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Jul 1, 2018 at 20:18 comment added user105178 @dhy I see. Thank you for clarifying!
Jul 1, 2018 at 19:43 comment added dhy 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.
Jul 1, 2018 at 19:22 comment added user105178 @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.
Jul 1, 2018 at 19:06 comment added user105178 @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.
Jul 1, 2018 at 18:46 comment added dhy 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.)
Jul 1, 2018 at 17:43 comment added meh 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.
Jul 1, 2018 at 3:43 history asked user105178 CC BY-SA 4.0