Section of a vector bundle

This is a section of a fiber bundle whose fibers are vector spaces.

A section of a vector bundle $\pi :E\to B$ is a map $s:B\to E$ satisfying $\pi \circ s={\mathrm{id}}_B$.

The zero locus of a holomorphic section of a vector bundle of a rank-$r$ bundle is regular when its derivative has rank $r$ along the locus. Its kernel is then the tangent bundle of the smooth zero set. Smoothness of the underlying set alone is insufficient: the square of a reduced defining function has the same set of zeros but zero derivative there.

The smooth sections of a vector bundle form a module over the commutative ring $C^{\infty }(M)$ by pointwise multiplication. A vector bundle morphism over the identity induces a $C^{\infty }(M)$-linear map between these modules. Bundle morphisms from maps of smooth sections gives the converse, without assuming the bundle has a global frame.

A nowhere-zero section of a vector bundle is a section $s$ with $s(x)\ne 0$ in every fiber. Every trivial vector bundle of positive rank has one.

Identify ${\mathbb{R}}^{2m+2}$ with ${\mathbb{C}}^{m+1}$ and let $J$ be multiplication by $i$. On the unit sphere $S^{2m+1}$, the vector field $V(x)=Jx$ has norm one and satisfies $⟨x,Jx⟩=0$. It is therefore a continuous tangent nowhere-zero section, including on $S^1$.

A unit nowhere-zero section $\sigma$ of a real vector bundle with a fiber metric defines a rank-one trivial vector bundle. Its fiberwise orthogonal complement is a subbundle: locally complete $\sigma$ to an orthonormal frame. The vector bundle isomorphism $(v,t)\mapsto v+t\sigma$ has inverse $w\mapsto (w-⟨w,\sigma ⟩\sigma ,⟨w,\sigma ⟩)$. For an oriented bundle, orient the complement so that its positive frame followed by $\sigma$ is positive.

The hairy ball theorem says that every continuous tangent vector field on an even-dimensional sphere vanishes somewhere. Equivalently, the tangent bundle of an even-dimensional sphere has no nowhere-zero section.

A nowhere-zero tangent vector field on $S^n$ can be normalized to a unit field $w(x)$ perpendicular to $x$. Then $\cos (\pi t)x+\sin (\pi t)w(x)$ is a homotopy from the identity to the antipodal map. Their mapping degrees are $1$ and $(-1{)}^{n+1}$, respectively, contradicting homotopy invariance of mapping degree when $n$ is positive and even. For $S^0$ the tangent spaces are zero, so every tangent field is zero directly.