# Line subbundles of maximal degree

I want to know whether it is possible to bound the degree of line subbundles of certain holomorphic line bundles over Riemann surfaces.

Even more concretely, consider the complex projective line $$\mathbb{P}_{\mathbb{C}}^1$$ and the rank-2 vector bundle $$E=\mathcal{O}(a)\oplus \mathcal{O}(b)$$ for $$a,b\in\mathbb{Z}$$. Let $$L\subset E$$ be a holomorphic line subbundle. Thanks to the classification of line bundles over the Riemann sphere, this is some $$\mathcal{O}(d)$$ for $$d\in\mathbb{Z}$$.

Question: is necessarily $$d$$ bounded by $$a,b$$ or $$a+b=deg(E)$$? How can I prove that? Is there a maximal-degree subbundle of $$E$$?

• $d$ is bounded by $\max\{a,b\}$. Jan 14, 2022 at 12:40
• Any sketch of a proof? I know Chern classes are obstructions for the rank of trivial subbundles to exist, but this doesn't say much unless I demand $a+b=0$. Jan 14, 2022 at 12:50

Suppose we have

$$0 \to \mathcal{O}(d) \to \mathcal{O}(a)\oplus\mathcal{O}(b) \to \mathcal{O}(c)\to 0$$

for some $$c \in \mathbb{Z}$$. Tensoring through by $$\mathcal{O}(-d)$$ gives the short exact sequence

$$0 \to \mathcal{O} \to \mathcal{O}(a - d)\oplus\mathcal{O}(b - d) \to \mathcal{O}(c-d) \to 0.$$

Therefore the bundle $$\mathcal{O}(a - d)\oplus\mathcal{O}(b - d)$$ has a non-zero section, so either $$\mathcal{O}(a - d)$$ has a section (in which case $$a - d \geq 0$$), or $$\mathcal{O}(b - d)$$ has a section (in which case $$b - d \geq 0$$). That is, $$a \geq d$$ or $$b \geq d$$, so $$d \leq \max\{a, b\}$$ as pointed out by Sasha in the comments.

• Thank you! Now I see this has a nice generalization for $E=\bigoplus_{i} \mathcal{O}(a_i)$ vector bundle over $\mathbb{P}^1$ of arbitrary rank: necessarily $d\leq \max_i a_i$. Jan 14, 2022 at 16:05
• Note that the quotient is not necessarily a line bundle. For example an ideal sheaf $\mathcal O(-1) \subset \mathcal O$ has a torsion quotient. Doesn't change the argument though. Jan 14, 2022 at 20:58
• @red_trumpet: The quotient of a vector bundle by a subbundle is always a vector bundle. Note that $\mathcal{O}(-1)$ is not a subbundle of $\mathcal{O}$ (but it is a subbundle of $\mathcal{O}\oplus\mathcal{O}$). Jan 15, 2022 at 17:44
• @MichaelAlbanese Isn't the ideal sheaf $\mathcal I \subset \mathcal O$ of a point $p \in \mathbb P^1$ isomorphic to $\mathcal O(-1)$, and the quotient $\mathcal O / \mathcal I$ is the structure sheaf of $p$, hence torsion? Jan 15, 2022 at 17:50
• @red_trumpet: That's a subsheaf, not a subbundle. Jan 15, 2022 at 18:51