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Jul 30, 2012 at 21:37 comment added Dox @paul-reynolds Thank you for the comment, it's a very sketchy proof of the statement. I'll look Lawson & Michelsohn's and Nakahara's books for more details.
Jul 30, 2012 at 21:27 vote accept Dox
Jul 30, 2012 at 20:37 answer added José Figueroa-O'Farrill timeline score: 6
Jul 30, 2012 at 20:22 comment added Paul Reynolds On page 20 of Lawson & Michelsohn it says "It is a classical fact (cf. Helgason [1]) that $SO(n)$ is connected and that $SO(r,s)$, for $r,s \geq 1$, has exactly two connected components. It is also a classical fact that $\pi_1SO(n) \cong \mathbb{Z}_2$ for $n \geq 3$ and $\pi_1SO(r,s) \cong \pi_1SO(r) \times \pi_1SO(s)$ for all $r,s$. Hence, $\pi_1SO^0(1,r) = \pi_1SO^0(r,1) = \mathbb{Z}_2$ and $\pi_1SO^0(r,s) = \mathbb{Z}_2 \times \mathbb{Z}_2$ for all $r,s \geq 3$. (Here $SO^0(r,s)$ denotes the connected component of the identity.)". I didn't see a proof in Helgason but I didn't look too hard.
Jul 30, 2012 at 19:48 history asked Dox CC BY-SA 3.0