Subgroups and temperaments, part 1: tmonzos, tvals

Some of this I posted before, albeit in a form lacking somewhat in clarity. I'll try to tighten things up a bit here. View this email in fixed width.

First, some simple terminology: I'm calling a "tmonzo" a element in a tempered group of intervals, by analogy with Gene's "smonzo" concept. For example, in terms of meantone intervals, things like |1 0> and |0 1> are tmonzos. A tval is then an element in the dual group to the group of tmonzos.

These are interesting for a number of reasons, and I'll go into detail about them below.

All of this fits nicely into Gene's homomorphism paradigm, so I'll start where he left off:
- A temperament is a group epimorphism H: A -> B between free abelian groups.
- where A is the group representing JI intervals, or monzos,
- and where B is the group representing tempered intervals, or tmonzos.
- The group A*, the dual group to A, is the group of linear functionals on those JI intervals, or vals.
- The group B*, which is dual to B, is then the group of linear functionals on the -tempered- intervals, which makes them the "tvals" I talked about above.

The first very important thing to note is that, for free abelian groups A, B and homomorphism H: A -> B, a dual homomorphism H*: B* -> A* also exists, which has some interesting implications.

Let's make this more concrete by giving an example. For instance, say you're mapping from the 5-limit JI group A to the porcupine group B by tempering out 250/243. Then B* is the set of "tempered vals", or tvals, acting on the group of porcupine intervals. For instance, if |1 0> is the porcupine octave and |0 1> is the porcupine 10/9, then the tval <15 2| represents the usual mapping for 15-EDO, given in terms of porcupine intervals rather than in terms of JI ratio mappings - something Keenan first noted to me offlist about a year ago.

The dual homomorphism immediately gives us a way to turn <15 2| back into <15 24 35|. Furthermore, this dual homomorphism has the property that if the original homomorphism has the property that rank(B) <= rank(A), the dual homomorphism will be injective.

Let's run away from abstract algebra now and get back to linear algebra. The dual transformation I'm talking about is also given by the mapping matrix, but this time being multiplied on the other side. Left-multiply the mapping matrix for porcupine by <15 2|, and I will use the convention throughout that rows are vals and cols are monzos:

[1 2 3]
[15 2] * [0 -3 -5]

and you get <15 24 35| as a result.

Note that we're multiplying the matrix on the left this time, so the answer is going to be a linear combination of rows in the mapping. This shows that the mapping from tvals onto regular vals is one-to-one but not onto, a property which was first noticed by Keenan. The image of this map is only going to be the set of vals supporting the temperament in question.

One huge implication of the dual homomorphism not being surjective in general is that the question of "which porcupine tval does <12 19 28| map to" is undecidable: there's no tval mapping to it, so it doesn't "map back to" anything. You can (usually) come up with a map from A* -> B* if you like that satisfies the matter if you like, but you'll immediately run into the following theorem:

Theorem: for free abelian groups A, B and group homomorphism H*: B* -> A*, there does not in general exist a -unique- map H*^-1: A* -> B* which respects the property that, for all b in B*, H*^-1(H*(b)) = b.

For homomorphisms which aren't surjective (e.g. are contorted), you don't get a map at all. For homomorphisms which are surjective but with nontrivial kernel, you get an infinite number of maps. You only get a unique map if the homomorphism is an automorphism.

IOW, if you want to take your favorite val and find the "tempered equivalent" of it in your favorite temperament, you're SOL in general unless that val supports the temperament in question. You're either going to have an infinite amount of potential tvals you can map it to, or none. Allowing for rational coefficients in your monzos doesn't really fix the problem either. Sorry.

Much of the above stuff may seem obvious to you, which is fine: this is a precursor to the next part, which will apply the same principles in reverse. Next time we'll look at applying group homomorphisms to the group of -vals-, and see what that looks like. The same intuitive properties that we can see here will also apply over there, which will help us make sense out of what it means to "temper out a val."

-Mike

--- In tuning-math@yahoogroups.com, "battaglia01" <battaglia01@...> wrote:
>
> Some of this I posted before, albeit in a form lacking somewhat in clarity. I'll try to tighten things up a bit here. View this email in fixed width.
>
> First, some simple terminology: I'm calling a "tmonzo" a element in a tempered group of intervals, by analogy with Gene's "smonzo" concept. For example, in terms of meantone intervals, things like |1 0> and |0 1> are tmonzos. A tval is then an element in the dual group to the group of tmonzos.
>
> These are interesting for a number of reasons, and I'll go into detail about them below.
>
> All of this fits nicely into Gene's homomorphism paradigm, so I'll start where he left off:
> - A temperament is a group epimorphism H: A -> B between free abelian groups.
> - where A is the group representing JI intervals, or monzos,
> - and where B is the group representing tempered intervals, or tmonzos.
> - The group A*, the dual group to A, is the group of linear functionals on those JI intervals, or vals.
> - The group B*, which is dual to B, is then the group of linear functionals on the -tempered- intervals, which makes them the "tvals" I talked about above.
>
> The first very important thing to note is that, for free abelian groups A, B and homomorphism H: A -> B, a dual homomorphism H*: B* -> A* also exists, which has some interesting implications.
>
> Let's make this more concrete by giving an example. For instance, say you're mapping from the 5-limit JI group A to the porcupine group B by tempering out 250/243. Then B* is the set of "tempered vals", or tvals, acting on the group of porcupine intervals. For instance, if |1 0> is the porcupine octave and |0 1> is the porcupine 10/9, then the tval <15 2| represents the usual mapping for 15-EDO, given in terms of porcupine intervals rather than in terms of JI ratio mappings - something Keenan first noted to me offlist about a year ago.
>
> The dual homomorphism immediately gives us a way to turn <15 2| back into <15 24 35|. Furthermore, this dual homomorphism has the property that if the original homomorphism has the property that rank(B) <= rank(A), the dual homomorphism will be injective.
>
> Let's run away from abstract algebra now and get back to linear algebra. The dual transformation I'm talking about is also given by the mapping matrix, but this time being multiplied on the other side. Left-multiply the mapping matrix for porcupine by <15 2|, and I will use the convention throughout that rows are vals and cols are monzos:
>
> [1 2 3]
> [15 2] * [0 -3 -5]
>
> and you get <15 24 35| as a result.
>
> Note that we're multiplying the matrix on the left this time, so the answer is going to be a linear combination of rows in the mapping. This shows that the mapping from tvals onto regular vals is one-to-one but not onto, a property which was first noticed by Keenan. The image of this map is only going to be the set of vals supporting the temperament in question.
>
> One huge implication of the dual homomorphism not being surjective in general is that the question of "which porcupine tval does <12 19 28| map to" is undecidable: there's no tval mapping to it, so it doesn't "map back to" anything. You can (usually) come up with a map from A* -> B* if you like that satisfies the matter if you like, but you'll immediately run into the following theorem:
>
> Theorem: for free abelian groups A, B and group homomorphism H*: B* -> A*, there does not in general exist a -unique- map H*^-1: A* -> B* which respects the property that, for all b in B*, H*^-1(H*(b)) = b.
>
> For homomorphisms which aren't surjective (e.g. are contorted), you don't get a map at all. For homomorphisms which are surjective but with nontrivial kernel, you get an infinite number of maps. You only get a unique map if the homomorphism is an automorphism.
>
> IOW, if you want to take your favorite val and find the "tempered equivalent" of it in your favorite temperament, you're SOL in general unless that val supports the temperament in question. You're either going to have an infinite amount of potential tvals you can map it to, or none. Allowing for rational coefficients in your monzos doesn't really fix the problem either. Sorry.
>
> Much of the above stuff may seem obvious to you, which is fine: this is a precursor to the next part, which will apply the same principles in reverse. Next time we'll look at applying group homomorphisms to the group of -vals-, and see what that looks like. The same intuitive properties that we can see here will also apply over there, which will help us make sense out of what it means to "temper out a val."
>
> -Mike
>

Is the mapping matrix for porcupine just the regular temperament mapping? I think I understand most of this, in terms of the theorem you state, what can you say about H^-1:B->A. Illuminating that piece would help me understand why the dual situation is one-to-one but not onto. (Maybe I'm just being lazy?) I like the tmonzo and tval idea.

PGH

--- In tuning-math@yahoogroups.com, "Paul" <phjelmstad@...> wrote:

>I like the tmonzo and tval idea.

I called them smonzos and svals. I don't know why Mike wants to make it "t"--the "s" stands for "subgroup", but what does the "t" mean?

On Fri, Jul 6, 2012 at 12:02 AM, Paul <phjelmstad@msn.com> wrote:
>
> Is the mapping matrix for porcupine just the regular temperament mapping?

Right.

> I think I understand most of this, in terms of the theorem you state, what
> can you say about H^-1:B->A. Illuminating that piece would help me
> understand why the dual situation is one-to-one but not onto. (Maybe I'm
> just being lazy?) I like the tmonzo and tval idea.

Think about the mapping matrix for 5-limit porcupine. If you
right-multiply this mapping matrix by a 1x3 matrix which is a single
monzo, your result will lie in the column space of the matrix. This
implies a linear transformation from Z^3 -> Z^2, which won't be one to
one (intuitively because intervals are getting tempered together). In
the case of porcupine, which isn't contorted, it will be onto.

On the other hand, if you left-multiply the mapping matrix by a 1x2
matrix which is a single tval, your result will now lie in the row
space of the mapping (read: set of all vals supporting the
temperament). You'll get a mapping from Z^2 -> Z^3. This couldn't
possibly be surjective, since you're gaining a dimension. It will be
injective, however.

-Mike

On Sat, Jul 7, 2012 at 1:26 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:
>
> --- In tuning-math@yahoogroups.com, "Paul" <phjelmstad@...> wrote:
>
> >I like the tmonzo and tval idea.
>
> I called them smonzos and svals. I don't know why Mike wants to make it
> "t"--the "s" stands for "subgroup", but what does the "t" mean?

Tmonzos and tvals aren't smonzos and svals. The "t" stands for tempered.

-Mike