$S^2 \times R$ geometry

There are exactly four compact manifolds with $$S^2 \times R$$ geometry. They are $$S^2 \times S^1 , \mathbb{RP}_2 \times S^1, M_2, \mathbb{RP}_3\# \mathbb{RP}_3$$ where $$M_2$$ denotes the mapping torus of an orientation reversing isometry of the sphere $$S^2$$.

I am curious about these manifolds:

• $$\mathbb{RP}_2 \times S^1, M_2$$ are both nonorientable. And $$\mathbb{RP}_2 \times S^1$$ has orientable double cover $$S^2 \times S^1$$. What is the orientable double cover of $$M_2$$?

• In www2.math.umd.edu/~wmg/icm.pdf page 8 claims that all these manifolds are quotients of $$S^2 \times S^1$$. How can I see that $$\mathbb{RP}_3\# \mathbb{RP}_3$$ and $$M_2$$ are quotients of $$S^2 \times S^1$$?

• $$\mathbb{RP}_2\# \mathbb{RP}_2$$ (the klein bottle) has a transitive action by the group $$E_2$$ of isometries of the flat plane . Does $$\mathbb{RP}_3\# \mathbb{RP}_3$$ (the 3d Klein bottle) also admit a transitive action by some non compact group? This article https://link.springer.com/article/10.1007/BF00967152 seems to be claiming that the group $$E_3$$ of isometries of flat 3 space acts transitively on $$\mathbb{RP}_3\# \mathbb{RP}_3$$. Can someone describe this action? What closed subgroup of $$E_3$$ can I quotient by to get $$\mathbb{RP}_3\# \mathbb{RP}_3$$?

• I think $E_3$ could not act transitively on $\mathbb{RP}^3 \# \mathbb{RP}^3$ by Thurston's geometrization. Commented Jan 4, 2022 at 23:32
• Thurston Geometrization just says that it can't act with compact stabilizers. But I agree that intuitively it seems wrong. The result is very strange to me and I don't know what the action would be. You can see in the second to last entry of table 1 (the table is at the top of page 2) they list a circle bundle over $RP_2$, Gorbatsevich calls it $S(RP_2)$ and claim it is diffeomorphic to $RP_3 \# RP_3$. Gorbatsevich also claims a transitive action by $SU_2$ semi direct $R^3$ on this manifold $S(RP_2)$ . And $SU_2$ semi direct $R^3$ is basically the universal cover of $E_3$. Commented Jan 4, 2022 at 23:46
• The author seemed to have explained the quotient after Theorem $2.4$. Commented Jan 4, 2022 at 23:56
• $M_2$ is the underlying space of the conformally flat Lorentzian manifold called the Einstein universe, which conformally compactifies Minkowski space $\mathbb{E}^{2, 1}$ by adding a light cone at infinity (i.e. the complement of any light cone in $\mathrm{Ein}^3$ is conformally equivalent to $\mathbb{E}^{2, 1}$). You can see it in $\mathbb{E}^{2, 1}$ by imagining a vertex added at infinity for each tangent plane to any null cone, and one extra point serving as the infinity cone's vertex. See A primer on the (2 + 1) Einstein universe Commented Dec 4, 2023 at 0:04

$$\newcommand{\Number}[1]{\mathbf{#1}}\newcommand{\Reals}{\Number{R}}\newcommand{\Cpx}{\Number{C}}\newcommand{\Proj}{\mathbf{P}}\newcommand{\CSum}{\mathop{\#}}$$All four manifolds are double-covered by the product $$S^{2} \times S^{1}$$. These coverings may be represented conveniently by writing $$(x, z)$$ for the general element of $$S^{2} \times S^{1} \subset \Reals^{3} \times \Cpx$$:

• The quotient by $$(x, z) \mapsto (-x, z)$$ is $$\Reals\Proj^{2} \times S^{1}$$. A fundamental domain is a closed hemisphere times the circle.
• The quotient by $$(x, z) \mapsto (-x, -z)$$ is the mapping cylinder of the antipodal map of $$S^{2}$$. A fundamental domain is the sphere cross half the circle, i.e., $$S^{2} \times [0, 1]$$, and $$(x, 0) \sim (-x, 1)$$.
• The quotient by $$(x, z) \mapsto (-x, \bar{z})$$ is the connected sum $$\Reals\Proj^{3} \CSum\Reals\Proj^{3}$$. A fundamental domain is the sphere cross half the circle, this time with boundary identification $$(x, 0) \sim (-x, 0)$$ and $$(x, 1) \sim (-x, 1)$$. To see this is a connected sum of two $$\Reals\Proj^{3}$$s, note that projective space itself may be viewed as a closed ball in $$\Reals^{3}$$ with antipodal identification on the boundary. Removing a ball from this amounts to removing a concentric ball, i.e., identifying $$(x, 0) \sim (-x, 0)$$ in $$S^{2} \times [0, 1]$$.
• Just for completeness, $$(x, z) \mapsto (x, z^{2})$$ is a double-covering of $$S^{2} \times S^{1}$$ over itself.

Incidentally, the isometry group of a flat Klein bottle does not act transitively: A flat Klein bottle may be viewed as a quotient of the flat cylinder $$\Reals \times S^{1}$$ under the mapping $$(t, z) \mapsto (t + 1, \bar{z})$$. Consequently, an isometry of the Klein bottle lifts to an invariant isometry of the cylinder, and conversely every invariant isometry of the cylinder descends. Translations along the $$\Reals$$ factor descend, as does the reflection $$(t, z) \mapsto (t, \bar{z})$$, but rotations of the $$S^{1}$$ factor do not. Geometrically, there are two distinguished circles on a flat Klein bottle coming from central circles on two flat Möbius strips. The "nearby" local sections of the non-trivial circle bundle over $$S^{1}$$ are topologically circles mapping $$2$$-to$$1$$ to the "central" circles.

The isometries of $$\Reals\Proj^{3} \CSum\Reals\Proj^{3}$$ may be analyzed similarly; here we seek isometries of $$\Reals \times S^{2}$$ commuting with $$(t, x) \mapsto (t + 1, -x)$$. Here, every orthogonal transformation of Euclidean three-space descends to an isometry of the quotient.

• Didn't see Jason's (+1) answer until I posted this; leaving this anyway in case a more geometric (less topological) accounting is of interest. Commented Jan 4, 2022 at 23:36
• Jason's answer was super awesome too but I'm more of a geometry person so this is exactly what I was looking for. Thanks so much Andrew! Maybe some of the insights here would carry over to my question about mapping tori of isometries of spheres math.stackexchange.com/questions/4348711/… Commented Jan 4, 2022 at 23:50
• For what it's worth, I'd also select this answer as best ;-). Mine was way too complicated! Commented Jan 4, 2022 at 23:55
• If I'm reading this correctly you are saying that the flat Klein bottle is basically a bundle of round circles (by round I just mean the isometry group is $O_2$ obviously the curvature is still 0) over a distorted base circle and so the isometry group of the flat Klein bottle is exactly the isometry group $O_2$ of the circle fiber? And then $RP_3 \# RP_3$ is basically a bundle of round spheres over a distorted base circle so the isometry group of $RP_3 \# RP_3$ is exactly the isometry group $O_3$ of the round sphere $S^2$ fiber? Commented Jan 6, 2022 at 12:54
• I think of it as: For each of these spaces $M$, we have a Thurston geometry on the universal cover, every Thurston metric on $M$ lifts, and every isometry of $M$ lifts to an isometry commuting with deck transformations. I'd be hesitant to speak of a distorted circle because intrinsically all circles are flat. <> Incidentally, remarkable features about flat Klein bottles (which I hadn't thought about until this question) include that (i) the "central circles" exist, (ii) they're necessarily orthogonal to the circle fibers. (Flat tori are not all Riemannian products.) Commented Jan 6, 2022 at 13:53

I don't know much about non-compact groups, so I'll leave your last question to someone else.

I claim that $$S^1\times S^2$$ not only covers both $$M_2$$ and $$X:=\mathbb{R}P^3\sharp \mathbb{R}P^3$$, but that it double covers them. In particular, since $$\pi_1(M_2) \cong \mathbb{Z}$$, it has a unique double cover, so $$S^1\times S^2$$ is the orientation covering of $$M_2$$.

You can see that $$M_2$$ is double covered by $$S^1\times S^2$$ directly. In fact, if you consider the $$\mathbb{Z}_2$$ action on $$S^1\times S^2$$ which maps $$(x,y)$$ to $$(-x,-y)$$, then $$M_2$$ is the quotient by this free action. The map $$M_2\rightarrow S^1$$ given by mapping $$[(x,y)]$$ to $$x^2$$ is a fiber bundle map with fiber $$S^2$$. This bundle structure allows you to identify $$M_2$$ as the mapping torus of the antipodal map on $$S^2$$ and also tells you that $$\pi_1(M_2)\cong \mathbb{Z}$$ (via the long exact sequence in homotopy groups.)

For $$X$$, we'll argue as follows. I'm going to view $$\mathbb{R}P^3$$ as $$S^2\times [0,1]/\sim$$ where $$S^2\times \{0\}$$ is identified to a single point, and $$(x,1) \sim (-x,1)$$. That is, we do the usual antipodal identification on the boundary. (This is precisely the ball model of $$\mathbb{R}P^3$$). To form $$X$$ we take two copies of $$\mathbb{R}P^3$$, with $$S^2\times \{0\}$$ (which is a single point) removed, and glue. Thus, we can view $$X$$ as $$S^2\times [-1,1]/\sim$$ where antipodal points in $$S^2\times \{-1\}$$ are identified and likewise antipodal points in $$S^2\times \{1\}$$ are identified.

Now, define $$\pi:X\rightarrow \mathbb{R}P^2$$ by making $$\pi$$ to be the identity on the end points and the usual double cover $$S^2\rightarrow \mathbb{R}P^2$$ on each slice $$S^2\times \{t\}$$ with $$t\in (-1,1)$$. This is a bundle map with fiber $$S^1$$.

This shows that we have a bundle $$S^1\rightarrow X\rightarrow \mathbb{R}P^2$$. If we pull this bundle back along the double cover $$S^2\rightarrow \mathbb{R}P^2$$, we obtain a bundle $$S^1\rightarrow \tilde{X}\rightarrow S^2$$ where $$\tilde{X}$$ is a double cover of $$X$$. It remains to determine the diffeomorphism type of $$\tilde{X}$$. To do this, simply note that the map $$\pi_1(X)\rightarrow \pi_1(\mathbb{R})^2$$ is a group homomorphism from an infinite group to a finite one, so it must have an infinite kernel. Thus, the map $$\pi_1(S^1)\rightarrow \pi_1(X)$$ must be injective.

This, in turn implies that the map $$\pi_1(S^1)\rightarrow \pi_1(\tilde{X})$$ is injective, so $$\tilde{X}$$ has infinite fundamental group. From, say, the Gysin sequence, this implies that the bundle $$S^1\rightarrow \tilde{X}\rightarrow S^2$$ has trivial Euler class, so the bundle is trivial. That is, $$\tilde{X}$$ is diffeomorphic to $$S^1\times S^2$$.

• I suspect there is a significantly easier proof that $\mathbb{R}P^3\sharp\mathbb{R}P^3$ is double covered by $S^2\times S^1$, but this is what I came up with... Commented Jan 4, 2022 at 22:54
• $\mathbb{R}P^3 \# \mathbb{R}P^3$ can be seen as the quotient of $S^2 \times S^1$ under the equivalence $(x,y) \sim (-x,\bar{y})$, where $x \in S^2$ and $y \in U(1) \cong S^1$. Commented Jan 4, 2022 at 23:11
• @Zerox: Oh, duh, thanks. I knew the map had to be easy to write down since it had to be an isometry. Not sure why I didn't think of complex conjugation on the $S^1$ factor... Commented Jan 4, 2022 at 23:49