Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I wish to know if there is a rank 2 vector bundle $E$ on $\mathbb{P}^1 \times \mathbb{P}^1$ such that $\mathbb{P}(E)$ when restricted to $\mathbb{P}^1 \times [0:1]$ is the $n$th Hirzebruch surface and when restricted to $\mathbb{P}^1 \times [x:y]$ is the $(n-2)$th Hirzebruch surface.

share|improve this question

1 Answer 1

up vote 4 down vote accepted

Let $F = O \oplus O(n-2,0)$ and denote the line $P^1\times[0:1]$ by $L$. Note that $F_{|L} = O \oplus O(n-2)$. Consider any surjective map $O_L \oplus O_L(n-2) \to O_L(n-1)$ (for example the one given by $u^{n-1}$ on the first summand and by $v$ on the second, where $(u:v)$ are the homogeneous coordinates on $L$). Consider the composition $F \to F_{|L} \to O_L(n-1)$ and let $E$ be its kernel. Note that for any line $L' = P^1\times[x:y]$ we have $E_{|L'} = F_{|L'} = O \oplus O(n-2)$. On the other hand, restricting to $L$ we obtain an exact quadruple $$ 0 \to L_1i^*O_L(n-1) \to E_{|L} \to O_L \oplus O_L(n-2) \to O_L(n-1) \to 0 $$ where $i:L \to P^1\times P^1$ is the embedding. Note that $L_1i^*O_L(n-1) = O_L(n-1)$, and the kernel of the rightmost map is $O_L(-1)$. Hence we have an exact triple $$ 0 \to O_L(n-1) \to E_{|L} \to O_L(-1) \to 0. $$ Since there are no notrivial extensions, we see that $E_{|L} = O_L(-1) \oplus O_L(n-1)$. So, the bundle $E$ gives what you need.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.