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What is known about isometric actions on $\mathbb S^n$ such that the quotient space is homeomorphic to $\mathbb S^n$?


  • I am mostly interested in (maybe trivial) properties of such actions for large $n$.
  • I see that the orientation preserving part of Coxeter's group has this property.
  • Originally I thought that any such action is generated by rotations around $\mathbb S^{n-2}$'s; now I see that there are other examples for $\mathbb S^3$; thanks to Lee Mosher.
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Your $S^n$'s didn't come out well; I put them into TeX. – Lee Mosher Jul 25 '12 at 13:24
I prefer this way, it visible on the question list. – Anton Petrunin Jul 25 '12 at 13:43
It's probably browser dependent. On my browser it looks the same in the question list as on the actual page, which is to say, indecipherable, like a little box with six binary digits. – Lee Mosher Jul 25 '12 at 13:55
That's why, as you'll see, I reverted it briefly, I thought it might look better in the question list. But it didn't. – Lee Mosher Jul 25 '12 at 13:56
I just learned that this question is very close to conjecture on p.9 of Lectures on orbifolds and reflection groups by Michael W. Davis, – Anton Petrunin Jul 30 '12 at 12:20
up vote 11 down vote accepted

I think I have completely answered the question in the following form:

Theorem. For a finite subgroup $\Gamma < O(n)$ the quotient space $S^{n-1}/\Gamma$ is homeomorphic to $S^{n-1}$ if and only if $\Gamma$ has the form \begin{eqnarray*} \Gamma = \Gamma_{ps} \times P_1 \times \ldots \times P_k \end{eqnarray*} for a pseudoreflection group $\Gamma_{ps}$ and Poincaré groups $P_i<SO(4)$, $i=1,\ldots,k$, such that the factors act in pairwise orthogonal spaces and such that $n>5$ if $k=1$.


Here, a pseudoreflection group is understood in the sense of Mikhailova and a Poincaré $P$ group comes from the binary icosahedral group in $SU(2)$, i.e. $S^3/P$ is Poincaré's homology sphere.

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Your question translates into the language of orbifolds as saying: what is known about spherical $n$-orbifolds with underlying space homeomorphic to $S^n$?

In $S^2$, the examples you give are all there are.

Orbifolds with the geometry of $S^3$ were enumerated by William Dunbar in his thesis. His published paper MR1118824 contains the enumeration of the 21 oriented $S^3$-orbifolds which do not have a circle fibration over a 2-orbifold. The equivalence relation here is up to orientation preserving isometry; if you allow orientation reversing isometry then the list is cut down somewhat. Each of the 21 has underlying space homeomorphic to $S^3$. At the end of Dunbar's paper you will see that exactly 8 of the 21 are orientable double covers of Coxeter group quotients, with the corresponding Dynkin diagrams listed out explicitly. That leaves 13 examples as you ask for in $S^3$.

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PDF download link for Dunbar's Transactions paper:… – Joseph O'Rourke Jul 25 '12 at 18:12

In the following article of M.A.Mikhailova (М.А. Михайлова)

Изв. АН СССР. Сер. матем., 48:1 (1984)


it is proven:

Theorem. The quotient $\mathbb R^n/G$ by a linear action of a finite group $G$ is homeomorphic to $\mathbb R^n$ if and only if $G$ is generated by pseudo-reflections (i.e, rotations of $\mathbb R^n$ that fix a subspace of codimension 2).

The proof relies on a complete classification of finite groups generated by pseudo-reflections (there is a reference to this classification at the end of the article)

(there should be of course an English translation of this article, but I can not find it now).

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She probably wanted to say "PL-homeomorphic" otherwise cone over double suspension over Poincare sphere would be a counterexample. – Anton Petrunin Jul 29 '12 at 14:45
The article is available at for real money. I was able to get a copy through the proxy server of my university library. – Lee Mosher Jul 29 '12 at 19:26

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