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

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!

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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. – user52865 Dec 12 '12 at 20:54
@dan: Thanks for pointing this out. – Sam Dec 14 '12 at 18:12

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 $n \ge 2$.
@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 $O(n-1,m-1)$? – Sam Dec 14 '12 at 18:22