# abelianization of adelic points of an algebraic group

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.

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This is related to the Kneser-Tits conjecture --- see the Wikipedia article and its references. –  anon Feb 24 '13 at 20:46

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.
This is not quite correct. You can take $G=SL_1(D)$ over $K_v$ where $D$ is a central division algebra over $K_v$. In general, it is not its own commutator; however, the algebraic group is semi-simple and simply connected. –  Venkataramana Feb 24 '13 at 14:06
@Aakumadula: thanks for the correction. I should have assumed the group has no $K$-anisotropic factor; I edit accordingly. I don't know much about the anisotropic case so I'm not able to give an explicit example of a division algebra over a number field for which $SL_1(D)$ is not a perfect group but I'm ready to believe it indeed exists. –  YCor Feb 24 '13 at 15:25
Over a local field, take a quaternionic division algebra $D$. Then $D$ contains every quadratic subfield of $K$ (this is known; Platonov-Rapinchuk also state this). In particular, $D$ contains the unique unramified quadratic extension $E/K$. Moreover, the nontirvial Galois element of $E/K$ also lies in $D$. If $\pi$ is the uniformising parameter of $K$, it is also a uniformising parameter for $E$ and one can see that $SL_1(D)$ modulo the congruence subgroup of level one with respect to $\pi$ is the group of norm one elements of $E/K$ plus the Galois automorphism; this is solvable. –  Venkataramana Feb 24 '13 at 16:14
For a reference, there is a paper by Raghunathan and Gopal Prasad on $SL_1(D)$. –  Venkataramana Feb 24 '13 at 16:15