There are different descriptions of the category $Coh(\mathbf{P}^{n})$. One can either describe it as modules over Beilinson's quiver algebra (Let us denote it by $A$) using the exceptional collection $O$, $O(1)$,$\cdots O(n)$, or one can identify it with graded modules over the exterior algebra (Denote it by $B$). My question is the following: Is it true that the algebra $A$ is Koszul and that the Koszul dual of $A$ is equal to $B$? If not, what is the precise relationship between $A$ and $B$? Moreover, it seems that the resolution of the diagonal of $\mathbf{P}^{n}$ should produce a complex of $A$-$B$ bimodules, does this has anything to do with the Koszul complex of algebras?
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1$\begingroup$ I'm not sure what the Beilinson quiver module might be, but you should look at Eisenbud's book on syzygies in which there is a duality established, one side of which is what you call B. $\endgroup$– mehCommented Jul 1, 2018 at 17:43
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1$\begingroup$ I don't think there is any relationship between $A$ and $B$. The Koszul dual of $B$ is the symmetric algebra, which looks completely different from $A$ (in particular, it is not finite dimensional.) $B$ is also very canonical in a sense, while $A$ is not (it depends highly on an exceptional collection.) $\endgroup$– dhyCommented Jul 1, 2018 at 18:46
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$\begingroup$ @dhy The Koszul dual of B is indeed a polynomial algebra. However, since A an inhomogeneous quadratic algebra, there is still a chance that the Koszul dual of A is B, just like the relationship between the enveloping algebra of a Lie algebra and the Chevalley complex. $\endgroup$– user105178Commented Jul 1, 2018 at 19:06
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$\begingroup$ @dhy Also, I seem to remember that the exterior algebra B can be obtained from an exceptional collection, which also comes from the resolution of the diagonal. $\endgroup$– user105178Commented Jul 1, 2018 at 19:22
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$\begingroup$ What version of inhomogeneous Koszul duality are you using? The only version I know of requires $A$ to be a filtered deformation of a quadratic algebra, in which case what is your filtration on $A$? The only one that comes to mind for me makes $A$ a homogeneous quadratic algebra. Also I was under the impression that in this case you still necessarily have the identity that the power series of dimensions of grade components of $A$ and $B$ multiplied to $1$, which is evidently false here. $\endgroup$– dhyCommented Jul 1, 2018 at 19:43
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