Temperament addition: Difference between revisions

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In order to understand how to do temperament arithmetic on <math>g_{\text{min}}>1</math> temperaments, we must first understand addability.

~~===Verbal explanation===~~

In order to understand addability, we must work up to it, understanding these concepts in this order:

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#linear independence between temperaments by only one basis vector (that's addability)

====1. Linear dependence====

===1. Linear dependence===

This is explained here: [[linear dependence]].

====2. Linear dependence between temperaments====

===2. Linear dependence between temperaments===

Linear dependence has been defined for the matrices and multivectors that represent temperaments, but it can also be defined for temperaments themselves. The conditions of temperament arithmetic motivate a definition of linear dependence for temperaments whereby temperaments are considered linearly dependent if ''either of their mappings or their comma bases are linearly dependent''<ref>or — equivalently, in EA — either their multimaps or their multicommas are linearly dependent</ref>.

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To make this point visually, we could say that two temperaments are linearly dependent if they intersect in one or the other of tone space and tuning space. So you have to check both views.<ref>You may be wondering — what about two temperaments which are parallel in tone or tuning space, e.g. compton and blackwood in tuning space? Their comma bases are each <math>n=1</math>, and they meet to give a <math>n=2</math> [[comma basis]], which corresponds to a <math>r=1</math> mapping, which means it should appear as an ET point on the PTS diagram. But how could that be? Well, here's their meet: {{bra|{{vector|1 0 0}} {{vector|0 1 0}}}}, and so that corresponding mapping is {{ket|{{map|0 0 1}}}}. So it's some degenerate ET. I suppose we could say it's the point at infinity away from the center of the diagram.</ref>

====3. Linear independence between temperaments====

===3. Linear independence between temperaments===

Linear dependence may be considered as a boolean (yes/no, linearly dependent/independent) or it may be considered as an integer count of linearly dependent basis vectors. In other words, it is the dimension of the linear-dependence basis <math>\dim(L_{\text{dep}})</math>. To refer to this count, we may hyphenate it as '''linear-dependence''', and use the variable <math>l_{\text{dep}}</math>. For example, 5-ET and 7-ET, per the example in the previous section, are <math>l_{\text{dep}}=1</math> (read "linear-dependence-1") temperaments.

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A proof of this conjecture is given here: [[Temperament arithmetic#Sintel's proof of the linear-independence conjecture]].

====4. Linear independence between temperaments by only one basis vector (i.e. addability)====

===4. Linear independence between temperaments by only one basis vector (i.e. addability)===

Two temperaments are addable if they are <math>l_{\text{ind}}=1</math>. In other words, both their mappings and their comma bases share all but one basis vector.

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And so this is why <math>g_{\text{min}}=1</math> temperaments are all addable. Because if <math>g_{\text{min}}=1</math>, and the temperaments are different from each other so <math>l_{\text{ind}}</math> is at least 1, and <math>l_{\text{ind}}</math> can't be greater than <math>g_{\text{min}}</math>, so then necessarily <math>l_{\text{ind}}</math> <math>= 1</math> exactly, and therefore the temperaments are addable.

===~~Diagrammatic explanation~~===

==Multivector approach==

The simplest approach to <math>g_{\text{min}}>1</math> temperament arithmetic is to use multivectors. This is discussed in more detail here: [[Douglas Blumeyer and Dave Keenan's Intro to exterior algebra for RTT#Temperament arithmetic]].

==Matrix approach==

~~====Introduction====~~

Temperament arithmetic for <math>g_{\text{min}}>1</math> temperaments (again, that's with both <math>r>1</math> and <math>n>1</math>) can also be done using matrices, and it works in essentially the same way — entry-wise addition or subtraction — but there are some complications that make it significantly more involved than it is with multivectors. There's essentially five steps:

The diagrams used for this explanation were inspired in part by [[Kite Giedraitis|Kite]]'s [[gencom]]s, and specifically how in his "twin squares" matrices — which have dimensions <math>d×d</math> — one can imagine shifting a bar up and down to change the ~~boundary between vectors that form a~~ basis ~~for the commas and those that form a basis for preimage intervals (this basis is typically called~~ "~~the [[generator]]s~~"~~). The count of the former is the nullity~~ <math>n</math>~~, and the count of the latter is the rank <math>r~~</~~math~~>~~, and~~ the ~~shifting of the boundary bar between them~~ with the ~~total~~ <~~math~~>d</math> ~~vectors corresponds to the insight of the rank-nullity theorem, which states that~~ <math>~~r + n=d~~</~~math~~>~~. And so this diagram's square grid has just the right amount of room to portray both the mapping~~ and ~~the comma basis~~ for ~~a given temperament~~ (~~with the comma basis's vectors rotated 90 degrees to appear as rows~~, ~~to match up with the rows of the mapping).~~

# Find the linear-dependence basis <math>L_{\text{dep}}</math>

#Put the matrices in a form with the <math>L_{\text{dep}}</math>

#Check for enfactoring, and perform an addabilization defactor (if necessary)

#Check for negation, and change negation (if necessary)

#Entry-wise add, and canonicalize

~~So consider this first example of such a diagram:~~

===The steps===

~~{| class~~=~~"wikitable"~~

==== 1. Find the <math>L_{\text{dep}}</math>====

|+

~~| rowspan~~=~~"4" |<math>d~~=~~4</math>~~

~~| style="border-bottom: 3px solid black;"|<math>g_{\text{min}}~~=1~~</math>~~

~~| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |  ↑  ~~

~~| style="border-top: 1px solid #B6321C;" |<math>l_{\text{ind}}=1</math>~~

|-

~~| rowspan="3" |<math>g_{\text{max}}=3</math>~~

~~| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓  ~~

~~| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |~~  ↓  

~~| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓  ~~

~~| style="border-bottom: 1px solid #B6321C;" |<math>l_{\text{~~ind}}=1</math>

|-

~~|style~~=~~"background-color: #BED5BA;"|~~      

~~|style="background-color: #BED5BA;"|~~      

~~| rowspan="2" |~~

|-

~~|style="background-color: #BED5BA;"|~~      

~~|style~~=~~"background-color: #BED5BA;"|~~      

~~|style="background-color: #BED5BA;"|~~      

|}

~~This represents a~~ <~~math~~>~~d=4~~</~~math~~> ~~temperament~~. ~~These diagrams are grade-agnostic~~, ~~which is to say~~ that ~~they are agnostic as to which side counts~~ the ~~<math>r</math>~~ and ~~which side counts~~ the ~~<math>n</math>~~. ~~So we~~ are ~~showing them as~~ <~~math>g_{\text{min}}</math> and <math>g_{\text{max}}</math~~> ~~instead. We could say there's a variation on the rank~~-~~nullity theorem whereby~~ <math>~~g_{\text{min}} + g_~~{\text{~~max~~}}=d</math>~~, just as <math>r + n=d~~</~~math>. So we can then say that this diagram represents either a <math>r=1</math>, <math>n=3</math> temperament, or perhaps a <math>n=1</math>, <math>r=3</math~~> ~~temperament~~.

For matrices, it is necessary to make explicit the basis for the linearly dependent vectors shared between the involved matrices before performing the arithmetic. In other words, any vectors that can be found through linear combinations of any of the involved matrices' basis vectors must appear explicitly and in the same position of each matrix before the sum or difference is taken. These vectors are called the linear-dependence basis, or <math>L_{\text{dep}}</math>.

~~But actually,~~ this ~~diagram represents more than just a single temperament. It represents a relationship between a pair of temperaments (which have the same [[dimensions]]~~, ~~non-grade-agnostically, i.e. not a pairing~~ of ~~a <math>r=1</math>, <math>n=3</math> temperament with a <math>r=3</math>, <math>n=1</math> temperament). As elsewhere~~, ~~green coloration indicates the linearly dependent basis vectors~~ <math>L_{\text{dep}}</math> ~~between this pair~~ of ~~temperaments~~, ~~and red coloration indicates~~ linearly ~~''in''~~dependent ~~basis~~ vectors ~~<math>L_{\text{ind}}</math> between the same pair of temperaments.~~

Before this can be done, of course, we need to actually find the <math>L_{\text{dep}}</math>. This can be done using the technique described here: [[Linear dependence#For a given set of basis matrices, how to compute a basis for their linearly dependent vectors]]

~~So, in this case, the two ET maps are linearly independent~~. ~~This should be unsurprising; because ET maps are constituted by only~~ a ~~single vector (they're <math>r=1</math> by definition), if they ''were''~~ ~~linearly dependent, then they'd necessarily be the ''same'' exact ET! Temperament arithmetic on two of the same ET is never interesting;~~ <math>~~T_1~~</math> ~~plus <math>T_1~~</~~math~~> simply equals <math>T_1</math> again, and <math>T_1</math> minus <math>T_1</math> is undefined. That said, if we ''were'' to represent temperament arithmetic between two of the same temperament on such a diagram as this, then every cell would be green. And this is true regardless whether <math>r=~~1</math> or otherwise.~~

====2. Put the matrices in a form with the <math>L_{\text{dep}}</math>====

~~From this information~~, ~~we can see that~~ the ~~comma bases~~ of ~~any randomly selected pair of ''different''~~ <math>d=4</math> ~~ETs are going to~~ ~~share 2 vectors~~, ~~or in other words, their~~ <math>L_{\text{dep}}</math> ~~will have two basis~~ vectors. ~~In terms~~ of the ~~diagram,~~ we'~~re saying~~ that ~~they~~'~~ll always~~ have ~~two~~ ~~green-colored~~ vectors ~~under the black bar~~.

The <math>L_{\text{dep}}</math> will always have one less vector than the original matrix, by the definition of addability as <math>L_{\text{ind}}=1</math>. And the <math>L_{\text{dep}}</math> is not a full recreation of the original temperament; it needs that one extra vector to get back to representing it.

So a next step, we need pad out the <math>L_{\text{dep}}</math> by drawing from vectors from the original matrices. We can start from their first vectors. But if that vector happens to be linearly dependent on the <math>L_{\text{dep}}</math>, then it won't result in a representation of the original matrix. Otherwise we'll produce a [[rank-deficient]] matrix that doesn't still represent the same temperament as we started with. So we just have to keep going until we get it.

====3. Addabiliziation defactoring====

But it is not quite as simple as determining the <math>L_{\text{dep}}</math> and then supplying the remaining vectors necessary to match the grade of the original matrix, because the results may then be [[enfactored]]. And defactoring them without compromising the explicit <math>L_{\text{dep}}</math> cannot be done using existing [[defactoring algorithms]]; it's a tricky process, or at least computationally intensive. This is called '''addabilization defactoring'''.

~~These diagrams are a good way to understand which temperament relationships are possible and which aren't, where by a "relationship" here we mean a particular combination~~ of ~~their matching dimensions and their linear-independence integer count. A good way to use these diagrams for this purpose is to imagine~~ the ~~red coloration emanating away from the black bar in both directions simultaneously, one pair~~ of ~~vectors at~~ a ~~time~~. ~~Doing it like this captures the fact~~, ~~as previously stated~~, ~~that the~~ <math>l_{\text{~~ind~~}}</math> ~~on either side of duality is always equal~~. ~~There~~'~~s no notion of a max or min here, as there is with <math>g</math> or~~ <math>l_{\text{dep}}</math>; the <math>l_{\text{~~ind~~}}</math> on either side is always the same, so we can capture it with a single number, which counts the red vectors on just one half (that is, half of the total count of red vectors, or half of the width of the red band in the middle of the grid).

Most established defactoring algorithms will alter any or all of the entries of a matrix. This is not an option if we still want to be able to do temperament arithmetic, however, because these matrices must retain their explicit <math>L_{\text{dep}}</math>. And we can't defactor and then paste the <math>L_{\text{dep}}</math> back over the first vector or something, because then we might just be enfactored again. We need to find a defactoring algorithm that manages to work without altering any of the vectors in the <math>L_{\text{dep}}</math>.

~~There's no need~~ to ~~look at diagrams like this where the black bar~~ is ~~below the center. This is because, even though for convenience~~ we~~'re currently treating~~ the ~~top half as <math>r</math> and~~ the ~~bottom half as <math>n</math>, these diagrams are ultimately grade-agnostic. So we could say that each one essentially represents not just one possibility for~~ the ~~relationship between two temperaments' dimensions and <math>l_{\text{ind}}</math>~~, but ''~~two~~'' ~~such possibilities. Again~~, this ~~diagram equally represents both <math>d=4~~, r=1, ~~n=3, </math><math>l_{\text{ind}}=1</math> as well as <math>d=4~~, ~~r=3, n=1, </math><math>l_{\text{ind}}=1</math>. Which is another way of saying~~ we ~~could vertically mirror it without changing it~~.

The first step to addabilization defactoring is inspired by [[Pernet-Stein defactoring]]: we find the value of the enfactoring factor (the "greatest factor") by following this algorithm until the point where we have a square transformation matrix, but instead of inverting it and multiplying by it to ''remove'' the defactoring, we simply take this square matrix's determinant, which is the factor we were about to remove. If that determinant is 1, then we're already defactored; if not, then we need to take do some additional steps.

~~With~~ the ~~black bar always either~~ in the ~~top half or exactly~~ in the ~~center, we can see that~~ the ~~emanating~~ ~~red band~~ ~~will always either hit the top edge~~ of the ~~square grid first, or they will hit both the top and bottom edges of it simultaneously. So this is how these diagrams visually convey the fact that~~ the <math>l_{\text{~~ind~~}}</math> between two temperaments will always be less than or equal to their <math>g_{\text{min}}</math>: because a situation where <math>g_{\text{min}}>l</math> would visually look like the red band spilling past the edges of the square grid.

It turns out that you can always<ref>This conjecture was first suggested by Mike Battaglia, but it has not yet been mathematically proven. Sintel and Tom Price have done some experiments but nothing complete yet. Douglas Blumeyer's test cases in the [[RTT library in Wolfram Language]] suggest this is true, though.</ref> isolate the greatest factor in the single final vector of the matrix — the linearly independent vector — through linear combinations of the vectors in the <math>L_{\text{dep}}</math>.

~~We could also say~~ that ~~two temperaments are~~ linearly dependent ~~on each other when~~ ~~<math>l_{\text{ind}}~~</~~math~~><math>~~<g_~~{\text{~~max~~}}</math>, ~~that~~ is, ~~their linear-independence<~~/~~span> is less than their ''max''-grade~~.

The example that will be worked through in this section below is as simple as it can get: the <math>L_{\text{dep}}</math> consists of only a single vector, so we simply add some number of this single linearly dependent vector to the linearly independent vector. However, if there are multiple vectors in the <math>L_{\text{dep}}</math>, the linear combination which surfaces the greatest factor may involve just one or potentially all of those vectors, and the best approach to finding this combination is simply an automatic solver. An example of this approach is demonstrated in the [[RTT library in Wolfram Language]], here: https://github.com/cmloegcmluin/RTT/blob/main/main.m#L477

~~Perhaps more importantly~~, ~~we can also see from these diagrams that any pair of <math>g_{\text{min}}=1</math> temperaments will~~ be ~~addable~~. ~~Because if they are <math>g_{\text{min}}=1</math>~~, ~~then~~ the ~~furthest~~ the ~~red band can extend from the black bar~~ is ~~1 vector~~, ~~and 1 mirrored set~~ of ~~red vectors means <math>l_{\text{ind}}=1</math>~~, and ~~that's~~ the ~~definition of addability~~.

Another complication is that the greatest factor may be very large, or be a highly composite number. In this case, searching for the linear combination that isolates the greatest factor in its entirety directly may be intractable; it is better to eliminate it piecemeal, i.e., whenever the solver finds a factor of the greatest factor, eliminate it, and repeat until the greatest factor is fully eliminated. The RTT library code linked to above works in this way.

====~~A simple <math>d=3</math> example~~====

====4. Negation====

~~Let's back-pedal to <math>d=3</math>~~ for ~~a simple illustrative example~~.

Temperament negation is more complex with matrices, both in terms of checking for it, as well as changing it.

~~{| class="wikitable"~~

|+

~~| rowspan="3" |<math>d=3</math>~~

~~|style="border-bottom: 3px solid black;"|<math>g_{\text{min}}=1</math>~~

~~| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |  ↑  ~~

~~| style="border-top: 1px solid #B6321C;" |<math>l_{\text{ind}}=1</math>~~

|-

~~| rowspan="2" |<math>g_{\text{max}}=2</math>~~

~~| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓  ~~

~~| style="border-bottom: 1px solid #B6321C;" |<math>l_{\text{ind}}=1</math>~~

|-

~~|style="background-color: #BED5BA;"|~~      

|

|}

~~This diagram shows us that any two <math>d=3</math>~~, ~~<math>g_{\text{min}}=1</math> temperaments (like 5-limit ETs) will be linearly dependent, i~~.~~e. their~~ comma bases ~~will share one~~ vector~~. You may already know this intuitively if you are familiar with the 5-limit [[projective tuning space]] diagram from the [[Paul_Erlich#Papers|Middle Path]] paper~~, ~~which shows how we can draw~~ a ~~line through any two ETs and that line will represent~~ a ~~temperament~~, ~~and the single comma that temperament tempers out~~ is ~~this shared vector. The diagram also tells us that any two 5-limit temperaments that temper out only~~ a ~~single comma will also be linearly dependent, for the opposite reason: their ''mappings'' will always share one vector~~.

For matrices, negation is accomplished by choosing a single vector and changing the sign of every entry in it. In the case of comma bases, a vector is a column, whereas in a mapping a vector (technically a row vector, or covector) is a row.

~~And we can see that there~~ are ~~no other diagrams of interest~~ for ~~<math>d=3</math>, because there's no sense in looking at diagrams with no red band, but~~ we ~~can't extend~~ the ~~red band any further than 1 vector on each side without going over~~ the ~~edge~~, and ~~we can't lower~~ the ~~black bar any further without going below~~ the ~~center. So we're done. And our conclusion~~ is ~~that any pair of different <math>d=3</math> temperaments that are nontrivial (<math>0 < n < d=3</math>~~ and ~~<math>0 < r < d=3</math>) will be addable~~.

For matrices, the check for negation is related to canonicalization of multivectors as are used in exterior algebra for RTT. Essentially we take the minors of the matrix, and then look at their leading or trailing entry (leading in the case of a covariant matrix, like a mapping; trailing in the case of a contravariant matrix, like a comma basis): if this entry is positive, so is the temperament, and vice versa.

====~~Completing the suite of <math>d=4</math> examples~~====

====5. Entry-wise add====

~~Okay, back to~~ <~~math~~>~~d=4~~</~~math~~>. We~~'ve already looked at~~ the <~~math>g_{\text{min}}~~=~~1</math> possibility (which, for any <math~~>d</math>~~, there will only ever be one of). So let's start looking at the possibilities where <math>g_~~{\text{~~min~~}}=2</math>~~, which in the case of <math>d=4~~</~~math> leaves us only one pair of values for <math>r</math~~> and ~~<math>n</math>: both being 2~~.

The entry-wise addition of elements works mostly the same as for vectors. But there's one catch: we only do it for the pair of linearly independent vectors. We set the <math>L_{\text{dep}}</math> aside, and reintroduce it at the end.

~~{| class="wikitable"~~

When taking the sum, this is just for simplicity's sake. There's no sense in adding the two copies of the <math>L_{\text{dep}}</math> together, as we'll just get the same vector but 2-enfactored. So we may as well set it aside, and deal only with the linearly independent vectors, and put it back at the end.

|+

~~| rowspan="4" |~~<~~math>d=4</math>~~

When taking the difference, it's essential that we set the <math>L_{\text{dep}}</math> aside before entry-wise arithmetic, though, because if we were to subtract it from itself, we'd end up with all zeros. Unlike the case of the sum, where we'd just end up with an enfactored version of the starting vectors, we couldn't even defactor to get back to where we started if we completely wiped out the relevant information by sending it all to zeros.

~~| rowspan="2"~~ style="~~border-bottom~~: ~~3px solid black~~;" |<math>g_{\text{~~min~~}}=2</math>

~~|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|~~      

As a final step, as is always good to do when concluding temperament operations, put the result in [[canonical form]].

~~|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|~~      

===Example===

~~|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|~~      

~~|style="border-bottom: 1px solid #B6321C;"|~~

For our example, let’s look at septimal meantone plus flattone. The canonical forms of these temperaments are {{ket|{{map|1 0 -4 -13}} {{map|0 1 4 10}}}} and {{ket|{{map|1 0 -4 17}} {{map|0 1 4 -9}}}}.

|-

~~|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑~~

'''0. Counterexample.''' Before we try following the detailed instructions just described above, let's do the counterexample, to illustrate why we have to follow them at all. Simple entry-wise addition of these two mapping matrices gives {{ket|{{map|2 0 -8 4}} {{map|0 2 8 1}}}}, which is not the correct answer:

~~|style="background~~-~~color: #E7BBB3; border-bottom: 3px solid black;"|~~  ↑

~~|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑  ~~

<math>\left[ \begin{array} {rrr}

~~| style="border-top: 1px solid #B6321C;" |~~<math>l_{\text{~~ind~~}}=1</math>

|-

1 & 0 & -4 & -13 \\

~~| rowspan~~=~~"2"~~ |~~<math>g_~~{~~\text~~{~~max~~}}~~=2</math>~~

0 & 1 & 4 & 10 \\

| ~~style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"~~ |~~  ↓  ~~

| ~~style="background~~-~~color: #E7BBB3; border~~-~~bottom: 1px solid #B6321C;" |  ↓  ~~

~~| style="background~~-~~color: #E7BBB3; border-bottom: 1px solid #B6321C;"~~ |~~  ↓  ~~

| ~~style="background~~-~~color: #E7BBB3; border-bottom: 1px solid #B6321C;"~~ |~~  ↓  ~~

~~| style="border-bottom: 1px solid #B6321C;" |~~<math>~~l_{~~\~~text~~{~~ind~~}}~~=1</math>~~

|-

~~|style="background~~-~~color: #BED5BA;"|~~      

~~|style="background-color: #BED5BA;"|~~      

|

|}

~~But even with <math>d</math>, <math>r</math>, and <math>n</math> fixed, we still have more than one possibility for <math>L_{~~\~~text~~{~~dep~~}~~}</math>. The above diagram shows <math>l_{~~\~~text~~{~~ind~~}~~}=1</math>. The below diagram shows <math>l_{\text~~{~~ind}~~}~~=2</math>.~~

\end{array} \right]

+

\left[ \begin{array} {rrr}

~~{| class="wikitable"~~

1 & 0 & -4 & 17 \\

|+

0 & 1 & 4 & -9 \\

~~| rowspan="4" |<math>d=~~4~~</math>~~

~~| rowspan="2" style="border-bottom: 3px solid black;" |<math>g_{~~\~~text{min}}=2</math>~~

~~| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |  ↑  ~~

~~| rowspan="2" style="border-top: 1px solid #B6321C;" |<math>l_{~~\~~text{ind}}=2</math>~~

|-

~~|style="background~~-~~color: #E7BBB3; border-bottom: 3px solid black;"|  ↑  ~~

~~|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑  ~~

|-

~~| rowspan="2" |<math>g_{~~\~~text{max}}=2</math>~~

~~| style="background-color: #E7BBB3;" |  ↓  ~~

~~| rowspan="2" style="border-bottom: 1px solid #B6321C;" |<math>l_{~~\~~text{ind}}=2</math>~~

|-

~~|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|  ↓  ~~

|}

In the ~~former possibility~~, where <math>l_{\text{~~ind~~}}=1</math> (and ~~therefore the temperaments are addable)~~, ~~we have a pair of different~~ <math>~~d=4~~</math>, <~~math~~>r=2</~~math~~> ~~temperaments where we can find a single comma that both temperaments temper out, and — equivalently — we can find one ET that supports both temperaments~~.

\end{array} \right]

=

\left[ \begin{array} {rrr}

2 & 0 & -8 & 4 \\

0 & 2 & 8 & 1 \\

\end{array} \right]</math>

And it's wrong not only because it is clearly enfactored (at least one factor of 2, that is visible in the first vector). The full explanation of why this is the wrong answer is beyond the scope of this example. However, if we now follow through with the instructions described above, we can find the correct answer.

'''1. Find the linear-dependence basis.''' We know where to start: first find the <math>L_{\text{dep}}</math> and put each of these two mappings into a form that includes it explicitly. In this case, their <math>L_{\text{dep}}</math> consists of a single vector: {{ket|{{map|19 30 44 53}}}}.

In the ~~latter possibility~~, ~~where~~ ~~<math>l_~~{~~\text~~{~~ind~~}}~~=2</math>~~~~, neither side of duality~~ ~~shares~~ any vectors ~~in common. And so we've encountered our first example that is not addable. In other words~~, ~~if the~~ ~~red band~~ ~~ever extends more than 1 vector away~~ from the ~~black bar, temperament arithmetic~~ is not ~~possible~~. ~~So <math>d=4</math> is~~ the first ~~time we had enough room (half of <math>d</math>) to support that condition~~.

'''2. Reproduce the original temperament.''' The original matrices had two vectors, so as our second step, we pad out these matrices by drawing from vectors from the original matrices, starting from their first vectors, so now we have [{{map|19 30 44 53}} {{map|1 0 -4 -13}}⟩ and [{{map|19 30 44 53}} {{map|1 0 -4 17}}⟩. We could choose any vectors from the original matrices, as long as they are linearly independent from the ones we already have; if one is not, skip it and move on. In this case the first vectors are both fine, though.

~~We have now exhausted the possibility space for <math>d=4</math>. We can't extend either the red band or the black bar any further.~~

~~====~~<math>~~d=5</math> diagrams finally reveal important relationships====~~

<math>\left[ \begin{array} {rrr}

~~So how about we go to <math>d=5</math> (such as the 11~~-~~limit). As usual, starting with <math>g_~~{\~~text~~{~~min~~}}=1</math>:

\color{OliveGreen}19 & \color{OliveGreen}30 & \color{OliveGreen}44 & \color{OliveGreen}53 \\

~~{| class="wikitable"~~

1 & 0 & -4 & -13 \\

|+

~~| rowspan="5" |~~<~~math~~>d=5</~~math~~>

\end{array} \right]

|style="~~border-bottom~~: ~~3px solid black~~;" |<math>g_{\text{~~min~~}}=1</math>

+

~~| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |  ↑~~

\left[ \begin{array} {rrr}

~~| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |~~  ↑

~~| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |~~  ↑

\color{OliveGreen}19 & \color{OliveGreen}30 & \color{OliveGreen}44 & \color{OliveGreen}53 \\

~~| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |~~  ↑

1 & 0 & -4 & 17 \\

| ~~style="background~~-~~color: #E7BBB3; border~~-~~bottom: 3px solid black; border-top: 1px solid #B6321C;" |~~  ~~↑  ~~

~~| style="border-top: 1px solid #B6321C;" |<math>l_~~{~~\text~~{~~ind~~}}~~=1</math>~~

\end{array} \right]</math>

|-

~~| rowspan~~=~~"4" |<math>g_~~{~~\text~~{~~max~~}}=~~4</math>~~

~~| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |~~  ↓

'''3. Defactor.''' Next, verify that both matrices are defactored. In this case, both matrices ''are'' enfactored, each by a factor of 30<ref>or you may prefer to think of this as three different (prime) factors: 2, 3, 5 (which multiply to 30)</ref>. So we'll use addabilization defactoring. Since there's only a single vector in the <math>L_{\text{dep}}</math>, therefore all we need to do is repeatedly add that one linearly dependent vector to the linearly independent vector until we find a vector with the target GCF, which we can then simply divide out to defactor the matrix. Specifically, we add 11 times the linearly dependent vector. For the first matrix, {{map|1 0 -4 -13}} + 11⋅{{map|19 30 44 53}} = {{map|210 330 480 570}}, whose entries have a GCF = 30, so we can defactor the matrix by dividing that vector by 30, leaving us with {{map|7 11 16 19}}. Therefore the final matrix here is [{{map|19 30 44 53}} {{map|7 11 16 19}}⟩. The other matrix matrix happens to defactor in the same way: {{map|1 0 -4 17}} + 11⋅{{map|19 30 44 53}} = {{map|210 330 480 600}} whose GCF is also 30, reducing to {{map|7 11 16 20}}, so the final matrix is [{{map|19 30 44 53}} {{map|7 11 16 20}}⟩.

~~| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |~~  ↓

| ~~style="background-color~~: ~~#E7BBB3; border~~-~~bottom: 1px solid #B6321C;" |~~  ↓

'''4. Check negativity.''' The next thing we need to do is check the negativity of these two temperaments. If either of the matrices we're performing arithmetic on is negative, then we'll have to change it (if both are negative, then the problem cancels out, and we go back to being right). We check negativity by using the minors of these matrices. The first matrix's minors are (-1, -4, -10, -4, -13, -12) and the second matrix's minors are (-1, -4, 9, -4, 17, 32). What we're looking for here are their leading entries, because these are minors of a mapping (if we were looking at minors of comma bases, we'd be looking at the trailing entries instead). Specifically, we're looking to see if the leading entries are positive. They're not. Which tells us these matrices are both negative! But again, since they were ''both'' negative, the effect cancels out; no need to change anything (but if we wanted to, we could just take the linearly independent vector for each matrix and negate every entry in it).

~~| style~~=~~"background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"~~ |  ↓

~~| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |~~  ↓

~~| style="border-bottom: 1px solid #B6321C;" |~~~~<math>l_~~{~~\text~~{~~ind~~}}~~=1</math>~~

|-

~~|style="background~~-~~color: #BED5BA;"|~~      

~~| rowspan="3" |~~

|-

~~|style="background~~-~~color: #BED5BA;"|~~      

~~|style="background~~-~~color: #BED5BA~~;"|      

|style="~~background-~~color: #~~BED5BA~~;"|      

~~|style="background-color: #BED5BA;"|~~      

|-

~~|style="background-color: #BED5BA;"|~~      

|}

~~Just as with~~ the ~~<math>l_{\text{ind}}=1</math> diagrams given for <math>d=3</math> and <math>d=5</math>, we can see these are addable temperaments.~~

'''5. Add.''' Now the matrices are ready to add:

~~Now let's look at <math>d=5</math> but with <math>g_{\text{min}}=2</math>. This presents two possibilities. First, <math>l_{\text{ind}}=1</math>:~~

~~{| class="wikitable"~~

<math>\left[ \begin{array} {rrr}

|+

~~| rowspan="5" |~~<math>~~d=5</math>~~

\color{OliveGreen}19 & \color{OliveGreen}30 & \color{OliveGreen}44 & \color{OliveGreen}53 \\

~~| rowspan="2" style="border-bottom: 3px solid black;" |<math>g_{~~\~~text~~{~~min~~}}~~=2</math>~~

\color{BrickRed}7 & \color{BrickRed}11 & \color{BrickRed}16 & \color{BrickRed}19 \\

~~|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|~~      

~~|style="background-~~color~~: #BED5BA; border-bottom: 1px solid #B6321C;"|~~      

\end{array} \right]

~~|style="background-~~color~~: #BED5BA; border-bottom: 1px solid #B6321C;"|~~      

+

~~|style="background-~~color~~: #BED5BA; border-bottom: 1px solid #B6321C;"|~~      

\left[ \begin{array} {rrr}

~~|style="background-~~color~~: #BED5BA; border-bottom: 1px solid #B6321C;"|~~      

~~| style="border-bottom: 1px solid #B6321C;" |~~

\color{OliveGreen}19 & \color{OliveGreen}30 & \color{OliveGreen}44 & \color{OliveGreen}53 \\

|-

\color{BrickRed}7 & \color{BrickRed}11 & \color{BrickRed}16 & \color{BrickRed}20 \\

~~|style="background-~~color~~: #E7BBB3; border-bottom: 3px solid black;"|  ↑  ~~

\end{array} \right]</math>

~~|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑  ~~

~~| style="border-top: 1px solid #B6321C;" |<math>l_{~~\~~text~~{~~ind~~}}~~=1</math>~~

|-

~~| rowspan="3" |<math>g_~~{\~~text~~{~~max~~}}~~=3</math>~~

~~| style="background-~~color~~: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓  ~~

~~| style="background-~~color~~: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓  ~~

~~| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓  ~~

~~| style="border-bottom: 1px solid #B6321C;" |<math>l_{~~\~~text~~{~~ind~~}~~}=1~~</math>~~~~

|-

~~|style="background-color: #BED5BA;"|~~      

|

|-

~~|style="background-color: #BED5BA;"|~~      

|

|}

~~And second,~~ <math>l_{\text{~~ind~~}}=2</math>:

We set the <math>L_{\text{dep}}</math> aside, and deal only with the linearly independent vectors:

<math>\left[ \begin{array} {rrr}

\color{BrickRed}7 & \color{BrickRed}11 & \color{BrickRed}16 & \color{BrickRed}19 \\

\end{array} \right]

+

\left[ \begin{array} {rrr}

\color{BrickRed}7 & \color{BrickRed}11 & \color{BrickRed}16 & \color{BrickRed}20 \\

{~~| class="wikitable"~~

\end{array} \right]

|+

=

~~| rowspan="5" |<math>d~~=~~5</math>~~

\left[ \begin{array} {rrr}

~~| rowspan="2" style="border-bottom: 3px solid black;" |<math>g_{~~\~~text~~{~~min~~}}~~=2</math>~~

~~| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |  ↑  ~~

\color{BrickRed}14 & \color{BrickRed}22 & \color{BrickRed}32 & \color{BrickRed}39 \\

~~| style="background-~~color~~: #E7BBB3; border-top: 1px solid #B6321C;" |  ↑  ~~

\end{array} \right]</math>

~~| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |  ↑  ~~

~~| rowspan="2" style="border-top: 1px solid #B6321C;" |<math>l_{~~\~~text~~{~~ind~~}~~}=2~~</math~~>~~

Then we can reintroduce the <math>L_{\text{dep}}</math> afterwards:

|-

~~|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑  ~~

~~|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑  </span~~>

<math>\left[ \begin{array} {rrr}

~~|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑  ~~

\color{OliveGreen}19 & \color{OliveGreen}30 & \color{OliveGreen}44 & \color{OliveGreen}53 \\

~~|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|~~  ~~↑  ~~

\color{BrickRed}14 & \color{BrickRed}22 & \color{BrickRed}32 & \color{BrickRed}39 \\

|-

~~| rowspan="3" |~~<math>g_{\text{~~max~~}}=3</math>

\end{array} \right]</math>

~~| style="background-color: #E7BBB3;" |  ↓~~

~~| style="background-color: #E7BBB3;" |  ↓  ~~

~~| style="background-color: #E7BBB3;" |  ↓  ~~

~~| style="background-color: #E7BBB3;" |  ↓  ~~

~~| rowspan="2" style="border-bottom: 1px solid #B6321C;" |~~<math>~~l_{~~\~~text~~{~~ind~~}}~~=2</math>~~

|-

~~|style="background-~~color~~: #E7BBB3; border-bottom: 1px solid #B6321C;"|  ↓  ~~

~~|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|  ↓~~  </~~span~~>

|-

~~|style="background-color: #BED5BA;"|~~      

|

|}

Here's where things really get interesting. Because in both of these cases, the pairs of temperaments represented are linearly dependent on each other (i.e. either their mappings are linearly dependent, their comma bases are linearly dependent, or both). And so far, every possibility where temperaments have been linearly dependent, they have also been <math>l_{\text{ind}}=1</math>, and therefore addable. But if you look at the second case here, we are <math>l_{\text{ind}}=2</math>, but since <math>d=5</math>, the temperaments still manage to be linearly dependent. So this is the first example of a linearly dependent temperament pairing which is not addable.

~~====Back to <math>d=2</math>, for a surprisingly tricky example====~~

And finally [[canonical form|canonicalize]]:

Beyond <math>d=5</math>, these diagrams get cumbersome to prepare, and cease to reveal further insights. But if we step back down to <math>d=2</math>, a place simpler than anywhere we've looked so far, we actually find another surprisingly tricky example, which is hopefully still illuminating.

So <math>~~d=2</math> (such as the 3-limit) presents another case — similar to the <math>d=5</math>, <math>g_{~~\~~text{min}}=2</math>, <math>l_{~~\~~text~~{~~ind}~~}=2</math> case explored most recently above — where the properties of linearly dependence and addability do not match each other. But while in the other case, we had a temperament pair that was linearly dependent yet not addable, in this <math>d=2</math> (and therefore <math>g_{\text{min}}=1</math>, <math>l_{\text{~~ind}~~}~~=1</math>) case, it is the other way around: addable yet linearly independent!~~

<math>\left[ \begin{array} {rrr}

~~{| class="wikitable"~~

1 & 0 & -4 & 2 \\

|+

0 & 2 & 8 & 1 \\

~~| rowspan="~~2~~" |<math>d=2</math>~~

~~|style="border-bottom: 3px solid black;" |<math>g_{~~\~~text{min}}=1</math>~~

~~|style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;"|  ↑  ~~

~~| style="border-top: 1px solid #B6321C;" |<math>l_{\text{ind}}=~~1~~</math>~~

|-

~~|<math>g_{~~\~~text{max}}=1</math>~~

~~|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|  ↓  ~~

~~| style="border-bottom: 1px solid #B6321C;" |<math>l_{~~\~~text{ind}}=1</math>~~

|}

~~Basically, in the case of <math>d=2</math>, <math>g_{~~\~~text~~{~~max~~}~~}=1~~</math> (in non-trivial cases, i.e. not JI or the unison temperament), so any two different ETs or commas you pick are going to be linearly independent (because the only way they could be linearly dependent would be to be the same temperament). And yet we know we can still entry-wise add them to new vectors that are [[Douglas_Blumeyer_and_Dave_Keenan%27s_Intro_to_exterior_algebra_for_RTT#Decomposability|decomposable]], because they're already vectors (decomposing means to express a [[Douglas_Blumeyer_and_Dave_Keenan%27s_Intro_to_exterior_algebra_for_RTT#From_vectors_to_multivectors|multivector]] in the form of a list of monovectors, so decomposing a multivector that's already a monovector like this is tantamount to merely putting array braces around it.)

\end{array} \right]</math>

~~====Conclusion====~~

This explanation has hopefully helped get a grip on how addability AKA <math>l_{\text{ind}}=1</math> behaves given various temperament dimensions. But it still hasn't quite explained why <math>l_{\text{ind}}=1</math> is ~~one and the same thing as addability. We will look at this in another section soon~~.

so we can now see that meantone plus flattone is [[godzilla]].

==~~=Geometric explanation===~~

As long as we've done all this work to set these matrices up for arithmetic, let's check their difference as well. Again, set aside the <math>L_{\text{dep}}</math>, and just entry-wise subtract the two linearly independent vectors:

We've presented a diagrammatic illustration of the behavior of linear-independence <math>l_{\text{ind}}</math> with respect to temperament dimensions. But some of the results might have seemed surprising. For instance, when looking at the diagram for <math>d=4, g_{\text{min}}=1, g_{\text{max}}=3</math>, it might have seemed intuitive enough that the the red band could not extend beyond the square grid, but then again, why shouldn't it be possible to have, say, two 7-limit ETs which temper out only a single comma in common? Perhaps it doesn't seem clear that this is impossible, and that they must temper out two commas in common (and of course the infinitude of combinations of these two commas). If this is as unclear to you as it was to the author when exploring this topic, then this explanatory section is for you! Here, we will use geometrical representations of temperaments to hone our intuitions about the possible combinations of dimensions and linear-independence <math>l_{\text{ind}}</math> of temperaments.

~~In this approach, we’re actually not going to focus directly on the linear-independence~~ <math>~~l_{~~\~~text{ind}}</math> of temperaments. Instead, we're going to look at the linear-''de''pendence <math>l_{~~\~~text~~{~~dep}~~}~~</math> of matrices representing temperaments such as mappings and comma bases, and then compute the linear-independence <math>l_~~{~~\text{ind}}</math> from it and the grade <math>g</math>. As we’ve established, the linear-dependence <math>l_{\text{dep}~~}~~</math> differs from one side of duality to the other, so we’ll only be looking at one side of duality at a time.~~

<math>\left[ \begin{array} {rrr}

=~~===Introduction====~~

\color{BrickRed}7 & \color{BrickRed}11 & \color{BrickRed}16 & \color{BrickRed}19 \\

\end{array} \right]</math>

-

<math>\left[ \begin{array} {rrr}

\color{BrickRed}7 & \color{BrickRed}11 & \color{BrickRed}16 & \color{BrickRed}20 \\

\end{array} \right]

=

\left[ \begin{array} {rrr}

\color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}-1 \\

\end{array} \right]</math>

In this geometric approach, we'll be imagining individual vectors as points (0D), sets of two vectors as lines (1D), sets of three as planes (2D), four as volumes (3D), and so forth, as according to this table:

~~{| class="wikitable center-all"~~

|+

~~!vector~~

~~count~~

~~!geometric~~

~~dimension~~

~~!form~~

|-

|0

~~|undefined~~

~~|(emptiness)~~

|-

|1

|0

~~|point~~

|-

|2

|1

~~|line~~

|-

|3

|2

~~|plane~~

|-

|4

|3

~~|volume~~

|-

|5

|4

~~|hypervolume~~

|-

~~| ⋮~~

|⋮

|}

~~This is a~~ "~~vector space~~", and these geometric dimensions are consistent with how temperaments represented by these counts of vectors appear in ''projective'' vector space, which reduces geometric dimensions by 1. For example, a vector has a geometric interpretation as a directed line segment, which is 1D, but a point is 0D, which is one dimension lower. Essentially what we~~'re doing is ''assuming the origin''.~~

And so, reintroducing the <math>L_{\text{dep}}</math>, we have:

Think of it this way: geometric points are zero-dimensional, simply representing a position in space, whereas linear algebra vectors are one-dimensional, representing both a magnitude and direction; the way vectors manage to encode this extra dimension without providing any additional information is by being understood to describe this position in space ''relative to an origin''. Well, so we'll now switch our interpretation of these objects to the geometric one, where the vector's entries are nothing more than a coordinate for a point in space. And the "projection" involved in projective vector space essentially positions us at this discarded origin, looking out from it upon every individual point, which accomplishes the same feat, in a visual way.

Perhaps an example may help clarify this setup. Suppose we've got an (x,y,z) space, and two coordinates (5,8,12) and (7,11,16). You should recognize these as the simple maps for 5-ET and 7-ET, usually written as {~~{map|5 8 12}~~} ~~and {~~{~~map|7 11 16}~~}, respectively. Ask for the equation of the plane defined by the three points (5,8,12), (7,11,16), and the origin (0,0,0) and you'll get -4x + 4y -1z = 0, which clearly shows us the entries of the meantone comma. That's because meantone temperament can be defined by these two maps. 5-limit JI is a 3D space, and meantone temperament, as a rank-2 temperament, would be a 2D plane. But we don't normally need to think of the map corresponding to the origin, where everything is tempered out, including meantone. So we can just assume it, and think of a 2D plane as being defined by only 2 points, which in a view projected (from the origin) will look like just the line connecting (5,8,12) and (7,11,16).

<math>\left[ \begin{array} {rrr}

So, we've set the stage for our projective vector spaces. We will now be looking at representations of temperaments as counts of vector sets, and then using this scheme to convert them to primitive geometric forms. We'll place two of each form into the space, representing the two temperaments having the possibility of arithmetic performed on them checked. Then we will observe their possible ''intersections'' depending on how they're oriented in space, and it's these intersections that represent their linear-dependence. When the dimension of the intersection is then converted back to a vector set count, then we have their linear-dependence <math>l_~~{\~~text~~{~~dep~~}}~~</math>, for this side of duality, anyway (remember, unlike the linear-independence <math>l_~~{\~~text~~{~~ind}~~}~~</math>, this value isn't necessarily the same on both sides of duality). We can finally subtract the linear~~-~~dependence from the grade (vector count) to get the linear-indepedence, in order to determine if the two temperaments are addable.

\color{OliveGreen}19 & \color{OliveGreen}30 & \color{OliveGreen}44 & \color{OliveGreen}53 \\

\color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}-1 \\

In these examples, we'll be assuming that no two temperaments being compared are the same, because performing temperament arithmetic on copies of the same temperament is not interesting. The other things we'll be assuming is that no lines, planes, etc. are parallel to each other; this is due to a strange effect touched upon in footnote 4 whereby temperament geometry that appears parallel in projective space actually still intersects; the present author asks that if anyone is able to demystify this situation, that they please do!

\end{array} \right]</math>

~~====At <math>d=3</math>====~~

~~First, let's establish the geometric dimension of the space. With <math>d=3</math>, we've got a 2D space (one less than 3), so the entire space can be visualized on a plane.~~

Which canonicalizes to:

Our only possible values for <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> here are 1 and 2, respectively. So these are the two possible counts of vectors <math>g</math> possessed by matrices representing temperaments here.

~~So let's look at temperaments represented by matrices with 1 vector first (~~<math>g=1</math>). In this case, each of the two temperaments is a point in the plane. Unless these two temperaments are the same temperament, the intersection of these two points is empty. Emptiness isn't even 0D! So that tells us that these temperaments have 0 vectors worth of linear dependence. With <math>g=1</math>, that gives us a <math>l_{\~~text~~{~~ind}~~}~~</math><math> = g</math> <math> - </math> <math>l_~~{~~\text{dep~~}~~}</math> <math>= 1 -</math> <math>0</math> <math>= 1</math>:~~

<math>\left[ \begin{array} {rrr}

~~[[File:D3 g1 dep0.png|200px|none]]~~

19 & 30 & 44 & 0 \\

0 & 0 & 0 & 1 \\

Next, let's look at temperaments represented by matrices with 2 vectors (<math>g=2</math>). In this case, each of the two temperaments is a line in the plane. Again, assuming the two lines are not the same line or parallel, their intersection is a point. Being 0D, that tells us that the linear-dependence of these matrices is 1. So that gives us an <math>l_{\~~text~~{~~ind~~}~~}</math> <math>= g</math> <math>-</math> <math>l_{~~\~~text{dep}}</math> <math>= 2 -</math> <math>1</math> <math>= 1</math>. This matches the value we found via the <math>g=1~~</math>~~, so we've effectively checked our work:~~

\end{array} \right]</math>

~~[[File:D3 g2 dep1.png|200px|none]]~~

~~====At <math>d=2</math>====~~

And so we can see that meantone minus flattone is [[meanmag]].

~~Let's step back to <math>d~~=~~2</math>. Here we've got a 2 minus 1 equals 1D space, so the entire space can be visualized~~ on ~~a single line (one direction corresponds to an increasing ratio between the two coordinates, and the other to a decreasing ratio).~~

== Arithmetic on non-addable temperaments ==

We know our only possible value for <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> here is 1. So in either case, each of the two temperaments is a point on the line. As with two points in a plane — when <math>d=~~3</math> — unless these two temperaments are the same temperament, the intersection of these two points is empty. So again the <math>l_{\text{ind}}</math><math>~~ = g = ~~1</math>:~~

== Further explanations ==

~~[[File:D2 g1 dep0.png|200px|none]]~~

===Diagrammatic explanation===

====~~At <math>d=4</math>~~====

====Introduction====

~~First~~, ~~let'~~s ~~establish~~ the ~~geometric dimension~~ of the ~~space. With~~ <math>d=4</math>~~, we~~'~~ve got~~ a ~~3D space~~ (~~one less than 4)~~, so the ~~entire space can be visualized in a volume~~.

The diagrams used for this explanation were inspired in part by [[Kite Giedraitis|Kite]]'s [[gencom]]s, and specifically how in his "twin squares" matrices — which have dimensions <math>d×d</math> — one can imagine shifting a bar up and down to change the boundary between vectors that form a basis for the commas and those that form a basis for preimage intervals (this basis is typically called "the [[generator]]s"). The count of the former is the nullity <math>n</math>, and the count of the latter is the rank <math>r</math>, and the shifting of the boundary bar between them with the total <math>d</math> vectors corresponds to the insight of the rank-nullity theorem, which states that <math>r + n=d</math>. And so this diagram's square grid has just the right amount of room to portray both the mapping and the comma basis for a given temperament (with the comma basis's vectors rotated 90 degrees to appear as rows, to match up with the rows of the mapping).

~~At <math>d=4</math>, we have~~ a ~~couple options for the grade~~: ~~either <math>g_{\text{min}}=1</math> and <math>g_{\text{max}}=3</math>, or both <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> equal 2.~~

So consider this first example of such a diagram:

~~Let's look at temperaments represented by matrices with 1 vector first (~~<math>g=1</math>~~). Yet again, we find ourselves with two separate points, but now we find them in a space that's not a line, not a plane, but a volume. This doesn't change <span~~ style="~~color~~: ~~#B6321C~~;"><math>l_{\text{~~ind~~}}~~</math><math> = g = 1</math>, so we're not even going to show it, or any further cases of <math>g~~=1</math>~~. These are all addable.~~

{| class="wikitable"

|+

~~And when <math>g~~=3<~~/math>, because this is paired with <math>g~~=~~1</math> from the min and max values, we should expect to get the same answer as with <math~~>~~g=1~~</~~math~~>~~. And indeed, it will check out that way. Because two <math>g~~=~~3</math> temperaments will be planes in this volume, and the intersection of two planes is a line. Which means that~~ ~~<math>l_{\text{dep}}</math>~~~~<math>~~ = ~~2</math>. And so~~ ~~<math>l_{\text{ind}}</math>~~~~<math>~~ = ~~g</math> <math>~~ - ~~</math>~~ ~~<math>l_{\text{dep}}</math>~~ ~~<math>~~= 3 -~~</math>~~ <math>2</math> <math>= 1</math>. And here's where our geometric approach begins to pay off! This was the example given at the beginning that might seem unintuitive when relying only on the diagrammatic approach. But here we can see clearly that there would be no way for two planes in a volume to intersect only at a point, which proves the fact that two 7-~~limit ETs could never only temper out a single comma in common.~~

| rowspan="4" |<math>d=4</math>

| style="border-bottom: 3px solid black;"|<math>g_{\text{min}}=1</math>

~~[[File:D4 g3 dep2.png~~|~~200px]]~~

| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |  ↑

~~Next let's look at temperaments represented by matrices with 2 vectors, that is, when both <math>g_{\text{min}}</math> and~~ <math>g_{\text{max}}~~</math> are equal to 2. What are the possible ways lines can occupy a volume together? In a plane, as it was with <math>d~~=3</math> ~~(and again assuming no parallel objects in these examples), they must intersect. But in a volume, here in <math>d~~=~~4</math>, this is possible. So, with <math>g=2</math>, it is possible to have a~~ ~~<math>l_{\text{dep}}</math>~~ ~~<math>~~= ~~0</math>, which leads to~~ ~~<math>l_{\text{ind}}</math>~~~~<math>~~ = ~~g</math> <math>~~ - ~~</math>~~ ~~<math>l_{\text{dep}}</math>~~ ~~<math>~~= 2 -~~</math>~~ ~~<math>0</math>~~ ~~<math>~~= ~~2</math>. Not addable in this case.~~

| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |  ↑

~~[[File~~:~~D4 g2 dep0.png~~|~~200px]]~~

| style="border-top: 1px solid #B6321C;" |<math>l_{\text{ind}}=1</math>

|-

~~But we can also imagine two lines in a volume that do intersect at a point. This is the case where~~ <math>l_{\text{ind}}</math>~~<math>~~ = ~~g</math> <math>~~ - ~~</math> <span~~ style="color: #~~3C8031~~;"~~><math>l_{\text{dep}}</math> <math>~~= 2 -~~</math> <span~~ style="color: #~~3C8031~~;"~~><math>1</math> <math>~~= ~~1</math>~~: ~~addable!~~

| rowspan="3" |<math>g_{\text{max}}=3</math>

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

| style="border-bottom: 1px solid #B6321C;" |<math>l_{\text{ind}}=1</math>

|-

|style="background-color: #BED5BA;"|      

| rowspan="2" |

|-

|style="background-color: #BED5BA;"|      

|}

~~[[File:D4 g2 dep1~~.~~png|200px]]~~

This represents a <math>d=4</math> temperament. These diagrams are grade-agnostic, which is to say that they are agnostic as to which side counts the <math>r</math> and which side counts the <math>n</math>. So we are showing them as <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> instead. We could say there's a variation on the rank-nullity theorem whereby <math>g_{\text{min}} + g_{\text{max}}=d</math>, just as <math>r + n=d</math>. So we can then say that this diagram represents either a <math>r=1</math>, <math>n=3</math> temperament, or perhaps a <math>n=1</math>, <math>r=3</math> temperament.

==== At <math>d=5</math>=~~===~~

But actually, this diagram represents more than just a single temperament. It represents a relationship between a pair of temperaments (which have the same [[dimensions]], non-grade-agnostically, i.e. not a pairing of a <math>r=1</math>, <math>n=3</math> temperament with a <math>r=3</math>, <math>n=1</math> temperament). As elsewhere, green coloration indicates the linearly dependent basis vectors <math>L_{\text{dep}}</math> between this pair of temperaments, and red coloration indicates linearly ''in''dependent basis vectors <math>L_{\text{ind}}</math> between the same pair of temperaments.

~~First~~, ~~let's establish~~ the ~~geometric dimension of the space~~. ~~With~~ <math>d=5</math>, we'~~ve got a 4D space (one less than 5)~~, so the ~~entire space can be visualized in a hypervolume. We~~'~~ve now gone beyond~~ the ~~dimensionality of physical reality~~, ~~so things get a little harder to conceptualize unfortunately. But~~ <math>~~d=5~~</math> is the ~~first~~ <math>d</math> ~~where we can make an important point about addability, so please bear with!~~

So, in this case, the two ET maps are linearly independent. This should be unsurprising; because ET maps are constituted by only a single vector (they're <math>r=1</math> by definition), if they ''were'' linearly dependent, then they'd necessarily be the ''same'' exact ET! Temperament arithmetic on two of the same ET is never interesting; <math>T_1</math> plus <math>T_1</math> simply equals <math>T_1</math> again, and <math>T_1</math> minus <math>T_1</math> is undefined. That said, if we ''were'' to represent temperament arithmetic between two of the same temperament on such a diagram as this, then every cell would be green. And this is true regardless whether <math>r=1</math> or otherwise.

~~At <math>d=5</math>~~, we ~~also have a couple options for~~ the ~~grade: either~~ <math>~~g_{\text{min}}~~=1</math> ~~and~~ <~~math~~>~~g_{\text{max}}=4~~</~~math~~>, or <math>g_{\text{~~min~~}}=2</math> ~~and~~ <~~math~~>~~g_{\text{max}}~~=3</~~math~~>.

From this information, we can see that the comma bases of any randomly selected pair of ''different'' <math>d=4</math> ETs are going to share 2 vectors, or in other words, their <math>L_{\text{dep}}</math> will have two basis vectors. In terms of the diagram, we're saying that they'll always have two green-colored vectors under the black bar.

~~First~~ we~~'ll look~~ at <math>g_{\text{~~min~~}}=1</math> ~~and~~ <math>g_{\text{~~max~~}}=4</math>~~. Temperament matrices with~~ <math>~~g=1~~</math> ~~are still addable. And temperament matrices~~ with <~~math~~>g=4</~~math~~> ~~should be too. We can see this visually as how two volumes~~ in ~~a hypervolume together will have an intersection~~ the ~~shape~~ of ~~a plane~~. ~~We can now see that there~~'s ~~a generalizable principal~~ that ~~any~~ two <math>~~(d-1)~~</math>~~-dimensional objects will necessarily have a~~ <math>(d~~-2)~~</math>~~-dimensional intersection, and thus have~~ <math>l_{\text{ind}}</math> <math>= 1</math> ~~and be addable~~. So we ~~won't need to show this one or any further like~~ it~~, either~~.

These diagrams are a good way to understand which temperament relationships are possible and which aren't, where by a "relationship" here we mean a particular combination of their matching dimensions and their linear-independence integer count. A good way to use these diagrams for this purpose is to imagine the red coloration emanating away from the black bar in both directions simultaneously, one pair of vectors at a time. Doing it like this captures the fact, as previously stated, that the <math>l_{\text{ind}}</math> on either side of duality is always equal. There's no notion of a max or min here, as there is with <math>g</math> or <math>l_{\text{dep}}</math>; the <math>l_{\text{ind}}</math> on either side is always the same, so we can capture it with a single number, which counts the red vectors on just one half (that is, half of the total count of red vectors, or half of the width of the red band in the middle of the grid).

There's no need to look at diagrams like this where the black bar is below the center. This is because, even though for convenience we're currently treating the top half as <math>r</math> and the bottom half as <math>n</math>, these diagrams are ultimately grade-agnostic. So we could say that each one essentially represents not just one possibility for the relationship between two temperaments' dimensions and <math>l_{\text{ind}}</math>, but ''two'' such possibilities. Again, this diagram equally represents both <math>d=4, r=1, n=3, </math><math>l_{\text{ind}}=1</math> as well as <math>d=4, r=3, n=1, </math><math>l_{\text{ind}}=1</math>. Which is another way of saying we could vertically mirror it without changing it.

~~So let's look at temperament matrices with <math>g_{\text{min}}=2</math> and <math>g_{\text{max}}=3</math>. For <math>g=2</math>~~, we ~~have two possible values for~~ ~~<math>l_{\text{dep}}</math>~~~~: 0 or 1. Meaning that~~ either the ~~two lines through this hypervolume do not intersect (0)~~, or they ~~intersect at a point (1). These diagrams would look very much like~~ the corresponding diagrams for <math>d=4</math>, so we will not be showing them. But what about when <math>g=3</math>? We can certainly imagine two planes in a hypervolume intersecting at a line, just as they do in an ordinary volume — they're just not taking advantage of ~~the additional geometric dimension~~. So ~~we won't show that example either. But where it gets really interesting~~ is ~~imagining then taking one of~~ these ~~two planes and rotating it in~~ the ~~fifth dimension; this causes the intersection between~~ the ~~two planes to be reduced down to a single point. And this corresponds with the case of~~ <math>l_{\text{~~dep~~}}=1</math> ~~here, which means ~~<math>l_{\text{~~ind~~}}</math>~~<math> = g</math> <math> - </math> ~~<math>l_{\text{~~dep~~}}~~</math~~>~~ <math>= 3 -~~</math> ~~<math>1</math>~~ ~~<math>= 2</math>, so therefore not addable:~~

With the black bar always either in the top half or exactly in the center, we can see that the emanating red band will always either hit the top edge of the square grid first, or they will hit both the top and bottom edges of it simultaneously. So this is how these diagrams visually convey the fact that the <math>l_{\text{ind}}</math> between two temperaments will always be less than or equal to their <math>g_{\text{min}}</math>: because a situation where <math>g_{\text{min}}>l</math> would visually look like the red band spilling past the edges of the square grid.

~~[[File~~:~~D5 g3 dep1~~.~~png|300px]]~~

We could also say that two temperaments are linearly dependent on each other when <math>l_{\text{ind}}</math><math><g_{\text{max}}</math>, that is, their linear-independence is less than their ''max''-grade.

~~So for~~ <math>g_{\text{min}}=2</math> ~~and~~ <math>g_{\text{~~max~~}}=3</math> ~~we got two different possibilities for~~ ~~<math>l_{\text{ind}}</math>~~: 1 ~~and 2~~, and ~~for each~~ of ~~these two possibilities, we found it twice. We can see then that these match up, that is, that the <math>g_{\text{min}}=2</math> case with~~ ~~<math>l_{\text{ind}}=1</math>~~ ~~matches with the <math>g_{\text{max}}=3</math> case with~~ <math>l_{\text{ind}}=1</math>, and the ~~<math>l_{\text{ind}}=2</math> cases match in the same way~~.

Perhaps more importantly, we can also see from these diagrams that any pair of <math>g_{\text{min}}=1</math> temperaments will be addable. Because if they are <math>g_{\text{min}}=1</math>, then the furthest the red band can extend from the black bar is 1 vector, and 1 mirrored set of red vectors means <math>l_{\text{ind}}=1</math>, and that's the definition of addability.

==== ~~Conclusion~~====

====A simple <math>d=3</math> example====

~~Here~~'s a ~~summary table of our geometric findings so far:~~

Let's back-pedal to <math>d=3</math> for a simple illustrative example.

{| class="wikitable ~~center-all~~"

{| class="wikitable"

|+

! rowspan="2" |<math>d</math> ( <math>= g_{\text{min}} ~~+ g_{\text{max}}~~</math>)

| rowspan="3" |<math>d=3</math>

~~! colspan~~="2" |<~~math~~>~~g_{\text{min}}~~</~~math~~>

|style="border-bottom: 3px solid black;"|<math>g_{\text{min}}=1</math>

~~! colspan~~="2" |<~~math~~>~~g_{\text{max}}~~</~~math~~>

| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |  ↑

~~! rowspan~~="2" |~~<math>l_{\text{ind}}</math>~~ ~~( <math>~~= g -~~</math>~~ <math>l_{\text{~~dep~~}}</math>)

| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |  ↑

| style="border-top: 1px solid #B6321C;" |<math>l_{\text{ind}}=1</math>

|-

!<math>g</math>

| rowspan="2" |<math>g_{\text{max}}=2</math>

!<~~math~~>~~l_{\text{dep}}~~<~~/math~~>

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

!<~~math~~>g</~~math~~>

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

!<math>l_{\text{~~dep~~}}</math>

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

| style="border-bottom: 1px solid #B6321C;" |<math>l_{\text{ind}}=1</math>

|-

|2

|style="background-color: #BED5BA;"|      

|1

|style="background-color: #BED5BA;"|      

|0

|style="background-color: #BED5BA;"|      

|1

|

|0

|}

|1

|-

This diagram shows us that any two <math>d=3</math>, <math>g_{\text{min}}=1</math> temperaments (like 5-limit ETs) will be linearly dependent, i.e. their comma bases will share one vector. You may already know this intuitively if you are familiar with the 5-limit [[projective tuning space]] diagram from the [[Paul_Erlich#Papers|Middle Path]] paper, which shows how we can draw a line through any two ETs and that line will represent a temperament, and the single comma that temperament tempers out is this shared vector. The diagram also tells us that any two 5-limit temperaments that temper out only a single comma will also be linearly dependent, for the opposite reason: their ''mappings'' will always share one vector.

|3

|1

And we can see that there are no other diagrams of interest for <math>d=3</math>, because there's no sense in looking at diagrams with no red band, but we can't extend the red band any further than 1 vector on each side without going over the edge, and we can't lower the black bar any further without going below the center. So we're done. And our conclusion is that any pair of different <math>d=3</math> temperaments that are nontrivial (<math>0 < n < d=3</math> and <math>0 < r < d=3</math>) will be addable.

|0

|2

====Completing the suite of <math>d=4</math> examples====

|1

Okay, back to <math>d=4</math>. We've already looked at the <math>g_{\text{min}}=1</math> possibility (which, for any <math>d</math>, there will only ever be one of). So let's start looking at the possibilities where <math>g_{\text{min}}=2</math>, which in the case of <math>d=4</math> leaves us only one pair of values for <math>r</math> and <math>n</math>: both being 2.

{| class="wikitable"

|+

| rowspan="4" |<math>d=4</math>

| rowspan="2" style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=2</math>

|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|      

|style="border-bottom: 1px solid #B6321C;"|

|-

| ~~rowspan~~="3" |4

|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑

| 1

|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑

|0

|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑

|3

|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑

|2

| style="border-top: 1px solid #B6321C;" |<math>l_{\text{ind}}=1</math>

|1

|-

|2

| rowspan="2" |<math>g_{\text{max}}=2</math>

|0

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

|2

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

|0

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

|2

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

| style="border-bottom: 1px solid #B6321C;" |<math>l_{\text{ind}}=1</math>

|-

|2

|style="background-color: #BED5BA;"|      

|1

|style="background-color: #BED5BA;"|      

|2

|style="background-color: #BED5BA;"|      

|1

|style="background-color: #BED5BA;"|      

|1

|

|}

But even with <math>d</math>, <math>r</math>, and <math>n</math> fixed, we still have more than one possibility for <math>L_{\text{dep}}</math>. The above diagram shows <math>l_{\text{ind}}=1</math>. The below diagram shows <math>l_{\text{ind}}=2</math>.

{| class="wikitable"

|+

| rowspan="4" |<math>d=4</math>

| rowspan="2" style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=2</math>

| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |  ↑

| rowspan="2" style="border-top: 1px solid #B6321C;" |<math>l_{\text{ind}}=2</math>

|-

| ~~rowspan~~="3" | 5

|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑

|1

|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑

|0

|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑

|4

|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑

|3

|1

|-

|2

| rowspan="2" |<math>g_{\text{max}}=2</math>

|0

| style="background-color: #E7BBB3;" |  ↓

|3

| style="background-color: #E7BBB3;" |  ↓

|1

| style="background-color: #E7BBB3;" |  ↓

|2

| style="background-color: #E7BBB3;" |  ↓

| rowspan="2" style="border-bottom: 1px solid #B6321C;" |<math>l_{\text{ind}}=2</math>

|-

|2

|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|  ↓

|1

|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|  ↓

|3

|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|  ↓

|2

|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|  ↓

|1

|}

~~The geometric explanation still hasn't answered~~ the ~~question as to why~~ <math>l_{\text{ind}}=1</math> is the ~~condition on addability. It just increased our intuitions about its relationship with temperament dimensions. We'll still look to~~ a ~~later section for an answer on this~~.

In the former possibility, where <math>l_{\text{ind}}=1</math> (and therefore the temperaments are addable), we have a pair of different <math>d=4</math>, <math>r=2</math> temperaments where we can find a single comma that both temperaments temper out, and — equivalently — we can find one ET that supports both temperaments.

=== ~~Algebraic explanation=~~==

In the latter possibility, where <math>l_{\text{ind}}=2</math>, neither side of duality shares any vectors in common. And so we've encountered our first example that is not addable. In other words, if the red band ever extends more than 1 vector away from the black bar, temperament arithmetic is not possible. So <math>d=4</math> is the first time we had enough room (half of <math>d</math>) to support that condition.

~~This explanation relies on comparing~~ the ~~results of the multivector and matrix approaches to temperament arithmetic, and showing algebraically how the matrix approach~~ can ~~only achieve the same answer as the multivector approach on the condition that it keeps all but one vector between the added matrices the same, that is, not only are~~ the ~~temperaments addable, but their~~ ~~<math>L_{\text{dep}}</math>~~ ~~appears explicitly in~~ the ~~added matrices~~.

We have now exhausted the possibility space for <math>d=4</math>. We can't extend either the red band or the black bar any further.

To compare results, we eventually get both approaches into a multivector form. With the multivector approach, we wedge the vector set first and then add the resultant multivectors to get a new multivector. With the matrix approach, we treat the vector set as a matrix and add first, then treat the resultant matrix as a vector set and wedge those vectors to get a new multivector.

====<math>d=5</math> diagrams finally reveal important relationships====

~~The diagrams below are organized into a 2×2 layout. The left part shows the multivector approach, and the right part shows the matrix approach. The top part shows~~ how ~~the results of two approaches match when the ~~<math>~~L_{\text{dep}}~~</math>~~ is successfully explicit~~ (~~and in these cases~~, ~~the ~~<math>L_{\text{~~dep~~}}</math> vectors are highlighted in green and the <math>L_{\text{ind}}</math> vectors are highlighted in red), and the bottom part shows how the results fail to match when it is not. Successful matches are highlighted in yellow and failures to match are highlighted in blue.

So how about we go to <math>d=5</math> (such as the 11-limit). As usual, starting with <math>g_{\text{min}}=1</math>:

{| class="wikitable"

~~This first diagram demonstrates this situation for a <math>d=3, g=2</math> case.~~

{| class="wikitable ~~center-all~~"

|+

!

| rowspan="5" |<math>d=5</math>

!

|style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=1</math>

!

| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |  ↑

~~! colspan~~="11" |

| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |  ↑

!

| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |  ↑

~~! colspan~~="11" |

| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |  ↑

!

| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |  ↑

| style="border-top: 1px solid #B6321C;" |<math>l_{\text{ind}}=1</math>

|-

!

| rowspan="4" |<math>g_{\text{max}}=4</math>

!

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

!

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

| ~~colspan~~="11" ~~rowspan~~="1" |~~'''multivector approach'''~~

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

!

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

| ~~colspan~~="11" ~~rowspan~~="1~~" |'''matrix approach'''~~

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

!

| style="border-bottom: 1px solid #B6321C;" |<math>l_{\text{ind}}=1</math>

|-

!

|style="background-color: #BED5BA;"|      

!

|style="background-color: #BED5BA;"|      

!

|style="background-color: #BED5BA;"|      

~~! colspan~~="11" |

|style="background-color: #BED5BA;"|      

!

|style="background-color: #BED5BA;"|      

~~! colspan~~="11" |

| rowspan="3" |

!

|-

~~! rowspan="5" |~~

|style="background-color: #BED5BA;"|      

~~| colspan="1" rowspan="5" |explicit <math>L_{\text{dep}}</math>~~

|style="background-color: #BED5BA;"|      

~~{{bra|{{vector|<math>a</math> <math>b</math> <math>c</math>}}}}~~

|style="background-color: #BED5BA;"|      

~~! rowspan="5" |~~

|style="background-color: #BED5BA;"|      

| style="background-color: #BED5BA;"|~~<math>a</math>~~

| style="background-color: #BED5BA;"|~~<math>b</math>~~

| style="background-color: #BED5BA;"|~~<math>c</math>~~

~~| rowspan="2" |~~

| style="background-color: #BED5BA;"|~~<math>a</math>~~

| style="background-color: #BED5BA;"|~~<math>b</math>~~

~~| style="background-color: #BED5BA;"|<math>c</math>~~

~~| rowspan="2" |~~

~~| colspan="3" rowspan="2" |~~

~~! rowspan="5" |~~

~~| style="background-color: #BED5BA;"|<math>a</math>~~

~~| style="background-color: #BED5BA;"|<math>b</math>~~

~~| style="background-color: #BED5BA;"|<math>c</math>~~

~~| colspan="1" rowspan="2" |<math>+</math>~~

~~| style="background-color: #BED5BA;"|<math>a</math>~~

~~| style="background-color: #BED5BA;"|<math>b</math>~~

~~| style="background-color: #BED5BA;"|<math>c</math>~~

~~| colspan="1" rowspan="2" |<math>=</math>~~

~~|<math>2a</math>~~

~~|<math>2b</math>~~

~~|<math>2c</math>~~

~~! rowspan="5" |~~

|-

| style="background-color: #~~E7BBB3~~;"|~~<math>d</math>~~

|style="background-color: #BED5BA;"|      

| style="background-color: #~~E7BBB3~~;"|~~<math>e</math>~~

|style="background-color: #BED5BA;"|      

| style="background-color: #~~E7BBB3~~;"|~~<math>f</math>~~

|style="background-color: #BED5BA;"|      

| style="background-color: #~~E7BBB3~~;"|~~<math>g</math>~~

|style="background-color: #BED5BA;"|      

| style="background-color: #~~E7BBB3~~;"|<math>h</math>

|style="background-color: #BED5BA;"|      

| style="~~background-~~color: #~~E7BBB3~~;"|<math>i</math>

|}

| ~~style~~="~~background-color: #E7BBB3;~~"|<math>d</math>

| style="~~background~~-~~color~~: ~~#E7BBB3~~;"|<math>e</math>

Just as with the <math>l_{\text{ind}}=1</math> diagrams given for <math>d=3</math> and <math>d=5</math>, we can see these are addable temperaments.

| style="background-color: #~~E7BBB3~~;"|~~<math>f</math>~~

| style="background-color: #~~E7BBB3~~;"|~~<math>g</math>~~

Now let's look at <math>d=5</math> but with <math>g_{\text{min}}=2</math>. This presents two possibilities. First, <math>l_{\text{ind}}=1</math>:

| style="background-color: #~~E7BBB3~~;"|~~<math>h</math>~~

| style="background-color: #~~E7BBB3~~;"|~~<math>i</math>~~

{| class="wikitable"

|~~<math>d+g</math>~~

|+

|~~<math>e+h</math>~~

| rowspan="5" |<math>d=5</math>

|~~<math>f+i</math>~~

| rowspan="2" style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=2</math>

|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|      

| style="border-bottom: 1px solid #B6321C;" |

|-

| ~~colspan~~="3" ~~rowspan~~="1" ~~|<math~~>∧</~~math~~>

|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑

|

|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑

| ~~colspan~~="3" ~~rowspan~~="1" |<~~math~~>∧</~~math~~>

|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑

|

|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑

| ~~colspan~~="3" |

|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑

| ~~colspan~~="3" |

| style="border-top: 1px solid #B6321C;" |<math>l_{\text{ind}}=1</math>

|

~~| colspan~~="3" |

|

| ~~colspan~~="3" ~~rowspan~~="1" |<math>∧</math>

|-

|<math>~~bf-ce~~</math>

| rowspan="3" |<math>g_{\text{max}}=3</math>

|~~<math>af~~-~~cd</math>~~

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

~~|<math>ae~~-~~bd</math>~~

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

|<~~math~~> +</~~math~~>

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

|~~<math>bi~~-~~ch</math>~~

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

|<~~math~~>~~ai-cg~~</~~math~~>

| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |  ↓

|~~<math>ah~~-~~bg</math>~~

| style="border-bottom: 1px solid #B6321C;" |<math>l_{\text{ind}}=1</math>

|<~~math>~~=~~</math>~~

~~|<math~~>~~bf-ce+bi-ch~~</~~math~~>

|~~<math>af~~-~~cd+ai~~-~~cg</math>~~

|<~~math~~>~~ae-bd+ah-bg~~</~~math~~>

| ~~colspan~~="~~3" rowspan="2~~" |

~~| rowspan~~="2" |

| ~~colspan~~="~~3" rowspan="2~~" |

~~| rowspan~~="2" |

~~|<math~~>~~2b(f+i)-2c(e+h)~~</math>

~~|<math>2a(f+i)-2c(d+g)~~</math>

~~|<math>2a(e+h)-2b(d+g)~~</~~math~~>

|-

| ~~colspan~~="3" |

|style="background-color: #BED5BA;"|      

|

| ~~colspan~~="3" |

|-

|style="background-color: #BED5BA;"|      

|

| style="~~background-~~color: ~~LightYellow~~;"|<math>~~b(f+i)~~-~~c(e+h)~~</math>

|}

| style="background-color: ~~LightYellow~~;"|<~~math~~>~~a(f+i)-c(d+g)~~</~~math~~>

| style="background-color: ~~LightYellow~~;"|<~~math~~>~~a(e+h)-b(d+g)~~</~~math~~>

And second, <math>l_{\text{ind}}=2</math>:

| style="background-color: ~~LightYellow~~;"|<~~math~~>~~b(f+i)-c(e+h)~~</~~math~~>

| style="background-color: ~~LightYellow~~;"|<~~math~~>~~a(f+i)-c(d+g)~~</~~math~~>

{| class="wikitable"

| style="background-color: ~~LightYellow~~;"|<math>~~a(e+h)-b(d+g)~~</math>

|+

|-

| rowspan="5" |<math>d=5</math>

!

| rowspan="2" style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=2</math>

!

| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |  ↑

!

| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |  ↑

~~! colspan~~="11" |

| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |  ↑

!

| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |  ↑

~~! colspan~~="11" |

| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |  ↑

!

| rowspan="2" style="border-top: 1px solid #B6321C;" |<math>l_{\text{ind}}=2</math>

|-

|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|  ↑

|-

! rowspan="5" |

| rowspan="3" |<math>g_{\text{max}}=3</math>

| ~~rowspan~~="5" |~~hidden~~ <~~math~~>~~L_{\text{dep}}~~</~~math~~>

| style="background-color: #E7BBB3;" |  ↓

~~! rowspan~~="5" |

| style="background-color: #E7BBB3;" |  ↓

|<~~math~~>a</~~math~~>

| style="background-color: #E7BBB3;" |  ↓

|<math>b</math>

| style="background-color: #E7BBB3;" |  ↓

|<~~math~~>c</~~math~~>

| style="background-color: #E7BBB3;" |  ↓

| ~~rowspan~~="2" |

| rowspan="2" style="border-bottom: 1px solid #B6321C;" |<math>l_{\text{ind}}=2</math>

|<~~math~~>j</~~math~~>

|-

|<~~math~~>k</~~math~~>

|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|  ↓

|<~~math~~>l</~~math~~>

|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|  ↓

|-

|style="background-color: #BED5BA;"|      

|

| ~~colspan~~="3" ~~rowspan~~="2" |

|}

~~! rowspan~~="5" |

|<~~math~~>a</~~math~~>

Here's where things really get interesting. Because in both of these cases, the pairs of temperaments represented are linearly dependent on each other (i.e. either their mappings are linearly dependent, their comma bases are linearly dependent, or both). And so far, every possibility where temperaments have been linearly dependent, they have also been <math>l_{\text{ind}}=1</math>, and therefore addable. But if you look at the second case here, we are <math>l_{\text{ind}}=2</math>, but since <math>d=5</math>, the temperaments still manage to be linearly dependent. So this is the first example of a linearly dependent temperament pairing which is not addable.

|<~~math~~>b</math>

|<math>c</~~math~~>

====Back to <math>d=2</math>, for a surprisingly tricky example====

~~| colspan~~="1" ~~rowspan~~="2~~" |~~<math>+</~~math~~>

|<math>j</math>

Beyond <math>d=5</math>, these diagrams get cumbersome to prepare, and cease to reveal further insights. But if we step back down to <math>d=2</math>, a place simpler than anywhere we've looked so far, we actually find another surprisingly tricky example, which is hopefully still illuminating.

|<~~math~~>k</~~math~~>

|<~~math~~>l</~~math~~>

So <math>d=2</math> (such as the 3-limit) presents another case — similar to the <math>d=5</math>, <math>g_{\text{min}}=2</math>, <math>l_{\text{ind}}=2</math> case explored most recently above — where the properties of linearly dependence and addability do not match each other. But while in the other case, we had a temperament pair that was linearly dependent yet not addable, in this <math>d=2</math> (and therefore <math>g_{\text{min}}=1</math>, <math>l_{\text{ind}}=1</math>) case, it is the other way around: addable yet linearly independent!

~~| colspan~~=~~"1" rowspan~~=~~"2" |~~<math>=</math>

|<math>~~a+j~~</math>

{| class="wikitable"

|<math>~~b+k~~</math>

|+

|<math>~~c+l~~</math>

| rowspan="2" |<math>d=2</math>

~~! rowspan="5" |~~

|style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=1</math>

|-

|style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;"|  ↑

|<math>d</math>

|style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;"|  ↑

|<math>e</math>

| style="border-top: 1px solid #B6321C;" |<math>l_{\text{ind}}=1</math>

|<math>f</math>

|<~~math~~>g</~~math~~>

|<~~math~~>h</~~math~~>

|<math>i</math>

|<math></math>

|<~~math~~>d</math>

|<math>e</~~math~~>

|<~~math~~>f</~~math~~>

|<math>g</math>

|<math>h</math>

|<~~math~~>i</~~math~~>

|<~~math~~>~~d+g~~</~~math~~>

|<~~math~~>~~e+h~~</math>

|<math>~~f+i~~</~~math~~>

|-

~~| colspan="3" rowspan="1"~~ |<math>∧</math>

|<math>g_{\text{max}}=1</math>

|

|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|  ↓

~~| colspan~~="3" ~~rowspan~~="1" ~~|<math~~>∧</~~math~~>

|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|  ↓

|

| style="border-bottom: 1px solid #B6321C;" |<math>l_{\text{ind}}=1</math>

~~| colspan~~="3" |

|}

~~| colspan~~="3" |

|

Basically, in the case of <math>d=2</math>, <math>g_{\text{max}}=1</math> (in non-trivial cases, i.e. not JI or the unison temperament), so any two different ETs or commas you pick are going to be linearly independent (because the only way they could be linearly dependent would be to be the same temperament). And yet we know we can still entry-wise add them to new vectors that are [[Douglas_Blumeyer_and_Dave_Keenan%27s_Intro_to_exterior_algebra_for_RTT#Decomposability|decomposable]], because they're already vectors (decomposing means to express a [[Douglas_Blumeyer_and_Dave_Keenan%27s_Intro_to_exterior_algebra_for_RTT#From_vectors_to_multivectors|multivector]] in the form of a list of monovectors, so decomposing a multivector that's already a monovector like this is tantamount to merely putting array braces around it.)

| ~~colspan~~="3" |

|

====Conclusion====

~~| colspan~~="3" ~~rowspan~~="1~~" |~~<math>∧</~~math~~>

|-

This explanation has hopefully helped get a grip on how addability AKA <math>l_{\text{ind}}=1</math> behaves given various temperament dimensions. But it still hasn't quite explained why <math>l_{\text{ind}}=1</math> is one and the same thing as addability. We will look at this in another section soon.

|<math>~~bf-ce~~</math>

|<math>~~af-cd~~</math>

===Geometric explanation===

|<~~math~~>~~ae-bd~~</~~math~~>

|<~~math~~>+</~~math~~>

We've presented a diagrammatic illustration of the behavior of linear-independence <math>l_{\text{ind}}</math> with respect to temperament dimensions. But some of the results might have seemed surprising. For instance, when looking at the diagram for <math>d=4, g_{\text{min}}=1, g_{\text{max}}=3</math>, it might have seemed intuitive enough that the the red band could not extend beyond the square grid, but then again, why shouldn't it be possible to have, say, two 7-limit ETs which temper out only a single comma in common? Perhaps it doesn't seem clear that this is impossible, and that they must temper out two commas in common (and of course the infinitude of combinations of these two commas). If this is as unclear to you as it was to the author when exploring this topic, then this explanatory section is for you! Here, we will use geometrical representations of temperaments to hone our intuitions about the possible combinations of dimensions and linear-independence <math>l_{\text{ind}}</math> of temperaments.

|<~~math~~>~~ki-lh~~</math>

|<math>~~ji-lg~~</~~math~~>

In this approach, we’re actually not going to focus directly on the linear-independence <math>l_{\text{ind}}</math> of temperaments. Instead, we're going to look at the linear-''de''pendence <math>l_{\text{dep}}</math> of matrices representing temperaments such as mappings and comma bases, and then compute the linear-independence <math>l_{\text{ind}}</math> from it and the grade <math>g</math>. As we’ve established, the linear-dependence <math>l_{\text{dep}}</math> differs from one side of duality to the other, so we’ll only be looking at one side of duality at a time.

|<~~math~~>~~jh-kg~~</math>

|<math>=</~~math~~>

====Introduction====

|<~~math~~>bf-~~ce+ki-lh~~</math>

|<math>~~af-cd+ji-lg~~</~~math~~>

In this geometric approach, we'll be imagining individual vectors as points (0D), sets of two vectors as lines (1D), sets of three as planes (2D), four as volumes (3D), and so forth, as according to this table:

|<math>~~ae-bd+jh-kg~~</math>

{| class="wikitable center-all"

~~| colspan~~="3" ~~rowspan~~="2" |

|+

~~| rowspan="2" |~~

!vector

~~| colspan~~="3" ~~rowspan~~="2" |

count

~~| rowspan~~="2" |

!geometric

|<math>~~(b+k)(f+i)-(c+l)(e+h)~~</math>

dimension

|<math>~~(a+j)(f+i)-(c+l)(d+~~g)</math>

!form

|<math>(~~a+j~~)(~~e+h~~)-(~~b+k~~)(d+~~g)</math>~~

|-

| ~~colspan="3"~~ |

|0

|

|undefined

| ~~colspan="3" |~~

|(emptiness)

|

|-

| ~~style="background-color: LightBlue;"|<math>bf-ce+ki-lh</math>~~

|1

| ~~style="background-color: LightBlue;"|<math>af-cd+ji-lg</math>~~

|0

| ~~style="background-color: LightBlue;"|<math>ae-bd+jh~~-~~kg</math>~~

|point

| ~~style="background-color: LightBlue;"|<math>bf+bi+kf+ki-ce-ch-le-lh</math>~~

|-

| ~~style="background-color: LightBlue;"|<math>af+ai+jf+ji-cd-cg-ld-lg</math>~~

|2

| ~~style="background-color: LightBlue;"|<math>ae+ah+je+jh-bd-bg-kd-kg</math>~~

|1

|line

|-

!

|3

!

|2

!

|plane

~~! colspan="11" |~~

!

~~! colspan="11"~~ |

!

|}

~~This second diagram demonstrates this situation for a <math>d=5, g=~~3~~</math> case. One pair of the <math>L_{\text{dep}}</math> vectors are explicitly matching, but not the other, which isn't enough.~~

~~{| class="wikitable center-all"~~

|+

!

~~! colspan="~~2~~" |~~

!

~~! colspan="32"~~ |

!

~~! colspan="22" |~~

!

|-

!

|4

~~! colspan="2"~~ |

|3

!

|volume

~~| colspan="32" rowspan="1"~~ |~~'''multivector approach'''~~

!

| ~~colspan="22" rowspan="1" |'''matrix approach'''~~

!

|-

!

|5

~~! colspan="2"~~ |

|4

!

|hypervolume

~~! colspan="32"~~ |

!

~~! colspan="22"~~ |

!

|-

~~! rowspan="7"~~ |

| ⋮

| ~~colspan=~~"1" ~~rowspan=~~"7" |~~explicit <math>L_~~{~~\text~~{~~dep~~}}~~</math>~~

|⋮

|}

This is a "vector space", and these geometric dimensions are consistent with how temperaments represented by these counts of vectors appear in ''projective'' vector space, which reduces geometric dimensions by 1. For example, a vector has a geometric interpretation as a directed line segment, which is 1D, but a point is 0D, which is one dimension lower. Essentially what we're doing is ''assuming the origin''.

Think of it this way: geometric points are zero-dimensional, simply representing a position in space, whereas linear algebra vectors are one-dimensional, representing both a magnitude and direction; the way vectors manage to encode this extra dimension without providing any additional information is by being understood to describe this position in space ''relative to an origin''. Well, so we'll now switch our interpretation of these objects to the geometric one, where the vector's entries are nothing more than a coordinate for a point in space. And the "projection" involved in projective vector space essentially positions us at this discarded origin, looking out from it upon every individual point, which accomplishes the same feat, in a visual way.

Perhaps an example may help clarify this setup. Suppose we've got an (x,y,z) space, and two coordinates (5,8,12) and (7,11,16). You should recognize these as the simple maps for 5-ET and 7-ET, usually written as {{map|5 8 12}} and {{map|7 11 16}}, respectively. Ask for the equation of the plane defined by the three points (5,8,12), (7,11,16), and the origin (0,0,0) and you'll get -4x + 4y -1z = 0, which clearly shows us the entries of the meantone comma. That's because meantone temperament can be defined by these two maps. 5-limit JI is a 3D space, and meantone temperament, as a rank-2 temperament, would be a 2D plane. But we don't normally need to think of the map corresponding to the origin, where everything is tempered out, including meantone. So we can just assume it, and think of a 2D plane as being defined by only 2 points, which in a view projected (from the origin) will look like just the line connecting (5,8,12) and (7,11,16).

~~⟨{{~~vector|<~~math~~>a</~~math~~> <~~math~~>b</~~math~~> <~~math~~>c</~~math~~> <~~math~~>d</math> <math>e</~~math~~>}}

So, we've set the stage for our projective vector spaces. We will now be looking at representations of temperaments as counts of vector sets, and then using this scheme to convert them to primitive geometric forms. We'll place two of each form into the space, representing the two temperaments having the possibility of arithmetic performed on them checked. Then we will observe their possible ''intersections'' depending on how they're oriented in space, and it's these intersections that represent their linear-dependence. When the dimension of the intersection is then converted back to a vector set count, then we have their linear-dependence <math>l_{\text{dep}}</math>, for this side of duality, anyway (remember, unlike the linear-independence <math>l_{\text{ind}}</math>, this value isn't necessarily the same on both sides of duality). We can finally subtract the linear-dependence from the grade (vector count) to get the linear-indepedence, in order to determine if the two temperaments are addable.

~~{{vector|~~<~~math~~>f</math> <math>g</~~math~~> <~~math~~>h</~~math> <math>i</math~~> <~~math>j</math>}}]~~

| style="~~background-~~color: #~~BED5BA~~;"~~|<math~~>~~r_1~~</~~math~~>

In these examples, we'll be assuming that no two temperaments being compared are the same, because performing temperament arithmetic on copies of the same temperament is not interesting. The other things we'll be assuming is that no lines, planes, etc. are parallel to each other; this is due to a strange effect touched upon in footnote 4 whereby temperament geometry that appears parallel in projective space actually still intersects; the present author asks that if anyone is able to demystify this situation, that they please do!

! ~~rowspan~~=~~"7" |~~

~~| style~~=~~"background-color: #BED5BA;" colspan~~=~~"2" rowspan~~=~~"1" |~~<math>a</math>

====At <math>d=3</math>====

~~| style~~=~~"background-color: #BED5BA;" colspan~~=~~"2" rowspan~~=~~"1" |~~<math>b</math>

~~| style="background-color: #BED5BA;" colspan="~~2~~" rowspan="1" |~~<math>c</math>

First, let's establish the geometric dimension of the space. With <math>d=3</math>, we've got a 2D space (one less than 3), so the entire space can be visualized on a plane.

| style="~~background-~~color: #~~BED5BA~~;" ~~colspan~~="2" ~~rowspan="1" |~~<math>d</math>

| style="~~background-~~color: #~~BED5BA~~;" ~~colspan="2" rowspan="1" |~~<math>e</math>

Our only possible values for <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> here are 1 and 2, respectively. So these are the two possible counts of vectors <math>g</math> possessed by matrices representing temperaments here.

~~| rowspan~~=~~"3" |~~

| style="~~background-~~color: #~~BED5BA~~;" ~~colspan="2" rowspan="1" |~~<math>a</math>

So let's look at temperaments represented by matrices with 1 vector first (<math>g=1</math>). In this case, each of the two temperaments is a point in the plane. Unless these two temperaments are the same temperament, the intersection of these two points is empty. Emptiness isn't even 0D! So that tells us that these temperaments have 0 vectors worth of linear dependence. With <math>g=1</math>, that gives us a <math>l_{\text{ind}}</math><math> = g</math> <math> - </math> <math>l_{\text{dep}}</math> <math>= 1 -</math> <math>0</math> <math>= 1</math>:

| style="~~background-~~color: #~~BED5BA~~;" ~~colspan="2" rowspan="1" |~~<math>b</math>

| ~~style~~="~~background-~~color: #~~BED5BA~~;" ~~colspan="2" rowspan="1" |<math~~>c</~~math~~>

[[File:D3 g1 dep0.png|200px|none]]

| style="~~background-~~color: #~~BED5BA~~;" ~~colspan="2" rowspan="1" |<math~~>d</~~math~~>

| style="~~background-~~color: #~~BED5BA~~;" ~~colspan="2" rowspan="1" |~~<math>e</math>

Next, let's look at temperaments represented by matrices with 2 vectors (<math>g=2</math>). In this case, each of the two temperaments is a line in the plane. Again, assuming the two lines are not the same line or parallel, their intersection is a point. Being 0D, that tells us that the linear-dependence of these matrices is 1. So that gives us an <math>l_{\text{ind}}</math> <math>= g</math> <math>-</math> <math>l_{\text{dep}}</math> <math>= 2 -</math> <math>1</math> <math>= 1</math>. This matches the value we found via the <math>g=1</math>, so we've effectively checked our work:

~~| rowspan~~=~~"3" |~~

~~| colspan~~="10" ~~rowspan="3" |~~

[[File:D3 g2 dep1.png|200px|none]]

~~! rowspan="7" |~~

~~| style="background-color: #BED5BA;"|~~<math>a</math>

====At <math>d=2</math>====

~~| style="background~~-~~color: #BED5BA;"|~~<math>b<~~/math>~~

| style="~~background-~~color: #~~BED5BA~~;"|<math>c</math>

Let's step back to <math>d=2</math>. Here we've got a 2 minus 1 equals 1D space, so the entire space can be visualized on a single line (one direction corresponds to an increasing ratio between the two coordinates, and the other to a decreasing ratio).

~~| style="background-color: #BED5BA;"|<~~math>d</math>

~~| style="background-color: #BED5BA;"|~~<math>e</math>

We know our only possible value for <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> here is 1. So in either case, each of the two temperaments is a point on the line. As with two points in a plane — when <math>d=3</math> — unless these two temperaments are the same temperament, the intersection of these two points is empty. So again the <math>l_{\text{ind}}</math><math> = g = 1</math>:

| ~~colspan~~=~~"1" rowspan~~=~~"3" |+~~

~~| style~~=~~"background-color: #BED5BA;"|<math~~>a</math>

[[File:D2 g1 dep0.png|200px|none]]

~~| style="background-color: #BED5BA;"|~~<math>b</math>

~~| style="background-color: #BED5BA;"|~~<math>c</math>

====At <math>d=4</math>====

~~| style="background-color: #BED5BA;"|~~<math>d</math>

| style="~~background-~~color: #~~BED5BA~~;"|<math>e</math>

First, let's establish the geometric dimension of the space. With <math>d=4</math>, we've got a 3D space (one less than 4), so the entire space can be visualized in a volume.

~~| colspan="1" rowspan="3" |~~<math>=</math>

| ~~colspan~~=~~"2" rowspan~~=~~"1" |~~<math>2a</math>

At <math>d=4</math>, we have a couple options for the grade: either <math>g_{\text{min}}=1</math> and <math>g_{\text{max}}=3</math>, or both <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> equal 2.

~~| colspan~~=~~"2" rowspan~~=~~"1" |~~<math>2b</math>

~~| colspan="2" rowspan="1" |~~<math>2c</math>

Let's look at temperaments represented by matrices with 1 vector first (<math>g=1</math>). Yet again, we find ourselves with two separate points, but now we find them in a space that's not a line, not a plane, but a volume. This doesn't change <math>l_{\text{ind}}</math><math> = g = 1</math>, so we're not even going to show it, or any further cases of <math>g=1</math>. These are all addable.

~~| colspan~~=~~"2" rowspan="~~1~~" |~~<math>2d</math>

~~| colspan~~=~~"2" rowspan="1" |~~<math>2e</math>

And when <math>g=3</math>, because this is paired with <math>g=1</math> from the min and max values, we should expect to get the same answer as with <math>g=1</math>. And indeed, it will check out that way. Because two <math>g=3</math> temperaments will be planes in this volume, and the intersection of two planes is a line. Which means that <math>l_{\text{dep}}</math><math> = 2</math>. And so <math>l_{\text{ind}}</math><math> = g</math> <math> - </math> <math>l_{\text{dep}}</math> <math>= 3 -</math> <math>2</math> <math>= 1</math>. And here's where our geometric approach begins to pay off! This was the example given at the beginning that might seem unintuitive when relying only on the diagrammatic approach. But here we can see clearly that there would be no way for two planes in a volume to intersect only at a point, which proves the fact that two 7-limit ETs could never only temper out a single comma in common.

~~! rowspan~~=~~"7" |~~

|-

[[File:D4 g3 dep2.png|200px]]

| style="~~background-~~color: #~~BED5BA~~;"|<math>~~r_2~~</math>

~~| style~~=~~"background-color: #BED5BA;" colspan~~=~~"2" rowspan~~="1~~" |~~<math>f</math>

Next let's look at temperaments represented by matrices with 2 vectors, that is, when both <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> are equal to 2. What are the possible ways lines can occupy a volume together? In a plane, as it was with <math>d=3</math> (and again assuming no parallel objects in these examples), they must intersect. But in a volume, here in <math>d=4</math>, this is possible. So, with <math>g=2</math>, it is possible to have a <math>l_{\text{dep}}</math> <math>= 0</math>, which leads to <math>l_{\text{ind}}</math><math> = g</math> <math> - </math> <math>l_{\text{dep}}</math> <math>= 2 -</math> <math>0</math> <math>= 2</math>. Not addable in this case.

~~| style~~=~~"background-color: #BED5BA;" colspan="2" rowspan="~~1~~" |~~<math>g</math>

| style="~~background-~~color: #~~BED5BA~~;" ~~colspan="2" rowspan="1" |~~<math>h</math>

[[File:D4 g2 dep0.png|200px]]

| style="~~background-~~color: #~~BED5BA~~;" ~~colspan="2" rowspan="1" |<math~~>i</math>

~~| style="background-color: #BED5BA;" colspan="2" rowspan="1" |~~<math>j</~~math~~>

But we can also imagine two lines in a volume that do intersect at a point. This is the case where <math>l_{\text{ind}}</math><math> = g</math> <math> - </math> <math>l_{\text{dep}}</math> <math>= 2 -</math> <math>1</math> <math>= 1</math>: addable!

| style="~~background-~~color: #~~BED5BA~~;" ~~colspan="2" rowspan="1" |~~<math>f</math>

| style="~~background-~~color: #~~BED5BA~~;" ~~colspan="~~2~~" rowspan="1" |~~<math>g</math>

[[File:D4 g2 dep1.png|200px]]

~~| style~~=~~"background-color: #BED5BA;" colspan="~~2~~" rowspan="1" |~~<math>h</math>

~~| style="background-color: #BED5BA;" colspan="2" rowspan="1" |~~<math>i</math>

==== At <math>d=5</math>====

~~| style="background-color: #BED5BA;" colspan="~~2~~" rowspan="1" |~~<math>j</math>

~~| style~~=~~"background-color: #BED5BA;" |~~<math>f</math>

First, let's establish the geometric dimension of the space. With <math>d=5</math>, we've got a 4D space (one less than 5), so the entire space can be visualized in a hypervolume. We've now gone beyond the dimensionality of physical reality, so things get a little harder to conceptualize unfortunately. But <math>d=5</math> is the first <math>d</math> where we can make an important point about addability, so please bear with!

| style="~~background-~~color: #~~BED5BA~~;" |<math>g</math>

| style="~~background-~~color: #~~BED5BA~~;" |<math>h</math>

At <math>d=5</math>, we also have a couple options for the grade: either <math>g_{\text{min}}=1</math> and <math>g_{\text{max}}=4</math>, or <math>g_{\text{min}}=2</math> and <math>g_{\text{max}}=3</math>.

~~| style~~=~~"background-color: #BED5BA;" |~~<math>i</math>

~~| style~~="~~background-~~color: #~~BED5BA~~;" |<math>j</math>

First we'll look at <math>g_{\text{min}}=1</math> and <math>g_{\text{max}}=4</math>. Temperament matrices with <math>g=1</math> are still addable. And temperament matrices with <math>g=4</math> should be too. We can see this visually as how two volumes in a hypervolume together will have an intersection the shape of a plane. We can now see that there's a generalizable principal that any two <math>(d-1)</math>-dimensional objects will necessarily have a <math>(d-2)</math>-dimensional intersection, and thus have <math>l_{\text{ind}}</math> <math>= 1</math> and be addable. So we won't need to show this one or any further like it, either.

| style="~~background-~~color: #~~BED5BA~~;" |<math>f</math>

| style="~~background-~~color: #~~BED5BA~~;" ~~|<math~~>g</math>

So let's look at temperament matrices with <math>g_{\text{min}}=2</math> and <math>g_{\text{max}}=3</math>. For <math>g=2</math>, we have two possible values for <math>l_{\text{dep}}</math>: 0 or 1. Meaning that either the two lines through this hypervolume do not intersect (0), or they intersect at a point (1). These diagrams would look very much like the corresponding diagrams for <math>d=4</math>, so we will not be showing them. But what about when <math>g=3</math>? We can certainly imagine two planes in a hypervolume intersecting at a line, just as they do in an ordinary volume — they're just not taking advantage of the additional geometric dimension. So we won't show that example either. But where it gets really interesting is imagining then taking one of these two planes and rotating it in the fifth dimension; this causes the intersection between the two planes to be reduced down to a single point. And this corresponds with the case of <math>l_{\text{dep}}=1</math> here, which means <math>l_{\text{ind}}</math><math> = g</math> <math> - </math> <math>l_{\text{dep}}</math> <math>= 3 -</math> <math>1</math> <math>= 2</math>, so therefore not addable:

~~| style="background-color: #BED5BA;" |~~<math>h</math>

~~| style~~=~~"background-color: #BED5BA;" |~~<math>i</math>

[[File:D5 g3 dep1.png|300px]]

| style="~~background-~~color: #~~BED5BA~~;" |<math>j</math>

~~| colspan~~="2" ~~rowspan=~~"1~~" |~~<math>2f</math>

So for <math>g_{\text{min}}=2</math> and <math>g_{\text{max}}=3</math> we got two different possibilities for <math>l_{\text{ind}}</math>: 1 and 2, and for each of these two possibilities, we found it twice. We can see then that these match up, that is, that the <math>g_{\text{min}}=2</math> case with <math>l_{\text{ind}}=1</math> matches with the <math>g_{\text{max}}=3</math> case with <math>l_{\text{ind}}=1</math>, and the <math>l_{\text{ind}}=2</math> cases match in the same way.

~~| colspan~~=~~"2" rowspan="1" |<math>2g~~</math>

| ~~colspan~~=~~"2" rowspan~~=~~"1" |~~<math>2h</math>

==== Conclusion====

~~| colspan~~="~~2" rowspan="1" |<math>2i~~</math>

~~| colspan="2" rowspan="1" |<math>2j~~</~~math~~>

Here's a summary table of our geometric findings so far:

{| class="wikitable center-all"

|+

! rowspan="2" |<math>d</math> ( <math>= g_{\text{min}} + g_{\text{max}}</math>)

! colspan="2" |<math>g_{\text{min}}</math>

! colspan="2" |<math>g_{\text{max}}</math>

! rowspan="2" |<math>l_{\text{ind}}</math> ( <math>= g -</math> <math>l_{\text{dep}}</math>)

|-

~~| style="background-color: #E7BBB3;"|<math>r_3</math>~~

!<math>g</math>

~~| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>k</math>~~

!<math>l_{\text{dep}}</math>

~~| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>l</math>~~

!<math>g</math>

~~| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>m</math>~~

!<math>l_{\text{dep}}</math>

~~| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>n</math>~~

~~| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>o</math>~~

~~| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>p</math>~~

~~| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>q</math>~~

~~| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>r</math>~~

~~| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>s</math>~~

~~| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>t</math>~~

~~| style="background-color: #E7BBB3;"|<math>k</math>~~

~~| style="background-color: #E7BBB3;"|~~<math>l</math>

~~| style="background-color: #E7BBB3;"|<math>m~~<~~/math>~~

| style="~~background-~~color: #~~E7BBB3~~;"~~|<math>n</math~~>

~~| style="background-color: #E7BBB3;"|<math>o</math>~~

~~| style="background-color: #E7BBB3;"|~~<math>p</math>

~~| style="background-color: #E7BBB3;"|<math>q~~</~~math~~>

~~| style="background-color: #E7BBB3;"|~~<math>r</math>

~~| style="background-color: #E7BBB3;"|~~<~~math>s</math>~~

| style="~~background-~~color: #~~E7BBB3~~;"~~|<math>t</math>~~

~~| colspan="2" rowspan="1" |<math>k+p</math>~~

~~| colspan="2" rowspan="1" |<math~~>~~l+q~~</math>

~~| colspan="2" rowspan="1" |<math>m+r~~</math>

~~| colspan="2" rowspan="1" |<math>n+s~~</~~math>~~

~~| colspan="2" rowspan="1" |<math>o+t</math~~>

|-

|

|2

| ~~colspan="10" rowspan="~~1~~" |<math>∧</math>~~

|1

|

|0

| ~~colspan="10" rowspan="~~1~~" |<math>∧</math>~~

|1

|

|0

~~| colspan="10" |~~

|1

~~| colspan="5" |~~

|

~~| colspan="5" |~~

|

| ~~colspan="10" rowspan="~~1~~" |<math>∧</math>~~

|-

|~~<math>r_1∧r_2</math>~~

|3

| ~~rowspan="2" |<math>ag-bf</math>~~

|1

| ~~rowspan="2" |<math>ah-cf</math>~~

|0

| ~~rowspan="~~2~~" |<math>ai-df</math>~~

|2

| ~~rowspan="2" |<math>aj-ef</math>~~

|1

| ~~rowspan="2" |<math>bh-cg</math>~~

|1

| ~~rowspan="2" |<math>bi~~-~~dg</math>~~

|-

| rowspan="2" |~~<math>bj-eg</math>~~

| rowspan="3" |4

| ~~rowspan="2" |<math>ci-dh</math>~~

| 1

| ~~rowspan="2" |<math>cj-eh</math>~~

|0

| ~~rowspan="2" |<math>dj-ei</math>~~

|3

| ~~rowspan="~~2~~" |~~

|2

~~| rowspan="2" |<math>ag-bf</math>~~

|1

~~| rowspan="2" |<math>ah-cf</math>~~

|-

~~| rowspan="2" |<math>ai-df</math>~~

|2

~~| rowspan="2" |<math>aj-ef</math>~~

|0

~~| rowspan="2" |<math>bh-cg</math>~~

|2

| ~~rowspan="2" |<math>bi-dg</math>~~

|0

| ~~rowspan="2" |<math>bj~~-~~eg</math>~~

|2

| ~~rowspan="~~2~~" |<math>ci-dh</math>~~

| ~~rowspan="2" |<math>cj-eh</math>~~

| ~~rowspan="~~2~~" |<math>dj-ei</math>~~

| ~~rowspan="2" |~~

| ~~colspan="10" rowspan="~~2~~" |~~

~~| colspan="5" rowspan="3" |~~

~~| rowspan="3" |~~

~~| colspan="5" rowspan="3" |~~

~~| rowspan="3" |~~

~~|<math>4ag-4bf</math>~~

~~|<math>4ah-4cf</math>~~

~~|<math>4ai-4df</math>~~

~~|<math>4aj-4ef</math>~~

~~|<math>4bh-4cg</math>~~

~~|<math>4bi-4dg</math>~~

~~|<math>4bj-4eg</math>~~

~~|<math>4ci-4dh</math>~~

~~|<math>4cj-4eh</math>~~

~~|<math>4dj-4ei</math>~~

|-

|~~simplify <math>r_1∧r_2</math> if necessary~~

|2

|~~<math>ag-bf</math>~~

|1

|~~<math>ah-cf</math>~~

|2

|~~<math>ai-df</math>~~

|1

|~~<math>aj~~-~~ef</math>~~

|1

|~~<math>bh-cg</math>~~

|-

|~~<math>bi-dg</math>~~

| rowspan="3" | 5

|~~<math>bj-eg</math>~~

|1

|~~<math>ci-dh</math>~~

|0

|~~<math>cj-eh</math>~~

|4

|~~<math>dj-ei</math>~~

|3

|1

|-

|~~<math>(r_1∧r_2)∧r_3</math>~~

|2

~~|<math>k(bh-cg)\\-l(ah-cf)\\+m(ag-bf)</math>~~

|0

~~|<math>k(bi-dg)\\-l(ai-df)\\+n(ag-bf)</math>~~

|3

~~|<math>k(bj-eg)\\-l(aj-ef)\\+o(ag-bf)</math>~~

|1

~~|<math>k(ci-dh)\\-m(ai-df)\\+n(ah-cf)</math>~~

|2

~~|<math>k(cj-eh)\\-m(aj-ef)\\+o(ah-cf)</math>~~

|-

~~|<math>k(dj-ei)\\-n(aj-ef)\\+o(ai-df)</math>~~

|2

~~|<math>l(ci-dh)\\-m(bi-dg)\\+n(bh-cg)</math>~~

|1

|~~<math>l(cj-eh)\\-m(bj-eg)\\+o(bh-cg)</math>~~

|3

|~~<math>l(dj-ei)\\-n(bj-eg)\\+o(bi-dg)</math>~~

|2

|~~<math>m(dj-ei)\\-n(cj-eh)\\+o(ci-dh)</math>~~

|1

|~~<math>+</math>~~

|}

|~~<math>p(bh-cg)\\~~-~~q(ah-cf)\\+r(ag-bf)</math>~~

|~~<math>p(bi-dg)\\-q(ai-df)\\+s(ag-bf)</math>~~

The geometric explanation still hasn't answered the question as to why <math>l_{\text{ind}}=1</math> is the condition on addability. It just increased our intuitions about its relationship with temperament dimensions. We'll still look to a later section for an answer on this.

~~|<math>p(bj-eg)\\-q(aj-ef)\\+t(ag-bf)</math>~~

|~~<math>p(ci-dh)\\-r(ai-df)\\+s(ah-cf)</math>~~

=== Algebraic explanation===

|~~<math>p(cj-eh)\\-r(aj-ef)\\+t(ah-cf)</math>~~

|~~<math>p(dj-ei)\\-s(aj-ef)\\+t(ai-df)</math>~~

This explanation relies on comparing the results of the multivector and matrix approaches to temperament arithmetic, and showing algebraically how the matrix approach can only achieve the same answer as the multivector approach on the condition that it keeps all but one vector between the added matrices the same, that is, not only are the temperaments addable, but their <math>L_{\text{dep}}</math> appears explicitly in the added matrices.

|~~<math>q(ci-dh)\\-r(bi-dg)\\+s(bh-cg)</math>~~

|~~<math>q(cj-eh)\\-r(bj-eg)\\+t(bh-cg)</math>~~

To compare results, we eventually get both approaches into a multivector form. With the multivector approach, we wedge the vector set first and then add the resultant multivectors to get a new multivector. With the matrix approach, we treat the vector set as a matrix and add first, then treat the resultant matrix as a vector set and wedge those vectors to get a new multivector.

~~|<math>q(dj-ei)\\-s(bj-eg)\\+t(bi-dg)</math>~~

~~|<math>r(dj-ei)\\-s(cj-eh)\\+~~t~~(ci-dh)</math>~~

The diagrams below are organized into a 2×2 layout. The left part shows the multivector approach, and the right part shows the matrix approach. The top part shows how the results of two approaches match when the <math>L_{\text{dep}}</math> is successfully explicit (and in these cases, the <math>L_{\text{dep}}</math> vectors are highlighted in green and the <math>L_{\text{ind}}</math> vectors are highlighted in red), and the bottom part shows how the results fail to match when it is not. Successful matches are highlighted in yellow and failures to match are highlighted in blue.

|<~~math>=</math>~~

| style="~~background-~~color: ~~LightYellow~~;"|<math>~~(k+p)(bh-cg)~~\~~\-(l+q)(ah-cf)\\+(m+r)(ag-bf)~~</math>

This first diagram demonstrates this situation for a <math>d=3, g=2</math> case.

~~| style="background-color: LightYellow;"|<math>(k+p)(bi-dg)\\-(l+q)(ai-df)\\+(n+s)(ag-bf)~~</~~math~~>

{| class="wikitable center-all"

~~| style~~=~~"background-color: LightYellow;"|~~<~~math>(k+p)(bj-eg)\\-(l+q)(aj-ef)\\+(o+t)(ag-bf)</math>~~

|+

| style="~~background-~~color: ~~LightYellow~~;"|<math>~~(k+p)(ci-dh)~~\~~\-(m+r)(ai-df)\\+(n+s)(ah-cf)~~</math>

!

~~| style="background-color: LightYellow;"|<math>(k+p)(cj-eh)\\-(m+r)(aj-ef)\\+(o+t)(ah-cf)~~</~~math~~>

!

| style="~~background-~~color: ~~LightYellow~~;"|<math>~~(k+p)(dj-ei)~~\~~\-(n+s)(aj-ef)\\+(o+t)(ai-df)~~</math>

!

~~| style="background-color: LightYellow;"|~~<~~math~~>(~~l+q)(ci-dh)\\-(m+r)(bi-dg)\\+(n+s)(bh-cg)~~<~~/math>~~

! colspan="11" |

| style="~~background-~~color: ~~LightYellow~~;"|<math>~~(l+q)(cj-eh)\\-(m+r)(bj-eg)~~\~~\+(o+t)(bh-cg)~~</math>

!

~~| style="background-color: LightYellow;"|<math>(l+q)(dj-ei)\\-(n+s)(bj-eg)\\+(o+t)(bi-dg)~~</~~math~~>

! colspan="11" |

| style="~~background-~~color: ~~LightYellow~~;"|<math>~~(m+r)(dj-ei)\\-(n+s)(cj-eh)~~\~~\+(o+t)(ci-dh)~~</math>

!

~~| style="background-color: LightYellow;"|~~<~~math~~>~~(k+p)(bh-cg)\\-(l+q)(ah-cf)\\+(m+r)(ag-bf~~)</math>

~~| style~~=~~"background-color: LightYellow;"|<math>(k+p)(bi-dg)\\-(l+q)(ai-df)\\+(n+s)(ag-bf)~~</math>

| ~~style~~="~~background~~-~~color: LightYellow;~~"|~~<math>(k~~+~~p)(bj-eg)\\-(l+q)(aj-ef)\\+(o+t)(ag-bf)</math>~~

~~| style="background-color: LightYellow;"|<math>(k+p)(ci-dh)\\-(m+r)(ai-df)\\+(n+s)(ah-cf)</math>~~

~~| style="background-color: LightYellow;"|<math>(k+p)(cj-eh)\\-(m+r)(aj-ef)\\+(o+t)(ah-cf)</math>~~

~~| style="background-color: LightYellow;"|<math>(k+p)(dj-ei)\\-(n+s)(aj-ef)\\+(o+t)(ai-df)</math>~~

~~| style~~="~~background-color: LightYellow;~~"|~~<math>(l+q)(ci-dh)\\-(m+r)(bi-dg)\\+(n+s)(bh-cg)</math>~~

~~| style="background-color: LightYellow;"|<math>(l+q)(cj-eh)\\-(m+r)(bj-eg)\\+(o+t)(bh-cg)</math>~~

~~| style~~="~~background-color: LightYellow;~~"|~~<math>(l+q)(dj-ei)\\-(n+s)(bj-eg)\\+(o+t)(bi-dg)</math>~~

~~| style="background-color: LightYellow;"|<math>(m+r)(dj-ei)\\-(n+s)(cj-eh)\\+(o+t)(ci-dh)</math>~~

|-

!

| colspan="11" rowspan="1" |'''multivector approach'''

!

| colspan="11" rowspan="1" |'''matrix approach'''

!

|-

!

~~! colspan="32" |~~

!

~~! colspan="22" |~~

!

! colspan="11" |

!

! colspan="11" |

!

|-

! rowspan="5" |

| colspan="1" rowspan="5" |explicit <math>L_{\text{dep}}</math>

{{bra|{{vector|<math>a</math> <math>b</math> <math>c</math>}}}}

! rowspan="5" |

| style="background-color: #BED5BA;"|<math>a</math>

| style="background-color: #BED5BA;"|<math>b</math>

| style="background-color: #BED5BA;"|<math>c</math>

| rowspan="2" |

| style="background-color: #BED5BA;"|<math>a</math>

| style="background-color: #BED5BA;"|<math>b</math>

| style="background-color: #BED5BA;"|<math>c</math>

| rowspan="2" |

| colspan="3" rowspan="2" |

! rowspan="5" |

| style="background-color: #BED5BA;"|<math>a</math>

| style="background-color: #BED5BA;"|<math>b</math>

| style="background-color: #BED5BA;"|<math>c</math>

| colspan="1" rowspan="2" |<math>+</math>

| style="background-color: #BED5BA;"|<math>a</math>

| style="background-color: #BED5BA;"|<math>b</math>

| style="background-color: #BED5BA;"|<math>c</math>

| colspan="1" rowspan="2" |<math>=</math>

|<math>2a</math>

|<math>2b</math>

|<math>2c</math>

! rowspan="5" |

|-

~~! rowspan="7" |~~

| style="background-color: #E7BBB3;"|<math>d</math>

~~| rowspan="7"~~ |~~hidden <span~~ style="color: #~~3C8031~~;~~"><math>L_{\text{dep}}</math>~~

| style="background-color: #E7BBB3;"|<math>e</math>

~~|<math>r_1</math>~~

| style="background-color: #E7BBB3;"|<math>f</math>

~~! rowspan="7" |~~

| style="background-color: #E7BBB3;"|<math>g</math>

~~| colspan="2" rowspan="1" |<math>a</math>~~

| style="background-color: #E7BBB3;"|<math>h</math>

~~| colspan="2" rowspan="1" |<math>b</math>~~

| style="background-color: #E7BBB3;"|<math>i</math>

~~| colspan="2" rowspan="1" |<math>c</math>~~

| style="background-color: #E7BBB3;"|<math>d</math>

~~| colspan="2" rowspan="1~~" |<math>d</math>

| style="background-color: #E7BBB3;"|<math>e</math>

| ~~colspan~~="~~2" rowspan="1~~" |<math>e</math>

| style="background-color: #E7BBB3;"|<math>f</math>

| ~~rowspan~~="~~3" |~~

| style="background-color: #E7BBB3;"|<math>g</math>

~~| colspan="2" rowspan="1~~" |<math>a</math>

| style="background-color: #E7BBB3;"|<math>h</math>

| ~~colspan~~="~~2" rowspan="1~~" |<math>b</math>

| style="background-color: #E7BBB3;"|<math>i</math>

| ~~colspan~~="~~2" rowspan="1~~" |<math>c</math>

|<math>d+g</math>

| ~~colspan~~="~~2" rowspan="1~~" |<math>d</math>

|<math>e+h</math>

| ~~colspan~~="~~2" rowspan="1~~" ~~|<math>e</math>~~

|<math>f+i</math>

~~| rowspan="3" |~~

|-

~~| colspan="10" rowspan="3" |~~

| colspan="3" rowspan="1" |<math>∧</math>

~~! rowspan="7" |~~

|

~~|<math>a</math>~~

| colspan="3" rowspan="1" |<math>∧</math>

~~|<math>b</math>~~

|

~~|<math>c</math>~~

| colspan="3" |

|<math>d</math>

| colspan="3" |

|~~<math>e</math>~~

|

~~| colspan~~="1" ~~rowspan="3" |<math>+</math>~~

| colspan="3" |

~~|<math>a</math>~~

|

~~|<math>b</math>~~

| colspan="3" rowspan="1" |<math>∧</math>

~~|<math>c</math>~~

~~|<math>d</math>~~

|<math>e</math>

| ~~colspan="1" rowspan="3" |<math>=</math>~~

~~| colspan="2" rowspan="1" |<math>2a</math>~~

~~| colspan="2" rowspan="1" |<math>2b</math>~~

~~| colspan="2" rowspan="1" |<math>2c</math>~~

~~| colspan="2" rowspan="1" |<math>2d</math>~~

~~| colspan~~="~~2" rowspan="1" |<math>2e</math>~~

~~! rowspan="7" |~~

|-

~~|<math>r_2</math>~~

~~| colspan="2" rowspan="1~~" |<math>f</math>

| ~~colspan~~="~~2" rowspan="1~~" |<math>g</math>

| ~~colspan~~="~~2" rowspan="1~~" |<math>h</math>

| ~~colspan~~="~~2" rowspan="1~~" |<math>i~~</math>~~

~~| colspan="2" rowspan="1" |<math>j</math>~~

~~| colspan="2" rowspan="1" |<math>u</math>~~

~~| colspan="2" rowspan="1" |<math>v</math>~~

~~| colspan="2" rowspan="1" |<math>w</math>~~

~~| colspan="2" rowspan="1" |<math>x</math>~~

~~| colspan="2" rowspan="1" |<math>y</math>~~

~~|<math>f~~</math>

|<math>g</math>

|<math>h</math>

~~|<math>i</math>~~

~~|<math>j</math>~~

~~|<math>u</math>~~

~~|<math>v</math>~~

~~|<math>w</math>~~

~~|<math>x</math>~~

~~|<math>y</math>~~

~~| colspan="2" rowspan="1"~~ |<math>f+~~u</math>~~

~~| colspan="2" rowspan="1" |<math>g+v</math>~~

~~| colspan="2" rowspan="1" |<math>w+h</math>~~

~~| colspan="2" rowspan="1" |<math>~~i~~+x</math>~~

~~| colspan="2" rowspan="1" |<math>j+y~~</math>

|-

~~|<math>r_3</math>~~

| colspan="2" rowspan="1" |<math>k</math>

| ~~colspan="2" rowspan="1" |<math>l</math>~~

| colspan="2" rowspan="1" |<math>m</math>

| ~~colspan="2" rowspan="1" |<math>n</math>~~

| colspan="~~2" rowspan="1~~" |~~<math>o</math>~~

| colspan="~~2" rowspan="1~~" |~~<math>p</math>~~

| ~~colspan="2" rowspan="1" |<math>q</math>~~

| colspan="~~2" rowspan="1~~" |~~<math>r</math>~~

| ~~colspan="2" rowspan="1" |<math>s</math>~~

| colspan="~~2" rowspan="1" |<math>t</math>~~

~~|<math>k</math>~~

~~|<math>l</math>~~

~~|<math>m</math>~~

~~|<math>n</math>~~

~~|<math>o</math>~~

~~|<math>p</math>~~

~~|<math>q</math>~~

~~|<math>r</math>~~

~~|<math>s</math>~~

~~|<math>t</math>~~

~~| colspan="2~~" rowspan="1" |<math>~~k+p</math>~~

~~| colspan="2" rowspan="1" |<math>l+q</math>~~

~~| colspan="2" rowspan="1" |<math>m+r</math>~~

~~| colspan="2" rowspan="1" |<math>n+s</math>~~

~~| colspan="2" rowspan="1" |<math>o+t~~</math>

|-

|<math>bf-ce</math>

|<math>af-cd</math>

|<math>ae-bd</math>

|<math> +</math>

|<math>bi-ch</math>

|<math>ai-cg</math>

|<math>ah-bg</math>

|<math>=</math>

|<math>bf-ce+bi-ch</math>

|<math>af-cd+ai-cg</math>

|<math>ae-bd+ah-bg</math>

| colspan="3" rowspan="2" |

| rowspan="2" |

| colspan="3" rowspan="2" |

| rowspan="2" |

|<math>2b(f+i)-2c(e+h)</math>

|<math>2a(f+i)-2c(d+g)</math>

|<math>2a(e+h)-2b(d+g)</math>

|-

| colspan="3" |

|

| colspan="~~10" rowspan="1~~" |~~<math>∧</math>~~

| colspan="3" |

|

| ~~colspan~~="10" ~~rowspan~~="1" |<math>∧</math>

| style="background-color: LightYellow;"|<math>b(f+i)-c(e+h)</math>

|

| style="background-color: LightYellow;"|<math>a(f+i)-c(d+g)</math>

~~| colspan~~="10" |

| style="background-color: LightYellow;"|<math>a(e+h)-b(d+g)</math>

| ~~colspan~~="5" |

| style="background-color: LightYellow;"|<math>b(f+i)-c(e+h)</math>

|

| style="background-color: LightYellow;"|<math>a(f+i)-c(d+g)</math>

~~| colspan~~="5" |

| style="background-color: LightYellow;"|<math>a(e+h)-b(d+g)</math>

|

~~| colspan="10" rowspan~~="1" |<math>∧</math>

|-

~~|<math>r_1∧r_2</math>~~

!

~~| rowspan~~="2" |~~<math>ag-bf</math>~~

!

~~| rowspan~~="2" |~~<math>ah~~-~~cf</math>~~

!

| rowspan="2" |~~<math>ai-df</math>~~

! colspan="11" |

| rowspan="2" |<~~math>aj-ef</math>~~

!

~~| rowspan~~="2" |<math>~~bh-cg~~</math>

! colspan="11" |

~~| rowspan="2" |<math>bi-dg~~</~~math~~>

!

| rowspan="2" |~~<math>bj-eg</math>~~

|-

~~| rowspan="2"~~ |<math>~~ci-dh~~</math>

! rowspan="5" |

~~| rowspan="2"~~ |<math>~~cj-eh~~</math>

| rowspan="5" |hidden <math>L_{\text{dep}}</math>

~~| rowspan="2"~~ |<math>~~dj-ei~~</math>

! rowspan="5" |

|<math>a</math>

|<math>b</math>

|<math>c</math>

| rowspan="2" |

~~| rowspan="2"~~ |<math>~~av-bu~~</math>

|<math>j</math>

~~| rowspan="2"~~ |<math>~~aw-cu~~</math>

|<math>k</math>

~~| rowspan="2"~~ |<math>~~ax-du~~</math>

|<math>l</math>

| ~~rowspan="2" |<math>ay-eu</math>~~

|

| ~~rowspan~~="2" ~~|<math>bw-cv</math>~~

| colspan="3" rowspan="2" |

| rowspan="2" |~~<math>bx-dv</math>~~

! rowspan="5" |

| rowspan="2" |~~<math>by-ev</math>~~

|<math>a</math>

~~| rowspan="2"~~ |<math>~~cx-dw~~</math>

|<math>b</math>

~~| rowspan="2"~~ |<math>~~cy-ew~~</math>

|<math>c</math>

~~| rowspan="2"~~ |<math>~~dy-ex~~</math>

| colspan="1" rowspan="2" |<math>+</math>

~~| rowspan="2" |~~

|<math>j</math>

| colspan="10" rowspan="2" |

|<math>k</math>

~~| colspan="5" rowspan="3" |~~

|<math>l</math>

~~| rowspan="3" |~~

| colspan="1" rowspan="2" |<math>=</math>

~~| colspan="5" rowspan="3" |~~

|<math>a+j</math>

~~| rowspan="3" |~~

|<math>b+k</math>

|<math>~~2a(g~~+~~v)\\-2b(f+u)~~</math>

|<math>c+l</math>

|<math>~~2a(w+h)\\-2c(f+u)~~</math>

! rowspan="5" |

|<math>~~2a(i+x)\\-2d(f+u)~~</math>

|<math>~~2a(j+y)\\-2e(f+u)~~</math>

|<math>~~2b(w+h)\\-2c(g+v)~~</math>

|<math>~~2b(i~~+~~x)\\-2d(g+v)</math>~~

~~|<math>2b(~~j~~+y)\\-2e(g+v)~~</math>

|<math>~~2c(i+x)\\-2d(w~~+h)</math>

|<math>~~2c(j~~+~~y)\\-2e(w+h)~~</math>

|~~<math>2d(j+y)\\-2e(i+x)</math>~~

|-

|~~simplify~~ <math>~~(r_1∧r_2)~~</math> ~~if necessary~~

|<math>d</math>

|<math>a(g~~+v)\\-b(f+u)~~</math>

|<math>e</math>

|<math>~~a(w+~~h~~)\\-c(f+u)~~</math>

|<math>f</math>

|<math>a(i~~+x)\\-~~d~~(f+u)~~</math>

|<math>g</math>

|<math>~~a(j+y)\\-~~e(f~~+u)~~</math>

|<math>h</math>

|<math>~~b(w+h)\\-c(~~g~~+v)~~</math>

|<math>i</math>

|<math>~~b(i+x)\\-d(g+v)~~</math>

|<math></math>

|<math>~~b(j+y)\\-e(g+v)~~</math>

|<math>d</math>

|<math>~~c(i+x)\\-~~d(w+h)</math>

|<math>e</math>

|<math>~~c(j+y)\\-~~e(w+h)</math>

|<math>f</math>

|<math>~~d(j~~+~~y)\\-e(~~i~~+x)~~</math>

|<math>g</math>

|<math>h</math>

|<math>i</math>

|<math>d+g</math>

|<math>e+h</math>

|<math>f+i</math>

|-

|<math>~~(r_1∧r_2)∧r_3~~</math>

| colspan="3" rowspan="1" |<math>∧</math>

|<math>~~k(bh-cg)\\-l(ah-cf)\\+m(ag-bf)~~</math>

|

|~~<math>k(bi-dg)\\-l(ai-df)\\+n(ag-bf)</math>~~

| colspan="3" rowspan="1" |<math>∧</math>

|~~<math>k(bj-eg)\\-l(aj-ef)\\+o(ag-bf)</math>~~

|

|~~<math>k(ci-dh)\\-m(ai-df)\\+n(ah-cf)</math>~~

| colspan="3" |

|~~<math>k(cj-eh)\\-m(aj-ef)\\+o(ah-cf)</math>~~

| colspan="3" |

|<math>~~k(dj-ei)\\-n(aj-ef)\\+o(ai-df)~~</math>

|

|~~<math>l(ci~~-~~dh)\\-m(bi-dg)\\+n(bh-cg)</math>~~

| colspan="3" |

|<math>~~l(cj~~-~~eh)\\-m(bj-eg)\\+o(bh-cg)~~</math>

|

|<math>~~l(dj-ei)\\-n(bj~~-~~eg)\\+o(bi-dg)~~</math>

| colspan="3" rowspan="1" |<math>∧</math>

|<math>~~m(dj-ei)\\-n(cj-eh)\\+o(ci~~-~~dh)~~</math>

|-

|<math>bf-ce</math>

|<math>af-cd</math>

|<math>ae-bd</math>

|<math>+</math>

|<math>~~p(bw~~-~~cv)\\-q(aw-cu)\\+r(av-bu)~~</math>

|<math>ki-lh</math>

|<math>~~p(bx-dv)\\-q(ax-du)\\+s(av~~-~~bu)~~</math>

|<math>ji-lg</math>

|<math>~~p(by~~-~~ev)\\-q(ay-eu)\\+t(av-bu)~~</math>

|<math>jh-kg</math>

|<math>~~p(cx-dw)\\-r(ax-du)\\+s(aw-cu)~~</math>

|<math>=</math>

|<math>~~p(cy-ew)\\-r(ay~~-~~eu)\\~~+~~t(aw~~-~~cu)~~</math>

|<math>bf-ce+ki-lh</math>

|<math>~~p(dy~~-~~ex)\\-s(ay-eu)\\~~+~~t(ax~~-~~du)~~</math>

|<math>af-cd+ji-lg</math>

|<math>~~q(cx~~-~~dw)\\-r(bx-dv)\\~~+~~s(bw~~-~~cv)</math>~~

|<math>ae-bd+jh-kg</math>

~~|<math>q(cy-ew)\\-r(by-ev)\\+t(bw-cv)</math>~~

| colspan="3" rowspan="2" |

~~|<math>q(dy-ex)\\-s(by-ev)\\+t(bw-cv)</math>~~

| rowspan="2" |

~~|<math>r(dy-ex)\\-s(cy-ew)\\+t(cx-dw)</math>~~

| colspan="3" rowspan="2" |

~~|<math>=~~</math>

| rowspan="2" |

| ~~style~~="~~background-color: LightBlue;~~"~~|<math>k(bh-cg)\\-l(ah-cf)\\+m(ag-bf)\\+p(bw-cv)\\-q(aw-cu)\\+r(av-bu)</math>~~

|<math>(b+k)(f+i)-(c+l)(e+h)</math>

~~| style~~="~~background-color: LightBlue;~~"|~~<math>k(bi-dg)\\-l(ai-df)\\+n(ag-bf)\\+p(bx-dv)\\-q(ax-du)\\+s(av-bu)</math>~~

|<math>(a+j)(f+i)-(c+l)(d+g)</math>

| ~~style~~="~~background-color: LightBlue;~~"|~~<math>k(bj-eg)\\-l(aj-ef)\\+o(ag-bf)\\+p(by-ev)\\-q(ay-eu)\\+t(av-bu)</math>~~

|<math>(a+j)(e+h)-(b+k)(d+g)</math>

| ~~style~~="~~background-color: LightBlue;~~"~~|<math>k(ci-dh)\\-m(ai-df)\\+n(ah-cf)\\+p(cx-dw)\\-r(ax-du)\\+s(aw-cu)</math>~~

|-

~~| style~~="~~background-color: LightBlue;~~"|~~<math>k(cj-eh)\\-m(aj-ef)\\+o(ah-cf)\\+p(cy-ew)\\-r(ay-eu)\\+t(aw-cu)</math>~~

| colspan="3" |

| ~~style~~="~~background-color: LightBlue;~~"|~~<math>k(dj-ei)\\-n(aj-ef)\\+o(ai-df)\\+p(dy-ex)\\-s(ay-eu)\\+t(ax-du)</math>~~

|

~~| style="background-color: LightBlue;"~~|<math>l(~~ci-dh)\\-m(bi-dg)\\~~+~~n(bh-cg~~)~~\\+q~~(~~cx-dw)\\-r(bx-dv)\\~~+~~s(bw-cv)</math>~~

| colspan="3" |

~~| style="background-color: LightBlue;"|<math>l(cj-eh~~)\\-m(~~bj-eg)\\+o(bh-cg)\\+q(cy-ew)\\-r(by-ev)\\~~+~~t(bw-cv)</math>~~

|

~~| style="background-color: LightBlue;"|<math>~~l~~(dj-ei~~)~~\\-n~~(~~bj-eg)\\~~+~~o(bi-dg)\\+q(dy-ex)\\-s(by-ev)\\+t(bw-cv~~)</math>

| style="background-color: LightBlue;"|<math>bf-ce+ki-lh</math>

~~| style="background-color: LightBlue;"~~|<math>m(~~dj-ei)\\-n(cj-eh)\\~~+~~o(ci-dh~~)~~\\+r~~(~~dy-ex)\\-s(cy-ew)\\~~+~~t(cx-dw~~)~~</math>~~

| style="background-color: LightBlue;"|<math>af-cd+ji-lg</math>

~~| style="background~~-~~color: LightBlue;"|<math>(k+p)\\~~(~~b(w+h)-~~c(g+~~v))\\-(~~l+q)\\(~~a(w~~+~~h)-c(f+u))\\+(m+r)\\(a(~~g~~+v)-b(f+u)~~)</math>

| style="background-color: LightBlue;"|<math>ae-bd+jh-kg</math>

~~| style="background-color: LightBlue;"~~|<math>(k+p)\\(~~b(i~~+x)-d(g+~~v))\\-(l+q~~)\\(~~a(i+x)-~~d~~(f+u))\\+(n~~+~~s)\\(a(~~g~~+v)-b(f+u)~~)</math>

| style="background-color: LightBlue;"|<math>bf+bi+kf+ki-ce-ch-le-lh</math>

| ~~style~~="~~background-color: LightBlue;~~"|~~<math>(k+p)\\(b(j+y)-e(g+v))\\-(l+q)\\(a(j+y)-e(f+u))\\+(o+t)\\(a(g+v)-b(f+u))</math>~~

| style="background-color: LightBlue;"|<math>af+ai+jf+ji-cd-cg-ld-lg</math>

| ~~style~~="~~background-color: LightBlue;~~"|~~<math>(k+p)\\(c(i+x)-d(w+h))\\-(m+r)\\(a(i+x)-d(f+u))\\+(n+s)\\(a(w+h)-c(f+u))</math>~~

| style="background-color: LightBlue;"|<math>ae+ah+je+jh-bd-bg-kd-kg</math>

| style="background-color: LightBlue;"|<math>~~(k+p)\\(c(j+y)~~-~~e(w~~+~~h))\\~~-~~(m+r)\\(a(j+y)-e(f+u))\\+(o+t)\\(a(w+h)-c(f+u))~~</math>

| style="background-color: LightBlue;"|<math>~~(k+p)\\(d(j+y)~~-~~e(i+x))\\-(n+s)\\(a(j~~+y)-~~e(f+u))\\+(o+t)\\(a(i+x)-d(f+u))~~</math>

| style="background-color: LightBlue;"|<math>~~(l+q)\\(c(i+x)~~-~~d(w~~+~~h))\\~~-~~(m+r)\\(b(i+x)-d(g+v))\\+(n+s)\\(b(w+h)-c(g+v))~~</math>

| style="background-color: LightBlue;"|<math>(l+~~q)\\(c(j~~+y)-~~e(w+h))\\~~-~~(m+r)\\(b(j+y)~~-~~e(g+v))\\+(o+t)\\(b(w+h)~~-~~c(g+v))~~</math>

| style="background-color: LightBlue;"|<math>(l+~~q)\\(d(j~~+y)-~~e(i+x))\\~~-~~(n+s)\\(b(j+y)~~-~~e(g+v))\\+(o+t)\\(b(i+x)~~-~~d(g+v))~~</math>

| style="background-color: LightBlue;"|<math>(m+~~r)\\(d(j~~+y)-~~e(i+x))\\~~-~~(n+s)\\(c(j+y)~~-~~e(w+h))\\+(o+t)\\(c(i+x)~~-~~d(w+h))~~</math>

|-

!

~~! colspan="2" |~~

!

~~! colspan="32" |~~

!

! colspan="22" |

! colspan="11" |

!

! colspan="11" |

!

|}

~~These two examples are by no means~~ a ~~proof~~, ~~but meditation on the patterns in the variables is at least fairly convincing~~.

This second diagram demonstrates this situation for a <math>d=5, g=3</math> case. One pair of the <math>L_{\text{dep}}</math> vectors are explicitly matching, but not the other, which isn't enough.

{| class="wikitable center-all"

~~===Sintel's proof~~ of the ~~linear-independence~~</~~span~~> ~~conjecture===~~

|+

!

~~====Sintel's original text====~~

! colspan="2" |

!

<~~nowiki~~>~~If A and B~~ are ~~mappings from Z^n to Z^m~~, ~~with n > m~~, ~~A, B full rank (using A and B as their rowspace equivalently):~~

! colspan="32" |

!

~~dim(A~~ + ~~B) - m~~ = ~~dim(ker(A) + ker(B)) - (n-m)~~

! colspan="22" |

!

~~>> dim(A)+dim(B)~~=~~dim(A+B)+dim(A∩B)~~ =~~> dim(A + B) = dim(A) + dim(B)~~ - ~~dim(A∩B)~~

|-

!

~~dim(A) + dim(B) - dim(A∩B) - m~~ = ~~dim(ker(A) + ker(B)) - (n-m)~~

! colspan="2" |

!

~~>> by duality of kernel, dim(ker(A) + ker(B))~~ = ~~dim(ker(A ∩ B))~~

| colspan="32" rowspan="1" |'''multivector approach'''

!

~~dim(A) + dim(B) - dim(A∩B) - m~~ = ~~dim(ker(A ∩ B))~~ - ~~(n-m)~~

| colspan="22" rowspan="1" |'''matrix approach'''

!

~~>> rank nullity: dim(ker(A ∩ B)) + dim(A ∩ B)~~ = n

|-

!

~~dim(A) + dim(B) - dim(A∩B) - m~~ = ~~n - dim(A ∩ B) - (n-m)~~

! colspan="2" |

!

~~m + m - dim(A∩B) - m~~ = ~~n - dim(A ∩ B) - (n-m)~~

! colspan="32" |

!

~~m + m~~ - m = ~~n - n + m~~

! colspan="22" |

!

m = m</~~nowiki~~>

|-

! rowspan="7" |

| colspan="1" rowspan="7" |explicit <math>L_{\text{dep}}</math>

~~====Douglas Blumeyer's interpretation====~~

⟨{{vector|<math>a</math> <math>b</math> <math>c</math> <math>d</math> <math>e</math>}}

{{vector|<math>f</math> <math>g</math> <math>h</math> <math>i</math> <math>j</math>}}]

~~We're going to take the strategy of beginning with what we're trying to prove, then reducing it to an obvious equivalence, which will show that our initial statement must be just as true.~~

| style="background-color: #BED5BA;"|<math>r_1</math>

! rowspan="7" |

~~So here's the statement we're trying to prove:~~

| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>a</math>

| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>b</math>

| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>c</math>

~~<math>\text{rank}(\text{union}(M_1, M_2)) - r = \text{nullity}(\text{union}(C_1, C_2)) - n</math>~~

| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>d</math>

| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>e</math>

| rowspan="3" |

<math>M_1</math> and <math>M_2</math> are mappings which both have dimensionality <math>d</math>, rank <math>r</math>, nullity <math>n</math>, and are full-rank, and <math>C_1</math> and <math>C_2</math> are their comma bases, respectively.

| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>a</math>

| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>b</math>

Technically since these matrices are representing subspace bases, the correct operation here is "sumset", not "union", but because "union" is a more commonly known opposite of intersection and would work for plain matrices, I've decided to stick with it here.

| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>c</math>

| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>d</math>

~~So, the left-hand side of this equation is~~

| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>e</math>

| rowspan="3" |

| colspan="10" rowspan="3" |

! rowspan="7" |

| style="background-color: #BED5BA;"|<math>a</math>

| style="background-color

~~\color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}-1 \\~~

~~\end{array} \right]~~</math>

~~And so, reintroducing the <math>L_{\text{dep}}</math>,~~ we ~~have:~~

So we know this is true.

~~<math>\left[ \begin{array} {rrr}~~

~~\color{OliveGreen}19 & \color{OliveGreen}30 & \color{OliveGreen}44 & \color{OliveGreen}53 \\~~

~~\color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}-1 \\~~

~~\end{array} \right]</math>~~

~~Which canonicalizes to:~~

~~<math>\left[ \begin{array} {rrr}~~

~~19 & 30 & 44 & 0 \\~~

~~0 & 0 & 0 & 1 \\~~

~~\end{array} \right]</math>~~

~~And so we can see that meantone minus flattone~~ is ~~[[meanmag]]~~.

=Glossary=

@@ Line 236: / Line 236: @@
 In order to understand how to do temperament arithmetic on <math>g_{\text{min}}>1</math> temperaments, we must first understand addability.
-===Verbal explanation===
 In order to understand addability, we must work up to it, understanding these concepts in this order:
@@ Line 245: / Line 243: @@
 #<span style="color: #B6321C;">linear independence</span> between temperaments by only one basis vector (that's addability)
-====1. <span style="color: #3C8031;">Linear dependence</span>====
+===1. <span style="color: #3C8031;">Linear dependence</span>===
 This is explained here: [[linear dependence]].
-====2. <span style="color: #3C8031;">Linear dependence</span> between temperaments====
+===2. <span style="color: #3C8031;">Linear dependence</span> between temperaments===
 <span style="color: #3C8031;">Linear dependence</span> has been defined for the matrices and multivectors that represent temperaments, but it can also be defined for temperaments themselves. The conditions of temperament arithmetic motivate a definition of <span style="color: #3C8031;">linear dependence</span> for temperaments whereby temperaments are considered <span style="color: #3C8031;">linearly dependent</span> if ''either of their mappings or their comma bases are <span style="color: #3C8031;">linearly dependent</span>''<ref>or — equivalently, in EA — either their multimaps or their multicommas are <span style="color: #3C8031;">linearly dependent</span></ref>.
@@ Line 257: / Line 255: @@
 To make this point visually, we could say that two temperaments are <span style="color: #3C8031;">linearly dependent</span> if they intersect in one or the other of tone space and tuning space. So you have to check both views.<ref>You may be wondering — what about two temperaments which are parallel in tone or tuning space, e.g. compton and blackwood in tuning space? Their comma bases are each <math>n=1</math>, and they meet to give a <math>n=2</math> [[comma basis]], which corresponds to a <math>r=1</math> mapping, which means it should appear as an ET point on the PTS diagram. But how could that be? Well, here's their meet: {{bra|{{vector|1 0 0}} {{vector|0 1 0}}}}, and so that corresponding mapping is {{ket|{{map|0 0 1}}}}. So it's some degenerate ET. I suppose we could say it's the point at infinity away from the center of the diagram.</ref>
-====3. <span style="color: #B6321C;">Linear independence</span> between temperaments====
+===3. <span style="color: #B6321C;">Linear independence</span> between temperaments===
 <span style="color: #3C8031;">Linear dependence</span> may be considered as a boolean (yes/no, linearly <span style="color: #3C8031;">dependent</span>/<span style="color: #B6321C;">independent</span>) or it may be considered as <span style="color: #3C8031;">an integer count of linearly dependent basis vectors</span>. In other words, it is the dimension of <span style="color: #3C8031;">the linear-dependence basis <math>\dim(L_{\text{dep}})</math></span>. To refer to this count, we may hyphenate it as <span style="color: #3C8031;">'''linear-dependence'''</span>, and use the variable <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span>. For example, 5-ET and 7-ET, per the example in the previous section, are <span style="color: #3C8031;"><math>l_{\text{dep}}=1</math></span> (read <span style="color: #3C8031;">"linear-dependence-1"</span>) temperaments.
@@ Line 267: / Line 265: @@
 A proof of this conjecture is given here: [[Temperament arithmetic#Sintel's proof of the linear-independence conjecture]].
-====4. <span style="color: #B6321C;">Linear independence</span> between temperaments by only one basis vector (i.e. addability)====
+===4. <span style="color: #B6321C;">Linear independence</span> between temperaments by only one basis vector (i.e. addability)===
 Two temperaments are addable if they are <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>. In other words, both their mappings and their comma bases <span style="color: #3C8031;">share</span> all but one basis vector.
@@ Line 273: / Line 271: @@
 And so this is why <math>g_{\text{min}}=1</math> temperaments are all addable. Because if <math>g_{\text{min}}=1</math>, and the temperaments are different from each other so <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span> is at least 1, and <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span> can't be greater than <math>g_{\text{min}}</math>, so then necessarily <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span> <math>= 1</math> exactly, and therefore the temperaments are addable.
-===Diagrammatic explanation===
+==Multivector approach==
+The simplest approach to <math>g_{\text{min}}>1</math> temperament arithmetic is to use multivectors. This is discussed in more detail here: [[Douglas Blumeyer and Dave Keenan's Intro to exterior algebra for RTT#Temperament arithmetic]].
+==Matrix approach==
-====Introduction====
+Temperament arithmetic for <math>g_{\text{min}}>1</math> temperaments (again, that's with both <math>r>1</math> and <math>n>1</math>) can also be done using matrices, and it works in essentially the same way — entry-wise addition or subtraction — but there are some complications that make it significantly more involved than it is with multivectors. There's essentially five steps:
-The diagrams used for this explanation were inspired in part by [[Kite Giedraitis|Kite]]'s [[gencom]]s, and specifically how in his "twin squares" matrices — which have dimensions <math>d×d</math> — one can imagine shifting a bar up and down to change the boundary between vectors that form a basis for the commas and those that form a basis for preimage intervals (this basis is typically called "the [[generator]]s"). The count of the former is the nullity <math>n</math>, and the count of the latter is the rank <math>r</math>, and the shifting of the boundary bar between them with the total <math>d</math> vectors corresponds to the insight of the rank-nullity theorem, which states that <math>r + n=d</math>. And so this diagram's square grid has just the right amount of room to portray both the mapping and the comma basis for a given temperament (with the comma basis's vectors rotated 90 degrees to appear as rows, to match up with the rows of the mapping).
+# Find the linear-dependence basis <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>
+#Put the matrices in a form with the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>
+#Check for enfactoring, and perform an addabilization defactor (if necessary)
+#Check for negation, and change negation (if necessary)
+#Entry-wise add, and canonicalize
-So consider this first example of such a diagram:
+===The steps===
-{| class="wikitable"
+==== 1. Find the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>====
-|+
-| rowspan="4" |<math>d=4</math>
-| style="border-bottom: 3px solid black;"|<math>g_{\text{min}}=1</math>
-| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
-|-
-| rowspan="3" |<math>g_{\text{max}}=3</math>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C; "><math>l_{\text{ind}}=1</math></span>
-|-
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-| rowspan="2" |
-|-
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|}
-This represents a <math>d=4</math> temperament. These diagrams are grade-agnostic, which is to say that they are agnostic as to which side counts the <math>r</math> and which side counts the <math>n</math>. So we are showing them as <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> instead. We could say there's a variation on the rank-nullity theorem whereby <math>g_{\text{min}} + g_{\text{max}}=d</math>, just as <math>r + n=d</math>. So we can then say that this diagram represents either a <math>r=1</math>, <math>n=3</math> temperament, or perhaps a <math>n=1</math>, <math>r=3</math> temperament.
+For matrices, it is necessary to make explicit <span style="color: #3C8031;">the basis for the linearly dependent vectors shared</span> between the involved matrices before performing the arithmetic. In other words, any vectors that can be found through linear combinations of any of the involved matrices' basis vectors must appear explicitly and in the same position of each matrix before the sum or difference is taken. These vectors are called the <span style="color: #3C8031;">linear-dependence basis, or <math>L_{\text{dep}}</math></span>.
-But actually, this diagram represents more than just a single temperament. It represents a relationship between a pair of temperaments (which have the same [[dimensions]], non-grade-agnostically, i.e. not a pairing of a <math>r=1</math>, <math>n=3</math> temperament with a <math>r=3</math>, <math>n=1</math> temperament). As elsewhere, <span style="color: #3C8031;">green coloration indicates the linearly dependent basis vectors <math>L_{\text{dep}}</math></span> between this pair of temperaments, and <span style="color: #B6321C;">red coloration indicates linearly ''in''dependent basis vectors <math>L_{\text{ind}}</math></span> between the same pair of temperaments.
+Before this can be done, of course, we need to actually find the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>. This can be done using the technique described here: [[Linear dependence#For a given set of basis matrices, how to compute a basis for their linearly dependent vectors]]
-So, in this case, the two ET maps are <span style="color: #B6321C;">linearly independent</span>. This should be unsurprising; because ET maps are constituted by only a single vector (they're <math>r=1</math> by definition), if they ''were'' <span style="color: #3C8031;">linearly dependent</span>, then they'd necessarily be the ''same'' exact ET! Temperament arithmetic on two of the same ET is never interesting; <math>T_1</math> plus <math>T_1</math> simply equals <math>T_1</math> again, and <math>T_1</math> minus <math>T_1</math> is undefined. That said, if we ''were'' to represent temperament arithmetic between two of the same temperament on such a diagram as this, then every cell would be green. And this is true regardless whether <math>r=1</math> or otherwise.
+====2. Put the matrices in a form with the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>====
-From this information, we can see that the comma bases of any randomly selected pair of ''different'' <math>d=4</math> ETs are going to <span style="color: #3C8031;">share 2 vectors</span>, or in other words, their <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> will have two basis vectors. In terms of the diagram, we're saying that they'll always have two <span style="color: #3C8031;">green-colored vectors</span> under the black bar.
+The <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> will always have one less vector than the original matrix, by the definition of addability as <span style="color: #B6321C;"><math>L_{\text{ind}}=1</math></span>. And the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> is not a full recreation of the original temperament; it needs that one extra vector to get back to representing it.
+So a next step, we need pad out the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> by drawing from vectors from the original matrices. We can start from their first vectors. But if that vector happens to be linearly dependent on the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>, then it won't result in a representation of the original matrix. Otherwise we'll produce a [[rank-deficient]] matrix that doesn't still represent the same temperament as we started with. So we just have to keep going until we get it.
+====3. Addabiliziation defactoring====
+But it is not quite as simple as determining the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> and then supplying the remaining vectors necessary to match the grade of the original matrix, because the results may then be [[enfactored]]. And defactoring them without compromising the explicit <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> cannot be done using existing [[defactoring algorithms]]; it's a tricky process, or at least computationally intensive. This is called '''addabilization defactoring'''.
-These diagrams are a good way to understand which temperament relationships are possible and which aren't, where by a "relationship" here we mean a particular combination of their matching dimensions and their linear-independence integer count. A good way to use these diagrams for this purpose is to imagine the <span style="color: #B6321C;">red coloration</span> emanating away from the black bar in both directions simultaneously, one pair of vectors at a time. Doing it like this captures the fact, as previously stated, that the <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span> on either side of duality is always equal. There's no notion of a max or min here, as there is with <math>g</math> or <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span>; the <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span> on either side is always the same, so we can capture it with a single number, which counts the <span style="color: #B6321C;">red vectors</span> on just one half (that is, half of the total count of <span style="color: #B6321C;">red vectors</span>, or half of the width of the <span style="color: #B6321C;">red band</span> in the middle of the grid).
+Most established defactoring algorithms will alter any or all of the entries of a matrix. This is not an option if we still want to be able to do temperament arithmetic, however, because these matrices must retain their explicit <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>. And we can't defactor and then paste the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> back over the first vector or something, because then we might just be enfactored again. We need to find a defactoring algorithm that manages to work without altering any of the vectors in the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>.
-There's no need to look at diagrams like this where the black bar is below the center. This is because, even though for convenience we're currently treating the top half as <math>r</math> and the bottom half as <math>n</math>, these diagrams are ultimately grade-agnostic. So we could say that each one essentially represents not just one possibility for the relationship between two temperaments' dimensions and <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span>, but ''two'' such possibilities. Again, this diagram equally represents both <math>d=4, r=1, n=3, </math><span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span> as well as <math>d=4, r=3, n=1, </math><span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>. Which is another way of saying we could vertically mirror it without changing it.
+The first step to addabilization defactoring is inspired by [[Pernet-Stein defactoring]]: we find the value of the enfactoring factor (the "greatest factor") by following this algorithm until the point where we have a square transformation matrix, but instead of inverting it and multiplying by it to ''remove'' the defactoring, we simply take this square matrix's determinant, which is the factor we were about to remove. If that determinant is 1, then we're already defactored; if not, then we need to take do some additional steps.
-With the black bar always either in the top half or exactly in the center, we can see that the emanating <span style="color: #B6321C;">red band</span> will always either hit the top edge of the square grid first, or they will hit both the top and bottom edges of it simultaneously. So this is how these diagrams visually convey the fact that the <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span> between two temperaments will always be less than or equal to their <math>g_{\text{min}}</math>: because a situation where <math>g_{\text{min}}>l</math> would visually look like the <span style="color: #B6321C;">red band</span> spilling past the edges of the square grid.
+It turns out that you can always<ref>This conjecture was first suggested by Mike Battaglia, but it has not yet been mathematically proven. Sintel and Tom Price have done some experiments but nothing complete yet. Douglas Blumeyer's test cases in the [[RTT library in Wolfram Language]] suggest this is true, though.</ref> isolate the greatest factor in the single final vector of the matrix — the <span style="color: #B6321C;">linearly independent vector</span> — through linear combinations of the vectors in the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>.
-We could also say that two temperaments are <span style="color: #3C8031;">linearly dependent</span> on each other when <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math><g_{\text{max}}</math>, that is, their <span style="color: #B6321C;">linear-independence</span> is less than their ''max''-grade.
+The example that will be worked through in this section below is as simple as it can get: the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> consists of only a single vector, so we simply add some number of this <span style="color: #3C8031;">single linearly dependent vector</span> to the <span style="color: #B6321C;">linearly independent vector</span>. However, if there are multiple vectors in the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>, the linear combination which surfaces the greatest factor may involve just one or potentially all of those vectors, and the best approach to finding this combination is simply an automatic solver. An example of this approach is demonstrated in the [[RTT library in Wolfram Language]], here: https://github.com/cmloegcmluin/RTT/blob/main/main.m#L477
-Perhaps more importantly, we can also see from these diagrams that any pair of <math>g_{\text{min}}=1</math> temperaments will be addable. Because if they are <math>g_{\text{min}}=1</math>, then the furthest the <span style="color: #B6321C;">red band</span> can extend from the black bar is 1 vector, and 1 mirrored set of <span style="color: #B6321C;">red vectors</span> means <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>, and that's the definition of addability.
+Another complication is that the greatest factor may be very large, or be a highly composite number. In this case, searching for the linear combination that isolates the greatest factor in its entirety directly may be intractable; it is better to eliminate it piecemeal, i.e., whenever the solver finds a factor of the greatest factor, eliminate it, and repeat until the greatest factor is fully eliminated. The RTT library code linked to above works in this way.
-====A simple <math>d=3</math> example====
+====4. Negation====
-Let's back-pedal to <math>d=3</math> for a simple illustrative example.
+Temperament negation is more complex with matrices, both in terms of checking for it, as well as changing it.
-{| class="wikitable"
-|+
-| rowspan="3" |<math>d=3</math>
-|style="border-bottom: 3px solid black;"|<math>g_{\text{min}}=1</math>
-| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
-|-
-| rowspan="2" |<math>g_{\text{max}}=2</math>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
-|-
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|
-|}
-This diagram shows us that any two <math>d=3</math>, <math>g_{\text{min}}=1</math> temperaments (like 5-limit ETs) will be <span style="color: #3C8031;">linearly dependent</span>, i.e. their comma bases will <span style="color: #3C8031;">share</span> one vector. You may already know this intuitively if you are familiar with the 5-limit [[projective tuning space]] diagram from the [[Paul_Erlich#Papers|Middle Path]] paper, which shows how we can draw a line through any two ETs and that line will represent a temperament, and the single comma that temperament tempers out is <span style="color: #3C8031;">this shared vector</span>. The diagram also tells us that any two 5-limit temperaments that temper out only a single comma will also be <span style="color: #3C8031;">linearly dependent</span>, for the opposite reason: their ''mappings'' will always <span style="color: #3C8031;">share</span> one vector.
+For matrices, negation is accomplished by choosing a single vector and changing the sign of every entry in it. In the case of comma bases, a vector is a column, whereas in a mapping a vector (technically a row vector, or covector) is a row.
-And we can see that there are no other diagrams of interest for <math>d=3</math>, because there's no sense in looking at diagrams with no <span style="color: #B6321C;">red band</span>, but we can't extend the <span style="color: #B6321C;">red band</span> any further than 1 vector on each side without going over the edge, and we can't lower the black bar any further without going below the center. So we're done. And our conclusion is that any pair of different <math>d=3</math> temperaments that are nontrivial (<math>0 < n < d=3</math> and <math>0 < r < d=3</math>) will be addable.
+For matrices, the check for negation is related to canonicalization of multivectors as are used in exterior algebra for RTT. Essentially we take the minors of the matrix, and then look at their leading or trailing entry (leading in the case of a covariant matrix, like a mapping; trailing in the case of a contravariant matrix, like a comma basis): if this entry is positive, so is the temperament, and vice versa.
-====Completing the suite of <math>d=4</math> examples====
+====5. Entry-wise add====
-Okay, back to <math>d=4</math>. We've already looked at the <math>g_{\text{min}}=1</math> possibility (which, for any <math>d</math>, there will only ever be one of). So let's start looking at the possibilities where <math>g_{\text{min}}=2</math>, which in the case of <math>d=4</math> leaves us only one pair of values for <math>r</math> and <math>n</math>: both being 2.
+The entry-wise addition of elements works mostly the same as for vectors. But there's one catch: we only do it for the pair of <span style="color: #B6321C;">linearly independent vectors</span>. We set the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> aside, and reintroduce it at the end.
-{| class="wikitable"
+When taking the sum, this is just for simplicity's sake. There's no sense in adding the two copies of the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> together, as we'll just get the same vector but 2-enfactored. So we may as well set it aside, and deal only with the <span style="color: #B6321C;">linearly independent vectors</span>, and put it back at the end.
-|+
-| rowspan="4" |<math>d=4</math>
+When taking the difference, it's essential that we set the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> aside before entry-wise arithmetic, though, because if we were to subtract it from itself, we'd end up with all zeros. Unlike the case of the sum, where we'd just end up with an enfactored version of the starting vectors, we couldn't even defactor to get back to where we started if we completely wiped out the relevant information by sending it all to zeros.
-| rowspan="2" style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=2</math>
-|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
+As a final step, as is always good to do when concluding temperament operations, put the result in [[canonical form]].
-|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
-|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
+===Example===
-|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
-|style="border-bottom: 1px solid #B6321C;"|
+For our example, let’s look at septimal meantone plus flattone. The canonical forms of these temperaments are {{ket|{{map|1 0 -4 -13}} {{map|0 1 4 10}}}} and {{ket|{{map|1 0 -4 17}} {{map|0 1 4 -9}}}}.
-|-
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
+'''0. Counterexample.''' Before we try following the detailed instructions just described above, let's do the counterexample, to illustrate why we have to follow them at all. Simple entry-wise addition of these two mapping matrices gives {{ket|{{map|2 0 -8 4}} {{map|0 2 8 1}}}}, which is not the correct answer:
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
+<math>\left[ \begin{array} {rrr}
-| style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
-|-
+& 0 & -4 & -13 \\
-| rowspan="2" |<math>g_{\text{max}}=2</math>
+& 1 & 4 & 10 \\
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
-|-
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|
-|}
-But even with <math>d</math>, <math>r</math>, and <math>n</math> fixed, we still have more than one possibility for <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>. The above diagram shows <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>. The below diagram shows <span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>.
+\end{array} \right]
++
+\left[ \begin{array} {rrr}
-{| class="wikitable"
+& 0 & -4 & 17 \\
-|+
+& 1 & 4 & -9 \\
-| rowspan="4" |<math>d=4</math>
-| rowspan="2" style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=2</math>
-| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| rowspan="2" style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>
-|-
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|-
-| rowspan="2" |<math>g_{\text{max}}=2</math>
-| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-| rowspan="2" style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>
-|-
-|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|}
-In the former possibility, where <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span> (and therefore the temperaments are addable), we have a pair of different <math>d=4</math>, <math>r=2</math> temperaments where we can find a single comma that both temperaments temper out, and — equivalently — we can find one ET that supports both temperaments.
+\end{array} \right]
+=
+\left[ \begin{array} {rrr}
+& 0 & -8 & 4 \\
+& 2 & 8 & 1 \\
+\end{array} \right]</math>
+And it's wrong not only because it is clearly enfactored (at least one factor of 2, that is visible in the first vector). The full explanation of why this is the wrong answer is beyond the scope of this example. However, if we now follow through with the instructions described above, we can find the correct answer.
+'''1. Find the linear-dependence basis.''' We know where to start: first find the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> and put each of these two mappings into a form that includes it explicitly. In this case, their <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> consists of a single vector: <span style="color: #3C8031;">{{ket|{{map|19 30 44 53}}}}</span>.
-In the latter possibility, where <span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>, neither side of duality <span style="color: #3C8031;">shares</span> any vectors in common. And so we've encountered our first example that is not addable. In other words, if the <span style="color: #B6321C;">red band</span> ever extends more than 1 vector away from the black bar, temperament arithmetic is not possible. So <math>d=4</math> is the first time we had enough room (half of <math>d</math>) to support that condition.
+'''2. Reproduce the original temperament.''' The original matrices had two vectors, so as our second step, we pad out these matrices by drawing from vectors from the original matrices, starting from their first vectors, so now we have [<span style="color: #3C8031;">{{map|19 30 44 53}}</span> {{map|1 0 -4 -13}}⟩ and [<span style="color: #3C8031;">{{map|19 30 44 53}}</span> {{map|1 0 -4 17}}⟩. We could choose any vectors from the original matrices, as long as they are <span style="color: #B6321C;">linearly independent</span> from the ones we already have; if one is not, skip it and move on. In this case the first vectors are both fine, though.
-We have now exhausted the possibility space for <math>d=4</math>. We can't extend either the <span style="color: #B6321C;">red band</span> or the black bar any further.
-====<math>d=5</math> diagrams finally reveal important relationships====
+<math>\left[ \begin{array} {rrr}
-So how about we go to <math>d=5</math> (such as the 11-limit). As usual, starting with <math>g_{\text{min}}=1</math>:
+\color{OliveGreen}19 & \color{OliveGreen}30 & \color{OliveGreen}44 & \color{OliveGreen}53 \\
-{| class="wikitable"
+& 0 & -4 & -13 \\
-|+
-| rowspan="5" |<math>d=5</math>
+\end{array} \right]
-|style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=1</math>
++
-| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
+\left[ \begin{array} {rrr}
-| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
+\color{OliveGreen}19 & \color{OliveGreen}30 & \color{OliveGreen}44 & \color{OliveGreen}53 \\
-| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
+& 0 & -4 & 17 \\
-| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
+\end{array} \right]</math>
-|-
-| rowspan="4" |<math>g_{\text{max}}=4</math>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
+'''3. Defactor.''' Next, verify that both matrices are defactored. In this case, both matrices ''are'' enfactored, each by a factor of 30<ref>or you may prefer to think of this as three different (prime) factors: 2, 3, 5 (which multiply to 30)</ref>. So we'll use addabilization defactoring. Since there's only a single vector in the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>, therefore all we need to do is repeatedly add that <span style="color: #3C8031;">one linearly dependent vector</span> to the <span style="color: #B6321C;">linearly independent vector</span> until we find a vector with the target GCF, which we can then simply divide out to defactor the matrix. Specifically, we add 11 times the <span style="color: #3C8031;">linearly dependent vector</span>. For the first matrix, {{map|1 0 -4 -13}} + 11⋅<span style="color: #3C8031;">{{map|19 30 44 53}}</span> = {{map|210 330 480 570}}, whose entries have a GCF = 30, so we can defactor the matrix by dividing that vector by 30, leaving us with <span style="color: #B6321C;">{{map|7 11 16 19}}</span>. Therefore the final matrix here is [<span style="color: #3C8031;">{{map|19 30 44 53}}</span> {{map|7 11 16 19}}⟩. The other matrix matrix happens to defactor in the same way: {{map|1 0 -4 17}} + 11⋅<span style="color: #3C8031;">{{map|19 30 44 53}}</span> = {{map|210 330 480 600}} whose GCF is also 30, reducing to <span style="color: #B6321C;">{{map|7 11 16 20}}</span>, so the final matrix is [<span style="color: #3C8031;">{{map|19 30 44 53}}</span> <span style="color: #B6321C;">{{map|7 11 16 20}}</span>⟩.
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
+'''4. Check negativity.''' The next thing we need to do is check the negativity of these two temperaments. If either of the matrices we're performing arithmetic on is  negative, then we'll have to change it (if both are negative, then the problem cancels out, and we go back to being right). We check negativity by using the minors of these matrices. The first matrix's minors are (-1, -4, -10, -4, -13, -12) and the second matrix's minors are (-1, -4, 9, -4, 17, 32). What we're looking for here are their leading entries, because these are minors of a mapping (if we were looking at minors of comma bases, we'd be looking at the trailing entries instead). Specifically, we're looking to see if the leading entries are positive. They're not. Which tells us these matrices are both negative! But again, since they were ''both'' negative, the effect cancels out; no need to change anything (but if we wanted to, we could just take the <span style="color: #B6321C;">linearly independent vector</span> for each matrix and negate every entry in it).
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
-|-
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-| rowspan="3" |
-|-
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|-
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|}
-Just as with the <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span> diagrams given for <math>d=3</math> and <math>d=5</math>, we can see these are addable temperaments.
+'''5. Add.''' Now the matrices are ready to add:
-Now let's look at <math>d=5</math> but with <math>g_{\text{min}}=2</math>. This presents two possibilities. First, <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>:
-{| class="wikitable"
+<math>\left[ \begin{array} {rrr}
-|+
-| rowspan="5" |<math>d=5</math>
+\color{OliveGreen}19 & \color{OliveGreen}30 & \color{OliveGreen}44 & \color{OliveGreen}53 \\
-| rowspan="2" style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=2</math>
+\color{BrickRed}7 & \color{BrickRed}11 & \color{BrickRed}16 & \color{BrickRed}19 \\
-|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
-|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
+\end{array} \right]
-|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
++
-|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
+\left[ \begin{array} {rrr}
-|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
-| style="border-bottom: 1px solid #B6321C;" |
+\color{OliveGreen}19 & \color{OliveGreen}30 & \color{OliveGreen}44 & \color{OliveGreen}53 \\
-|-
+\color{BrickRed}7 & \color{BrickRed}11 & \color{BrickRed}16 & \color{BrickRed}20 \\
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
+\end{array} \right]</math>
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-| style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
-|-
-| rowspan="3" |<math>g_{\text{max}}=3</math>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
-|-
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|
-|-
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|
-|}
-And second, <span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>:
+We set the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> aside, and deal only with the <span style="color: #B6321C;">linearly independent vectors</span>:
+<math>\left[ \begin{array} {rrr}
+\color{BrickRed}7 & \color{BrickRed}11 & \color{BrickRed}16 & \color{BrickRed}19 \\
+\end{array} \right]
++
+\left[ \begin{array} {rrr}
+\color{BrickRed}7 & \color{BrickRed}11 & \color{BrickRed}16 & \color{BrickRed}20 \\
-{| class="wikitable"
+\end{array} \right]
-|+
+=
-| rowspan="5" |<math>d=5</math>
+\left[ \begin{array} {rrr}
-| rowspan="2" style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=2</math>
-| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
+\color{BrickRed}14 & \color{BrickRed}22 & \color{BrickRed}32 & \color{BrickRed}39 \\
-| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
+\end{array} \right]</math>
-| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-| rowspan="2" style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>
+Then we can reintroduce the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> afterwards:
-|-
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
+<math>\left[ \begin{array} {rrr}
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
+\color{OliveGreen}19 & \color{OliveGreen}30 & \color{OliveGreen}44 & \color{OliveGreen}53 \\
-|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
+\color{BrickRed}14 & \color{BrickRed}22 & \color{BrickRed}32 & \color{BrickRed}39 \\
-|-
-| rowspan="3" |<math>g_{\text{max}}=3</math>
+\end{array} \right]</math>
-| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-| rowspan="2" style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>
-|-
-|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|-
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|style="background-color: #BED5BA;"|
-|
-|}
-Here's where things really get interesting. Because in both of these cases, the pairs of temperaments represented are <span style="color: #3C8031;">linearly dependent</span> on each other (i.e. either their mappings are <span style="color: #3C8031;">linearly dependent</span>, their comma bases are <span style="color: #3C8031;">linearly dependent</span>, or both). And so far, every possibility where temperaments have been <span style="color: #3C8031;">linearly dependent</span>, they have also been <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>, and therefore addable. But if you look at the second case here, we are <span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>, but since <math>d=5</math>, the temperaments still manage to be <span style="color: #3C8031;">linearly dependent</span>. So this is the first example of a <span style="color: #3C8031;">linearly dependent</span> temperament pairing which is not addable.
-====Back to <math>d=2</math>, for a surprisingly tricky example====
+And finally [[canonical form|canonicalize]]:
-Beyond <math>d=5</math>, these diagrams get cumbersome to prepare, and cease to reveal further insights. But if we step back down to <math>d=2</math>, a place simpler than anywhere we've looked so far, we actually find another surprisingly tricky example, which is hopefully still illuminating.
-So <math>d=2</math> (such as the 3-limit) presents another case — similar to the <math>d=5</math>, <math>g_{\text{min}}=2</math>, <span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span> case explored most recently above — where the properties of <span style="color: #3C8031;">linearly dependence</span> and addability do not match each other. But while in the other case, we had a temperament pair that was <span style="color: #3C8031;">linearly dependent</span> yet not addable, in this <math>d=2</math> (and therefore <math>g_{\text{min}}=1</math>, <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>) case, it is the other way around: addable yet <span style="color: #B6321C;">linearly independent</span>!
+<math>\left[ \begin{array} {rrr}
-{| class="wikitable"
+& 0 & -4 & 2 \\
-|+
+& 2 & 8 & 1 \\
-| rowspan="2" |<math>d=2</math>
-|style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=1</math>
-|style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↑  </span>
-|style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↑  </span>
-| style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
-|-
-|<math>g_{\text{max}}=1</math>
-|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-| style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
-|}
-Basically, in the case of <math>d=2</math>, <math>g_{\text{max}}=1</math> (in non-trivial cases, i.e. not JI or the unison temperament), so any two different ETs or commas you pick are going to be <span style="color: #B6321C;">linearly independent</span> (because the only way they could be <span style="color: #3C8031;">linearly dependent</span> would be to be the same temperament). And yet we know we can still entry-wise add them to new vectors that are [[Douglas_Blumeyer_and_Dave_Keenan%27s_Intro_to_exterior_algebra_for_RTT#Decomposability|decomposable]], because they're already vectors (decomposing means to express a [[Douglas_Blumeyer_and_Dave_Keenan%27s_Intro_to_exterior_algebra_for_RTT#From_vectors_to_multivectors|multivector]] in the form of a list of monovectors, so decomposing a multivector that's already a monovector like this is tantamount to merely putting array braces around it.)
+\end{array} \right]</math>
-====Conclusion====
-This explanation has hopefully helped get a grip on how addability AKA <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span> behaves given various temperament dimensions. But it still hasn't quite explained why <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span> is one and the same thing as addability. We will look at this in another section soon.
+so we can now see that meantone plus flattone is [[godzilla]].
-===Geometric explanation===
+As long as we've done all this work to set these matrices up for arithmetic, let's check their difference as well. Again, set aside the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>, and just entry-wise subtract the two <span style="color: #B6321C;">linearly independent vectors</span>:
-We've presented a diagrammatic illustration of the behavior of <span style="color: #B6321C;">linear-independence <math>l_{\text{ind}}</math></span> with respect to temperament dimensions. But some of the results might have seemed surprising. For instance, when looking at the diagram for <math>d=4, g_{\text{min}}=1, g_{\text{max}}=3</math>, it might have seemed intuitive enough that the the <span style="color: #3C8031;">red band</span> could not extend beyond the square grid, but then again, why shouldn't it be possible to have, say, two 7-limit ETs which temper out only a single comma in common? Perhaps it doesn't seem clear that this is impossible, and that they must temper out two commas in common (and of course the infinitude of combinations of these two commas). If this is as unclear to you as it was to the author when exploring this topic, then this explanatory section is for you! Here, we will use geometrical representations of temperaments to hone our intuitions about the possible combinations of dimensions and <span style="color: #B6321C;">linear-independence <math>l_{\text{ind}}</math></span> of temperaments.
-In this approach, we’re actually not going to focus directly on the <span style="color: #B6321C;">linear-independence <math>l_{\text{ind}}</math></span> of temperaments. Instead, we're going to look at the <span style="color: #3C8031;">linear-''de''pendence <math>l_{\text{dep}}</math></span> of matrices representing temperaments such as mappings and comma bases, and then compute the <span style="color: #B6321C;">linear-independence <math>l_{\text{ind}}</math></span> from it and the grade <math>g</math>. As we’ve established, the <span style="color: #3C8031;">linear-dependence <math>l_{\text{dep}}</math></span> differs from one side of duality to the other, so we’ll only be looking at one side of duality at a time.
+<math>\left[ \begin{array} {rrr}
-====Introduction====
+\color{BrickRed}7 & \color{BrickRed}11 & \color{BrickRed}16 & \color{BrickRed}19 \\
+\end{array} \right]</math>
+-
+<math>\left[ \begin{array} {rrr}
+\color{BrickRed}7 & \color{BrickRed}11 & \color{BrickRed}16 & \color{BrickRed}20 \\
+\end{array} \right]
+=
+\left[ \begin{array} {rrr}
+\color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}-1 \\
+\end{array} \right]</math>
-In this geometric approach, we'll be imagining individual vectors as points (0D), sets of two vectors as lines (1D), sets of three as planes (2D), four as volumes (3D), and so forth, as according to this table:
-{| class="wikitable center-all"
-|+
-!vector
-count
-!geometric
-dimension
-!form
-|-
-|0
-|undefined
-|(emptiness)
-|-
-|1
-|0
-|point
-|-
-|2
-|1
-|line
-|-
-|3
-|2
-|plane
-|-
-|4
-|3
-|volume
-|-
-|5
-|4
-|hypervolume
-|-
-| ⋮
-|⋮
-|⋮
-|}
-This is a "vector space", and these geometric dimensions are consistent with how temperaments represented by these counts of vectors appear in ''projective'' vector space, which reduces geometric dimensions by 1. For example, a vector has a geometric interpretation as a directed line segment, which is 1D, but a point is 0D, which is one dimension lower. Essentially what we're doing is ''assuming the origin''.
+And so, reintroducing the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>, we have:
-Think of it this way: geometric points are zero-dimensional, simply representing a position in space, whereas linear algebra vectors are one-dimensional, representing both a magnitude and direction; the way vectors manage to encode this extra dimension without providing any additional information is by being understood to describe this position in space ''relative to an origin''. Well, so we'll now switch our interpretation of these objects to the geometric one, where the vector's entries are nothing more than a coordinate for a point in space. And the "projection" involved in projective vector space essentially positions us at this discarded origin, looking out from it upon every individual point, which accomplishes the same feat, in a visual way.
-Perhaps an example may help clarify this setup. Suppose we've got an (x,y,z) space, and two coordinates (5,8,12) and (7,11,16). You should recognize these as the simple maps for 5-ET and 7-ET, usually written as {{map|5 8 12}} and {{map|7 11 16}}, respectively. Ask for the equation of the plane defined by the three points (5,8,12), (7,11,16), and the origin (0,0,0) and you'll get -4x + 4y -1z = 0, which clearly shows us the entries of the meantone comma. That's because meantone temperament can be defined by these two maps. 5-limit JI is a 3D space, and meantone temperament, as a rank-2 temperament, would be a 2D plane. But we don't normally need to think of the map corresponding to the origin, where everything is tempered out, including meantone. So we can just assume it, and think of a 2D plane as being defined by only 2 points, which in a view projected (from the origin) will look like just the line connecting (5,8,12) and (7,11,16).
+<math>\left[ \begin{array} {rrr}
-So, we've set the stage for our projective vector spaces. We will now be looking at representations of temperaments as counts of vector sets, and then using this scheme to convert them to primitive geometric forms. We'll place two of each form into the space, representing the two temperaments having the possibility of arithmetic performed on them checked. Then we will observe their possible <span style="color: #3C8031;">''intersections''</span> depending on how they're oriented in space, and it's these <span style="color: #3C8031;">intersections that represent their linear-dependence</span>. When the dimension of the <span style="color: #3C8031;">intersection</span> is then converted back to a vector set count, then we have their <span style="color: #3C8031;">linear-dependence <math>l_{\text{dep}}</math></span>, for this side of duality, anyway (remember, unlike the <span style="color: #B6321C;">linear-independence <math>l_{\text{ind}}</math></span>, this value isn't necessarily the same on both sides of duality). We can finally subtract the <span style="color: #3C8031;">linear-dependence</span> from the grade (vector count) to get the <span style="color: #B6321C;">linear-indepedence</span>, in order to determine if the two temperaments are addable.
+\color{OliveGreen}19 & \color{OliveGreen}30 & \color{OliveGreen}44 & \color{OliveGreen}53 \\
+\color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}-1 \\
-In these examples, we'll be assuming that no two temperaments being compared are the same, because performing temperament arithmetic on copies of the same temperament is not interesting. The other things we'll be assuming is that no lines, planes, etc. are parallel to each other; this is due to a strange effect touched upon in footnote 4 whereby temperament geometry that appears parallel in projective space actually still intersects; the present author asks that if anyone is able to demystify this situation, that they please do!
+\end{array} \right]</math>
-====At <math>d=3</math>====
-First, let's establish the geometric dimension of the space. With <math>d=3</math>, we've got a 2D space (one less than 3), so the entire space can be visualized on a plane.
+Which canonicalizes to:
-Our only possible values for <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> here are 1 and 2, respectively. So these are the two possible counts of vectors <math>g</math> possessed by matrices representing temperaments here.
-So let's look at temperaments represented by matrices with 1 vector first (<math>g=1</math>). In this case, each of the two temperaments is a point in the plane. Unless these two temperaments are the same temperament, the <span style="color: #3C8031;">intersection</span> of these two points is empty. Emptiness isn't even 0D! So that tells us that these temperaments have 0 vectors worth of <span style="color: #3C8031;">linear dependence</span>. With <math>g=1</math>, that gives us a <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math> = g</math> <math> - </math> <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span> <math>= 1 -</math> <span style="color: #3C8031;"><math>0</math></span> <math>= 1</math>:
+<math>\left[ \begin{array} {rrr}
-[[File:D3 g1 dep0.png|200px|none]]
+& 30 & 44 & 0 \\
+& 0 & 0 & 1 \\
-Next, let's look at temperaments represented by matrices with 2 vectors (<math>g=2</math>). In this case, each of the two temperaments is a line in the plane. Again, assuming the two lines are not the same line or parallel, their <span style="color: #3C8031;">intersection</span> is a point. Being 0D, that tells us that the <span style="color: #3C8031;">linear-dependence</span> of these matrices is 1. So that gives us an <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span> <math>= g</math> <math>-</math> <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span> <math>= 2 -</math> <span style="color: #3C8031;"><math>1</math></span> <math>= 1</math>. This matches the value we found via the <math>g=1</math>, so we've effectively checked our work:
+\end{array} \right]</math>
-[[File:D3 g2 dep1.png|200px|none]]
-====At <math>d=2</math>====
+And so we can see that meantone minus flattone is [[meanmag]].
-Let's step back to <math>d=2</math>. Here we've got a 2 minus 1 equals 1D space, so the entire space can be visualized on a single line (one direction corresponds to an increasing ratio between the two coordinates, and the other to a decreasing ratio).
+== Arithmetic on non-addable temperaments ==
-We know our only possible value for <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> here is 1. So in either case, each of the two temperaments is a point on the line. As with two points in a plane — when <math>d=3</math> — unless these two temperaments are the same temperament, the intersection of these two points is empty. So again the <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math> = g = 1</math>:
+== Further explanations ==
-[[File:D2 g1 dep0.png|200px|none]]
+===Diagrammatic explanation===
-====At <math>d=4</math>====
+====Introduction====
-First, let's establish the geometric dimension of the space. With <math>d=4</math>, we've got a 3D space (one less than 4), so the entire space can be visualized in a volume.
+The diagrams used for this explanation were inspired in part by [[Kite Giedraitis|Kite]]'s [[gencom]]s, and specifically how in his "twin squares" matrices — which have dimensions <math>d×d</math> — one can imagine shifting a bar up and down to change the boundary between vectors that form a basis for the commas and those that form a basis for preimage intervals (this basis is typically called "the [[generator]]s"). The count of the former is the nullity <math>n</math>, and the count of the latter is the rank <math>r</math>, and the shifting of the boundary bar between them with the total <math>d</math> vectors corresponds to the insight of the rank-nullity theorem, which states that <math>r + n=d</math>. And so this diagram's square grid has just the right amount of room to portray both the mapping and the comma basis for a given temperament (with the comma basis's vectors rotated 90 degrees to appear as rows, to match up with the rows of the mapping).
-At <math>d=4</math>, we have a couple options for the grade: either <math>g_{\text{min}}=1</math> and <math>g_{\text{max}}=3</math>, or both <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> equal 2.
+So consider this first example of such a diagram:
-Let's look at temperaments represented by matrices with 1 vector first (<math>g=1</math>). Yet again, we find ourselves with two separate points, but now we find them in a space that's not a line, not a plane, but a volume. This doesn't change <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math> = g = 1</math>, so we're not even going to show it, or any further cases of <math>g=1</math>. These are all addable.
+{| class="wikitable"
+|+
-And when <math>g=3</math>, because this is paired with <math>g=1</math> from the min and max values, we should expect to get the same answer as with <math>g=1</math>. And indeed, it will check out that way. Because two <math>g=3</math> temperaments will be planes in this volume, and the intersection of two planes is a line. Which means that <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span><math> = 2</math>. And so <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math> = g</math> <math> - </math> <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span> <math>= 3 -</math> <span style="color: #3C8031;"><math>2</math></span> <math>= 1</math>. And here's where our geometric approach begins to pay off! This was the example given at the beginning that might seem unintuitive when relying only on the diagrammatic approach. But here we can see clearly that there would be no way for two planes in a volume to intersect only at a point, which proves the fact that two 7-limit ETs could never only temper out a single comma in common.
+| rowspan="4" |<math>d=4</math>
+| style="border-bottom: 3px solid black;"|<math>g_{\text{min}}=1</math>
-[[File:D4 g3 dep2.png|200px]]
+| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
+| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-Next let's look at temperaments represented by matrices with 2 vectors, that is, when both <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> are equal to 2. What are the possible ways lines can occupy a volume together? In a plane, as it was with <math>d=3</math> (and again assuming no parallel objects in these examples), they must intersect. But in a volume, here in <math>d=4</math>, this is possible. So, with <math>g=2</math>, it is possible to have a <span style="color: #B6321C;"><math>l_{\text{dep}}</math></span> <math>= 0</math>, which leads to <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math> = g</math> <math> - </math> <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span> <math>= 2 -</math> <span style="color: #3C8031;"><math>0</math></span> <math>= 2</math>. Not addable in this case.
+| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
+| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-[[File:D4 g2 dep0.png|200px]]
+| style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
+|-
-But we can also imagine two lines in a volume that do intersect at a point. This is the case where <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math> = g</math> <math> - </math> <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span> <math>= 2 -</math> <span style="color: #3C8031;"><math>1</math></span> <math>= 1</math>: addable!
+| rowspan="3" |<math>g_{\text{max}}=3</math>
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
+| style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C; "><math>l_{\text{ind}}=1</math></span>
+|-
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
+| rowspan="2" |
+|-
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
+|}
-[[File:D4 g2 dep1.png|200px]]
+This represents a <math>d=4</math> temperament. These diagrams are grade-agnostic, which is to say that they are agnostic as to which side counts the <math>r</math> and which side counts the <math>n</math>. So we are showing them as <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> instead. We could say there's a variation on the rank-nullity theorem whereby <math>g_{\text{min}} + g_{\text{max}}=d</math>, just as <math>r + n=d</math>. So we can then say that this diagram represents either a <math>r=1</math>, <math>n=3</math> temperament, or perhaps a <math>n=1</math>, <math>r=3</math> temperament.
-==== At <math>d=5</math>====
+But actually, this diagram represents more than just a single temperament. It represents a relationship between a pair of temperaments (which have the same [[dimensions]], non-grade-agnostically, i.e. not a pairing of a <math>r=1</math>, <math>n=3</math> temperament with a <math>r=3</math>, <math>n=1</math> temperament). As elsewhere, <span style="color: #3C8031;">green coloration indicates the linearly dependent basis vectors <math>L_{\text{dep}}</math></span> between this pair of temperaments, and <span style="color: #B6321C;">red coloration indicates linearly ''in''dependent basis vectors <math>L_{\text{ind}}</math></span> between the same pair of temperaments.
-First, let's establish the geometric dimension of the space. With <math>d=5</math>, we've got a 4D space (one less than 5), so the entire space can be visualized in a hypervolume. We've now gone beyond the dimensionality of physical reality, so things get a little harder to conceptualize unfortunately. But <math>d=5</math> is the first <math>d</math> where we can make an important point about addability, so please bear with!
+So, in this case, the two ET maps are <span style="color: #B6321C;">linearly independent</span>. This should be unsurprising; because ET maps are constituted by only a single vector (they're <math>r=1</math> by definition), if they ''were'' <span style="color: #3C8031;">linearly dependent</span>, then they'd necessarily be the ''same'' exact ET! Temperament arithmetic on two of the same ET is never interesting; <math>T_1</math> plus <math>T_1</math> simply equals <math>T_1</math> again, and <math>T_1</math> minus <math>T_1</math> is undefined. That said, if we ''were'' to represent temperament arithmetic between two of the same temperament on such a diagram as this, then every cell would be green. And this is true regardless whether <math>r=1</math> or otherwise.
-At <math>d=5</math>, we also have a couple options for the grade: either <math>g_{\text{min}}=1</math> and <math>g_{\text{max}}=4</math>, or <math>g_{\text{min}}=2</math> and <math>g_{\text{max}}=3</math>.
+From this information, we can see that the comma bases of any randomly selected pair of ''different'' <math>d=4</math> ETs are going to <span style="color: #3C8031;">share 2 vectors</span>, or in other words, their <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> will have two basis vectors. In terms of the diagram, we're saying that they'll always have two <span style="color: #3C8031;">green-colored vectors</span> under the black bar.
-First we'll look at <math>g_{\text{min}}=1</math> and <math>g_{\text{max}}=4</math>. Temperament matrices with <math>g=1</math> are still addable. And temperament matrices with <math>g=4</math> should be too. We can see this visually as how two volumes in a hypervolume together will have an intersection the shape of a plane. We can now see that there's a generalizable principal that any two <math>(d-1)</math>-dimensional objects will necessarily have a <math>(d-2)</math>-dimensional intersection, and thus have <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span> <math>= 1</math> and be addable. So we won't need to show this one or any further like it, either.
+These diagrams are a good way to understand which temperament relationships are possible and which aren't, where by a "relationship" here we mean a particular combination of their matching dimensions and their linear-independence integer count. A good way to use these diagrams for this purpose is to imagine the <span style="color: #B6321C;">red coloration</span> emanating away from the black bar in both directions simultaneously, one pair of vectors at a time. Doing it like this captures the fact, as previously stated, that the <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span> on either side of duality is always equal. There's no notion of a max or min here, as there is with <math>g</math> or <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span>; the <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span> on either side is always the same, so we can capture it with a single number, which counts the <span style="color: #B6321C;">red vectors</span> on just one half (that is, half of the total count of <span style="color: #B6321C;">red vectors</span>, or half of the width of the <span style="color: #B6321C;">red band</span> in the middle of the grid).
+There's no need to look at diagrams like this where the black bar is below the center. This is because, even though for convenience we're currently treating the top half as <math>r</math> and the bottom half as <math>n</math>, these diagrams are ultimately grade-agnostic. So we could say that each one essentially represents not just one possibility for the relationship between two temperaments' dimensions and <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span>, but ''two'' such possibilities. Again, this diagram equally represents both <math>d=4, r=1, n=3, </math><span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span> as well as <math>d=4, r=3, n=1, </math><span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>. Which is another way of saying we could vertically mirror it without changing it.
-So let's look at temperament matrices with <math>g_{\text{min}}=2</math> and <math>g_{\text{max}}=3</math>. For <math>g=2</math>, we have two possible values for <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span>: 0 or 1. Meaning that either the two lines through this hypervolume do not intersect (0), or they intersect at a point (1). These diagrams would look very much like the corresponding diagrams for <math>d=4</math>, so we will not be showing them. But what about when <math>g=3</math>? We can certainly imagine two planes in a hypervolume intersecting at a line, just as they do in an ordinary volume — they're just not taking advantage of the additional geometric dimension. So we won't show that example either. But where it gets really interesting is imagining then taking one of these two planes and rotating it in the fifth dimension; this causes the intersection between the two planes to be reduced down to a single point. And this corresponds with the case of <span style="color: #3C8031;"><math>l_{\text{dep}}=1</math></span> here, which means <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math> = g</math> <math> - </math> <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span> <math>= 3 -</math> <span style="color: #3C8031;"><math>1</math></span> <math>= 2</math>, so therefore not addable:
+With the black bar always either in the top half or exactly in the center, we can see that the emanating <span style="color: #B6321C;">red band</span> will always either hit the top edge of the square grid first, or they will hit both the top and bottom edges of it simultaneously. So this is how these diagrams visually convey the fact that the <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span> between two temperaments will always be less than or equal to their <math>g_{\text{min}}</math>: because a situation where <math>g_{\text{min}}>l</math> would visually look like the <span style="color: #B6321C;">red band</span> spilling past the edges of the square grid.
-[[File:D5 g3 dep1.png|300px]]
+We could also say that two temperaments are <span style="color: #3C8031;">linearly dependent</span> on each other when <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math><g_{\text{max}}</math>, that is, their <span style="color: #B6321C;">linear-independence</span> is less than their ''max''-grade.
-So for <math>g_{\text{min}}=2</math> and <math>g_{\text{max}}=3</math> we got two different possibilities for <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span>: 1 and 2, and for each of these two possibilities, we found it twice. We can see then that these match up, that is, that the <math>g_{\text{min}}=2</math> case with <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span> matches with the <math>g_{\text{max}}=3</math> case with <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>, and the <span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span> cases match in the same way.
+Perhaps more importantly, we can also see from these diagrams that any pair of <math>g_{\text{min}}=1</math> temperaments will be addable. Because if they are <math>g_{\text{min}}=1</math>, then the furthest the <span style="color: #B6321C;">red band</span> can extend from the black bar is 1 vector, and 1 mirrored set of <span style="color: #B6321C;">red vectors</span> means <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>, and that's the definition of addability.
-==== Conclusion====
+====A simple <math>d=3</math> example====
-Here's a summary table of our geometric findings so far:
+Let's back-pedal to <math>d=3</math> for a simple illustrative example.
- {| class="wikitable center-all"
+{| class="wikitable"
 |+
-! rowspan="2" |<math>d</math> ( <math>= g_{\text{min}} + g_{\text{max}}</math>)
+| rowspan="3" |<math>d=3</math>
-! colspan="2" |<math>g_{\text{min}}</math>
+|style="border-bottom: 3px solid black;"|<math>g_{\text{min}}=1</math>
-! colspan="2" |<math>g_{\text{max}}</math>
+| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-! rowspan="2" |<span style="color: #B6321C;"><math>l_{\text{ind}}</math></span> ( <math>= g -</math> <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span>)
+| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
+| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
+| style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
 |-
-!<math>g</math>
+| rowspan="2" |<math>g_{\text{max}}=2</math>
-!<span style="color: #3C8031;"><math>l_{\text{dep}}</math></span>
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-!<math>g</math>
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-!<span style="color: #3C8031;"><math>l_{\text{dep}}</math></span>
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
+| style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
 |-
-|2
+|style="background-color: #BED5BA;"|
-|1
+|style="background-color: #BED5BA;"|
-|0
+|style="background-color: #BED5BA;"|
-|1
+|
-|0
+|}
-|1
-|-
+This diagram shows us that any two <math>d=3</math>, <math>g_{\text{min}}=1</math> temperaments (like 5-limit ETs) will be <span style="color: #3C8031;">linearly dependent</span>, i.e. their comma bases will <span style="color: #3C8031;">share</span> one vector. You may already know this intuitively if you are familiar with the 5-limit [[projective tuning space]] diagram from the [[Paul_Erlich#Papers|Middle Path]] paper, which shows how we can draw a line through any two ETs and that line will represent a temperament, and the single comma that temperament tempers out is <span style="color: #3C8031;">this shared vector</span>. The diagram also tells us that any two 5-limit temperaments that temper out only a single comma will also be <span style="color: #3C8031;">linearly dependent</span>, for the opposite reason: their ''mappings'' will always <span style="color: #3C8031;">share</span> one vector.
-|3
-|1
+And we can see that there are no other diagrams of interest for <math>d=3</math>, because there's no sense in looking at diagrams with no <span style="color: #B6321C;">red band</span>, but we can't extend the <span style="color: #B6321C;">red band</span> any further than 1 vector on each side without going over the edge, and we can't lower the black bar any further without going below the center. So we're done. And our conclusion is that any pair of different <math>d=3</math> temperaments that are nontrivial (<math>0 < n < d=3</math> and <math>0 < r < d=3</math>) will be addable.
-|0
-|2
+====Completing the suite of <math>d=4</math> examples====
-|1
-|1
+Okay, back to <math>d=4</math>. We've already looked at the <math>g_{\text{min}}=1</math> possibility (which, for any <math>d</math>, there will only ever be one of). So let's start looking at the possibilities where <math>g_{\text{min}}=2</math>, which in the case of <math>d=4</math> leaves us only one pair of values for <math>r</math> and <math>n</math>: both being 2.
+{| class="wikitable"
+|+
+| rowspan="4" |<math>d=4</math>
+| rowspan="2" style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=2</math>
+|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
+|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
+|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
+|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
+|style="border-bottom: 1px solid #B6321C;"|
 |-
-| rowspan="3" |4
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-| 1
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|0
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|3
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|2
+| style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
-|1
 |-
-|2
+| rowspan="2" |<math>g_{\text{max}}=2</math>
-|0
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-|2
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-|0
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-|2
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
+| style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
 |-
-|2
+|style="background-color: #BED5BA;"|
-|1
+|style="background-color: #BED5BA;"|
-|2
+|style="background-color: #BED5BA;"|
-|1
+|style="background-color: #BED5BA;"|
-|1
+|
+|}
+But even with <math>d</math>, <math>r</math>, and <math>n</math> fixed, we still have more than one possibility for <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>. The above diagram shows <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>. The below diagram shows <span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>.
+{| class="wikitable"
+|+
+| rowspan="4" |<math>d=4</math>
+| rowspan="2" style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=2</math>
+| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
+| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
+| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
+| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
+| rowspan="2" style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>
 |-
-| rowspan="3" | 5
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|1
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|0
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|4
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|3
-|1
 |-
-|2
+| rowspan="2" |<math>g_{\text{max}}=2</math>
-|0
+| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-|3
+| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-|1
+| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-|2
+| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
+| rowspan="2" style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>
 |-
-|2
+|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|1
+|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|3
+|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|2
+|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|1
 |}
-The geometric explanation still hasn't answered the question as to why <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span> is the condition on addability. It just increased our intuitions about its relationship with temperament dimensions. We'll still look to a later section for an answer on this.
+In the former possibility, where <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span> (and therefore the temperaments are addable), we have a pair of different <math>d=4</math>, <math>r=2</math> temperaments where we can find a single comma that both temperaments temper out, and — equivalently — we can find one ET that supports both temperaments.
-=== Algebraic explanation===
+In the latter possibility, where <span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>, neither side of duality <span style="color: #3C8031;">shares</span> any vectors in common. And so we've encountered our first example that is not addable. In other words, if the <span style="color: #B6321C;">red band</span> ever extends more than 1 vector away from the black bar, temperament arithmetic is not possible. So <math>d=4</math> is the first time we had enough room (half of <math>d</math>) to support that condition.
-This explanation relies on comparing the results of the multivector and matrix approaches to temperament arithmetic, and showing algebraically how the matrix approach can only achieve the same answer as the multivector approach on the condition that it keeps all but one vector between the added matrices the same, that is, not only are the temperaments addable, but their <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> appears explicitly in the added matrices.
+We have now exhausted the possibility space for <math>d=4</math>. We can't extend either the <span style="color: #B6321C;">red band</span> or the black bar any further.
-To compare results, we eventually get both approaches into a multivector form. With the multivector approach, we wedge the vector set first and then add the resultant multivectors to get a new multivector. With the matrix approach, we treat the vector set as a matrix and add first, then treat the resultant matrix as a vector set and wedge those vectors to get a new multivector.
+====<math>d=5</math> diagrams finally reveal important relationships====
-The diagrams below are organized into a 2×2 layout. The left part shows the multivector approach, and the right part shows the matrix approach. The top part shows how the results of two approaches match when the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> is successfully explicit (and in these cases, the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> vectors are highlighted in green and the <span style="color: #B6321C;"><math>L_{\text{ind}}</math></span> vectors are highlighted in red), and the bottom part shows how the results fail to match when it is not. Successful matches are highlighted in yellow and failures to match are highlighted in blue.
+So how about we go to <math>d=5</math> (such as the 11-limit). As usual, starting with <math>g_{\text{min}}=1</math>:
+{| class="wikitable"
-This first diagram demonstrates this situation for a <math>d=3, g=2</math> case.
-{| class="wikitable center-all"
 |+
-!
+| rowspan="5" |<math>d=5</math>
-!
+|style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=1</math>
-!
+| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-! colspan="11" |
+| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-!
+| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-! colspan="11" |
+| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-!
+| style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
+| style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
 |-
-!
+| rowspan="4" |<math>g_{\text{max}}=4</math>
-!
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-!
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| colspan="11" rowspan="1" |'''multivector approach'''
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-!
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-| colspan="11" rowspan="1" |'''matrix approach'''
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-!
+| style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
 |-
-!
+|style="background-color: #BED5BA;"|
-!
+|style="background-color: #BED5BA;"|
-!
+|style="background-color: #BED5BA;"|
-! colspan="11" |
+|style="background-color: #BED5BA;"|
-!
+|style="background-color: #BED5BA;"|
-! colspan="11" |
+| rowspan="3" |
-!
 |-
-! rowspan="5" |
+|style="background-color: #BED5BA;"|
-| colspan="1" rowspan="5" |explicit <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
-{{bra|{{vector|<math>a</math> <math>b</math> <math>c</math>}}}}
+|style="background-color: #BED5BA;"|
-! rowspan="5" |
+|style="background-color: #BED5BA;"|
-| style="background-color: #BED5BA;"|<math>a</math>
-| style="background-color: #BED5BA;"|<math>b</math>
-| style="background-color: #BED5BA;"|<math>c</math>
-| rowspan="2" |
-| style="background-color: #BED5BA;"|<math>a</math>
-| style="background-color: #BED5BA;"|<math>b</math>
-| style="background-color: #BED5BA;"|<math>c</math>
-| rowspan="2" |
-| colspan="3" rowspan="2" |
-! rowspan="5" |
-| style="background-color: #BED5BA;"|<math>a</math>
-| style="background-color: #BED5BA;"|<math>b</math>
-| style="background-color: #BED5BA;"|<math>c</math>
-| colspan="1" rowspan="2" |<math>+</math>
-| style="background-color: #BED5BA;"|<math>a</math>
-| style="background-color: #BED5BA;"|<math>b</math>
-| style="background-color: #BED5BA;"|<math>c</math>
-| colspan="1" rowspan="2" |<math>=</math>
-|<math>2a</math>
-|<math>2b</math>
-|<math>2c</math>
-! rowspan="5" |
 |-
-| style="background-color: #E7BBB3;"|<math>d</math>
+|style="background-color: #BED5BA;"|
-| style="background-color: #E7BBB3;"|<math>e</math>
+|style="background-color: #BED5BA;"|
-| style="background-color: #E7BBB3;"|<math>f</math>
+|style="background-color: #BED5BA;"|
-| style="background-color: #E7BBB3;"|<math>g</math>
+|style="background-color: #BED5BA;"|
-| style="background-color: #E7BBB3;"|<math>h</math>
+|style="background-color: #BED5BA;"|
-| style="background-color: #E7BBB3;"|<math>i</math>
+|}
-| style="background-color: #E7BBB3;"|<math>d</math>
-| style="background-color: #E7BBB3;"|<math>e</math>
+Just as with the <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span> diagrams given for <math>d=3</math> and <math>d=5</math>, we can see these are addable temperaments.
-| style="background-color: #E7BBB3;"|<math>f</math>
-| style="background-color: #E7BBB3;"|<math>g</math>
+Now let's look at <math>d=5</math> but with <math>g_{\text{min}}=2</math>. This presents two possibilities. First, <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>:
-| style="background-color: #E7BBB3;"|<math>h</math>
-| style="background-color: #E7BBB3;"|<math>i</math>
+{| class="wikitable"
-|<math>d+g</math>
+|+
-|<math>e+h</math>
+| rowspan="5" |<math>d=5</math>
-|<math>f+i</math>
+| rowspan="2" style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=2</math>
+|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
+|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
+|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
+|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
+|style="background-color: #BED5BA; border-bottom: 1px solid #B6321C;"|
+| style="border-bottom: 1px solid #B6321C;" |
 |-
-| colspan="3" rowspan="1" |<math>∧</math>
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-| colspan="3" rowspan="1" |<math>∧</math>
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-|
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-| colspan="3" |
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
-| colspan="3" |
+| style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
-|
-| colspan="3" |
-|
-| colspan="3" rowspan="1" |<math>∧</math>
 |-
-|<math>bf-ce</math>
+| rowspan="3" |<math>g_{\text{max}}=3</math>
-|<math>af-cd</math>
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-|<math>ae-bd</math>
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-|<math> +</math>
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-|<math>bi-ch</math>
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-|<math>ai-cg</math>
+| style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↓  </span>
-|<math>ah-bg</math>
+| style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
-|<math>=</math>
-|<math>bf-ce+bi-ch</math>
-|<math>af-cd+ai-cg</math>
-|<math>ae-bd+ah-bg</math>
-| colspan="3" rowspan="2" |
-| rowspan="2" |
-| colspan="3" rowspan="2" |
-| rowspan="2" |
-|<math>2b(f+i)-2c(e+h)</math>
-|<math>2a(f+i)-2c(d+g)</math>
-|<math>2a(e+h)-2b(d+g)</math>
 |-
-| colspan="3" |
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
 |
-| colspan="3" |
+|-
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
 |
-| style="background-color: LightYellow;"|<math>b(f+i)-c(e+h)</math>
+|}
-| style="background-color: LightYellow;"|<math>a(f+i)-c(d+g)</math>
-| style="background-color: LightYellow;"|<math>a(e+h)-b(d+g)</math>
+And second, <span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>:
-| style="background-color: LightYellow;"|<math>b(f+i)-c(e+h)</math>
-| style="background-color: LightYellow;"|<math>a(f+i)-c(d+g)</math>
+{| class="wikitable"
-| style="background-color: LightYellow;"|<math>a(e+h)-b(d+g)</math>
+|+
-|-
+| rowspan="5" |<math>d=5</math>
-!
+| rowspan="2" style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=2</math>
-!
+| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-!
+| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-! colspan="11" |
+| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-!
+| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-! colspan="11" |
+| style="background-color: #E7BBB3; border-top: 1px solid #B6321C;" |<span style="color: #B6321C;">  ↑  </span>
-!
+| rowspan="2" style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>
+|-
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
+|style="background-color: #E7BBB3; border-bottom: 3px solid black;"|<span style="color: #B6321C;">  ↑  </span>
 |-
-! rowspan="5" |
+| rowspan="3" |<math>g_{\text{max}}=3</math>
-| rowspan="5" |hidden <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>
+| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-! rowspan="5" |
+| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-|<math>a</math>
+| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-|<math>b</math>
+| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-|<math>c</math>
+| style="background-color: #E7BBB3;" |<span style="color: #B6321C;">  ↓  </span>
-| rowspan="2" |
+| rowspan="2" style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>
-|<math>j</math>
+|-
-|<math>k</math>
+|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|<math>l</math>
+|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
+|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
+|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
+|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
+|-
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
+|style="background-color: #BED5BA;"|
 |
-| colspan="3" rowspan="2" |
+|}
-! rowspan="5" |
-|<math>a</math>
+Here's where things really get interesting. Because in both of these cases, the pairs of temperaments represented are <span style="color: #3C8031;">linearly dependent</span> on each other (i.e. either their mappings are <span style="color: #3C8031;">linearly dependent</span>, their comma bases are <span style="color: #3C8031;">linearly dependent</span>, or both). And so far, every possibility where temperaments have been <span style="color: #3C8031;">linearly dependent</span>, they have also been <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>, and therefore addable. But if you look at the second case here, we are <span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span>, but since <math>d=5</math>, the temperaments still manage to be <span style="color: #3C8031;">linearly dependent</span>. So this is the first example of a <span style="color: #3C8031;">linearly dependent</span> temperament pairing which is not addable.
-|<math>b</math>
-|<math>c</math>
+====Back to <math>d=2</math>, for a surprisingly tricky example====
-| colspan="1" rowspan="2" |<math>+</math>
-|<math>j</math>
+Beyond <math>d=5</math>, these diagrams get cumbersome to prepare, and cease to reveal further insights. But if we step back down to <math>d=2</math>, a place simpler than anywhere we've looked so far, we actually find another surprisingly tricky example, which is hopefully still illuminating.
-|<math>k</math>
-|<math>l</math>
+So <math>d=2</math> (such as the 3-limit) presents another case — similar to the <math>d=5</math>, <math>g_{\text{min}}=2</math>, <span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span> case explored most recently above — where the properties of <span style="color: #3C8031;">linearly dependence</span> and addability do not match each other. But while in the other case, we had a temperament pair that was <span style="color: #3C8031;">linearly dependent</span> yet not addable, in this <math>d=2</math> (and therefore <math>g_{\text{min}}=1</math>, <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>) case, it is the other way around: addable yet <span style="color: #B6321C;">linearly independent</span>!
-| colspan="1" rowspan="2" |<math>=</math>
-|<math>a+j</math>
+{| class="wikitable"
-|<math>b+k</math>
+|+
-|<math>c+l</math>
+| rowspan="2" |<math>d=2</math>
-! rowspan="5" |
+|style="border-bottom: 3px solid black;" |<math>g_{\text{min}}=1</math>
-|-
+|style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↑  </span>
-|<math>d</math>
+|style="background-color: #E7BBB3; border-bottom: 3px solid black; border-top: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↑  </span>
-|<math>e</math>
+| style="border-top: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
-|<math>f</math>
-|<math>g</math>
-|<math>h</math>
-|<math>i</math>
-|<math></math>
-|<math>d</math>
-|<math>e</math>
-|<math>f</math>
-|<math>g</math>
-|<math>h</math>
-|<math>i</math>
-|<math>d+g</math>
-|<math>e+h</math>
-|<math>f+i</math>
 |-
-| colspan="3" rowspan="1" |<math>∧</math>
+|<math>g_{\text{max}}=1</math>
-|
+|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-| colspan="3" rowspan="1" |<math>∧</math>
+|style="background-color: #E7BBB3; border-bottom: 1px solid #B6321C;"|<span style="color: #B6321C;">  ↓  </span>
-|
+| style="border-bottom: 1px solid #B6321C;" |<span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>
-| colspan="3" |
+|}
-| colspan="3" |
-|
+Basically, in the case of <math>d=2</math>, <math>g_{\text{max}}=1</math> (in non-trivial cases, i.e. not JI or the unison temperament), so any two different ETs or commas you pick are going to be <span style="color: #B6321C;">linearly independent</span> (because the only way they could be <span style="color: #3C8031;">linearly dependent</span> would be to be the same temperament). And yet we know we can still entry-wise add them to new vectors that are [[Douglas_Blumeyer_and_Dave_Keenan%27s_Intro_to_exterior_algebra_for_RTT#Decomposability|decomposable]], because they're already vectors (decomposing means to express a [[Douglas_Blumeyer_and_Dave_Keenan%27s_Intro_to_exterior_algebra_for_RTT#From_vectors_to_multivectors|multivector]] in the form of a list of monovectors, so decomposing a multivector that's already a monovector like this is tantamount to merely putting array braces around it.)
-| colspan="3" |
-|
+====Conclusion====
-| colspan="3" rowspan="1" |<math>∧</math>
-|-
+This explanation has hopefully helped get a grip on how addability AKA <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span> behaves given various temperament dimensions. But it still hasn't quite explained why <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span> is one and the same thing as addability. We will look at this in another section soon.
-|<math>bf-ce</math>
-|<math>af-cd</math>
+===Geometric explanation===
-|<math>ae-bd</math>
-|<math>+</math>
+We've presented a diagrammatic illustration of the behavior of <span style="color: #B6321C;">linear-independence <math>l_{\text{ind}}</math></span> with respect to temperament dimensions. But some of the results might have seemed surprising. For instance, when looking at the diagram for <math>d=4, g_{\text{min}}=1, g_{\text{max}}=3</math>, it might have seemed intuitive enough that the the <span style="color: #3C8031;">red band</span> could not extend beyond the square grid, but then again, why shouldn't it be possible to have, say, two 7-limit ETs which temper out only a single comma in common? Perhaps it doesn't seem clear that this is impossible, and that they must temper out two commas in common (and of course the infinitude of combinations of these two commas). If this is as unclear to you as it was to the author when exploring this topic, then this explanatory section is for you! Here, we will use geometrical representations of temperaments to hone our intuitions about the possible combinations of dimensions and <span style="color: #B6321C;">linear-independence <math>l_{\text{ind}}</math></span> of temperaments.
-|<math>ki-lh</math>
-|<math>ji-lg</math>
+In this approach, we’re actually not going to focus directly on the <span style="color: #B6321C;">linear-independence <math>l_{\text{ind}}</math></span> of temperaments. Instead, we're going to look at the <span style="color: #3C8031;">linear-''de''pendence <math>l_{\text{dep}}</math></span> of matrices representing temperaments such as mappings and comma bases, and then compute the <span style="color: #B6321C;">linear-independence <math>l_{\text{ind}}</math></span> from it and the grade <math>g</math>. As we’ve established, the <span style="color: #3C8031;">linear-dependence <math>l_{\text{dep}}</math></span> differs from one side of duality to the other, so we’ll only be looking at one side of duality at a time.
-|<math>jh-kg</math>
-|<math>=</math>
+====Introduction====
-|<math>bf-ce+ki-lh</math>
-|<math>af-cd+ji-lg</math>
+In this geometric approach, we'll be imagining individual vectors as points (0D), sets of two vectors as lines (1D), sets of three as planes (2D), four as volumes (3D), and so forth, as according to this table:
-|<math>ae-bd+jh-kg</math>
+{| class="wikitable center-all"
-| colspan="3" rowspan="2" |
+|+
-| rowspan="2" |
+!vector
-| colspan="3" rowspan="2" |
+count
-| rowspan="2" |
+!geometric
-|<math>(b+k)(f+i)-(c+l)(e+h)</math>
+dimension
-|<math>(a+j)(f+i)-(c+l)(d+g)</math>
+!form
-|<math>(a+j)(e+h)-(b+k)(d+g)</math>
 |-
-| colspan="3" |
+|0
-|
+|undefined
-| colspan="3" |
+|(emptiness)
-|
+|-
-| style="background-color: LightBlue;"|<math>bf-ce+ki-lh</math>
+|1
-| style="background-color: LightBlue;"|<math>af-cd+ji-lg</math>
+|0
-| style="background-color: LightBlue;"|<math>ae-bd+jh-kg</math>
+|point
-| style="background-color: LightBlue;"|<math>bf+bi+kf+ki-ce-ch-le-lh</math>
+|-
-| style="background-color: LightBlue;"|<math>af+ai+jf+ji-cd-cg-ld-lg</math>
+|2
-| style="background-color: LightBlue;"|<math>ae+ah+je+jh-bd-bg-kd-kg</math>
+|1
+|line
 |-
-!
+|3
-!
+|2
-!
+|plane
-! colspan="11" |
-!
-! colspan="11" |
-!
-|}
-This second diagram demonstrates this situation for a <math>d=5, g=3</math> case. One pair of the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> vectors are explicitly matching, but not the other, which isn't enough.
-{| class="wikitable center-all"
-|+
-!
-! colspan="2" |
-!
-! colspan="32" |
-!
-! colspan="22" |
-!
 |-
-!
+|4
-! colspan="2" |
+|3
-!
+|volume
-| colspan="32" rowspan="1" |'''multivector approach'''
-!
-| colspan="22" rowspan="1" |'''matrix approach'''
-!
 |-
-!
+|5
-! colspan="2" |
+|4
-!
+|hypervolume
-! colspan="32" |
-!
-! colspan="22" |
-!
 |-
-! rowspan="7" |
+| ⋮
-| colspan="1" rowspan="7" |explicit <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>
+|⋮
+|⋮
+|}
+This is a "vector space", and these geometric dimensions are consistent with how temperaments represented by these counts of vectors appear in ''projective'' vector space, which reduces geometric dimensions by 1. For example, a vector has a geometric interpretation as a directed line segment, which is 1D, but a point is 0D, which is one dimension lower. Essentially what we're doing is ''assuming the origin''.
+Think of it this way: geometric points are zero-dimensional, simply representing a position in space, whereas linear algebra vectors are one-dimensional, representing both a magnitude and direction; the way vectors manage to encode this extra dimension without providing any additional information is by being understood to describe this position in space ''relative to an origin''. Well, so we'll now switch our interpretation of these objects to the geometric one, where the vector's entries are nothing more than a coordinate for a point in space. And the "projection" involved in projective vector space essentially positions us at this discarded origin, looking out from it upon every individual point, which accomplishes the same feat, in a visual way.
+Perhaps an example may help clarify this setup. Suppose we've got an (x,y,z) space, and two coordinates (5,8,12) and (7,11,16). You should recognize these as the simple maps for 5-ET and 7-ET, usually written as {{map|5 8 12}} and {{map|7 11 16}}, respectively. Ask for the equation of the plane defined by the three points (5,8,12), (7,11,16), and the origin (0,0,0) and you'll get -4x + 4y -1z = 0, which clearly shows us the entries of the meantone comma. That's because meantone temperament can be defined by these two maps. 5-limit JI is a 3D space, and meantone temperament, as a rank-2 temperament, would be a 2D plane. But we don't normally need to think of the map corresponding to the origin, where everything is tempered out, including meantone. So we can just assume it, and think of a 2D plane as being defined by only 2 points, which in a view projected (from the origin) will look like just the line connecting (5,8,12) and (7,11,16).
-⟨{{vector|<math>a</math> <math>b</math> <math>c</math> <math>d</math> <math>e</math>}}
+So, we've set the stage for our projective vector spaces. We will now be looking at representations of temperaments as counts of vector sets, and then using this scheme to convert them to primitive geometric forms. We'll place two of each form into the space, representing the two temperaments having the possibility of arithmetic performed on them checked. Then we will observe their possible <span style="color: #3C8031;">''intersections''</span> depending on how they're oriented in space, and it's these <span style="color: #3C8031;">intersections that represent their linear-dependence</span>. When the dimension of the <span style="color: #3C8031;">intersection</span> is then converted back to a vector set count, then we have their <span style="color: #3C8031;">linear-dependence <math>l_{\text{dep}}</math></span>, for this side of duality, anyway (remember, unlike the <span style="color: #B6321C;">linear-independence <math>l_{\text{ind}}</math></span>, this value isn't necessarily the same on both sides of duality). We can finally subtract the <span style="color: #3C8031;">linear-dependence</span> from the grade (vector count) to get the <span style="color: #B6321C;">linear-indepedence</span>, in order to determine if the two temperaments are addable.
-{{vector|<math>f</math> <math>g</math> <math>h</math> <math>i</math> <math>j</math>}}]
-| style="background-color: #BED5BA;"|<math>r_1</math>
+In these examples, we'll be assuming that no two temperaments being compared are the same, because performing temperament arithmetic on copies of the same temperament is not interesting. The other things we'll be assuming is that no lines, planes, etc. are parallel to each other; this is due to a strange effect touched upon in footnote 4 whereby temperament geometry that appears parallel in projective space actually still intersects; the present author asks that if anyone is able to demystify this situation, that they please do!
-! rowspan="7" |
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>a</math>
+====At <math>d=3</math>====
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>b</math>
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>c</math>
+First, let's establish the geometric dimension of the space. With <math>d=3</math>, we've got a 2D space (one less than 3), so the entire space can be visualized on a plane.
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>d</math>
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>e</math>
+Our only possible values for <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> here are 1 and 2, respectively. So these are the two possible counts of vectors <math>g</math> possessed by matrices representing temperaments here.
-| rowspan="3" |
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>a</math>
+So let's look at temperaments represented by matrices with 1 vector first (<math>g=1</math>). In this case, each of the two temperaments is a point in the plane. Unless these two temperaments are the same temperament, the <span style="color: #3C8031;">intersection</span> of these two points is empty. Emptiness isn't even 0D! So that tells us that these temperaments have 0 vectors worth of <span style="color: #3C8031;">linear dependence</span>. With <math>g=1</math>, that gives us a <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math> = g</math> <math> - </math> <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span> <math>= 1 -</math> <span style="color: #3C8031;"><math>0</math></span> <math>= 1</math>:
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>b</math>
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>c</math>
+[[File:D3 g1 dep0.png|200px|none]]
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>d</math>
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>e</math>
+Next, let's look at temperaments represented by matrices with 2 vectors (<math>g=2</math>). In this case, each of the two temperaments is a line in the plane. Again, assuming the two lines are not the same line or parallel, their <span style="color: #3C8031;">intersection</span> is a point. Being 0D, that tells us that the <span style="color: #3C8031;">linear-dependence</span> of these matrices is 1. So that gives us an <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span> <math>= g</math> <math>-</math> <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span> <math>= 2 -</math> <span style="color: #3C8031;"><math>1</math></span> <math>= 1</math>. This matches the value we found via the <math>g=1</math>, so we've effectively checked our work:
-| rowspan="3" |
-| colspan="10" rowspan="3" |
+[[File:D3 g2 dep1.png|200px|none]]
-! rowspan="7" |
-| style="background-color: #BED5BA;"|<math>a</math>
+====At <math>d=2</math>====
-| style="background-color: #BED5BA;"|<math>b</math>
-| style="background-color: #BED5BA;"|<math>c</math>
+Let's step back to <math>d=2</math>. Here we've got a 2 minus 1 equals 1D space, so the entire space can be visualized on a single line (one direction corresponds to an increasing ratio between the two coordinates, and the other to a decreasing ratio).
-| style="background-color: #BED5BA;"|<math>d</math>
-| style="background-color: #BED5BA;"|<math>e</math>
+We know our only possible value for <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> here is 1. So in either case, each of the two temperaments is a point on the line. As with two points in a plane — when <math>d=3</math> — unless these two temperaments are the same temperament, the intersection of these two points is empty. So again the <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math> = g = 1</math>:
-| colspan="1" rowspan="3" |+
-| style="background-color: #BED5BA;"|<math>a</math>
+[[File:D2 g1 dep0.png|200px|none]]
-| style="background-color: #BED5BA;"|<math>b</math>
-| style="background-color: #BED5BA;"|<math>c</math>
+====At <math>d=4</math>====
-| style="background-color: #BED5BA;"|<math>d</math>
-| style="background-color: #BED5BA;"|<math>e</math>
+First, let's establish the geometric dimension of the space. With <math>d=4</math>, we've got a 3D space (one less than 4), so the entire space can be visualized in a volume.
-| colspan="1" rowspan="3" |<math>=</math>
-| colspan="2" rowspan="1" |<math>2a</math>
+At <math>d=4</math>, we have a couple options for the grade: either <math>g_{\text{min}}=1</math> and <math>g_{\text{max}}=3</math>, or both <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> equal 2.
-| colspan="2" rowspan="1" |<math>2b</math>
-| colspan="2" rowspan="1" |<math>2c</math>
+Let's look at temperaments represented by matrices with 1 vector first (<math>g=1</math>). Yet again, we find ourselves with two separate points, but now we find them in a space that's not a line, not a plane, but a volume. This doesn't change <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math> = g = 1</math>, so we're not even going to show it, or any further cases of <math>g=1</math>. These are all addable.
-| colspan="2" rowspan="1" |<math>2d</math>
-| colspan="2" rowspan="1" |<math>2e</math>
+And when <math>g=3</math>, because this is paired with <math>g=1</math> from the min and max values, we should expect to get the same answer as with <math>g=1</math>. And indeed, it will check out that way. Because two <math>g=3</math> temperaments will be planes in this volume, and the intersection of two planes is a line. Which means that <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span><math> = 2</math>. And so <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math> = g</math> <math> - </math> <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span> <math>= 3 -</math> <span style="color: #3C8031;"><math>2</math></span> <math>= 1</math>. And here's where our geometric approach begins to pay off! This was the example given at the beginning that might seem unintuitive when relying only on the diagrammatic approach. But here we can see clearly that there would be no way for two planes in a volume to intersect only at a point, which proves the fact that two 7-limit ETs could never only temper out a single comma in common.
-! rowspan="7" |
-|-
+[[File:D4 g3 dep2.png|200px]]
-| style="background-color: #BED5BA;"|<math>r_2</math>
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>f</math>
+Next let's look at temperaments represented by matrices with 2 vectors, that is, when both <math>g_{\text{min}}</math> and <math>g_{\text{max}}</math> are equal to 2. What are the possible ways lines can occupy a volume together? In a plane, as it was with <math>d=3</math> (and again assuming no parallel objects in these examples), they must intersect. But in a volume, here in <math>d=4</math>, this is possible. So, with <math>g=2</math>, it is possible to have a <span style="color: #B6321C;"><math>l_{\text{dep}}</math></span> <math>= 0</math>, which leads to <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math> = g</math> <math> - </math> <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span> <math>= 2 -</math> <span style="color: #3C8031;"><math>0</math></span> <math>= 2</math>. Not addable in this case.
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>g</math>
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>h</math>
+[[File:D4 g2 dep0.png|200px]]
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>i</math>
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>j</math>
+But we can also imagine two lines in a volume that do intersect at a point. This is the case where <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math> = g</math> <math> - </math> <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span> <math>= 2 -</math> <span style="color: #3C8031;"><math>1</math></span> <math>= 1</math>: addable!
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>f</math>
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>g</math>
+[[File:D4 g2 dep1.png|200px]]
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>h</math>
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>i</math>
+==== At <math>d=5</math>====
-| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>j</math>
-| style="background-color: #BED5BA;" |<math>f</math>
+First, let's establish the geometric dimension of the space. With <math>d=5</math>, we've got a 4D space (one less than 5), so the entire space can be visualized in a hypervolume. We've now gone beyond the dimensionality of physical reality, so things get a little harder to conceptualize unfortunately. But <math>d=5</math> is the first <math>d</math> where we can make an important point about addability, so please bear with!
-| style="background-color: #BED5BA;" |<math>g</math>
-| style="background-color: #BED5BA;" |<math>h</math>
+At <math>d=5</math>, we also have a couple options for the grade: either <math>g_{\text{min}}=1</math> and <math>g_{\text{max}}=4</math>, or <math>g_{\text{min}}=2</math> and <math>g_{\text{max}}=3</math>.
-| style="background-color: #BED5BA;" |<math>i</math>
-| style="background-color: #BED5BA;" |<math>j</math>
+First we'll look at <math>g_{\text{min}}=1</math> and <math>g_{\text{max}}=4</math>. Temperament matrices with <math>g=1</math> are still addable. And temperament matrices with <math>g=4</math> should be too. We can see this visually as how two volumes in a hypervolume together will have an intersection the shape of a plane. We can now see that there's a generalizable principal that any two <math>(d-1)</math>-dimensional objects will necessarily have a <math>(d-2)</math>-dimensional intersection, and thus have <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span> <math>= 1</math> and be addable. So we won't need to show this one or any further like it, either.
-| style="background-color: #BED5BA;" |<math>f</math>
-| style="background-color: #BED5BA;" |<math>g</math>
+So let's look at temperament matrices with <math>g_{\text{min}}=2</math> and <math>g_{\text{max}}=3</math>. For <math>g=2</math>, we have two possible values for <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span>: 0 or 1. Meaning that either the two lines through this hypervolume do not intersect (0), or they intersect at a point (1). These diagrams would look very much like the corresponding diagrams for <math>d=4</math>, so we will not be showing them. But what about when <math>g=3</math>? We can certainly imagine two planes in a hypervolume intersecting at a line, just as they do in an ordinary volume — they're just not taking advantage of the additional geometric dimension. So we won't show that example either. But where it gets really interesting is imagining then taking one of these two planes and rotating it in the fifth dimension; this causes the intersection between the two planes to be reduced down to a single point. And this corresponds with the case of <span style="color: #3C8031;"><math>l_{\text{dep}}=1</math></span> here, which means <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span><math> = g</math> <math> - </math> <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span> <math>= 3 -</math> <span style="color: #3C8031;"><math>1</math></span> <math>= 2</math>, so therefore not addable:
-| style="background-color: #BED5BA;" |<math>h</math>
-| style="background-color: #BED5BA;" |<math>i</math>
+[[File:D5 g3 dep1.png|300px]]
-| style="background-color: #BED5BA;" |<math>j</math>
-| colspan="2" rowspan="1" |<math>2f</math>
+So for <math>g_{\text{min}}=2</math> and <math>g_{\text{max}}=3</math> we got two different possibilities for <span style="color: #B6321C;"><math>l_{\text{ind}}</math></span>: 1 and 2, and for each of these two possibilities, we found it twice. We can see then that these match up, that is, that the <math>g_{\text{min}}=2</math> case with <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span> matches with the <math>g_{\text{max}}=3</math> case with <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span>, and the <span style="color: #B6321C;"><math>l_{\text{ind}}=2</math></span> cases match in the same way.
-| colspan="2" rowspan="1" |<math>2g</math>
-| colspan="2" rowspan="1" |<math>2h</math>
+==== Conclusion====
-| colspan="2" rowspan="1" |<math>2i</math>
-| colspan="2" rowspan="1" |<math>2j</math>
+Here's a summary table of our geometric findings so far:
+ {| class="wikitable center-all"
+|+
+! rowspan="2" |<math>d</math> ( <math>= g_{\text{min}} + g_{\text{max}}</math>)
+! colspan="2" |<math>g_{\text{min}}</math>
+! colspan="2" |<math>g_{\text{max}}</math>
+! rowspan="2" |<span style="color: #B6321C;"><math>l_{\text{ind}}</math></span> ( <math>= g -</math> <span style="color: #3C8031;"><math>l_{\text{dep}}</math></span>)
 |-
-| style="background-color: #E7BBB3;"|<math>r_3</math>
+!<math>g</math>
-| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>k</math>
+!<span style="color: #3C8031;"><math>l_{\text{dep}}</math></span>
-| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>l</math>
+!<math>g</math>
-| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>m</math>
+!<span style="color: #3C8031;"><math>l_{\text{dep}}</math></span>
-| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>n</math>
-| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>o</math>
-| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>p</math>
-| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>q</math>
-| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>r</math>
-| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>s</math>
-| style="background-color: #E7BBB3;" colspan="2" rowspan="1" |<math>t</math>
-| style="background-color: #E7BBB3;"|<math>k</math>
-| style="background-color: #E7BBB3;"|<math>l</math>
-| style="background-color: #E7BBB3;"|<math>m</math>
-| style="background-color: #E7BBB3;"|<math>n</math>
-| style="background-color: #E7BBB3;"|<math>o</math>
-| style="background-color: #E7BBB3;"|<math>p</math>
-| style="background-color: #E7BBB3;"|<math>q</math>
-| style="background-color: #E7BBB3;"|<math>r</math>
-| style="background-color: #E7BBB3;"|<math>s</math>
-| style="background-color: #E7BBB3;"|<math>t</math>
-| colspan="2" rowspan="1" |<math>k+p</math>
-| colspan="2" rowspan="1" |<math>l+q</math>
-| colspan="2" rowspan="1" |<math>m+r</math>
-| colspan="2" rowspan="1" |<math>n+s</math>
-| colspan="2" rowspan="1" |<math>o+t</math>
 |-
-|
+|2
-| colspan="10" rowspan="1" |<math>∧</math>
+|1
-|
+|0
-| colspan="10" rowspan="1" |<math>∧</math>
+|1
-|
+|0
-| colspan="10" |
+|1
-| colspan="5" |
-|
-| colspan="5" |
-|
-| colspan="10" rowspan="1" |<math>∧</math>
 |-
-|<math>r_1∧r_2</math>
+|3
-| rowspan="2" |<math>ag-bf</math>
+|1
-| rowspan="2" |<math>ah-cf</math>
+|0
-| rowspan="2" |<math>ai-df</math>
+|2
-| rowspan="2" |<math>aj-ef</math>
+|1
-| rowspan="2" |<math>bh-cg</math>
+|1
-| rowspan="2" |<math>bi-dg</math>
+|-
-| rowspan="2" |<math>bj-eg</math>
+| rowspan="3" |4
-| rowspan="2" |<math>ci-dh</math>
+| 1
-| rowspan="2" |<math>cj-eh</math>
+|0
-| rowspan="2" |<math>dj-ei</math>
+|3
-| rowspan="2" |
+|2
-| rowspan="2" |<math>ag-bf</math>
+|1
-| rowspan="2" |<math>ah-cf</math>
+|-
-| rowspan="2" |<math>ai-df</math>
+|2
-| rowspan="2" |<math>aj-ef</math>
+|0
-| rowspan="2" |<math>bh-cg</math>
+|2
-| rowspan="2" |<math>bi-dg</math>
+|0
-| rowspan="2" |<math>bj-eg</math>
+|2
-| rowspan="2" |<math>ci-dh</math>
-| rowspan="2" |<math>cj-eh</math>
-| rowspan="2" |<math>dj-ei</math>
-| rowspan="2" |
-| colspan="10" rowspan="2" |
-| colspan="5" rowspan="3" |
-| rowspan="3" |
-| colspan="5" rowspan="3" |
-| rowspan="3" |
-|<math>4ag-4bf</math>
-|<math>4ah-4cf</math>
-|<math>4ai-4df</math>
-|<math>4aj-4ef</math>
-|<math>4bh-4cg</math>
-|<math>4bi-4dg</math>
-|<math>4bj-4eg</math>
-|<math>4ci-4dh</math>
-|<math>4cj-4eh</math>
-|<math>4dj-4ei</math>
 |-
-|simplify <math>r_1∧r_2</math> if necessary
+|2
-|<math>ag-bf</math>
+|1
-|<math>ah-cf</math>
+|2
-|<math>ai-df</math>
+|1
-|<math>aj-ef</math>
+|1
-|<math>bh-cg</math>
+|-
-|<math>bi-dg</math>
+| rowspan="3" | 5
-|<math>bj-eg</math>
+|1
-|<math>ci-dh</math>
+|0
-|<math>cj-eh</math>
+|4
-|<math>dj-ei</math>
+|3
+|1
 |-
-|<math>(r_1∧r_2)∧r_3</math>
+|2
-|<math>k(bh-cg)\\-l(ah-cf)\\+m(ag-bf)</math>
+|0
-|<math>k(bi-dg)\\-l(ai-df)\\+n(ag-bf)</math>
+|3
-|<math>k(bj-eg)\\-l(aj-ef)\\+o(ag-bf)</math>
+|1
-|<math>k(ci-dh)\\-m(ai-df)\\+n(ah-cf)</math>
+|2
-|<math>k(cj-eh)\\-m(aj-ef)\\+o(ah-cf)</math>
+|-
-|<math>k(dj-ei)\\-n(aj-ef)\\+o(ai-df)</math>
+|2
-|<math>l(ci-dh)\\-m(bi-dg)\\+n(bh-cg)</math>
+|1
-|<math>l(cj-eh)\\-m(bj-eg)\\+o(bh-cg)</math>
+|3
-|<math>l(dj-ei)\\-n(bj-eg)\\+o(bi-dg)</math>
+|2
-|<math>m(dj-ei)\\-n(cj-eh)\\+o(ci-dh)</math>
+|1
-|<math>+</math>
+|}
-|<math>p(bh-cg)\\-q(ah-cf)\\+r(ag-bf)</math>
-|<math>p(bi-dg)\\-q(ai-df)\\+s(ag-bf)</math>
+The geometric explanation still hasn't answered the question as to why <span style="color: #B6321C;"><math>l_{\text{ind}}=1</math></span> is the condition on addability. It just increased our intuitions about its relationship with temperament dimensions. We'll still look to a later section for an answer on this.
-|<math>p(bj-eg)\\-q(aj-ef)\\+t(ag-bf)</math>
-|<math>p(ci-dh)\\-r(ai-df)\\+s(ah-cf)</math>
+=== Algebraic explanation===
-|<math>p(cj-eh)\\-r(aj-ef)\\+t(ah-cf)</math>
-|<math>p(dj-ei)\\-s(aj-ef)\\+t(ai-df)</math>
+This explanation relies on comparing the results of the multivector and matrix approaches to temperament arithmetic, and showing algebraically how the matrix approach can only achieve the same answer as the multivector approach on the condition that it keeps all but one vector between the added matrices the same, that is, not only are the temperaments addable, but their <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> appears explicitly in the added matrices.
-|<math>q(ci-dh)\\-r(bi-dg)\\+s(bh-cg)</math>
-|<math>q(cj-eh)\\-r(bj-eg)\\+t(bh-cg)</math>
+To compare results, we eventually get both approaches into a multivector form. With the multivector approach, we wedge the vector set first and then add the resultant multivectors to get a new multivector. With the matrix approach, we treat the vector set as a matrix and add first, then treat the resultant matrix as a vector set and wedge those vectors to get a new multivector.
-|<math>q(dj-ei)\\-s(bj-eg)\\+t(bi-dg)</math>
-|<math>r(dj-ei)\\-s(cj-eh)\\+t(ci-dh)</math>
+The diagrams below are organized into a 2×2 layout. The left part shows the multivector approach, and the right part shows the matrix approach. The top part shows how the results of two approaches match when the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> is successfully explicit (and in these cases, the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> vectors are highlighted in green and the <span style="color: #B6321C;"><math>L_{\text{ind}}</math></span> vectors are highlighted in red), and the bottom part shows how the results fail to match when it is not. Successful matches are highlighted in yellow and failures to match are highlighted in blue.
-|<math>=</math>
-| style="background-color: LightYellow;"|<math>(k+p)(bh-cg)\\-(l+q)(ah-cf)\\+(m+r)(ag-bf)</math>
+This first diagram demonstrates this situation for a <math>d=3, g=2</math> case.
-| style="background-color: LightYellow;"|<math>(k+p)(bi-dg)\\-(l+q)(ai-df)\\+(n+s)(ag-bf)</math>
+{| class="wikitable center-all"
-| style="background-color: LightYellow;"|<math>(k+p)(bj-eg)\\-(l+q)(aj-ef)\\+(o+t)(ag-bf)</math>
+|+
-| style="background-color: LightYellow;"|<math>(k+p)(ci-dh)\\-(m+r)(ai-df)\\+(n+s)(ah-cf)</math>
+!
-| style="background-color: LightYellow;"|<math>(k+p)(cj-eh)\\-(m+r)(aj-ef)\\+(o+t)(ah-cf)</math>
+!
-| style="background-color: LightYellow;"|<math>(k+p)(dj-ei)\\-(n+s)(aj-ef)\\+(o+t)(ai-df)</math>
+!
-| style="background-color: LightYellow;"|<math>(l+q)(ci-dh)\\-(m+r)(bi-dg)\\+(n+s)(bh-cg)</math>
+! colspan="11" |
-| style="background-color: LightYellow;"|<math>(l+q)(cj-eh)\\-(m+r)(bj-eg)\\+(o+t)(bh-cg)</math>
+!
-| style="background-color: LightYellow;"|<math>(l+q)(dj-ei)\\-(n+s)(bj-eg)\\+(o+t)(bi-dg)</math>
+! colspan="11" |
-| style="background-color: LightYellow;"|<math>(m+r)(dj-ei)\\-(n+s)(cj-eh)\\+(o+t)(ci-dh)</math>
+!
-| style="background-color: LightYellow;"|<math>(k+p)(bh-cg)\\-(l+q)(ah-cf)\\+(m+r)(ag-bf)</math>
-| style="background-color: LightYellow;"|<math>(k+p)(bi-dg)\\-(l+q)(ai-df)\\+(n+s)(ag-bf)</math>
-| style="background-color: LightYellow;"|<math>(k+p)(bj-eg)\\-(l+q)(aj-ef)\\+(o+t)(ag-bf)</math>
-| style="background-color: LightYellow;"|<math>(k+p)(ci-dh)\\-(m+r)(ai-df)\\+(n+s)(ah-cf)</math>
-| style="background-color: LightYellow;"|<math>(k+p)(cj-eh)\\-(m+r)(aj-ef)\\+(o+t)(ah-cf)</math>
-| style="background-color: LightYellow;"|<math>(k+p)(dj-ei)\\-(n+s)(aj-ef)\\+(o+t)(ai-df)</math>
-| style="background-color: LightYellow;"|<math>(l+q)(ci-dh)\\-(m+r)(bi-dg)\\+(n+s)(bh-cg)</math>
-| style="background-color: LightYellow;"|<math>(l+q)(cj-eh)\\-(m+r)(bj-eg)\\+(o+t)(bh-cg)</math>
-| style="background-color: LightYellow;"|<math>(l+q)(dj-ei)\\-(n+s)(bj-eg)\\+(o+t)(bi-dg)</math>
-| style="background-color: LightYellow;"|<math>(m+r)(dj-ei)\\-(n+s)(cj-eh)\\+(o+t)(ci-dh)</math>
 |-
 !
 !
 !
+| colspan="11" rowspan="1" |'''multivector approach'''
+!
+| colspan="11" rowspan="1" |'''matrix approach'''
+!
+|-
 !
-! colspan="32" |
 !
-! colspan="22" |
 !
+! colspan="11" |
+!
+! colspan="11" |
+!
+|-
+! rowspan="5" |
+| colspan="1" rowspan="5" |explicit <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>
+{{bra|{{vector|<math>a</math> <math>b</math> <math>c</math>}}}}
+! rowspan="5" |
+| style="background-color: #BED5BA;"|<math>a</math>
+| style="background-color: #BED5BA;"|<math>b</math>
+| style="background-color: #BED5BA;"|<math>c</math>
+| rowspan="2" |
+| style="background-color: #BED5BA;"|<math>a</math>
+| style="background-color: #BED5BA;"|<math>b</math>
+| style="background-color: #BED5BA;"|<math>c</math>
+| rowspan="2" |
+| colspan="3" rowspan="2" |
+! rowspan="5" |
+| style="background-color: #BED5BA;"|<math>a</math>
+| style="background-color: #BED5BA;"|<math>b</math>
+| style="background-color: #BED5BA;"|<math>c</math>
+| colspan="1" rowspan="2" |<math>+</math>
+| style="background-color: #BED5BA;"|<math>a</math>
+| style="background-color: #BED5BA;"|<math>b</math>
+| style="background-color: #BED5BA;"|<math>c</math>
+| colspan="1" rowspan="2" |<math>=</math>
+|<math>2a</math>
+|<math>2b</math>
+|<math>2c</math>
+! rowspan="5" |
 |-
-! rowspan="7" |
+| style="background-color: #E7BBB3;"|<math>d</math>
-| rowspan="7" |hidden <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>
+| style="background-color: #E7BBB3;"|<math>e</math>
-|<math>r_1</math>
+| style="background-color: #E7BBB3;"|<math>f</math>
-! rowspan="7" |
+| style="background-color: #E7BBB3;"|<math>g</math>
-| colspan="2" rowspan="1" |<math>a</math>
+| style="background-color: #E7BBB3;"|<math>h</math>
-| colspan="2" rowspan="1" |<math>b</math>
+| style="background-color: #E7BBB3;"|<math>i</math>
-| colspan="2" rowspan="1" |<math>c</math>
+| style="background-color: #E7BBB3;"|<math>d</math>
-| colspan="2" rowspan="1" |<math>d</math>
+| style="background-color: #E7BBB3;"|<math>e</math>
-| colspan="2" rowspan="1" |<math>e</math>
+| style="background-color: #E7BBB3;"|<math>f</math>
-| rowspan="3" |
+| style="background-color: #E7BBB3;"|<math>g</math>
-| colspan="2" rowspan="1" |<math>a</math>
+| style="background-color: #E7BBB3;"|<math>h</math>
-| colspan="2" rowspan="1" |<math>b</math>
+| style="background-color: #E7BBB3;"|<math>i</math>
-| colspan="2" rowspan="1" |<math>c</math>
+|<math>d+g</math>
-| colspan="2" rowspan="1" |<math>d</math>
+|<math>e+h</math>
-| colspan="2" rowspan="1" |<math>e</math>
+|<math>f+i</math>
-| rowspan="3" |
+|-
-| colspan="10" rowspan="3" |
+| colspan="3" rowspan="1" |<math>∧</math>
-! rowspan="7" |
+|
-|<math>a</math>
+| colspan="3" rowspan="1" |<math>∧</math>
-|<math>b</math>
+|
-|<math>c</math>
+| colspan="3" |
-|<math>d</math>
+| colspan="3" |
-|<math>e</math>
+|
-| colspan="1" rowspan="3" |<math>+</math>
+| colspan="3" |
-|<math>a</math>
+|
-|<math>b</math>
+| colspan="3" rowspan="1" |<math>∧</math>
-|<math>c</math>
-|<math>d</math>
-|<math>e</math>
-| colspan="1" rowspan="3" |<math>=</math>
-| colspan="2" rowspan="1" |<math>2a</math>
-| colspan="2" rowspan="1" |<math>2b</math>
-| colspan="2" rowspan="1" |<math>2c</math>
-| colspan="2" rowspan="1" |<math>2d</math>
-| colspan="2" rowspan="1" |<math>2e</math>
-! rowspan="7" |
-|-
-|<math>r_2</math>
-| colspan="2" rowspan="1" |<math>f</math>
-| colspan="2" rowspan="1" |<math>g</math>
-| colspan="2" rowspan="1" |<math>h</math>
-| colspan="2" rowspan="1" |<math>i</math>
-| colspan="2" rowspan="1" |<math>j</math>
-| colspan="2" rowspan="1" |<math>u</math>
-| colspan="2" rowspan="1" |<math>v</math>
-| colspan="2" rowspan="1" |<math>w</math>
-| colspan="2" rowspan="1" |<math>x</math>
-| colspan="2" rowspan="1" |<math>y</math>
-|<math>f</math>
-|<math>g</math>
-|<math>h</math>
-|<math>i</math>
-|<math>j</math>
-|<math>u</math>
-|<math>v</math>
-|<math>w</math>
-|<math>x</math>
-|<math>y</math>
-| colspan="2" rowspan="1" |<math>f+u</math>
-| colspan="2" rowspan="1" |<math>g+v</math>
-| colspan="2" rowspan="1" |<math>w+h</math>
-| colspan="2" rowspan="1" |<math>i+x</math>
-| colspan="2" rowspan="1" |<math>j+y</math>
-|-
-|<math>r_3</math>
-| colspan="2" rowspan="1" |<math>k</math>
-| colspan="2" rowspan="1" |<math>l</math>
-| colspan="2" rowspan="1" |<math>m</math>
-| colspan="2" rowspan="1" |<math>n</math>
-| colspan="2" rowspan="1" |<math>o</math>
-| colspan="2" rowspan="1" |<math>p</math>
-| colspan="2" rowspan="1" |<math>q</math>
-| colspan="2" rowspan="1" |<math>r</math>
-| colspan="2" rowspan="1" |<math>s</math>
-| colspan="2" rowspan="1" |<math>t</math>
-|<math>k</math>
-|<math>l</math>
-|<math>m</math>
-|<math>n</math>
-|<math>o</math>
-|<math>p</math>
-|<math>q</math>
-|<math>r</math>
-|<math>s</math>
-|<math>t</math>
-| colspan="2" rowspan="1" |<math>k+p</math>
-| colspan="2" rowspan="1" |<math>l+q</math>
-| colspan="2" rowspan="1" |<math>m+r</math>
-| colspan="2" rowspan="1" |<math>n+s</math>
-| colspan="2" rowspan="1" |<math>o+t</math>
 |-
+|<math>bf-ce</math>
+|<math>af-cd</math>
+|<math>ae-bd</math>
+|<math> +</math>
+|<math>bi-ch</math>
+|<math>ai-cg</math>
+|<math>ah-bg</math>
+|<math>=</math>
+|<math>bf-ce+bi-ch</math>
+|<math>af-cd+ai-cg</math>
+|<math>ae-bd+ah-bg</math>
+| colspan="3" rowspan="2" |
+| rowspan="2" |
+| colspan="3" rowspan="2" |
+| rowspan="2" |
+|<math>2b(f+i)-2c(e+h)</math>
+|<math>2a(f+i)-2c(d+g)</math>
+|<math>2a(e+h)-2b(d+g)</math>
+|-
+| colspan="3" |
 |
-| colspan="10" rowspan="1" |<math>∧</math>
+| colspan="3" |
 |
-| colspan="10" rowspan="1" |<math>∧</math>
+| style="background-color: LightYellow;"|<math>b(f+i)-c(e+h)</math>
-|
+| style="background-color: LightYellow;"|<math>a(f+i)-c(d+g)</math>
-| colspan="10" |
+| style="background-color: LightYellow;"|<math>a(e+h)-b(d+g)</math>
-| colspan="5" |
+| style="background-color: LightYellow;"|<math>b(f+i)-c(e+h)</math>
-|
+| style="background-color: LightYellow;"|<math>a(f+i)-c(d+g)</math>
-| colspan="5" |
+| style="background-color: LightYellow;"|<math>a(e+h)-b(d+g)</math>
-|
-| colspan="10" rowspan="1" |<math>∧</math>
 |-
-|<math>r_1∧r_2</math>
+!
-| rowspan="2" |<math>ag-bf</math>
+!
-| rowspan="2" |<math>ah-cf</math>
+!
-| rowspan="2" |<math>ai-df</math>
+! colspan="11" |
-| rowspan="2" |<math>aj-ef</math>
+!
-| rowspan="2" |<math>bh-cg</math>
+! colspan="11" |
-| rowspan="2" |<math>bi-dg</math>
+!
-| rowspan="2" |<math>bj-eg</math>
+|-
-| rowspan="2" |<math>ci-dh</math>
+! rowspan="5" |
-| rowspan="2" |<math>cj-eh</math>
+| rowspan="5" |hidden <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>
-| rowspan="2" |<math>dj-ei</math>
+! rowspan="5" |
+|<math>a</math>
+|<math>b</math>
+|<math>c</math>
 | rowspan="2" |
-| rowspan="2" |<math>av-bu</math>
+|<math>j</math>
-| rowspan="2" |<math>aw-cu</math>
+|<math>k</math>
-| rowspan="2" |<math>ax-du</math>
+|<math>l</math>
-| rowspan="2" |<math>ay-eu</math>
+|
-| rowspan="2" |<math>bw-cv</math>
+| colspan="3" rowspan="2" |
-| rowspan="2" |<math>bx-dv</math>
+! rowspan="5" |
-| rowspan="2" |<math>by-ev</math>
+|<math>a</math>
-| rowspan="2" |<math>cx-dw</math>
+|<math>b</math>
-| rowspan="2" |<math>cy-ew</math>
+|<math>c</math>
-| rowspan="2" |<math>dy-ex</math>
+| colspan="1" rowspan="2" |<math>+</math>
-| rowspan="2" |
+|<math>j</math>
-| colspan="10" rowspan="2" |
+|<math>k</math>
-| colspan="5" rowspan="3" |
+|<math>l</math>
-| rowspan="3" |
+| colspan="1" rowspan="2" |<math>=</math>
-| colspan="5" rowspan="3" |
+|<math>a+j</math>
-| rowspan="3" |
+|<math>b+k</math>
-|<math>2a(g+v)\\-2b(f+u)</math>
+|<math>c+l</math>
-|<math>2a(w+h)\\-2c(f+u)</math>
+! rowspan="5" |
-|<math>2a(i+x)\\-2d(f+u)</math>
-|<math>2a(j+y)\\-2e(f+u)</math>
-|<math>2b(w+h)\\-2c(g+v)</math>
-|<math>2b(i+x)\\-2d(g+v)</math>
-|<math>2b(j+y)\\-2e(g+v)</math>
-|<math>2c(i+x)\\-2d(w+h)</math>
-|<math>2c(j+y)\\-2e(w+h)</math>
-|<math>2d(j+y)\\-2e(i+x)</math>
 |-
-|simplify <math>(r_1∧r_2)</math> if necessary
+|<math>d</math>
-|<math>a(g+v)\\-b(f+u)</math>
+|<math>e</math>
-|<math>a(w+h)\\-c(f+u)</math>
+|<math>f</math>
-|<math>a(i+x)\\-d(f+u)</math>
+|<math>g</math>
-|<math>a(j+y)\\-e(f+u)</math>
+|<math>h</math>
-|<math>b(w+h)\\-c(g+v)</math>
+|<math>i</math>
-|<math>b(i+x)\\-d(g+v)</math>
+|<math></math>
-|<math>b(j+y)\\-e(g+v)</math>
+|<math>d</math>
-|<math>c(i+x)\\-d(w+h)</math>
+|<math>e</math>
-|<math>c(j+y)\\-e(w+h)</math>
+|<math>f</math>
-|<math>d(j+y)\\-e(i+x)</math>
+|<math>g</math>
+|<math>h</math>
+|<math>i</math>
+|<math>d+g</math>
+|<math>e+h</math>
+|<math>f+i</math>
 |-
-|<math>(r_1∧r_2)∧r_3</math>
+| colspan="3" rowspan="1" |<math>∧</math>
-|<math>k(bh-cg)\\-l(ah-cf)\\+m(ag-bf)</math>
+|
-|<math>k(bi-dg)\\-l(ai-df)\\+n(ag-bf)</math>
+| colspan="3" rowspan="1" |<math>∧</math>
-|<math>k(bj-eg)\\-l(aj-ef)\\+o(ag-bf)</math>
+|
-|<math>k(ci-dh)\\-m(ai-df)\\+n(ah-cf)</math>
+| colspan="3" |
-|<math>k(cj-eh)\\-m(aj-ef)\\+o(ah-cf)</math>
+| colspan="3" |
-|<math>k(dj-ei)\\-n(aj-ef)\\+o(ai-df)</math>
+|
-|<math>l(ci-dh)\\-m(bi-dg)\\+n(bh-cg)</math>
+| colspan="3" |
-|<math>l(cj-eh)\\-m(bj-eg)\\+o(bh-cg)</math>
+|
-|<math>l(dj-ei)\\-n(bj-eg)\\+o(bi-dg)</math>
+| colspan="3" rowspan="1" |<math>∧</math>
-|<math>m(dj-ei)\\-n(cj-eh)\\+o(ci-dh)</math>
+|-
+|<math>bf-ce</math>
+|<math>af-cd</math>
+|<math>ae-bd</math>
 |<math>+</math>
-|<math>p(bw-cv)\\-q(aw-cu)\\+r(av-bu)</math>
+|<math>ki-lh</math>
-|<math>p(bx-dv)\\-q(ax-du)\\+s(av-bu)</math>
+|<math>ji-lg</math>
-|<math>p(by-ev)\\-q(ay-eu)\\+t(av-bu)</math>
+|<math>jh-kg</math>
-|<math>p(cx-dw)\\-r(ax-du)\\+s(aw-cu)</math>
+|<math>=</math>
-|<math>p(cy-ew)\\-r(ay-eu)\\+t(aw-cu)</math>
+|<math>bf-ce+ki-lh</math>
-|<math>p(dy-ex)\\-s(ay-eu)\\+t(ax-du)</math>
+|<math>af-cd+ji-lg</math>
-|<math>q(cx-dw)\\-r(bx-dv)\\+s(bw-cv)</math>
+|<math>ae-bd+jh-kg</math>
-|<math>q(cy-ew)\\-r(by-ev)\\+t(bw-cv)</math>
+| colspan="3" rowspan="2" |
-|<math>q(dy-ex)\\-s(by-ev)\\+t(bw-cv)</math>
+| rowspan="2" |
-|<math>r(dy-ex)\\-s(cy-ew)\\+t(cx-dw)</math>
+| colspan="3" rowspan="2" |
-|<math>=</math>
+| rowspan="2" |
-| style="background-color: LightBlue;"|<math>k(bh-cg)\\-l(ah-cf)\\+m(ag-bf)\\+p(bw-cv)\\-q(aw-cu)\\+r(av-bu)</math>
+|<math>(b+k)(f+i)-(c+l)(e+h)</math>
-| style="background-color: LightBlue;"|<math>k(bi-dg)\\-l(ai-df)\\+n(ag-bf)\\+p(bx-dv)\\-q(ax-du)\\+s(av-bu)</math>
+|<math>(a+j)(f+i)-(c+l)(d+g)</math>
-| style="background-color: LightBlue;"|<math>k(bj-eg)\\-l(aj-ef)\\+o(ag-bf)\\+p(by-ev)\\-q(ay-eu)\\+t(av-bu)</math>
+|<math>(a+j)(e+h)-(b+k)(d+g)</math>
-| style="background-color: LightBlue;"|<math>k(ci-dh)\\-m(ai-df)\\+n(ah-cf)\\+p(cx-dw)\\-r(ax-du)\\+s(aw-cu)</math>
+|-
-| style="background-color: LightBlue;"|<math>k(cj-eh)\\-m(aj-ef)\\+o(ah-cf)\\+p(cy-ew)\\-r(ay-eu)\\+t(aw-cu)</math>
+| colspan="3" |
-| style="background-color: LightBlue;"|<math>k(dj-ei)\\-n(aj-ef)\\+o(ai-df)\\+p(dy-ex)\\-s(ay-eu)\\+t(ax-du)</math>
+|
-| style="background-color: LightBlue;"|<math>l(ci-dh)\\-m(bi-dg)\\+n(bh-cg)\\+q(cx-dw)\\-r(bx-dv)\\+s(bw-cv)</math>
+| colspan="3" |
-| style="background-color: LightBlue;"|<math>l(cj-eh)\\-m(bj-eg)\\+o(bh-cg)\\+q(cy-ew)\\-r(by-ev)\\+t(bw-cv)</math>
+|
-| style="background-color: LightBlue;"|<math>l(dj-ei)\\-n(bj-eg)\\+o(bi-dg)\\+q(dy-ex)\\-s(by-ev)\\+t(bw-cv)</math>
+| style="background-color: LightBlue;"|<math>bf-ce+ki-lh</math>
-| style="background-color: LightBlue;"|<math>m(dj-ei)\\-n(cj-eh)\\+o(ci-dh)\\+r(dy-ex)\\-s(cy-ew)\\+t(cx-dw)</math>
+| style="background-color: LightBlue;"|<math>af-cd+ji-lg</math>
-| style="background-color: LightBlue;"|<math>(k+p)\\(b(w+h)-c(g+v))\\-(l+q)\\(a(w+h)-c(f+u))\\+(m+r)\\(a(g+v)-b(f+u))</math>
+| style="background-color: LightBlue;"|<math>ae-bd+jh-kg</math>
-| style="background-color: LightBlue;"|<math>(k+p)\\(b(i+x)-d(g+v))\\-(l+q)\\(a(i+x)-d(f+u))\\+(n+s)\\(a(g+v)-b(f+u))</math>
+| style="background-color: LightBlue;"|<math>bf+bi+kf+ki-ce-ch-le-lh</math>
-| style="background-color: LightBlue;"|<math>(k+p)\\(b(j+y)-e(g+v))\\-(l+q)\\(a(j+y)-e(f+u))\\+(o+t)\\(a(g+v)-b(f+u))</math>
+| style="background-color: LightBlue;"|<math>af+ai+jf+ji-cd-cg-ld-lg</math>
-| style="background-color: LightBlue;"|<math>(k+p)\\(c(i+x)-d(w+h))\\-(m+r)\\(a(i+x)-d(f+u))\\+(n+s)\\(a(w+h)-c(f+u))</math>
+| style="background-color: LightBlue;"|<math>ae+ah+je+jh-bd-bg-kd-kg</math>
-| style="background-color: LightBlue;"|<math>(k+p)\\(c(j+y)-e(w+h))\\-(m+r)\\(a(j+y)-e(f+u))\\+(o+t)\\(a(w+h)-c(f+u))</math>
-| style="background-color: LightBlue;"|<math>(k+p)\\(d(j+y)-e(i+x))\\-(n+s)\\(a(j+y)-e(f+u))\\+(o+t)\\(a(i+x)-d(f+u))</math>
-| style="background-color: LightBlue;"|<math>(l+q)\\(c(i+x)-d(w+h))\\-(m+r)\\(b(i+x)-d(g+v))\\+(n+s)\\(b(w+h)-c(g+v))</math>
-| style="background-color: LightBlue;"|<math>(l+q)\\(c(j+y)-e(w+h))\\-(m+r)\\(b(j+y)-e(g+v))\\+(o+t)\\(b(w+h)-c(g+v))</math>
-| style="background-color: LightBlue;"|<math>(l+q)\\(d(j+y)-e(i+x))\\-(n+s)\\(b(j+y)-e(g+v))\\+(o+t)\\(b(i+x)-d(g+v))</math>
-| style="background-color: LightBlue;"|<math>(m+r)\\(d(j+y)-e(i+x))\\-(n+s)\\(c(j+y)-e(w+h))\\+(o+t)\\(c(i+x)-d(w+h))</math>
 |-
 !
-! colspan="2" |
 !
-! colspan="32" |
 !
-! colspan="22" |
+! colspan="11" |
+!
+! colspan="11" |
 !
 |}
-These two examples are by no means a proof, but meditation on the patterns in the variables is at least fairly convincing.
+This second diagram demonstrates this situation for a <math>d=5, g=3</math> case. One pair of the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span> vectors are explicitly matching, but not the other, which isn't enough.
+{| class="wikitable center-all"
-===Sintel's proof of the <span style="color: #B6321C;">linear-independence</span> conjecture===
+|+
+!
-====Sintel's original text====
+! colspan="2" |
+!
- <nowiki>If A and B are mappings from Z^n to Z^m, with n > m, A, B full rank (using A and B as their rowspace equivalently):
+! colspan="32" |
+!
-  dim(A + B) - m = dim(ker(A) + ker(B)) - (n-m)
+! colspan="22" |
+!
-  >> dim(A)+dim(B)=dim(A+B)+dim(A∩B) => dim(A + B) = dim(A) + dim(B) - dim(A∩B)
+|-
+!
-  dim(A) + dim(B) - dim(A∩B) - m = dim(ker(A) + ker(B)) - (n-m)
+! colspan="2" |
+!
-  >> by duality of kernel, dim(ker(A) + ker(B))  = dim(ker(A ∩ B))
+| colspan="32" rowspan="1" |'''multivector approach'''
+!
-  dim(A) + dim(B) - dim(A∩B) - m = dim(ker(A ∩ B))  - (n-m)
+| colspan="22" rowspan="1" |'''matrix approach'''
+!
-  >> rank nullity: dim(ker(A ∩ B)) + dim(A ∩ B) = n
+|-
+!
-  dim(A) + dim(B) - dim(A∩B) - m = n -  dim(A ∩ B)  - (n-m)
+! colspan="2" |
+!
-  m + m - dim(A∩B) - m = n -  dim(A ∩ B)  - (n-m)
+! colspan="32" |
+!
-  m + m - m = n - n + m
+! colspan="22" |
+!
-  m = m</nowiki>
+|-
+! rowspan="7" |
+| colspan="1" rowspan="7" |explicit <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>
-====Douglas Blumeyer's interpretation====
+⟨{{vector|<math>a</math> <math>b</math> <math>c</math> <math>d</math> <math>e</math>}}
+{{vector|<math>f</math> <math>g</math> <math>h</math> <math>i</math> <math>j</math>}}]
-We're going to take the strategy of beginning with what we're trying to prove, then reducing it to an obvious equivalence, which will show that our initial statement must be just as true.
+| style="background-color: #BED5BA;"|<math>r_1</math>
+! rowspan="7" |
-So here's the statement we're trying to prove:
+| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>a</math>
+| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>b</math>
+| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>c</math>
-<math>\text{rank}(\text{union}(M_1, M_2)) - r = \text{nullity}(\text{union}(C_1, C_2)) - n</math>
+| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>d</math>
+| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>e</math>
+| rowspan="3" |
-<math>M_1</math> and <math>M_2</math> are mappings which both have dimensionality <math>d</math>, rank <math>r</math>, nullity <math>n</math>, and are full-rank, and <math>C_1</math> and <math>C_2</math> are their comma bases, respectively.
+| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>a</math>
+| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>b</math>
-Technically since these matrices are representing subspace bases, the correct operation here is "sumset", not "union", but because "union" is a more commonly known opposite of intersection and would work for plain matrices, I've decided to stick with it here.
+| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>c</math>
+| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>d</math>
-So, the left-hand side of this equation is
+| style="background-color: #BED5BA;" colspan="2" rowspan="1" |<math>e</math>
+| rowspan="3" |
+| colspan="10" rowspan="3" |
+! rowspan="7" |
+| style="background-color: #BED5BA;"|<math>a</math>
+| style="background-color
-\color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}-1 \\
-\end{array} \right]</math>
+<math>r = r</math>
-And so, reintroducing the <span style="color: #3C8031;"><math>L_{\text{dep}}</math></span>, we have:
+So we know this is true.
-<math>\left[ \begin{array} {rrr}
-\color{OliveGreen}19 & \color{OliveGreen}30 & \color{OliveGreen}44 & \color{OliveGreen}53 \\
-\color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}0 & \color{BrickRed}-1 \\
-\end{array} \right]</math>
-Which canonicalizes to:
-<math>\left[ \begin{array} {rrr}
-& 30 & 44 & 0 \\
-& 0 & 0 & 1 \\
-\end{array} \right]</math>
-And so we can see that meantone minus flattone is [[meanmag]].
 =Glossary=