abelianization of adelic points of an algebraic group

2012-06-28T18:20:08Z

Let $G$ be a connected reductive group defined over a number field $K$ and $G^{der}$ its derived subgroup.
Let $\mathbb{A}_K$ denote the adeles of $K$.

Then for $G=GL_n$ we have $[GL_n(\mathbb{A}_K),GL_n(\mathbb{A}_K)]=SL_n(\mathbb{A}_K)=GL_n^{der}(\mathbb{A}_K)$. I'm interested in what generality this holds, in other words I'd like to ask:

Question 1: When is the commutator subgroup $[G(\mathbb{A}_K),G(\mathbb{A}_K)]$ equal to $G^{der}(\mathbb{A}_K)$?

As I think this question is really a local one, so let me put it this way:

Let $K_v$ be a local field of char 0. $G$ a reductive group over $K_v$, $G^{der}$ its derived subgroup.

Question 2: Is the commutator subgroup $[G(K_v),G(K_v)]$ equal to $G^{der}(K_v)$?

These questions came up when I wanted to understand 1-dimensional automorphic representations of unitary groups coming from a division algebra with an involution of the second kind and I realized I didn't know what the abelianizations of the adelic points of the groups in question were.

Answer by Yves Cornulier for abelianization of adelic points of an algebraic group

2012-06-28T19:53:30Z

I think question 2 has a positive answer when $G^{der}$ is simply connected [EDIT: and without anisotropic factor], but not in general. If $G=PGL_d$ (so $G=G^{der}$) then $G(K)/[G(K),G(K)]=M/M^d$ where $M=K^*$, this is not a trivial group in general.

Relevant refences should be the preliminary chapters in book by Margulis "discrete subgroups of semisimple Lie groups", as well as Platonov-Rapinchuk's book.

abelianization of adelic points of an algebraic group - MathOverflow

abelianization of adelic points of an algebraic group

Answer by Yves Cornulier for abelianization of adelic points of an algebraic group