Conjugation of root subgroups by the Weyl group

Question

Fix a field $k$ of characteristic zero, and let $G$ be a connected reductive algebraic $k$-group of isotropic rank $\ge 1$. Fix a maximal $k$-split torus $S$, and let $\Phi_k$ be the relative root system of $G$ with respect to $S$. Assume that $\Phi_k$ is reduced and irreducible.

By Theorem 2 of Petrov and Stavrova, for every root $\alpha \in \Phi_k$ there is an embedding $X_\alpha:V_\alpha \to G$ where $V_\alpha$ is a vector group (EDIT: vector group scheme), and the image of $X_\alpha$ is the root subgroup $U_\alpha$. Given $v \in V_\alpha(k)$, $v \neq 0$, associated to $X_\alpha(v) \in U_\alpha(k)$ is a "Weyl group element" $w_\alpha(v)$, using Lemma 1.3 of Deodhar. If $G$ is a Chevalley group, these are the elements denoted $w_\alpha(t)$ in Steinberg, and $w_\alpha(1)$ is literally a representative of a Weyl group element. Steinberg lists some relations, of interest here is (R7), which says that for any two roots $\alpha,\beta \in \Phi_k$ and $v \in k$, $$ w_\alpha(1) \cdot X_\beta(v) \cdot w_\alpha(1)^{-1} = X_{w_\alpha \beta} \Big( c_{\alpha,\beta} v\Big) $$ This says that the action of the (lifts of) Weyl group elements on $G(k)$ by conjugation corresponds to the action of the Weyl group on $\Phi_k$; conjugating the root subgroup $U_\beta$ by a Weyl group representative $w_\alpha(1)$ takes it to the root subgroup $U_{w_\alpha \beta}$. The coefficient $c_{\alpha,\beta}$ is just a sign $\pm 1$ depending on $\alpha$ and $\beta$.

I have done a variety of explicit computations in quasi-split groups, and found in every case that $$ w_\alpha(u) \cdot X_\beta(v) \cdot w_\alpha(u)^{-1} = X_{w_\alpha \beta} \Big( c_{\alpha,\beta}(u,v) \Big) $$ for some function $c_{\alpha,\beta}:V_\alpha(k) \times V_\beta(k) \to V_{w_\alpha \beta}(k)$. Steinberg's (R7) says that in the split case $c_{\alpha,\beta}(1,v) = \pm v$. In the preprint (reference 3, equation 3) we give a version of this for a class of quasi-split special unitary groups, where $c_{\alpha,\beta}(1,v) = \pm v$ or $\pm \overline{v}$, with the bar denoting a Galois automorphism of order 2. I have also done computations in a quasi-split special orthogonal group, where some more complicated functions $c_{\alpha,\beta}$ arise.

Question 1: Why does the conjugation on the LHS above always end up in $U_{w_\alpha \beta}$, even in non-split cases? I suspect this is an obvious consequence of a Bruhat decomposition, but I don't understand that as well as I should.

Question 2: Is there a known generalization of this relation to reductive isotropic groups, or perhaps just for quasi-split groups?

1. Deodhar, Vinay V., On central extensions of rational points of algebraic groups, Am. J. Math. 100, 303-386 (1978). ZBL0392.20027.2.
2. Petrov, V.; Stavrova, A., Elementary subgroups of isotropic reductive groups., St. Petersbg. Math. J. 20, No. 4, 625-644 (2009); translation from Algebra Anal. 20, No. 4, 160-188 (2008). ZBL1206.20053.
3. Rapinchuk, I.; Ruiter, J., On abstract homomorphisms of some special unitary groups, arXiv:2107.07351, preprint.
4. Steinberg, Robert, Lectures on Chevalley groups, University Lecture Series 66. Providence, RI: American Mathematical Society (AMS) (ISBN 978-1-4704-3105-1/pbk; 978-1-4704-3631-5/ebook). xi, 160 p. (2016). ZBL1361.20003.

Answer 1

Let me use $a$ and $b$ for relative roots, so that I can later switch to $\alpha$ and $\beta$ for absolute roots.

If $b$ is a non-multipliable root, then, as you have said, $V_b$ is the $b$-root space, and $X_b$ is an exponential-type map. Specifically, it is the unique group homomorphism $V_b \to G$ whose derivative at the identity is the inclusion of the $b$-root subspace of $\operatorname{Lie}(G)$. It can be described ‘explicitly’, for small values of explicitly, as $v \mapsto \prod X_\beta(v_\beta)$, where $\beta$ runs over the absolute roots whose restriction to $S$ is $b$, and $v = \sum v_\beta$.

If $a$ is also non-multipliable, then we have that $w_a(u)$ is the commuting product $\prod w_\alpha(u_\alpha)$, where $\alpha$ runs over the absolute roots whose restriction to $S$ is $a$, and $u = \sum u_\alpha$. (Actually, now that I think about it, I might already be assuming $G$ quasi-split here.) In particular, $w_a(u)w_a(1)^{-1}$ equals $\prod \alpha^\vee(u_\alpha)$ (or maybe the inverse of this, depending how things are normalised; I didn't check).

You have already observed that $\operatorname{Int}(w_a(1))\bigl(X_b(v)\bigr)$ equals $\prod X_{w_a\beta}\bigl(c_{a\beta}\operatorname{Ad}(w_a(1))v_\beta\bigr)$, where $c_{a\beta} = \prod c_{\alpha\beta}$. Now suppose that $G$ is quasi-split. Then the set of absolute roots $\beta$ restricting to $b$ is a Galois orbit, and it is clear that $\beta \mapsto c_{a\beta}$ is constant on Galois orbits, so $\operatorname{Int}(w_a(1))\bigl(X_b(v)\bigr)$ equals $X_{w_a b}\bigl(c_{a b}\operatorname{Ad}(w_a(1)v)\bigr)$, where $c_{a b}$ is the common value of $c_{a\beta}$. Now just conjugate by $\prod \alpha^\vee(u_\alpha)$ to finish.