Proving that $O(n,m)$ is simply connected.

Question

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!

Answer 1

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.