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Let $P$ be a finite connected poset with incidence algebra $A_P$. For the definition and results on Koszul algebras for incidence algebras, see for example here

Question: Which posets have the property that $A_P$ is a Koszul algebra such that the Koszul dual of $A_P$ is isomorphic to $A_P$?

It seems this property is extremely rare and only the finest posets are in this class such as Boolean lattices (and no other distributive lattices ?!) or the strong Bruhat order for the symmetric group.

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  • $\begingroup$ Is the Eulerian property (Enumerative Combinatorics, vol. 1, second ed., Section 3.16) necessary and/or sufficient for your property? $\endgroup$
    – Richard Stanley
    Commented Sep 28, 2021 at 14:54
  • $\begingroup$ @RichardStanley I dont know, it might be good to know more example of posets with this Koszul-selfduality property. $\endgroup$
    – Mare
    Commented Sep 28, 2021 at 16:03
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    $\begingroup$ A speculative conjecture would be that Koszul self-duality is equivalent to the order complex being Gorenstein, since Gorenstein complexes (or their face rings) are self-dual in a certain homological sense. $\endgroup$
    – Richard Stanley
    Commented Sep 28, 2021 at 18:00
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    $\begingroup$ Perhaps this is a silly question, but does it have anything to do with the poset being self-dual in the order sense (isomorphic to its own dual poset)? $\endgroup$
    – Sam Hopkins
    Commented Sep 29, 2021 at 19:15
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    $\begingroup$ @SamHopkins Yes, I think this implies it is self-dual. $\endgroup$
    – Mare
    Commented Sep 29, 2021 at 19:20

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Elaborating on Richard's suggestion, I think that it is necessary (but don't know about sufficiency) for the poset $P$ to be Gorenstein* over the field $k$ used in defining the incidence algebra $A_P$. To see this, note that Koszulity of $A_P$ is equivalent to Cohen-Macaulay-ness of $P$ over $k$, by the results of Polo [23], Woodcock [29] (reference numbers from the paper that you quoted). Then Eulerian-ness is required by the numerology relating Hilbert series of a Koszul algebra to that of its quadratic dual, as in the reference [1, Lemma 2.11.1] by Beilinson, Ginzburg, and Soergel. And as Richard said, the Gorenstein* property is defined as being both Cohen-Macaulay and Eulerian.

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    $\begingroup$ The lattices of faces of a convex polytope (or more generally, any regular CW-complex whose geometric realization is homeomorphic to a sphere) is Gorenstein*. I don't know about Sage, but the Macaulay2 software package can check the Gorenstein property by inputting the face ring of the order complex of $P$. $\endgroup$
    – Richard Stanley
    Commented Sep 29, 2021 at 17:50
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    $\begingroup$ In Sage a poset has "is_eulerian" and also "order_complex" which returns a simplicial complex. Then a simplicial complex has "is_cohen_macaulay" so we can check for Gorenstein* with built-in functions. $\endgroup$
    – John Machacek
    Commented Sep 29, 2021 at 18:00
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    $\begingroup$ Most of the other examples that I know that haven't already been mentioned are variations on Bruhat intervals, such as electrical network posets, or Bruhat orders on involutions or twisted involutions. $\endgroup$
    – Vic Reiner
    Commented Sep 29, 2021 at 18:16
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    $\begingroup$ For $n \leq 9$ there are no other bounded posets with $n$ elements other than Boolean or Bruhat order of Coxeter groups. The sequence starts for $n \geq 1$ with 1,1,0,1,0,1,0,2,0. $\endgroup$
    – Mare
    Commented Sep 29, 2021 at 18:37
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    $\begingroup$ @Mare: I think that every bounded Gorenstein* poset with at most nine elements is a Bruhat order or a boolean algebra. (Actually, a boolean algebra is a Bruhat order for a reducible root system.) However, there are three bounded Gorenstein* posets (one just the dual of another) with ten elements that are not Bruhat orders. See Figure 3.17 of Enumerative Combinatorics, vol. 1, second ed. Can you test them? $\endgroup$
    – Richard Stanley
    Commented Oct 7, 2021 at 14:56

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