# If $M$ and $N$ are closed and $M\times S^1$ is diffeomorphic to $N\times S^1$, is it true that $M$ and $N$ are diffeomorphic?

If $$M$$ and $$N$$ are closed smooth manifolds, and $$M\times S^1$$ is diffeomorphic to $$N\times S^1$$, is it true that $$M$$ and $$N$$ are diffeomorphic?

• This has to be in the literature somewhere... Internet suggests it's true for simply-connected n-manifolds with n > 4. Jun 25, 2020 at 3:02
• Definitely not true if M is of dim 4. Jun 25, 2020 at 3:04
• @ Chris: I am sure it is. I don't know where to find it. I googled but did not find anything. I asked our topologists; no response yet. The one with S^2 instead of S^1 is (negative) pretty famous and important. Jun 25, 2020 at 3:04
• @IgorBelegradek: Sorry, I didn't see your comment when I posted my answer which is effectively your first sentence. Jun 25, 2020 at 14:04
• @MichaelAlbanese: no worries. I was too lazy to post an answer, and it is good that you did that, since the answers get more visibility. Jun 25, 2020 at 16:00

If $$M$$ is of dimension $$<4$$ then the answer is YES because there are no exotic structures on $$M$$ and there are full classification results.[EDIT: In case of 3-manifolds this is true except some surface bundles over $$S^1$$ with fiber genus $$>1$$ and periodic monodromy, Stability of 3-manifolds.]

It is not true in dimension 4. For example, any closed simply-connected 4-manifold $$M$$ and an exotic copy $$M'$$ are h-cobordant by a theorem of Wall. Thus $$M\times S^1$$ is h-cobordant to $$M'\times S^1$$ (as one can extend the previus h-cobordsim trivially on the $$S^1$$ component). This is then a trivial cobordism by the high-dimensional s-cobordism theorem which says that such an h-cobordism is trivial if the Whitehead torsion of $$\pi_1(M\times S^1)$$ vanishes and indeed $$Wh(\pi_1(M\times S^1))= Wh(\mathbb Z)=0$$ by a result of Bass. So they are in fact diffeomorphic.

When the dimension is $$>4$$ the answer is YES if $$M$$ is simply-connected. To see this, notice that it is enough to show that $$M$$ and $$M'$$ are h-cobordant. Since $$M\times S^1$$ is diffeomorphic to $$M'\times S^1$$ there is a map $$f:M \to M'\times S^1$$. Since $$M$$ is simply-connected $$f$$ has a lift $$\bar{f}$$ to the universal cover $$M'\times \mathbb R$$. We claim that the image of $$\bar{f}$$ separates $$M'\times \mathbb R$$. Otherwise we can cut $$M'\times \mathbb R$$ along $$Im(\bar{f})$$ and connect the two boundary components by an arc $$\gamma$$. This arc in the original manifold $$M'\times \mathbb R$$ gives rise to a closed curve $$\gamma'$$ which transversally intersects $$Im(\bar{f})$$ at a single point. But $$M'\times \mathbb R$$ is simply-connected and thus $$\gamma'$$ is homotopic to a point disjoint from $$Im(\bar{f})$$, contradicting the homotopy-invariant count of transverse intersection points. Now, since $$M$$ is compact we can find a cobordism from $$Im(\bar{f})\approx M$$ to $$M'\times \{t\}$$ for some sufficiently large $$t\in \mathbb R$$. Since everything is simply-connected and the projection map induces isomorphisms on homologies, by Hurewitz' theorem we can conclude that this is an h-cobordism and so $$M$$ is diffeomorphic to $$M'$$.

• Thanks, that was quick, though I am not familiar with some of the content. Probably you can expand the second and third sentences of your answer a little bit. Jun 25, 2020 at 3:17
• Sorry if I am being daft, but why is the image of $\bar{f}$ separating? Jun 25, 2020 at 15:34
The manifolds $$M$$ and $$N$$ may not even be homotopy equivalent!
In Compact Flat Riemannian Manifolds: I, Charlap showed that there are two closed flat manifolds $$M$$ and $$N$$ of the same dimension which are not homotopy equivalent (this is equivalent to $$\pi_1(M) \not\cong \pi_1(N)$$ as $$M$$ and $$N$$ are aspherical), such that $$M\times S^1$$ and $$N\times S^1$$ are diffeomorphic (which is equivalent to $$\pi_1(M\times S^1) \cong \pi_1(N\times S^1)$$ as closed flat manifolds are determined up to diffeomorphism by their fundamental group).
• Thanks for the answer. I was thinking about that too. In your response, $M$ and $N$ are compact but not closed, correct? Jun 25, 2020 at 13:26