"Sane" and "insane" temperaments

(Related thread: /tuning/topicId_101888.html#101888 )

I was trying to revise the conditions for temperaments to be allowed to have the same name, and I wanted to prove some mathematical statements, but they turned out to be false. However, they're only false for some pathological temperaments that no one would ever use. Here's an example of such a pathological temperament:

http://x31eq.com/cgi-bin/rt.cgi?ets=22_27&limit=2_9_5_7

This is a real, non-contorted temperament, because 5 is mapped to an odd number of generators. The problem with this temperament is that it's mapping the 2.9.5.7 subgroup, but 9 has a perfectly good square root that is not mapped to 3. So this is an "insane" version of superpyth where you forget about 3 even though the interval representing it is still there (in this case it's a generator). The following rigorous definition is intended to reflect this intuition.

Definition: A temperament is "sane" if, for every rational number in the subgroup it's mapping, represented by a monzo v in the *standard* basis (the primes), the rational number represented by v / gcd(v[0],v[1],...,m[1],m[2],...) is also in the subgroup. Here v[0],v[1],... are the monzo entries of v and m[1],m[2],... are the entries of its mapping in the temperament. A temperament is "insane" iff it is not sane.

To repeat, a temperament is "sane" if I can always divide any monzo in the subgroup by the gcd of {all its entries together with the entries of its mapping} to get another JI interval that is *also* in the subgroup. The example above is "insane" because 9 has the monzo |2, 0, 0, 0> and the mapping [4,-2], so the gcd of everything is 2, but the monzo |1, 0, 0, 0> is not in the subgroup. This formalizes the notion that "9 has a square root in the temperament, but that square root isn't mapped to 3".

Why did I bother to make this definition? Because it allows me to say this:

If I have two (uncontorted) temperaments A and B, where
* A and B have the same rank
* B is the restriction of A to a smaller subgroup, having exactly the same mapping matrix as A on that subgroup, and
* A and B are both "sane"
then the quotient group of (A's subgroup) / (B's subgroup) is free.

Proof: Assume the quotient group has a torsion element, x. Let v be any element of A's subgroup that is in the coset corresponding to x. Because x is not the identity, v is not in B's subgroup, but because x has finite order, some power of v is in B's subgroup, let's say v^n. The GCD of the monzo entries of v^n is a multiple of n, and the GCD of the mapping entries (in both A and B) is also a multiple of n. Therefore, because B is a sane temperament, v must belong to B's subgroup. But this contradicts the assumption that the quotient group has torsion, so the quotient group must be torsion-free.

As everybody knows, every finitely-generated torsion-free abelian group is a free abelian group, therefore the quotient qroup (A's subgroup) / (B's subgroup) is free.

Thus I can limit myself to thinking about sane temperaments (as every sane person should do), and avoid certain kinds of silly things happening.

No 2.9.5.7 temperament can share the name "meantone", because it can't agree with the mapping matrix without being contorted. This has nothing to do with "sane" or "insane" temperaments.

No 2.9.5.7 temperament can share the name "superpyth", because although there does exist an uncontorted temperament with the same mapping matrix (exhibited above), that temperament is insane. (If anybody wants to start naming insane temperaments, they'll just have to choose unique names for them.)

Keenan

On Thu, Nov 3, 2011 at 7:50 PM, Keenan Pepper <keenanpepper@gmail.com> wrote:
>
> Why did I bother to make this definition? Because it allows me to say this:
>
> If I have two (uncontorted) temperaments A and B, where
> * A and B have the same rank
> * B is the restriction of A to a smaller subgroup, having exactly the same mapping matrix as A on that subgroup, and
> * A and B are both "sane"
> then the quotient group of (A's subgroup) / (B's subgroup) is free.

What do you mean by "having exactly the same mapping matrix?" This
would seem at first glance to fit your definition above:

A = Meantone
2 3 5
[<1 0 0],
<0 1 4]>

B = WTF
2 9 5
[<1 0 0],
<0 1 4]>

I assume this isn't what you meant. Can you give an example of two
temperaments that would fit your description of A and B above?

-Mike

On Fri, Nov 4, 2011 at 2:03 AM, Mike Battaglia <battaglia01@gmail.com> wrote:
>
> A = Meantone
>   2 3 5
> [<1 0 0],
>  <0 1 4]>
>
> B = WTF
>   2 9 5
> [<1 0 0],
>  <0 1 4]>

Typo - that should be <1 1 0] for the top row above.

Also this -

B = Less WTF, but still WTF
2 5 9
[<1 0 0],
<0 1 4]>

B = FTW
25 81 2
[<1 0 0],
<0 1 4]>

-Mike

Keenan wrote:

>Why did I bother to make this definition? Because it allows me to say this:
>
>If I have two (uncontorted) temperaments A and B, where
>* A and B have the same rank
>* B is the restriction of A to a smaller subgroup, having exactly the
>same mapping matrix as A on that subgroup, and
>* A and B are both "sane"
>then the quotient group of (A's subgroup) / (B's subgroup) is free.

That all looks good, but what's the significance of this
quotient group being a free group?

-Carl

Mike wrote:
>What do you mean by "having exactly the same mapping matrix?" This
>would seem at first glance to fit your definition above:
>
>A = Meantone
> 2 3 5
>[<1 0 0],
> <0 1 4]>
>
>B = WTF
> 2 9 5
>[<1 0 0],
> <0 1 4]>
>
>I assume this isn't what you meant.

I took

>> * B is the restriction of A to a smaller subgroup, having
>> exactly the same mapping matrix as A on that subgroup,

to mean that B's subgroup is a subset of A's. Maybe Keenan
can clarify.

-Carl

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
> What do you mean by "having exactly the same mapping matrix?" This
> would seem at first glance to fit your definition above:

Sorry, sorry, I should have written "having exactly the same mapping *function* restricted to the smaller subgroup". You just take each generator of your smaller subgroup, map it with the same mapping function, and there you go.

The mapping matrix will not be an identical matrix of integers, sorry about that.

> A = Meantone
> 2 3 5
> [<1 0 0],
> <0 1 4]>
>
> B = WTF
> 2 9 5
> [<1 0 0],
> <0 1 4]>
>
> I assume this isn't what you meant. Can you give an example of two
> temperaments that would fit your description of A and B above?

In meantone what you'd get is

Meantone:
2 3 5
[<1 1 0],
<0 1 4]>

Attempted 2.9.5 meantone:
2 9 5
[<1 2 0],
<0 2 4]>

but this isn't an actual temperament because it's contorted. For 5-limit superpyth you get

Superpyth:
2 3 5
[<1 1 -3],
<0 1 9]>

Insane 2.9.5 superpyth:
2 9 5
[<1 2 -3],
<0 2 9]>

This is not contorted, but it is insane.

Keenan

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
> I took
>
> >> * B is the restriction of A to a smaller subgroup, having
> >> exactly the same mapping matrix as A on that subgroup,
>
> to mean that B's subgroup is a subset of A's. Maybe Keenan
> can clarify.

I guess you're asking if the generators of B's subgroup must form a subset of the generators of A's subgroup. That is not what I meant.

For example, 2.9.5 is a subgroup of 2.3.5. Why? Because 2.3.5 represents

{2^x 3^y 5^z | x,y,z in Z}

and 2.9.5 represents

{2^x 3^(2y) 5^z | x,y,z in Z}

So the set of rational numbers represented by 2.9.5 is a subset of that represented by 2.3.5 (consisting exactly of the elements where y is even), and in fact forms a subgroup under multiplication.

Keenan

--- In tuning-math@yahoogroups.com, Carl Lumma <carl@...> wrote:
>
> Keenan wrote:
>
> >Why did I bother to make this definition? Because it allows me to say this:
> >
> >If I have two (uncontorted) temperaments A and B, where
> >* A and B have the same rank
> >* B is the restriction of A to a smaller subgroup, having exactly the
> >same mapping matrix as A on that subgroup, and
> >* A and B are both "sane"
> >then the quotient group of (A's subgroup) / (B's subgroup) is free.
>
> That all looks good, but what's the significance of this
> quotient group being a free group?

Basically Gene mentioned it would be desirable, lol. Then I thought about it a little and decided it should be true pretty much all the time, but when I thought about it a little more I realized it wasn't true for insane temperaments.

In really vague intuitive terms, the quotient group being free means that the larger subgroup is simply adding some new, independent stuff to the smaller subgroup, and not screwing around with anything that's already there.

For example, going from 2.5/3.7/3 to 2.3.5.7 has a free quotient group, which makes sense because I'm just adding in 3 and not disturbing 5/3 or 7/3 at all. 5/3 and 7/3 are still technically generators of the subgroup, just not the particular generators I'm listing.

On the other hand, going from 2.9.5 from 2.3.5 has a non-free quotient group (it's cyclic of order 2), which means that adding 3 didn't leave 9 alone, instead it split 9 in half. So now 9 isn't a generator anymore.

Keenan

Keenan wrote:
>So the set of rational numbers represented by 2.9.5 is a subset of
>that represented by 2.3.5 (consisting exactly of the elements where y
>is even), and in fact forms a subgroup under multiplication.

Ok, understood.

>On the other hand, going from 2.9.5 from 2.3.5 has a non-free
>quotient group (it's cyclic of order 2), which means that adding
>3 didn't leave 9 alone, instead it split 9 in half. So now 9 isn't
>a generator anymore.

It sounds like you've just shifted the work to your second
condition about the mapping matrix. That is, you've shown Gene's
suggestion is almost always equivalent to your "distinction" we
were discussing on tuning.

-Carl