Assume $X$ is a Riemann surface and $\pi: E \to X$ is a holomorphic vector bundle of rank $n$ on $X$. Let's say $f$ is a holomorphic section of $E$ over $X$.

It is clear that if $f$ is a nonvanishing holomorphic function then $F := \bigcup_{x \in X} F_x \subset E$ (where $F_x = \mathbb{C} \cdot f(x)$) is a holomorphic subbundle of $E$ of rank 1 for which $f$ is a holomorphic section.

But I do not want to assume that our $f$ is nonvanishing. Then how do I construct a subbundle $F' \subset E$ of rank 1 such that $f$ is a holomorphic section of the subbundle...in other words, what do I do at the zeros? Obviously $\mathbb{C} \cdot f(x)$ is zero dimensional (not 1-dimensional) when $f(x)$ is the zero element of the vector space $\pi^{-1}(x)$. How do I "extend" the line bundle to these vanishing points to get a subbundle of $F' \subset E$ of rank 1 such that $f$ is a holomorphic section of $F'$? I would appreciate some help, I totally don't know how I could do it.

This is a special case of exercise 29.1(b) from Otto Forster's Lectures on Riemann Surfaces.


Here's one way to do it. Take a trivialization $\phi$ of $E$ over a neighborhood $U$ of $x$. For convenience, think of $z\in\Bbb C$ as a holomorphic coordinate on $U$, with $x$ corresponding to $0$. If $\phi\colon \pi^{-1}(U)\to U\times \Bbb C^n$ is the trivialization, write $\phi(f(z)) = (z,\psi(z))$. Now we use the standard trick. Expand (the holomorphic component functions of) $\psi$ in power series centered at $0$ and factor out the largest power of $z$ possible. The remaining vector function has a nonzero limit as $z\to 0$, say $\ell\in\Bbb C^n-\{0\}$, and we can define the fiber $F'_x$ to be the span of $\phi^{-1}(x,\ell)$.

  • $\begingroup$ Thank you for your explanation. But I think that any trivialisation of a holomorphic subbundle must be the restriction of some trivialisation of the original vector bundle...and in your case, that trivialisation on the original bundle $E$ would look like $\phi': \pi^{-1}(U) \to U \times \mathbb{C}^n, t_z \mapsto (z, t_1/z^k, \dots , t_n/z^k)$ (where $t_z$ is any element in the fibre of $z \in U$, $k$ is the largest power of z possible, and $\phi(t_z) = (z, t_1, \dots t_n)$ defines $t_1, \dots t_n$. But then the map $\phi'$ is not well defined on $\pi^{-1}(U)$! Sorry, am I making sense? $\endgroup$ – Acton Jul 9 '18 at 4:13
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    $\begingroup$ I'm not changing the trivialization. I'm just finding the limiting position of the line spanned by $f(z)$ by using the local power series expansions. This is a variant of the usual argument used when working with holomorphic maps $U\to\Bbb P^n$. $\endgroup$ – Ted Shifrin Jul 9 '18 at 5:18
  • $\begingroup$ Thanks again, Ted. I don't have much experience working with holomorphic maps $U \to \mathbb{P}^n$, but I think I'm a little lost because my definition of a holomorphic subbundle $F \subset E$ of rank 1 is: for each $x \in X$ there is a nhd $U \subset X$ and a trivialisation $h: E_U \to U \times \mathbb{C}^n$ such that $h(F_U) = U \times (\mathbb{C} \times 0)$. That's why I'm trying to see how your trivialisation can be modified to show that your $F$ is a subbundle of rank 1 by my definition. Sorry, am I just not seeing something obvious? $\endgroup$ – Acton Jul 9 '18 at 6:30
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    $\begingroup$ @Acton, you should be able to see that a nowhere-vanishing section gives you a trivialization of a line bundle. (To fit your definition, you need locally to extend that to a basis ...) Using our notation, the obvious extension of $f(z)/z^k$ across $0$ gives that nowhere-zero holomorphic section on $U$. $\endgroup$ – Ted Shifrin Jul 9 '18 at 6:48

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