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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?

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  • $\begingroup$ This has to be in the literature somewhere... Internet suggests it's true for simply-connected n-manifolds with n > 4. $\endgroup$ – Chris Gerig Jun 25 at 3:02
  • $\begingroup$ Definitely not true if M is of dim 4. $\endgroup$ – Anubhav Mukherjee Jun 25 at 3:04
  • $\begingroup$ @ 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. $\endgroup$ – Mohammad Farajzadeh-Tehrani Jun 25 at 3:04
  • $\begingroup$ @ Mukherjee: Definitely means you know a result, please let me know. $\endgroup$ – Mohammad Farajzadeh-Tehrani Jun 25 at 3:06
  • $\begingroup$ There are closed flat manifolds that are not homotopy equivalent but become diffeomorphic after product with $S^1$, see math.stackexchange.com/questions/396608/…. Also if $M$ is closed manifold of dimension $\ge 6$ (maybe $5$?) such that the Whitehead group of $\pi_1(M)$ is nonzero, then there is a $h$-cobordism $W$ from $M$ to a manifold $L$ not diffeomorphic to $M$, and $W\times S^1$ is trivial, so $M\times S^1$, $L\times S^1$ are diffeomorphic. $\endgroup$ – Igor Belegradek Jun 25 at 11:26
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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.

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 (http://www.numdam.org/item/?id=PMIHES_1964__22__61_0). 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 $M'$.

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  • $\begingroup$ 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. $\endgroup$ – Mohammad Farajzadeh-Tehrani Jun 25 at 3:17
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    $\begingroup$ @MohammadFarajzadeh-Tehrani I tried to add a few references. Please let me know if it is helpful or not. $\endgroup$ – Anubhav Mukherjee Jun 25 at 3:38
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    $\begingroup$ very good. thanks $\endgroup$ – Mohammad Farajzadeh-Tehrani Jun 25 at 3:39
  • $\begingroup$ @MohammadFarajzadeh-Tehrani I made a few changes in my ans. $\endgroup$ – Anubhav Mukherjee Jun 25 at 4:06
  • $\begingroup$ Sorry if I am being daft, but why is the image of $\bar{f}$ separating? $\endgroup$ – Michael Albanese Jun 25 at 15:34
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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).

For a more explicit example, see this excellent answer by George Lowther.

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  • $\begingroup$ Thanks for the answer. I was thinking about that too. In your response, $M$ and $N$ are compact but not closed, correct? $\endgroup$ – Mohammad Farajzadeh-Tehrani Jun 25 at 13:26
  • $\begingroup$ They are closed, I will edit to make this clear. $\endgroup$ – Michael Albanese Jun 25 at 13:50

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