$\pi_n(SU(2)/Z_N)\simeq?$, $\pi_n(SO(3)/Z_N)\simeq?$, $\pi_n(U(1)/Z_N)\simeq?$

Question

So based on the tool, I have attempted to compute the following threes homotopy groups.

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

$\pi_n(SO(3)/Z_N)\simeq?$

$\pi_n(U(1)/Z_N)\simeq?$

With a fixed positive integer $N$, and for any positive integer $n\geq1$.

Right now I would appreciate someone stated explicitly what is the exact answer. Many thanks.

Answer 1

The answer is trivial in the following sense. Denote ${\bf Z} /n {\bf Z}$ by $H$ and let $G$ be some Lie group. Because $H$ is discrete, the homotopy groups of $G$ and $G/H$ are the same for all $n \geq 2$ (you can get this from the long exact sequence since all homotopy groups of $H$ are trivial). On the other hand there is a short exact sequence $$0 \to \pi_1(G) \to \pi_1(G/H) \to H \to 0.$$ When $G$ is simply connected, as for $G = SU(2)$ this gives an isomorphism $\pi_1(G/H) \cong H$ (this can be obtained also from the standard covering space theory).

Now, for your specific $G$. First, $U(1)$ is trivial, it has no higher homotopy groups and has $\pi_1(U(1)) \cong {\bf Z}$.

The other two groups one can study together since $SU(2)$ is a double cover of $SO(3)$, i.e. $SO(3)/({\bf Z}/n{\bf Z}) \cong SU(2) / ({\bf Z} / 2n {\bf Z})$. Given all the information I've given in the first paragraph, you should be able to figure out most of the homotopy groups.

The only part left to be done is $\pi_n(SU(2))$. But $SU(2)$ has homotopy type of a sphere $S^3$, so you actually want homotopy groups of spheres. These are notoriously hard to compute, but you can find low degree ones in literature.

Answer 2

Let $X$ be a Lie group and $\mathbb{Z}_N$ be embedded as a closed subgroup. For $n\geq 2$, $\pi_n(X/\mathbb{Z}_N)$ is isomorphic to $\pi_n(X)$ because we get a short exact sequence (part of the long exact sequence in homotopy) $$\pi_n(\mathbb{Z}_N)\rightarrow\pi_n(X)\rightarrow\pi_n(X/\mathbb{Z}_N)\rightarrow\pi_{n-1}(\mathbb{Z}_N)$$ which is then $$0\rightarrow\pi_n(X)\rightarrow\pi_n(X/\mathbb{Z}_N)\rightarrow 0$$ for all $n\geq 2$ and hence there is an induced isomorphism between the groups.

For $n=1$ you get the short exact sequence $$\pi_1(\mathbb{Z}_N)\rightarrow\pi_1(X)\rightarrow\pi_1(X/\mathbb{Z}_N)\rightarrow\pi_{0}(\mathbb{Z}_N)\rightarrow \pi_0(X)$$ which is then $$0\rightarrow\pi_1(X)\rightarrow\pi_1(X/\mathbb{Z}_N)\rightarrow\mathbb{Z}_N\to \pi_0(X)$$ and so if $X$ is connected (so $\pi_0(X)=0$) then you have an extension problem to solve (I'll leave that to you).

Note that $U(1)\cong S^1$ the circle and so $\pi_n(U(1))$ is trivial for $n\geq 2$ and isomorphic to the integers for $n=1$.

Note also that $SU(2)\cong S^3$ and so $\pi_n(SU(2))$ is only actually known for low values of $n$ (can be looked up in most texts which mention the homotopy groups of the spheres).

Note also that $SO(3)$ has a double cover by $SU(2)$ so shares its homotopy groups for all $n\geq 2$ and will have a subgroup of index $2$ isomorphic to $\pi_1(SU(2))=0$ for $n=1$. Hence $\pi_1(SO(3))\cong\mathbb{Z}_2$.