16
$\begingroup$

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

Comments.

  • 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.
$\endgroup$
5
  • 2
    $\begingroup$ Your $S^n$'s didn't come out well; I put them into TeX. $\endgroup$
    – Lee Mosher
    Jul 25, 2012 at 13:24
  • 2
    $\begingroup$ I prefer this way, it visible on the question list. $\endgroup$ Jul 25, 2012 at 13:43
  • $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Jul 25, 2012 at 13:55
  • $\begingroup$ 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. $\endgroup$
    – Lee Mosher
    Jul 25, 2012 at 13:56
  • $\begingroup$ 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, math.osu.edu/~davis.12/papers/lectures%20on%20orbifolds.pdf $\endgroup$ Jul 30, 2012 at 12:20

3 Answers 3

12
$\begingroup$

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 $k=1$.

cf. http://arxiv.org/abs/1307.4875.

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.

$\endgroup$
20
$\begingroup$

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 $S^3$.

$\endgroup$
1
7
$\begingroup$

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

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

О ФАКТОРПРОСТРАНСТВЕ ПО ДЕЙСТВИЮ КОНЕЧНОЙ ГРУППЫ, ПОРОЖДЕННОЙ ПСЕВДООТРАЖЕНИЯМИ.

http://www.mathnet.ru/links/33220b8c84645bec685e85bf17d65994/im1420.pdf

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).

$\endgroup$
2
  • $\begingroup$ She probably wanted to say "PL-homeomorphic" otherwise cone over double suspension over Poincare sphere would be a counterexample. $\endgroup$ Jul 29, 2012 at 14:45
  • 1
    $\begingroup$ The article is available at iopscience.iop.org/0025-5726/24/1 for real money. I was able to get a copy through the proxy server of my university library. $\endgroup$
    – Lee Mosher
    Jul 29, 2012 at 19:26

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.