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If $D={\bigcap }_{i=1}^q\ker {\alpha }^i$ has constant rank, the Frobenius theorem says that $D$ is integrable exactly when its annihilator is closed under the exterior derivative: equivalently,This is also equivalent to involutivity of $D$.

For a nowhere-zero 1-form $\alpha$, the hyperplane distribution $\ker \alpha$ is tangent to hypersurfaces exactly when $\alpha \wedge d\alpha =0$. In general relativity this characterizes hypersurface-orthogonal vector fields and implies vanishing twist.

The differential-forms version of the Frobenius theorem says that a constant-rank distribution $D={\bigcap }_i\ker {\alpha }^i$ is integrable exactly whenfor suitable 1-forms ${\beta }^i{}_j$. For a plane distribution $D=\ker \alpha$ on ${\mathbb{R}}^3$, this reduces to the integrability criterion for a plane distribution $\alpha \wedge d\alpha =0$.

For example, $\ker dz$ is integrable: its integral surfaces are the horizontal planes $z=\text{constant}$. In contrast, for $\alpha =dz-xdy$,Thus $\ker (dz-xdy)$ is not integrable; it is the standard contact structure on ${\mathbb{R}}^3$.