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.
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$\begingroup$ Probably it's enough to deal with Chevalley groups (split case): when $G$ is $k$-isotropic, $\S7$ of the Borel-Tits paper Groupes reductifs constructs a split subgroup having as maximal torus a maximal $k$-split torus of $G$. (When $G$ is simply connected this subgroup is also of simply connected type, but that doesn't seem to be relevant here.) $\endgroup$– Jim HumphreysCommented May 15, 2014 at 19:47
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1$\begingroup$ If G=PGL_p and k has char p, the commutator subgroup only contains elements whose determinant is a p-th power and so doesn't contain any open neighbourhood of 1. If G is split, semisimple and simply connected then G(k) is perfect for all fields k with |k|>3 by the Steinberg presentation so you're fine there. The most likely general criterion seems to be if the kernel of the canonical map from the simply connected cover is a group scheme of order prime to p=char(k). $\endgroup$– Peter McNamaraCommented May 17, 2014 at 0:06
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$\begingroup$ and I don't understand your comment about the anisotropic case since if you take an anisotropic form of PGL_p, the commutator subgroup is not open either. Does strong topology mean what I think it does (usual topology from topology of k)? I've never heard the words strong topology in this context before. $\endgroup$– Peter McNamaraCommented May 17, 2014 at 8:53
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$\begingroup$ Many thanks for your help. Yes, strong topology does mean the topology coming from the topology on $k$. I think I heard that phrase used somewhere in an algebraic geometry textbook but I'm not sure. Do you know of anywhere where I can read more about the Steinberg presentation? $\endgroup$– RupertCommented May 19, 2014 at 11:25
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1$\begingroup$ Steinberg: Lectures on Chevalley Groups contains the presentation $\endgroup$– Peter McNamaraCommented May 20, 2014 at 11:27
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I claim the following is true:
Theorem: Let G be an isotropic (definition: contains a subgroup isomorphic to 𝔾m) 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.
answered May 20, 2014 at 11:50