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Consider a triangulated ball $D$, and assume that $\omega$ is an assignment of real weights to the simplices of $D$, including the empty one, such that for every maximal simplex $F$ and every simplex $G$ in $F$, we have

$\sum_{x \in [G,F]} \omega(x) \ge 0.$

Is it true that

$\sum_{x \in D} \omega(x) \ge 0$?

Here $[G,F]$ is the interval of $G$ to $F$, that is, all simplices containing $G$ and contained in $F$.

Edit: for Cohen-Macaulay complexes (which are more general than balls) the answer is no. Raman Sanyal showed that the dual linear program is infeasible for an example of nonpartitionable complexes. I will leave it open since my original question is unsolved, but that provides an indication.

Accepting Raman's answer even though the question for disks is open

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    $\begingroup$ An equivalent reformulation: can we assign to each pair $(F,G)$ of a maximal simplex $F$ and a simplex $G$ in $F$ a nonnegative weight $u(F,G)$ such that $\sum_{x\in [G,F]} u(F,G)=1$ for all $x\in D$? $\endgroup$
    – Ilya Bogdanov
    Commented Sep 14, 2023 at 6:14
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    $\begingroup$ Thanks Ilya. Let me add that I also don't know this to be false for any Cohen-Macaulay poset. $\endgroup$
    – Karim Adiprasito
    Commented Sep 14, 2023 at 23:06
  • $\begingroup$ @IlyaBogdanov: your summand does not depend on $x$. $\endgroup$
    – Richard Stanley
    Commented Sep 16, 2023 at 1:37
  • $\begingroup$ @RichardStanley yes that is the point. You sum over those pairs $G$ and $F$ whose interval contains $x$ $\endgroup$
    – Karim Adiprasito
    Commented Sep 16, 2023 at 2:42
  • $\begingroup$ You can think of this as a fractional partitioning $\endgroup$
    – Karim Adiprasito
    Commented Sep 16, 2023 at 2:44

1 Answer 1

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SAGE says it's not true for non-partionable but Cohen-Macaulay complexes. Problem of finding $\omega$ can be reformulated as a LP feasibility problem (as explained by Ilya) and for an example from "A non-partitionable Cohen–Macaulay simplicial complex" the dual problem was infeasible.

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    $\begingroup$ Thanks. Can you add the weight that provides the counterexample? $\endgroup$
    – Karim Adiprasito
    Commented Sep 17, 2023 at 11:29
  • $\begingroup$ Checked as well, seems correct! $\endgroup$
    – Karim Adiprasito
    Commented Sep 17, 2023 at 14:22
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    $\begingroup$ I'm confused as to how this answers the question as literally stated. The example under discussion is not a triangulated ball, is it? $\endgroup$
    – Sam Hopkins
    Commented Sep 17, 2023 at 14:33
  • $\begingroup$ @SamHopkins I would assume that the example is realizablo as such. But, anyway, it would be much more pleasant if such details appeared here, in order not to urge each reader to find those by themselves. $\endgroup$
    – Ilya Bogdanov
    Commented Sep 17, 2023 at 22:26
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    $\begingroup$ @IlyaBogdanov: I doubt this, because if you look at the paper (at least, the arXiv version arxiv.org/abs/1504.04279), you'll see "Question 4.1. Is every simplicial ball partitionable?" $\endgroup$
    – Sam Hopkins
    Commented Sep 17, 2023 at 22:30

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