# Pushforward of a very ample line bundle on a curve to $\mathbb{P}^1$

Let $$p:C\to\mathbb{P}^1$$ be a degree $$k$$ morphism from a smooth projective curve $$C$$ to the projective line and $$L$$ a very ample line bundle on $$C$$. We know that $$p_*\mathcal{O}_C(L)$$ is a rank $$k$$ locally free sheaf on $$\mathbb{P}^1$$ and hence is in the form $$\mathcal{O}(e_1)\oplus\cdots\oplus\mathcal{O}(e_k)$$ by Birkhoff–Grothendieck theorem. In this case, are all $$e_1,...,e_k$$ positive?

No, not in general. Take $$C=\mathbb{P}^1$$, $$L=\mathcal{O}(1)$$, $$p$$ to be map $$x\mapsto x^2$$ in affine coordinates. Then $$p_*L$$ has rank $$2$$, but
$$2=h^0(L)=h^0(p_*L)=h^0(\mathcal{O}(e_1))+h^0(\mathcal{O}(e_2))$$ If $$e_1$$ and $$e_2$$ were both positive, then term on the right would be at least $$4$$. So this is impossible.

Added in response to comment. If you are allowed to pick $$\deg L\gg 0$$ relative to $$k$$, then I think it's probably true. Here's a result in that direction.

Lemma. If $$\deg L\gg 0$$ relative to $$k$$, then all $$e_i\ge 0$$.

Sketch. We can assume $$L=\omega_{C/\mathbb{P}^1}(M)$$ with $$M$$ globally generated. By a standard trick, we can find a cyclic cover $$\pi:\tilde C\to C$$ such that $$L$$ is a direct summand of $$\pi_*\omega_{\tilde C/\mathbb{P}^1}$$. Then $$p_*L$$ is a summand of $$(p\circ \pi)_*\omega_{\tilde C/\mathbb{P}^1}$$. The last sheaf is semipositive by a theorem of Fujita.

I suspect with more work, you can make the $$e_i$$ positive, but I leave that to you.

• Thanks for your answering. But I'm still wondered whether this is true if the condition is strenghened such that the degree of $L$ is sufficiently large. – Li Li Aug 11 '20 at 11:10
• A vector bundle $E$ on $\mathbb{P}^1$ is of the form $\ \bigoplus \mathscr{O}_{\mathbb{P}^1}(e_i)\,$ with the $e_i$ positive if and only if $H^1(E(-2))=0$. If $E=p_*L$, this is equivalent to $H^1(L(-2p^*[0]))=0$. This will hold in particular as soon as $\deg L > 2g(C)-2+2k$. – abx Aug 11 '20 at 14:39
• Good. That does it. – Donu Arapura Aug 11 '20 at 15:00