Line subbundles of vector bundles on smooth curves

Let $V$ be a vector bundle of rank at least 2 on a smooth (integral, projective) curve $C$. We know that a global section of $V$ is the same as a morphism $\mathcal{O}_C \to \mathcal{V}$ (letting $\mathcal{V}$ denote the sheaf of sections), and that this section is nonzero if and only if the morphism of sheaves is injective.

On the other hand, let $\mathbb{P}(V)$ be the associated projective bundle. A nonzero section $s : C \to V$ defines a rational map to $\mathbb{P}(V)$. Since $C$ is a smooth curve and $\mathbb{P}(V)$ is proper, this extends to an actual morphism $C \to \mathbb{P}(V)$, and so yields an actual line-subbundle $L \subset V$, and (again) an injective morphism of sheaves $\mathcal{L} \to \mathcal{V}$.

Question: How can we obtain the line bundle $L$ (and its sheaf of sections) in sheaf language? e.g. is there an algebraic way of converting the morphism $\mathcal{O}_C \to \mathcal{V}$ into the morphism $\mathcal{L} \to \mathcal{V}$ ? Also, what is $c_1(L)$ in this case? (edit: I guess $c_1(L) = Z(s)$, the section's vanishing locus?)

Ok, I have a somewhat ad-hoc construction. The key algebraic fact about smooth curves (to use instead of the fact that rational morphisms from smooth curves to proper varieties extend to genuine morphisms) is that a coherent sheaf on $C$ is locally free if and only if it is torsionfree. This is because $C$ is covered by affine charts which are the spectra of Dedekind domains.

Let $\varphi_s : \mathcal{O}_C \to \mathcal{V}$ be the section and let $\mathcal{F} = \mathrm{coker}(\varphi_s)$. Let $\mathcal{F}_{tors}$ be the torsion subsheaf, which is supported on $Z(s)$. So

$$0 \to \mathcal{F}_{tors} \to \mathcal{F} \to \mathcal{F}' \to 0$$

and $\mathcal{F}'$ is torsionfree, hence locally free. Let $\mathcal{L} \subset \mathcal{V}$ be the preimage of $\mathcal{F}_{tors}$ (note that $\mathrm{im}(\varphi_s)$ is a subsheaf of $\mathcal{L}$). We have

$$0 \to \mathcal{L} \to \mathcal{V} \to \mathcal{F}' \to 0,$$

which shows that $\mathcal{L}$ is locally free, and

$$0 \to \mathcal{O}_C \to \mathcal{L} \to \mathcal{F}_{tors} \to 0,$$

which shows that $\mathcal{L}$ has rank 1 and that $c_1(L) = Z(s)$.

I think what is going on is the following. The nonzero section $s$ determines a morphism $\varphi_s : \mathcal{O} \to \mathcal{V}$, while the subbundle $\mathcal{L} \subset \mathcal{V}$ is the image of $\varphi_s$ (see new edit).

The point is that scaling $s$ by $\mathcal{O}^*$ changes the particular morphism $\varphi_s$ but leaves the image of $\varphi_s$ invariant. So the induced map $C \to \mathbb{P}(\mathcal{V})$ only depends on $s$ modulo the action of $\mathcal{O}^*$ and so only sees the image of $\varphi_s$ and not $\varphi_s$ itself.

The local picture over a point is the fact that if I have a vector space $V$ as a bundle over a point $p$, a nonzero section is a choice of vector $0 \neq v \in V$ which determines a morphism $k \to V$ given by $1 \mapsto v$. The induced section of $\mathbb{P}(V)$ is a point corresponding to the line $l \subset V$ spanned by $v$. Multiplying $v$ by some $c \in k^*$ gives the same point in $\mathbb{P}(V)$ and so induces the same line $l \subset V$ by $cv$ corresponds to a different morphism $k \to V$ with image $l$.

$\textbf{EDIT 2}$ I realized from the comment discussions there was a big gap in my understanding of the difference between locally free sheaf and vector bundle. The map $\varphi_s$ is indeed injective as a map of sheaves so that $\mathcal{O}$ is a subsheaf of $\mathcal{V}$. However, $\varphi_s$ does not induce an embedding of associated bundles whenever $s$ vanishes.

If $s$ vanishes at $p$, this means that $s_p \in \mathfrak{p}$ in the local ring $\mathcal{O}_{p}$ and so even though the map on stalks is injective, the map on fibers of the associated line bundle is not.

A characterization of $\mathcal{L}$ that agrees with the construction in your answer is that $\varphi_s$ factors as $\mathcal{O} \to \mathcal{L} \subset \mathcal{V}$ AND where the inclusion $\mathcal{L} \subset \mathcal{V}$ is not just an injective map of locally free sheaves but also induces an embedding of associated vector bundles.

Another algebraic way to construct $\mathcal{L}$ is to look at $Z(s)$. Since $s$ is not identically zero and $C$ is a curve, then $Z(s)$ is a divisor and $\mathcal{L}$ is its associated line bundle. From this point of view, the fact that $C$ is a curve comes up because in higher dimensions, $Z(s)$ will have higher codimension. I'm pretty sure this is just a rephrasing of what you said in your answer though.

• Hmm, unfortunately it is not true that $\varphi_s$ fails to be injective wherever the section vanishes. Rather $\varphi_s$ fails to be injective at $p \in C$ if $s$ vanishes identically on a neighborhood of $p$, and since $C$ is integral, this never happens (unless $s$ is the zero section). In particular, $\mathcal{O}_C$ is isomorphic to its image. – Jake Levinson May 17 '15 at 2:20
• (Simple proof: the map of sheaves $\varphi_s : \mathcal{O}_C \to \mathcal{V}$ is generically injective, so $\ker \varphi$ is not supported at the generic point, i.e. it is a torsion sheaf. But the only torsion subsheaf of a torsionfree sheaf (like $\mathcal{O}_C$) is zero.) – Jake Levinson May 17 '15 at 2:24
• Yes you're right. Somehow I knew this as I was writing my original answer, and then confused myself and thought it was wrong when I reread it and made the edit... Sorry about that! Anyway, I think the original answer is still correct, the the difference between the two bundles you were asking about is that one comes with a particular map induced by the section, while the pullback from the projective bundle is an honest to god subbundle and not an injective map from $\mathcal{O}_C$. Though I will think about it some more as now I don't trust myself! – Dori Bejleri May 17 '15 at 2:33
• Perhaps my phrasing was unclear. I am asking for a way of proving or computing that $L$ exists, directly in terms of sheaves of modules, rather than by the geometrical argument about projective bundles. Also, both $\mathcal{O}_C$ and $\mathcal{L}$ have a section -- that part of your post is true, ultimately the map $\mathcal{O}_C \to V$ factors through $\mathcal{L}$. But what, algebraically, is $\mathcal{L}$? – Jake Levinson May 17 '15 at 2:37
• For sure at least $\mathcal{L}$ is isomorphic to the image of $\varphi_s$ over the locus where $s$ is nonzero. – Dori Bejleri May 17 '15 at 2:37