Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 115 / 4

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.

The vertical tangent space of the trivial principal $\mathbb{R}$-bundle is spanned by ${\partial }_z$. Sinceprojects isomorphically onto the tangent space of $S^1\times \mathbb{R}$, it is always complementary to the vertical direction. It is the horizontal distribution of a principal connection precisely when it is invariant under the principal translations $z\mapsto z+a$. The horizontal lifts of ${\partial }_{\theta }$ and ${\partial }_t$ are unique, so this invariance is equivalent toThe functions must also be smooth and periodic in $\theta$, as is already required for them to be functions on the cylinder.

Under these conditions the connection form isIt sends ${\partial }_z$ to $1$, is translation-invariant, and has kernel $D$, proving sufficiency as well. Since the structure group $\mathbb{R}$ is abelian, the bracket term vanishes and

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.