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Let $D$ be a connected Dynkin diagram with an automorphism $\nu$ of order 2. Let $Q=Q(D)$ denote the root lattice of $D$. Let $W=W(D)$ denote the Weyl group, it acts effectively on $Q$ and it is generated by reflections $r_\alpha$ for $\alpha\in D$. The automorphism $\nu$ acts on $Q$. Let $W_0$ denote the centralizer of $\nu$ in $W\subset {\rm Aut}\, Q$.

I want to understand this group $W_0$. Let $D^\nu$ denote the subset of $\nu$-fixed vertices in $D$. For $\beta\in D^\nu$ we have $r_\beta\in W_0$. I assume that for all $\gamma\in D\smallsetminus D^\nu$, the vertices $\gamma$ and $\nu(\gamma)$ are not connected by an edge (thus I exclude the case $D={\bf A}_{2n}$). Then $r_\gamma$ and $r_{\nu(\gamma)}$ commute, and we have $r_\gamma r_{\nu(\gamma)}\in W_0$.

Question. Is it true that $W_0$ is generated by $r_\beta$ for $\beta\in D^\nu$ and by $r_\gamma r_{\nu(\gamma)}\in W_0$ for $\gamma\in D\smallsetminus D^\nu$?

I am interested in the case $D={\bf D}_n$, but I would prefer to get a classification-free answer.

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    $\begingroup$ This set-up is well-studied in connection with the construction of quasi-split groups over various fields such as finite fields. I think Carter's old book Simple Groups of Lie Type may be a good source for an affirmative answer to your question, but I'd need to check more carefully. There are also accounts by Tits, Satake, etc. $\endgroup$
    – Jim Humphreys
    Commented Aug 23, 2013 at 13:20
  • $\begingroup$ P.S. The early sections of Chapter 13 in Carter's book deal with your situation, though his notation differs from yours. Some of the arguments are classification-free, involving generation of a reflection subgroup of the Weyl group. Then there is case-by-case discussion of the actual possibilities. He is of course treating all possible diagram symmetries. $\endgroup$
    – Jim Humphreys
    Commented Aug 23, 2013 at 13:30
  • $\begingroup$ @Jim Humphreys: Thank you, it was very helpful. I am interested in twisting compact groups over $\mathbb{R}$, rather than split groups over finite fields. $\endgroup$
    – Mikhail Borovoi
    Commented Aug 23, 2013 at 18:17
  • $\begingroup$ For the Weyl group it doesn't really make any difference. I think Helgason's book deals more directly with your issue, but the formalism for roots and Weyl groups doesn't change. Satake and Tits also have lecture note treatments covering the Lie groups, as do some textbooks I don't have at hand. Carter's treatment is quite concrete in any case. $\endgroup$
    – Jim Humphreys
    Commented Aug 23, 2013 at 20:37
  • $\begingroup$ @Jim Humphreys: Sure! Carter's Proposition 13.1.2 gives the affirmative answer to my question. $\endgroup$
    – Mikhail Borovoi
    Commented Aug 23, 2013 at 21:20

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As indicated in my comments, the 1968 book Simple Groups of Lie Type by R.W. Carter has a good elementary treatment of your question (to which the answer is yes) in Chapter 13. All of this goes back pretty far in the history of Lie theory, with the Weyl group and root system arising from a simple Lie algebra (or Lie group) or from a simple algebraic group. Typically the discussion of symmetries of the Dynkin diagram occurs along with specific constructions in the Lie algebra or associated group. But Carter's exposition is clear and detailed, showing how to pass from the original Weyl group to its subgroup commuting with the given diagram symmetry. In particular, this subgroup has a natural set of generators as in your question.

While there are many treatments in textbooks or lecture notes, one online source may be useful: the 1967-68 Yale lecture notes on Chevalley groups by Steinberg here. See his section 11, where he starts with standard examples and then treats the general theory. In all such sources, notation varies quite a bit but the ideas are pretty much the same.

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