Let $X$ be a compact manifold with $\pi_1(W)=S_3$, $T$ be an $n$-dimensional complex torus. Denote the universal cover of $W$ by $\tilde{W}$. Let $S_3$ act on $(T\times T\times T) \times \tilde{W}$ by permutation on the first part and by deck transformation on the second. Now we consider $Y:=((T\times T\times T) \times \tilde{W})/S_3$.

I feel confused with the following:

We can take $\tilde{Y}$ as the $6$-fold covering of $Y$ corresponding to the surjection $\pi_1(Y)=\mathbb{Z}^{2n}\times \mathbb{Z}^{2n}\times \mathbb{Z}^{2n}⋊ S_3\to S_3$.

What I know is a covering map $p:\tilde{Y}\to Y$ induces an injection $p^*:\pi_1(\tilde{Y})\to\pi_1(Y)$, I don't understand what exactly the $6$-fold covering corresponding to the surjection from is.

Also is it obvious that $\pi_1(Y)=\mathbb{Z}^{2n}\times \mathbb{Z}^{2n}\times \mathbb{Z}^{2n}⋊ S_3$?

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    $\begingroup$ Consider the fibration $T \times T \times T \to (T \times T \times T \times \widetilde{W})/S_3 = Y \stackrel{\pi}{\to} W$. Note that diagonal $\Delta \subset T \times T \times T$ is fixed under this $S_3$-action, so pick any point $(x, x, x) \in T \times T \times T$, and note that the quotient $T \times T \times T \times \widetilde{W} \to Y$ restricted to $(x, x, x) \times \widetilde{W}$ is the covering projection $\widetilde{W} \to W$, so this gives a section $W \to Y$ to $\pi$ by including this copy of $W$ in $Y$. Run the homotopy exact sequence to get a split short sequence of $\pi_1$'s $\endgroup$ – Balarka Sen Dec 29 '18 at 16:09

If $X$ is a $G$-space and $G$ acts properly discontinuously and freely on a path-connected space $Y$, then we have the fiber bundle $X \to (X \times Y)/G \to Y/G$ on which running the homotopy long exact sequence gives $$\cdots \to \pi_{n+1}(Y/G) \to \pi_n X \to \pi_n ((X \times Y)/G) \to \pi_n(Y/G) \to \cdots $$ If $x \in X$ is a fixed point under the action of $G$, then given the quotient map $q : X \times Y \to (X \times Y)/G$, $q|_{x \times Y}$ is the covering projection $Y \to Y/G$. This gives a canonical copy of $Y/G$ sitting inside $(X \times Y)/G$, and the inclusion map $s : Y/G \hookrightarrow (X \times Y)/G$ constitutes a section of the said fiber bundle.

Then $s_* : \pi_n(Y/G) \to \pi_n((X \times Y)/G)$ is a right-inverse to $\pi_n((X \times Y)/G) \to \pi_n(Y/G)$ for every $n$, forcing these maps to be surjection, and the long exact sequence above to break into short exact pieces $0 \to \pi_n(X) \to \pi_n((X \times Y)/G) \to \pi_n(Y/G) \to 0$ for every $n$. If $n = 1$, then we have a split short exact sequence

$$0 \to \pi_1(X) \to \pi_1((X \times Y)/G) \to \pi_1(Y/G) \to 0$$

This establishes $\pi_1((X \times Y)/G)$ as a semidirect product $\pi_1(X) \rtimes \pi_1(Y/G)$.

In this particular case, take $X = T \times T \times T$ and $Y = \widetilde{W}$, letting $G = S_3$ act on the former by permuting the factors and on the latter by deck transformations. This $S_3$-action has a fixed point in $X$, just choose any point $(x, x, x) \in T \times T \times T$ in the diagonal. Then $\pi_1((T \times T \times T \times \widetilde{W})/S_3)$ is isomorphic to $(\Bbb Z^{2n})^3 \rtimes S_3$, as required. The semidirect product is taken with respect to the homomorphism $S_3 \to \text{Aut}((\Bbb Z^{2n})^3)$ permuting the three coordinate $\Bbb Z^{2n}$s, induced from the $S_3$-action on $T \times T \times T$ as described before.

  • $\begingroup$ Can you explain why $(X\times Y)/G\to Y/G$ is a fiber bundle with fiber $X$? $\endgroup$ – 6666 Dec 30 '18 at 13:01
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    $\begingroup$ The projection map is obtained from $X \times Y \to Y/G$ which first projects to $Y$, then projects to $Y/G$ by the covering map. This is a fiber bundle with fiber $X \times G$. By universal property, this map factors through $(X \times Y)/G \to Y/G$ which has fibers $(X \times G)/G = X$; that it is locally trivial follows from the properly discontinuous action of $G$. It's a good exercise to check all the details here. $\endgroup$ – Balarka Sen Dec 30 '18 at 13:20
  • $\begingroup$ $\pi_1(Y)=\mathbb{Z}^{2n}\times \mathbb{Z}^{2n}\times \mathbb{Z}^{2n}⋊ S_3\to S_3$ in your sense is corresponding to the induced map of $(X\times Y)/G\to Y/G$, however, why is this $6$-fold? $\endgroup$ – 6666 Jan 1 at 17:01
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    $\begingroup$ @6666 Your questions do not make sense to me. Whenever you have a space $Y$, equipped with a surjective homomorphism $\rho: \pi_1 Y \to G$, there is a unique covering space $Y' \to Y$ so that $\text{ker}(\rho) = \pi_1 Y'$, and furnishes us with an isomorphism $\text{Deck}(Y') \cong G$. This is the main theorem of covering space theory, see Hatcher ch 1.3. One explicit formula is that if $\tilde Y$ is the universal cover, then $\tilde Y \times_{\pi_1 Y} G \cong Y'$, where we use $\rho$ to give an action of $\pi_1 Y$ on $G$. This is all one is expected to be able to say about $Y'$. $\endgroup$ – user98602 Jan 1 at 17:18
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    $\begingroup$ @6666 All they are doing is defining $Y'$ (what they call $\tilde Y$, but I reserve that notation for universal cover), and they are defining it by the property I quote. $\endgroup$ – user98602 Jan 1 at 17:29

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