Let $N$ be a smooth manifold. Let $\Delta$ be a $C^{\infty}$ distribution on $N$. Suppose we have for all $q \in U$, an open set, $$ \Delta_q = Span ( X'_1(q), ..., X'_r(q) ) $$ for $X'_j \in X(N)$ ($C^{\infty}$ vector fields).

Is it then the case that given any $X \in X(N)$ which belong to $\Delta$
we have $$ X|_{U} = \sum_{j=1}^r a_j X'_j $$ where each $a_j \in C^{\infty}(N)$?
I think this is the case but I was wondering how can I show this? Thank you.


You may as well assume that $U=N$, otherwise restrict to $U$.

Thus the question is, given a manifold $N$, and a distribution $\Delta$ such that $\Delta_q= \newcommand\Span{\operatorname{Span}}\Span(X_{1,q},\ldots,X_{i,q},\ldots)$ for $X_i$ global vector fields on $N$, then for any $X\in \Delta$, do there exist locally finite smooth functions $a_i$ such that $X=\sum_i a_iX_i$. (I've generalized slightly by allowing the index set to be infinite).

The answer is yes. (Admittedly I'm a little new to differential geometry, but this should be correct).


Let $k = \dim \Delta$.

For any $q\in N$, $\Delta_q$ is the span of the $X_{i,q}$, so there are indices $i_1,\ldots,i_k$ such that $X_{i_1},\ldots X_{i_k}$ form a basis for $\Delta_q$ at $q$. Since being a basis for the distribution is an open condition ($\det\ne 0$), there is some neighborhood of $q$, $U_q$, on which $X_{i_1}|_{U_q},\ldots,X_{i_k}|_{U_q}$ form a local basis. On $U_q$, we can write $X|_{U_q}$ as $X|_{U_q} = \sum_{j=1}^k a_{q,i_j}X_{i_j}|_{U_q}$ for some smooth functions $a_{q,i_j}:U_q\to \Bbb{R}$.

Then for the other indices, define $a_{q,i} : U_q\to \Bbb{R}$ by $a_{q,i}=0$ when $i\ne i_j$ for some $j$.

On $U_q$, we now have $$X=\sum_i a_{q,i} X_i |_{U_q}.$$

To extend this to the whole manifold, observe that since we constructed these functions for some neighborhood of every point $q$, we can cover $N$ by open sets $U_q$.

Taking a smooth partition of unity $\{\rho_q\}$, where $\operatorname{supp}\rho_q \subseteq U_q$, we now have $$ X = \sum_q \rho_q \sum_i a_{q,i}X_i =\sum_i \sum_q \rho_q a_{q,i} X_i.$$

Thus if we define $a_i = \sum_q a_{q,i}$, we obtain smooth functions $a_i$ such that $$X=\sum_i a_iX_i,$$ as desired.

Point of clarification

When I say that being a basis is an open condition ($\det\ne 0$), what I mean precisely is the following.

There is some open neighborhood of $q$ on which $\Delta$ has a local basis, $e_1,\ldots,e_k$. With respect to this basis, we can write our vector fields $$X_{i_j} =\sum_{\ell} a_{j\ell} e_\ell$$ for some smooth functions $a_{j\ell}$ defined on this neighborhood. Taking the determinant of the matrix $(a_{j\ell})$, we have that the vector fields $X_{i_j}$ are a local basis when the determinant of this matrix is nonzero. This is true on a neighborhood of $q$, since it is true at $q$.

Side note

Note that this proof didn't require the index set to have any particular cardinality, nor did it require $\Delta$ to be a distribution specifically rather than a general vector bundle. Also, it's fairly clear that the $a_{q,i}$s we construct, if we regard them as functions of $X$ will be linear in $X$. Thus from some slight modifications of this proof we obtain the following result.

Proposition: Let $E$ be a vector bundle on a manifold $N$. Given global sections $\{X_i\}\subseteq \Gamma(E)$ such that $\{X_{i,q}\}$ span $E_q$ at every point $q$ of $N$, there are global sections $a_i \in \Gamma(E^*)$ such that for every $X\in \Gamma(E)$, $$X=\sum_i a_i(X)X_i.$$

An algebraist will recognize this as roughly saying that $\Gamma(E)$ is a projective module over $C^\infty(N)$. This statement is a little stronger, at least when $N$ is compact. When $N$ isn't compact, the proof given here doesn't quite show that it's projective, since our sum is only locally finite. Compare with this characterization of projective modules.

  • $\begingroup$ I realised there was one bit I didn't understand even though it seemed clear a month ago... When you say "Since being a basis for the distribution is an open condition ($\det\ne 0$), there is some neighborhood of $q$, $U_q$, on which $X_{i_1}|_{U_q},\ldots,X_{i_k}|_{U_q}$ form a local basis." I get the first part but don't see how $X_{i_1}|_{U_q},\ldots,X_{i_k}|_{U_q}$ form a local basis part follows... I would appreciate if you could you possibly explain me how this works? $\endgroup$ – Johnny T. Jun 28 at 8:41

I know a partial result.

Theorem: If the linearly independent vector fields $X’_1,...,X’_r$ on a smooth manifold $N$ are such that their pairwise brackets equal $0$, then for any vector field $X$ whose values are in their span we have $X=\sum_{i} a_i X’_i$ with $a_i$’s being smooth maps on $N$.

  • $\begingroup$ Can you possibly give me a reference for this? $\endgroup$ – Johnny T. Jun 28 at 8:11

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