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Oct 10, 2023 at 7:56 history edited Karim Adiprasito CC BY-SA 4.0
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Oct 10, 2023 at 7:54 vote accept Karim Adiprasito
Sep 18, 2023 at 13:25 comment added Sam Hopkins As you mention, Karim, the problem (in Ilya's dual reformulation) is a kind of fractional partitioning: specifically, if we have a partitioning then we can set $u(F,G)=1$ if $F$ and $G$ are the bottom and top elements in an interval in our partitioning, and $0$ otherwise. Since the question of whether every triangulated ball is partitionable appears to be open, this would suggest the answer to your question is also not known. (Of course in theory it might be possible to show that a fractional partitioning always exists without showing a non-fractional partitioning does.)
Sep 18, 2023 at 4:57 history edited Karim Adiprasito CC BY-SA 4.0
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Sep 17, 2023 at 21:12 comment added Sam Hopkins @IlyaBogdanov: it might be worth pointing out that you are using duality of linear programming, or more precisely Farkas' lemma (en.wikipedia.org/wiki/Farkas%27_lemma)
Sep 17, 2023 at 14:22 vote accept Karim Adiprasito
Sep 18, 2023 at 3:35
Sep 17, 2023 at 10:20 answer added Raman Sanyal timeline score: 4
Sep 16, 2023 at 2:44 comment added Karim Adiprasito You can think of this as a fractional partitioning
Sep 16, 2023 at 2:42 comment added Karim Adiprasito @RichardStanley yes that is the point. You sum over those pairs $G$ and $F$ whose interval contains $x$
Sep 16, 2023 at 1:37 comment added Richard Stanley @IlyaBogdanov: your summand does not depend on $x$.
Sep 14, 2023 at 23:06 comment added Karim Adiprasito Thanks Ilya. Let me add that I also don't know this to be false for any Cohen-Macaulay poset.
Sep 14, 2023 at 6:14 comment added Ilya Bogdanov 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 $x\in D$?
Sep 14, 2023 at 1:49 history asked Karim Adiprasito CC BY-SA 4.0