isometries + fixed point = 3-sphere

[Graven Water]

I came up with a nice proof. Or possibly it's an embarassing poof. I
skimmed over a bunch of articles on conjugate loci and didn't see this
mentioned.

Suppose there's a complete Riemannian 3-manifold M, with a lot of
symmetry: there's an isometry group G(p) which fixes a point p and is
transitive on the unit
tangent vectors at p, and another point q which is a fixed point for
G(p). Then, M is homeomorphic to the 3-sphere.

Let q be the closest fixed point for G(p) other than p. There's a
closest point because the set of fixed points of G(p) is closed, and close
enough to p, the exponential map is a diffeomorphism.

All the geodesics going through p have to go through q - it's a complete
manifold and so there IS a geodesic from p to q. They all have the same
length because G(p) is transitive on the geodesics.

But it also seems to me then that all the geodesics going through q have
to go through p.

There's an exponential map on the tangent space at p, and it would have
a singular 2-sphere of radius d, d being the distance to q.

There's a neighborhood V of q, in which the exponential map at q is a
diffeomorphism.

You could find a 2-sphere S- in the tangent space at p, close enough to the
singular 2-sphere radius d, that the whole sphere is mapped into V. Since
the exponential map at q is a diffeomorphism, you'd have a continuous
(smooth, I guess) map from S- to the tangent space at q.

If you mapped this image of S- by radial projection onto the unit 2-sphere
in the tangent
space at q, the projection is well defined because the image of S- doesn't
include the 0 vector. Also the projection has to be injective, because if
not, two different geodesics starting from p going to q are crossing or
merging inside V, which they can't.

So you'd have a 1-1 continuous map from a 2-sphere to a proper subset of
it, which can't happen.

This would mean that any point on the Riemannian manifold would be on a
geodesic from p to q, since the manifold is complete. So it would be
compact. Also G(p)=G(q), and G(p) has no fixed points besides p and q.

Can there be points other than q in the tangent space at p where the
exponential map is singular? They would be singularities of order 1,
which means that the kernel of the jacobian of the exponential map is
dimension 1. There would be a spherical shell of singularities in the
tangent space at p. But an isometry of the tangest space is a rotation
of the spherical shell, and it must have a fixed point. So a singularity
at that point has to have rotational symmetry around that point -so it
can't be of dimension 1.

So the space must be homeomorphic to the 3-sphere.

Laura