Cross-Section of Vector Bundle

Definition

A cross-section vector bundle $\xi$ is a section of $\pi :E\to B$ such that takes each $b\in B$ into the corresponding fiber $F_b(\xi )$.

Zero Section

The zero section of the vector bundle $\xi$ is a map $z:B\to E$ such that $b\mapsto 0\in {\pi }^{-1}(b)$.

The zero section is an homotopy inverse of $\pi$ for each fiber $F_b$ because $\pi \circ z=I{d}_B$ and $z\circ \pi =0_{\pi }(v)$ for $v\in E$.

Note that we can find the homotopy between $0_{\pi (v)}$ and $I{d}_E$ such that

$$H:E\times I\to E,H(v,t)=(1-t)v.$$

Example

Section of $TM$ is a vector field $X:M\to TM$. As a space of vector fields on a manifold, we denote $\mathfrak{X}(M)$.

Now we show there is some twisting.

$\xi$ trivial bundle implies

$$E_0\cong B\times ({\mathbb{R}}^k\setminus e)\simeq B\times S^{k-1}.$$

We have the tautological line bundle ${\gamma }_1^1$:

We can think of it like $(\text{line},\text{point on line})\in E$. This gives us the open mobius strip.

As we see for $n=1$ is non trivial bundle, it still holds for $n\ge 1$.

Theorem ${\gamma }_n^1$ is nontrivial for $n\ge 1$.

Proof

$E_0$ is homotopic to $E\cap {\mathbb{RP}}^n\times {\mathbb{S}}^n$ which is homeomorphic to ${\mathbb{S}}^n$. So it is connected. However, if it is trivial bundle, $E_0$ is homotopic to ${\mathbb{RP}}^n\times {\mathbb{S}}^0$ that disconnected. Therefore tautological line bundle is nontrivial

Linearly Independent Section

Definition

For sections $s_1,\cdots ,s_n$ in rank $n$ vector bundle,

are linearly independent if $\forall b\in B,\left\{s_1(b),\cdots ,s_n(b)\right\}$ is linearly independent in ${\pi }^{-1}(b)$.

Theorem

Rank $n$ vector bundle $\xi$ is trivial $\iff$ $\xi$ has $n$ linearly independent sections.

Proof

Suppose that $\xi$ is a trivial bundle. Then canonically, we can choice a linearly independent section as $(b,e_i)$.

On the other hand, $f:B\times {\mathbb{R}}^n\to E$ as $(b,\Sigma a_i{s}_i)\mapsto \Sigma a_i{s}_i(b)$ is an global bundle map.

Example

$T{T}^2$ has $2$ linearly independent sections because $T{T}^2=T(S^1\times S^1)\cong T{S}^1\oplus T{S}^1$, thus $T{T}^2\cong T^2\times {\mathbb{R}}^2$.