Suppose $\mathbf{G}$ is a connected reductive (possibly non-split!) group over a field $F$, $\mathbf{S} \leq \mathbf{G}$ a maximal split subtorus and $\mathbf{Z} \leq \mathbf{G}$ its centralizer. For a root $\alpha \in \Phi(\mathbf{G},\mathbf{S})$ let $\mathbf{U}\_\alpha \leq \mathbf{G}$ denote the associated root subgroup (possibly nonabelian, if $\mathbf{G}$ is not split). Let $\mathbf{G}\_\alpha$ denote smallest closed subgroup of $\mathbf{G}$ containing $\mathbf{U}\_\alpha$ and $\mathbf{U}\_{-\alpha}$.


Question:


1. What can one say about $\mathbf{G}\_\alpha$ in the general non-split case?
2. More specifically, is it true that $\mathbf{G}\_\alpha \cap \mathbf{G}\_\beta = \{1\}$ if $\alpha,\beta$ are not rational multiples of each other?
3. What about questions 1 and 2 if one replaces $\mathbf{G}\_\alpha$ by $\mathbf{Z}\_\alpha := \mathbf{Z}\cap \mathbf{G}\_\alpha$ ?


Edit: Replaced 'local field' by 'field' in order to make the question less confusing. Also, the closest thing to an answer of question 2 I could find in Borel-Tits is Proposition 3.22 where they show that $G\_\psi \cap G\_\eta = G\_{\psi \cap \eta}$. Here $G\_\psi$ is a subgroup defined for a quasi-closed set $\psi$ of roots. However my questions concerns the subgroups $G^\ast\_\psi \leq G\_\psi$ (in the notation of Borel-Tits), and so this does not help me at all.



Subgroups generated by opposite root groups
I believe that (2) and (3) will often fail. For example, let $\mathbf{G} = \mathrm{SL}(3,\mathbb{H})$, where $\mathbb{H}$ is the algebra of quaternions. This is an almost-simple algebraic group over $\mathbb{R}$. Consider 
$$\mathbf{G}\_\alpha = \begin{bmatrix} \* & \* & 0 \\ \* & \* & 0 \\ 0 & 0 & 1 \end{bmatrix}, 
\ \mathbf{G}\_\beta = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \* & \* \\ 0 & \* & \* \end{bmatrix}, 
\ t\_\omega = \begin{bmatrix} 1 & 0 & 0 \\ 0 & \omega & 0 \\ 0 & 0 & 1 \end{bmatrix}, $$
where $\omega$ is a unit quaternion. Then $t\_\omega$ is in $\mathbf{G}\_\alpha \cap \mathbf{G}\_\beta$, so (2) fails. In fact, $t\_\omega$ is in $\mathbf{Z}\_\alpha \cap \mathbf{Z}\_\beta$, since it centralizes the maximal $\mathbb{R}$-split torus consisting of real diagonal matrices, so (3) fails.




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Here is a guess at the answer to (1) when $F$ has characteristic zero. (I do not have the expertise to speculate about fields of positive characteristic.)


Let $\mathbf{T}$ be a maximal torus that contains $\mathbf{S}$. It is easy to see that $\mathbf{U}\_\alpha$ is normalized by $\mathbf{T}$, so the Lie algebra $\mathfrak{u}\_\alpha$ is a sum of root spaces for $\mathbf{T}$. Let $\psi$ be the smallest quasi-closed set of roots that contains all of the roots that occur in either $\mathfrak{u}\_\alpha$ or $\mathfrak{u}\_{-\alpha}$. It seems to me that $\mathbf{G}\_\alpha$ will usually, if not always, be the corresponding almost simple subgroup $\mathbf{G}^\*\_\psi$.


As a special case, $\alpha$ could be a circled root in the Tits-Satake diagram. Removing all of the other circled roots yields a diagram that may be disconnected, and I suspect that $\mathbf{G}\_\alpha$ may be the almost-simple group corresponding to the connected component that contains $\alpha$. 


At least, this works for $\mathrm{SL}(n,\mathbb{H})$. Take, for example, $n = 3$. The Tits-Satake diagram is 
${\bullet}{-}{\circ}{-}{\bullet}{-}{\circ}{-}{\bullet}$.
Let $\alpha$ and $\beta$ be the two circled vertices. After deleting $\beta$ (the rightmost circled vertex), there are two connected components. One, which we will ignore, is an isolated black vertex. The other, which consists of the first three vertices (${\bullet}{-}{\circ}{-}{\bullet}$), and represents $\mathbf{G}\_\alpha$, corresponds to the copy of $\mathrm{SL}(2,\mathbb{H})$ in the top left corner. This agrees with the above description of $\mathbf{G}\_\alpha$.


Note that, in this example, $\mathbf{G}\_\alpha$ is the subgroup corresponding to the first three vertices, and $\mathbf{G}\_\beta$ corresponds to the last three vertices. Their overlap is the middle vertex, which corresponds to the set of elements $t\_\omega$ mentioned above. In general, I think that if $\alpha$ and $\beta$ are adjacent to the same component of black vertices, then $\mathbf{G}\_\alpha \cap \mathbf{G}\_\beta$ will contain the anisotropic group corresponding to these black vertices.


