# “Basis extension theorem” for local smooth vector fields

Let $\pi: E \to M$ be a smooth vector bundle of rank $n$, and suppose $s_1, \ldots, s_m$ are independent smooth local sections over an open subset $U \subset M$.

Can I prove the “basis extension theorem” for smooth sections? That is, for each $p \in U$, there are smooth sections $s_{m+1},\ldots,s_n$ defined on some neighborhood of $V$ of $p$ such that $(s_1, \ldots, s_n)$ is a smooth local frame for $E$ over $U \cap V$.

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Yes. Let $p:E\to M$ denote the fibration. Let $W\to U$ be the subbundle of $p^{-1}(U)\to U$ given by $W=\coprod _{p\in U}span\{s_{1}(p),…,s_{m}(p)\}$. Since the secions are smooth, this does indeed define a subbundle. Now let me give you an idea of how to proceed. You want to look at the normal bundle. If you fix a metric $g$ on $M$, there is a natural choice of such a bundle. Namely you can define $N=\coprod_{p\in U}N_{p}$ where $N_{p}$ is the orthogonal complement of $W_{p}$. Check (using the metric) that this is indeed a vector bundle. Then $N$ must be trivial, since $U$ is contractible. Choose a global section of the frame bundle of $N$ and your done!. You can do this without fixing a metric on $M$ too, you just need to use the quotient bundle $p^{-1}(U)/W$.

Let $W$ be a neighborhood of $p$ in $M$ such that there is a local trivialization $\varPhi:\pi^{-1}(W)\to W\times \mathbb{R}^n$. Then $v_i(q)=(\pi_{\mathbb{R^n}}\circ\varPhi\circ s_i)(q)$ are linearly independent in $\mathbb{R^n}$ for each $q\in W$. We can find $v_m, \ldots,v_n \in \mathbb{R}$ such that $\{v_1(p), \ldots,v_m(p),v_{m+1},\ldots,v_n\}$ is a basis for $\mathbb{R^n}$. So $\det [v_1(p), \ldots,v_m(p),v_{m+1},\ldots,v_n]\neq0$, because the $\det$ is continuous there is a neighborhood $V$ of $p$ such that $\det [v_1(q), \ldots,v_m(q),v_{m+1},\ldots,v_n]\neq0$ for each $q\in V$. Then $(s_1, \ldots s_m,s_{m+1}=\varPhi^{-1}(\cdot,v_{m+1}), \ldots, s_{n}=\varPhi^{-1}(\cdot,v_{n}))$ is a frame over $U\cap V$.