# Notion of isomorphism of lattices / intrinsic definition of lattices

I'm trying to construct the "correct" (read: a good) notion of moduli space of $$n$$-dimensional lattices. Here, an $$n$$-dimensional lattice is the $$\mathbb{Z}$$-span of an $$\mathbb{R}$$-basis of $$\mathbb{R}^n$$. For this, one has to find the "correct" (read: a good) notion of isomorphism between two lattices.

The first immediate thought is that the orthogonal transformations $$O_n(\mathbb{R})$$ should definitely be (or rather "induce") isomorphisms between lattices. So for example, when columns of a matrix are considered to be bases of lattices, then $$\pmatrix{1 & 0 \\ 0 &1}$$ and $$\pmatrix{1/\sqrt{2} & 1/\sqrt{2} \\ -1/\sqrt{2} & 1/\sqrt{2}}$$ should be isomorphic, because they are just rotated versions of each other. That's because I really only care about intrinsic properties of lattices such as density, shortest vector, etc., and all of this is preserved under orthogonal transformations.

A second thing one might want to mod out is the scaling action of $$\mathbb{R}^\times$$, so that $$\lambda I$$ is the same as $$I$$, but I guess that depends on personal preferences.

Then I thought, fine, I'm done. However, after thinking about it, I don't see a clear reason why this must be everything. Why shouldn't there be more transformations which preserve all the intrinsic properties of a lattice? Please convince me that there can't be!

Precisely: Can you come up with a property/invariant $$\Phi$$ of lattices, such that for any $$M \in \text{GL}_n(\mathbb{R})$$ which is not in $$\mathbb{R}^\times \cdot O_n(\mathbb{R})$$, there exists a lattice $$L$$ with $$\Phi(M L) \neq \Phi(L)$$?

Ideally, one could even find an sort-of intrinsic definition of lattice, so please let me know if you have ideas for that.

EDIT: I think it's more complicated, namely when we are working with matrices then we have to consider left actions by the orthogonal stuff and then right actions by $$\text{SL}_n(\mathbb{Z})$$ or something like this. (Or the other way around if you consider row instead of column vectors.)

• Perhaps the notion of similitude is helpful. Maybe you can consider a lattice as something like "a (free) abelian group equipped with an inner product" - then the notion of isomorphism this corresponds to is "a group isomorphism which is also a similitude wrt the induced metric" (possibly..). I am spitballing; I haven't checked the details (and I know nothing about moduli spaces!!). Similitude is a bit weird - I don't know if there's a nice category that gives rise to it. If you replace "similitude" with "isometry" then there is.. Commented Mar 5 at 2:07

By definition, a lattice $$L$$ is an additive discrete subgroup of $$\mathbb{R}^n$$. An additive subgroup $$L$$ of $$\mathbb{R}^n$$ is discrete iff $$L$$ is the $$\mathbb{Z}-$$span of some $$t$$ linearly independent vectors over $$\mathbb{R}^n$$. Let's assume that $$L$$ is full-rank, i.e., $$t=n$$. Thus, when we say "given" $$L$$, we mainly mean given its basis, which can be represented by an invertible real matrix in $$\mathbb{R}^{n\times n}$$. To "compare" two lattices with respect to some invariant, e.g., shortest vector, covering radius, it is natural to scale both lattices to the same volume, say volume 1. Hence, $$\mathcal{L}_n$$, the space of all possible $$n-$$dimensional lattice bases (of volume 1) is identified with $$SL_n(\mathbb{R})$$. As any change of basis generates the same lattice, that is multiplying your basis matrix by a matrix in $$SL_n(\mathbb(R))$$ (right multiplication if you write your basis as column vectors), then $$\mathcal{L}_n$$ is in fact $$SL_n(\mathbb{R})/SL_n(\mathbb{Z}).$$ Now back to your question. If you want to study lattices w.r.t some Euclidean invariant, i.e., a property depending on the Euclidean distance which is invariant under rotation, such as short vectors...etc., then it is natural to consider that two lattices in $$\mathcal{L}_n$$ are equivalent if we can obtain one from another via rotation, i.e, they belong to the same orbit of the action of $$O_n(\mathbb{R})$$. So you work on $$L_n=O_n(\mathbb{R})\backslash SL_n(\mathbb{R})/SL_n(\mathbb{Z}).$$ In the above, I restricted to the Euclidean distance point of view, but we can get to the same conclusion if we see lattices from the point of view of positive definite quadratic forms. Anyway, let us now consider a "non-Euclidean" invariant. For $$x=(x_1,\dots,x_n)\in\mathbb{R}^n$$ we define $$N(x)=\prod_{i=1}^n |x_i|.$$ Define a notion of "shortest vector" w.r.t $$N$$ on $$L$$, namely $$N_{\min}(L)=\min_{x\in L} N(x).$$ It is possible to obtain two lattices $$L$$ and $$L'$$ that are in the same class in $$L_n$$ but with $$N_{\min}(L)\neq N_{\min}(L')$$. One strategy, is to take any lattice containing a point with some coordinate equals to zero, i.e., $$N_{\min}(L)=0$$, and try to rotate it to avoid points with zero coordinates, i.e., $$N_{\min}(L)\neq 0$$. For example, the standard lattice $$\mathbb{Z}^n$$ has $$N_{\min}=0$$, but one can construct rotations of $$\mathbb{Z}^n$$ with non-vanishing N_{\min}. Such rotations are interesting in coding theory, see for instance this paper. Note that $$N(x)$$ is invariant under multiplication by elements in $$A_n$$, the set of diagonal matrices of determinant one. So when we deal with problems related to $$N(L)$$, it is more suitable to consider the action of $$A_n$$ rather than $$O_n(\mathbb{R})$$, thus it is better to work on the space $$A_n(\mathbb{R})\backslash SL_n(\mathbb{R})/SL_n(\mathbb{Z}).$$