Timeline for Sets M,N with G action such that C[M] = C[N] as G modules, how are they related ?
Current License: CC BY-SA 3.0
9 events
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| Sep 19, 2012 at 15:48 | comment | added | Alexander Chervov | @Andreas Blass May I ask another question. Consider GL_n(F_q) and its action of it on V=F_q^n and on V^* . This is an example of the setup we need X=V and Y = V^*. How to construct P(V) and P(V^*) isomorphism ? Many many many thanks in advance. | |
| Sep 18, 2012 at 13:55 | comment | added | Alexander Chervov | @Andreas Blass Thank you very much ! Probably there should not be unique correspondence (e.g. if G is trivial - that would be any two sets canonically isomorphic - nonsense), however might be not everything is that much bad, in some specific situations, although it is far from being clear for me at the moment, what are these "specific situations"... | |
| Sep 18, 2012 at 13:20 | comment | added | Andreas Blass | @Alexander Chernov: The proof is not directly constructive. It involved comparing the characters of the linear representations over $\mathbb C$ with the marks (in Burnside's sense) of the power-set permutation actions. I haven't really considered the question whether one can convert it into something more constructive, setting up a specific, canonical correspondence between the two sorts of representations. | |
| Sep 18, 2012 at 11:59 | comment | added | Alexander Chervov | @Andreas Blass May I ask you ? Vaguely: is the proof constructive ? More precisely may be there is something like "correspondence" between X,Y which is G-ivariant ? Can the isomrphism between P(x) P(Y) be chosen uniquely ? or "almost" uniequely up to something... | |
| Sep 17, 2012 at 8:02 | vote | accept | Alexander Chervov | ||
| Sep 12, 2012 at 6:02 | comment | added | Alexander Chervov | @Andreas Blass Thank you very much ! Very nice result ! | |
| Sep 12, 2012 at 3:01 | comment | added | Benjamin Steinberg | This is of course the same as P(X) and P(Y) being isomorphic as modules of the semiring P(G). What other idempotent semirings R have the property that $\mathbb CX\cong \mathbb CY$ iff $RX\cong RY$. | |
| Sep 12, 2012 at 2:59 | comment | added | Benjamin Steinberg | This seems to me to really answer the question. | |
| Sep 11, 2012 at 21:31 | history | answered | Andreas Blass | CC BY-SA 3.0 |