User david eisenbud - MathOverflow

Symmetric powers and duals of vector bundles in char p

2010-08-03T23:27:47Z

Suppose that $X$ is a smooth projective variety (eg $P^n$) and $E$ is a vector bundle (eg the tangent bundle). If the characteristic is zero, then taking symmetric powers "commutes" with taking duals: $ Sym_m(E)^* $ and $Sym_m(E^*)$ are canonically isomorphic. This is not true in characteristic $p>0$ (one has a canonical isomorphism with a divided power, instead.) But even in characteristic $p$, if the bundle is trivial, then there are non-canonical isomorphisms, and it is not hard to show that the Chern classes are the same.

Question: Is $Sym_m(E)^*$ isomorphic to $Sym_m(E^*)$ (non-canonically) in general?

Answer by David Eisenbud for Vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$

2010-05-02T05:27:12Z

There is a precise sense in which the theory of vector bundles on $P^1\times P^1$ is exactly as complicated as that for $P^2$ (and this is true for any hypersurface in $P^3$, and conjecturally for all surfaces):

For any bundle $E$ on $P^1\times P^1$, the (singly graded) cohomology table of $E$ is the collection of numbers $$ h^i(E(p)); \quad p,q\in Z\times Z, \quad i=0,1,2. $$ Let $C(P^1\times P^1)$ be the positive rational convex cone generated by the Betti tables of bundles on $P^1\times P^1$. By a Theorem of mine with Schreyer (soon to be on the arXiv), the cone $C(P^1\times P^1)$ is identical to the corresponding cone for $P^2$.

The idea of the proof is simple: there exist Ulrich sheaves on $P^1\times P^1$---these are sheaves (bundles in this case) with the property that under a finite map to $P^2$ they push forward to a direct sum of copies of the structure sheaf. Pulling back a vector bundle from $P^2$ and tensoring with an Ulrich sheaf only multiplies the cohomology table by the rank of the Ulrich sheaf. This gives one inclusion. For the other, observe that pushing a bundle forward by a finite linear projection $P^1\times P^1 \to P^2$ preserves the cohomology table of the bundle.

Answer by David Eisenbud for Simple example of a ring which is normal but not CM

2009-10-23T21:36:27Z

Another family of examples is given by the homogeneous coordinate rings of irregular surfaces (ie 2-dimensional $X$ such that $H^1({\mathcal O}_X) \neq 0$); these surfaces cannot be embedded in any way so that their homogeneous coordinate rings become Cohen-Macaulay. Elliptic scrolls (such as the one in the previous answer) and Abelian surfaces in P4, made from the sections of the Horrocks-Mumford bundle, are such examples.

The point is that sufficiently positive, complete embeddings of any smooth variety (or somewhat more generally) will have normal homogeneous coordinate rings, and they will be Cohen-Macaulay iff the intermediate cohomology of the variety vanishes. All the examples above fall into this category. It's an interesting general question to ask how positive is "sufficiently positive".