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

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    $\begingroup$ What does "bounded generation" / "boundedly generated" mean? $\endgroup$
    – LSpice
    Sep 17 at 20:05
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    $\begingroup$ (Also, your $P$ and $P^-$ must not contain any simple factor of $G$.) $\endgroup$
    – LSpice
    Sep 17 at 20:11
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    $\begingroup$ @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. $\endgroup$
    – BharatRam
    Sep 17 at 20:13
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    $\begingroup$ You need $G$ to be $k$-isotropic. If $G$ is $k$-anisotropic, this is clearly false. $\endgroup$
    – YCor
    Sep 17 at 21:40
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    $\begingroup$ @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 $x\neq x_1\dots x_N$. (...) $\endgroup$
    – YCor
    Sep 18 at 7:16

1 Answer 1

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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.

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    $\begingroup$ 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? $\endgroup$ Sep 17 at 21:42
  • $\begingroup$ @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! $\endgroup$
    – LSpice
    Sep 17 at 22:56
  • $\begingroup$ @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? $\endgroup$
    – BharatRam
    Sep 20 at 6:40
  • $\begingroup$ @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. $\endgroup$
    – LSpice
    Sep 20 at 13:31
  • $\begingroup$ @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. $\endgroup$
    – LSpice
    Sep 23 at 18:49

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