Mapping torus of the antipodal map of $ S^2 $

Question

Let $ M $ be the mapping torus for the antipodal map of $ S^2 $. I know the following things:

• $ M $ is a bundle $$ S^2 \to M \to S^1 $$ in particular $ M $ is a compact connected 3 manifold

• By LES homotopy $$ \pi_1(M) \cong \pi_1(S^1) \cong \mathbb{Z} $$ (since $ S^2 $ is connected simply connected)

• Similarly by LES homotopy $$ \pi_n(M) \cong \pi_n(S^2) $$ all $ n \geq 2 $.

• $ M $ is non orientable since the antipodal map on $ S^2 $ is orientation reversing

• $ M $ admits $ S^2 \times E^1 $ geometry (in other words the universal cover of $ M $ is $ S^2 \times E^1 $ and this cover is Riemannian)

• $ Iso(S^2) \times Iso(E^1) $ acts transitively on $ S^2 \times E^1 $. So the connected component of the identity, namely $ G:=SO_3(\mathbb{R}) \times \mathbb{R} $, also acts transitively on $ S^2 \times E^1 $. The action by an isometry $ (R,t) $ on a point is $$ (v,x) \mapsto (Rv,x+t) $$

• Since the antipodal map is central in $ Iso(S^2)=O_3(\mathbb{R}) $ the transitive action of $ G $ on the cover descends to a transitive action of $ G $ on $ M $. (according to Natural group action on mapping torus)

• As a result $ M $ is diffeomorphic to $ G/H $ where again $ G $ is the four dimensional noncompact group of isometries $ SO_3(\mathbb{R}) \times \mathbb{R} $ and $ H $ is the closed subgroup consisting of $$ \{ (\begin{bmatrix} R & 0 \\ 0 & 1 \end{bmatrix},2n):R \in SO_2 , n \in \mathbb{Z} \} \cup \{ (\begin{bmatrix} J & 0 \\ 0 & -1 \end{bmatrix},2n+1): J \in O_2 \setminus SO_2 , n \in \mathbb{Z} \} $$

I would like to better understand the action of $ G $ on $ M $. Is it faithful? Is it an action by isometries? Mostly I am interested in determining whether or not $ M $ is Riemannian homogeneous. What is the isometry group of $ M $?

I also know the following:

Given a Riemannian manifold $ (M,g) $, the universal cover $ \tilde{M} $ can be equipped with a metric $ \tilde{g} $ such that the covering map $ \tilde{M} \to M $ coincides with the quotient map $ \tilde{M}↦\tilde{M}/Γ $ where $ Γ $ is a discrete subgroup of isometries of $ \tilde{M} $. Using among other things the lifting property of universal coverings, you can show that the isometry groups are related by the following expression: $$ Isom(M)≅N_{Isom(\tilde{M})}(Γ)/Γ $$ I believe in this case $ \Gamma $ is the group generated by the isometry $ (f,1) $ where the $ f $ represents the antipodal map as an element of $ O_3 $ (just the negative identity matrix).Note that the generator of $ \Gamma $ is not in $ G $ the identity component of the isometry group.

Anyway, I know all these things but somehow I still can't quite put it all together. So my question is:

What is the isometry group of the the mapping torus of the antipodal map of $ S^2 $ and is it transitive?

Answer 1

The mapping torus of the antipodal map of $ S^2 $ is diffeomorphic to $$ SO_3(\mathbb{R})\times SO_2(\mathbb{R})/H $$ where $ H $ is the closed subgroup consisting of $$ \{ (\begin{bmatrix} R & 0 \\ 0 & 1 \end{bmatrix},I):R \in SO_2 \} \cup \{ (\begin{bmatrix} J & 0 \\ 0 & -1 \end{bmatrix},-I): J \in O_2 \setminus SO_2 \} $$ and $ I $ is the $ 2 \times 2 $ identity matrix.

To calculate the isometry group observe that the mapping torus is a quotient of $ S^2 \times S^1 $ by diagonal antipodal action, in other words the isometry $ (-1,-1) \in \text{Iso}(S^2 \times S^1)\cong O_3(\mathbb{R}) \times O_2(\mathbb{R}) $. In general you can calculate the isometry group of the quotient of a manifold by a group of discrete isometries $ \Gamma $ using the formula $$ \text{Iso}(M/\Gamma) \cong N_{\text{Iso}(M)}(\Gamma)/\Gamma $$ where $ N $ is the normalizer (here $ \Gamma $ must act freely so that the quotient $ M/ \Gamma $ is a manifold). In this case $ \Gamma $ is just the two element group generated by $ (-1,-1) $ and so it normal, indeed it is even central. Thus we can apply the formula and conclude that the isometry group of the mapping torus of the antipodal map of $ S^2 $ is $$ O_3(\mathbb{R}) \times O_2(\mathbb{R})/(-1,-1) $$ In particular the isometry group has two connected components and the connected component of the identity is isomorphic to $ SO_3(\mathbb{R}) \times SO_2(\mathbb{R}) $.

For more related material and the source of all these ideas see

Which 3 manifolds admit transitive action by compact group?

and also for the isometry group formula

Isometry group(s) of flat surfaces