Let $f:X\rightarrow Y$ be smooth family of complex curves over $Y\backslash S$, S-finite set, Y-smooth complex curve. Let $Z \rightarrow Y $ be a ramified covering, ramified over points of $S .$ Let $W$ be a normalization of $X\times_{Y} Z,$ and we have the induced map $g:W\rightarrow X.$ If $\mathcal{F},\mathcal{G}$ are locally free sheaves on $W$ and $g_{*}(\mathcal{F})= g_{*}(\mathcal{G}),$ do we have that $\mathcal{F}=\mathcal{G} ?$


In general no. For instance take $Y = P^1$, $X = P^1 \times Y$, and take $S$ to be 4 different points. Then $Z$ is an elliptic curve and $W = P^1 \times Z$. Take $F$ to be (the pullback to $W$ of) a nontrivial line bundle on $Z$ of degree $0$. Then $g_*F$ is (the pullback to $X$ of) the pushforward of that line bundle w.r.t. $Z \to Y$. It is easy to see that this pushforward is isomorphic to $O(-1) \oplus O(-1)$, independently of the choice of $F$. In particular, if $G$ corresponds to another nontrivial line bundle of degree $0$, then $g_*(F) \cong g_*(G)$, but $F \not\cong G$.

  • $\begingroup$ is there any condition that we can add to the map $f$, in order that this holds ?If the map $f$ is a semistable non isotrivial fibration , does it change the situation? $\endgroup$
    – And Rub
    Sep 1 '16 at 17:12
  • $\begingroup$ @AndRub: As you see, the problem is not with the map $f$ (you can even take $f$ to be the identity map, still the same counterexample works), but with the map $Z \to Y$ that takes non-isomorphic objects to isomorphic objects. $\endgroup$
    – Sasha
    Sep 1 '16 at 18:09

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.