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Given a distribution $L^r=\{X_1,...X_r\}$, that is $\{X_1(p),...X_r(p)\}$ linearly independent on each point $p$ on a manifold $M$ of dimension $n$. At a point $p \in M$, the system $\{X_1(p),...X_r(p)\}$ can be extended to a basis $B$, and since the condition of being linearly independent is open we can extend it in a differentiable way to a neighborhood $U$ of $p$. If the set of 1-forms $\{\phi_1,...,\phi_n\}$ evaluated at each point of $U$ is its dual basis, then locally the distribution $L^r$ is given by the vanishing of $\{\phi_{r+1},...,\phi_n\}$. I don't understand this last statement.

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up vote 1 down vote accepted

By duality we have $\phi_i\,X_j =\begin{cases} 1 \text{ if } i=j\text{ and }\\ 0 \text{ otherwise}\end{cases}$

so the linear span of $X_1,...,X_r$ can be represented as: $$\left\{ Z \in T_p M \text{ such that } \phi_j Z = 0 \text{ for all } j=r+1,...,n\right\}.$$

It's linear algebra, think of the $X_s$ and $\phi_s$ as just "vectors".

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