TITLE: Vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$
QUESTION [19 upvotes]: I have a question about vector bundles on the algebraic surface $\mathbb{P}^1\times\mathbb{P}^1$.  My motivation is the splitting theorem of Grothendieck, which says that every algebraic vector bundle $F$ on the projective line $\mathbb{P}^1$ is a direct sum of $r$ line bundles, where $r$ is the rank of $F$.
My question is: to what extent is this true for $\mathbb{P}^1\times\mathbb{P}^1$, if at all?  At first glance I imagine this is classical, but I haven't had luck working it out, nor do I have a good reference for where this might be done.
Irrespective of the content of the answer, would it follow that the answer would be the same if the question were asked for $(\mathbb{P}^1)^k$?
Thanks!

REPLY [23 votes]: 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.