My question is this. Suppose we are given some smooth vector fields $X_1, X_2,…, X_k$ which are linearly independent at all points in a neighborhood $U$ (EDIT: diffeomorphic to a ball) of $R^n$. Do there necessarily exist smooth vector fields $X_{k+1},…,X_n$ such that $X_1,…,X_n$ are linearly independent on $U$? Of course at any point $p$, one can complete $X_1(p),…,X_k(p)$ to a basis for $R^n$, but there are infinitely many choices, and one can't hope to get smooth vector fields choosing arbitrarily at each point. If the result is true, there must be some algorithm by which to assign vectors in a smooth way, and this algorithm is what I would like to know. I should add that I am in the situation where $X_1,…,X_n$ do not necessarily commute, so they cannot be said to be coordinate fields. Thanks.
[Math] Extending a Set of Linearly Independent Vector Fields to a Basis
differential-geometrymanifoldsvector-bundles
Best Answer
The answer to your revised question is yes, but with the caveat that we may have to shrink $U$.
Suppose $X_1, \ldots, X_k$ are pointwise linearly independent vector fields on an open subset $U$ of $\mathbb{R}^n$. Pick a point $p$ in $U$, and choose vectors $v_{k+1}, \ldots, v_n$ so that $X_1 (p), \ldots, X_k (p), v_{k+1}, \ldots, v_n$ forms a basis for the tangent space at $p$. Since $U$ is an open subset of $\mathbb{R}^n$, we can use the standard frame to extend each $v_i$ to a (smooth) vector field $X_i$. Now consider $\det (X_1, \ldots, X_n)$: this is a smooth function on $U$, and by construction it is non-zero at $p$, so by continuity it is non-zero on a neighbourhood $V$ of $p$. Thus we have a frame $X_1, \ldots, X_n$ on $V$ extending $X_1, \ldots, X_k$.