Is there a classification of vector bundles on $\mathbb P^1 \times \mathbb P^1$? I know that the analogue of Grothendieck's splitting theorem is not true for $\mathbb P^1 \times \mathbb P^1$.
Is it true that any line bundle on $\mathbb P^1 \times \mathbb P^1$ is of the form $\mathcal O(m,n) = p_1^*(\mathcal O(m)) \otimes p_2^*(\mathcal O(n))$, where $p_1$ and $p_2$ denote the projections on the two factors?
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3$\begingroup$ 2 is true by the exponential exact sequence $ H^1(X,\mathcal O_X) \to H^1(X,\mathcal O_X^\times) \to H^2(X,\mathbb Z)$, plus a computation of $H^1(\mathbb P^1 \times \mathbb P^1,\mathcal O_X)$ and $H^2(\mathbb P^1 \times \mathbb P^1,\mathbb Z)$, via the Kunneth formula, from the well known-cohomology of $\mathbb P^1$. $\endgroup$– Will SawinCommented Jan 28, 2013 at 6:50
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1$\begingroup$ For each triple $(r,c_1,c_2)$ there is a moduli space of stable vector bundles, which is not well understood. There is a lot of literature on the subject. $\endgroup$– SashaCommented Jan 28, 2013 at 7:06
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$\begingroup$ For some further references, see <mathoverflow.net/questions/21854/…>. $\endgroup$– AngeloCommented Jan 28, 2013 at 8:34
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$\begingroup$ Angelo was trying to link to mathoverflow.net/questions/21854/… $\endgroup$– Allen KnutsonCommented Jan 28, 2013 at 18:41
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$\begingroup$ @Sasha "There is a lot of literature on the subject." Can you give examples? I find nothing specific for $\mathbb{P}^1 \times \mathbb{P}^1$ except Huh: "Moduli of stable sheaves on a smooth quadric and a Brill–Noether locus" and Huh: "Cubic symmetroids and vector bundles on a quadric surface". I do find much work on complex surfaces that is not specific to $\mathbb{P}^1 \times \mathbb{P}^1$. $\endgroup$– user505117Commented Mar 20, 2023 at 21:01
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1 Answer
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As Will Sawin remarks in the comments, the answer to your question (2) is "yes," either via the exponential exact sequence or by an easy application of the "theorem of the square," (which works e.g. over a general field).
The answer to your question (1) is rather more interesting--while there is no classification of vector bundles on $\mathbb{P}^1\times \mathbb{P}^1$ (that I know of), there is a splitting theorem analogous to Grothendieck's result. First, it's worth noting that there is a result of Horrocks generalizing Grothendieck's theorem to projective spaces of arbitrary dimension:
Theorem. (Horrocks) Let $E$ be a vector bundle on $\mathbb{P}^n$. Then $E$ is a direct sum of line bundles if and only if $H^i(\mathbb{P}^n, E(r))=0$ for $0<i<n$ and all $r$.
Note that if $n=1$, this is precisely Grothendieck's splitting theorem--a good place to learn about this result and its cousins is Vector Bundles on Complex Projective Spaces by Okonek et al. It is this theorem which generalizes to $\mathbb{P}^1\times \mathbb{P}^1$.
Indeed, a literally identical theorem to Horrocks' result above works for a quadric surface $\mathbb{P}^1\times \mathbb{P}^1$ in $\mathbb{P}^3$ (a vector bundle splits if and only if the middle cohomology of all of its twists $E(n)$ vanishes). One reference is this paper by Buchweitz, Greuel, and Schreyer; the relevant comments are Conjecture B and Remark 2 on page 169. Conjecture B gives a (conjectural) generalization to smooth hypersurfaces; I don't know too much about this subject, so I have no idea what the status of this conjecture is.
Of course, this is far from a classification of vector bundles on $\mathbb{P}^1\times \mathbb{P}^1$--that said, I think that such a classification (at anywhere near the level of completeness of the classification on $\mathbb{P}^1$) is well beyond current technology. Vector bundles on $\mathbb{P}^2$ are already quite interesting and complicated.
answered Jan 28, 2013 at 8:14
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1$\begingroup$ The updated link for the "theorem of the square" is math.ru.nl/~bmoonen/BookAV/LineBund.pdf $\endgroup$ Commented Nov 6, 2017 at 13:18
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$\begingroup$ Is the statement about $\mathbb P^1\times\mathbb P^1$ correct? The line bundle $E=\mathcal O(0,-2) $ has cohomology in degree $1$ since $H^1(\mathcal O(m,n))=H^0(\mathcal O(m))\otimes H^1(\mathcal O(n))\oplus H^1(\mathcal O(m))\otimes H^0(\mathcal O(n))$. $\endgroup$ Commented Apr 1 at 0:45