Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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!

share|cite|improve this question
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
up vote 3 down vote accepted

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.

share|cite|improve this answer
@SamL. Indeed, these spaces are special cases of Stiefel manifolds, which should give you something to search for. – Miha Habič Dec 12 '12 at 22:33
@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
@MihaHabič: Thank you very much for pointing me to Stiefel manifolds! :) – Sam Feb 9 '13 at 8:11

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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