Timeline for Why is the symmetric monoidal structure on invertible modules strict?
Current License: CC BY-SA 2.5
6 events
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| Sep 15, 2010 at 20:18 | comment | added | Charles Rezk | It is a curious diagram chase. I do it this way: if $\chi: R\to R$ is the map which represents the euler characteristic of $X$, then show that $1_X\otimes \chi \otimes 1_X : X\otimes X\to X\otimes X$ is equal to the braid map. | |
| Sep 15, 2010 at 20:00 | history | edited | Charles Rezk | CC BY-SA 2.5 |
tau is automorphism of I tensor I
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| Sep 15, 2010 at 19:44 | comment | added | Eric Wofsey | Hmmm...the idea of identifying $\eta$ with the Euler characteristic seems promising (though there's a bit of a diagram to chase to verify that they really are the same--the subtlety is that $\eta$ is defined in terms of the unit map and its inverse, while the Euler characteristic is defined in terms of the unit and the counit, which is NOT the same). This lets you really recast the question: why is the Euler characteristic of any dualizable module (locally) a nonnegative integer, rather than being an arbitrary element of $R$? | |
| Sep 15, 2010 at 15:05 | comment | added | Charles Rezk | Martin: if an object X in C has a dual Y, then the "Euler characteristic" is a map $R\to R$; the definition is basically the same as one way to define "trace of the identity map on a vector space". "Dual" means you have maps $\eta:R\to X\otimesY$ and $\epsilon:Y\otimes X\to R$ which encode the idea that "$Y\otimes{−}$ and $X\otimes{−}$" are adjoint functors on C; then $\chi(X)$ is $R\to X\otimes Y\approx Y\otimes X\to R$, where the map in the middle is the "braid map". | |
| Sep 15, 2010 at 8:07 | comment | added | Martin Brandenburg | What is the Euler characteristic of an invertible object? | |
| Sep 15, 2010 at 5:01 | history | answered | Charles Rezk | CC BY-SA 2.5 |