Timeline for on the Springer sheaf
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Sep 18, 2016 at 1:08 | comment | added | Geordie Williamson | The sentence starting "We know that..." is false: small tells us that it is the IC extension of the local system on the regular semi-simple locus. This representation is the regular representation and is not irreducible. Also, the answer to "Is this sheaf..." is no: over the regular (not necessarily semi-simple) elements the dimension is constantly zero, however the number of points in the fibre can change, varying between order of Weyl group and 1, thus the sheaf is not locally constant over this locus. | |
| Sep 17, 2016 at 20:07 | comment | added | prochet | the dimension is related to centralizer and ofr unipotent elements of $GL_{n}$, I think this dimension characterizes the conjugacy class, no? | |
| Sep 17, 2016 at 19:51 | comment | added | Sasha | Why do you think so? The stalks of this sheaf should reflect the topology of the fibers which is more rich than just their dimension... | |
| Sep 17, 2016 at 17:38 | history | edited | prochet | CC BY-SA 3.0 |
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| Sep 17, 2016 at 17:31 | comment | added | prochet | $\tilde{\mathfrak{g}}=\{(g,B)\in\mathfrak{g}\times G/B\vert g\in Lie(B)\}$ | |
| Sep 17, 2016 at 17:27 | review | Suggested edits | |||
| Sep 17, 2016 at 17:37 | |||||
| Sep 17, 2016 at 17:19 | history | edited | Ben Webster♦ | CC BY-SA 3.0 |
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| Sep 17, 2016 at 16:49 | comment | added | SashaP | Probably $\mathfrak{g}$ should be replaced by the nilpotent cone of it. | |
| Sep 17, 2016 at 16:02 | comment | added | abx | What do you call the Springer resolution of $\mathfrak{g}$? $\mathfrak{g}$ is smooth. | |
| Sep 17, 2016 at 15:56 | comment | added | SashaP | Don't you need to shift by $-\dim\mathfrak g$ to make it perverse? | |
| Sep 17, 2016 at 15:34 | history | asked | prochet | CC BY-SA 3.0 |