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Suppose I have a root system $\Phi$ (of a semisimple Lie algebra) with a set of simple roots $\Delta$. I am interested in describing Weyl group elements $w$ preserving a given subset $\Delta'$ in the sense that $w(\Delta')=\Delta'$.

Denote by $\Delta''$ the largest subset of $\Delta$ such that $\Delta'\cap\Delta''=\emptyset$ and all roots in $\Delta'$ are orthogonal to roots in $\Delta''$. Obviously, any product of simple reflections corresponding to roots of $\Delta''$ fixes $\Delta'$. For example, one may hope that this gives me all such $w$ if $\Delta'$ is a connected subset of $\Delta$.

As far as I understand, my question is equivalent to asking which Weyl group elements fix a given Levi subalgebra.

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    $\begingroup$ This does not give you all such w. For example take the root system $A_3$ and $\Delta'$ equal to the two orthogonal simple roots. Then $\Delta'' = \emptyset$ but the product of the longest element with the two simple reflections preserves $\Delta'$. $\endgroup$
    – Arkandias
    Commented Nov 22, 2016 at 13:02
  • $\begingroup$ Thank you Arkandias! I have modified the question to exclude simple counterexamples :-). $\endgroup$
    – Roman Fedorov
    Commented Nov 22, 2016 at 13:38
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    $\begingroup$ I think Arkandias' counterexample is easily modified to a counterexample when $\Delta'$ is connected: Staying in $A_3$, take $\Delta'$ to be the singleton containing the middle root in the diagram. Then $\Delta''$ is still empty, but $s w_0$ still fixes $\Delta'$, where $s$ is the reflection for that middle root and $w_0$ is the longest element. $\endgroup$
    – Nathan Reading
    Commented Nov 22, 2016 at 15:00
  • $\begingroup$ It seems to me that this question is essentially equivalent to the following: Given a subset $J$ of the simple generators $S$ for $W$, find the subgroup of $W$ consisting of elements that fix the parabolic subgroup $W_J$ under conjugation. (These would be equivalent in the sense that if you solve one, you easily solve the other. There would be some "translation" needed from one answer to the other.) $\endgroup$
    – Nathan Reading
    Commented Nov 22, 2016 at 15:10

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I'm not sure whether there is an efficient way to answer your question (or a written reference), but it's possible to analyze the situation case-by-case.

It's probably best to start with an irreducible root system $\Phi$ (reduced in the Bourbaki sense), which can be asociated with a simple Lie algebra over $\mathbb{C}$ but can just as well be studied abstractly. Fix a simple system $\Delta$ and the corresponding Weyl group $W$ generated by the simple reflections $s_\alpha$ (with $\alpha \in \Delta$). Now choose a (proper) subset $\Delta' \subset \Delta$ with a corresponding root subsystem $\Phi'$ (not necessarily irreducible) and Weyl subgroup $W'$.

It's important to keep in mind that the automorphism group Aut($\Phi$) is the semidirect product of the normal subgroup $W$ (which acts simply transitively on simple systems in $\Phi$) and the possibly trivial group $\Gamma$ of graph automorphisms. An example is given by Nathan Reading for type $A_3$, where the nontrivial graph automorphism has order 2. Similarly, Aut($\Phi'$) is the product of such automorphism groups (and a permutation group on irreducible components if there is more than one), which might or might not be realized within $W$ and might or might not involve graph automorphisms. So a certain amount of case-by-case description could be needed.

[EDIT] With this set-up, suppose $1 \neq w \in W$ stabilizes $\Delta'$. To summarize the possibilities for $w$ based on the above description of Aut($\Phi'$): (1) $w \notin W'$ by the simple transitivity of $W'$ on bases such as $\Delta'$. (2) If there exist simple roots orthogonal to all $\alpha \in \Delta'$ (i.e. $\Delta'' \neq \emptyset$), then $w$ can be any nontrivial element of $W''$, the subgroup of $W$ generated by $\Delta''$. (3) As in the comments, $w$ might involve both reflections $s_\alpha$ for $\alpha \in \Delta'$ and elements outside $W'$ such as the longest element $w_0$. (Note that any reduced expression for $w_0$ involves all simple roots. Also, $w_0$ takes $\Delta$ to $-\Delta$ but might also involve a Dynkin diagram automorphism. In either case, if $w_0 \alpha = -\alpha$ in the case $\Delta' =\{\alpha\}$, then $s_\alpha w_0$ fixes $\Delta'$. Similarly for larger, or disconnected, $\Delta'$ as suggested by Arkandias.) As indicated above, case-by-case work should determine all such possibilities.

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