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As we know, an exotic $\mathbb{R}^4$ is a manifold which is homeomorphic, but not diffeomorphic to the standard $(\mathbb{R}^4,id)$, and there are even very explicit descriptions of them (Kirby diagrams, etc). Being descriptions from the "outside" (forgive the less exact tone) here, I wonder if there are "inside" descriptions, i.e.:

When you are sitting inside a four-manifold which have the topological properties of being an $\mathbb{R}^4$ (for example the right homology and homotopy properties), can you decide (and if yes, how) that it is not the standard $(\mathbb{R}^4,id)$, but an exotic one ?

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One possible formalization of "inside" is "local," and in that case it might be possible to have an exotic $\mathbb{R}^4$ which is locally, but not globally, diffeomorphic to standard $\mathbb{R}^4$. (I don't know if this is true or not.) –  Qiaochu Yuan Jul 10 '11 at 22:39
@Qiaochu: By definition of "manifold" an exotic $\mathbb{R}^4$ is locally diffeomorphic to the standard $\mathbb{R}^4$. –  Jason DeVito Jul 10 '11 at 22:52
@Jason: ha. Right. So one can't perform local experiments on an exotic $\mathbb{R}^4$ to distinguish it from standard $\mathbb{R}^4$, then, right? –  Qiaochu Yuan Jul 10 '11 at 23:03

1 Answer 1

Here is one way of possibly deciding which is a mixture of local and global, but is all internal. It "only" requires knowledge of every complete Riemannian metric on your manifold (so is completely unpractical, but is still completely internal).

It's local in the sense that you can look at your manifold pointwise and get the answer (which is nice because, as covered in the comments, an exotic $\mathbb{R}^4$ is locally diffeomorphic to the standard $\mathbb{R}^4$), but it's global in that you really do have to look at all of the points.

Let $M^4$ be a manifold which has the same topological type as $\mathbb{R}^4$. For every complete Riemannian metric $g$ on $M$, compute the sectional curvature. I'll use the notation "$sec_g \geq 0$" to mean that the sectional curvatures of all 2-planes at all points are greater than or equal to $0$ in the (complete) metric $g$.

Your manifold $M^4$ is an exotic $\mathbb{R}^4$ iff it is never the case that $sec_g\geq 0$ or $sec_g\leq 0$.

Here's the idea of the proof. If your metric $g$ on your manifold happens to satisfy $sec_g \leq 0$, then by the Cartan-Hadamard theorem, $M^4$ is covered by the standard $\mathbb{R}^4$. Since $M^4$ is simply connected, this implies it's diffeomorphic to the standard $\mathbb{R}^4$.

On the other hand, if your metric $g$ happens to satisfy $sec_g\geq 0$, then by the Soul Theorem, $M$ is diffeomorphic to the normal bundle over some compact connected submanifold $N$ inside of it. In particular, the topology of $M^4$ is that of a compact manifold. But $M^4$ is simply connected, so $N$ must also be simply connected. This implies $N$ is orientable. But then this implies $H_{\text{dim }N}(N)\neq 0$. But since $N$ has the homotopy type of $M^4$, the only nontrivial homology it has is in dimension $0$. Thus $N$ has dimension $0$, i.e., $N$ is a point. But the normal bundle of a point in any manifold is diffeomorphic to the standard $\mathbb{R}^n$, showing that $M^4$ is diffeomorphic to the standard $\mathbb{R}^4$.

Conversely, the standard Euclidean metric on the standard $\mathbb{R}^4$ has $sec = 0$

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