Timeline for Intuition behind the age grading in quantum cohomology of orbifolds
Current License: CC BY-SA 3.0
6 events
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| Apr 13, 2017 at 12:58 | history | edited | CommunityBot |
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| Oct 4, 2011 at 14:20 | vote | accept | Dan Petersen | ||
| Sep 25, 2011 at 18:13 | comment | added | euklid345 | It is very nice that the age grading turns out to be compatible with the crepant resolution conjecture, but I don't think this answers the question. There must be some geometric sense in which the twisted sectors of an orbifold have fractional dimension. | |
| Sep 22, 2011 at 9:50 | comment | added | Paul Johnson | You are correct. The point is that $X$ will not have a crepant resolution unless $H^{orb}(X)$ is actually integrally graded. In fact, for $X$ to have a crepant resolution, all the degree shifting numbers should be even. To see that this is reasonable, note that over a point $x$ in $X$, the isotropy group $G_x$ will act on $K_X$. To be crepant, we want $f^*(K_X)=K_Y$ -- since $K_Y$ has no orbifold structure, it seems that we should have that $K_X$ is the trivial representation of $G_x$, which means that $G_x$ acts on $T_xX$ with determinant 1. | |
| Sep 22, 2011 at 7:45 | comment | added | Dan Petersen | Thanks a lot for the answer. I'm going to reveal my ignorance here but I am slightly confused about the statement that $H^{\mathrm{orb}}(X)$ is supposed to coincide with $H^\ast(Y)$ as graded vector spaces -- isn't the former $\mathbf Q$-graded and the latter $\mathbf Z$-graded? | |
| Sep 21, 2011 at 12:46 | history | answered | Paul Johnson | CC BY-SA 3.0 |