$ S^2 \times R $ geometry

Question

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 $?

Answer 1

$\newcommand{\Number}[1]{\mathbf{#1}}\newcommand{\Reals}{\Number{R}}\newcommand{\Cpx}{\Number{C}}\newcommand{\Proj}{\mathbf{P}}\newcommand{\CSum}{\mathop{\#}}$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.

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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.

Answer 2

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 $S^1\times S^2$.