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Hi,

Let $S=G/K$ be a sphere written as coset space. I know there are just few possibilities for $G$, and $K$ due to the classification of compact connected groups that can be transitive on a sphere.

If $F \subset G$ is a finite fixed point free subgroup, can we write $S/F = G / (F \times K)$??

This should be true for real projective spaces, but for example how about lens spaces $S^{2n-1}/\mathbb Z_q$?

thanks

David

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    $\begingroup$ No, generally you can't do this. It's easy enough to work out in the case of lens spaces using linear algebra -- elements of $K$ do not commute with most of the elements of $F$. $\endgroup$
    – Ryan Budney
    Commented Nov 26, 2011 at 21:54
  • $\begingroup$ And in the general case what if $F$ and $K$ commute? For example for finite subgroups of $SU(2)$ $\endgroup$
    – David P
    Commented Nov 26, 2011 at 23:23
  • $\begingroup$ Sorry for adding a comment as answer. Anyway it is enough to have $F$ (finite) normal (hence cenral) in $G$, or just central to well-define a $G$-action on $M/F$, but what about other sufficient cases? For instance, if we see $S^{2n-1} = U(n)/U(n-1)$ and $F$ is the cyclic group acting by multiplying by $e^{2 \pi i/q}$ on each complex coordinates, of course $F$ is central in $G$ and we can say that $S^{2n-1}/F$ is homogeneous under $U(n)$. How about a $SO(2n)$-action? $\endgroup$
    – David P
    Commented Nov 28, 2011 at 21:37

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so what are the spherical space forms homogeneous under? Starting from various ways to write the spheres themselves $S^{n-1} = SO(n)/SO(n-1)$, $S^{2n-1} = SU(n)/SU(n-1)$ and $S^{4n-1} = Sp(n)/Sp(n-1)$ ??

Is it enough to have $F$ commute with $K$? (notation as in the first post)

thanks

D

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    $\begingroup$ Please don't leave further questions as "answers" - instead, you can add things to your earlier question, as edits or updates $\endgroup$
    – Yemon Choi
    Commented Nov 27, 2011 at 22:17
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    $\begingroup$ Also, you should register your account, or find a way to use the same cookie, so you don't create new accounts with the same name. $\endgroup$
    – S. Carnahan ♦
    Commented Nov 28, 2011 at 2:39

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