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I'm looking to compute the fundamental group of a couple of different quotients of the $n$-torus. The first of these I'm interested is the space $\mathbb{T}^n/S_n$ where the symmetric group $S_n$ acts on $\mathbb{T}^n$ by permuting the elements: If $\sigma \in S_n$ then $\sigma\cdot(x_1, \ldots, x_n) = (x_{\sigma(1)}, \ldots, x_{\sigma(n)})$. The other spaces are $\mathbb{T}^{n-1}/S_n$ and $\mathbb{T}^{n-1}/(S_n \times \mathbb{Z}_2)$. These have all come from a couple papers I'm reading and the explanation for the latter two seems a little more cryptic. Here's an excerpt:

The $\mathbb{T}^{n-1}/S_n$ is the quotient of an $(n-1)$-simplex whose boundary is singular, by the rigid transformation cyclically permuting its vertices. The space $\mathbb{T}^{n-1}/(S_n \times \mathbb{Z}_2)$ is the quotient of the [preceding] space by an additional reflection.

Unfortunately my skill in computing the fundamental groups of spaces is not very good at the moment. I'm pretty sure that $S_n$ is a covering space action on $\mathbb{T}^n$ and that the quotient map $q:\mathbb{T}^n \to \mathbb{T}^n/S_n$ sending each point to its orbit is a covering map of the resulting space. Furthermore, I know there needs to be a relationship between the automorphism group of $\mathbb{T}^n$ with respect to $q$, and the fundamental group $\mathbb{Z}^n$ of $\mathbb{T}^n$ in relation to the quotient space.

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  • $\begingroup$ The quotient map is not a covering map. It fails to be a cover on the locus where some of the coordinates of a point in the $n$-torus are the same. $\endgroup$ – Qiaochu Yuan Apr 22 '15 at 6:13
  • $\begingroup$ @QiaochuYuan Ok, what you're saying makes sense. Is there an overall strategy you would recommend to approaching this problem? What facts/constructions regarding covering spaces and group actions would you say are most relevant? $\endgroup$ – Mnifldz Apr 22 '15 at 6:23
  • $\begingroup$ Taking quotients by anything other than a free action is a delicate procedure and I don't know if there's anything that can be said in general about how it interacts with taking the fundamental group. In this particular case you could try looking at the locus on which the action is free and then trying to figure out what the non-free bits add in. $\endgroup$ – Qiaochu Yuan Apr 22 '15 at 6:29
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    $\begingroup$ @QiaochuYuan: Chapter 11 of my book Topology and Groupoids is on orbit spaces and orbit groupoids. In particular, it proves that for a discontinuous action on a Hausdorff space which is locally nice, then the fundamental groupoid of the orbit space is isomorphic to the orbit groupoid of the fundamental groupoid. The one calculation given is that of the fundamental group of the symmetric square of a space. So I am hoping for more applications! $\endgroup$ – Ronnie Brown Apr 22 '15 at 9:15
  • $\begingroup$ An older version of this chapter on orbit grouopids is available at . arXiv:math/0212271. $\endgroup$ – Ronnie Brown Apr 22 '15 at 9:38
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Related to QiaochuYuan's comment. It might seem a little tricky but maybe you can use something out of it. If you do not want to go through orbifold maybe this discussion could help you :

$\pi_1$ and $H_1$ of Symmetric Product of surfaces

Instead of looking topologically $\mathbb{T}^n/S_n$ you could say it is an orbifold (i.e. almost everywhere a manifold but with some points being singular : their neighborhood appears as finite quotient $V/G$ where $V$ is a usual chart and $G$ a finite group acting on it, see Thurston : http://library.msri.org/books/gt3m/PDF/13.pdf or this paper https://people.math.osu.edu/davis.12/papers/lectures%20on%20orbifolds.pdf). The good thing about it, is that when $\Gamma$ acts properly (and not only freely) on a manifold $M$ then $M/\Gamma$ is an orbifold (ok we do not win anything here). Using some Galois theory of orbifold coverings you get that :

$$1\rightarrow \pi_1^{Orb}(\mathbb{T}^n)\rightarrow \pi_1^{Orb}(\mathbb{T}^n/S_n)\rightarrow Aut(q)\rightarrow 1 $$

Moreover $\pi_1^{Orb}$ is constructed as the automorphism group of the universal orbifold cover so that it coincides with the topological notion on manifold. It is easy to see that $Aut(q)$ will still be $S_n$ so that :

$$1\rightarrow \mathbb{Z}^n\rightarrow \pi_1^{Orb}(\mathbb{T}^n/S_n)\rightarrow S_n\rightarrow 1 $$

So the $\underline{\text{orbifold}}$ fundamental group of $\mathbb{T}^n/S_n$ appears as an extension of $S_n$ by $\mathbb{Z}^n$.

I truly think that its orbifold fundamental group will actually be :

$$\mathbb{Z}^n\rtimes_{\phi} S_n $$

Where $\phi(\sigma)(a_1,...,a_n)=(a_{\sigma(1)},...,a_{\sigma(n)})$. It is not so hard to see why. I claim that the extension I wrote above is splitted.That is I can find a subgroup of $\pi_1^{Orb}(\mathbb{T}^n/S_n)$ isomorphic to $S_n$ whose intersection with $\mathbb{Z}^n$ is trivial. Indeed, by definition :

$$\pi_1^{Orb}(\mathbb{T}^n/S_n)=Aut(\text{universal cover of }\mathbb{T}^n/S_n) $$

But the universal cover (from a universal property) of $\mathbb{T}^n/S_n$ must be the same as the universal cover of $\mathbb{T}^n$. Hence to look at $\pi_1^{Orb}(\mathbb{T}^n/S_n)$ it suffices to understand the automorphism group of this universal cover :

$$p: \mathbb{R}^n\rightarrow \mathbb{T}^n/S_n $$

$$(x_1,...,x_n)\mapsto (x_1\text{ mod } \mathbb{Z},...,x_n\text{ mod } \mathbb{Z})\text{ mod } S_n $$

I claim that for $\sigma\in S_n$ :

$$\phi_{\sigma}: \mathbb{R}^n\rightarrow\mathbb{R}^n$$

$$(x_1,...,x_n)\mapsto (x_{\sigma(1)},...,x_{\sigma(n)})$$

is in $Aut(p)$ and that $\sigma\mapsto \phi_{\sigma}$ is an injective function which is a morphism (or maybe an antimorphism but it does not change anything here) and gives a subgroup $H$ of $\pi_1^{Orb}(\mathbb{T}^n/S_n)$ isomrphic to $S_n$. Clearly $\phi(\sigma)$ (unless $\sigma$ is trivial) won't fix fibers over $\mathbb{T}^n$ hence $H\cap\mathbb{Z}^n$ is trivial. It is not hard to see that the action by conjugation of $H$ over $\mathbb{Z}^n$ is the one I gave above and we are done.

What remains is to relate $\pi_1^{Orb}(\mathbb{T}^n/S_n)$ and $\pi_1(\mathbb{T}^n/S_n)$. There is a well known surjection of $\pi_1^{Orb}(\mathbb{T}^n/S_n)$ on $\pi_1(\mathbb{T}^n/S_n)$ (essentially, looking at orbifold curves and forgetting that they are orbifold to just look at them as topological curves of the underlying topological space). The kernel of this map can be seen as the subgroup generated by local isotropies (to do this you must know the interpetation of $\pi_1^{Orb}$ as orbifold loops around a point modulo some homotopy relation). Using this we get a surjective group morphism :

$$\psi : \pi_1^{Orb}(\mathbb{T}^n/S_n)\rightarrow \pi_1(\mathbb{T}^n/S_n) $$

And a good understanding of this morphism gives you that :

$$Ker(\psi)=<<H>> $$

The group generated by all the conjugates of $H$. It is then easily seen (working in the semi-direct product structure) that :

$$Ker(\psi)=<((a_1-a_{\sigma(1)},...,a_n-a_{\sigma(n)}),\sigma)> $$

To finish if we denote $K$ to be the kernel of :

$$\mathbb{Z}^n\rightarrow \mathbb{Z} $$

$$(a_1,...,a_n)\mapsto \sum_{i=1}^na_i $$

I claim that $Ker(\psi)=K\rtimes_{\phi} S_n$. First $Ker(\psi)\cap \mathbb{Z}^n$ is easily seen to be in $K$, secondly $Ker(\psi)$ surjects on $\mathfrak{S}_n$. So the non-trivial thing to do is just to show that $K\subseteq Ker(\psi)\cap \mathbb{Z}^n$ and this because $Ker(\psi)\cap \mathbb{Z}^n$ contains $(1,0,...,0,-1,0,...,0)$ where the $-1$ is at the $i$-th place for all $i$. To do this just realize that :

$$((1,0...,0,-1,0,...,0),(1,i))\text{ and } ((0,...,0),(1,i))\text{ are in } Ker(\psi)$$

Hence the product is in $Ker(\psi)$ so :

$$((1,0...,0,-1,0,...,0),(1,i)).((0,...,0),(1,i))=((1,0...,0,-1,0,...,0),(1,i)^2)=((1,0,...,0,-1,0,...,0),Id)\in Ker(\psi)\cap \mathbb{Z}^n$$

This shows that :

$$Ker(\psi)=K\rtimes_{\phi}S_n $$

Finally the quotient of $\pi_1^{Orb}(\mathbb{T}^n/S_n)$ by $Ker(\psi)$ is easily seen to be isomorphic to $\mathbb{Z}^n/K=\mathbb{Z}$ hence :

$$\pi_1(\mathbb{T}^n/S_n)=\mathbb{Z} $$

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  • $\begingroup$ Thank you for your incredibly detailed answer. I'm tempted to accept this as an answer, but my issue is that I don't understand all of the constructions you presented: I'm very new to covering spaces, haven't learned homology yet, and only have a minimal grasp on orbifolds. This question has a much more involved answer than I anticipated. I will definitely take the time to read the resources you presented as well as your proof. Thanks! $\endgroup$ – Mnifldz Apr 22 '15 at 17:22
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Let us denote the orbit space obtained from the $S_n$-action on $X \times X \times \cdots \times X$ as $SP^n(X)$.

Given a continuous map $f : X \to Y$, $SP^n$ induces a map $SP^n(f) : SP^n(X) \to SP^n(Y)$. In particular, $SP^n : \mathbf{Top} \to \mathbf{Top}$ is a functor. It can be proved that $SP^n$ takes homotopic maps to homotopic maps, which means if $X \simeq Y$ then $SP^n(X) \simeq SP^n(Y)$.

Claim : $SP^n(S^1)$ is homotopy equivalent to $S^1$.

Consider the map $f : SP^n(\Bbb C^*) \to \Bbb C^n$ given by $[z_1, z_2, \cdots, z_n] \mapsto (\sigma_1, \sigma_1, \cdots, \sigma_n)$, where $\sigma_i$s are the elementary symmetric polynomials of $z_1, \cdots, z_n \in \Bbb C^*$.

We see that the map is surjective away from the codimension $1$ subspace of points in $\Bbb C^n$ with last coordinate zero. The map $SP^n(\Bbb C^*) \to \text{im} f \cong \Bbb C^n - \Bbb C^{n-1}$, however, is provably a homeomorphism. Thus, $SP^n(S^1) \simeq SP^n(\Bbb C^*) \cong \Bbb C^n - \Bbb C^{n-1} \cong\Bbb C^{n-1} \times \Bbb C^*$ which is homotopy equivalent to $S^1$.

Thus, $\pi_1(SP^n(S^1)) \cong \Bbb Z \;\; \blacksquare$

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