Timeline for Weyl group elements fixing a set of simple roots
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Dec 11, 2016 at 13:58 | vote | accept | Roman Fedorov | ||
| Dec 11, 2016 at 14:01 | |||||
| Nov 22, 2016 at 20:37 | answer | added | Jim Humphreys | timeline score: 4 | |
| Nov 22, 2016 at 15:10 | comment | added | Nathan Reading | 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.) | |
| Nov 22, 2016 at 15:00 | comment | added | Nathan Reading | 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. | |
| Nov 22, 2016 at 13:38 | comment | added | Roman Fedorov | Thank you Arkandias! I have modified the question to exclude simple counterexamples :-). | |
| Nov 22, 2016 at 13:35 | history | edited | Roman Fedorov | CC BY-SA 3.0 |
added 66 characters in body
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| Nov 22, 2016 at 13:02 | comment | added | Arkandias | 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 $\Delta'$. | |
| Nov 22, 2016 at 10:26 | history | asked | Roman Fedorov | CC BY-SA 3.0 |