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I learned that the fundamental group of $O(n)$ is $\Bbb{Z}/2\Bbb{Z}$ (for $n>2$). What is the explicit expression for its topological invariant? To be specific:

Given a smooth path $\{M(t):T^1\rightarrow{O(n)}\}$, find the functional that maps each path to its corresponding element in $\pi_1[O(n)]$.

Chen Fang
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  • Possible duplicate: http://math.stackexchange.com/questions/123650/fundamental-group-of-the-special-orthogonal-group-son – Sammy Black Dec 02 '14 at 02:59
  • You may look at Chapter 6.6 of http://www.amazon.de/Geometrie-Mathematik-Physikstudierende-Aufbaukurs-Mathematik/dp/3834802107 which explains in very much detail the nontrivial element of $\pi_1(SO(3))$ (which is of course the same as $\pi_1(O(3))$). – user39082 Dec 02 '14 at 11:09
  • @Sammy Black: The answer explains in detail which path corresponds to the generator, which I find helpful. But I need an explicit isomorphism: given any smooth path, is it trivial or nontrivial? Or: find a well-defined process by which any smooth path can shrink to a point or a known generator. – Chen Fang Dec 02 '14 at 12:02
  • @user39082 I do not have access to the book at the moment but I will check in the library. But as I said in my other reply, I was not asking for the generator, but an explicit isomorphism. – Chen Fang Dec 02 '14 at 12:06
  • Since $pi_1(SO(n))$ and $H_1$ agree, it suffices to check that the loop is not trivial in homology. That should be easier. You could also pass to cohomology, check whether the nontrivial element of $H^1$ vanishes. Perhaps use (twisted) de Rham cohomology and Poincare duality to compute it as an integral. – ziggurism Dec 03 '14 at 04:55

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