5
$\begingroup$

My question is the following: Under which conditions on given integers $n\le m$ is $$O(n,m) = \{A \in \mathbb R^{m\times n} : A^TA = \mathbf 1\}$$ simply connected? Does anyone know a reference for this?

Simple connectedness of this for certain $n,m$ (I believe it should be ok for $m-n\ge 2$) was used in a proof, I'm currently trying to understand. But I haven't been able to fill the gap myself.

Thanks a lot!

  • 5
    From the definition you give you probably mean what you write, however in that case $O(n,m)$ is a rather [unfortunate choice of notation](http://en.wikipedia.org/wiki/Indefinite_orthogonal_group).2012-12-12
  • 0
    @dan: Thanks for pointing this out.2012-12-14

1 Answers 1

3

Here's something that should get you started. Motivated by the standard way of computing fundamental groups of Lie groups, consider the fibration $$ O(n-1,m-1) \to O(n,m) \to S^n $$ from the map $O(n,m) \ni A \mapsto A(1,0,\cdots,0) \in \mathbb R^n$. For $n \ge 2$ $S^n$ is simply connected so the long exact sequence of homotopy groups gives $\pi_1 O(n,m) = \pi_1 O(n-1,m-1)$ for $n \ge 2$.

Now this just leaves a few cases which I think are doable.

  • 3
    @SamL. Indeed, these spaces are special cases of [Stiefel manifolds](http://en.wikipedia.org/wiki/Stiefel_manifold), which should give you something to search for.2012-12-12
  • 0
    @Eric: Thanks for the hint! Unfotunately, I only have only very basic knowledge of Lie groups and even less knowledge of fibrations. It may be a bit much to ask for, but if it's not too much trouble, could you perhaps indicate how one might obtain an explicit isomorhpism $\pi_1 O(n,m) = \pi_1 O(n-1,m-1)$? I guess the idea is to take a loop, project it down to the sphere where we know there is a homotopy to a point, and then somehow lift all of this back up to $O(n,m)$ (but how?) to get a homotopy which homotops the initial loop into a single fiber $O(n-1,m-1)$?2012-12-14
  • 0
    @MihaHabič: Thank you very much for pointing me to Stiefel manifolds! :)2013-02-09