0
$\begingroup$

Suppose we are given a scheme $S$ and two vector bundles $V$ and $W$ over $S$. Is it always true that $\mathbb{P}(V)\cong \mathbb{P}(W)$ implies that $V\cong W$ as $S$-schemes?

If the statement is false then what is the most general condition on $S$ for which it becomes true?

$\endgroup$

1 Answer 1

7
$\begingroup$

This is not true as stated, because for any line bundle $L$ on $S$ one has $$ \mathbb{P}(V \otimes L) \cong \mathbb{P}(V), $$ but this is the only issue. Indeed, if $X = \mathbb{P}(V) \stackrel{p}\to S$ and $S$ is connected, the relative Picard group $\mathrm{Pic}(X/S)$ is cyclic, and if $H$ is a lift to $\mathrm{Pic}(X)$ of its relatively ample generator, then $$ p_*\mathcal{O}_X(H) \cong V^\vee, $$ and since $H$ is uniquely defined up to twist by $p^*(\mathrm{Pic}(S))$, it follows that $V$ is uniquely defined up to twist by $\mathrm{Pic}(S)$.

$\endgroup$
2
  • $\begingroup$ Thanks! Does this require any condition on S? $\endgroup$
    – Arun Kumar
    Feb 27, 2023 at 18:15
  • $\begingroup$ Not that I know (but you need, of course, to modify the argument slightly when $S$ is not connected). $\endgroup$
    – Sasha
    Feb 28, 2023 at 6:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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