Sections of line bundles

A section of a line bundle $L$ is like a vector field. That is it is a map $\varphi : M \to L$ such that $\varphi (m) \in L_{m}$ for all $m$ or more succinctly $\pi\circ \varphi = id_{m}$.

Example 1.5 (The trivial bundle.)   $L = \mathbb{C}\times M.$ Every section $\varphi$ looks like $\varphi (x) = (f (x), x)$ for some function $f$.

Example 1.6 (The tangent bundle to $S^2$.)   $TS^2$. Sections are vector fields. Alternatively because each $T_{xS^{2}}\subset \mathbb{R}^{3}$ we can think of a section $s$ as a map $s \colon S^{2}\to \mathbb{R}^{3}$ such that $\langle s(x) , x \rangle = 0$ for all $x \in S^{2}$.

Example 1.7 (The Hopf bundle)   By definition a section $s \colon \mathbb{C}P_{1} \to H$ is a map

$$s \colon \mathbb{C}P_{1} \to H \subset \mathbb{C}^{2} \times \mathbb{C}P_1$$

which must have the form $[z] \mapsto ([z], w)$. For convenience we will write it as $s([z]) = ([z], s(z))$ where, for any $[z]$ $s \colon \mathbb{C}P_{1} \to \mathbb{C}^{2}$ satisfies $s([z]) = \lambda z$ for some $\lambda \in \mathbb{C}^{\times}$.

The set of all sections, denoted by $\Gamma (M,L)$, is a vector space under pointwise addition and scalar multiplication. I like to think of a line bundle as looking like Figure 1.

Figure 1: A line bundle.

Here $O$ is the set of all zero vectors or the image of the zero section. The curve $s$ is the image of a section and thus generalises the graph of a function.

We have the following result:

Proposition 1.1   A line bundle $L$ is trivial if and only if it has a nowhere vanishing section.

Proof. Let $\varphi\colon L \to M \times \mathbb{C}$ be the trivialisation then $\varphi^{-1} ( m, 1)$ is a nowhere vanishing section.

Conversely if $s$ is a nowhere vanishing section then define a trivialisation $M \times \mathbb{C}\to L$ by $(m, \lambda) \mapsto \lambda s (m).$ This is an isomorphism. $\qedsymbol$

Note 1.2   . The condition of local triviality in the definition of a line bundle could be replaced by the existence of local nowhere vanishing sections. This shows that $TS^2$ is locally trivial as it clearly has local nowhere-vanishing vector fields. Recall however the so called `hairy-ball theorem' from topology which tells us that $S^2$ has no global nowhere vanishing vector fields. Hence $TS^2$ is not trivial. We shall prove this result a number of times.