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(?).

How about:

(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.

  • 2
    $\begingroup$ 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. $\endgroup$ Dec 17 '13 at 23:58
  • $\begingroup$ 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.) $\endgroup$
    – wonderich
    Dec 18 '13 at 0:01
  • $\begingroup$ This question does not appear to be about research level mathematics, and may be more appropriate at math.stackexchange. $\endgroup$
    – Ricardo Andrade
    Dec 18 '13 at 0:06
  • $\begingroup$ that is fine. I can ask at math.stackexchange. $\endgroup$
    – wonderich
    Dec 18 '13 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.

  • $\begingroup$ 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? $\endgroup$
    – wonderich
    Dec 18 '13 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 $\pi_3(SU(2)/Z_N)\cong\mathbb{Z}$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.