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

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?

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 $$SO_3(\mathbb{R}) \times SO_2(\mathbb{R})$$.