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Show that $\mathcal{O}(p)$ on $\mathbb{P}^1$ is $\mathcal{O}(1)$.

I think I would like to show that there exist a rational section of this sheaf with order equal to $1$.

I am splitting $\mathbb{P}^1$ into its affine covering $U_0(z_0); \ x_0 \neq 0$ and $U_1(y_1); \ x_1 \neq 0$.

Suppose $p=[0:1] \in U_1(y_1)$.

Let us look at the rational section $y_1$ which has a zero at $[0:1]$, thus I imagine it is a perfectly valid section for $\mathcal{O}(p)(U_1)$. Let us look at what happens on $U_1 \cap U_0$, we then have that $y_1=1/z_0$.

We would now say that the section $y_1$ has order zero. However, this can't be correct. Is $1/z_0$, with having a pole at $[1:0]$ not a valid rational section?

If not, how do I find a rational section of this sheaf?

user7090
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  • What is your definition of $\mathcal O(1)$? – Alex Mathers Jun 16 '17 at 00:35
  • Rigorously, $\mathcal{O}(1) \cong S(1)^{\sim}$ – user7090 Jun 16 '17 at 01:32
  • Recall that $\mathcal O(p)$ is a line bundle, in particular you should have some cocycles $\varphi_{ij}$ such that for all sections $s_i,s_j$ one has $s_i = \varphi_{ij} s_j$. If you simply do a change of variable $z = 1/w$, you will find nothing as the only holomorphic functions defined on $\Bbb P^1$ are the constant. –  Jun 16 '17 at 07:14

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Here is the description of a rational section of $\mathcal O_{\mathbb P^1}(p)$ which is even regular.
The sheaf $\mathcal O_{\mathbb P^1}(p)$ is the subsheaf $\mathcal O_{\mathbb P^1}(p)\subset \mathcal K$ of the constant sheaf $\mathcal K$ of rational functions defined by the specification: $$\Gamma(U,\mathcal O_{\mathbb P^1}(p))=\{g\in \Gamma(U,\mathcal K)\vert \operatorname {div}(g)+1\cdot p\geq 0\}$$ where $U\subset \mathbb P^1$ is open and $\operatorname {div}(g)$ is the divisor associated to $g$.
The constant function $1$ is clearly a global regular section $f=1 \in \Gamma(\mathbb P^1,\mathcal O_{\mathbb P^1}(p))$ since $$ \operatorname {div}(f)+1\cdot p=0+1\cdot p\geq 0$$ Notice that the divisor of $f$ seen as a section of $\mathcal O_{\mathbb P^1}(p)$ is $$\operatorname {div}^{\mathcal O_{\mathbb P^1}(p)}(f)=\operatorname {div}(f)+1.p=1.p$$ The subtle fact that for $n\gt 0$ the constant function $1$ is a regular section of $\mathcal O_{\mathbb P^1}(n\cdot p)$, and that this section has a zero of order $n$ at $p$ (and no other zero) is unfortunately not sufficiently emphasized in textbooks.

Georges Elencwajg
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