Fundamental group Pi1(SU(n)) and Pi2(SU(n))

Question

I need to find the fundamental group $\pi_1(SU(n))$ and $\pi_2(SU(n))$ for all $n$. I don't have any idea.

Answer 1

Since $SU(1)$ is a singleton, $\pi_1(SU(1)) = 0 = \pi_2(SU(1))$. Now let $n \ge 2$. There is a fiber bundle $SU(n-1) \to SU(n) \to S^{2n-1}$, which induces homotopy exact sequences $\pi_1(SU(n-1)) \to \pi_1(SU(n)) \to \pi_1(S^{2n-1})$ and $\pi_2(SU(n-1)) \to \pi_2(SU(n)) \to \pi_2(S^{2n-1})$. Since $n \ge 2$, $2n - 1 \ge 3$, and thus $\pi_1(S^{2n-1}) = 0 = \pi_2(S^{2n-1})$. Therefore, the maps $\pi_1(SU(n-1)) \to \pi_1(SU(n))$ and $\pi_2(SU(n-1)) \to \pi_2(SU(n))$ are onto. Using this fact and the fact that $SU(2)$ is homeomorphic to $S^3$, argue inductively to show that $\pi_1(SU(n)) = 0 = \pi_2(SU(n))$.