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For the quotient space $G=G_0/G_1$, knowing the homotopy groups of $G_0$ and $G_1$, one can determine homotopy groups from the long exact sequence

$$ ... \to \pi_n(G_1) \to \pi_n(G_0) \to \pi_n(G_0/G_1) \to \pi_{n-1}(G_1) \to \pi_{n-1}(G_0) \to \pi_{n-1}(G_0/G_1) \to .... $$ But in practice, there are some subtleties.

Consider this particular group:

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

We define $G_0 \equiv SU(N)_A \times SU(N)_{B} \times U(1)$ and $G_1 \equiv {(\mathbb{Z}_N)^2}$.

Question: What is the fundamental group $\pi_1(G)=\pi_1(G_0/G_1)=?$


This group can be understood as a triplet $$(g_A, g_B, e^{i \theta}) \in SU(N)_A \times SU(N)_B \times U(1),$$ such that $(e^{i \frac{2\pi}{N}} \mathbb{I}_N, \mathbb{I}_N, e^{-i \frac{2\pi}{N}})$ is the first $\mathbb{Z}_N$ generator mod out in $G$, while $(\mathbb{I}_N,e^{i \frac{2\pi}{N}} \mathbb{I}_N, e^{-i \frac{2\pi}{N}})$ is the second $\mathbb{Z}_N$ generator mod out in $G$.

The $\mathbb{I}_N$ means the rank-N identity matrix.

Namely (1) the center of $SU(N)_A$, (2) the center of $SU(N)_B$ and (3) the $e^{i \frac{2\pi}{N}}\in U(1)$ overlap, thus we only mod out the twice redundant $(\mathbb{Z}_N)^2$.

In other words, the following three $\mathbb{Z}_N$ generators within $G_0$ are identified: $$ (e^{i \frac{2\pi}{N}} \mathbb{I}_N, \mathbb{I}_N, 1) \simeq ( \mathbb{I}_N, e^{i \frac{2\pi}{N}} \mathbb{I}_N, 1) \simeq ( \mathbb{I}_N, \mathbb{I}_N, e^{i \frac{2\pi}{N}}), $$ that is why we mod out $G_1={(\mathbb{Z}_N)^2}$ out of $G_0$, when we define ${G} \equiv \frac{G_0}{G_1}$ earlier.

Attempt:

One can see that $$ ... \to \pi_1(G_1) =0 \to \pi_1(G_0) = \mathbb{Z} \to \pi_1(G_0/G_1)$$ $$\to \pi_{0}(G_1)=(\mathbb{Z}_N)^2 \to \pi_{0}(G_0)=0 \to \pi_{0}(G_0/G_1)=0 $$ so to speak, $$ 0 \to \mathbb{Z} \to \pi_1(G_0/G_1)\to (\mathbb{Z}_N)^2 \to 0 \to 0, \tag{i} $$ so there is a short exact sequence determine the fundamental group $\pi_1(G_0/G_1)$, this is possible a non-Abelian group? And this extension can be classified by the cohomology group $H^2(B(\mathbb{Z}_N)^2, \mathbb{Z} )=H^1(B(\mathbb{Z}_N)^2, U(1) )=(\mathbb{Z}_N)^2$?

Attempt 2: By the earlier definition, one can rewrite

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

by the fact that $U(N)_A \equiv \frac{SU(N)_A\times U(1)}{(\mathbb{Z}_N)}$, $G_0' \equiv U(N)_A \times SU(N)_{B}$, $G_1' \equiv \mathbb{Z}_N$. Now compute again $$ ... \to \pi_1(G_1') =0 \to \pi_1(G_0') = \mathbb{Z} \to \pi_1(G_1'/G_0') $$ $$\to \pi_{0}(G_1')=\mathbb{Z}_N \to \pi_{0}(G_0')=0 \to \pi_{0}(G)=0 $$ so to speak, $$ 0 \to \mathbb{Z} \to \pi_1(G)=\pi_1(G_1'/G_0')\to \mathbb{Z}_N \to 0 \to 0. \tag{ii} $$ And this distinct extension can be classified by a distinct cohomology group $H^2(B\mathbb{Z}_N, \mathbb{Z} )=H^1(B\mathbb{Z}_N, U(1) )=\mathbb{Z}_N$?

Since

$\pi_1(G)=\pi_1(G_1/G_0)=\pi_1(G_1'/G_0')$,

so the answers from my own Attempt 1 and Attempt 2 must be the same, from two different short exact sequences (i) and (ii). What is the unique answer?

Is $\pi_1(G)=\mathbb{Z}$ or $(\mathbb{Z} \rtimes \mathbb{Z}_N)$, or something else? We can take $N=3$ as a specific example.

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  • $\begingroup$ +1, thanks for the insightful ref - my example shall be a much easier example! $\endgroup$ – wonderich Sep 3 '18 at 20:29
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    $\begingroup$ You mod out by a central subgroup, so the quotient is a Lie group, and in particular has an abelian fundamental group. $\endgroup$ – YCor Sep 3 '18 at 20:40
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    $\begingroup$ As Ycor explained, general theory says the group in question is indeed Abelian. While there may be non-Abelian groups that can fit into the sequence (i), the group you're looking for is Abelian. $\endgroup$ – Dan Ramras Sep 4 '18 at 3:51
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    $\begingroup$ The fundamental group is isomorphic to the preimage $\tilde{G}_1$ of $G_1$ in the universal covering of $G_0$. This is indeed isomorphic to $Z\times (Z/nZ)$. $\endgroup$ – YCor Sep 4 '18 at 6:29
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    $\begingroup$ $A\rtimes B$ means nothing if you don't specify the action of $B$ on $A$. Anyway, since $G$ is abelian, the only possibility is the direct product (which is one instance of semidirect product). By the way, if $N$ is odd, the only semidirect product $Z\rtimes (Z/NZ)$ is the direct product... $\endgroup$ – YCor Sep 4 '18 at 16:06

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