4
$\begingroup$

Let $G$ be a reductive group over a number field $K$. Let $\Gamma\subset G(K)$ be a subgroup.

My extremely naive question is - When can you deduce that $\Gamma$ is Zariski-dense? I'm looking for criteria for Zariski-density, possibly making additional assumptions on $G,K,\Gamma$.

To my surprise I've found it to be very difficult to find information on this question. References would also be appreciated!

As a special case of this, let $R_K\subset K$ be the ring of integers. Fix an embedding of algebraic $K$-groups $i : G\hookrightarrow \text{GL}_{n,K}$, and define $G(R_K) := \text{GL}_n(R_K)\cap i(G(K))$. When is $G(R_K)$ Zariski-dense? Does it depend on the embedding $i$?

It might be appropriate to make this a community wiki

$\endgroup$

1 Answer 1

7
$\begingroup$

The groups $G(R_K)$ you are interested in are arithmetic groups. If $G$ is semisimple, then a theorem of Borel and Harish-Chandra says that $G(R_K)$ is a lattice in $G(K \otimes \mathbb{R})$. Modulo some other minor hypotheses (connected, no compact factors) you can then appeal to the Borel Density Theorem to see that $G(R_K)$ is Zariski dense in $G(K \otimes \mathbb{R})$. A nice source that discusses this kind of thing is Dave Witte-Morris's book "Introduction to Arithmetic groups", available here

For this, you definitely need $G$ to be semisimple. For instance, $GL(n,\mathbb{Z})$ is not Zariski dense in $GL(n,\mathbb{R})$ (and is also not a lattice).

$\endgroup$
2
  • $\begingroup$ Do you mean to say that $G(R_K)$ is Zariski dense in $G$? The Borel density theorem seems to only address the case that $K$ can be embedded in $\mathbb{R}$. What if $K$ is imaginary? $\endgroup$ Jan 5, 2022 at 5:46
  • 1
    $\begingroup$ @stupid_question_bot: Use Weil Restriction is to reduce to the case $K=\mathbb{Q}$. en.m.wikipedia.org/wiki/Weil_restriction $\endgroup$ Jan 5, 2022 at 11:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy