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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 $M$?

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    $\begingroup$ 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. $\endgroup$
    – Ryan Budney
    Commented May 14, 2022 at 7:09
  • $\begingroup$ @RyanBudney Snap! $\endgroup$
    – Jonny Evans
    Commented May 14, 2022 at 7:12
  • $\begingroup$ @RyanBudney I think Freedman proved that every homology 3-sphere bounds a contractible topological 4-manifold. $\endgroup$
    – Zerox
    Commented May 14, 2022 at 11:33
  • $\begingroup$ @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. $\endgroup$
    – Ryan Budney
    Commented May 14, 2022 at 23:45
  • $\begingroup$ @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). $\endgroup$
    – Danny Ruberman
    Commented May 15, 2022 at 2:28

2 Answers 2

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There are exotic spheres (which are null cobordant) which do not bound a parallelisable manifold. Since the convex hull is contractible, it would be parallelisable if it were a manifold, so these guys do not admit embeddings like you want.

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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.

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