sc2

van den Bergh

Here is a slightly different proof I think (once you know something is true it becomes easier to prove).

(0) I will use the “quotient Grassmannian”. This is of course equivalent.

(1) It is sufficient to construct a quiver Grassmannian for a quiver with relations with the desired property. Indeed if a representation satisfies certain relations then so does any sub or quotient representation.

(2) Assume we want to do P^n (projective space). Then we take the Beilinson quiver with 3 vertices and 2 x n+1 arrows and we impose the commutativity relations.

The moduli space of representations of dimension vector (1,1,1) generated in the first vertex is P^n.

This moduli space is also the quiver Grassmannian of quotients with dimension vector (1,1,1) of the projective representation corresponding to the first vertex.

(3) For an arbitrary projective variety we use the fact that it can be defined by quadratic relations in some P^n and impose these relations on the corresponding Beilinson quiver from (2) (which already has the commutativity relations). Then we use again the quiver Grassmannian of quotients with dimension vector (1,1,1) of the projective representation corresponding to the first vertex.

(4) If we do not want to use the fact that we can use quadratic relations then we can use larger Beilinson quivers.

Created on May 22, 2012 at 20:40:28. See the history of this page for a list of all contributions to it.