Not much is known about vector bundles on $\mathbb{P}^2$ but I wonder if the following is a tractable question:

If $E,E'$ are non-isomorphic vector bundles on $\mathbb{P}^2$, then is there always a smooth curve $C \subset \mathbb{P}^2$ such that $E|_C$ and $E'|_C$ are still non isomorphic?

A related and perhaps easier question: Can a vector bundle restrict to the trivial bundles on every curve without itself being trivial on $\mathbb{P}^2$?

  • 2
    $\begingroup$ To add to Angelo's answer and to more directly address your second question, something even stronger is true: If E is trivial on every line in P^n then E is trivial. In fact, something stronger still is true: if E is trivial on every line through a fixed point of P^n, then E is trivial. $\endgroup$ – mdeland Mar 30 '11 at 15:44

Any curve of large enough degree will do. Set $F:= E'\otimes E^{\vee}$; if $d$ is a very large integer, then $\mathrm H^1(F(-d)) = 0$. Take any curve $C$ of degree $d$, and suppose that $E\mid_C$ and $E'\mid_C$ are isomorphic; this isomorphism is given by a section of $F\mid_C$. Since $\mathrm H^1(F(-d)) = 0$, this section extends to a global section of $F$, which yields a homomorphism $f \colon E'\to E$ which is an isomorphism when restricted to $C$. This $f$ is injective. Obviously $E$ and $E'$ have the same rank, so the cokernel of $f$ is torsion, and the same degree, so it in fact concentrated in finitely many points. But then the cokernel must be 0, because a sheaf concentrated in codimension 2 can't have projective dimension 1, and this concludes the proof.


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.