Grassmannians of submodules of a module

Paper here: http://arxiv.org/pdf/1010.4761v2.pdf

Can anyone tell me if this actually defines some sort of object which
is analogous to a Grassmannian, but which allows us to distinguish
between all the submodules of a module and not just the subspaces of a
vector space?

If so, how the hell does it parameterize it? 5-limit JI is a Z-module,
how do I represent the 2.9.5 submodule using this thing?

Anyone have any insight? And by anyone I'm quite certain I mean Gene?

-Mike

Also, the reason I care about this is I wish I had something like the
wedge product for "subgroup mappings," or whatever the hell I'm
supposed to call linear transformations on val space.

For instance, 2.9.5/3.11 is the same as 2.9.15.11. I wish I had some
operator like ^ such that I could just do 2/1 ^ 9/1 ^ 5/3 ^ 11/1 and
get the same thing as if I did 2/1 ^ 9/1 ^ 15/1 ^ 11/1. The wedge
product does the trick, but still fails catastrophically because it
also makes 2/1 ^ 9/1 ^ 5/3 ^ 11/1 the same as 4/1 ^ 3/1 ^ 5/3 ^ 11/1.
Both are just |||0 2 0 0 0>>>.

This is because a^2b^c^d = 2a^b^c^d = 2(a^b^c^d), which is why this
all gets screwed up. This is Bad. I tried telling this to Gene on
Facebook, but I was afraid if I typed too much he'd run away, so we
ended off with him getting the impression that I just hate bilinear
maps for no reason at all. Bilinear is actually my middle name, but in
this case it means you equate subgroups that Must Not Be Equated.

The whole thing is the same situation as that, in the 5-limit, 7p^24p
= 14p^12p: both are just <<2 8 8||. You can claim that this isn't a
problem because contorted temperaments are evil, but subgroups with
prime powers are definitely not evil.

Obviously, representing the subgroup as a matrix and putting it into
Hermite form fits the bill, but it would be nice to have something as
powerful as exterior algebra at ones disposal. This is why this
Grassmannian of submodules thing is interesting to me. Is it really a
coarse moduli space for the submodules of any module, including a
Z-module?

-Mike

On Wed, Jun 20, 2012 at 2:38 AM, Mike Battaglia <battaglia01@gmail.com> wrote:
> Paper here: http://arxiv.org/pdf/1010.4761v2.pdf
>
> Can anyone tell me if this actually defines some sort of object which
> is analogous to a Grassmannian, but which allows us to distinguish
> between all the submodules of a module and not just the subspaces of a
> vector space?
>
> If so, how the hell does it parameterize it? 5-limit JI is a Z-module,
> how do I represent the 2.9.5 submodule using this thing?
>
> Anyone have any insight? And by anyone I'm quite certain I mean Gene?
>
> -Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> Paper here: http://arxiv.org/pdf/1010.4761v2.pdf
>
> Can anyone tell me if this actually defines some sort of object which
> is analogous to a Grassmannian, but which allows us to distinguish
> between all the submodules of a module and not just the subspaces of a
> vector space?

Where do you think it might be doing that? And are you talking modules in general, or just injective modules, eg a vector space over Q considered as a Z-module?

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> Also, the reason I care about this is I wish I had something like the
> wedge product for "subgroup mappings," or whatever the hell I'm
> supposed to call linear transformations on val space.
>
> For instance, 2.9.5/3.11 is the same as 2.9.15.11. I wish I had some
> operator like ^ such that I could just do 2/1 ^ 9/1 ^ 5/3 ^ 11/1 and
> get the same thing as if I did 2/1 ^ 9/1 ^ 15/1 ^ 11/1.

What's wrong with the normal list?

On Wed, Jun 20, 2012 at 12:10 PM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...>
> wrote:
> >
> > Also, the reason I care about this is I wish I had something like the
> > wedge product for "subgroup mappings," or whatever the hell I'm
> > supposed to call linear transformations on val space.
> >
> > For instance, 2.9.5/3.11 is the same as 2.9.15.11. I wish I had some
> > operator like ^ such that I could just do 2/1 ^ 9/1 ^ 5/3 ^ 11/1 and
> > get the same thing as if I did 2/1 ^ 9/1 ^ 15/1 ^ 11/1.
>
> What's wrong with the normal list?

There's nothing wrong with the normal list, just like there's nothing
wrong with representing temperaments in RREF form. But it'd be simpler
and more powerful to have something like the wedge product to use, and
might open up new creative ways of doing things.

-Mike

On Wed, Jun 20, 2012 at 12:09 PM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> Where do you think it might be doing that? And are you talking modules in
> general, or just injective modules, eg a vector space over Q considered as a
> Z-module?

I don't care about injective modules; I only care about things applying to Z^n.

The paper I linked might be out of context. It's just one result out of this

https://www.google.com/search?q=%22Grassmannian%20of%20submodules%22

There's a number of results referencing this concept, of which the
above paper I linked is one. This might be a better one

http://arxiv.org/pdf/math/0611074.pdf

So it says this:
"Let M be a finite dimensional space on a field k. The Grassmannian
Gre(M, k) of M
is the set of subspaces of dimension e. It is well known that Gre(M,
k) is an algebraic
variety with nice properties. For instance, the linear group GLe(M, k)
acts transitively on
Gre(M, k) with parabolic stabilizer, hence the variety Gre(M, k) is
smooth and projective."

OK, got that. But now it says this

"Suppose now that M has a structure of A-module, where A is a finitely generated
k-algebra. It is natural to define the Grassmannian Gre(M,A) of
A-submodules of M of
given dimension e. It is a closed subvariety of Gre(M, k), hence it is
a projective variety.
But in general this variety is not smooth."

This I don't get. They're talking about vector spaces which are also
structured to be modules over algebras over a field? Would this apply
to Z^n? To Q^n?

I'm trying to figure out if there's something analogous to a
Grassmannian which would apply to the submodules of Z^n, so I can have
a parameter space handling all of the "lattices" in it.

-Mike

On Wed, Jun 20, 2012 at 12:09 PM, genewardsmith
<genewardsmith@sbcglobal.net> wrote:
>
> --- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...>
> wrote:
> >
> > Paper here: http://arxiv.org/pdf/1010.4761v2.pdf
> >
> > Can anyone tell me if this actually defines some sort of object which
> > is analogous to a Grassmannian, but which allows us to distinguish
> > between all the submodules of a module and not just the subspaces of a
> > vector space?
>
> Where do you think it might be doing that? And are you talking modules in
> general, or just injective modules, eg a vector space over Q considered as a
> Z-module?

By the way, I asked here:

http://mathoverflow.net/questions/100217/generalized-grassmannians-that-parameterize-the-submodules-of-a-module

Got some interesting responses, most of which involve, as you'd might
expect, algebraic geometry. I have a semi-decent grasp on commutative
algebra these days, so maybe it's about time to dive into Eisenbud and
Harris.

Again, my hope is that this will lead to some operation analogous to
the wedge product, but which a) allows for a nice way to handle
subgroups which trace out unsaturated lattices in the larger JI group
(slash the quotient group is nonfree), and b) allows for a nice way to
handle temperaments with torsion.

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> By the way, I asked here:
>
> http://mathoverflow.net/questions/100217/generalized-grassmannians-that-parameterize-the-submodules-of-a-module

Good idea, and interesting responses.

Some of this stuff I've never thought about before, like the quotient group
of a subgroup of Z^n.

In music theory terms, if we're working in the 2.3 group, would the
quotient group of the 2.9 subgroup be a torsion group of order 2 with only
1/1 and 3/1 in it or something?

What would the quotient group of the 4.6 subgroup be?

-Mike

On Jun 21, 2012, at 8:28 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> By the way, I asked here:
>
>
http://mathoverflow.net/questions/100217/generalized-grassmannians-that-parameterize-the-submodules-of-a-module

Good idea, and interesting responses.

-Mike

On Jun 21, 2012, at 11:19 PM, Mike Battaglia <battaglia01@gmail.com> wrote:

Some of this stuff I've never thought about before, like the quotient group
of a subgroup of Z^n.

In music theory terms, if we're working in the 2.3 group, would the
quotient group of the 2.9 subgroup be a torsion group of order 2 with only
1/1 and 3/1 in it or something?

Oops, this is wrong; the quotient group of a subgroup is just the same as
the resulting tempered space you get if you treat the subgroup as the
kernel of a temperament. So the quotient group for 2.9.5 would be an order
two cyclic group for which a -transversal- of the non-identity element is
3/1, but which also represents 3/2, 6/1, 3/5, etc.

Still not sure wtf the structure of the (2.3)/(4.6) quotient group is. The
4.6 subgroup traces out an XXXXXX crisscrossing diamond pattern in the 2.3
group, so the quotient group has one element in it for which 2/1 is a
transversal, and another element for which 3/1 is a transversal. (2/1)^2 =
1/1 in this group and (3/1)^2 = 1/1 as well, but 2/1 * 3/1 = 1/1 as well.
What does one call this sort of group?

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> Some of this stuff I've never thought about before, like the quotient group
> of a subgroup of Z^n.

I think I mentioned it some time back when talking about the differnce between finite and infinite index.

> In music theory terms, if we're working in the 2.3 group, would the
> quotient group of the 2.9 subgroup be a torsion group of order 2 with only
> 1/1 and 3/1 in it or something?

More or less.

> What would the quotient group of the 4.6 subgroup be?

Similar.

On Jun 22, 2012, at 12:13 AM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:
>
> Some of this stuff I've never thought about before, like the quotient
group
> of a subgroup of Z^n.

I think I mentioned it some time back when talking about the differnce
between finite and infinite index.

Index of what?

> In music theory terms, if we're working in the 2.3 group, would the
> quotient group of the 2.9 subgroup be a torsion group of order 2 with only
> 1/1 and 3/1 in it or something?

More or less.

But the thing I called 3/1 would also represent 3/2, 6/1, etc, right?

> What would the quotient group of the 4.6 subgroup be?

Similar.

Wouldn't it be order 3?

How does the 4.6 quotient group differ from the 4.9 quotient group?

-Mike

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> I think I mentioned it some time back when talking about the differnce
> between finite and infinite index.
>
> Index of what?

http://en.wikipedia.org/wiki/Index_of_a_subgroup

> > In music theory terms, if we're working in the 2.3 group, would the
> > quotient group of the 2.9 subgroup be a torsion group of order 2 with only
> > 1/1 and 3/1 in it or something?
>
> More or less.
>
> But the thing I called 3/1 would also represent 3/2, 6/1, etc, right?

Right.

> > What would the quotient group of the 4.6 subgroup be?
>
> Similar.
>
> Wouldn't it be order 3?

No, still order 2.

> How does the 4.6 quotient group differ from the 4.9 quotient group?

They're not even the same size.

On Jun 22, 2012, at 1:59 PM, genewardsmith <genewardsmith@sbcglobal.net>
wrote:

--- In tuning-math@yahoogroups.com, Mike Battaglia <battaglia01@...> wrote:

> I think I mentioned it some time back when talking about the differnce
> between finite and infinite index.
>
> Index of what?

http://en.wikipedia.org/wiki/Index_of_a_subgroup

Ok.

> > What would the quotient group of the 4.6 subgroup be?
>
> Similar.
>
> Wouldn't it be order 3?

No, still order 2.

Ok yeah, I see why it's order 2.

Muchas gracias!

-Mike