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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.

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It's not obvious to me why existence would imply an algorithm in this case. If you just said that on general principle, note that that general principle is false; for instance, we can prove the existence of uncomputable numbers, but there is by definition no algorithm for finding one. –  joriki Nov 4 '12 at 10:42
    
Yes, sorry, I exaggerate when I say "must." –  Cass Nov 4 '12 at 10:58
    
For some basic information about writing math at this site see e.g. here, here, here and here. –  Julian Kuelshammer Nov 4 '12 at 11:44

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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 $X_1, \ldots, X_k$.

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Perfect. Thanks! –  Cass Nov 5 '12 at 8:01

In general, the answer is no. Here's a counterexample. Let $U = \mathbb R^3\smallsetminus\{0\}$, and let $X_1$ be the unit outward-pointing radial vector field on $U$. If we could complete $X_1$ to a global frame $(X_1,X_2,X_3)$, then we could apply the Gram-Schmidt algorithm to it and produce a global orthonormal frame $(X_1,\tilde X_2,\tilde X_3)$. At points of the unit sphere, because $X_1$ is normal to the sphere, it follows that $\tilde X_2$ and $\tilde X_3$ are tangent to it. Thus we would have a global frame $(\tilde X_2,\tilde X_3)$ for $S^2$, which is impossible.

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Thanks Jack (also, thanks Julian for tidying up my post). Unfortunately, I phrased the question poorly. I should have said a neighborhood diffeomorphic to a ball in $R^n$ (or just take an actual ball in $R^n$). I basically want to know if this result is even true locally on a manifold. –  Cass Nov 5 '12 at 6:39

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