Vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$

Question

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!

Answer 1

The splitting theorem is most certainly false for vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$. In fact, the theory of vector bundles on quadric surfaces is probably as complicated as the theory of vector bundles on $\mathbb{P}^2$ (that is, very complicated).

Here is a simple example of an indecomposable rank 2 bundle. By the Künneth formula, we see that $\mathrm H^1(\mathcal O(1,-2))$ is not 0; hence there is a non-split extension $$ 0 \to \mathcal O(1,-2) \to E \to \mathcal O \to 0. $$ I claim that $E$ is not split. Suppose that it splits as the sum $L_1 \oplus L_2$ of two line bundles. If $L_1$ were to map to 0 in $\mathcal O$, the map $E \to \mathcal O$ would factor through $L_2$, and then the map $L_2 \to \mathcal O$, being surjective, would have to be an isomorphism, and the sequence would split. Hence both $L_1$ and $L_2$ admit a non-zero map to $\mathcal O$; this means that they have to be of the form $\mathcal O(m,n)$ with $m,n \leq 0$. But $L_1\otimes L_2$, which is the determinant of $E$, is $\mathcal O(1,-2)$, and this gives a contradiction.

I know that there have been people studying the moduli theory of vector bundles on $\mathbb{P}^1\times\mathbb{P}^1$(see for example arXiv:0810.4392; but there must be lots of earlier papers).

Answer 2

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

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

The idea of the proof is simple: there exist Ulrich sheaves on $\mathbb{P}^1\times \mathbb{P}^1$---these are sheaves (bundles in this case) with the property that under a finite map to $\mathbb{P}^2$ they push forward to a direct sum of copies of the structure sheaf. Pulling back a vector bundle from $\mathbb{P}^2$ 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 $\mathbb{P}^1\times \mathbb{P}^1 \to \mathbb{P}^2$ preserves the cohomology table of the bundle.

Answer 3

It may be a bit unfair to compare $X=\mathbb P^1 \times \mathbb P^1$ to $\mathbb P^1$. EDIT: I removed a too optimistic statement about restricting vector bundles on $\mathbb P^3$ to a smooth hypersurface, which works for $\mathbb P^4$ but not always for $\mathbb P^3$.

On the other hand, if we compare $X$ with the smooth surface $\mathbb P^2$, then there is this famous result by Horrocks, which says that a vector bundle $E$ on $\mathbb P^2$ splits if

$$\oplus_{i\in \mathbb Z} H^1(\mathbb P^2,E(i)) =0 $$

Interestingly, over $X$, the same result works. In other words, $E$ on $X$ splits if: $$\oplus_{i\in \mathbb Z} H^1(X,E(i)) =0 $$

the reason is such $E$ corresponds to a (graded) maximal Cohen-Macaulay (MCM) module over the cone of $X$, namely the hypersurface $R=k[x,y,u,v]/(xu-yv)$. But $R$ has finite MCM type and all the indecomposables have rank one (but note that they will not always be twists of the trivial line bundle). A lot more details and references are available in this paper by Buchweitz, Greuel and Schreyer.