What is an example of a Frobenius algebra that is not Koszul? Are there reasonable requirements for a Frobenius to be Koszul?
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asked Nov 2, 2022 at 12:59
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5$\begingroup$ Surely you can pick a non-quadratic Frobenius algebra? $\endgroup$– PedroCommented Nov 2, 2022 at 13:14
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2 Answers
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If an algebra A is Koszul with Koszul daul B, one has the formula gldim A=Loewy Length B-1.
Thus when a Frobenius algebra is Koszul (and not semisimple), the quadratic dual must be infinite dimensional.
For example for preprojective algebras of Dynkin type, the quadratic dual is finite dimensional and thus they can not be Koszul, see https://link.springer.com/article/10.1023/A:1020146502185 .
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Probably the simplest example is $k[x|x^3=0]$. It is not Koszul since it is not quadratic.