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<h2>IMPORTED REVISION FROM WIKISPACES</h2>
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One interesting way to think of covectors, since they're these dual vectors that "act on" normal vectors, is thus as functions - they take in a vector as input, multiply each coefficient of the vector by the corresponding coefficient of the covector, sum them up, and spit out a number. In other words, you know that the action of the covector \((12,19,28)^*\) on any arbitrary vector \((a,b,c)\) is going to be \(12a + 19b + 28c\). So, you can think of \((12,19,28)^*\) itself as a function looking something like \(f(\v{v}) = 12a + 19b + 28c\) for some vector of the form (a, b, c). Note that the formatting on \(\v{v}\) specifies that \(\v{v}\) is a vector that's being taken in as input.
One interesting way to think of covectors, since they're these dual vectors that "act on" normal vectors, is thus as functions - they take in a vector as input, multiply each coefficient of the vector by the corresponding coefficient of the covector, sum them up, and spit out a number. In other words, you know that the action of the covector \((12,19,28)^*\) on any arbitrary vector \((a,b,c)\) is going to be \(12a + 19b + 28c\). So, you can think of \((12,19,28)^*\) itself as a function looking something like \(f(\v{v}) = 12a + 19b + 28c\) for some vector of the form (a, b, c). Note that the formatting on \(\v{v}\) specifies that \(\v{v}\) is a vector that's being taken in as input.


Before we go on, however, let's clean up the notation a bit. In physics, the notation commonly used is to notate covectors \(\bratext{like this}\) and to notate vectors \(\kettext{like this}\). Physicists call this "bra-ket" notation, or sometimes "Dirac" notation. So instead of writing covectors as \((x,y,z)^*\), I'll just write \(\bra{x \s y \s z}\) from now on. And instead of writing vectors as \((a,b,c)), I'll just write \(\ket{|a \s b \s c&gt;}\) from now on.
Before we go on, however, let's clean up the notation a bit. In physics, the notation commonly used is to notate covectors \(\bratext{like this}\) and to notate vectors \(\kettext{like this}\). Physicists call this "bra-ket" notation, or sometimes "Dirac" notation. So...
* Instead of writing covectors as \((x,y,z)^*\), I'll just write \(\bra{x \s y \s z}\) from now on.
* Instead of writing vectors as \((a,b,c)\), I'll just write \(\ket{a \s b \s c}\) from now on.
* When I want to denote the dot product of a covector and a vector, I'll write it as \braket{x \s y \s z}{a \s b \s c}.
 
Now then, let's say you're going to ask a harmlessly ordinary question like "does 81/80 vanish in 12-EDO?" But let's think: when you ask that question, what are you really asking? Another way to rephrase that question is to ask this: "if I go up four tempered 3/2's, then go down a
tempered 5/1, thus putting me at what's supposed to be 81/80 - how many steps in 12-EDO do I arrive at? Is it 0?"
 
Or, to represent everything in terms of primes, that's also the same as saying "if I go up four tempered 3/1's, then down a tempered 5/1, and then down four (possibly tempered) 2/1's to reduce within the octave, thus putting me at what's supposed to be 81/80 - how many steps in 12-EDO do I arrive at? Is it 0?"
 
And, if you want to put that in prime order, so that first you move around by all a bunch of 2/1's, then you move around by a bunch of 3/1's then you move around by a bunch of 5/1's, we arrive at yet another equivalent expression: "if I first go down four possibly tempered 2/1's, then up four tempered 3/1's, then down a tempered 5/1, thus putting me at what's supposed to be 81/80 - how many steps in 12-EDO do I arrive at? Is it 0?"
 
You may want to take a second to confirm for yourself that all of these expressions are the same thing - they'll all move you up by exactly 81/80.
 
This last formulation of the question is especially easy to figure out if we know how we want to map the primes in 12-EDO. Which, hopefully, we do: in 12-EDO, 2/1 is most obviously mapped to 12 steps (duh), 3/1 is best mapped to 19 steps, and 5/1 is best mapped to 28 steps. Let's see what happens, then, if we mechanically solve the above problem by stacking and removing intervals from one another like lego pieces or something, in exactly the fashion I mentioned above, and see what we're left with. Let's see what we get:


Technically, the application of &lt;x y z| to |a b c&gt; isn't called the dot product, for obscure mathematical reasons. It's sometimes called the "bracket product." But I've seen even Gene call it the "dot product" before, so I'm just going to informally use that usage for now because it's something everyone's familiar with (and it's basically the same exact thing).
Now then, let's say you're going to ask a question like "does 81/80 vanish in 12-EDO?" But when you ask that question, what are you really asking?
Well, you're asking the question - "if I go up four 3/2's, then go down a 5/1, thus putting me at the syntonic comma - how many steps in 12-EDO do I arrive at?"
Or, to represent everything in terms of primes, that's the same as saying "if I go up four 3/1's, then down a 5/1, and then down four 2/1's to reduce within the octave - how many steps in 12-EDO do I arrive at?"
So, in 12-EDO, 2/1 is best mapped to 12 steps, 3/1 is best mapped to 19 steps, and 5/1 is best mapped to 28 steps. Let's see what happens if we do it mechanically out by stacking and removing intervals from one another like lego pieces or something.
1) first you go down 4 octaves, at a rate of 12 steps per octave, putting you underwater at -48 steps.
1) first you go down 4 octaves, at a rate of 12 steps per octave, putting you underwater at -48 steps.
2) then, you go up 4 tritaves, times 19 steps per tritave, giving you 76 steps. This lands you at a net of -48 + 76 = 28 steps.
3) finally, you go down one 5/1, times 28 steps per 5/1, putting you down 28 more steps. This lands you at a net of 28 - 28 steps = 0.
So, if you mechanically work out the way that you'd compute how many steps 81/80 is in 12-EDO, you get 0 steps, meaning you're back at 1/1 and hence tempered out. No surprise there.
But, if you really think about it, what you've also just done is evaluate the expression \(12*\-4 + 19*4 + 28*\-1): down four (possibly tempered) octaves, up four tempered tritaves, down a tempered 5/1, see what the result is. This is the same exact thing as multiplying out \(\braket{12 \s 19 \s 28}{\-4 \s 4 \s -1}\). Looks like you've just seemingly applied \(\bra{12 \s 19 \s 18}\) to \(\ket{\-4 \s 4 \s \-1}\).
OK, so that's kind of neat. Now what?
==1.3: Why the fact that covectors mean stuff matters. (OR: PREPARE FOR WEDGIE)==
Assuming you've understood my exposition thus far, you now hopefully see where all things like monzos and vals come from.
* - Monzos, straightforwardly, are elements in a vector space - they're vectors - like the spaces you learned in high school algebra.
* - Vals, straightforwardly, are elements in the "dual space" to that space - they're covectors - which is the new thing you just learned about and which you'd end up learning in college linear algebra.
Also, hopefully, now you see that this whole \(\brakettext{covector}{vector}\) thing is just a mathematical representation of the mechanical process you'd be doing if you sat down in scala for a while and stacked and subtracted intervals and figured out what you end up arriving at in some EDO. In other words, \(\bra{12 \s 19 \s 28}\) is just a little machine that takes in JI intervals and spits out steps, and the mathematical name for this sort of little machine is "covector." Very nice!
Now, historically speaking, the concept of taking JI intervals and treating them like vectors isn't new. Theorists have been doing that for years and years. Neither Paul nor Gene nor Graham nor Joe Monzo nor anyone I know first came up with the concept of plotting JI intervals as vectors on a JI lattice.
The real quantum leap in thought, here, is to consider the musical interpretation of the elements in the -dual space- - the covectors. It's is one of those things that appears to be completely meaningless from a musical standpoint unless you really think about it. Someone at some point figured out all of the above and I think it's the one of the best insights ever made in music theory. Looking at the dual space is, in a sense, the logical completion of the idea of putting JI into a "lattice," as per Tenney/Fokker/Wilson/etc.
Now, where do we go from here?
Well, if there's one thing mathematicians know a lot of random crap about, it's vectors and covectors. They've been exploring manipulations of these sorts of objects for literally thousands of years under the moniker of "linear algebra." These objects may be new to music theory, but they're definitely not new to math. And since mathematics has concerned itself for such a long time with manipulating covectors and vectors, we can simply take some of the techniques that mathematicians have developed to do so, and apply them here, in a musical context. Some of these linear-algebraic manipulations may seem to have no musical purpose at all. Others do. Others may seem to have no purpose at all, at first, and then end up having a purpose that strikes you in a flash after a little bit of thought.
One of the manipulations that would appear to be totally useless is a deceptively simple extension of all of this called **exterior algebra**. It introduces a single product, called the **wedge product**, and you can multiply vectors together using this in a similar sort of way that you'd multiply polynomials on paper in high school algebra or something. It also introduces a new type of object, called a "multivector" - defined as the product of vectors. This is one of those things that seems totally useless unless you're [[Gene Ward Smith|Gene Smith]], at which point you realize that multivals represent higher-rank temperaments, and that this random field of mathematics has a rather musical interpretation.
So then, how do we continue? Obviously higher-rank temperaments are things we care about: we can't just stop at vals and leave it at that. Now that you're here, there are basically two different paths you can go down, both of which are basically the same exact thing:
1) The exterior algebra/"wedgie" route, a la Gene.
2) The matrix algebra/"mapping" route, a la Graham Breed.


2) then, you go up 4 tritaves, times 19 steps per tritave, giving you 76 steps. This lands you at a net of -48 + 76 = 28 steps.
Both of these have different strengths. Gene's approach I find to be remarkably elegant in a certain kind of way, in that multivectors are very intuitive objects to think about. Graham's approach, on the other hand, isn't quite as "elegant," but may be a bit simpler to work out and immediately glean information from - mapping matrices are a bit easier to understand than wedgies.
 
We'll cover both of these eventually. But first, we'll need to dig deeper into what, exactly, temperaments "are," conceptually. Stay tuned for that...


3) finally, you go down one 5/1, times 28 steps per 5/1, putting you down 28 more steps. This lands you at a net of 28 - 28 steps = 0.
So, if you mechanically work out the way that you'd compute how many steps 81/80 is in 12-EDO, you get 0 steps, meaning it's "equated with 1/1" and hence tempered out. No surprise there.
But, if you really think about it, you'd also find that you've just seemingly applied &lt;12 19 28| to |-4 4 -1&gt;. What you've just done is evaluate the expression 12*-4 + 19*4 + 28*-1: down four octaves, up four tritaves, down a 5/1. This is the same as multiplying &lt;12 19 28|-4 4 -1&gt;.
And, to answer Mike S's question, note also that 12*-4 + 19*4 + 28*-1 is also the dot product of the vectors (12, 19, 28) and (-4, 4, -1). Or it's the bracket product of &lt;12 19 28|-4 4 -1&gt; - same basic thing.
So that's what the interpretation of the dot product is.
For some of you, this may be review, but it's meant to give a basic foundation of the mathematical reasoning underpinning some of these objects. Stay tuned for more...


[[#ref1]][1] - Note that some have raised technical concerns about this operation being called the "dot product," insisting that the dot product is something that's only done between two vectors, or two covectors, but never between one covector and one vector. Another term that's sometimes been used for this product in the "**bracket product**", for reasons we don't need to get into here. However, confusingly, the term bracket product has also been used for the ordinary dot product, and it's also very common to hear people call the thing I'm calling the dot product above. It's best at this point to just know that the two terms are out there. I'm going to continue calling it the dot product since its' something more people are familiar with.
[[#ref1]][1] - Note that some have raised technical concerns about this operation being called the "dot product," insisting that the dot product is something that's only done between two vectors, or two covectors, but never between one covector and one vector. Another term that's sometimes been used for this product in the "**bracket product**", for reasons we don't need to get into here. However, confusingly, the term bracket product has also been used for the ordinary dot product, and it's also very common to hear people call the thing I'm calling the dot product above. It's best at this point to just know that the two terms are out there. I'm going to continue calling it the dot product since its' something more people are familiar with.
Line 150: Line 188:
One interesting way to think of covectors, since they're these dual vectors that &amp;quot;act on&amp;quot; normal vectors, is thus as functions - they take in a vector as input, multiply each coefficient of the vector by the corresponding coefficient of the covector, sum them up, and spit out a number. In other words, you know that the action of the covector \((12,19,28)^*\) on any arbitrary vector \((a,b,c)\) is going to be \(12a + 19b + 28c\). So, you can think of \((12,19,28)^*\) itself as a function looking something like \(f(\v{v}) = 12a + 19b + 28c\) for some vector of the form (a, b, c). Note that the formatting on \(\v{v}\) specifies that \(\v{v}\) is a vector that's being taken in as input.&lt;br /&gt;
One interesting way to think of covectors, since they're these dual vectors that &amp;quot;act on&amp;quot; normal vectors, is thus as functions - they take in a vector as input, multiply each coefficient of the vector by the corresponding coefficient of the covector, sum them up, and spit out a number. In other words, you know that the action of the covector \((12,19,28)^*\) on any arbitrary vector \((a,b,c)\) is going to be \(12a + 19b + 28c\). So, you can think of \((12,19,28)^*\) itself as a function looking something like \(f(\v{v}) = 12a + 19b + 28c\) for some vector of the form (a, b, c). Note that the formatting on \(\v{v}\) specifies that \(\v{v}\) is a vector that's being taken in as input.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Before we go on, however, let's clean up the notation a bit. In physics, the notation commonly used is to notate covectors \(\bratext{like this}\) and to notate vectors \(\kettext{like this}\). Physicists call this &amp;quot;bra-ket&amp;quot; notation, or sometimes &amp;quot;Dirac&amp;quot; notation. So instead of writing covectors as \((x,y,z)^*\), I'll just write \(\bra{x \s y \s z}\) from now on. And instead of writing vectors as \((a,b,c)), I'll just write \(\ket{|a \s b \s c&amp;gt;}\) from now on.&lt;br /&gt;
Before we go on, however, let's clean up the notation a bit. In physics, the notation commonly used is to notate covectors \(\bratext{like this}\) and to notate vectors \(\kettext{like this}\). Physicists call this &amp;quot;bra-ket&amp;quot; notation, or sometimes &amp;quot;Dirac&amp;quot; notation. So...&lt;br /&gt;
&lt;ul&gt;&lt;li&gt;Instead of writing covectors as \((x,y,z)^*\), I'll just write \(\bra{x \s y \s z}\) from now on.&lt;/li&gt;&lt;li&gt;Instead of writing vectors as \((a,b,c)\), I'll just write \(\ket{a \s b \s c}\) from now on.&lt;/li&gt;&lt;li&gt;When I want to denote the dot product of a covector and a vector, I'll write it as \braket{x \s y \s z}{a \s b \s c}.&lt;/li&gt;&lt;/ul&gt;&lt;br /&gt;
Now then, let's say you're going to ask a harmlessly ordinary question like &amp;quot;does 81/80 vanish in 12-EDO?&amp;quot; But let's think: when you ask that question, what are you really asking? Another way to rephrase that question is to ask this: &amp;quot;if I go up four tempered 3/2's, then go down a&lt;br /&gt;
tempered 5/1, thus putting me at what's supposed to be 81/80 - how many steps in 12-EDO do I arrive at? Is it 0?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
Or, to represent everything in terms of primes, that's also the same as saying &amp;quot;if I go up four tempered 3/1's, then down a tempered 5/1, and then down four (possibly tempered) 2/1's to reduce within the octave, thus putting me at what's supposed to be 81/80 - how many steps in 12-EDO do I arrive at? Is it 0?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
And, if you want to put that in prime order, so that first you move around by all a bunch of 2/1's, then you move around by a bunch of 3/1's then you move around by a bunch of 5/1's, we arrive at yet another equivalent expression: &amp;quot;if I first go down four possibly tempered 2/1's, then up four tempered 3/1's, then down a tempered 5/1, thus putting me at what's supposed to be 81/80 - how many steps in 12-EDO do I arrive at? Is it 0?&amp;quot;&lt;br /&gt;
&lt;br /&gt;
You may want to take a second to confirm for yourself that all of these expressions are the same thing - they'll all move you up by exactly 81/80.&lt;br /&gt;
&lt;br /&gt;
This last formulation of the question is especially easy to figure out if we know how we want to map the primes in 12-EDO. Which, hopefully, we do: in 12-EDO, 2/1 is most obviously mapped to 12 steps (duh), 3/1 is best mapped to 19 steps, and 5/1 is best mapped to 28 steps. Let's see what happens, then, if we mechanically solve the above problem by stacking and removing intervals from one another like lego pieces or something, in exactly the fashion I mentioned above, and see what we're left with. Let's see what we get:&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Technically, the application of &amp;lt;x y z| to |a b c&amp;gt; isn't called the dot product, for obscure mathematical reasons. It's sometimes called the &amp;quot;bracket product.&amp;quot; But I've seen even Gene call it the &amp;quot;dot product&amp;quot; before, so I'm just going to informally use that usage for now because it's something everyone's familiar with (and it's basically the same exact thing).&lt;br /&gt;
Now then, let's say you're going to ask a question like &amp;quot;does 81/80 vanish in 12-EDO?&amp;quot; But when you ask that question, what are you really asking?&lt;br /&gt;
Well, you're asking the question - &amp;quot;if I go up four 3/2's, then go down a 5/1, thus putting me at the syntonic comma - how many steps in 12-EDO do I arrive at?&amp;quot;&lt;br /&gt;
Or, to represent everything in terms of primes, that's the same as saying &amp;quot;if I go up four 3/1's, then down a 5/1, and then down four 2/1's to reduce within the octave - how many steps in 12-EDO do I arrive at?&amp;quot;&lt;br /&gt;
So, in 12-EDO, 2/1 is best mapped to 12 steps, 3/1 is best mapped to 19 steps, and 5/1 is best mapped to 28 steps. Let's see what happens if we do it mechanically out by stacking and removing intervals from one another like lego pieces or something.&lt;br /&gt;
1) first you go down 4 octaves, at a rate of 12 steps per octave, putting you underwater at -48 steps.&lt;br /&gt;
1) first you go down 4 octaves, at a rate of 12 steps per octave, putting you underwater at -48 steps.&lt;br /&gt;
2) then, you go up 4 tritaves, times 19 steps per tritave, giving you 76 steps. This lands you at a net of -48 + 76 = 28 steps.&lt;br /&gt;
3) finally, you go down one 5/1, times 28 steps per 5/1, putting you down 28 more steps. This lands you at a net of 28 - 28 steps = 0.&lt;br /&gt;
&lt;br /&gt;
So, if you mechanically work out the way that you'd compute how many steps 81/80 is in 12-EDO, you get 0 steps, meaning you're back at 1/1 and hence tempered out. No surprise there.&lt;br /&gt;
&lt;br /&gt;
But, if you really think about it, what you've also just done is evaluate the expression \(12*\-4 + 19*4 + 28*\-1): down four (possibly tempered) octaves, up four tempered tritaves, down a tempered 5/1, see what the result is. This is the same exact thing as multiplying out \(\braket{12 \s 19 \s 28}{\-4 \s 4 \s -1}\). Looks like you've just seemingly applied \(\bra{12 \s 19 \s 18}\) to \(\ket{\-4 \s 4 \s \-1}\).&lt;br /&gt;
&lt;br /&gt;
OK, so that's kind of neat. Now what?&lt;br /&gt;
&lt;br /&gt;
&lt;!-- ws:start:WikiTextHeadingRule:14:&amp;lt;h2&amp;gt; --&gt;&lt;h2 id="toc5"&gt;&lt;a name="LECTURE 1: Vector Spaces and Dual Spaces-1.3: Why the fact that covectors mean stuff matters. (OR: PREPARE FOR WEDGIE)"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextHeadingRule:14 --&gt;1.3: Why the fact that covectors mean stuff matters. (OR: PREPARE FOR WEDGIE)&lt;/h2&gt;
&lt;br /&gt;
Assuming you've understood my exposition thus far, you now hopefully see where all things like monzos and vals come from.&lt;br /&gt;
&lt;br /&gt;
&lt;ul&gt;&lt;li&gt;- Monzos, straightforwardly, are elements in a vector space - they're vectors - like the spaces you learned in high school algebra.&lt;/li&gt;&lt;li&gt;- Vals, straightforwardly, are elements in the &amp;quot;dual space&amp;quot; to that space - they're covectors - which is the new thing you just learned about and which you'd end up learning in college linear algebra.&lt;/li&gt;&lt;/ul&gt;&lt;br /&gt;
Also, hopefully, now you see that this whole \(\brakettext{covector}{vector}\) thing is just a mathematical representation of the mechanical process you'd be doing if you sat down in scala for a while and stacked and subtracted intervals and figured out what you end up arriving at in some EDO. In other words, \(\bra{12 \s 19 \s 28}\) is just a little machine that takes in JI intervals and spits out steps, and the mathematical name for this sort of little machine is &amp;quot;covector.&amp;quot; Very nice!&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
Now, historically speaking, the concept of taking JI intervals and treating them like vectors isn't new. Theorists have been doing that for years and years. Neither Paul nor Gene nor Graham nor Joe Monzo nor anyone I know first came up with the concept of plotting JI intervals as vectors on a JI lattice.&lt;br /&gt;
&lt;br /&gt;
The real quantum leap in thought, here, is to consider the musical interpretation of the elements in the -dual space- - the covectors. It's is one of those things that appears to be completely meaningless from a musical standpoint unless you really think about it. Someone at some point figured out all of the above and I think it's the one of the best insights ever made in music theory. Looking at the dual space is, in a sense, the logical completion of the idea of putting JI into a &amp;quot;lattice,&amp;quot; as per Tenney/Fokker/Wilson/etc.&lt;br /&gt;
&lt;br /&gt;
Now, where do we go from here?&lt;br /&gt;
&lt;br /&gt;
Well, if there's one thing mathematicians know a lot of random crap about, it's vectors and covectors. They've been exploring manipulations of these sorts of objects for literally thousands of years under the moniker of &amp;quot;linear algebra.&amp;quot; These objects may be new to music theory, but they're definitely not new to math. And since mathematics has concerned itself for such a long time with manipulating covectors and vectors, we can simply take some of the techniques that mathematicians have developed to do so, and apply them here, in a musical context. Some of these linear-algebraic manipulations may seem to have no musical purpose at all. Others do. Others may seem to have no purpose at all, at first, and then end up having a purpose that strikes you in a flash after a little bit of thought.&lt;br /&gt;
&lt;br /&gt;
One of the manipulations that would appear to be totally useless is a deceptively simple extension of all of this called &lt;strong&gt;exterior algebra&lt;/strong&gt;. It introduces a single product, called the &lt;strong&gt;wedge product&lt;/strong&gt;, and you can multiply vectors together using this in a similar sort of way that you'd multiply polynomials on paper in high school algebra or something. It also introduces a new type of object, called a &amp;quot;multivector&amp;quot; - defined as the product of vectors. This is one of those things that seems totally useless unless you're &lt;a class="wiki_link" href="/Gene%20Ward%20Smith"&gt;Gene Smith&lt;/a&gt;, at which point you realize that multivals represent higher-rank temperaments, and that this random field of mathematics has a rather musical interpretation.&lt;br /&gt;
&lt;br /&gt;
So then, how do we continue? Obviously higher-rank temperaments are things we care about: we can't just stop at vals and leave it at that. Now that you're here, there are basically two different paths you can go down, both of which are basically the same exact thing:&lt;br /&gt;
&lt;br /&gt;
1) The exterior algebra/&amp;quot;wedgie&amp;quot; route, a la Gene.&lt;br /&gt;
2) The matrix algebra/&amp;quot;mapping&amp;quot; route, a la Graham Breed.&lt;br /&gt;
&lt;br /&gt;
Both of these have different strengths. Gene's approach I find to be remarkably elegant in a certain kind of way, in that multivectors are very intuitive objects to think about. Graham's approach, on the other hand, isn't quite as &amp;quot;elegant,&amp;quot; but may be a bit simpler to work out and immediately glean information from - mapping matrices are a bit easier to understand than wedgies.&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
2) then, you go up 4 tritaves, times 19 steps per tritave, giving you 76 steps. This lands you at a net of -48 + 76 = 28 steps.&lt;br /&gt;
We'll cover both of these eventually. But first, we'll need to dig deeper into what, exactly, temperaments &amp;quot;are,&amp;quot; conceptually. Stay tuned for that...&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
3) finally, you go down one 5/1, times 28 steps per 5/1, putting you down 28 more steps. This lands you at a net of 28 - 28 steps = 0.&lt;br /&gt;
So, if you mechanically work out the way that you'd compute how many steps 81/80 is in 12-EDO, you get 0 steps, meaning it's &amp;quot;equated with 1/1&amp;quot; and hence tempered out. No surprise there.&lt;br /&gt;
But, if you really think about it, you'd also find that you've just seemingly applied &amp;lt;12 19 28| to |-4 4 -1&amp;gt;. What you've just done is evaluate the expression 12*-4 + 19*4 + 28*-1: down four octaves, up four tritaves, down a 5/1. This is the same as multiplying &amp;lt;12 19 28|-4 4 -1&amp;gt;.&lt;br /&gt;
And, to answer Mike S's question, note also that 12*-4 + 19*4 + 28*-1 is also the dot product of the vectors (12, 19, 28) and (-4, 4, -1). Or it's the bracket product of &amp;lt;12 19 28|-4 4 -1&amp;gt; - same basic thing.&lt;br /&gt;
So that's what the interpretation of the dot product is.&lt;br /&gt;
For some of you, this may be review, but it's meant to give a basic foundation of the mathematical reasoning underpinning some of these objects. Stay tuned for more...&lt;br /&gt;
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&lt;!-- ws:start:WikiTextAnchorRule:14:&amp;lt;img src=&amp;quot;/i/anchor.gif&amp;quot; class=&amp;quot;WikiAnchor&amp;quot; alt=&amp;quot;Anchor&amp;quot; id=&amp;quot;wikitext@@anchor@@ref1&amp;quot; title=&amp;quot;Anchor: ref1&amp;quot;/&amp;gt; --&gt;&lt;a name="ref1"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextAnchorRule:14 --&gt;[1] - Note that some have raised technical concerns about this operation being called the &amp;quot;dot product,&amp;quot; insisting that the dot product is something that's only done between two vectors, or two covectors, but never between one covector and one vector. Another term that's sometimes been used for this product in the &amp;quot;&lt;strong&gt;bracket product&lt;/strong&gt;&amp;quot;, for reasons we don't need to get into here. However, confusingly, the term bracket product has also been used for the ordinary dot product, and it's also very common to hear people call the thing I'm calling the dot product above. It's best at this point to just know that the two terms are out there. I'm going to continue calling it the dot product since its' something more people are familiar with.&lt;br /&gt;
&lt;!-- ws:start:WikiTextAnchorRule:16:&amp;lt;img src=&amp;quot;/i/anchor.gif&amp;quot; class=&amp;quot;WikiAnchor&amp;quot; alt=&amp;quot;Anchor&amp;quot; id=&amp;quot;wikitext@@anchor@@ref1&amp;quot; title=&amp;quot;Anchor: ref1&amp;quot;/&amp;gt; --&gt;&lt;a name="ref1"&gt;&lt;/a&gt;&lt;!-- ws:end:WikiTextAnchorRule:16 --&gt;[1] - Note that some have raised technical concerns about this operation being called the &amp;quot;dot product,&amp;quot; insisting that the dot product is something that's only done between two vectors, or two covectors, but never between one covector and one vector. Another term that's sometimes been used for this product in the &amp;quot;&lt;strong&gt;bracket product&lt;/strong&gt;&amp;quot;, for reasons we don't need to get into here. However, confusingly, the term bracket product has also been used for the ordinary dot product, and it's also very common to hear people call the thing I'm calling the dot product above. It's best at this point to just know that the two terms are out there. I'm going to continue calling it the dot product since its' something more people are familiar with.&lt;br /&gt;
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