# Conditions for underlying space of an orbifold $\Bbb T^n/\Gamma$ to be a sphere?

Given a $n$-dimensional torus, is it always possible to find a discrete action to produce an orbifold such that its underlying space is the $n$-dimensional sphere? Or does it only happens for specific dimensions? Is there some particular property of the action $\Gamma$ -which I expect to be some realization of $\Bbb Z_n$- that I should check?

Bonus question: And to be a projective plane? I am thinking that according Arnold-Kuiper-Massey theorem one can see $\Bbb CP^2$ as a branched covering of $S^4$, so if $T^4$ could be a branched covering of $\Bbb CP^2$ we would have an interesting way to cover $S^4$

The personal motivation for this is, as usual, more fuzzy that the question itself but it could add some heat to the comments: I am interested on how to sequence manifolds in a way that increases some continuous symmetry. For manifolds with a metric the two extreme cases in a given dimension would be the n-Torus, where you can act with group $U(1)^n$ and the n-sphere, where you act with $SO(n+1)$. Of course the sphere that comes out from orbifolding is not exactly the sphere, one must consider the conical singularities and all the metric properties go amok, but still it is, to me, a hint of how do symmetries appear and disappear. For physics this is specially funny in dimension seven because you have also the the manifold $CP^2\times S^3$ where the isometry group has 14 generators and then you could try to produce a sequence 7 -> 14 -> 28

EDIT: It seems than for n=2 the physicists call "pillows" to the spheres, in order to stress the existence of singular points. So the question is, when is the orbifold a n-pillow?

• $T^{2n}$ can be realized as a branched covering of $\mathbb{C}\mathbb{P}^n$ for all $n$: take a $n$-dimensional abelian variety embedded in some $\mathbb{C}\mathbb{P}^N$ and a generic projection to $\mathbb{C}\mathbb{P}^n$. – abx Aug 4 '15 at 13:27
• @abx nice, so with $T^7 = T^4 \times T^3$ mi "motivation" has at least one answer if one can cover $S^3$; it seems I should look for generic families of actions. – arivero Aug 4 '15 at 14:03
• What about the action of $A_n$ on $T^{n-1}$ by its $n-1$-dimensional standard representation. – Will Sawin Aug 8 '15 at 14:26
• @WillSawin I am a bit out of shape now ans I can not see how to prove that such action produces an sphere... could you expand? – arivero Aug 11 '15 at 2:26

Here is a theorem due to Armstrong:

A group $G$ of homeomorphisms of a topological space $X$is called discontinuous if (1) the stabilizer of each point of $X$ is finite, and (2) each point $\chi \in X$ has a neighborhood $U$ such that any element of $G$ not in the stabilizer of $\chi$ maps $U$ outside itself. Theorem: Let $G$ be a discontinuous group of homeomorphisms of a path connected, simply connected, locally compact metric space $X,$ and let $H$ be the normal subgroup of $G$ generated by those elements which have fixed points. Then the fundamental group of the orbit space $X/G$ is isomorphic to the factor group $G/H.$

(the above is actually the math review of

Armstrong, M. A. The fundamental group of the orbit space of a discontinuous group. Proc. Cambridge Philos. Soc. 64 1968 299–30 )

So, in order for the (underlying topological space of the) quotient of $\mathbb{R}^n$ to be simply connected, it must be generated by elements with fixed points (and vice versa). Of course, a quotient of $\mathbb{T}^n$ is also a quotient of $\mathbb{R}^n,$ so this applies to your question. On the other hand, simply connected does not a sphere make, but figuring out the higher homology groups of the quotient should be relatively easy, and if they are all trivial, then the quotient is a sphere by the Poincare conjecture.

(By the way, the reason I know of Armstrong's theorem is that it was used by Colin Maclachlan to show that the (underlying topological space of the) moduli space of Riemann surfaces is simply connected (since the mapping class group acts discontinuously, and is generated by torsion).

A little example Of course, in two dimension, the standard example is the quotient of the torus $\mathbb{T}^2$ by the elliptic involution, which is the sphere with four distinguishe (Weierstrass) points.

A bigger example If I understand correctly, any projective variety has a branch covering map to $\mathbb{P}^n.$ (see, e.g., https://math.stackexchange.com/questions/679768/projective-noether-normalization). Since there are plenty of projective tori running around, you always have branched cover maps to $\mathbb{CP}^n.$

Three dimensions This case was completely analyzed by Bill Dunbar in his thesis. It seems that most torus quotients with singular locus have $S^3$ as their underlying topological space.

more examples Will Sawin's example for $n=2$ gives $S^1.$ For $n=3$ it gives $S^2,$ with three singular points of cone angle $2\pi/3,$ so a doubled regular triangle. In dimension $3,$ the nicest example of a branched cover of $S^3$ by $T^3$ is the orbifold whose branch locus is the Borromean rings (see page 81 in this 2008 paper by Montesinos, which is highly recommended in general), with branching index $2.$ This might be the Sawin example for $n=4,$ but I am not completely certain.

Is it a manifold? In a comment discussion with Will Sawin, we agreed that a key part of the question is whether the underlying topological space is actually a manifold. This is equivalent to the quotient of the link of every branch point being a topological sphere (it is always a rational homology sphere). Now, the dimension of the link is $n-1,$ so since every rational homology sphere in dimensions $0, 1, 2$ is an actual sphere, the question is vacuous for $n=1, 2, 3.$ For $n=4,$ it becomes nontrivial. Luckily, spherical $3$-orbifolds have been studied by Bill Dunbar. Most are, indeed, topological spheres, see:

MR1118824 (94c:57022) Reviewed Dunbar, William D.(1-PASE) Nonfibering spherical 3-orbifolds. Trans. Amer. Math. Soc. 341 (1994), no. 1, 121–142.

but there is more Notice that every element of $SO(2n-1)$ has fixed points. Therefore, for a five dimensional orbifold (assuming the point stabilizer action is linear, which is certainly true for a crystallographic action on $\mathbb{T}^5,$ the vertex links are simply connected. They are not necessarily manifolds, but if they are, they are simply connected rational homology spheres of dimension 4. Does this mean they are spheres? This merits anther question...

• Hmm having it for n=1,2,3 and possibly 4 starts to signal that dimension is not a main worry. It could have still some argument about having only conical singularities or more extended ones. – arivero Aug 8 '15 at 23:46
• well the underlying manifold must be a manifold, yep, but for the orbifold of course I am happy to allow for conical singularities and usual stuff (I just mention it for benefit of the casual passer-by) – arivero Aug 12 '15 at 17:21
• I think you could be overlooking one of the conditions of Dunbar when you say 'most' quotients of $T^3$ have underlying space $S^3$. In fact, there are several quotients of $T^3$ with underlying space $S^1\times S^2$, for example the product of the pillowcase $S^2(2,2,2,2)$ and $S^1$. These all turn out to be fibered orbifolds, so Dunbar doesn't focus on them. However, it should be pointed out that $S^2(2,2,2,2) \times S^1$ can cover itself, so 'most' perhaps might be 'most (up to homeomorphism)'. – Neil Hoffman Oct 14 '15 at 0:14

I am just expanding abx's comment: it is possible to realize this branched cover as a group quotient. If abx prefers to explain this, I am happy to delete this answer.

Denote by $\mu_2$ the $2$-element group with an identity element $1$ and a non-identity element $-1$. There is an action of $\mu_2$ on $T^2$ by $-1\cdot (a,b) = (-a,-b)$. The quotient is $\mathbb{CP}^1$ considered as the projective space of nonzero, degree $1$, homogeneous polynomials in variables $x$ and $y$ up to scaling, $$q:T^2 \to \mathbb{CP}^1, \ \ q(a,b) = [r(a,b)y-s(a,b)x].$$ Now, for arbitrary $n$, let $\Gamma$ be the wreath product group $$\Gamma = \left( \mu_2 \right)^n \rtimes \mathfrak{S}_n,$$ where $\mathfrak{S}_n$ is the symmetric group on $n$ letters. For an element $(g_1,\dots,g_n)\in (\mu_2)^n$ and for $\sigma\in \mathfrak{S}_n$, the element $\left((g_1,\dots,g_n),\sigma\right)\in \Gamma$ acts on $(T^2)^n$ in the evident manner, $$((g_1,\dots,g_n),\sigma)\cdot ((a_1,b_1),\dots,(a_n,b_n)) = \left(g_1\cdot (a_{\sigma(1)},b_{\sigma(1)}),\dots, g_n\cdot(a_{\sigma(n)},b_{\sigma(n)})\right).$$ The quotient of this group action is $\mathbb{CP}^n$ considered as the projective space of nonzero, degree $n$, homogeneous polynomials in variables $x$ and $y$ up to scaling , $$q_n:T^{2n} \to \mathbb{CP}^n, \ \ q_n\left((a_1,b_1),\dots,(a_n,b_n)\right) =$$ $$\left[(r(a_1,b_1)y-s(a_1,b_1)x)\cdots(r(a_n,b_n)y-s(a_n,b_n)x)\right].$$

• Thanks! I am still looking forward for answers on the spheres, but this $\mathbb CP^n$ stuff is interesting too. I wonder if both this quotient and the A-K-M theorem could also be understood in the context of orbifolds. – arivero Aug 4 '15 at 14:37
• This quotient is an orbifold structure on $\mathbb{CP}^n$. What would you like to know about this orbifold? – Jason Starr Aug 4 '15 at 14:41
• @Json Ah, the usual stuff: singularities, conic points... and same for $\mathbb C \mathbb P^2$ down to $S^4$. General understanding :-) – arivero Aug 4 '15 at 14:45
• This construction is neat. In particular, if I understand well, for $n=2$ and combined with the fact that the 4-sphere is a $\mathbb{Z}_2$-quotient of $\mathbb{C}P^2$ (ncatlab.org/nlab/show/…) this gives a flat orbifold whose coarse space the the 4-sphere. This must be in textbooks, in citable form. Does anyone have a pointer? – Urs Schreiber Jul 28 '20 at 13:34
• @UrsSchreiber Unfortunately I do not know a citation for this. – Jason Starr Jul 29 '20 at 8:08

Let me clarify my comment. $S_n$ acts on $T^n$ in the obvious way, and preserves the multiplication map $T^n \to T$, so acts on the fiber $T^{n-1}$ in a natural way. I think the quotient by this action is a simplex, the Weyl alcove of type $A_{n-1}$.

The quotient by $A_n$ is a double cover of this. So it is two simplices glued together. I hope that they are glued together in the obvious way, to form a sphere. But I don't see immediately how to see this.

• Thanks, it is more clear now :-) At least it could work for S^3; I will try to visualize the general case. – arivero Aug 11 '15 at 10:02
• Why do you want to go to the fiber? – Igor Rivin Aug 11 '15 at 18:59
• @Igor Rivin otherwise the quotient has nontrivial fundamental group and thus is not a simplex. – Will Sawin Aug 11 '15 at 19:51
• @IgorRivin For instance because $\mathbb Z^n \rtimes S_n$ is not generated by elements with fixed points, because there is a homomorphism to $\mathbb Z$ where you sum up the $n$ elements of $S^n$ and everything with a fixed point is in the kernel. – Will Sawin Aug 11 '15 at 20:14
• Fair enough. How do you know that once you go to the fiber, this problem goes away? – Igor Rivin Aug 11 '15 at 20:28