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I read that $O(M)$ is connected if and only if $M$ is not orientable. Is it true? If there is an answer, I would very much like to see a proof. Thank you.

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up vote 6 down vote accepted

If you assume that $M$ itself is connected, then it is true.

First, if $M$ is orientable and connected, then $O(M)$ has (at least) two connected components, one for each orientation of frame, either agreeing with a specified orientation or not.

If $M$ is not orientable, then, using the Levi-Civita connection for the Riemannian metric and parallel translating a frame around a non-orientable loop in the manifold will show that any two frames at a point can be connected, since the orthogonal group in any dimension has exactly two components. Then any two fibers of $O(M)\to M$ can be connected since $M$ is connected, so it follows that $O(M)$ itself is connected.

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Sorry,I forgot to mention that M is connected. I was quite surprised because by using a connection one can connect 2 points in the bundle by lifting the curve connecting the projections(of course,only differentiable curves can be lifted) so I do not understand where this reasoning is fallacious. Anyway,thank you very much for your answer,Mr. Bryant,it was most useful. – Codan Acelaire Jun 13 '13 at 15:01
OK,now I understand.The problem consists in connecting frames over the same point,and it can be done if M is not orientable. – Codan Acelaire Jun 13 '13 at 15:59

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