Smooth Vector Bundles and Sections - Examples

If $s:M\to M\times \mathbb{R}$ is a section, then ${\mathrm{pr}}_1(s(p))=p$, so $s(p)=(p,f(p))$ for a unique scalar function $f:M\to \mathbb{R}$.

The map $s$ is smooth exactly when its second component $f$ is smooth. Conversely every smooth $f$ gives a smooth section $p\mapsto (p,f(p))$. Thus sections of the trivial line bundle are the same as smooth functions on $M$.

The overlap $U_0\cap U_1$ has two connected components. Defining the rank-one transition function to be $1$ on the upper component and $-1$ on the lower one gives a locally constant smooth cocycle, so [L1] constructs a smooth line bundle $L\to S^1$.

The refutation recorded in [L2] is exactly the argument that this line bundle admits no global frame, so $L$ is not trivial. This smooth nontrivial line bundle is the Mobius bundle.

On $U_i$, every line $[x]$ has a unique representative with $i$-th coordinate $1$. Let $s_i([x])$ be that representative. Then $s_i([x])$ spans the fibre of ${\gamma }^1$ over $[x]$, so $s_i$ is a local frame of the tautological line bundle on $U_i$.

On $U_i\cap U_j$, the two generators satisfy $s_j=\frac{x_i}{x_j}{s}_i$, because both are rescalings of the same representative vector $x$. Therefore a vector with $s_i$-coordinate $a$ has $s_j$-coordinate $(x_j\mathrm{/}{x}_i)a$, so the chart transition $g_{ji}$ is the nonzero smooth scalar $x_j\mathrm{/}{x}_i$. By [L1], these transition functions define a smooth line bundle, namely ${\gamma }^1$.

The earlier tangent-bundle construction gives local coordinates of the form $(p,v)\in U\times {\mathbb{R}}^n$, and the fibre over $p$ is the vector space $T_{p}M$. Thus $TM\to M$ is a smooth rank-$n$ vector bundle.

The cotangent-bundle construction gives the same kind of local description with fibres $T_p^{\ast }M$, so $T^{\ast }M\to M$ is likewise a smooth rank-$n$ vector bundle.

Write $S^n$ as the regular level set of $F(z)=⟨z,z⟩$. Since $d{F}_x(v)=2⟨x,v⟩$, [L3] gives $T_x{S}^n=x^{\perp }$. Its Euclidean orthogonal complement is therefore the one-dimensional space $\mathbb{R}x$, so [L1] identifies the normal bundle with a line bundle spanned by the radial vector.

The smooth section $x\mapsto x$ is nowhere zero and spans that line at every point, so it is a global frame. Therefore [L2] implies that the normal bundle of $S^n$ is trivial.

Define $X:S^1\to T{S}^1$ by $X(x,y)=(-y,x)$. This vector is tangent to $S^1$ at $(x,y)$ because it is orthogonal to the radial vector $(x,y)$, and it is never zero on the circle.

Because $T{S}^1$ has rank $1$, the nowhere-zero tangent field $X$ is a global frame. Hence [L2] shows that $T{S}^1$ is trivial.

A point of $q^{\ast }{\gamma }^1$ is a pair $(x,([x],v))$ with $v\in \mathbb{R}x$. Define a section $s:S^n\to q^{\ast }{\gamma }^1$ by $s(x)=(x,([x],x))$. This is well defined because $x$ lies in the line represented by $[x]$.

The section $s$ is nowhere zero. Since $q^{\ast }{\gamma }^1$ has rank $1$, it is a global frame, so [L2] implies that the pulled-back tautological bundle is trivial.

Define $G:E\to E\oplus F$ by $G(e)=(e,\mathrm{\Phi }(e))$. This is a smooth bundle map over ${\mathrm{id}}_M$, and each fibre map $G_p(v)=(v,{\mathrm{\Phi }}_p(v))$ is injective, hence has constant rank $\mathrm{rank}E$.

The image of $G$ is exactly the graph ${\mathrm{\Gamma }}_{\mathrm{\Phi }}$. Therefore [L2] implies that ${\mathrm{\Gamma }}_{\mathrm{\Phi }}$ is a smooth vector subbundle of $E\oplus F$.

On the trivial line bundle $\mathbb{R}\times \mathbb{R}\to \mathbb{R}$, define the smooth bundle map $\mathrm{\Phi }(x,v)=(x,xv)$. For $x\mathrm{\ne }0$, the fibre map is injective, so $\ker {\mathrm{\Phi }}_x=\{0\}$. At $x=0$, the fibre map is zero, so $\ker {\mathrm{\Phi }}_0=\mathbb{R}$.

The fibre dimensions of $\ker \mathrm{\Phi }$ jump from $0$ to $1$, so the kernel is not locally trivial and therefore not a smooth vector subbundle. This is exactly why [L1] requires constant rank.