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Suppose that $M(\Bbb R,n)$ is the set of all $n \times n$ square real matrices. The special linear group $\text{SL}(\Bbb R,n)$ acts on this group via right-multiplication.

It is easy to see that the determinant will be preserved under the group action, so that for any $A \in M(\Bbb R,n)$ and $B \in \text{SL}(\Bbb R,n)$, we have $\det(AB) = \det(A)$.

Additionally, on this Wikipedia page, it claims that this is basically the only polynomial on the coefficients of $A \in M(\Bbb R,n)$ that is preserved for arbitrary $B \in \text{SL}(\Bbb R,n)$, in that any other polynomial preserved is just a linear combination of powers of the determinant. That is, the ring of invariants of the special linear group is generated entirely from the determinant.

A fairly important subgroup of $\text{SL}(\Bbb R,n)$ is $\text{SL}(\Bbb Z,n)$, the group of integer unimodular matrices with determinant equal to +1. This is just the subgroup of the special linear group whose coefficients are integers.

Although the coefficients of $\text{SL}(\Bbb Z,n)$ are integers, we will again view it as acting on the real square matrices $M(\Bbb R,n)$. Then it is easy to see that this subgroup again preserves the determinant. However, does it now also preserve anything else? Or do we get the same invariant ring as $\text{SL}(\Bbb Z,n)$?

(Also, is there some methodology for computing invariant rings for groups like these?)

EDIT: doing a brute force search of polynomials has yielded no invariants (so far) for $2 \times 2$ matrices except the determinant, so it may well be only that.

Mike Battaglia
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  • Try computing the Zariski closure? – Qiaochu Yuan Jun 12 '19 at 19:05
  • How does one do that? – Mike Battaglia Jun 12 '19 at 20:52
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    What you want to show is that $SL_n(\mathbb{Z})$ is Zariski dense in $SL_n(\mathbb{R})$, which would imply that they have the same polynomial invariants. So, given some nonzero polynomial function on $SL_n(\mathbb{R})$, you want to show that it doesn't vanish on $SL_n(\mathbb{Z})$, which means finding some element in $SL_n(\mathbb{Z})$ on which it doesn't vanish. This shouldn't be too hard. As a warmup, start by showing that $\mathbb{Z}$ is Zariski dense in $\mathbb{R}$, then that $\mathbb{Z}^n$ is Zariski dense in $\mathbb{R}^n$. – Qiaochu Yuan Jun 13 '19 at 01:25
  • OK, it's simple to see what the Zariski topologies are on $\Bbb Z$ and $\Bbb R$, but in what sense is $\Bbb Z$ dense in $\Bbb R$? $\Bbb Z$ isn't an ideal in $\Bbb R$, so I don't know how to represent it in $\text{Spec}(\Bbb R)$ to determine if it's dense or not. (Also, do you mean $\Bbb Z[x_1, ... x_n^2]$ rather than $\Bbb Z^n$?) – Mike Battaglia Jun 14 '19 at 02:07
  • @Qiaochu Yuan - when you say showing that $\Bbb Z$ is Zariski dense in $\Bbb R$, do I just look at the set of all prime ideals in $\Bbb R$ supersetting $\Bbb Z$? I get that $\Bbb R$ has only two ideals, but I don't know how to represent an arbitrary (non-ideal) subset in the spectrum. – Mike Battaglia Jun 16 '19 at 22:14

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