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I'm looking for something like a Grassmannian, but which parameterizes the submodules of a module rather than the subspaces of a vector space. Most specifically, I'm looking for something which parameterizes the submodules of specifically $\mathbb{Z}^n$. So another way to say it is that I'm looking for a space parameterizing for the subgroups of a free abelian group. (A moduli space?)

I've seen some references to the concept of a "Grassmannian of submodules" here and there (like the papers on the first page of https://www.google.com/search?q=%22grassmannian+of+submodules%22) but can't figure out if this handles modules like $\mathbb{Z}^n$ or not.

Does anyone know if such an object exists and if so, how to construct it? Where I can get more information on this?

EDIT: to give a bit more info, the only specific application I really care about is parameterizing the free subgroups of a free abelian group, or the "lattices" in the $\mathbb{Z}$-module $\mathbb{Z}^n$, etc. A solution which works only for that, but which doesn't handle more exotic modules would be fine for my purposes. (And if it doesn't work out for all free abelian groups in general, then having a solution for at least finitely generated free abelian groups would even be a great start.)

I framed the question in terms of the "submodules of a module" in general just because I saw some references to there being a "Grassmannian of submodules" before, so I thought such a construction might be widely known.

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  • $\begingroup$ Some thoughts. $\text{GL}_n(\mathbb{Z})$ naturally acts on the set of such submodules $L$. This action preserves the isomorphism type of the f.g. abelian group $A = \mathbb{Z}^n/L$, so you might as well fix this isomorphism type in advance. Given $A$ you now want to classify the possible ways of choosing $n$ generators of it. You could try to classify the orbits of this set under the action of $\text{GL}_n(\mathbb{Z})$ for some simple $A$ and see if you can reach any general conclusions. $\endgroup$
    – Qiaochu Yuan
    Commented Jun 21, 2012 at 13:25
  • $\begingroup$ For example if $A = (\mathbb{Z}/p\mathbb{Z})^n$ then $n$ generators are precisely a basis and the resulting set is a torsor over $\text{GL}_n(\mathbb{F}_p)$ (with the action of $\text{GL}_n(\mathbb{Z})$ factoring through this). $\endgroup$
    – Qiaochu Yuan
    Commented Jun 21, 2012 at 13:30

2 Answers 2

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Building on Donu Arapura's answer and Qiaochu Yuan's comment:

If the quotient is torsion then $k$ changes as a functioin of the characteristic, so no $Grass_{k,n}$ will do. If it is non-torsion then classifying the $\mathbb Z$-module is equivalent to classifying the $\mathbb Q$-subsoace it generates, so a point on the Grassmanian will be a fine enough invariant. It's also clear that every rational subspace contains a rank-k $\mathbb Z$-submodule whose quotient is torsion-free, so for the torsion-free case that is sufficient.

The next-simplest case is probably when the quotient is finite, since every other case can be viewed as a combination of this and the other one (by splitting the quotient into rank and torsion.) As Quiaochu points out, this is choosing a surjection $\mathbb Z^n\to A$, which is equivalent to choosing a set of $n$ generators of $A$. But since only the kernel of the map matters, and composing with an automorphism of $A$ doesn't change the kernel, you need to quotient out by the automorphism group of $A$. (The automorphism group of $A$ is not the image of $GL_n(\mathbb Z)$, for instance $GL_n(\mathbb Z) \to GL_n(\mathbb F_p)$ is not a surjection.)

For instance, there is only one sublattice of $\mathbb Z^n$ whose quotient is $(\mathbb Z/p)^n$: that being $p\mathbb Z^n$.

We can break this finite quotient into a product of $p$-torsion parts and consider those separately. If the $p$-torsion part is $(\mathbb Z/p)^k$, then $Grass_{k,n}(\mathbb F_P)$ will classify the choices of generators. If the torsion involves prime powers then I am not sure what to do.

But a deeper problem is that I don't think one can make a scheme over $\mathbb Z$ whose $\mathbb Z$-points have a natural bijection to the $\mathbb F_p$ points of an arbitrary scheme over $\mathbb F_p$ like $Grass_{k,n}(\mathbb F_p)$, other then silly stuff like the disjoint union of one copy of $\textrm{Spec} \mathbb Z$ for each point on the $\mathbb F_p$-scheme.

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  • $\begingroup$ Thanks for the answer - lots of interesting ideas here. I'm not sure I totally get the idea about using $Grass_{k,n}(\mathbb F_P)$ though. i get that $Grass_{k,n}(\mathbb F_P)$ can classify the choices of generator for $\mathbb F_P$, but how does that assist in representing the subgroup in question...? For instance, say I'm in $\mathbb Z^2$, and I want to represent the submodule $S$ generated by (1,0) and (0,3). $\mathbb Z^2/S$ is going to be a cyclic group of order 3. How were you thinking the Grassmannian over $\mathbb F_3$ would help represent this as a submodule of $\mathbb Z^2$? $\endgroup$
    – Mike Battaglia
    Commented Jun 25, 2012 at 4:18
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    $\begingroup$ look at the whole thing modulo $3$. since the submodule already contains $3\mathbb Z^2$, you're not losing any information. Mod $3$, it's the submodule generated by $(1,0)$ and $(0,0)$, or rather just the submodule generated by $(1,0)$. This makes it a $1$-dimensional subspace of a $2$-dimensional vector space, and these are classified by a Grassmanian. $\endgroup$
    – Will Sawin
    Commented Jun 25, 2012 at 4:25
  • $\begingroup$ Ah, OK. Interesting. So then you also mentioned that this wouldn't work for prime powers. It would also not work for something like the submodule generated by (1,0) and (0,6), right? Because then you don't have a finite field at all. $\endgroup$
    – Mike Battaglia
    Commented Jun 25, 2012 at 20:33
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    $\begingroup$ You can do different primes separately. This basically works the same as the canonical way to express an abelian group as the direct product of $p$-groups. So the mod $2$ data (in your case, the subgroup generated by $(1,0)$) and the mod $3$ data (in your case, the subgroup generated by $(1,0)$) are enough to tie down the lattice. If it was, on the other hand, $(1,0)$ mod $2$ and $(0,1)$ mod $3$, then the lattice would be generated by $(3,0)$ and $(0,2)$. $\endgroup$
    – Will Sawin
    Commented Jun 25, 2012 at 20:45
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Perhaps you can use the Grassmannian $Grass_{k,n}$ as a scheme over $Spec\, \mathbb{Z}$. Among other things, this means that you can plug in a commutative ring $R$ and get back the set $Grass_{k,n}(R)$ of rank $n-k$ direct summands of $R^n$. If this is too much of a restriction, then you might look at Grothendieck's Quot scheme. A gentle introduction to these ideas can be found in Eisenbud and Harris' The Geometry of Schemes.

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