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The question looks like an exercise in elementary algebraic topology, but I didn't manage to solve it. I am considering this question because it is a toy example in a problem I'm thinking about.

Let's consider the natural embedding $GL_n(\mathbb C) \to \mathbb C^{n^2} \backslash \{0\}$. As was discussed in this question, cohomology with rational coefficients of $GL_n(\mathbb C)$ is an exterior algebra on generators in degrees 1, 3, ..., 2n-1 (one generator in each degree), whereas $\mathbb C^{n^2} \backslash \{0\}$ is homotopy equivalent to a sphere $S^{n^2-1}$.

I'd like to prove that the map induced on cohomology of degree $n^2 - 1$ is a zero map. Any ideas?


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up vote 11 down vote accepted

$\mathbb C^{n^2} \smallsetminus {0}$ is homotopy equivalent to $S^{2n^2-1}$, not $S^{n^2-1}$, and $\mathrm H^{2n^2-1}(GL_n(\mathbb C)) = 0$.

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Just to be really pedantic, n should be at least 2. – damiano Apr 29 '10 at 14:56
That was quick. Thanks! – Evgeny Shinder Apr 29 '10 at 14:57
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 $n^2$. – Angelo Apr 29 '10 at 15:07
This matches with n^2=1+2+...+(2n-1), the degree of a top cohomology class in the exterior algebra H^*(GL_n(C)). – Evgeny Shinder Apr 29 '10 at 19:59
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 $\{(g,x) \in \mathbb{C}^{n \times n} \times \mathbb{C} : x \det g =1 \}$.) – David Speyer Apr 30 '10 at 11:29

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