1
$\begingroup$

I am trying to learn arithmetic groups, from a dynamical point of view. These questions (maybe silly) come to my mind, but I do not know the answer.

  1. How does $SL_2(\mathbb Z[\sqrt 2])$ embed into $SL(2,\mathbb R)\times SL(2,\mathbb R)$? (The former is an irreducible lattice in the latter.)

  2. What are the centralizers of $SL_2(\mathbb Z[\sqrt 2])$ in $SL(4,\mathbb R)$? Trivial?

===========================================================================

I notice that the above questions are rather silly in some sense. Let me include another two (maybe also silly) questions.

  1. Why $SL_2(\mathbb Z[\sqrt 2])$ is a lattice in $SL(2,\mathbb R)\times SL(2,\mathbb R)$? Why not in $SL(2,\mathbb R)$ as $SL(2,\mathbb Z)$?

  2. How does $SL_2(\mathbb Z[\sqrt 2])$ embed into $SL(4,\mathbb Z)$? So acts on $\mathbb T^4$.

=========================================================================== Add on Question 4.

The embedding can be constructed as follows. For any $A+B\sqrt 2\in SL_2(\mathbb Z[\sqrt 2])$ with $A,B$ integer matrices. Let $\tau:A+B\sqrt 2\mapsto \begin{bmatrix} A & 2B\\ B&A\end{bmatrix}\in SL(4,\mathbb Z)$. One can check $\tau$ is indeed a group homomorphism.

===========================================================================

Answers, references and comments are highly appreciated.

$\endgroup$
  • 4
    $\begingroup$ 1. You embed $\mathbb{Z}[\sqrt{2}]$ into $\mathbb{R}$ in two ways $\endgroup$ – Mikhail Borovoi Jun 5 '18 at 14:31
  • 3
    $\begingroup$ The identity map $\mathbb{Z} \mapsto \mathbb{Z}$ extends to two embeddings $\mathbb{Z}[\sqrt{2}] \mapsto \mathbb{R}$: one (the injection) taking $\sqrt{2} \to \sqrt{2}$; the other taking $\sqrt{2} \to -\sqrt{2}$. These two embeddings each induce two embeddings of $SL(2,\mathbb{Z}[\sqrt{2}] \to SL(2,\mathbb{R})$, which are the two coordinate maps of the embedding you desired in question 1. $\endgroup$ – Lee Mosher Jun 5 '18 at 14:33
  • 2
    $\begingroup$ It's Zariski dense in $SL_2\times SL_2$. So the centralizer is equal to the centralizer of $SL_2\times SL_2$ (and of the subalgebra $M_2\times M_2$ it generates), which you can compute (it's not reduced to scalars). $\endgroup$ – YCor Jun 5 '18 at 15:29
  • $\begingroup$ @YCor What if centralizer in $SL(4, \mathbb Z)$? $\endgroup$ – Changguang Jun 5 '18 at 15:42
  • 2
    $\begingroup$ Closely related to your question 3 is this: Why is $\mathbb{Z}[\sqrt{2}]$ a lattice in $\mathbb{R} \times \mathbb{R}$ under the two embeddings of my previous comment, but not a lattice in $\mathbb{R}$ itself as $\mathbb{Z}$ is? The general pattern is explained by number theory: for every number field $F$, its ring of integers $O$ has a lattice embedding in $\mathbb{R}^m \oplus \mathbb{C}^n$ where $m$ is the number of real embeddings and $n$ is the number of complex conjugate pairs of complex embeddings. You can learn this in most algebraic number theory books. $\endgroup$ – Lee Mosher Jun 5 '18 at 15:52

Your Answer

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

Browse other questions tagged or ask your own question.