Codex @codex  0

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.

A smooth curve tangent to a regular integrable distribution lies in one maximal leaf of a regular foliation. In a Frobenius theorem chart its transverse coordinates have zero derivative, so each short segment lies in one plaque. A finite chain of such segments covers any compact parameter interval and joins plaques in the same leaf. A single embedded local plaque only gives confinement while the curve remains in its chart. A dense immersed cylinder in a three-dimensional torus shows why global embedded confinement cannot generally replace the immersed statement.

An integral manifold of a rank-$r$ smooth distribution $D$ is an immersed $r$-dimensional submanifold with $T_{p}N=D_p$ at every point. For an integrable distribution, maximal connected integral manifolds form its leaves.

A smooth distribution is involutive if its local smooth sections are closed under the Lie bracket of vector fields. The Frobenius theorem identifies this algebraic condition with being an integrable distribution.

A leaf is a maximal connected integral manifold of the smooth constant-rank integrable distribution defining a regular foliation. It carries its intrinsic manifold topology and is immersed in the ambient manifold. It need not be embedded or closed. Local Frobenius theorem coordinates make each leaf segment a plaque with constant transverse coordinates; global confinement to an immersed leaf follows by joining such plaques along tangent curves.

A coorientation gives a globally nonvanishing $C^2$ one-form $\zeta$ with $E=\ker \zeta$. For tangent fields $A,B$ in the integrable distribution,since integrability makes their bracket tangent to $E$. Thus $\zeta \wedge d\zeta =0$. Choose a transverse field $Y$ with $\zeta (Y)=1$. The $C^1$ one-form $\eta =-{\iota }_{Y}d\zeta$ then satisfies $d\zeta =\eta \wedge \zeta$, as is checked on $E$ and $Y$.

The Godbillon-Vey invariant in the unnormalized convention isThe three-form is continuous at the specified regularity, so its integral is defined. We prove independence of both choices by producing exact differences with $C^1$ primitives, allowing ordinary Stokes theorem.

First, for a fixed $\zeta$, another admissible auxiliary form has the form ${\eta }^{\prime }=\eta +a\zeta$. From $d^2\zeta =0$ and $d\zeta =\eta \wedge \zeta$, one gets $d\eta \wedge \zeta =0$. Expanding the two three-forms, using this relation and $\zeta \wedge d\zeta =0$, givesIts integral on the closed manifold vanishes.

Next any other defining form is ${\zeta }^{\prime }=h\zeta$ with $h$ nowhere zero. Let $g=\log |h|$. Thenso one may use ${\eta }_0^{\prime }=\eta +dg$. Its three-form differs byAgain the difference integrates to zero; any subsequent auxiliary choice is handled by the first calculation. This includes reversing the chosen coorientation, since $h$ can be negative. Thus the invariant, and the corresponding de Rham cohomology class, depends only on the oriented manifold and the foliation defined by $E$. Reversing the orientation of $N$, unlike reversing the coorientation of $E$, changes the sign of its numerical value.

The printed hint is impossible: if ${\int }_0^{1}f=0$, thenChanging the order of integration gives the same zero, so the additional condition involving $\int tf$ cannot repair it. Using the same function in both proposed controls cannot generate an arbitrary third coordinate.

Instead, put $\Delta =P_z-P_x{P}_y/2$, $a=2(\Delta +P_x)$, and choose two different oscillatory controls:Their integrals give $x(t)=P_{x}t+a\sin (2\pi t)/(2\pi )$ and $y(t)=P_{y}t+1-\cos (2\pi t)$, hence $x(1)=P_x$ and $y(1)=P_y$. The third endpoint isThe resulting curve is smooth and horizontal at every time, proving smooth horizontal reachability in the real Heisenberg group.

If the Heisenberg horizontal distribution were an integrable distribution, every horizontal curve from the identity would remain in its maximal connected integral leaf. Reachability would force that two-dimensional leaf to be the whole three-dimensional group, contradicting the local leaf coordinates of the Frobenius theorem. The bracket computation in part (i) gives the same obstruction directly.

A smooth rank-$r$ smooth distribution $D\subset TM$ is involutive distribution if the Lie bracket of vector fields of two local sections remains a section. It is an integrable distribution if every point lies on an immersed $r$-dimensional integral manifold with tangent spaces equal to $D$. The Frobenius theorem says these conditions are equivalent, and gives local coordinates with $D=\mathrm{span}({\partial }_1,\dots ,{\partial }_r)$.

For necessity, fields tangent to an integral manifold have brackets tangent to it: they annihilate functions vanishing on the manifold, and so does their commutator. For sufficiency, induct on $r$. The rank-zero case is immediate. Straighten a nonvanishing local section using the flow-box theorem to obtain ${\partial }_1\in D$. Choose a frame ${\partial }_1,Y_2,\dots ,Y_r$ with the $Y_i$ having no ${\partial }_1$ component. Involutivity gives ${\partial }_{1}Y=AY$ for the column of these fields and a smooth matrix $A$. Solve the matrix ordinary differential equation ${\partial }_{1}B=-BA$, with $B=I$ on $x_1=0$. It stays invertible, and $Z=BY$ satisfies ${\partial }_{1}Z=0$. On the transverse slice, the $Z_i$ span an involutive rank-$(r-1)$ distribution. The induction hypothesis supplies adapted slice coordinates; extend them independently of $x_1$. These give the required rank-$r$ coordinate distribution and its integral manifolds.

For the real Heisenberg group, multiplication isDifferentiating left translation at the identity gives its left-invariant frame of the real Heisenberg group:Then $[E_1,E_2]=E_3$ and the other basis brackets vanish. Since $E_3\notin \mathrm{span}(E_1,E_2)$, the Heisenberg horizontal distribution is not involutive, hence not integrable by the Frobenius theorem.

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 ${\mathbb{R}}^3$.