Are exotic $\mathbb{R}^4$ flat?

Question

$\mathbb{R}^4$ can be given two different differential structures which are not diffeomorphic.

The standard $\mathbb{R}^4$ with its common differential structure can be given a metric in which its Riemann tensor is zero everywhere.

Do exotic differential structures also admit a metric with Riemann tensor vanishing everywhere?

Answer 1

Below, $M=M^4$ is a smooth contractible 4-manifold (for instance, an exotic $R^4$).

The answer to your question about existence of a flat metric on $M$ depends on how much do you assume about the flat metric:

1. If you assume that the metric on $M$ is complete, then $(M,g)$ is isometric to the Euclidean 4-space $E^4$, hence, $M$ cannot be exotic.

2. If you do not assume completeness then $M$ ideed admits a flat metric. The existence of such a metric is equivalent to the existence of an immersion $M^4\to E^4$. But, since $M$ is contractible, it is parallelizable. Every parallelizable $n$-manifold admits an immersion in $E^n$ by the Hirsch-Smale immersion theory. See this Mathoverflow discussion for a collection of references. The original reference is

M. Hirsch, Immersions of manifolds, Trans. Amer. Math. Soc. 93 (1959), 242–276.