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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.

Let $k$ generate the Killing horizon and let $\ell =fk$ be an affinely parametrized generator wherever $k\ne 0$. Choose an auxiliary null vector $n$ with $n\cdot \ell =-1$ and the screen-space projector $q_{ab}=g_{ab}+{\ell }_a{n}_b+n_a{\ell }_b$. The optical tensor is the screen projection $B_{ab}=q_a{}^c{q}_b{}^d{\nabla }_d{\ell }_c$.

Since the projector annihilates $k$, the symmetric part isby the Killing equation. Its trace is the null expansion and its trace-free symmetric part is the null shear, so both vanish. Also $\ell$ is normal to the null hypersurface. Frobenius theorem gives ${\ell }_{[a}{\nabla }_b{\ell }_{c]}=0$, which implies $q_a{}^c{q}_b{}^d{\nabla }_{[d}{\ell }_{c]}=0$: the null twist vanishes. These arguments use derivatives tangent to the horizon and hence apply to any smooth null geodesic congruence containing its generators, independently of its extension off the horizon. At a regular bifurcation surface, use the smooth affine generator and continuity. ThusThis is the vanishing of the optical scalars on a Killing horizon.

Write $V={\sum }_a{f}_a{Z}^{(a)}$ and $W={\sum }_b{g}_b{Z}^{(b)}$. Relation (2) reduces $[V,W]$ to brackets of individual summands. Applying relations (3), (1), and (3), in that order, givesThe middle term vanishes because the $Z^{(a)}$ commute, and every remaining term lies in their smooth span. ThereforeThis is the involutivity condition in the Frobenius theorem.

On the null hypersurface, the generator covector is proportional to a normal, locally $U_a=f{\nabla }_{a}u$ for a level-set function $u$. The Frobenius theorem therefore givesContracting once with $N^a$ and projecting the remaining indices with $P^a{}_b$ removes every term containing $U$ or $N$ and leaves $P_b{}^d{P}_c{}^e{\nabla }_{[d}{U}_{e]}=0$. Henceon the hypersurface.

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

For a principal connection with connection form $\mathcal{A}$, the horizontal distribution of a principal connection is $H=\ker \mathcal{A}$, the complement of the tangent spaces to the $G$-orbits. Its curvature of a principal connection isIf $X,Y$ are horizontal vector fields, then $\mathcal{A}(X)=\mathcal{A}(Y)=0$, and henceThe Frobenius theorem says that $H$ is integrable exactly when $[X,Y]$ is horizontal for all horizontal $X,Y$. The displayed identity makes this equivalent to the vanishing of the horizontal two-form $\mathcal{F}$, hence to $\mathcal{F}=0$. Thus the horizontal distribution is integrable exactly for a flat principal connection.

For $f=t\cos \theta +1$ and $g=\sin \theta$,The connection is therefore flat, so the Frobenius theorem gives horizontal sections locally.

A global section has the form $z=h(\theta ,t)$ and is horizontal exactly whenThe second equation gives $h=t\sin \theta +k(\theta )$, and the first then forces $k^{\prime }(\theta )=1$. No such $k$ is periodic on $S^1$, so no global horizontal section exists. Equivalently, the horizontal lift of one positive circuit in the $\theta$ direction changes $z$ bywhich is nontrivial holonomy.

A hypersurface normal is locally proportional to the gradient of a defining function. The Frobenius theorem therefore impliesTaking the antisymmetric part of the formula in part i gives $K_{[ab]}=0$, henceEquivalently, torsion freedom gives $K(X,Y)-K(Y,X)=-n\cdot [X,Y]=0$ because the bracket of tangent vector fields is tangent.

For generators of a null hypersurface, the Frobenius theorem gives ${\hat{\omega }}_{ab}=0$. The null energy condition and the Einstein field equations imply $R_{ab}{U}^a{U}^b\ge 0$, while the shear norm is nonnegative. With $A=D-2$, Raychaudhuri's equation givesAs long as $\theta <0$ remains finite,If $\theta (0)={\theta }_0<0$, integration givesThe right-hand side reaches zero at $\lambda =A/|{\theta }_0|$, which a finite negative $\theta$ cannot cross. Hence the null focusing theorem forcesprovided the generator extends that far.

Differentiate $B_{ab}={\nabla }_b{U}_a$ along an affinely parametrized generator. Commuting covariant derivatives and using ${\nabla }_{U}U=0$ gives the optical evolution equationTaking its screen trace and usingproduces the Null Raychaudhuri equationThe generators lie in a null hypersurface, so they are hypersurface orthogonal. The Frobenius theorem therefore gives ${\hat{\omega }}_{ab}=0$, leaving