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For some reason, I can't find a reference for $\pi_i GL(n,\mathbb C)$ nor can I figure what they are. For most Lie groups, you can get a nice fibration and use the long exact sequence in homotopy to inductively compute the homotopy groups (e.g. the fibration $SO(n-1) \to SO(n) \to S^{n-1}$). However, I can't think of a nice fibration; $GL(n)$ acts transitively on $\mathbb C^n$ but I don't know a nice description for the stabilizer subgroup.

This is motivated by understanding the statement that $GL(n)/GL(k)$ is $k-1$ connected (for the real and complex cases), so if there's an easy explanation for that without appealing to $\pi_1 GL(n)$, then that would also be appreciated.

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If you work through the iterated fibrations in Ryan's answer below to try and compute these groups, I think you'll find that knowing them is equivalent to knowing the (unstable) homotopy groups of spheres. –  Aaron Mazel-Gee Feb 3 '11 at 2:46
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up vote 12 down vote accepted

There's a fibration

$$GL(n, \mathbb C) \to GL(n+1, \mathbb C) \to \mathbb C^{n+1} \setminus \{0\}$$

By Gram-Schmidt, this fibration is fibre homotopy-equivalent to

$$U_n \to U_{n+1} \to S^{2n+1}$$

given by only remembering the 1st vector in the matrix, just as in your $SO(n)$ example.

The stable homotopy groups of the unitary groups are known. Google "Bott Periodicity". The unstable groups for $U_n$, just like for $SO_n$, are only known in a range.

I believe these fibrations are discussed in Bredon's book, as well as May, among others. This is example 4.55 in Section 4.2 of Hatcher's book.

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That first fibration is the one I was thinking about (but I forgot to exclude zero) but is that really right? The dimensions don't seem to add up... –  Eric O. Korman Feb 2 '11 at 19:34
You're right -- the first "fibration" I wrote down is really a "homotopy fibre sequence" in that the fibre contains $GL(n, \mathbb C)$ and is homotopy-equivalent to it, but it's a little larger than $GL(n, \mathbb C)$. Technically the fibre is all $(n+1)\times (n+1)$ invertible matrices such that the 1st column vector is $(1,0,\cdots,0)$. Projection into the orthogonal complement of $(1,0,\cdots,0)$ gives the homotopy-equivalence. –  Ryan Budney Feb 2 '11 at 19:40
ahh, ok. that makes sense, thanks! –  Eric O. Korman Feb 2 '11 at 19:42
Another way to see this, if you want to kill a fly with a sledgehammer, is a theorem of Iwasawa stating that any connected Lie group deformation retracts to any maximal compact subgroup. In particular $GL_n(\mathbf{C})$ is homotopy equivalent to $U(n)$. –  Dan Petersen Feb 2 '11 at 22:31
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