Fix an algebraically closed field $k$ of characteristic zero. I would like to know a classification of all vector bundles on $(\mathbb P_k^1)^r$, or at least of semistable vector bundles.

A theorem of Grothendieck states that a vector bundle $E$ of rank $m$ on $\mathbb P^1$ decomposes as a direct sum of line bundles: $E \cong \mathcal O(a_1) \oplus \dots \oplus \mathcal O(a_r)$.

Let $X$ be the $r$-fold fibre product of $\mathbb P^1$'s. We know (Hartshorne III Ex 12.6) that $\operatorname{Pic}(X) = \mathbb Z^r$, so every line bundle on $X$ is of the form $\pi_1^*\mathcal O_{\mathbb P^1}(a_1) \otimes \cdots \otimes \pi_r^*\mathcal O_{\mathbb P^1}(a_r)$, with $\pi_i$ the projection maps.

For $r=2$, we can find vector bundles of rank $2$ by computing $\operatorname{Ext}^1(\pi_i^*\mathcal O_{\mathbb P^1}(a_i), \pi_j^*\mathcal O_{\mathbb P^1}(a_j))$, which is non-trivial if $a_i \leq -2$ and $a_j \geq 0$ or if $a_i \geq 0$ and $a_j \leq -2$.

Does every rank $2$ vector bundle on $(\mathbb P^1)^2$ arise as such an extension? What about vector bundles of rank $2$ on $(\mathbb P^1)^r$ for $r> 2$? What about bundles of higher rank? What do vector bundles corresponding to non-trivial extensions look like in terms of transition functions?

By the Harder-Narasimhan filtration, every vector bundle on a projective scheme arises as an iterated extension of semistable vector bundles. Of the line bundles which arise as direct sums of tensor products of pullbacks of line bundles on $\mathbb P^1$ as above, it is not hard to show that the only semistable such are those of the form $\bigoplus_{i=1}^m (\pi_1^*\mathcal O_{\mathbb P^1}(a_1) \otimes \cdots \otimes \pi_r^*\mathcal O_{\mathbb P^1}(a_r))$. Are there any other semistable vector bundles on $X$?


1 Answer 1


Vector bundles on these are much more complicated. Consider $r=2$. We have a two to one map to the projective plane and we can pull back vector bundles from the plane. If $E,F$ are bundles on the plane and they become isomorphic when you pull back, then $E\oplus E(-1)$ and $F\oplus F(-1)$ must be isomorphic.


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