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.

  • $\begingroup$ 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. $\endgroup$ – Aaron Mazel-Gee Feb 3 '11 at 2:46

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.

  • $\begingroup$ 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... $\endgroup$ – Eric O. Korman Feb 2 '11 at 19:34
  • 1
    $\begingroup$ 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. $\endgroup$ – Ryan Budney Feb 2 '11 at 19:40
  • $\begingroup$ ahh, ok. that makes sense, thanks! $\endgroup$ – Eric O. Korman Feb 2 '11 at 19:42
  • 2
    $\begingroup$ 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 $U(n)$. $\endgroup$ – Dan Petersen Feb 2 '11 at 22:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.