# Are two homeomorphic hypersurfaces of the same smooth manifold also diffeomorphic?

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.

• 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
– BCLC
May 24, 2021 at 5:08

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 $$N_i$$.

• 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? Apr 17, 2021 at 8:07
• No, and in fact that was my original idea. I only chose $S^1$over $\mathbb{R}$ due to the compactness bit. 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.