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Inverse limits in the category of temperaments

🔗Mike Battaglia <battaglia01@gmail.com>

12/7/2012 11:57:23 PM

First, we formulate a temperament as an equivalence class of homomorphisms
in the usual way.

Let Q+ be the positive multiplicative group of rationals, and let G be the
set of all finitely generated subgroups of Q+. Note that G partially
ordered by subgroup inclusion, and that furthermore it is a directed
poset. Then, for each group g in G, let T_g be the set of rank-r
temperaments of that group, where r is constant, and that T is the union of
all T_g's. To keep the math simple for now, we exclude from T those T_g
where rank(g) < r.

We note that, for any groups g and h in G, where g is a subgroup of h, we
have a corresponding homomorphism: T_h -> T_g, which sends each temperament
in T_h to its restriction to subgroup g, which is a temperament in T_g.

It is extremely interesting, then, that T is an inverse system (in the
category of temperaments, I guess).

So then here's my question: what, exactly, are the elements in the inverse
limit of T?

I'd like to think that they're simply infinite-limit temperaments of Q+
itself, which temper out an infinite number of commas to bring you down to
a rank-r temperament. However, after seeing things like the ring of adeles,
I'm very suspicious that this may not be the case.

-Mike

🔗Mike Battaglia <battaglia01@gmail.com>

12/7/2012 11:59:21 PM

On Sat, Dec 8, 2012 at 2:57 AM, Mike Battaglia <battaglia01@gmail.com> wrote:
>
> It is extremely interesting, then, that T is an inverse system (in the
> category of temperaments, I guess).
>
> So then here's my question: what, exactly, are the elements in the inverse
> limit of T?
>
> I'd like to think that they're simply infinite-limit temperaments of Q+
> itself, which temper out an infinite number of commas to bring you down to a
> rank-r temperament. However, after seeing things like the ring of adeles,
> I'm very suspicious that this may not be the case.

Also related to this, here's a theorem for you to chew on: since
Cangwu badness is a norm if the free parameter > 0, then for any
temperament and choice of free parameter, there exists exactly one
extension of that temperament on any subgroup you like which has the
exact same Cangwu badness, and for which badness is minimal among all
extensions of that temperament.

(This may not be the case the way that Graham's defined it now because
there are probably some factors of the norm of the JIP floating
around, but it's trivial to change it to be this way, and then the
theorem above follows easily from a) the understanding that svals are
isomorphic to cosets of vals and b) the definition of the quotient
norm.)

-Mike

🔗Mike Battaglia <battaglia01@gmail.com>

12/10/2012 9:59:08 PM

bump for gene
-Mike

On Sat, Dec 8, 2012 at 2:57 AM, Mike Battaglia <battaglia01@gmail.com> wrote:
> First, we formulate a temperament as an equivalence class of homomorphisms
> in the usual way.
>
> Let Q+ be the positive multiplicative group of rationals, and let G be the
> set of all finitely generated subgroups of Q+. Note that G partially ordered
> by subgroup inclusion, and that furthermore it is a directed poset. Then,
> for each group g in G, let T_g be the set of rank-r temperaments of that
> group, where r is constant, and that T is the union of all T_g's. To keep
> the math simple for now, we exclude from T those T_g where rank(g) < r.
>
> We note that, for any groups g and h in G, where g is a subgroup of h, we
> have a corresponding homomorphism: T_h -> T_g, which sends each temperament
> in T_h to its restriction to subgroup g, which is a temperament in T_g.
>
> It is extremely interesting, then, that T is an inverse system (in the
> category of temperaments, I guess).
>
> So then here's my question: what, exactly, are the elements in the inverse
> limit of T?
>
> I'd like to think that they're simply infinite-limit temperaments of Q+
> itself, which temper out an infinite number of commas to bring you down to a
> rank-r temperament. However, after seeing things like the ring of adeles,
> I'm very suspicious that this may not be the case.
>
> -Mike