2

I want to prove that $\pi_1(SO(3))\cong \mathbb{Z}/2\mathbb{Z}$. I have already proved that there exists a surjection $\mathbb{Z}/2\mathbb{Z}\rightarrow \pi_1(SO(3))$.So I want to show that $\pi_1(SO(3))\neq 0$, but I still haven't come up with how to do it. Please give me some hints.

  • 4
    Are you working with the homotopy exact sequence? Alternatively, you might try to understand why $SO(3)\approx \mathbb RP^3$. – Ted Shifrin Jun 16 '13 at 13:51
  • No, I don't know exact sequence very well. I didn't know that, I try to understand it. Thank you. –  Jun 17 '13 at 01:58
  • OK ... Every element of $SO(3)$ is a rotation through some angle $0\le\theta\le\pi$ about some unit vector $v\in S^2$. – Ted Shifrin Jun 17 '13 at 02:12
  • 1
    Just extending Ted's comment, If you know that $So(3)\approx\mathbb{R}P^3$, then as $S^3$ is a double cover of $\mathbb{R}P^3$ and has trivial fundamental group with index 2 in $\pi _1(\mathbb{R}P^3)$, you are done. – tessellation Jun 17 '13 at 06:35

0 Answers0