User david eisenbud - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-18T05:20:48Z http://mathoverflow.net/feeds/user/5771 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/34452/symmetric-powers-and-duals-of-vector-bundles-in-char-p Symmetric powers and duals of vector bundles in char p David Eisenbud 2010-08-03T23:27:47Z 2010-08-05T19:47:17Z <p>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.</p> <p>Question: Is $Sym_m(E)^*$ isomorphic to $Sym_m(E^*)$ (non-canonically) in general?</p> http://mathoverflow.net/questions/21854/vector-bundles-on-mathbbp1-times-mathbbp1/23231#23231 Answer by David Eisenbud for Vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$ David Eisenbud 2010-05-02T05:27:12Z 2010-05-02T05:27:12Z <p>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):</p> <p>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 $P^2$. </p> <p>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.</p> http://mathoverflow.net/questions/1652/simple-example-of-a-ring-which-is-normal-but-not-cm/2194#2194 Answer by David Eisenbud for Simple example of a ring which is normal but not CM David Eisenbud 2009-10-23T21:36:27Z 2009-10-23T21:36:27Z <p>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. </p> <p>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".</p>