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I want to prove that the fundamental group of $GL(n, \mathbb{C}) $ is infinite.

I don't know how to proceed, any hint ?

WLOG
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  • One way might be $GL(1, \mathbb{C})$ is the circle. So you may want to prove that $GL(2,\mathbb{C})$ is $GL(1,\mathbb{C})$ plus a higher dimensional cell which does not alter its fundamental group. But I feel any explicit CW-decomposition is difficult. – Bombyx mori Oct 01 '14 at 16:45

1 Answers1

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The determinant maps it to $\mathbb C^*$, which has the homotopy type of a circle. So you just have to find a continuous map $\mathbb C^*\to GL(n,\mathbb C)$ such that the composition with the determinant is the identity, and apply the $\pi_1$ functor.

Cheerful Parsnip
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  • This is really nice! I did not thought about it before. – Bombyx mori Oct 01 '14 at 16:42
  • This nice argument is enough to conclude that $\pi_1({\rm GL}_n({\Bbb C})$ is infinite, but isn't enough to compute it. – Andrea Mori Oct 01 '14 at 16:57
  • @AndreaMori: agreed. Here's a link to that more general question. http://math.stackexchange.com/questions/3637/fundamental-group-of-gln-c-is-isomorphic-to-z-how-to-learn-to-prove-facts-lik – Cheerful Parsnip Oct 01 '14 at 17:15