Is a manifold Euclidean if its tangent bundle is Euclidean?

Question

I'm wondering whether an $n$-dimensional manifold diffeomorphic to $\mathbb{R}^n$ if its tangent bundle is diffeomorphic to $\mathbb{R}^{2n}$. Many thanks!

Answer 1

A $3$-dimensional counterexample is the Whitehead manifold $W$. It is known $W$ is a contractible manifold, and so its tangent bundle is trivial. As well, $W \times \mathbb{R}$ is homeomorphic to $\mathbb{R}^4$ (see here), and so $T(W)$ is homeomorphic to $\mathbb{R}^6$. However, $\mathbb{R}^6$ has a single smooth structure, so we conclude there is a diffeomorphism $T(W) \cong \mathbb{R}^6$, but of course $W \not \cong \mathbb{R}^3$.

Answer 2

Let $M$ be a smooth $n$-manifold. Its tangent bundle $TM$ is diffeomorphic to $\mathbb{R}^{2n}$ if and only if $M$ is open and contractible.

Proof: One direction is straightforward. Any vector bundle is homotopy equivalent to its base, so if $TM$ is diffeomorphic to $\mathbb{R}^{2n}$, then $M$ is contractible. If $M$ had boundary, then so too would $TM$, so we see that $M$ must be open.

For the converse direction, if $M$ is a contractible $n$-manifold, then its tangent bundle is trivial so $TM$ is diffeomorphic to $M\times\mathbb{R}^n$. If $n \leq 2$, then $M$ is diffeomorphic to $\mathbb{R}^n$ so $TM$ is diffeomorphic to $M\times\mathbb{R}^n = \mathbb{R}^n\times\mathbb{R}^n = \mathbb{R}^{2n}$. Suppose now that $n \geq 3$. The product $M\times\mathbb{R}$ is simply connected at infinity (see Proposition 2.2 of Stallings - The Piecewise Linear Structure of Euclidean Space, together with Proposition 16.4.1 of Geoghegan - Topological Methods in Group Theory). So $M\times\mathbb{R}^n = (M\times\mathbb{R})\times\mathbb{R}^{n-1}$ is homeomorphic to $\mathbb{R}^{n+1}\times\mathbb{R}^{n-1} = \mathbb{R}^{2n}$, see this answer. Since $\mathbb{R}^{2n}$ has a unique smooth structure up to diffeomorphism (because $n \geq 3$), we see that $M\times\mathbb{R}^n$ is diffeomorphic to $\mathbb{R}^{2n}$. $\square$

Therefore, there are two types of examples which provide a negative answer to your question:

• smooth open contractible manifolds which are homeomorphic but not diffeomorphic to $\mathbb{R}^n$, and
• smooth open contractible manifolds which are not homeomorphic to $\mathbb{R}^n$.

The former examples are necessarily exotic $\mathbb{R}^4$'s as mentioned in Aleksei Kulikov's comment. The latter examples can only exist for $n \geq 3$ and can be equivalently described as open contractible manifolds which are not simply connected at infinity. An example of such a manifold is the Whitehead manifold as in Connor Malin's answer. There are also examples for $n \geq 4$, see Theorems 5.a.1 and 5.b.1 of Davis & Januszkiewicz - Hyperbolization of Polyhedra - note, the manifolds they construct do not necessarily admit a smooth structure, but their universal covers do (see the corollary on page 91 of Lashof - The Immersion Approach to Triangulation and Smoothing).