For $x\in S^2\subset\mathbb R^3$, let $F(x)=x^T$, viewed as a nonzero $1$ by $3$ matrix. This defines a <smooth map between manifolds>[smooth map] $F:S^2\to X_{1,3}$. Its pulled-back fiber is
$$
(F^*E_{1,3})_x=\ker x^T=x^\perp=T_xS^2,
$$
so the evident fiberwise identity gives an isomorphism of <vector bundle>[vector bundles] $F^*E_{1,3}\cong TS^2$.

If $E_{1,3}$ were a <trivial vector bundle>, its <pullback vector bundle>[pullback] would be trivial. A trivial rank-two bundle has a <nowhere-zero section>, whereas the assumed form of the <Hairy ball theorem> says that the <tangent bundle> $TS^2$ does not. Hence $E_{1,3}$ is nontrivial.

Solution

Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 115 / 1 / d / Solution

A nowhere-zero section of a vector bundle is a section $s$ with $s(x)\ne0$ in every fiber. Every trivial vector bundle of positive rank has one.


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 Incoming links: Nowhere-zero section