The group G^+ of algebraic groups over local fields

Question

Let $G$ be an algebraic group defined over a char 0 local field $k$. Following Borel and Tits (73) we define the group $G^+(k)$ or $G^+$ by the subgroup of $G(k)$ generated by the unipotent elements of $G(k)$.

Suppose $G$ is generated by a finite set of unipotent $k$-subgroups, say $U_1,\cdots, U_n$. Is it true that the group generated by $U_1(k), \cdots, U_n(k)$ is $G^+$?

I feel the answer is positive but do not know how to prove it. It seems that the ideas of the original paper of Borel and Tits can help, but I still do not read French (which I always plan to learn) yet.

Answer 1

I haven't really checked the details, but I guess an argument works as follows. First, there exist $(k_i)$ such that the product map $\prod U_{k_i}\to G$ is onto. I think this implies that its differential is surjective on a Zariski open set. Therefore it is onto at some point in $\prod U_{k_i}(K)$; it follows (by the constant rank theorem, applies between $p$-adic manifolds) that the image of $\prod U_{k_i}(K)$ in $G(K)$ has non-empty interior for the norm topology. It follows that the subgroup $V$ generated by the $U_i(K)$ is open for the norm topology.

Now first assume $G$ is simple. Then a classical result (Prasad, Bull Soc Math France 1982, who attributes it to Tits) is that any open subgroup of $G(K)$ is either compact or contains $G^+$. Since $V$ is not compact (because any of the $U_i(K)$, if not trivial, is noncompact), it follows that $V=G^+$.

In the semisimple case, the same holds: since $V$ is an open subgroup and all its projection to $K$-simple factors are noncompact, $V=G^+$. Now since $G$ is generated by unipotents, it is semisimple modulo the unipotent radical. At this point the argument is complete for $G$ reductive but still needs a little more in general. If $H$ is the quotient by the unipotent radical $W$, it is probably true, that an open subgroup of $G(K)$ whose projection to the reductive quotient $H(K)$ contains $H(K)^+$, has to contain $N(K)^+$, where $N$ is the largest perfect normal subgroup in $G$. This should allow to conclude modulo $N$, i.e. reduce to the case when $G$ itself is unipotent, which is easy.

[Edit: I was more careful with the argument that the $U_i(K)$ generate an open subgroup]

Answer 2

First, the Borel-Tits paper you refer to is Homomorphismes "abstraits" de groupes algebriques simples, in Ann. of Math. 97 (1973), 499-571.

Their main concern is a connected reductive algebraic group $G$ defined over an arbitrary field $k$, but they pay special attention to the case of local fields later in the paper. In Section 6 they define the subgroup $H^+$ for any connected $k$-group $H$ to be the (normal) subgroup of $H(k)$ generated by the rational points of all $k$-split unipotent $k$-subgroups of $H$. (Here $k$-split is usually automatic, for instance when working over a perfect field.) The definition is sometimes given in a slightly different form in other papers but amounts to the same thing.

Assuming $H$ is actually generated by its connected unipotent subgroups, which often happens in the cases of interest such as semisimple groups, a finite number of such subgroups will be enough to generate the group (an elementary consequence of finite dimensionality). So the answer to your basic question is yes.

The group $G^+$ plays a natural role for Borel-Tits in the study of "abstract" homomorphisms of almost-simple algebraic groups, but it comes up more generally in older work of Tits where he conjectured that for simply connected groups it might coincide with the full group of rational points. This had already been suggested in special cases by Kneser and became known as the Kneser-Tits Conjecture. But Platonov found a counterexample, which led later to the label Kneser-Tits Problem in work of Prasad-Raghunathan and others (related to the "Whitehead group", etc.). So there is a further paper trail of interest here, much of it in English. However, the long paper by Borel-Tits is definitely in French.

EDIT: As Yves points out, I'm not directly answering the question asked. Probably I'm not understanding the motivation here, but the Borel-Tits subgroup is always required to be normal and in good cases is just the subgroup generated by all rational unipotent elements. Typically this comes up just for reductive groups, where the unipotent $k$-subgroups live inside the unipotent radicals of minimal $k$-parabolic subgroups. There are only finitely many conjugacy classes of the latter, so it's enough to start with finitely many unipotent groups (and take the normal closure). I've never seen any situation in which other finite choices of generating subgroups would be natural, but I suspect the answer to the (perhaps artificial) question asked is in fact yes. Is the question motivated somehow?