Douglas Blumeyer's RTT How-To: Difference between revisions

== multi(co)vectors ==

=== multicovectors ===

If you spend much time exploring information about RTT online, it won’t be long until you come across “wedgies”, which are an alternative way to represent temperaments. You may not need them to do RTT stuff; a lot of the information you can get from them can be gotten otherwise from the already-discussed matrices. But you may find they really appeal to you.

The mathematical structure used to represent a wedgie is called a '''multicovector'''. Only certain multicovectors technically qualify as wedgies<ref>Those without contorsion, etc.</ref>, so do not misunderstand me to be implying that multicovectors and wedgies are exactly equivalent. But for purposes of these materials, the distinction is not important, and even though multicovector has more syllables, I am doing what I can to lean away from RTT jargon in these materials, and toward generic mathematical concepts. So, “multicovector” it is.

 

Earlier we saw the term “covector”. This is the word for the mathematical structure that represents maps. Well, multicovectors and covectors have some things in common.

* They both exist on the left half of tuning duality, on the side built up out of ETs and concerns rank (not the side built up out of commas, which uses vectors/columns, and concerns nullity).

So we mentioned that a multicovector is an alternative way to represent a temperament. Let’s end the suspense. Here’s an example: meantone (in the 5-limit), as a multicovector, looks like this: {{multival|1 4 4}}. As you can see, it is a bicovector, because it has two of each bracket.

Why care about multicovectors? Well, a key reason is that they can serve the same purpose as the normal form of a temperament’s mapping: the process for converting a mapping to a multicovector will convert any equivalent mapping to the same exact multicovector. In other words, a multicovector can serve as a unique identifier for its temperament. And, unlike normal forms for matrices, there is no question about which normal form to use.

Alright, then, sounds great! But how do I convert a mapping to a multicovector? The process is doable. It’s closely related to the wedge product (hence the name “wedgie”), but better known as the '''exterior product'''. We write it with the symbol ∧, which is great because that’s the logical operator for “and”, forming a clear association with the unioning of two ETs.

First I’ll list the steps. Don’t worry if it doesn’t all make sense the first time. We’ll work through an example and go into more detail as we do. To be clear, what we're doing here is both more and less and different ways from the strict definition of the exterior product as you may see it elsewhere; I'm specifically here describing the process for finding the multicovector in the form you're going to be interested in for RTT purposes.

Take each combination of <span><math>r</math></span> primes where <span><math>r</math></span> is the rank, sorted in [https://en.wikipedia.org/wiki/Lexicographic_order lexicographic order], e.g. if we're in the 7-limit, we'd have <span><math>(2,3,5)</math></span>, <span><math>(2,3,7)</math></span>, <span><math>(2,5,7)</math></span>, and <span><math>(3,5,7)</math></span>. Now convert each of those combinations to a square <span><math>r×r</math></span> matrix by slicing a column for each prime out of the mapping and putting them together. Now take each matrix's determinant. Extract the GCD from the resulting sequence of scalars, then set them inside <span><math>r</math></span> brackets, and you've got your multicovector.

And so our results are <span><math>-2</math></span>, <span><math>3</math></span>, <span><math>1</math></span>, <span><math>-11</math></span>. At first glance we don’t have a GCD, but there is a slight trick here. We prefer for the first term to be positive. It doesn’t make a difference, but is a way to normalize things (we could have found the result where the first term came out positive by simply changing the order of the rows of our mapping, which doesn’t affect how the mapping works at all). So, we can simply think of this as extracting a GCD of -1, and our list ends up as <span><math>2</math></span>, <span><math>-3</math></span>, <span><math>-1</math></span>, <span><math>11</math></span>. Finally, set these inside triply-nested brackets, because it’s a tricovector for a rank-3 temperament, and we get {{multival|rank=3|2 -3 -1 11}}.

As for getting from the multicovector back to the mapping, you can solve a system of equations for that. Though it’s not easy and there may not be a unique solution. And you probably will never have the multicovector without the mapping anyway.

=== multivectors ===

You may have noticed that the multicovector for meantone, {{multival|1 4 4}}, looks really similar to the meantone comma, {{monzo|-4 4 -1}}. This is not a coincidence.

# Just as a vector is the dual of a covector, we also have a '''multivector''' which is the dual of a multicovector.

# We can calculate a multivector from a comma basis matrix much in the same way we can calculate a multicovector from a mapping matrix

# We can convert between multicovectors and multivectors using an operation called “taking the '''complement'''”<ref>Elsewhere on the wiki you may find the complement operation called "taking [[the dual]]", or even the dual of a multicovector being called simply "the dual". An alternate name for "taking the complement" may be "taking the Hodge dual", but in these materials, I am using the dual to refer to the general case, while the specific case of the dual of a multicovector is a multivector and the operation to get from one of these to its dual is called taking the complement (whereas to get to the dual of a mapping, which is a comma basis, the operation is called taking the null-space).</ref>, which basically involves reversing the order of terms and negating some of them.

To demonstrate these points, let’s first calculate the multivector from a comma basis, and then confirm it by calculating the same multivector as the complement of its dual multicovector.

Here’s the comma basis for meantone: {{val|{{monzo|-4 4 -1}}}}. Calculating the multivector is almost the same as calculating the multicovector. The only difference is that as a preliminary step you must transpose the matrix, or in other words, exchange rows and columns. In our bracket notation, that just looks like replacing {{val|{{monzo|-4 4 -1}}}} with {{monzo|{{val|-4 4 -1}}}}. Now we can see that this is just like our ET map example from the previous section: basically an identity operation, breaking the thing up into three 1×1 matrices <span><math>\begin{bmatrix}-4\end{bmatrix} \begin{bmatrix}4\end{bmatrix} \begin{bmatrix}-1\end{bmatrix}</math></span> which are their own determinants and then nesting back inside one layer of brackets because nullity is 1. So we have {{monzo|-4 4 -1}}. Except, be careful! We can’t skip the step where we extract the GCD, which in this case is -1 again, so the multivector (a monovector) is actually {{monzo|4 -4 1}}. By the way, we write this operation with a different symbol; it’s upside down from the exterior product, and so is called the interior product, and uses ∨, which is great because that’s also the logical operator for “or” which matches with the intersection of commas using the “|” operator which also means “or”.

# Multiply each term of the multicovector by these values. 1×1, 4×-1, 4×1, = 1, -4, 4.

What’s the proper count of brackets though? Well, the total count of brackets on the multivector and multicovector for a temperament must always sum to the dimensionality of the system from which you tempered. It’s the same thing as <span><math>d - n = r</math></span>, just phrased as <span><math>r + n = d</math></span>, and where <span><math>r</math></span> should be the bracket count for the multicovector and <span><math>n</math></span> should be the bracket count for the multivector. So with 5-limit meantone, with dimensionality 3, there should be 3 total pairs of brackets. If 2 are on the multicovector, then only 1 are on the multivector.

An important observation to make about multivectors and multicovectors is that — for a given temperament — they are always the same length. This may surprise you, since the rank and nullity for a temperament are often different, and the length of the multicovector comes from the rank while the length of the multivector comes from the nullity. But there’s a simple explanation for this. In either case, the length is not directly equal to the rank or nullity, but to the dimensionality choose the rank or nullity. And there’s a pattern to combinations that can be visualized in the symmetry of rows of Pascal’s triangle: <span><math>{d \choose n}</math></span> is always equal to <span><math>{d \choose {d - n}}</math></span>, or in other words, <span><math>{d \choose n}</math></span> is always equal to <span><math>{d \choose r}</math></span>. Here are some examples:

If you review the seven steps in the process for taking the complement, you may notice that a lot of it is busywork that will never change from one multicovector to another. It all amounts to a specific sequence of 1’s and -1’s corresponding to a given rank and dimensionality. In consideration of this, I have gone ahead and prepared a table with the sequences you need to multiply the terms of your multicovector by before flipping them:

If you’re instead converting a multivector to a multicovector, then you think of it a couple different ways. Either use <span><math>n</math></span> as <span><math>r</math></span> when looking up in this table, and then reverse the result, or find <span><math>r</math></span> by subtracting <span><math>n</math></span> from <span><math>d</math></span> and then look it up.

Note that multivectors could just as well serve the purpose of a unique identifier for temperaments. There’s no particular reason to prefer the multicovector to the multivector. By convention, however, the multicovector is the one that’s used.

|multimapping

|mapping

|?

So we now understand how to get to multicovectors. And we understand that they uniquely identify the temperament. But what about the individual terms — do they mean anything in and of themselves? It turns out: yes!

The first thing to understand is that each term of the multicovector pertains to a different combination of primes. We already know this: it’s how we calculated it from the mapping matrix. For example, in the multicovector for meantone, {{multival|1 4 4}}, the 1 is for <span><math>(2,3)</math></span>, the first 4 is for <span><math>(2,5)</math></span>, and the second 4 is for <span><math>(3,5)</math></span>.

Now, let’s convert every term of the multicovector by taking its absolute value and its inverse. In this case, each of our terms is already positive, so that has no effect. But taking the inverse converts us to <span><math>\frac 11</math></span>, <span><math>\frac 14</math></span>, <span><math>\frac 14</math></span>. These values tell us what fraction of the tempered lattice we can generate using the corresponding combination of primes.

One day you might come across a multicovector which has a term equal to zero. If you tried to interpret this term using the information here so far, you'd think it must generate <span><math>\frac 10</math></span>th of the tempered lattice. That's not easy to visualize or reason about. Does that mean it generates essentially infinity lattices? No, not really. More like the opposite. The question itself is somewhat undefined here. If anything, it's more like that combination of primes generates approximately ''none'' of the lattice. Because in this situation, the combination of primes whose multicovector term is zero generates so little of the tempered lattice that it's completely missing an entire dimension of it, so it's an infinitesimal amount of it that it generates. For example, the 11-limit temperament 7&12&31 has multicovector {{val|rank=3|0 1 1 4 4 -8 4 4 -12 -16}} and mapping {{monzo|{{val|1 0 -4 0 -12}} {{val|0 1 4 0 8}} {{val|0 0 0 1 1}}}}; we can see from this how primes <span><math>(2,3,5)</math></span> can only generate a rank-2 cross-section of the full rank-3 lattice, because while 2 and 3 do the trick of generating that rank-2 part (exactly as they do in 5-limit meantone), prime 5 doesn't bring anything to the table here so that's all we get.