Timeline for Euler Characteristic of General Linear Group
Current License: CC BY-SA 2.5
4 events
| when toggle format | what | by | license | comment | |
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| Aug 7, 2010 at 12:53 | comment | added | Donu Arapura | Karl: I realize now that you had thought it through and just wanted confirmation. Sorry if my comment seemed a little blunt. I also got zero using the same process. I guess you meant to write $SO(n)$ rather than $SL(n)$. | |
| Aug 7, 2010 at 10:33 | comment | added | Karl | Thank you all. My idea was to use Poincare Duality for $n>1$. Then using a homotopy equivalence of $GL(n)$ and $SL(n)$. Now, since Euler characteristic of a compact Lie group $Sl(n)$ for $n>1$ is zero. We will have $chi_c(Gl(n))=0.$ Which coincides with above answers. | |
| Aug 6, 2010 at 21:24 | comment | added | Theo Johnson-Freyd | For a complete answer, you should mention that $GL(0,\mathbb R)$ consists of a single point, or is empty, depending on the convention, and that $\chi(GL(1,\mathbb R)) = -2$. Note that the Euler characteristic you are using is the correct one --- it's additive on disjoint unions --- but is not a homotopy invariant. | |
| Aug 6, 2010 at 12:53 | history | answered | André Henriques | CC BY-SA 2.5 |