Let $M$ be a smooth (connected, without boundary) manifold and $N_1$, $N_2$ be two smooth (connected, without boundary) hypersurfaces of $M$. Suppose $N_1$ and $N_2$ are homeomorphic. Can $N_1$ and $N_2$ be non-diffeomorphic?

I am currently working on a problem where I have shown that two smooth, compact hypersurfaces $N_1$ and $N_2$ of the same manifold $M$ are both homeomorphic (even $C^{\alpha}$-homeomorphic for some $\alpha \in (0,1)$) to the same manifold $N$. However, I would like to use some differential properties of both $N_1$ and $N_2$ and it would be suitable that they have the same differential structure.

I know there exists many examples of non-diffeomorphic manifolds which are homeomorphic, such as exotic spheres. However, I don't know if one can realize two different differentiable spheres as hypersurfaces of the same smooth manifold.

  • $\begingroup$ wait technically your title question and question in the post actually ask the opposite things? but eh whatever i guess lol en.wikipedia.org/wiki/Tag_question $\endgroup$
    – BCLC
    May 24, 2021 at 5:08

2 Answers 2


Here is (what I believe is) a counterexample which works for any applicable $N_1,N_2$, which additionally is compact whenever $N_1$ and $N_2$ are:

Choose two homeomorphic, but not diffeomorphic manifolds $N_1,N_2$. Let $M=(N_1\times S^1)\#(N_2\times S^1)$, where $\#$ denotes a smooth connected sum (chosen with arbitrary orientation if applicable). We can always construct this sum by modifying a sufficiently small neighborhood from each $N_i\times S^1$ factor that $M$ retains an embedded hypersurface diffeomorphic to $N_i$.

  • $\begingroup$ Thank you very much for this beautifil coutner-example. I guess one could also construct two homeomorphic non-diffeomorphic compact hypersufaces of the same non-compact manifold with $(N_1\times \mathbb{R})\sharp (N_2\times \mathbb{R})$ by gluing on suitables neighbourhood, am I wrong? $\endgroup$
    – Didier
    Apr 17, 2021 at 8:07
  • 1
    $\begingroup$ No, and in fact that was my original idea. I only chose $S^1$over $\mathbb{R}$ due to the compactness bit. $\endgroup$
    – Kajelad
    Apr 17, 2021 at 16:28

In fact, this can fail spectacularly.

There is a submersion $\pi:\mathbb{R}^5\rightarrow \mathbb{R}$ with the property that for any $t\in \mathbb{R}$, $\pi^{-1}(t)$ is homeomorphic to $\mathbb{R}^4$, but for any $t\neq s\in\mathbb{R}$, the inverses images are not diffeomorphic.

I learned about this fact from Ryan Budney's answer to an MO question.

  • $\begingroup$ (Community Wiki since I didn't do anything other than link somewhere else) $\endgroup$ Apr 16, 2021 at 20:50
  • $\begingroup$ Thank you very much for this beautiful answer. It really is spectacular. I accepted the other answer because it is constructive and works for compact hypersurfaces, but yours is really helpful too. Thanks again. $\endgroup$
    – Didier
    Apr 17, 2021 at 8:04
  • 1
    $\begingroup$ I agree that Kajelad's answer should be accepted. But this is one of my favorite facts-I-have-no-idea-how-to-prove, so I felt I had to share it. $\endgroup$ Apr 17, 2021 at 16:21
  • $\begingroup$ And I am glad you shared it with me. It will certainly become one of my favorite facts-I-have-no-idea-how-to-prove too $\endgroup$
    – Didier
    Apr 17, 2021 at 16:23

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