Global section of the line bundle ${\cal O}_{{\Bbb P}_k}(1)$ over ${\Bbb P}_k^1$.

Question

The total space of the very ample line bundle ${\cal O}_{{\Bbb P}_k^1}(1)$ over ${\Bbb P}_k^1$ is the complement $T \colon= {\Bbb P}_k^2 \setminus {x}$ of a point $x$ in a projective surface, where $x = (1 \colon 0 \colon 0) \in {\Bbb P}_k^2$ and ${\Bbb P}_k^1 = (0 \,\colon k\,\colon k) \hookrightarrow {\Bbb P}_k^2 = (k \,\colon k \,\colon k)$. That is we have the fibration

\begin{equation*} \pi \colon T \to {\Bbb P}_k^1 \end{equation*}

Q. How can one visualise an everywhere holomorphic section of ${\cal O}_{{\Bbb P}_k^1}(1)$ as the section $s \colon {\Bbb P}_k^1 \to T$ to $\pi$ in the above context ?

I have difficulty to figure out the everywhere holomorphic section, i.e. locally holomorphic function which glues globally on ${\Bbb P}_k^1$. Any hint will be extremely helpful.

Answer 1

Answer: If $V:=k\{e_0,e_1\}$ is a 2-dimensional vector space over $k$ with dual space $V^*:=k\{x_0,x_1\}$ it follows there is a canonical isomorphism of $k$-vector spaces

$$H^0(\mathbb{P}(V^*), \mathcal{O}_{\mathbb{P}(V^*)}(1) ) \cong V^*.$$

Here $x_i:=e_i^*$ is the dual basis. Hence a global section $s$ of $\mathcal{O}_{\mathbb{P}(V^*)}(1)$ corresponds 1-1 to an element

$$s:=s_0x_0+s_1x_2 \in V^*$$

which is a map

$$s:V \rightarrow k.$$

The element $s$ may be zero, hence $s$ does not always span a "line" in $V^*$.

Note: You may construct the "geometric vector bundle" $\mathcal{L}_d:=\mathbb{V}(L_d^*)$ for any linebundle $L_d:=\mathcal{O}(d)$ on $C:=\mathbb{P}^1$ and by definition

$$\mathcal{L}_d:=Spec(Sym_{\mathcal{O}_{\mathbb{P}^1}}^*(L_d^*)) \cong Spec(Sym_{\mathcal{O}_{\mathbb{P}^1}}^*(L_{-d})) .$$

There is a canonical projection map $\pi_d: \mathcal{L}_d: \rightarrow \mathbb{P}^1$ and you must prove there is an "equality" between the set of sections $s:\mathbb{P}^1 \rightarrow \mathcal{L}_1$ of $\pi_1$ and the $k$-vector space $V^*$. More generally it follows the set of sections of the map $\pi_d: L_d \rightarrow \mathbb{P}^1_k$ equals $Sym^d(V^*)$. Note that the geometric vector bundle $L_d$ is locally trivial: $\pi^{-1}(D(x_i)) \cong D(x_i) \times \mathbb{A}^1_k$, where $\mathbb{A}^1_k \cong D(x_i)\subseteq \mathbb{P}^1_k$ is the standard open cover. If $x\in \mathbb{P}^1_k$ is a point and $\pi^{-1}(x) \cong \mathbb{A}^1_{\kappa(x)}$ is the fiber it follows the zero-section $s(0):=0\in Sym^d(V^*)$ corresponds to the zero section $s_0$ of $\pi_d$: It is the section that maps $x$ to the origin $(0)\in \mathbb{A}^1_{\kappa(x)}$ for all $x$.

Note: For a real smooth vector bundle $\pi: E \rightarrow M$ of a real smooth manifold $M$, two smooth sections $s,t$ may be "added": For any point $x\in M$ let $(s+t)(x):=s(x)+t(x)\in E_x$. The fiber $E_x$ is a real vector space and you may add sections fiberwise and you get a global section $s+t\in H^0(M,E)$. It follows the set of global section $H^0(M,E)$ is a real vector space. In the above situation the fiber $\pi_d^{-1}(x) \cong \mathbb{A}^1_{\kappa(x)}$ is not a vector space, but you still get a $k$-vector space structure on the set of global section of $L_d$.

In general if $X:=Spec(A)$ is an affine scheme and $E$ is a finite rank locally trivial $A$-module, it follows

$$\pi_E:\mathbb{V}(E^*):=Spec(Sym^*_A(E^*))\rightarrow X$$

is a "geometric vector bundle" on $X$. The set of global sections of $\pi_E$ equals $E$, hence the set of global sections is an $A$-module.

Answer 2

In this context, a section of $\pi$ is given by a line in $T$: the line given by vanishing of the first coordinate is the zero section, and any other line intersects the zero section in a unique point. To actually map a point $p \in \mathbb P^1$ to its image under $s$, draw a line between $p$ (considered as a point in $T$) and $x$; this is the fiber $\pi^{-1}(p)$, and its intersection with the line corresponding to $s$ is $s(p)$.