Past exam of the mathematics course of the University of Cambridge / 2018 / iii / Paper 115 / 2

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.

Using the left-invariant frame of the real Heisenberg group, horizontality givesFor an arbitrary starting point $(x_0,y_0,z_0)$,This is the control equation for the Heisenberg horizontal distribution. The final coordinate records a signed area-like integral, so changing the path in the $(x,y)$ plane can change the endpoint in the central direction.

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.