convex -> epimorphic?

In the archives the phrase "convex, epimorphic scale" is used
quite a bit. But it seems like convexity might almost imply
epimorphism. Is there a convex scale that isn't epimorphic?

Actually, didn't Keenan point out that all convex scales are
epimorphic with the <1 1 1...| val in the basis of their steps?

-Carl

On Sat, Mar 3, 2012 at 9:00 PM, Carl Lumma <carl@lumma.org> wrote:
>
> In the archives the phrase "convex, epimorphic scale" is used
> quite a bit. But it seems like convexity might almost imply
> epimorphism. Is there a convex scale that isn't epimorphic?

I think "convex but not epimorphic" would refer to scales that don't
tile the lattice completely. For example, consider a triangle or
something.

-Mike

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>
> In the archives the phrase "convex, epimorphic scale" is used
> quite a bit. But it seems like convexity might almost imply
> epimorphism. Is there a convex scale that isn't epimorphic?
>
> Actually, didn't Keenan point out that all convex scales are
> epimorphic with the <1 1 1...| val in the basis of their steps?

That's only true if the number of different steps is the same as the rank of the temperament. Otherwise the steps do not form a basis (they're linearly dependent).

As you know, this is not a general situation for rank > 2 (scales in a rank-3 temperament generally have 4 different steps, and only special cases have 3 different steps).

Keenan

>> In the archives the phrase "convex, epimorphic scale" is used
>> quite a bit. But it seems like convexity might almost imply
>> epimorphism. Is there a convex scale that isn't epimorphic?
>>
>> Actually, didn't Keenan point out that all convex scales are
>> epimorphic with the <1 1 1...| val in the basis of their steps?
>
>That's only true if the number of different steps is the same as the
>rank of the temperament. Otherwise the steps do not form a basis
>(they're linearly dependent).
>
>As you know, this is not a general situation for rank > 2 (scales in a
>rank-3 temperament generally have 4 different steps, and only special
>cases have 3 different steps).
>
>Keenan

Thanks for your response, but I confess I'm completely lost.
Can you... maybe expand on this a bit?

When you say the rank of the temperament -- what does temperament
have to do with it?

When you say steps, do you mean the variety of 2nds or the max
variety of the scale?

Can you exhibit a scale which is convex (in some basis) but not
epimorphic (in some basis)?

Do I remember right that if I'm convex in basis b, I'm still
convex after any change of basis?

Thanks,

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>
> In the archives the phrase "convex, epimorphic scale" is used
> quite a bit. But it seems like convexity might almost imply
> epimorphism. Is there a convex scale that isn't epimorphic?
>
> Actually, didn't Keenan point out that all convex scales are
> epimorphic with the <1 1 1...| val in the basis of their steps?

Since there's no reason to suppose the steps are independent generators, this doesn't seem to make sense. But I like the idea of showing a convex scale must be weakly epimorphic if it's actually true.

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
> Thanks for your response, but I confess I'm completely lost.
> Can you... maybe expand on this a bit?
>
> When you say the rank of the temperament -- what does temperament
> have to do with it?

It doesn't have to do with temperament per se; it has to do with the number of independent (incommensurate) generators of the tuning. Also known as its rank, or dimension.

Here's something I do claim:

Any scale with N different step sizes *that are all incommensurate* is epimorphic under the val <1 1 1...| (N 1's) in the basis of the steps.

However, if the step sizes are not incommensurate, but have nontrivial relationships between them, then they don't form a basis. For example, consider the scale:

9/8
5/4
4/3
3/2
5/3
9/5
2/1

This is a scale of 5-limit JI, which is a rank-3 tuning. But it contains 4 different kinds of steps: 9/8, 10/9, 16/15, and 27/25. Therefore those steps don't form a basis, because there's one too many of them. The set {9/8, 10/9, 16/15} is already a basis, and 27/25 can be expressed in terms of that basis, so it's redundant.

> When you say steps, do you mean the variety of 2nds or the max
> variety of the scale?

I'm talking about steps, a.k.a. 2nds.

> Can you exhibit a scale which is convex (in some basis) but not
> epimorphic (in some basis)?
>
> Do I remember right that if I'm convex in basis b, I'm still
> convex after any change of basis?

Yes, convexity is an intrinsic property of the scale and doesn't depend on a basis.

Here's a convex scale that I'm convinced is not even weakly epimorphic under any val:

25/24
9/8
6/5
5/4
4/3
3/2
25/16
8/5
5/3
9/5
15/8
2/1

It seems hard to prove that, but if it's epimorphic it ought to tile the octave-equivalent lattice, and I convinced myself it's impossible to tile it with no gaps.

Perhaps it's easier to see if you just consider a scale consisting of every note inside a huge *circle* on the 5-limit lattice. This scale would clearly be convex, because a circle is convex, but it wouldn't be epimorphic because you can't tile the plane with circles without leaving gaps.

Keenan

Keenan wrote:

>It doesn't have to do with temperament per se; it has to do with the
>number of independent (incommensurate) generators of the tuning. Also
>known as its rank, or dimension.

Right, that's why I said "in the ***basis*** of their steps"
(emphasis added). Steps aren't important of course, just often
convenient.

>Yes, convexity is an intrinsic property of the scale and doesn't
>depend on a basis.

Thanks.

>Here's a convex scale that I'm convinced is not even weakly epimorphic
>under any val:
> 25/24
> 9/8
> 6/5
> 5/4
> 4/3
> 3/2
> 25/16
> 8/5
> 5/3
> 9/5
> 15/8
> 2/1
>It seems hard to prove that, but if it's epimorphic it ought to tile
>the octave-equivalent lattice, and I convinced myself it's impossible
>to tile it with no gaps.
>Perhaps it's easier to see if you just consider a scale consisting of
>every note inside a huge *circle* on the 5-limit lattice. This scale
>would clearly be convex, because a circle is convex, but it wouldn't
>be epimorphic because you can't tile the plane with circles without
>leaving gaps.

Didn't you once post something about any convex scale being a
fundamental domain, because of hyperplanes separating them or
something? I must have misunderstood...

-Carl

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
> Didn't you once post something about any convex scale being a
> fundamental domain, because of hyperplanes separating them or
> something? I must have misunderstood...

The only post I can find related to this is /tuning-math/message/19632 , but that assumes the scales are epimorphic from the start. It doesn't say anything about "any convex scale", only "any convex epimorphic scale".

I'm sure that convex non-epimorphic scales exist. Just consider a huge circle of the octave-equivalent 5-limit lattice, containing, say, a million octave-equivalent pitch classes. It's convex but can't possibly tile the plane.

Keenan

At 05:34 PM 3/4/2012, you wrote:
>--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>> Didn't you once post something about any convex scale being a
>> fundamental domain, because of hyperplanes separating them or
>> something? I must have misunderstood...
>
>The only post I can find related to this is
>/tuning-math/message/19632 , but
>that assumes the scales are epimorphic from the start.

Ah, that makes more sense. Thanks. -Carl

Keenan Pepper wrote:

>Here's a convex scale that I'm convinced is not even weakly epimorphic
>under any val:
> 25/24
> 9/8
> 6/5
> 5/4
> 4/3
> 3/2
> 25/16
> 8/5
> 5/3
> 9/5
> 15/8
> 2/1
>It seems hard to prove that, but if it's epimorphic it ought to tile
>the octave-equivalent lattice, and I convinced myself it's impossible
>to tile it with no gaps.

It's funny, but earlier I mentioned on facebook a scale I
gave in 1999, that doesn't have constant structure

/tuning/topicId_5209.html#5209?var=0&l=1

It's clearly convex. Since epimorphic -> CS, it looks like
this disproves convex -> epimorphic straight away.

-Carl