Let $M$ be a smooth manifold of dimension $n$ and let $\pi: E \rightarrow M$ be a vector bundle of rank $k$. Then one can define the $C^{\infty}(M)$-module of smooth global sections of $E$, denoted by $\Gamma(M,E)$. Now my notes say that the rank of this module is at most $k$ and equal to $k$ if and only if the bundle is trivial. Does the second statement not need some extra assumptions? Maybe orientability of $E$ or maybe only consider nowhere vanishing sections?
Consider as a counterexample the Möbius strip $M$ as a non-trivial vector bundle of rank $1$ over $S^1$. However, any global section $s: S^1 \rightarrow M$ for itself will be linearly independent. Since it has to vanish at some point $x \in S^1$ (one can find a section that vanishes at only one point) the section cannot span $\Gamma(S^1,M)$, but it will be linearly independent since any smooth function that is non-zero at $x$ is also non-zero in some neighbourhood.
So is the counterexample false or are my notes missing extra assumptions?
Best Answer
In general we have:
Proposition
Let $M$ be a smooth manifold of dimension $n$ and let $E\longrightarrow M$ be a smooth vector bundle of rank $k$. Then $E$ is isomorphic to the trivial bundle if and only if there exist $k$ smooth linearly independent sections $s_1,\dots,s_k$, i.e. for every $x\in M$ the vectors $s_1(x),\dots,s_k(x)$ are linearly independent in $E_x$.
The proof is not difficult and it can be found in many books. Returning to your question, in light of this proposition, I think that the authors are saying that $\Gamma(M;E)$ has rank $r\le k$ if you can find $r$ linearly independent sections.
I hope this will help.