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.

  • $\begingroup$ 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. $\endgroup$ – Jake Levinson May 17 '15 at 2:20
  • $\begingroup$ (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.) $\endgroup$ – Jake Levinson May 17 '15 at 2:24
  • $\begingroup$ 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! $\endgroup$ – Dori Bejleri May 17 '15 at 2:33
  • $\begingroup$ 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 $\mathcal{L}$? $\endgroup$ – Jake Levinson May 17 '15 at 2:37
  • $\begingroup$ For sure at least $\mathcal{L}$ is isomorphic to the image of $\varphi_s$ over the locus where $s$ is nonzero. $\endgroup$ – Dori Bejleri May 17 '15 at 2:37

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.