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

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!

• First of all, I would notice the presence of the natural $2:1$ covering $\pi: ( \mathbb{P}^1 )^2 \rightarrow \operatorname{Sym}^2(\mathbb{P}^1) \cong \mathbb{P}^2$ ramified along the diagonal. On $\mathbb{P}^2$ there are certainly "non-splittable" vector bundles $E$ . Does the pullback $\pi^{*} E$ become splittable? – Qfwfq Apr 19 '10 at 17:16
• As others have pointed it's likely to be complicated. You can start with a paper of Brosius [Math. Ann. 1983] – Donu Arapura Apr 19 '10 at 17:58
• @Donu Arapura: Thank you for the reference, I'll have a look through it. – user5395 Apr 20 '10 at 11:25
• Alexander Klyachko has a number of papers studying equivariant vector bundles on toric varieties, which you can think of as the T-fixed-point locus inside the moduli space of all reasonable vector bundles. – Allen Knutson Apr 21 '10 at 15:29

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 http://front.math.ucdavis.edu/0810.4392; but there must be lots of earlier papers).

• Thank you, Angelo, for the counter-example and the reference! – user5395 Apr 20 '10 at 11:24

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.

• "By a Theorem of mine with Schreyer"... presumably it would helpful to know that this is David Eisenbud – Steven Sam May 2 '10 at 7:08
• – evgeny Nov 9 '16 at 8:06

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.

• Just a formal comment: since tensoring by a line bundle does not change the splitting type of a bundle, you can weaken your condition by saying that there exists an integer a such that E(0,a) satisfies the vanishing of H^1. With this change, all line bundles will satisfy the condition. – damiano Apr 19 '10 at 18:55
• @Hailong Dao: Thank you for the reference by Buchweitz et al. I was aware of the Horrocks splitting theorem for P^2, but I wasn't aware of its direct analogy for P^1 x P^1. – user5395 Apr 20 '10 at 11:22