Suppose that $G$ is an absolutely quasi-simple algebraic group defined over a non-archimedean local field $k$ of positive characteristic. Would there be any kind of reasonable sufficient condition for $[G(k),G(k)]$ to have nonempty interior in the strong topology? I have already asked this question in the anistropic case and the answer seems to be affirmative there. I thought that I had a proof in the isotropic case but it seems to be mistaken. It has been suggested to me that making the assumption that $G$ is simply connected might be enough. Any hints for an extra hypothesis that might do the trick would be very helpful.



when the derived group of the group of $k$-rational points has nonempty interior in the strong topology
I claim the following is true:


Theorem: Let G be an isotropic (definition: contains a subgroup isomorphic to m) semisimple algebraic group over the nonarchimedean local field k. Suppose that the characteristic of k does not divide the order of the fundamental group of G. Then the commutator subgroup [G(k),G(k)] is open (and of finite index).


This essentially boils down to the affirmative answer to the Kneser-Tits conjecture for local fields, which I think is due to Prasad and Raghunathan in all characteristics. This is the part where the hard work takes place. I have no idea how to proceed if G is anisotropic.


First consider the case when G is simply connected when I claim furthermore that G(k) is perfect. The Kneser-Tits problem asks if G(k) is generated by the (k-points of the) unipotent radicals of parabolic subgroups. So it suffices to show that root subgroups lie in the commutator subgroup.


So let S be a maximal split torus, a an indivisible root and Ua the corresponding root subgroup. Most of the time Ua is a vector group and it is easy to show Ua(k)⊆[G(k),G(k)] by computing the commutator of an element of S(k) and an element of Ua(k). In the remaining cases 2a is also a root, but Ua has a canonical two-step filtration with subquotients vector groups and we can proceed in much the same manner. (reference for the relevant structure theory: Section 3.3 in Pseudo-reductive Groups)


This deals with the simply-connected case. For the general case it suffices to show that H1(k,π1(G)) is finite. After passing to a finite Galois extension of k, the group scheme π1(G) becomes isomorphic to a product of μn's. So a spectral sequence argument reduces the finiteness to the finiteness of H^1(k,μn)=k\*/(k\*)n.


