Let $W$ be a $3$-dimensional $h$-cobordism of closed surfaces $M_0$ and $M_1$. Can we prove that $W$ is trivial? That is, $W$ is homeomorphic to $M_0 \times [0,1]$.
$\begingroup$
$\endgroup$
0
Add a comment
|
1 Answer
$\begingroup$
$\endgroup$
4
For $M_0=M_1=S^2$, this follows from the 3-dimensional Poincaré conjecture: glueing in 3-balls, you get a simply connected $3$-manifold, that has to be $S^3$, so $W=S^2\times\left[0,1\right]$.
For surfaces of higher genus, the result follows from Waldhausens rigidity theorem, which says that homotopy-equivalent Haken manifolds with homeomorphic boundaries are homeomorphic. (An irreducible $3$-manifold with non-spherical boundary is necessarily a Haken manifold.)
-
2$\begingroup$ This is morally correct but there are two points. First is that the result for an irreducible h-cobordism is due to Stallings (On fibering certain 3-manifolds) well before Waldhausen. To show irreducibility one also needs the Poincaré conjecture, proved subsequently by Perelman. $\endgroup$ Commented Jun 9, 2021 at 11:50
-
$\begingroup$ Does the same conclusion hold if the boundary is $P^2$? $\endgroup$– AdterramCommented Aug 29, 2021 at 6:07
-
$\begingroup$ @Adterram - Yes, this follows from a theorem of Livesay. - doi.org/10.2307/1970543 - in that case we get $P \times I$. $\endgroup$– Sam NeadCommented Sep 4, 2021 at 15:48
-
$\begingroup$ @Adterram - Also, if the answer answers your question, it is polite to accept it - you just tick the tick mark. best $\endgroup$– Sam NeadCommented Sep 4, 2021 at 15:51