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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 $E$?

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    $\begingroup$ $d$ is bounded by $\max\{a,b\}$. $\endgroup$
    – Sasha
    Jan 14, 2022 at 12:40
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    $\begingroup$ 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 $a+b=0$. $\endgroup$
    – topolosaurus
    Jan 14, 2022 at 12:50

1 Answer 1

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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.

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  • $\begingroup$ 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 $d\leq \max_i a_i$. $\endgroup$
    – topolosaurus
    Jan 14, 2022 at 16:05
  • $\begingroup$ 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. $\endgroup$
    – red_trumpet
    Jan 14, 2022 at 20:58
  • $\begingroup$ @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 $\mathcal{O}\oplus\mathcal{O}$). $\endgroup$
    – Michael Albanese
    Jan 15, 2022 at 17:44
  • $\begingroup$ @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? $\endgroup$
    – red_trumpet
    Jan 15, 2022 at 17:50
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    $\begingroup$ @red_trumpet: That's a subsheaf, not a subbundle. $\endgroup$
    – Michael Albanese
    Jan 15, 2022 at 18:51

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