Differential Geometry – Mapping Torus of Orientation Reversing Isometry of the Sphere

Question

$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$
Let $ f_n $ be an orientation reversing isometry of the round sphere $ S^n $. Let $ M_n $ be the mapping torus of $ f_n $. What can we say about $ M_n $?

Here are the things I think I know:

• $ M_n $ has dimension $ n+1 $
• $ M_n $ is a $ S^n $ bundle over $ S^1 $
• Applying LES homotopy to the fiber bundle be have
  $$
  1 \to \pi_1(S^n) \to \pi_1(M_n) \to \pi_1(S^1) \to \pi_0(S^n) \to \pi_0(M_n) \to 1
  $$
• For $ n=0 $, $ M_0 $ is the circle and the bundle map is just the standard map by which the circle double covers itself.
• For $ n \geq 1 $ the sphere is connected so the LES of homotopy simplifies to
  $$
  1 \to \pi_1(S^n) \to \pi_1(M_n) \to \pi_1(S^1) \to 1
  $$
• For $ n\geq 1 $ the sphere is connected so $ f_n $ orientation reversing implies $ M_n $ must be nonorientable (and moreover thanks to Zerox for pointing out that the orientable double cover will always be $ S^n \times S^1 $)
• $ M_1 $ is the Klein bottle
• For $ n \geq 2 $ then $ S^n $ is connected simply connected so the LES homotopy simplifies to
  $$
  \pi_1(M_n) \cong \pi_1(S^1) \cong \mathbb{Z}
  $$
• $ M_2 $ is a non orientable 3-manifold admitting $ S^2 \times R $ geometry. $ M_2 $ is the quotient of its orientable double cover $ S^2 \times S^1 $ by the free $ C_2 $ action given by $ (-x,-z) $ see this answer https://math.stackexchange.com/questions/4322584/s2-times-r-geometry.

(note that $ RP^2 \times S^1 $ is also a non orientable 3-manifold admitting $ S^2 \times R $ geometry whose orientable double cover is $ S^2 \times S^1 $, however it is not homeomorphic to $ M_2 $ since the mapping torus of a simply connected manifold has infinite cyclic fundamental group whereas $ RP^2 \times S^1 $ has fundamental group a direct product of infinite cyclic with 2 element cyclic)

I am interested in the geometry of this mapping torus $ M_n $. In particular, $ M_n $ always admits a Riemannian metric with respect to which it is locally isometric to the geometry of the universal cover of the trivial bundle $ S^n \times S^1 $. This geometry
$$
\widetilde{S^n \times S^1}
$$
is the product of a round geometry with a one dimensional flat
$$
S^n \times R
$$
for $ n \geq 2 $. For $ n=0,1 $ the geometry is just flat with universal cover $ \mathbb{R},\mathbb{R}^2 $ respectively. We can verify this in some examples by observing that $ S^1 $ and the Klein bottle both admit flat metrics. And $ M_2 $ is well known from Thurston geometrization as one of the exactly four compact 3-manifolds that admits $ S^2 \times R $ geometry.

Now to the question. Recall that $ M_n $ is the mapping torus of an orientation reversing isometry of $ S^n $. Let
$$
G_n:=\Iso(S^n \times R) \cong \O_{n+1} \times \Iso (R)
$$
For which $ n $ does there exists a transitive action of $ G_n $ on $ M_n $?

I'm also curious for which $ n $ the action factors through the compact group $ O_{n+1} \times \mathbb{R}/\mathbb{Z} $. Because then a transitive action by a compact group implies $ M_n $ admits the structure of a Riemannian homogeneous manifold. For example, there is a transitive action of $ G_n $ for both $ n=0,1 $. But that action can only factor though the action of a compact group in the case $ n=0 $, not the case $ n=1 $.

And I'm also curious how $ M_n $ might differ for odd and even $ n $, since odd and even orthogonal groups are significantly different.

This is cross posted from MSE: https://math.stackexchange.com/questions/4348711/mapping-torus-of-orientation-reversing-isometry-of-the-sphere?noredirect=1#comment9075822_4348711

Answer 1

For $n$ even, $M$ admits such an action. Indeed, the antipodal map of the even-dimensional sphere is orientation reversing, so you can realize $M$ as the quotient $$\langle \gamma \rangle \backslash \left(S^n\times \mathbb R\right)$$ where $\gamma = (-\mathrm{Id}, 1)\in O_{n+1} \times \mathbb R$. Since $\langle \gamma\rangle$ is central in $O_{n+1} \times \mathbb R$, the action of $O_{n+1}\times \mathbb R$ on $S_n\times \mathbb R$ factors to $M$.

For $n$ odd, orientation reversing isometries of $S_n$ are not central in $O_{n+1}$, and my guess would be that there is no such action.