Let $G$ be an almost $k$-simple group that is also simply connected (so that $G(k)^{+}=G(k)$). For opposite parabolic subgroups $P$ and $P^{-}$, it is known that $G(k)^{+}$ is generated by the unipotent radicals $R\_u(P)(k)$ and $R\_u(P^{-})(k)$ (Prop 1.5.4 in Margulis's book "Discrete Subgroups of Semisimple Lie groups).


Can this generation be extended to bounded generation? That is, is $G(k)=G(k)^{+}$ boundedly generated by $R\_u(P)(k)$ and $R\_u(P^{-})(k)$?



Bounded generation of group by unipotent radicals of opposite parabolic subgroups
I hope you will permit me to write $P^+$ in place of $P$. Put $U^\pm = R\_u(P^\pm)$.


Yes, at least in the split case.


Suppose first that $P^+$ and $P^-$ are minimal. Put $T = P \cap P^-$. By working in $\operatorname{SL}\_2(k)$ or $\operatorname{PGL}\_2(k)$, you see that, for all $t \in k^\times$ and all roots $\alpha$ of $T$ in $G$, you can write $\alpha^\vee(t)$ as a product of $6$ elements of $U^+(k)$ and $U^-(k)$. This covers $T(k)$, using $6r$ elements, where $r$ is the semisimple rank of the group. Once you have written, for each element $w$ of the (finite!) Weyl group of $T$ in $G$, a representative of $w$ as a product of elements of $U^+(k)$ and $U^-(k)$ (which, again by a rank-$1$ computation, can be done using at most $3\ell$ elements, where $\ell$ is the length of a minimal expression for $w$ as a product of reflections), you only need $2$ more elements to generate the corresponding Bruhat cell.


Now continue to suppose that $G$ is split, but drop the assumption that $P^+$ and $P^-$ are minimal. Put $M = P^+ \cap P^-$, and let $T$ be a split maximal torus in $M$.


Let $\alpha$ be a root of $T$ in $M$. Since we (should) have assumed that $P^+$ does not contain any isotropic factor of $G$, there is some root $\beta$ of $T$ in $U^+$ such that $\alpha + \beta$ is also a root of $T$ in $U^+$. Then, for a suitable Chevalley–Steinberg system $(u\_r : \operatorname{Add} \to U\_r)\_{\text{$r$ a root}}$, we have for all $t \in k$ that $[u\_{-\beta}(t), u\_{\alpha + \beta}(1)]$ lies in $u\_\alpha(t)U^-(k)$. That is, each element of $U\_\alpha(k)$ is a product of $4$ elements of $U\_{-\beta}(k) \subseteq U^-(k)$ and $U\_{\alpha + \beta}(k) \subseteq U^+(k)$ with an element of $U^-(k)$.


This shows that the group of $k$-rational points of every root subgroup of $M$ is boundedly generated by $U^+(k)$ and $U^-(k)$; so, if $B\_M^\pm$ are opposite Borel subgroups of $M$ containing $T$, then $R\_u(B\_M^\pm)(k)$ are boundedly generated by $U^+(k)$ and $U^-(k)$; so $R\_u(B\_M^\pm)(k)U^\pm(k) = R\_u(B\_M^\pm\cdot U^\pm)(k)$ is boundedly generated by $U^+(k)$ and $U^-(k)$. Since $B\_M^+\cdot U^+$ and $B\_M^-\cdot U^-$ are opposite Borel subgroups of $G$, we have reduced to the previous case.


