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Let $k$ be a nonarchimedean local field and $G$ a reductive $k$-group, which we assume to be semisimple and simply-connected. Recall that an abstract group $H$ is perfect if it is generated by commutators, that is, equals its derived subgroup.

Question: Is $G(k)$ perfect?

When $G$ is isotropic, $G(k)$ is known to be perfect. This is a consequence of the solution to the Kneser–Tits problem for $k$: the group $G(k)^+$ generated by (the rational points of) additive subgroups of $G$ is the same as the group generated by root subgroups of $G$ for a fixed maximal split torus, and the latter group is generated by commutators. It is known that $G(k)=G(k)^+$ when $G$ is simply-connected.

So the main thrust of my question is when $G$ is anistropic, or equivalently (hence the title), when $G(k)$ is compact in the analytic topology. In type $A$, Platonov and Jančevskiĭ have proved (On a conjecture of Harder) by computations with division rings that $G(k)$ is perfect. (Clarification: This result is for $D^\times$, not $\text{SL}_1(D)$.)

Related question: Are there any anisotropic simple $k$-groups not of type $A$?

I realize that if the answer to the related question is negative then this work of Platonov and Jančevskiĭ answers my first question. I once looked through the tables in Tits's Boulder notes Classification of semisimple algebraic groups and it seemed like there were compact simple reductive groups only in type $A$. But this result was so surprising, and the tables so hard for me to follow, that I was not confident in the correctness of my understanding.

Bonus points for questions that work with a more general $k$, say, a complete discretely valued field with residue field of dimension $\leq1$.

Clarification: I was too blithe about what happens in type $A$. As krl pointed out in the comments, $\operatorname{SL}_1(D)$ is not perfect, and in fact, its commutator subgroup is its pro-unipotent radical. Riehm's paper "The norm 1 group of a $\mathfrak P$-adic division algebra" carefully analyzes the situation. So in the end, the answer to my question is "no" overall, but "yes" in many cases.

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    $\begingroup$ The answer to the Related question is "no", there are no other such groups, but I know of no better reference than Tits's Boulder notes. $\endgroup$
    – LSpice
    Jul 5 at 14:40
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    $\begingroup$ That's really surprising: there are so many compact reductive groups over the reals. It would be nice to have a simpler explanation of this fact than a case-by-case analysis. Anyway, your comment completely resolves my question, so if you write it as an answer then I'll select it. $\endgroup$ Jul 5 at 16:12
  • $\begingroup$ The surprise works the other way for me: I'm always amazed by the abundance of compact groups in the real case! $\endgroup$
    – LSpice
    Jul 5 at 16:54
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    $\begingroup$ Maybe I'm missing something: isn't the answer to the question "no" in general? Can't one calculate that $[\textrm{SL}_1(D),\textrm{SL}_1(D)] \neq \textrm{SL}_1(D)$ (cf. Platonov-Rapinchuk, Thm. 1.9) ? $\endgroup$
    – krl
    Jul 6 at 2:05
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    $\begingroup$ Yes, that makes sense -- I appreciate your generosity. @krl, I will accept your comment as an answer if you post it. $\endgroup$ Jul 7 at 15:51

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It seems that the answer to the Question is "no" in general (in the anisotropic case): for example, if $D$ is a finite-dimensional central division algebra over $k$, then Theorem 1.9 in Platonov-Rapinchuk (based on the article of Riehm quoted above) implies that $$[\textrm{SL}_1(D), \textrm{SL}_1(D)] \neq \textrm{SL}_1(D).$$

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At your request, I post my comment as an answer: the answer to the Related question is "no", i.e., all simple, anisotropic groups over a non-Archimedean local field are of type $\mathsf A$; but I know of no better reference than Tits's Boulder notes Classification of algebraic semisimple groups.

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  • $\begingroup$ However, please see @krl's comment indicating that the answer to the question in the title is also "no". $\endgroup$
    – LSpice
    Jul 6 at 14:39

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