I'm interested in isometry groups of round 3 manifolds, especially Riemannian homogeneous ones.
Here are my thoughts so far:
-
The round three sphere $ S^3 $ has isometry group $ O_4 $.
-
Round projective space is the sphere mod the antipodal isometry (which is a rotation for odd spheres)
$$
\mathbb{R}P^3 \cong S^3/-I
$$
Using the normalizer formula for the isometry group of a quotient of a simply connected manifold we have
$$
\text{Iso}(\mathbb{R}P^3)\cong N_{\text{Iso}(S^3)}(-I)/-I=O_4/-I
$$
which further simplifies as
$$
\text{Iso}(\mathbb{R}P^3) \cong (O_1 \ltimes SO_4)/-I \cong O_1 \ltimes (SO_4/-I) \cong O_1 \ltimes (SO_3 \times SO_3)
$$ -
For a homogeneous lens space $ L_{p,1} \cong SU_2/C_p $ where $ C_p $ is $ p $ element cyclic subgroup the normalizer of $ C_p $ in $ SO_3 $ (or $ SU_2 $) is $ O_2 $ and $ O_2/C_p \cong O_2 $. So I think the isometry group should be something like
$$
O_2 \times O_3
$$ -
Another reason I'm guessing $ \text{Iso}(L_{p,1}) \cong O_3 \times O_2 $ is that from the Hopf fibration
$$
S^1 \to S^3 \to S^2
$$
we can mod out by $ C_p $
$$
S^1/C_p \to S^3/C_p \to S^2
$$
So we have
$$
S^1 \to L_{p,1} \to S^2
$$
and then I am just taking the the direct product of the isometry group of the fiber and of the base. (I know you can't generally do that for nontrivial fiber bundles for example no metric on the Klein bottle has isometries $ O_2 \times O_2 $ but just noting here that it sort of agrees with my earlier guess from the normalizer perspective). -
The normalizers of other groups are $ D_{2n} \trianglelefteq D_{4n} $ for $ n \geq 3 $, $ D_{4} \cong C_2 \times C_2 \trianglelefteq S_4 $, $ A_4 \trianglelefteq S_4 $, $ A_5 \trianglelefteq A_5 $. The point is that all other discrete subgroups of $ SO_3 $ (and $ SU_2 $) have discrete normalizer with finite index. So for the homogeneous prism manifolds and three exceptional manifolds $ SU_2/ \Gamma $ I think roughly the normalizer of $ \Gamma $ is $ \Gamma \times SU_2 $ so the isometry group is roughly $ (SU_2 \times \Gamma) /\Gamma $ or rather
$$
\text{Iso}(SU_2/\Gamma) \cong SU_2 \rtimes O_1
$$
to include orientation reversing isometries -
For round three manifolds that are not homogeneous I think they probably all have isometry group $ O_2 $ (just a guess)
Best Answer
The isometry groups of elliptic $3$-manifolds were computed by McCullough in
which can be found on the arxiv here.
(Community wiki because McCullough did all the work.)