# Question about distribution on a smooth manifold

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).

Proof.

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.

• 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? – 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 $$N$$.

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