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Preparation and my input:

For the quotient space $G/H$, knowing the homotopy groups of $G$ and $H$ one can determine homotopy groups from the long exact sequence

$$ ... \to \pi_n(H) \to \pi_n(G) \to \pi_n(G/H) \to \pi_{n-1}(H) \to \pi_{n-1}(G) \to \pi_{n-1}(G/H) \to .... $$ But in practice, this looks not so easy. I wonder can we use whatever tricks or methods to obtain the homopoty groups of the following quotient space $\pi_n(G/H)$? [With your patience, please see below two questions.]


Let us we take $N=3$ for all $N$.

  1. Consider the group:

$${G}=\frac{SU(N)_A \times SU(N)_{B_1} \times SU(N)_{B_2} \times U(1)}{(\mathbb{Z}_N)^2}\equiv \frac{G_0}{G_1},$$

The $SU(N)$ is a rank-$N$ special unitary matrix (in fundamental representation) containing the identity $\mathbb{I}_N$ (a rank-$N$ identity) matrix.

this group can be understood as a 4-multiplet $$(g_A, g_{B,1},g_{B,2}, e^{i \theta}) \in SU(N)_A \times SU(N)_{B_1} \times SU(N)_{B_2}\times U(1),$$ such that $(e^{i \frac{2\pi}{N}} \mathbb{I}_N, \mathbb{I}_N, \mathbb{I}_N, e^{-i \frac{2\pi}{N}}) \simeq 1$ is identified as a new identity element, thus the first $\mathbb{Z}_N$ generator mod out in $G$, while $( \mathbb{I}_N,e^{i \frac{2\pi}{N}} \mathbb{I}_N,e^{i \frac{2\pi}{N}} \mathbb{I}_N, e^{-i \frac{2\pi}{N}}) \simeq 1$ is identified as another new identity element, thus there is the second $\mathbb{Z}_N$ generator mod out in $G$. Namely (1) the center of $SU(N)_A$, (2) the center of $SU(N)_{B,1}$ together with the center of $SU(N)_{B,2}$, and (3) the $e^{i \frac{2\pi}{N}}\in U(1)$ overlap, thus we only mod out the twice redundant $(\mathbb{Z}_N)^2$. Another way to say this is that the following group elements ($\mathbb{Z}_N$ generator) in the original $G_0$ is identified to be the same element in $G$: $$(e^{i \frac{2\pi}{N}} \mathbb{I}_N, \mathbb{I}_N, \mathbb{I}_N, 1) \simeq (\mathbb{I}_N,e^{i \frac{2\pi}{N}} \mathbb{I}_N, e^{i \frac{2\pi}{N}} \mathbb{I}_N, 1) \simeq (\mathbb{I}_N, \mathbb{I}_N, \mathbb{I}_N,e^{i \frac{2\pi}{N}}).$$

Now consider the subgroup $H$ of the $G$ as:

$$H=\frac{SU(N)_{A,B}}{\mathbb{Z}_N}\times \mathbb{Z}_2=PSU(N)_{A,B} \times \mathbb{Z}_2,$$

where this group can be understood as a doublet $(g_{A,B}, g_1) \in SU(N)_{A,B}\times \mathbb{Z}_2= (SU(N)_{A,B}, \{\pm 1 \}) =H \subset G,$ where there is a one-to-one correspondence between the doublet $$(g_{A,B}, g_1) \in H$$ and the 4-multiplet $$(g_{A,B},g_{A,B}^*,g_{A,B}^*, g_1) \in G$$ with $g_{A,B}=g_A=g_{B,1}^*=g_{B,2}^*$, where $g^*$ means the complex conjugation (without the transpose $T$) of $g$.

Finally, we need $\frac{SU(N)_{A,B}}{\mathbb{Z}_N}$ to mod out ${\mathbb{Z}_N}$, where we consider $(g_{A,B}',g_{A,B}'^*)=(e^{i \frac{2\pi}{N}},e^{-i \frac{2\pi}{N}})\mathbb{I} \in {\mathbb{Z}_N}$; this particular rank-$N$ diagonal matrix is the ${\mathbb{Z}_N}$ we modded out.

We have that $$\frac{SU(N)_{A,B}}{\mathbb{Z}_N} \subset SU(N)_A \times SU(N)_{B_1} \times SU(N)_{B_2} , $$ $$ \mathbb{Z}_2 = \{\pm 1 \} \subset U(1). $$

So this explains how $H$ is embedded as a subgroup in $G$.

Question:

  • What is the precise space of the set of left cosets $$ G/H=? $$ Is this certain smooth homogeneous space like a sphere or a complex/real projective space?

  • What is the homotopy group? $$ \pi_j(G/H)=? $$ for $j=1,2,3,4,5$.


Some background info that I prepared for you:

(1). $\pi_i(U(N))=\pi_i(\frac{SU(N)\times U(1)}{\mathbb{Z}_N})$: $$\pi_m(U(N))=\pi_m(SU(N)), \text{ for } m \geq 2$$ $$\pi_1(U(N))=\mathbb{Z}, \;\;\pi_1(SU(N))=0,$$

(2). $\pi_0(\mathbb{Z}_N)=\mathbb{Z}_N$ and $\pi_i(\mathbb{Z}_N)=0$ for $i\geq 1$.

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  • $\begingroup$ Sorry, isn't $$\pi_0(H)=\pi_0(\frac{SU(N)_{A,B}}{\mathbb{Z}_N}\times \mathbb{Z}_2)={\mathbb{Z}_N}\times \mathbb{Z}_2?$$ Should some of your answer be corrected? $\endgroup$ – annie heart Sep 1 '18 at 3:20
  • $\begingroup$ @Mike Miller, "Note here that H acts trivially on π1G" why is that true? $\endgroup$ – annie heart Sep 1 '18 at 3:26
  • $\begingroup$ The $H$ looks complicated, it is not $G= (G/H) \times H$ for sure. $\endgroup$ – annie heart Sep 1 '18 at 3:26
  • $\begingroup$ I clearly misread some of the notation. I apologize and have deleted my comment. $\endgroup$ – Mike Miller Sep 1 '18 at 3:38
  • $\begingroup$ Dont we have $π1(G)=\mathbb{Z}$ and $\pi_0(H)=\mathbb{Z}_N \times \mathbb{Z}_2$? $\endgroup$ – annie heart Sep 1 '18 at 3:39

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