Contact manifold

Remarks

The condition does not depend on the choice of $\alpha$: if $\alpha' = h \alpha$ with $h$ nowhere zero, then $\alpha' \wedge (d\alpha')^n = h^{n+1} \alpha \wedge (d\alpha)^n$. Equivalently, $d\alpha$ restricted to each hyperplane $\xi_x$ is nondegenerate. A contact form exists globally exactly when $\xi$ is coorientable.

Example: on $\R^3$ with coordinates $(x, y, z)$, let $\alpha = dz - y\,dx$. Then $d\alpha = dx \wedge dy$ and $\alpha \wedge d\alpha = dz \wedge dx \wedge dy = dx \wedge dy \wedge dz \ne 0$, so $\ker \alpha$ is a contact structure, the standard one. The unit sphere $S^{2n+1} \subset \C^{n+1}$ with $\alpha = \sum_j (x_j\, dy_j - y_j\, dx_j)$ restricted to it is another standard example.

Non-example: on $\R^3$ the form $\alpha = dz$ has $d\alpha = 0$, so $\alpha \wedge d\alpha = 0$; its kernel is tangent to the planes $z = \text{const}$. By the Frobenius theorem, $\alpha \wedge d\alpha = 0$ is exactly the condition for $\ker \alpha$ to be tangent to a family of surfaces.