# Completion of local frames for the tangent bundle of a smooth manifold

In John M. Lee book Introductions to smooth manifolds Proposition 8.11 is left as exercise. Can anyone give me hints and suggestions to prove it?

In particular, i want to show:

Let $$M$$ be a smooth $$n$$-manifold with or without boundary and let $$(X_1,\dots,X_k)$$ be a linearly independent $$k$$-tuple of smooth vector fields on an open subset $$U$$ of $$M$$, with $$1\le k< n$$. Then for each $$p\in U$$ there exist smooth vector fields $$X_{k+1},\dots,X_n$$ in a neighborhood $$V$$ of $$p$$ such that $$(X_1,\dots,X_n)$$ is a smooth local frame for $$M$$ on $$U \cap V$$.

I can easily see that since $$X_1|_p,\dots,X_k|_p$$ are vectors linearly independent in $$T_pM$$ then there exist $$v_{k+1},\dots,v_n$$ in $$T_pM$$ such that $$\{X_1|_p,\dots,X_k|_p,v_{k+1},\dots,v_n\}$$ is a basis for $$T_pM$$.

I also know that I can extend each $$v_i$$ to a smooth global vector field on $$M$$, say $$X_i$$ (with $$i>k$$). But then, with this argument, it may be that $$(X_1,\dots,X_n)$$ is not a local frame for $$M$$ on $$U$$. I can only say that $$(X_1|_p,\dots,X_n|_p)$$ is a basis for $$T_pM$$.

EDIT Read this below as a (too long for a comment) comment for Sou's answer.

Thank you very much Sou! Please let me add some details to see if I have fully understood. Let be $$(W,(x^i))$$ a smooth chart for $$M$$ in $$p$$, then we have $$X_i|_q=X_i^j(q)\frac{\partial}{\partial x^j}|_q$$ for each $$q$$ in $$U \cap W$$ and each $$i\le k$$, and we have $$v_i=v_i^j\frac{\partial}{\partial x^j}|_p\in T_pM$$ for each $$k.

Now I define $$X_{k+i}|_q=v_i^j\frac{\partial}{\partial x^j}|_q$$ for each $$q\in W$$, and these are smooth vector fields on $$W$$.

Now the map $$G:U \cap W\to M(n\times n, \mathbb{R}), \quad q\mapsto (X_i^j(q))$$ is smooth and $$G^{-1}(GL(n,\mathbb{R}))$$ is the neighborhood of $$p$$ on which $$(X_1,\dots,X_n)$$ is a smooth local frame, right?

• Have you tried working in local coordinates around $p$, i.e., working in $\Bbb R^n$, and then going back to a neighborhood of $p$ in $M$? – Ted Shifrin Jan 22 '19 at 18:26
• Is there some result for $\mathbb{R}^n$ which may help? – Minato Jan 23 '19 at 6:57
• You may want to look here to compare your work. The accepted answer there by Diane is trustable since he is Geometer btw. – Si Kucing Jan 25 '19 at 8:46
• Thank- you Sou, you are a very kind person :) – Minato Jan 27 '19 at 13:38

$$\textbf{Hint: }$$ After you choose vectors $$v_{k+1},\dots,v_n$$, you have linearly independent vectors on $$T_pM$$. Extend $$\{v_{k+1},\dots,v_n\}$$ around a neighbourhood of $$p$$, say to constant local vector fields $$X_{k+1},\dots,X_n$$. Since $$\{X_1|_p,\dots,X_k|_p,X_{k+1}|_p,\dots,X_n|_p\}$$ linearly independent, the matrix $$[X_i^j(p)]$$ is invertible. Use continuity of determinant function to show that there is a smaller neighbourhood of $$p$$ such that $$\{X_1,\dots,X_n\}$$ is linearly independent there.