Timeline for Centralizer action on components of Springer fibers
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2016 at 15:47 | comment | added | Jim Humphreys | Note that Spaltenstein works in greater generality (with a reductive group that might not be connected), but his tables in Chapter IV may be useful. For exceptional types, some useful data is given in tables in Carter's 1985 book (based heavily on Spaltenstein's work), but note that several bonds were omitted in Carter's version of the partial ordering diagrams for nilpotent orbits. In both sources, good prime characteristic is allowed. | |
| Jun 14, 2016 at 11:11 | comment | added | Cheng-Chiang Tsai | Thank you very much! I will go check Spaltenstein's book. I should also have said that my $\mathrm{char}(\mathbb{F}_q)$ is large, so that this growth rate result is really equivalent to the original question. | |
| Jun 13, 2016 at 23:18 | comment | added | Jim Humphreys | I think your basic question has the answer yes, and in fact the entire group $A(u)$ should act trivially on at least one irredudible component of the Springer fiber. Have you looked at the work of Spaltenstein in Lect. Notes. in Math. 946 (1982)? This comes from his thesis work at Warwick supervised by Lusztig. | |
| Jun 13, 2016 at 18:12 | comment | added | Cheng-Chiang Tsai | Ouch - pardon for the language again. By "has the order of $q^d$" I meant to say things like there exist universal $C_1,C_2>0$ such that $C_1q^d<\#\mathcal{B}_u(\mathbb{F}_q)<C_2q^d$. | |
| Jun 13, 2016 at 17:35 | history | edited | Jim Humphreys | CC BY-SA 3.0 |
Edited to display question
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| Jun 13, 2016 at 17:33 | comment | added | Jim Humphreys | 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 $G = 1$.. | |
| Jun 13, 2016 at 5:46 | comment | added | Cheng-Chiang Tsai | 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 ... | |
| Jun 13, 2016 at 1:19 | comment | added | Allen Knutson | Argh, I confused this with the $W$-action, of course you're right. | |
| Jun 13, 2016 at 1:00 | history | edited | Cheng-Chiang Tsai | CC BY-SA 3.0 |
added 17 characters in body
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| Jun 13, 2016 at 0:50 | comment | added | Cheng-Chiang Tsai | 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... | |
| Jun 13, 2016 at 0:43 | comment | added | Allen Knutson | $A(u)$ 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. $g\cdot \vec C = \vec C$. | |
| Jun 13, 2016 at 0:14 | history | asked | Cheng-Chiang Tsai | CC BY-SA 3.0 |