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We define Riemann sphere as $S=\mathbb{C}^2-\{0\}/\sim$. Given a point $p$ over $S$, I have seen somewhere there exists a line bundle $L_p$ associated to $p$, and $L_p$ has a non-zero holomorphic section with only one zero at $p$.

I think this construction is well-known to most people, I want to know how to construct the line bundle associated to $p$, by definition of the line bundle, how to define the space $E$ and the map $E\rightarrow S$ such that each fiber is a complex vector space of dimension $1$?

Thanks!

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2 Answers 2

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For simplicity let's take $P = [0: 1]$.

Cover $S$ with the two opens $U_0 = S \setminus [1: 0]$ and $U_\infty = S \setminus [0:1]$.

Now take the disjoint union of two copies of the trivial bundle: $M = (U_0 \times \mathbb{C}) \sqcup(U_\infty \times \mathbb{C})$

We impose an equivalence relation on $M$ by the rule $([a:b], z)_0 \sim ([a:b], bz/a)_\infty$ when neither $a$ nor $b$ are zero, and set $L = M/\sim$.

Since $\sim$ respects the projection to $S$, there is a natural map $L \to S$. The fiber over a point is $\mathbb{C}$, and local triviality is clear (every point is in $U_0$ or $U_\infty$).

To give a section of $L$ is to give a map $f_0: U_0 \to \mathbb{C}$ and a map $f_\infty: U_\infty \to \mathbb{C}$ such that $$ bf_0([a: b]) = af_\infty([a:b]) $$ whenever neither $a$ nor $b$ are zero.

There is such a section: $f_0([a: b]) = a/b$ and $f_\infty([a: b]) = 1$. This is locally holomorphic, so it's holomorphic, and it has a unique zero of order one at $P$ as desired.

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  • $\begingroup$ (FWIW most algebraic geometers are not going to think about the concrete line bundle as spelled out here but just think about the invertible sheaf whose sections over an open are the degree one functions in the homogeneous coordinates. this answer just works thru the recipe that builds a physical line bundle if you know the invertible sheaf.) $\endgroup$
    – hunter
    Commented Jun 30, 2020 at 18:57
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    $\begingroup$ I know for a point $p$, then as a divisor, we can construct an invertible sheaf $\mathcal{O}(p)$, then by the correspondence between invertible sheaves and line bundles(for a line bundle $L$, the sheaf $\mathcal{O}{L}$ of regular sections of $L$ is invertible, this gives a group isomorphism), we have a line bundle associated to $p$. But I want to construct this line bundle directly from $p$ and I don't know how to get a line bundle directly from an invertible sheaf, so I asked this question. $\endgroup$
    – user685167
    Commented Jul 2, 2020 at 5:33
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    $\begingroup$ @Ang got it. The general recipe is the same as in this answer. given an invertible shefaf $\mathcal{F}$ take the disjoint union of the trivial bundle over the $U_i$ where $U_i$ is some trivializing cover and glue by $f_if_j^{-1}$ where $f_i$ and $f_j$ are choices of trivialization. $\endgroup$
    – hunter
    Commented Jul 2, 2020 at 5:38
  • $\begingroup$ Your answer is very clear and helpful! $\endgroup$
    – user685167
    Commented Jul 2, 2020 at 6:18
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This falls into the realm of the correspondence between divisors and holomorphic line bundles on a complex manifold. From this point of view, the line bundle you are describing is denoted $\mathcal{O}_{\mathbb{CP}^1}([p])$. However, this is a particularly special situation which allows for many different descriptions.

It turns out that the isomorphism type of $\mathcal{O}_{\mathbb{CP}^1}([p])$ doesn't depend on the choice of point $p$ (in the language of divisors, for any other point $q$ we have $[p] = [q]$), and is often denoted by $\mathcal{O}_{\mathbb{CP}^1}(1)$. One explicit model for the total space of $\mathcal{O}_{\mathbb{CP}^1}(1)$ is $\mathbb{CP}^2\setminus\{[0, 0, 1]\}$, where the projection map $\pi$ is given by $\pi([z_0, z_1, z_2]) \mapsto [z_0, z_1]$; geometrically, $\pi$ is the projection from the point $[0, 0, 1] \in \mathbb{CP}^2$ to the hyperplane given by $z_2 = 0$. For any choice of point $p = [a, b] \in \mathbb{CP}^1$, the bundle admits a section $\sigma$ which vanishes only at $p$, namely $\sigma([z_0, z_1]) = [z_0, z_1, bz_0 - az_1]$.

All of this generalises to $\mathbb{CP}^n$. The answers to this MathOverflow question give several other interpretations of the line bundle $\mathcal{O}_{\mathbb{CP}^n}(1)$.

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  • $\begingroup$ It's confusing to introduce both types of notation. The notation $\mathscr O(p)$ is used for functions with a pole at $p$, but you're correct, I guess. $\endgroup$ Commented Jun 30, 2020 at 14:17
  • $\begingroup$ Really? I have always thought that given an effective divisor $D$, the notation $\mathcal{O}(D)$ is the corresponding line bundle which admits a holomorphic section $s$ with vanishing locus $D$. $\endgroup$ Commented Jun 30, 2020 at 14:21
  • $\begingroup$ (Sheaf of germs of) functions versus sections. $\endgroup$ Commented Jun 30, 2020 at 14:23
  • $\begingroup$ Given a divisor $D$, the invertible sheaf $\mathcal{O}(D)$ which Shifrin talks about is defined on page 271 in Rick Miranda's book "Algebraic Curves and Riemann Surfaces". For any open subset $U$ of a Riemann surface $X$, $\mathcal{O}(D)(U)$ is the set of all meromorphic functions on $U$ which satisfy the condition that $\text{ord}_p(f)\geq-D(p)$ for all $p\in U$. $\endgroup$
    – user685167
    Commented Jul 2, 2020 at 5:24
  • $\begingroup$ Just a word of warning, the notation is not completely consistent among all references. For example, $\mathcal{O}(D)$ denotes a line bundle in Huybrechts' Complex Geometry: An Introduction rather than an invertible sheaf (I know the two notions are equivalent, but still). $\endgroup$ Commented Jul 2, 2020 at 10:56

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