# Bounded generation of group by unipotent radicals of opposite parabolic subgroups

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 $$R_u(P^{-})(k)$$?

• What does "bounded generation" / "boundedly generated" mean? Sep 17 at 20:05
• (Also, your $P$ and $P^-$ must not contain any simple factor of $G$.) Sep 17 at 20:11
• @LSpice: Bounded generation, in this case, would mean that there exists some constant $N$ such that any element $g \in G$ is of the form $g=a_1b_1a_2b_2\dots a_Nb_N$ where $a_i \in R_u(P)$ and $b_i \in R_u(P^{-})$. It is a strengthening of generation by insisting that lengths are bounded globally. Sep 17 at 20:13
• You need $G$ to be $k$-isotropic. If $G$ is $k$-anisotropic, this is clearly false.
– YCor
Sep 17 at 21:40
• @LSpice Write $G$ as closed $\mathbf{Q}$-subgroup of $\mathrm{SL}_n$. Write $U_1,U_2$ the two given unipotent subgroups, and write $U=U_1\cup U_2$ (it's not a subgroup). Let $G$ resp $U$ be the zero set of $P_G$, where $P_G$ resp $P_U$ is a tuple of polynomials. Suppose by contradiction for every $N$ there exists a field of char zero $k_N$ such that some element of $G(k_N)$ is not product of $N$ elements of $U(k_N)$. This can be written as 1st-order formula: $k_N$ satisfies the formula $F_N$: $\exists x: P_G(x)=0,\forall x_1,\dots x_N$ with $P_U(x_i)$, we have $x\neq x_1\dots x_N$. (...)
– YCor
Sep 18 at 7:16

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.

• Good answer! I have no idea, but is it possible that there's a sort of "purely Coxeter-group" reason for this? My own perception of the physical/mechanical causality is certainly in line with this answer, but it would be interesting if there were "yet another" reason for this working out. Perhaps the role of the little $SL(2)$'s is an indirect expression of something that could be said more abstractly? Sep 17 at 21:42
• @paulgarrett, re, I wouldn't know where to begin in giving a purely Coxeter-group-theoretic explanation of a fact about semisimple groups—I don't even know how the Coxeter group would "see" the unipotent radical of a parabolic subgroup; the closest I can come is observing that one can very crudely bound the length of a minimal expression for a representative of a Weyl-group element by $3\ell$, where $\ell$ is its minimal length as a product of reflections. However, I'd love to see whatever you come up with! Sep 17 at 22:56
• @LSpice: Hi, thanks for the answer. So basically, we first boundedly generate the maximal split torus $T$. Since the Weyl group is finite, these representatives too can be generated boundedly. And to generate the Bruhat cell, we use the Levy decomposition of $P$ (product of $T$ and $R_u(P)$). Does all of this also go through for non-minimal parabolics? Sep 20 at 6:40
• @BharatRam, re, to run the same argument, you'd need to replace the covering of the split maximal torus by a covering of a Levi component. I don't immediately see how to do that. In fact, I don't even immediately see how to handle minimal parabolics in the quasi-split but not split case, though I'm more confident that can be made to work. Sep 20 at 13:31
• @BharatRam, re, I still don't know how to handle the non-split case, but I have added an argument that I think covers the split but not-minimal case. Sep 23 at 18:49