Diffeomorphism of product implies diffeomorphism of each term?

Question

Suppose $M, N$ are manifolds and we know that $\mathbb{R}^n\times M$ is diffeomorphic to $\mathbb{R}^n\times N$, for the same natural number $n$.

Could it be possible that $M$ and $N$ are not diffeomorphic?

Answer 1

Suppose that $\Sigma^n$ is an exotic sphere, $T\Sigma^n\oplus\mathbb{R}\simeq \Sigma^n\times \mathbb{R}^{n+1}$ is trivial (see the reference). Since the tangent bundle of two exotic spheres are isomorphic, you deduce that if $S^n$ is another exotic sphere, $S^n\times\mathbb{R}^{n+1}$ is isomorphic to $\Sigma^n\times\mathbb{R}^{n+1}$ (see the reference). But $S^n$ and $\Sigma^n$ are not necessarily diffeomorphic.

See the first answer here.

https://mathoverflow.net/questions/58131/parallelizability-of-the-milnors-exotic-spheres-in-dimension-7

Answer 2

As explained by the existing answer, if $\Sigma$ is an exotic $n$-sphere, then $T\Sigma$ is vector bundle isomorphic to $TS^n$, which implies $T\Sigma \oplus \underline{\Bbb R}$ is vector bundle isomorphic to $TS^n \oplus \underline{\Bbb R}$. Since isomorphic vector bundles have diffeomorphic total spaces, that means $\Sigma \times \Bbb R^{n+1} \cong S^n \times \Bbb R^{n+1}$ but $\Sigma$ and $S^n$ are not diffeomorphic by definition.

For another example of a similar sort, if $M$ is an exotic $\Bbb R^4$, then $M \times \Bbb R$ is homeomorphic to $\Bbb R^5$. Since for all $n \neq 4$, topological $\Bbb R^n$'s are smooth $\Bbb R^n$'s (Stallings, 1962), you get than $M \times \Bbb R$ is diffeomorphic to $\Bbb R^4 \times \Bbb R$, but $M$ is not diffeomorphic to $\Bbb R^4$.

Notably this is not even true in the topological category; let $M$ be the Whitehead manifold, given by throwing away the Whitehead continuum (limit of a decreasing intersection of Whitehead links) out of $\Bbb R^3$. This is an open submanifold of $\Bbb R^3$, and $M \times \Bbb R$ is homeomorphic to $\Bbb R^4$ - but $M$ is not homeomorphic to $\Bbb R^3$; it's not simply connected at infinity.