# Mapping torus of orientation reversing isometry of the sphere

$$\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\SU{SU}\DeclareMathOperator\O{O}\DeclareMathOperator\Iso{Iso}$$ 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.

• The double cover of $M_2$ is not $\Bbb{RP}^3 \# \Bbb{RP}^3$, but $S^2 \times S^1$. The former is not simply connected by van Kampen. In fact the double cover of $M_n$ is always $S^n \times S^1$. Jan 8, 2022 at 15:38
• You should mention that this is a crosspost from MSE and include a link. Jan 8, 2022 at 15:56

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.
• $M_2$ is not listed in Gorbatsevich's article "Three-Dimensional Homogeneous Spaces". So your claim suggests an error in his article. That is certainly possible but I would like to better understand your reasoning about why the coset space you defined is homeomorphic to the mapping torus. For example a quick look at LES did not seem obvious to me that your space would have a particular fundamental group (since no group involved is simply connected or even connected computing the fundamental group of the quotient is not so obvious) Jan 12, 2022 at 14:52
• If I understand correctly you are claiming that $O_3×R/<(O_1 \oplus O_2)×Z>$ (here the direct sum means as a block diagonal matrix) is $M_2$ , the mapping torus of an orientation reversing isometry of the two sphere. Are you sure this is not just $RP_2×S^1$? Both spaces are non orientable with orientable double cover $S^2×S^1$. However $M_2$ has infinite cyclic fundamental group while $RP_2×S^1$ has fundamental group a direct product of infinite cyclic with two element cyclic. Jan 12, 2022 at 14:59
• No, I'm claiming that $M_2$ is the double quotient $<\gamma > \backslash O_3\times \mathbb R / O_1 \oplus O_2$, and that, since $\gamma$ is central in $O_3\times \mathbb R$, the left action of $O_3\times \mathbb R$ on $O_3\times \mathbb R /O_1\oplus O_2$ factors to an action on $M_2$. Jan 12, 2022 at 15:31
• Here is another way to see this: Consider the cover of degree 4 $S^2\times S^1 \to RP^2 \times S^1$, where the $S^1$ factor is mapped to itself by a double cover. This is a normal cover with Galois group $Z_2\times Z_2$, and $M_2$ is the intermediate cover corresponding to the diagonal $Z_2$ inside it. Now, the action of $O_3 \times R$ on $S^2 \times S^1$ clearly factors to $RP^2 \times S^1$, so it factors to the intermediate cover $M_2$. Jan 12, 2022 at 15:35
• Ok that sounds promising. Certainly $M_2$ is a double quotient. So you are claiming that $G=O_3 \times \mathbb{R}$ mod some closed subgroup is diffeomorphic to $M_2$. Could you point out which subgroup of $G$ you are considering? The closed subgroups of $O_3 \times \mathbb{R}$ have a very simple structure so it should be possible for you to be very explicit. The more specific you are about the closed subgroup the easier it will be for me to understand that your answer is correct. Jan 12, 2022 at 15:36