1
$\begingroup$

This seems like it should be easy, but unfortunately I don't see how to do it.

Let $X$ be a variety; I'm happy to assume that $X$ is quasiprojective. If $L_1$ and $L_2$ are two non-isomorphic line bundles on $X$, then can we find a curve $C$ in $X$ such that $L_1$ and $L_2$ restrict to non-isomorphic bundles on $C$? (In other words, is non-isomorphism of line bundles detected by curves?)

I think that for a general hyperplane section $H$ of $X$, the map $Pic(X) \rightarrow Pic(H)$ should be injective for some range of dimensions, which maybe breaks down for surfaces. So this reduces immediately to the case of $\dim X \approx 2$, but then...

$\endgroup$
  • $\begingroup$ You might look at the answer to the following MO question: mathoverflow.net/questions/233157/…. That question assumes that $X$ is a projective surface. Assuming $X$ is a normal, quasi-projective variety, there is always a normal projective compactification to which both $\mathcal{L}_1$ and $\mathcal{L}_2$ extend as invertible sheaves. $\endgroup$ – Jason Starr Aug 3 '16 at 18:01
4
$\begingroup$

Here is a proof for $\dim X=2$, projective and smooth. We may replace $L_1, L_2$ by $L_1\otimes L_2^{-1}=L$ and thus suffices to prove that if $L$ is not trivial, it is not trivial restricted to some curve. Take $H$ a large hypersurface section. Then $H^1(L-H)$ can be assumed to be zero and so if $L_{|H}$ is trivial,, we can assume that $H^0(L)\neq 0$ and thus $L=D$ for some effective curve and since $L\neq 0$, $D>0$. Now, restricted to $H$, it is clear that this line bundle has positive degree and can not be trivial.

$\endgroup$
  • $\begingroup$ Just to compare this proof with the proof in that other MO post, here the choice of sufficiently ample divisor $H$ depends, a priori, on $L$. In that other MO post, the goal was to prove that there exists a single ample hyperplane section (after base change to a bigger algebraically closed field, if the ground field is countable) that simultaneously works for all $L$. But that is not what user84144 asked, and Mohan's proof is definitely simpler. $\endgroup$ – Jason Starr Aug 3 '16 at 18:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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