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The following question arose from a survey paper I am writing on combinatorial reciprocity. Let $\mathcal{P}$ be a $d$-dimensional convex polytope. Let $\mathcal{Q}$ be a union of facets (codimension one faces) of $\mathcal{P}$. Suppose that $\mathcal{Q}$ has Euler characterisic $1$, and that the local Euler characteristic about any point $p$ of $\mathcal{Q}$ is either $1$ or $1+(-1)^d$. Does it follow that $\mathcal{Q}$ is homeomorphic to a ball?

If the answer is negative, what if we assume instead that $\mathcal{Q}$ is acyclic and the local homology at each point is the same as that of a $(d-2)$-ball or $(d-2)$-sphere?

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    $\begingroup$ If $P\subset \mathbb R^{n+1}$ is a simple polytope (en.wikipedia.org/wiki/Simple_polytope) then any union of facets will satisfy the conditions concerning local Euler characteristic – so we may essentially forget about those conditions. By taking $P$ with enough facets, one should be able to find examples of $Q$ that are homeomorphic to any $n$-manifold with boundary that embeds in the sphere $S^n$. So the condition about Euler characteristic being 1 sounds very much weaker than being homeomorphic to a ball. $\endgroup$
    – André Henriques
    Commented Dec 29, 2022 at 22:16
  • $\begingroup$ @AndréHenriques: what is a specific example? How can we be sure it embeds on the boundary of a convex polytope? $\endgroup$
    – Richard Stanley
    Commented Dec 29, 2022 at 22:50
  • $\begingroup$ A negative answer to my question follows from Theorem 2 of the paper services.math.duke.edu/~ezra/Reciprocal/reciprocal.pdf by Ezra Miller and Victor Reiner. $\endgroup$
    – Richard Stanley
    Commented Jan 12, 2023 at 15:26

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Here's an $(n+1)$-dimensional polytope that one can use to construct countereaxamples:
The convex hull of the points $$ \left(x_1,x_2,\ldots,x_n,\sum_{1=1}^nx_i^2\right) $$ where $x_i\in \{-N,\ldots,N-2,N-1,N\}$, for some fixed $N\gg1$.

This polytope's boundary contains an embedded copy of a large chunk of $\mathbb R^n$ with its standard cubulation.

Now fix any cell complex with Euler characteristic $1$ that embeds in $\mathbb R^n$. Thicken it so as to make another space that deformation retracts to it, and which is a union of unit cubes. If the thickening is an $n$-dimensional manifold with boundary, then you've got your counterexample.

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    $\begingroup$ This makes sense. Of course you don't want the thickened cell complex to be a ball, but this should be achievable from your comment above. $\endgroup$
    – Richard Stanley
    Commented Dec 31, 2022 at 19:44
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Here is a partial answer. There are finite acyclic 2-complexes that are not contractible. Any such 2-complex can be embedded in $\mathbb{R}^5$, and a regular neighbourhood of such a 2-complex will be a compact 5-manifold with boundary that is a subcomplex of (a finite piece of) $\mathbb{R}^5$ and that is acyclic and not contractible. It thus has the property that the local homology at each point is either that of a 4-ball or a 4-sphere. Take a larger piece of the given triangulated $\mathbb{R}^5$ that is homeomorphic to a 5-ball and that contains the given non-contractible piece as a subcomplex. Now project the larger ball-shaped piece homeomorphically onto a piece of a 5-sphere. Presumably (and this is why this is only a partial answer), one can do this in such a way that the given triangulation is part of a simplicial convex polytope structure on $S^5$. Note that there is a lot of freedom in how to choose the original triangulation of $\mathbb{R}^5$, so I don't see why this last step would fail.

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    $\begingroup$ I agree that it's not difficult to to find a negative answer to my question on the boundary of a sphere. The whole point of the question is to do it on the boundary of a convex polytope. Also, Kalai shows in link.springer.com/article/10.1007/BF02187893 that an overwhelming proportion of sphere triangulations (in dimension at least three) are not polytopal. This suggests that a "random" pure simplicial complex of dimension $d-1$ cannot realized as a a set of facets of a $d$-dimensional convex polytope, so it may not be so easy to find the necessary polytope structure. $\endgroup$
    – Richard Stanley
    Commented Dec 30, 2022 at 21:44
  • $\begingroup$ Is the barycentric subdivision of a polytopal triangulation also polytopal? I am not sure, but I suspect that any finite 2-complex will embed in a sufficiently fine subdivision of any triangulation of $S^5$. $\endgroup$
    – IJL
    Commented Jan 1, 2023 at 15:07

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