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Let $G$ be a complex adjoint group. Let $u\in G$ be unipotent. The group $A(u):=\pi_0(Z_G(u))$ acts on the set of components of the Springer fiber $\mathcal{B}_u$, the variety of Borel subgroups that contain $u$. We know $\mathcal{B}_u$ is equi-dimensional.

Could it be true that every element of $A(u)$ stabilizes at least one component of $\mathcal{B}_u$?

My motivation is the following: When $G$ is the corresponding split group over $\mathbb{F}_q$, I wish to say "$\#\mathcal{B}_u(\mathbb{F}_q)$ has the order of $q^{\dim\mathcal{B}_u}$." But then we need the above property to make sure the Frobenius stabilizes at least one component. Thank you!

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  • $\begingroup$ $A(u)$ doesn't act on the set of components. It acts on the vector space freely generated by them, and not as a permutation representation. I suppose your question still has sense, though, i.e. whether for each $g\in A(u)$ does there exist a component $C$ s.t. $g\cdot \vec C = \vec C$. $\endgroup$
    – Allen Knutson
    Commented Jun 13, 2016 at 0:43
  • $\begingroup$ The action of $Z_G(u)^o$ on the set of component has to be trivial, giving the $A(u)$-action on the components. Nope? Or maybe my language was too bad... $\endgroup$
    – Cheng-Chiang Tsai
    Commented Jun 13, 2016 at 0:50
  • $\begingroup$ Argh, I confused this with the $W$-action, of course you're right. $\endgroup$
    – Allen Knutson
    Commented Jun 13, 2016 at 1:19
  • $\begingroup$ In fact, I was confused for more than an hour before I realized I don't need the Springer action at all to ask this question. I did however rely on Pramod's Lusztig-Shoji algorithm package to check that this is true up to B4, C4 and G2 ... $\endgroup$
    – Cheng-Chiang Tsai
    Commented Jun 13, 2016 at 5:46
  • $\begingroup$ In your remark about motivation, it isn't clear to me what you mean by "has the order of". What does this mean when $u=1$, so $\mathcal{B}_u =\mathcal{B}$ (the full flag variety)? In type $A$ the flag variety is just projective space of the appropriate dimension and thus has for instance $q+1$ points over $\mathbb{F}_q$ if rank $G = 1$.. $\endgroup$
    – Jim Humphreys
    Commented Jun 13, 2016 at 17:33

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