\documentclass{amsart}
\newtheorem{theorem}{Theorem}
\newtheorem{definition}{Definition}
\newtheorem{proposition}[theorem]{Proposition}
\newtheorem{lemma}[theorem]{Lemma}
\newcommand{\zz}{{\mathbb Z}}
\newcommand{\re}{{\mathbb R}}
\begin{document}
\title{Three-manifolds with the sphere-separated property}
\author{Kevin Iga}
\date{May 6, 2002}
\maketitle
\begin{abstract}
We classify connected three-manifolds with the property that every
embedded sphere separates the manifold into two components.
\end{abstract}
\begin{definition}
A three-manifold $M^3$ has the {\bf sphere-separated property} if
every embedded two-sphere separates $M^3$ into two components.
\end{definition}
\begin{lemma}
Let $M^3$ be a 3-manifold and let $\Sigma^2$ be an embedded 2-sphere
in $M^3$. Then there is a neighborhood $N$ in $M^3$ diffeomorphic to
$\Sigma^2\times(-1,1)$ so that the image of $\Sigma^2\times\{0\}$ is
$\Sigma$.
\label{lem:annulus}
\end{lemma}
\begin{proof}
Let $N$ be a tubular neighborhood of $\Sigma^2$. Then $N$ is
diffeomorphic to the normal bundle of $\Sigma^2$ in $M^3$, where the
zero section gets mapped to $\Sigma$. This normal bundle is a line
bundle over a sphere. Since $\Sigma^2$ is simply-connected, every
line bundle over $\Sigma^2$ is trivial. Therefore, the normal bundle
of $\Sigma^2$ in $M^3$ is a trivial product $\Sigma^2\times(-1,1)$.
\end{proof}
\begin{lemma}
Let $M^3$ be a connected closed 3-manifold, and $\Sigma^2$ an embedded
2-sphere in $M^3$. The following are equivalent:
\begin{enumerate}
\item There is an open region $R\subset M^3$ which is not dense in
$M^3$ so that the closure of $R$ minus the interior of $R$ is
$\Sigma^2$.
\item $M^3-\Sigma^2$ is disconnected.
\item There does not exist a closed curve $\gamma:S^1\to M^3$ so that
$\gamma$ intersects $\Sigma^2$ transversally in one point.
\item $\Sigma^2$ represents zero in $H_2^t(M^3)$, the second homology in
twisted coefficients using the orientation bundle.
\end{enumerate}
\label{lem:equiv}
\end{lemma}
\begin{proof}
$3 \Rightarrow 2$
Suppose $M^3-\Sigma^2$ were connected. By the
Lemma~\ref{lem:annulus} there is a neighborhood $N$ in $M^3$
diffeomorphic to $\Sigma^2\times(-1,1)$ so that the image of
$\Sigma^2\times\{0\}$ is $\Sigma^2$. Let $x\in \Sigma^2$. Let $a$ be
the image of $(x,-1/2)$ in $N$, and let $b$ be the image of $(x,1/2)$
in $N$.
Since $M^3-\Sigma^2$ is connected, there is a path from $a$ to $b$ in
$M^3-\Sigma^2$ , which we call $\gamma$. We extend $\gamma$ using the
path $(x,t)$ from $a$ to $b$ through $N$. In this way, $\gamma$ is
now a closed loop in $M^3$ that intersects $\Sigma$ transversally
exactly once.
$2 \Rightarrow 1$
Suppose $M^3-\Sigma^2$ were disconnected. Then let $X$ and $Y$ be a
separation. Then $X$ and $Y$ are open and closed in $M^3-\Sigma^2$.
In $M^3$, $X$ and $Y$ are still open. But if either is closed, then
$M^3$ would be disconnected. But we have assumed that $M^3$ is connected.
So neither is closed.
By Lemma~\ref{lem:annulus}, there is a neighborhood $N$ of $\Sigma$ that is
diffeomorphic to $\Sigma\times(-1,1)$ in such a way that $\Sigma$ is
the image of $\Sigma\times\{0\}$. Then $X\cap N$ and $Y\cap N$, under this
diffeomorphism, get sent into $\Sigma\times ((-1,0)\cup (0,1))$.
Since $X\cup Y = M^3-\Sigma^2$, the image of $X\cap N$ and $Y\cap N$
must equal $\Sigma\times((-1,0)\cup(0,1))$, and since $X$ and $Y$ are
disjoint, $X\cap N$ and $Y\cap N$ must have as its image a partition of
$\Sigma\times ((-1,0)\cup (0,1))$. Since the connected components of
$\Sigma\times ((-1,0)\cup (0,1))$ are $\Sigma\times (-1,0)$ and $\Sigma
\times (0,1)$, we know that one is sent to $X\cap N$, and the other to
$Y\cap N$. Therefore the closure of $X$ includes all of $\Sigma^2$, and
similarly for the closure of $Y$.
Therefore $X$ is a region whose closure minus its interior is $\Sigma^2$.
$1\Rightarrow 4$
Suppose $R$ is an open region as described in condition 1. Let $C$ be
the closure of $R$ in $M^3$. Then $C$ is the union of $R$ and
$\Sigma^2$. Consider the following exact sequence:
\[H^t_3(C)\to H^t_3(C,\Sigma)\to H^t_2(\Sigma)\to H^t_2(C)\]
Since $C$ is a three-dimensional manifold with boundary, $C$ has
trivial $H^t_3$, but $H^t_3(C,\Sigma\cong\zz$. Since $\Sigma$ is a
sphere, $H^t_2(\Sigma)=\zz$. Therefore the sequence above is isomorphic to
the following sequence:
\[0 \to \zz \to \zz \to H^t_2(C)\]
By the definition of the map $H^t_3(C,\Sigma)\to H^t_2(\Sigma)$, it sends
the generator of the first to its boundary, which is $\Sigma$. So this
is an isomorphism. In particular, the map $H^t_2(\Sigma)\to H^t_2(C)$ is
zero. Therefore, the image of the class represented by $\Sigma$ is zero.
Now under the map $H^t_2(C)\to H^t_2(M^3)$ induced by inclusion, this
class is still zero.
$4\Rightarrow 3$
By intersection theory, $\Sigma\cap\gamma$ only depends on the
homology class of $\Sigma$, so $\Sigma$ cannot represent the zero
homology class. (Note: if $M^3$ is non-orientable, we can use
$\zz/2\zz$-coefficients and intersection theory is still valid.)
\end{proof}
\begin{proposition}
Let $A$ and $B$ be closed 3-manifolds.
If $M=A\#B$, then $M$ has the sphere-separated property if and only
if $A$ and $B$ have the sphere-separated property.
\label{prop:connsum}
\end{proposition}
\begin{proof}
Suppose $A$ and $B$ have the sphere-separated property. Then suppose
$\Sigma^2$ is an embedded two-sphere in $M=A\#B$. Let $\Lambda^2$ be
the sphere that $A$ and $B$ are connected along, and let $A'$ and $B'$
be the 3-manifolds with boundary that correspond to $A$ and $B$,
respectively, so that $M=A'\cup_\Lambda B'$.
By perturbing $\Sigma$ we can assume that $\Sigma$ and $\Lambda$
intersect transversely, which is to say, in a collection of closed
curves.
We now prove that $\Sigma$ is the boundary of a region. In $A'$,
$\Sigma\cap A'$ is a union of surfaces with boundary in $\Sigma\cap\Lambda$.
All of these surfaces must have genus 0, or else $\Sigma$ would not have
genus 0.
Now let W be a three-ball with boundary $\Lambda$ so that
$A=A'\cup_\Lambda W$. For each component of $\Sigma\cap\Lambda$, cap
it off with a disk in $W$, so that these disks are disjoint. Let $L$
be the union of these disks. Now $(\Sigma\cap A')\cup L$ is a union
of spheres. Each sphere has non-trivial intersection with $W$. If
there is more than one component, let $\gamma$ be a curve in $W$ that
connects two of the components. As $\gamma$ intersects $L$, along the
image of $\gamma$, there must be two consecutive intersections from
different components of $(\Sigma\cap A')\cup L$. Between these two,
use a tubular neighborhood $\gamma$ to connect these two. The result is
a union of spheres with one fewer component. Proceeding by induction,
we have an embedded sphere $S\subset A$ so that $S\cap A'=\Sigma\cap A'$.
Since $A$ has the sphere-separated property, we know that it bounds a
region $R_A$ in $A$. Its intersection with $A'$, $R'_A=R_A\cap A'$,
has as boundary $\Sigma\cap A'$ union with a subset of $\Lambda$.
Similarly, there is a region $R_B$ in $B$ so that $R'_B=R_B\cap B'$
has as boundary $\Sigma\cap B'$ union with a subset of $\Lambda$.
Now $\Sigma$ has trivial normal bundle because line bundles over two-spheres
must be trivial. Then a tubular neighborhood of $\Sigma$ is diffeomorphic
to the product $\Sigma\times[0,1]$. For both the $A'$ side and the $B'$
side, $R'_A$ must either intersect the image of $\Sigma\times\{0\}$ under
this diffeomorphism, or the image of $\Sigma\times\{1\}$. Similarly with
$R'_B$. If they both agree, then $R'_A\cup R'_B$ is a region whose boundary
is $\Sigma$. If they do not, then replace $R'_B$ with its complement in $B'$
and they will agree. Then proceed as before with their union.
Therefore $\Sigma$ bounds a region.
Conversely, suppose $M$ had the sphere-separated property. Then we
prove that $A$ and $B$ have the sphere-separated property. Let
$\Sigma$ be an embedded sphere in $A$. By perturbing slightly, we can
insist that $\Sigma$ be completely in $A'$. Then $\Sigma$ will be an
embedded sphere in $M$. Since $M$ has the sphere-separated property,
$\Sigma$ bounds a region. In fact, it separates $M$ into two regions.
Since $\Sigma\cap\Lambda=\emptyset$, and $\Lambda$ is connected,
$\Lambda$ must be in one of the regions. Consider the other region.
Since it does not intersect $\Lambda$, it lies completely inside $A'$.
Therefore it is in $A$, and its boundary is $\Sigma$. Therefore $A$
has the sphere-separated property. Similarly, B has the
sphere-separated property.
\end{proof}
\begin{theorem}
Suppose $M$ is a prime orientable closed manifold. Then $M$
has the sphere-separated property if and only if $\pi_2(M)=0$.
\label{thm:pi2}
\end{theorem}
\begin{proof}
Suppose $M$ has $\pi_2(M)=0$. Then for any embedded sphere, it is
homotopic to zero, so its homology class is zero. By Lemma~\ref{lem:equiv},
this means that it bounds a region. Note that this part does not require
that $M$ be orientable.
Suppose $M$ has $\pi_2(M)\not=0$. By the Sphere theorem \cite{Ro},
there exists an embedded sphere which is not null-homotopic. Since
$M$ is prime, this sphere either bounds a homotopy ball or does not
separate the space. If it bounded a homotopy ball it would be null-homotopic.
Therefore, this sphere does not bound a region; so $M$ does not have
the sphere-separated property.
\end{proof}
\begin{theorem}
Let $M$ be a 3-manifold, and $\tilde{M}$ a covering space of $M$.
Then if $\tilde{M}$ has the sphere-separated property, then $M$ has
the sphere-separated property.
\label{thm:covering}
\end{theorem}
\begin{proof}
Let $\pi:\tilde{M}\to M$ be projection.
Suppose $\tilde{M}$ has the sphere-separated property. Then let
$\Sigma$ be a 2-sphere embedded in $M$. We use basic convering space
theory \cite{Ma} to prove that $\pi$ is the trivial covering space
over $\Sigma$. Therefore there is a lift $\tilde{\Sigma}$ which is an
embedded sphere in $\tilde{M}$, on which $\pi:\tilde{\Sigma}\to\Sigma$
is a diffeomorphism.
Since $\tilde{M}$ has the sphere-separated property, there is a
connected region $C$ in $\tilde{M}$ with boundary $\tilde{\Sigma}$.
Now $C$ is compact, so if we modify $\tilde{M}$ outside $C$ by
intersecting it with a ball that contains $C$ and that intersects
$\tilde{M}$ transversally, then take the resulting boundary circles
and close them with disks, the resulting $\tilde{M}$ is a closed
3-manifold. We then apply Lemma~\ref{lem:equiv} to see that
$\tilde{\Sigma}$ represents zero in homology. Since $\pi$ is a
diffeomorphism on $\tilde{\Sigma}$, we see that
$\pi(\tilde{\Sigma})=\Sigma$, and $\pi_*([\tilde{\Sigma}])=0$ in
$H_2(M)$. Therefore there is a 3-chain in $M$ whose boundary is
$[\Sigma]$. The image of this 3-chain is a compact subset of $M$, so
by intersecting $M$ by a sufficiently large ball that contains this
3-chain and intersects $M$ transversally, then capping off the
resulting boundary circles with disks, we get a closed 3-manifold. We
take the connected component that contains $\Sigma$, and apply
Lemma~\ref{lem:equiv} and see that $\Sigma$ bounds a compact region in
$M$.
[Fix a few things: refer to connected components on $\tilde{M}$ part, and
also find out why $[\Sigma]$ is the same in $H_2(M)$ as it is on the
modified $M$, and similarly with $\tilde{M}$.]
\end{proof}
Now $\re P^2\times S^1$ has the sphere-separated property, but
$S^2\times S^1$ does not. So this is an example of where the converse
of the above is not true.
The following condition was noticed by Steven Kerkoff:
\begin{theorem}
Suppose $M^3$ is closed and orientable. If every homomorphism
$\phi:\pi_1(M^3)\to \zz$ is zero, then $M^3$ has the sphere-separated
property.
\end{theorem}
\begin{proof}
If $M^3$ has the property that every homomorphism
$\phi:\pi_1(M^3)\to\zz$ is zero, then it is also true for every
homomorphism $H_1(M^3)\to\zz$, since $H_1$ is the abelianization of
$\pi_1$ (by Hurewicz). By the universal coefficient theorem, this
means $H^1(M^3)=0$.
Since $M^3$ is closed and orientable, we can apply Poincar\'e duality,
and get that $H_2(M^3)=0$. This is the same as $H_2^t(M^3)$ since
$M^3$ is orientable. By Lemma~\ref{lem:equiv}, we know that any
embedded sphere separates $M^3$.
\end{proof}
\begin{theorem}
Suppose $M^3$ is closed and orientable. Then $M^3$ has the sphere-separated
property if and only if $M^3$ is a connected sum
\[M^3=A_1\#\dots\#A_k\]
so that the universal cover of each $A_i$ is either a
homotopy-$\re^3$, or a homotopy-$S^3$. So if the Poincar\'e
conjecture is true, the universal covers must be $\re^3$ or $S^3$.
\end{theorem}
\begin{proof}
By Proposition~\ref{prop:conn}, it suffices to prove this for $M^3$ prime.
Now suppose $M^3$ is prime, and its universal cover $\tilde{M}$ is a
homotopy-$\re^3$ or a homotopy-$S^3$. Suppose $\tilde{M}$ is a
homotopy-$S^3$. Then $\pi_2(\tilde{M})=0$, and $\tilde{M}$ is closed,
and orientable, so by Theorem~\ref{thm:pi2}, $\tilde{M}$ satisfies the
sphere-separated property.
If $\tilde{M}$ is a homotopy-$\re^3$, let $\hat{\tilde{M}}$ be the
one-point compactification of $\tilde{M}$. Now $\hat{\tilde{M}}$ is a
compact, orientable Poincar\'e Duality space, with
$\pi_2(\hat{\tilde{M}})=0$, so by Theorem~\ref{thm:pi2} (we need to
generalize this to Poincar\'e duality spaces, which can be done, but
requires us to go back to Lemma~\ref{lem:equiv}) $\hat{\tilde{M}}$
satisfies the sphere-separated property.
If an embedded sphere in $\tilde{M}$ separates $\hat{\tilde{M}}$, it
must separate $\tilde{M}$, since if $x$ and $y$ are points in
$\tilde{M}$, any path connecting $x$ and $y$ in $\tilde{M}$ also
connects $x$ and $y$ in $\hat{\tilde{M}}$. So $\tilde{M}$ satisfies
the sphere-separated property.
In either case, $\tilde{M}$ satisfies the sphere-separated property, so by
Theorem~\ref{thm:covering}, so does $M^3$.
Conversely, suppose $M^3$ is prime and has the sphere-separated
property. By Theorem~\ref{thm:pi2}, $\pi_2(M^3)=0$.
Let $\tilde{M}$ be the universal cover of $M^3$. Then
$\pi_2(\tilde{M})=\pi_2(M^3)=0$, and $\pi_1(\tilde{M})=0$. By the
Hurewicz theorem, $H_1(\tilde{M})=0$ and $H_2(\tilde{M})=0$.
Now since $\tilde{M}$ is a three-manifold, there are two cases: either
the covering space $\tilde{M}\to M$ is a finite covering, or it is an
infinite covering.
Case I: It is a finite covering. Then $\tilde{M}$ is a closed
orientable manifold, and simply-connected. Therefore it is either
$S^3$ or a counterexample to the Poincar\'e conjecture.
Case II: It is an infinite covering. Then $\tilde{M}$ is non-compact,
and $H_3(\tilde{M})=0$. Since $\tilde{M}$ is a three-manifold, its
homology vanishes in higher dimensions, too. By Hurewicz,
$\pi_k(\tilde{M})=0$ for all $k>0$, and by Whitehead's theorem,
$\tilde{M}$ is contractible. Therefore $\tilde{M}$ is a homotopy-$\re^3$.
\end{proof}
\begin{example}
$\re P^2\times S^1$ has the sphere-separated property, and is irreducible:
\end{example}
\begin{proof}
Let $\Sigma^2$ be any embedded two-sphere in $M^3 = \re P^2\times S^1$.
Let $\pi:S^2\times S^1\to \re P^2 \times S^1$ be the natural double cover of
$S^2$ to $\re P^2$, times the identity on $S^1$. Then $\Sigma^2$ lifts to
a sphere $\tilde{\Sigma}^2$ in $S^2\times S^1$. Now consider its
representative $[\tilde{\Sigma}^2]\in H_2(S^2\times S^1;\zz/2)$ in homology
in $\zz/2$ coefficients.
Now $H_2(S^2\times S^1;\zz/2)\cong \zz/2$, and
it is generated by the $S^2$ factor. The map
\[\pi_*:H_2(S^2\times S^1;\zz/2)\to H_2(\re P^2\times S^1;\zz/2)\]
sends this generator to twice itself, but in $\zz/2$ coefficients this is
zero. So $\pi_*$ is zero.
In particular, $[\Sigma^2]=\pi_*[\tilde{\Sigma}^2]=0$.
Now suppose $\Sigma^2$ does not separate $\re P^2\times S^1$. Then
by Lemma~\ref{lem:equiv}, there is a closed curve $\gamma:S^1\to
\re P^2\times S^1$ that intersects $\Sigma^2$ transversally in one point.
By intersection theory in $\zz/2$, $[\gamma]\cdot[\Sigma^2]=1$. But
$[\Sigma^2]=0$. This is a contradiction.
Therefore $\Sigma^2$ separates $\re P^2\times S^1$.
\end{proof}
\begin{thebibliography}{99}
\bibitem{Ma}W. Massey, {\em Algebraic Topology: an introduction},
Springer Verlag, 1967.
\bibitem{Ro}D. Rolfsen, {\em Knots and Links}, Publish or Perish Press,
Berkeley, 1976.
\end{thebibliography}
\end{document}
To fix:'
what if not compact?
Change open region -> compact region, manfld with boundary
Check Pi_2()=0 property holds up for poincare duality spaces (needed later)
Finish showing RP^2 x S^1 has sphere-separated property.
Classify what happens for non-orientable cases
Add bit about Hurewicz map on pi_2 being zero iff property? (Need orientable?)