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 $\mathbf{U}_{-\alpha}$.


  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.

  • $\begingroup$ Have you consulted the classic 1965 IHES paper by Borel-Tits on the structure of (isotropic) reductive groups over arbitrary fields, or the later papers by Bruhat-Tits specializing to local fields? These are freely available online at numdam.org and could help to answer your questions. (Also, your boldface notation here is unhelpful. It's better to follow Borel-Tits.) $\endgroup$ Mar 27, 2015 at 20:58
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    $\begingroup$ Sorry, I didn't mean to be so bold :-) I wouldn't be surprised if the answer was somewhere in Borel-Tits, but I rather hoped someone could quickly confirm question two to the affirmative or alternatively destroy all my hopes. $\endgroup$ Mar 27, 2015 at 21:01
  • $\begingroup$ If $G$ is not split, I think that $U_\alpha$ is not necessarily a subgroup; I'm not even sure how to define it, but say in char. 0, in the Lie algebra $\mathfrak{u}_\alpha$ makes sense but is not necessarily a subalgebra, if $2\alpha$ is a root, unless I misunderstand the context. $\endgroup$
    – YCor
    Mar 27, 2015 at 23:03
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    $\begingroup$ In general (I suppose this is a standard convention) one denotes by $U_\alpha$ the unique closed connected subgroup whose Lie algebra is given by the direct sum of all weight spaces $\text{Lie}(G)_\beta$ where $\beta$ runs over all positive integer multiples of $\alpha$. $\endgroup$ Mar 27, 2015 at 23:14
  • $\begingroup$ Just to have it written here, in response to @YCor's point, in the non-reduced setting it is no longer true that $\operatorname{Lie}(U_\alpha)$ is always the $\alpha$-root space; rather it is the sum of the $\alpha$-root space and the (possibly trivial) ($2\alpha$)-root space. $\endgroup$
    – LSpice
    Sep 8, 2021 at 0:46

1 Answer 1


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.

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.

  • $\begingroup$ Sorry could you explain why over $\mathbb R$ ${\mathbf G}_\alpha$ is the smallest closed subgroup containing a root and its opposite? $\endgroup$ Mar 28, 2015 at 7:49
  • $\begingroup$ For the root $\alpha(a_1,a_2,a_3) = a_1/a_2$, we have $$ \mathbf{U}_\alpha = \begin{bmatrix} 1 & * & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}, \ \mathbf{U}_{-\alpha} = \begin{bmatrix} 1 & 0 & 0 \\ * & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix}.$$ These generate the above-mentioned group $\mathbf{G}_\alpha$. $\endgroup$ Mar 28, 2015 at 17:06
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    $\begingroup$ Okay, I've convinced myself now by hand that such $t_\omega$ indeed exists, and so I believe your counterexample. Fortunately the statement that I was really after still seems to have a chance of being true, since it was much weaker. However this was a good opportunity for me to be reminded of the definition of $\text{SL}_n(D)$ and the fact that the Dieudonné determinant takes values in $D^\times/[D^\times, D^\times]$ and not $D^\times$. $\endgroup$ Mar 28, 2015 at 20:16

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