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$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$ 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 $ is a compact manifold of dimension $ n+1 $
  • $ M_n $ is an $ 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 $ M_0 \to S^1 $ is just the circle double covering 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
  • $ 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 $ S^2 \times R $ geometry.

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 $$ G_n:=\Iso(S^n \times R) \cong \O_{n+1} \times \mathbb{R} \times C_2 $$ 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} \times C_2 $. 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 in 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.

I'm especially curious about all this for the low dimensional examples $ M_2 $ and $ M_3 $.

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    You may already know this, but there are only two $S^n$-bundles over $S^1$ up to diffeomorphism: the trivial one $S^n\times S^1$, and a non-trivial non-orientable one. You are asking about the latter which is double covered by the former. In your setup, this can be seen by using the fact that the isomorphism class of the bundle only depends on the isotopy class of the diffeomorphism $f_n : S^n \to S^n$, but $f_n \in \operatorname{Isom}(S^n) \cong O(n+1)$ which has two connected components. In particular, the diffeomorphism type of $M_n$ is independent of the choice of $f_n$. – Michael Albanese Jan 04 '22 at 17:28
  • @MichaelAlbanese However your statement about coverings is new to me. That means the orientable double cover of $ M_2 $ is $ S^2 \times S^1 $, rather than $ RP_3 # RP_3 $ as I incorrectly claimed above. I'll edit that. In http://www2.math.umd.edu/~wmg/icm.pdf page 8 claims that every compact 3-manifold with $ S^2 \times R $ geometry is a quotient of $ S^2 \times S^1 $. Could you perhaps comment on that and clarify the covering relations between $ S^2 \times S^1 $ and the other manifolds $ RP_2 \times S^1, M_2, RP_3 # RP_3 $? – Ian Gershon Teixeira Jan 04 '22 at 17:45
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    I don't think every diffeomorphism is isotopic to an isometry of the round sphere, only homotopic. Regarding $M_2$, note that $M_2$ is non-orientable but $\mathbb{RP}^3#\mathbb{RP}^3$ is orientable, so they aren't the same. – Michael Albanese Jan 04 '22 at 18:27
  • Oh sorry homotopic . And about the other manifolds I don't think I was totally clear in my comment. I think I'm clearer in my question here https://math.stackexchange.com/questions/4322584/s2-times-r-geometry. Maybe you could take a look at it? – Ian Gershon Teixeira Jan 04 '22 at 21:56
  • It seems that Andrew D. Hwang and Jason DeVito have since cleared up your confusion on this front. – Michael Albanese Jan 05 '22 at 00:57
  • Haha ya everybody has been helping so much! Another question related to your previous comment: is every diffeomorphism of a real projective space homotopic to the identity? – Ian Gershon Teixeira Jan 05 '22 at 01:02
  • I'm not sure. I don't think so, but you should consider asking this as a new question. – Michael Albanese Jan 05 '22 at 01:11

1 Answers1

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The action of $ G_n $ on $ M_n $ is transitive for the even spheres ($ n $ even) since only for even $ n $ is the orientation reversing isometry central. See this answer:

Natural group action on mapping torus

credit to Kajelad and Moishe Kohan