Timeline for Algorithm for Brauer lifting via Brauer tree?
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| May 3, 2018 at 0:35 | vote | accept | Jim Humphreys | ||
| Oct 1, 2012 at 16:42 | comment | added | Jim Humphreys | Your comments make me more cautious about what can be expected in the way of a uniform algorithm. But leaving aside Green's earlier viewpoint for a moment, I'm still impressed by his ability to find all projective resolutions using the Brauer tree. Meanwhile I don't yet understand your Update and Example, since I'm not expecting symmetry but rather asymmetry in the choice of a lifting. In your example, I could lift the trivial Brauer character in two ways to an ordinary (not just virtual) character of degree 1, but prefer the lifting to the 1-character. | |
| Oct 1, 2012 at 0:34 | history | edited | fherzig | CC BY-SA 3.0 |
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| Sep 30, 2012 at 18:08 | comment | added | fherzig | ...and see what happens there. | |
| Sep 30, 2012 at 18:08 | comment | added | fherzig | You are welcome. i was able to find green's result in his 55 paper (in fact i had come across it as grad student but forgot). he constructs a character starting from a symmetric polynomial -- if you take the degree 1 symmetric polynomial you indeed get an ordinary character that extends the given Brauer character on p'-elements. it's not clear to me that this character can be read off purely from the tree, i.e. without using the structure of the group. if i had the time, i would take some (small) examples, e.g. open polygons that arise for general linear but also for alternating groups, ... | |
| Sep 30, 2012 at 14:56 | comment | added | Jim Humphreys | Thanks for the proposed algorithm, which I'll have to sort out. I've relied mainly on the general method of Green (and Lusztig) and will meanwhile add a link to Green's 1955 paper which gives his approach to Brauer lifting. | |
| Sep 28, 2012 at 17:13 | history | answered | fherzig | CC BY-SA 3.0 |