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](https://en.m.wikipedia.org/wiki/Kneser%E2%80%93Tits_conjecture) 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](https://mathscinet.ams.org/mathscinet-getitem?mr=379444)) 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](https://mathscinet.ams.org/mathscinet-getitem?mr=224710) 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](https://mathoverflow.net/questions/426066/is-every-compact-simply-connected-reductive-p-adic-group-perfect/426080#comment1095360_426066), $\operatorname{SL}\_1(D)$ 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](https://mathscinet.ams.org/mathscinet-getitem?mr=262250)" carefully analyzes the situation. So in the end, the answer to my question is "no" overall, but "yes" in many cases.



Is every compact simply-connected reductive p-adic group perfect?
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 $$[\textrm{SL}\_1(D), \textrm{SL}\_1(D)] \neq \textrm{SL}\_1(D).$$


