$\pi_j(G_{cont}/G_{finite})$

27 Views Asked by At

For the quotient group $$G'=G_{cont}/G_{finite},$$ knowing the homotopy groups of $G_{cont}$ and $G_{finite}$, one can determine homotopy groups from the long exact sequence

$$ ... \to \pi_n(G_{finite}) \to \pi_n(G_{cont}) \to \pi_n(G') \to \pi_{n-1}(G_{finite}) \to \pi_{n-1}(G_{cont}) \to \pi_{n-1}(G') \to .... $$ But in practice, one needs to be careful about the map.

Let us take one example,

$$G'=SU(N)/\mathbb{Z}_N,$$

where $\mathbb{Z}_N$ is a normal subgroup of $SU(N)$.

$$ ... \to \pi_1(\mathbb{Z}_N) \to \pi_1(SU(N)) \to \pi_1(G') \to \pi_{0}(\mathbb{Z}_N) \to \pi_{0}(SU(N)) \to \pi_{0}(G') $$

Are the following statements correct:

$$ ... \to \pi_1(\mathbb{Z}_N)=0 \to \pi_1(SU(N))=0 \to \pi_1(G') \to \pi_{0}(\mathbb{Z}_N)=\mathbb{Z}_N \to \pi_{0}(SU(N))=0 \to \pi_{0}(G') $$

so

$$ ... \to 0 \to 0 \to \pi_1(G') \to \mathbb{Z}_N \to 0 \to \pi_{0}(G') $$

Questions:

  1. $\pi_{0}(G')=0?$

  2. $\pi_{1}(G')=\mathbb{Z}_N?$

  3. $\pi_{2}(G')=\pi_{2}(SU(N))?$

  4. $\pi_{j}(G')=\pi_{j}(SU(N))?$, for $j\geq 2.$

Am I correct for the above equalities?