If the interior of a manifold with boundary is smooth, is the whole manifold smooth

Question

Let $M$ be a topological manifold with boundary. Let Int$M$, its interior, be a smooth manifold. Is it a known result that $M$ itself will be a smooth manifold with boundary? Can we extend a smooth structure on Int$M$ to include the boundary points of $M$?

I tried to come up with counterexamples, all the simple ones seem to work out. This feels like a natural result and it would be a useful lemma for a problem I'm working on.

If this isn't known, any hints on how to begin?

Answer 1

For instance, you can start with the $E8\oplus E8$ manifold $M$: This is a 4-dimensional closed simply connected manifold with the intersection form isomorphic to $E8\oplus E8$. This manifold has vanishing Kirby-Siebenmann invariant $k(M)$ (the signature is divisible by 16), but is not smoothable (this was first proven by Donaldson). The existence of such manifolds is nontrivial: Freedman proved that you can realize any unimodular integer quadratic form as the intersection form of a closed simply-connected topological 4-manifold. Since $k(M)=0$, the manifold $M\times (0,1)$ is smoothable. (This is again nontrivial and is due to Kirby and Siebenmann: For a closed $4$-manifold $k(M)=0$ if and only if $M\times (0,1)$ is smoothable.)

In particular, taking $W=M\times [0,1]$ we obtain a compact 5-dimensional manifold whose interior is smoothable but the boundary $M\sqcup M$ is not.

You can read much more at the manifold atlas.

Addendum. One can say precisely what happens for (closed) simply-connected 4-manifolds $M$ which bound 5-manifolds $N$ with smoothable interior. By looking at the collar neighborhood of $M$ in $N$, we see that $M\times (0,1)$ embeds in $int(N)$ and, hence, is smoothable. Thus, $k(M)=0$. The homeomorphism type of $M$ is then determined only by its intersection form (Freedman), more precisely, by the lattice $(H_2(M), Q)$ where $Q$ is the intersection form. The intersection form is unimodular (and even). There are two cases:

(a) $Q$ is definite. Then, by Donaldson's theorem, $Q$ is diagonal, i.e. is given by the rank $r$ identity matrix $I_r$ or by $-I_r$. This form $Q$ is realized by the smooth manifold which is the connected sum of $r$ copies of $CP^2$ or the same manifold with opposite orientation.

(b) $Q$ is indefinite. Each indefinite unimodular form is the $r$-fold direct sum of rank 2 hyperbolic quadratic forms. This is realized by the connected sum of $r$ copies of $S^2\times S^2$.

To conclude:

$M$ is smoothable if and only if its intersection form is indefinite or is the form $\pm I_r$, of rank $r$.

Once you drop simple connectivity, there are some known obstructions to smoothability (again, the intersection form is diagonal if definite). But the overall picture is totally unclear.