# The convex hull of a manifold whose cobordism class is trivial

Let $$M$$ be a compact orientable $$n$$ dimensional manifold. Assume that $$M$$ has trivial cobordism class.
Is there an embedding of $$M$$ in some Euclidean space $$\mathbb{R}^m$$ such that the convex hull of $$M$$ is a $$n+1$$ dimensional manifold whose boundary is $$M$$?

Here the image of $$M$$ under the embedding is denoted again by $$M$$.

Note: One can pose the same question in the following geometric manner:

Let $$M$$ be a compact Riemannian manifold with trivial cobordism class. Is there an isometric embedding of $$M$$ in some Euclidean space such that the convex hull of $$M$$ is a manifold whose boundary is $$M$$?

• Generally no. For example, plenty of homotopy spheres do not bound contractible manifolds. Many (?most?) homology 3-spheres do not bound contractible 4-manifolds, yet all 3-manifolds are null cobordant. May 14 at 7:09
• @RyanBudney Snap! May 14 at 7:12
• @RyanBudney I think Freedman proved that every homology 3-sphere bounds a contractible topological 4-manifold. May 14 at 11:33
• @Zerox, I doubt it, but if you can find such a claim somewhere, you will have found a contradiction. For example, the Poincare Dodecahedral Space does not bound a contractible 4-manifold, due to the Rochlin invariant. 2 days ago
• @RyanBudney I think you missed the word topological in Zerox's comment. On the other hand, the original question is a bit ambiguous about the category one is working in. The first part could be in the smooth/PL/topological category, and might have different answers. The second part presumably refers to smooth manifolds (because of the Riemannian metric). 2 days ago

Implicit in the other responses is the fact that if $$M$$ bounds a convex manifold $$W$$, then $$W$$ is contractible and so M has the homology of a sphere. So any null-cobordant manifold that is not a homology sphere is a counterexample. Eg take $$M = X \# -X$$ where $$X$$ is any orientable manifold with non-trivial homology.