Timeline for Counting cosets in the Quotient of Weyl groups
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
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| Jul 20, 2017 at 6:58 | comment | added | MIQ | @JimHumphreys The order of Weyl groups was never an issue but of parabolics, which you just cleared. Thanks | |
| Jul 19, 2017 at 17:21 | comment | added | Jim Humphreys | @MIQ: The orders of irreducible Weyl groups are well-known and given case-by-case near the end of the Bourbaki volume, as well as in the Wikipedia article linked, and in my books, etc. In turn, a proper parabolic subgroup is just a direct product of such Weyl groups, so the quotient giving the number of cosets is easy to calculate. (This can be done for any finite reflection group and is unrelated to representation theory.) | |
| Jul 19, 2017 at 14:00 | comment | added | Oliver | I don't know why you think tableaux should be involved, since they don't appear in the Grassmannian case that you seem to already understand. If you want an analogue of "partitions in a $k$ by $(n-k)$ rectangle" for general G/P, you can replace each element of $W^P$ by its set of inversions in the positive root poset. In the case of Grassmannian permutations, this is exactly giving you the partitions you expect. | |
| Jul 19, 2017 at 9:31 | answer | added | Vít Tuček | timeline score: 1 | |
| Jul 19, 2017 at 9:15 | comment | added | MIQ | @JimHumphreys I couldn't find that in the Bourbaki of Lie Algebras CH4-6 (English version, don't understand French). Could you please provide me with a more precise reference? Moreover, I have further made my question clearer by adding more details. Thanks | |
| Jul 19, 2017 at 9:12 | history | edited | MIQ | CC BY-SA 3.0 |
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| Jul 18, 2017 at 20:27 | comment | added | Jim Humphreys | As Hugh points out, you just need to know the order of $W$ and the orders of its various "parabolic" subgroups $W_P$, which are all known rather explicitly (cf. Bourbaki etc.). The answer is similar for any root system and its Weyl group, but of course it's more combinatorial-looking for type $A$: here a parabolic subgroup is a Young subgroup of the symmetric group, i.e., a product of smaller symmetric groups. | |
| Jul 18, 2017 at 20:02 | comment | added | Hugh Thomas | The number of elements in $W^P$ is the number of elements in $W$ divided by the number of elements in $W_P$. | |
| Jul 18, 2017 at 17:49 | history | edited | MIQ | CC BY-SA 3.0 |
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| Jul 18, 2017 at 15:50 | history | asked | MIQ | CC BY-SA 3.0 |