Let $F$ be a local field of characteristic $0$ and $G$ a connected split reductive group over $F$.

Let's look at the derived groups. We have $(G(F),G(F)) \subset (G,G)(F)$ and this inclusion is of finite index according to this MO question.

My question is : do we have (maybe under stronger assumptions) $U(F) \subset (G(F),G(F))$ for any unipotent subgroup $U \subset G$ ?

If no, same question for a given unipotent subgroup $U \subset G$.

  • 2
    $\begingroup$ The group $U(F)$ is an iterated extension of groups abstractly isomorphic to $\mathbf{Q}$-vector spaces and hence has no proper subgroup of finite index. Hence it is contained in any finite index subgroup of $(G,G)(F)$, hence is contained in $(G(F),G(F))$. $\endgroup$
    – YCor
    Nov 21, 2013 at 20:31
  • $\begingroup$ Thank you. Do you know if there is a more elementary proof (without proving the finiteness of the index, which is quite long) ? $\endgroup$
    – Arkandias
    Nov 21, 2013 at 20:54
  • $\begingroup$ A better proof should also carry over arbitrary fields of characteristic zero. $\endgroup$
    – YCor
    Nov 21, 2013 at 21:40

1 Answer 1


Yes, since in characteristic 0 all unipotent groups are connected and split. More generally, for any field $k \ne \mathbf{F}_2$ and any connected reductive $k$-group $G$ and split unipotent smooth connected closed $k$-subgroup $U \subset G$, necessarily $U(k) \subset (G(k), G(k))$. (This is false for ${\rm{SL}}_2(\mathbf{F}_2) = S_3$, whose commutator subgroup $A_3$ has order 3.)

In effect, this will reduce to the analogue for the non-reductive $H := \mathbf{G}_m \ltimes \mathbf{G}_{\rm{a}}$ with semi-direct product via either the usual scaling action or its precomposition with squaring on $\mathbf{G}_m$, taking $U$ to be $\mathbf{G}_{\rm{a}} = \mathscr{R}_u(H)$. In such cases, $(t,0)(1,x)(t,0)^{-1}(1,x)^{-1} = (1, (t^e-1)x)$ with $e = 1, 2$. But as $x$ varies through $k$ and $t$ varies through $k^{\times}$, $(t^e-1)x$ varies through $k$ when we can choose $t \in k^{\times}$ with $t^e - 1 \ne 0$; i.e., $k \ne \mathbf{F}_2$ when $e = 1$ and $k \ne \mathbf{F}_2, \mathbf{F}_3$ when $e = 2$. To deal with $\mathbf{F}_3$ below, we'll need to do a bit more work. (Note that ${\rm{SL}}_2(\mathbf{F}_3) = S_4$ does have commutator subgroup which contains the 3-Sylow subgroups.)

In the initial setup, every $U$ lies in $\mathscr{R}_u(P)$ for a minimal parabolic $k$-subgroup $P$ of $G$, so it suffices to treat such unipotent radicals (which are $k$-split). Of course, if $G$ has no proper parabolic $k$-subgroup (i.e., if $P = G$) then the only such $U$ is 1 and there is nothing to do, so we may assume $P \ne G$ and hence $G$ has a non-central split $k$-torus.

For a split maximal $k$-torus $S$ in $P$, the unipotent radical of $P$ is the direct product variety (under multiplication in $G$) given by $\prod_{c \in \Phi^+_{\rm{nd}}} U_c$ with the root groups associated to non-divisible elements in a positive system of roots in the relative root system $\Phi(G,S)$ (here ${\rm{Lie}}(U_c)$ is the $c$-weight space in ${\rm{Lie}}(G)$ unless $c$ is multipliable, in which case it is the span of the weight spaces for $c$ and $2c$). The hypothesis that $P \ne G$ ensures that the root system is non-empty (otherwise there is nothing to do). It suffices to treat the $U_c$'s separately, so by replacing $G$ with $\mathscr{D}(Z_G(S_c))$ for the codimension-1 subtorus $S_c = (\ker c)^0_{\rm{red}}$ killed by $c$ we may focus on the case of semisimple $G$ with $k$-rank 1.

For the purpose of handling $k = \mathbf{F}_3$, the following additional reduction steps are also convenient and make sense with arbitrary $k$. We may replace $G$ with its simply connected central cover $\widetilde{G}$ (as the preimage of $P$ in $\widetilde{G}$ is a minimal parabolic $k$-subgroup whose unipotent radical maps isomorphically onto $U$), so $G$ is simply connected. Hence, $G = \prod_i {\rm{R}}_{k_i/k}(G_i)$ for a finite separable extensions $k_i/k$ and absolutely simple connected semisimple $k_i$-groups $G_i$ that are simply connected. The $k$-rank 1 conditions forces all but one $G_i$ to be $k_i$-anisotropic, so we can drop those factors and thereby get to the case where there's a single $i = i_0$. Then $P = {\rm{R}}_{k_{i_0}/k}(P_{i_0})$ and $U = {\rm{R}}_{k_{i_0}/k}(U_{i_0})$, so on $k$-points we can pass to $(G_{i_0}, U_{i_0})$; i.e., we can assume $G$ is also absolutely simple over $k$.

If the relative root system is A$_1$ then $U_c$ is a direct sum of copies of the 1-dimensional representation $\mathbf{G}_{\rm{a}}(c)$ of $S = \mathbf{G}_m$ via a nontrivial character $c$ that is at worst 2-divisible in the character group (since $\langle c, c^{\vee} \rangle = 2$). If the relative root system is BC$_1$ then $U_c$ is an extension of a direct sum of copies $\mathbf{G}_{\rm{a}}(c)$ by a direct sum of copies of $\mathbf{G}_{\rm{a}}(2c)$ where $c$ is a basis of the character group, and this extension of $k$-groups remains short exact on $k$-points due to additive Hilbert 90.

Hence, it now suffices to show that if $H := \mathbf{G}_m \ltimes \mathbf{G}_{\rm{a}}(\chi)$ for the character $\chi$ of $\mathbf{G}_m$ given by $\chi(t) = t$ or $\chi(t) = t^2$ then $(H(k), H(k))$ contains $U(k)$ where $U = \mathscr{R}(H)$ is $\mathbf{G}_{\rm{a}}(\chi)$. The computation near the start handles this when $|k| > 3$. What if $k = \mathbf{F}_3$? This is where the above passage to the absolutely simple and simply connected case (not yet used) is useful. The only such cases over a finite field are ${\rm{SL}}_2$ and ${\rm{SU}}_3$. For the first case, by inspection ${\rm{SL}}_2(\mathbf{F}_3) = S_4$, whose commutator subgroup $A_4$ contains the 3-Sylow subgroups. In the other case, a direct argument shows that $G(k)$ is perfect whenever $|k| > 2$.

  • $\begingroup$ Is any of this discussion to be found in the older papers by Borel-Tits or Bruhat-Tits, or perhaps the book by Conrad-Gabber-Prasad? I recall that Tits was especially interested in subgroups generated by unipotents, in any characteristic (to include all local fields and residue fields). $\endgroup$ Nov 22, 2013 at 14:31

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