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**](http://dx.doi.org/10.2307/2373853), Am. J. Math. 100, 303-386 (1978). [ZBL0392.20027](https://zbmath.org/?q=an:0392.20027).2.
2. *Petrov, V.; Stavrova, A.*, [**Elementary subgroups of isotropic reductive groups.**](http://dx.doi.org/10.1090/S1061-0022-09-01064-4), St. Petersbg. Math. J. 20, No. 4, 625-644 (2009); translation from Algebra Anal. 20, No. 4, 160-188 (2008). [ZBL1206.20053](https://zbmath.org/?q=an:1206.20053).
3. *Rapinchuk, I.; Ruiter, J.*, On abstract homomorphisms of some special unitary groups, arXiv:2107.07351, preprint.
4. *Steinberg, Robert*, [**Lectures on Chevalley groups**](http://dx.doi.org/10.1090/ulect/066), 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](https://zbmath.org/?q=an:1361.20003).



Conjugation of root subgroups by the Weyl group
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](https://mathoverflow.net/questions/403345/conjugation-of-root-subgroups-by-the-weyl-group#comment1032860_403345), $V\_b$ 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.


