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Let $A,B$ be a pair of quasi-hereditary algebras and assume that $A$ and $B$ are both standard Koszul. Further assume that the graded decomposition matrices of $A$ and $B$ coincide (that is, the multiplicities of standard modules in projective modules coincide, as do their grading shifts). Under what circumstances could I conclude that these algebras $A$ and $B$ are graded Morita equivalent?

It feels to me that one should be able to make a general statement. If not, are there any counterexamples to the following claim? If $A$ and $B$ are standard Koszul and have the same decomposition matrices, then they are Morita equivalent.

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    $\begingroup$ Seems like there must be uncountably many standard Koszul algebras over an uncountable field, up to Morita equivalence, but only countably many possibilities for the decomposition matrix/grading shift data. I would imagine one could turn this into an explicit counterexample, and I’ll try when I have some time (unless somebody else does in the meantime … please do!) $\endgroup$
    – Jeremy Rickard
    Commented Mar 28, 2022 at 12:56

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Let $k$ be an infinite field. I will descibe an infinite family of standard Koszul algebras with the same graded decomposition matrix.

They will be algebras given by a quiver with relations. The quiver will be the same in all cases: nine vertices $a_{1}, a_{2}, a_{3}, b_{1}, b_{2}, b_{3}, c_{1}, c_{2}, c_{3}$, with an arrow from $a_{i}$ to $b_{j}$ and an arrow from $b_{i}$ to $c_{j}$ for all choices of $(i,j)$, so $18$ arrows in total.

For the relations, for each $(i,j)$ pick one of the three paths from $a_{i}$ to $c_{j}$ and for each of the other two paths, make it a scalar multiple of the first path. All $18$ scalars may be chosen independently. So this gives an $18$ parameter family of sets of relations.

Different choices may give isomorphic algebras. But since there is at most one arrow between each pair of vertices, the only ambiguity is that we can permute vertices (but only in finitely many ways) and multiply arrows by nonzero scalars. However, some choices of these scalars (for example, choosing them all to be equal) do not affect the relations, so we have at most $17$ parameter families of choices of relations giving the same algebra (in fact, less than $17$ parameters).

So we must have infinitely many nonisomorphic algebras.

Since the quiver is acyclic, we can order the vertices so that the standard modules are just the simple modules.

If we do this, it is straightforward to check that the algebras are all standard Koszul and have the same graded decomposition matrices.

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  • $\begingroup$ Hey Jeremy! I'm a bit terrible at general algebras via quivers, so sorry if this is stupid but here goes..... You say that your standard modules are all simple. If your algebra is quasi-hereditary this would imply that all these algebras are semisimple (for example, BGG reciprocity implies projectives are simple). This would imply that they are all Morita equivalent. Perhaps I've misunderstood, though. $\endgroup$
    – Chris Bowman
    Commented Mar 30, 2022 at 10:15
  • $\begingroup$ @ChrisBowman Without a duality, doesn't BGG reciprocity just imply that if standard modules are simple, then costandard modules are injective? But maybe your definition of "standard Koszul" assumes a duality? I was taking the definition in Ágoston, Dlab and Lukács' paper "Quasi-hereditary extension algebras", which doesn't seem to have such an assumption. $\endgroup$
    – Jeremy Rickard
    Commented Mar 30, 2022 at 10:34
  • $\begingroup$ Yes, you're quite right. I should have said that I wanted a BGG duality -- I was thinking of Lie theoretic example where this always happens. It's very interesting that you get counterexamples for quasi-hereditary algebras, thanks Jeremy. Do you think the duality is important here? Would you expect to be able to construct counter-examples in the case of ``BGG algebras"? $\endgroup$
    – Chris Bowman
    Commented Mar 31, 2022 at 11:20
  • $\begingroup$ @Chris I don’t know that much about BGG algebras, but I don’t see why there couldn’t be infinite familes of nonisomorphic examples depending on some choices of scalar parameters in the structure constants. If there weren’t, then this would be a surprising (to me) result about the rigidity of standard Koszul BGG algebras. $\endgroup$
    – Jeremy Rickard
    Commented Mar 31, 2022 at 13:20
  • $\begingroup$ Okay cool. Thanks Jeremy! I guess I can see your point... knowing (even graded) decomposition numbers and hence (by Koszulity) the radical layer structure of PIMs doesn't tell you the full submodule structure of the PIMs. So you can come up with quiver presentations which vary this submodule structure. $\endgroup$
    – Chris Bowman
    Commented Apr 1, 2022 at 10:41

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