Timeline for Natural embedding GL_n(C) -> C^{n^2} \ {0} induces zero on cohomology
Current License: CC BY-SA 2.5
6 events
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| Apr 30, 2010 at 11:29 | comment | added | David E Speyer |
Another reason that $GL_n(\mathbb{C})$ has top cohomology in dimension $n^2$ is that is an affine algebraic variety of dimension $n^2$. (To see that $GL_n(\mathbb{C})$ is affine, write it as $\{(g,x) \in \mathbb{C}^{n \times n} \times \mathbb{C} : x \det g =1 \}$.)
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| Apr 29, 2010 at 19:59 | comment | added | Evgeny Shinder | This matches with n^2=1+2+...+(2n-1), the degree of a top cohomology class in the exterior algebra H^*(GL_n(C)). | |
| Apr 29, 2010 at 15:07 | comment | added | Angelo | Yes, for $n=1$ the two spaces coincide. Let me add that the reason why $\mathrm H^{2n^2-1}(GL_n(\mathbb C)) = 0$ is that $GL_n(\mathbb C)$ is homotopy equivalent to the unitary group $U_n$, which has dimension $n^2$. Or, $\mathrm H^{i}(GL_n(\mathbb C)) = 0$ for $i > n^2$ because $GL_n(\mathbb C)$ is an affine variety of dimension $n^2$. | |
| Apr 29, 2010 at 14:58 | vote | accept | Evgeny Shinder | ||
| Apr 29, 2010 at 14:56 | comment | added | damiano | Just to be really pedantic, n should be at least 2. | |
| Apr 29, 2010 at 14:52 | history | answered | Angelo | CC BY-SA 2.5 |