Frobenius theorem

A constant-rank smooth distribution $D$ is an integrable distribution exactly when it is involutive. In differential-form language, let ${\alpha }^1,\dots ,{\alpha }^q$ be a local basis of its annihilator, so $D={\bigcap }_i\ker {\alpha }^i$. The equivalent condition isfor suitable local one-forms ${\beta }^i{}_j$. More precisely, the ideal generated by the annihilating forms inside the exterior algebra of differential forms is closed under the exterior derivative; this is an exterior differential ideal. It is not a claim that a two-form lies in the one-form annihilator itself.

For local vector fields $X,Y$ tangent to $D$, the definition of exterior derivative givesThus the displayed closure condition says exactly that the Lie bracket of vector fields remains tangent to $D$. The theorem turns this local algebraic condition into integral manifolds through every point.

For a nowhere-zero one-form $\alpha$, its kernel is tangent to local hypersurfaces precisely when $\alpha \wedge d\alpha =0$. In general relativity, applying this to a metric-dual one-form characterizes hypersurface orthogonality and vanishing twist.

An exterior differential ideal on a smooth manifold is a graded ideal $\mathcal{I}$ in the exterior algebra ${\Omega }^{\ast }(M)$ of differential forms that is closed under the exterior derivative:Being an ideal also requires $\alpha \wedge \beta \in \mathcal{I}$ for every $\beta \in {\Omega }^{\ast }(M)$. Thus stability under differentiation is an additional condition, not a consequence of wedge closure.

For example, the ideal generated by $dz$ in ${\mathbb{R}}^3$ is differential because $d(fdz)=df\wedge dz$. Its integral surfaces include the planes $z=\text{constant}$. In contrast, the ideal generated by $\alpha =dz-xdy$ is not differential: $d\alpha =-dx\wedge dy$ and $\alpha \wedge d\alpha =-dz\wedge dx\wedge dy\ne 0$, whereas wedging $\alpha$ with any element of its generated ideal gives zero. The Frobenius theorem interprets this obstruction as failure of the associated tangent-plane distribution to be integrable.

A nowhere-zero differential one-form $n$ is hypersurface orthogonal if locally $n=Fdt$ with $F\ne 0$. Its kernel consists of tangent vectors to the level hypersurfaces of $t$. Frobenius theorem makes this equivalent to $n\wedge dn=0$, where $d$ is the exterior derivative. With a torsion-free Levi-Civita connection, this is $n_{[\alpha }{\nabla }_{\beta }{n}_{\gamma ]}=0$. For a timelike unit normal, projecting the derivative on both indices gives zero antisymmetric part, which explains why hypersurface orthogonality implies symmetric extrinsic curvature.

For a non-null Killing vector field $V$ of a Levi-Civita connection, hypersurface orthogonality implies that $\eta =V^{♭}/F$, with $F=g(V,V)$, is a closed differential form. The Killing equation and $V^{♭}\wedge d{V}^{♭}=0$ give $2F{\nabla }_{\mu }{V}_{\nu }=V_{\nu }{\partial }_{\mu }F-V_{\mu }{\partial }_{\nu }F$, whence $d\eta =0$. The Poincaré lemma gives $\eta =d\varphi$ locally. For a timelike $V$, $\varphi$ supplies local static time coordinates. This does not assert global exactness.

For an involutive distribution of constant rank $r$, induct on $r$. Straighten a nonzero section by the flow-box theorem, and choose a frame ${\partial }_1,Y_2,\dots ,Y_r$ with no ${\partial }_1$ component in the remaining fields. Involutivity implies ${\partial }_{1}Y=AY$ for their column $Y$. The invertible solution of ${\partial }_{1}B=-BA$, initialized by $B=I$ on $x_1=0$, makes $Z=BY$ independent of $x_1$. Its restriction to the transverse slice is involutive of rank $r-1$, so induction provides coordinates spanning it. Extending these coordinates independently of $x_1$ gives $D=\mathrm{span}({\partial }_1,\dots ,{\partial }_r)$. Conversely, fields tangent to an integral manifold have tangent brackets because they preserve the ideal of smooth functions vanishing on it.