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?
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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.
answered Aug 10, 2020 at 18:55
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$\begingroup$ 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. $\endgroup$– Li LiCommented Aug 11, 2020 at 11:10
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1$\begingroup$ 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 $\deg L > 2g(C)-2+2k$. $\endgroup$– abxCommented Aug 11, 2020 at 14:39
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