Does every manifold with boundary admit a flat metric?

Question

There are several obstructions for a closed manifold $M$ admit a flat metric. for example Cartan-Hadamard Theorem implies the universal cover of $M$ must be $\mathbb R^n$. So if the manifold $M$ is allowed to have nonempty boundary and we do not put any restrictions on the boundary(say we do not ask it to be convex). Then can we say something?

Answer 1

You should explain what you mean by a flat metric is there is nonempty boundary. Whatever definition you have in mind, it should imply the existence of a flat metric on the interior of your manifold. You should also assume that the manifold is connected. For connected oriented manifolds of dimension $n\le 3$ there are no obstructions since every such manifold admits an immersion in $R^n$.

However, starting with dimension $n\ge 4$ there are obstructions. The simplest one is the following. Suppose that your manifold $M$ is also simply-connected. Then the existence of a flat metric on the interior of $M$ implies that $M$ admits an (isometric) immersion in $R^n$. This, in turn, implies that $M$ is parallelizable. (Actually, an open $n$-manifold is parallelizable if and only if it admits an immersion in $R^n$.) Now, take your favorite simply-connected non-parallelizable $n$-manifold with nonempty boundary, for instance, let $M$ be the complement to an open ball in $CP^2$. (It is not parallelizable because $CP^2$ is not a spin-manifold.) The interior of this manifold will not have a flat metric.