Homotopy groups of some magnetic monopoles

This is a list of homotopy groups which I (as a physics researcher) encounter when studying magnetic monopole under certain configuration of gauge field profiles.

\begin{gather} \pi_2(SU(2)/U(1)) \simeq \pi_2(S^2) \simeq Z.\\ \pi_1(SU(2)/U(1)) \simeq \pi_1(S^2) \simeq 0\\ \pi_1(U(1)/Z_N) \simeq \pi_1(S^1) \simeq Z\\ \pi_1(SU(2)/Z_N) \simeq Z_N ?\\ \pi_2(SU(2)/Z_N) \simeq ?\\ \pi_n(SU(2)/Z_N) \simeq ? \end{gather}

I suppose I can derive $(SU(2)/U(1))\simeq S^2$ and $U(1)/Z_N \simeq U(1) \simeq S^1$. So I can understand the first threes(?).

(a)$\pi_1(SU(2)/Z_N) \simeq Z_N$?

(b)$\pi_2(SU(2)/Z_N) \simeq$?

(b)$\pi_n(SU(2)/Z_N) \simeq$?

(is that $\pi_2(SU(2)/Z_N) \simeq Z \times Z_N$? is that $\pi_3(SU(2)/Z_N) \simeq 0$?)

Any explanation may help? Thank you.

• Have you looked in a textbook on homotopy theory? Very standard techniques can help you deal with this; in particular, your question is somewhat off-topic here, as explained in the FAQ. Dec 17, 2013 at 23:58
• would this story relate to the orbifolds we study? (yes I know basic homotopy theory, just like every mathematician knows basic quantum mechanics/quantum field theory.) Dec 18, 2013 at 0:01
• This question does not appear to be about research level mathematics, and may be more appropriate at math.stackexchange.
Dec 18, 2013 at 0:06
• that is fine. I can ask at math.stackexchange. Dec 18, 2013 at 0:07

Here is the basic tool. Let $F\rightarrow X\rightarrow B$ be a fibration with $X$, $B$, and $F$ manifolds. There is a long-exact sequence of homotopy groups $$\cdots\rightarrow\pi_n(F)\rightarrow\pi_n(X)\rightarrow\pi_n(B)\rightarrow\pi_{n-1}(F)\rightarrow\cdots.$$ Also, if $G$ is a Lie group and $H$ is a closed subgroup, then $G\rightarrow G/H$ is a fibration with fiber $H$. So, you should get $$\cdots\rightarrow\pi_n(H)\rightarrow\pi_n(G)\rightarrow\pi_n(G/H)\rightarrow\pi_{n-1}(H)\rightarrow\cdots.$$ In your examples, $H$ is a finite discrete subgroup, so $\pi_0(H)=H$ is its only non-zero homotopy group. Exactness considerations should allow you to compute your homotopy groups.
• Thanks Peter. Can you see'' the answer without doing computation? physicists rely a lot on intuition. Can you tell my answer is right or wrong? Dec 18, 2013 at 4:12
While the long exact sequence of homotopy groups for fibrations is the right tool in general, here we can use more elementary arguments. Note that $SU(2)$ is the universal cover of $SU(2)/Z_N$ (the quotient map is a covering map) so that
• $\pi_1(SU(2)/Z_N)=Z_N$
• $\pi_k(SU(2)/Z_N)\cong \pi_k(SU(2))$ (any continuous map $S^k\to SU(2)/Z_N$ lifts to one $S^k\to SU(2)$), for $k>1$.
Thus, $SU(2)$ being a copy of $S^3$, $\pi_2(SU(2)/Z_N)$ is trivial and $\pi_3(SU(2)/Z_N)\cong\mathbb{Z}$.