# Can curves differentiate vector bundles on P^2?

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 $\mathbb{P}^2$?

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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. – 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.