Why is this 3-manifold irreducible?

Question

Let $M = \mathbb{R}\mathbb{P}^2 \times \mathbb{S}^1$. It is a prime 3-manifold, but it cannot be reducible, since the only reducible prime connected 3-manifolds are the $\mathbb{S}^2$-bundles over $\mathbb{S}^1$. Thus, $M$ is irreducible, and so every smoothly embedded 2-sphere in $M$ must bound an embedded 3-ball. Is there a direct proof of this fact?

Answer 1

I'm going to extract the essential argument from the general theory of Seifert-fibered spaces. Let $M=\mathbb{R}\mathrm{P}^2\times S^1$, which as a Seifert-fibered space is very special since it is in fact a product $S^1$-bundle over $\mathbb{R}\mathrm{P}^2$.

Let $\alpha\subset\mathbb{R}\mathrm{P}^2$ be a non-separating simple closed curve (an image of a geodesic in $S^2$). Then, $T=\alpha\times S^1\subset M$ is a torus with $M-T$ a solid torus. Let $D=\mathbb{R}\mathrm{P}^2\times\{0\}-T$, which is a disk cross-section of this solid torus, and furthermore let $A=T-\mathbb{R}\mathrm{P}^2\times\{0\}$, which is an annulus. Then, $B=M-(A\cup D)$ is a ball. (Note: I am being cavalier with what I mean by a complement, where to be completely accurate I should be removing tubular neighborhoods.) We can use the sorts of techniques of normal surface theory, as seen in the proof of Kneser's prime decomposition theorem, to put a given surface in a nice position relative to $B$, $A$, and $D$.

Consider an embedded sphere $S\subset M$ that we presume does not bound a ball. By an isotopy, we can assume $S$ is transverse to $A$ and $D$. Suppose $S\cap A$ contains a closed loop that bounds a disk in $A$, and take the innermost such loop and disk. With it, we may do surgery on $S$ to get a pair of spheres. If both spheres bounded balls in $M$, then so too would $S$, so at least one does not; replace $S$ with this sphere. Hence, we may assume $S\cap A$ is a collection of arcs and essential loops in $A$. Similarly, we may assume $S\cap D$ is a collection of arcs.

Consider an innermost arc of $S\cap D$ that bounds a lune in $D$ such that the lune does not contain an antipodal pair of points in $\partial D$ (in the sense that $\partial D$ is a double cover of $\alpha$). We can isotope $S$ along this lune to remove a pair of points of intersection in $S\cap \alpha$. Hence, $S\cap D$ contains only arcs that connect antipodal points of $\partial D$. By considering the $\mathbb{Z}/2\mathbb{Z}$ intersection number, $S\cap D$ contains at most one such arc.

If there were such an arc, $S\cap \alpha$ would be a single point, and so the only arc in $S\cap A$ would be one that connects the two boundary components. This implies there are no essential loops, so $S\cap B$ contains a single component, which must be a disk since $S$ is a sphere. But such a disk in $B$ would imply that $S$ is a torus or Klein bottle, contrary to the fact that it is a sphere. Thus, it must be the case that $S\cap D$ is empty.

It follows that $S\cap A$ is a collection of essential loops. Again, we can assume $S\cap B$ is a collection of disks by performing surgery on $S$ if necessary and keeping one of the two components. Each loop in $S\cap A$, then, corresponds to an $\mathbb{R}\mathrm{P}^2$ component of $S$, so $S\cap A$ is empty.

Hence, $S\subset B$, but Alexander's theorem would imply $S$ bounds a ball, yet $S$ bounds no such thing.

Answer 2

The following argument, which is adapted from an argument Hatcher uses in his notes on $3$-manifolds to show $S^2\times S^1$ is prime, shows $\mathbb RP^2\times S^1$ is prime:

Let $S \subseteq S^1\times \mathbb RP^2$ be a separating sphere, so $(S^1\times\mathbb RP^2)|S$ consists of two compact $3$ manifolds $V$ and $W$ each with boundary a $2$-sphere. We have $\mathbb Z\times\mathbb Z_2 = \pi_1(S^1\times\mathbb RP^2) ≈\pi_1(V) \ast\pi_1(W)$, so either $V$ or $W$ must be simply connected, say $V$ is simply connected. The universal cover of $S^1\times \mathbb RP^2$ can be identified with $\mathbb R^3 − \{0\}$, and $V$ lifts to a diffeomorphic copy $\tilde V$ of itself in $\mathbb R^3\setminus\{0\}$.The sphere $\partial\tilde V$ bounds a ball in $\mathbb R^3$ by Alexander’s theorem. Since $\partial\tilde V$ also bounds $\tilde V$ in $\mathbb R^3$ we conclude that $\tilde V$ is a ball, hence also $V$ . Thus every separating sphere in $S^1\times\mathbb RP^2$ bounds a ball, so $S^1×\mathbb RP^2$ is prime.

The only part which remains to be justified is the part in italics. This can be done using the invariance of domain theorem for smooth functions:

Let us denote the lifting $V\rightarrow\mathbb R^3\setminus\{0\}$ by $f$, so that $\tilde V=f(V)$.

To prove $\tilde V$ is indeed a $3$-ball let us prove first that $\operatorname{Int}_{\mathbb R^3}(\tilde V)=f(\operatorname{Int}(V))=\operatorname{Int}(\tilde V)$.

We have that as $f$ is a diffeomorphism from $V$ onto $\tilde V$, $f(\operatorname{Int}(V))=\operatorname{Int}(\tilde V)$; $\operatorname{Int}(\tilde V)$ meaning the interior of $\tilde V$ as a manifold, since in particular $f$ is homeomorphism and the interior of a manifold with boundary is well defined topologically. Finally $\operatorname{Int}(\tilde V)=\operatorname{Int}_{\mathbb R^3}(\tilde V)$ because of the theorem of invariance of domain for smooth functions, as $\tilde V$ is a $3$-manifold smoothly embedded into $\mathbb R^3$.

Thus we have gotten $\operatorname{Int}_{\mathbb R^3}(\tilde V)=f(\operatorname{Int}(V))=\operatorname{Int}(\tilde V)$, and this in turn implies $f(\partial V)=\partial_{\mathbb R^3}\tilde V=\partial\tilde V$; in the last term we're taking the boundary of the manifold $\tilde V$

Now suppose $\tilde V$ is not a $3$-ball of $\mathbb R^3\setminus\{0\}$. Clearly, either $\operatorname{Int}(\tilde V)=\operatorname{Int}_{\mathbb R^3}(\tilde V)$ lies in the $3$-ball bounded by $\partial\tilde V$ or in its complement. The latter case cannot be as $\tilde V$ is compact and this would imply $\partial_{\mathbb R^3}\tilde V$ to intersect the complement of this $3$-ball as $\operatorname{Int}_{\mathbb R^3}(\tilde V)\neq\emptyset$ and this would be a contradiction.

$0$ cannot be in the $3$-ball bounded by $\partial\tilde V$, as otherwise either $\tilde V$ would be this ball minus $0$, which cannot be, or $\partial_{\mathbb R^3}\tilde V$ and the interior of this ball would intersect, which cannot be either.

Finally, $\tilde V$ must be the $3$-ball contained in $\mathbb R^3$, for otherwise as $\operatorname{Int}_{\mathbb R^3}(\tilde V)$ is contained in this $3$-ball we would have $\partial_{\mathbb R^3}\tilde V$ would intersect the interior of this ball, and this cannot be.