EDS vs Pfaffian systems

The question is: can a Pfaffian system describe the same EDS as a general EDS?

The answer depends on what is meant by "describe the same EDS":

1. The Geometric View (Yes)

If an EDS $\mathcal{I}$ on an $n$-dimensional manifold $M$ has exactly one $k$-dimensional integral element $E_p\subset T_{p}M$ at each point $p$, then the assignment $p\mapsto E_p$ defines a smooth $k$-dimensional distribution (a plane field) on $M$.

From linear algebra, any $k$-dimensional subspace in an $n$-dimensional tangent space can be uniquely cut out by the kernels of $n-k$ linearly independent 1-forms. Therefore, locally, you can always find smooth 1-forms ${\theta }^1,\dots ,{\theta }^{n-k}$ such that:

If you construct a Pfaffian system $\mathcal{J}=⟨{\theta }^1,\dots ,{\theta }^{n-k}⟩$, then by design:

• $\mathcal{I}$ and $\mathcal{J}$ have the identical $k$-dimensional integral elements at every point.
• $\mathcal{I}$ and $\mathcal{J}$ will yield the identical $k$-dimensional integral manifolds.

Nevertheless, while $\mathcal{I}$ and $\mathcal{J}$ share the exact same $k$-dimensional integral manifolds, they do not share the same lower-dimensional solutions. Specifically, $\mathcal{J}$ imposes strict restrictions on curves and lower-dimensional surfaces, whereas the lack of 1-forms in $\mathcal{I}$ allows for a much larger, less restricted set of lower-dimensional integral manifolds. Therefore, $\mathcal{J}$ only serves as a perfect geometric substitute if we are strictly looking for $k$-dimensional solutions.

2. The Algebraic View (No)

An Exterior Differential System is defined by its ideal of differential forms. Even if two systems share the same solutions, their underlying algebraic structures can be completely different.

Example:

• The Original EDS: ${\mathcal{I}}_1$ is generated entirely by 2-forms. It contains no algebraic 1-forms at all.
• The Pfaffian System: The unique 2-dimensional integral element is annihilated by the 1-form $\theta =dz+ydx$. The corresponding Pfaffian system is $\mathcal{J}=⟨dz+ydx⟩$.

Clearly, ${\mathcal{I}}_1\ne \mathcal{J}$ because $\mathcal{J}$ contains 1-forms, while ${\mathcal{I}}_1$ does not. They are completely different ideals in the algebra of forms.

Summary of the Relationship

While they are not equal, they are related by a strict containment. A foundational algebraic fact of differential forms states that if a form vanishes on a subspace defined by a set of 1-forms, it must lie in the algebraic ideal generated by those 1-forms.

Because every form in the original ideal $\mathcal{I}$ must vanish on its unique integral elements, it is guaranteed that:

The original EDS will always be a sub-ideal of the Pfaffian system that characterizes its distribution, meaning the Pfaffian system is a simplified, broader container holding the same geometric behavior.