Let $M$ be a contact manifold (with globally defined contact form, for simplicity) of dimension $2n+1 \ge 5$. So $H$ is a rank $2n$ subbundle of $TM$ and $[H,H] = TM$.
A partial connection with respect to $H$ on a vector bundle $E$ is a map $\Gamma(E) \to \Gamma(H^* \otimes E)$ satisfying the Leibniz rule in the sense
\begin{align}
\nabla fs = df|_H \otimes s + f\nabla s.
\end{align}
Let $L \le T^*M$ be the annihilator of the contact distribution. Then there is a canonical injective vector bundle homomorphism (the Levi map) $L \hookrightarrow \Lambda^2T^*M$ given by $\alpha \mapsto d\alpha|_H$ and the image consists of non-degenerate skew-forms. In particular there is a rank 3 subbundle of $\Lambda^2H^*$, call it $\Lambda^2_\perp H^*$ consisting of forms trace free with respect to the image of the Levi map.
There is an equivalence class of full connections $[\tilde{\nabla}]$ on E which extend $\nabla$.
The (partial) curvature of a partial connection can be defined by choosing such a representative and projecting the curvature $\Lambda^2 \otimes \operatorname{End}(E) \to \Lambda^2_\perp H^* \otimes \operatorname{End}(E)$. Furthermore in https://arxiv.org/abs/0910.5519 it is easily shown that there is a unique representative connection in $[\tilde{\nabla}]$ such that the projection $\Lambda^2 \otimes \operatorname{End}(E) \to L \otimes \operatorname{End}(E)$ vanishes. Accordingly a connection with vanishing partial curvature has a unique lift $\tilde{\nabla}$ such that
\begin{align}
\tilde{\nabla}_X\tilde{\nabla}_Ys – \tilde{\nabla}_Y\tilde{\nabla}_Xs – \tilde{\nabla}_{[X,Y]}s = 0 \ \ \ \forall X,Y \in H.
\end{align}
My instinct tells me the maximal non-integrability of H means that $\tilde{\nabla}$ must be flat, but I can't prove this for myself or find a source.
It is definitely sometimes true. For example one can define the obvious flat partial connection with respect to a trivialisation $\{e_i\}$ of $E$
\begin{align}
\nabla s = d s^{i}|_H \otimes e_i
\end{align}
and this has canonical representative
\begin{align}
\tilde{\nabla} s = d s^{i} \otimes e_i,
\end{align}
which is flat. So there is no general obstruction to this being true. So my question is: Is it possible to show that $\tilde{\nabla}$ must be flat, and if not, are there conditions one can enforce on $\nabla$ for this to be true?
Best Answer
It is true that $\tilde{\nabla}$ must be flat given the assumptions above.
First note that on a contact manifold we have an exact sequence $L \otimes T^*M|_H \to \Lambda^2T^*M \to \Lambda^2T^*M|_H $. The first map is given by $\alpha \otimes \rho \mapsto \alpha \wedge \tilde{\rho}$ where $\tilde{\rho}$ is any lift of $\rho$ and the second map is just the projection.
Accordingly we have an exact sequence $L \otimes T^*M|_H \otimes E \to \Lambda^2T^*M \otimes E \to \Lambda^2T^*M|_H \otimes E$ and writing $\kappa : E \to \Lambda^2T^*M \otimes E$ for the curvature we know that $\kappa(s)$ vanishes under the projection onto $ \Lambda^2T^*M|_H \otimes E$ for $s \in \Gamma(E)$. Accordingly write (over a trivialising neighbourhood) \begin{align} \kappa(s) = \sum_i \alpha \wedge \tilde{\rho}^{i} \otimes e_i \end{align} where $\{e_i\}_{i=1}^n$ is some local trivialisation of $E$. Now apply the exterior covariant derivative $\Lambda^2T^*M \otimes E \to \Lambda^3T^*M \otimes E$ and use the Bianchi identity to get. \begin{align} \kappa(\nabla s) = \sum_i d\alpha \wedge \tilde{\rho}^{i} \otimes e_i - \alpha \wedge d\tilde{\rho}^{i} \otimes e_i + \alpha \wedge \tilde{\rho}^{i} \wedge \nabla e_i \end{align} Projecting $\Lambda^3T^*M \otimes E \to \Lambda^3T^*M|_H \otimes E$, the left hand side vanishes since $\kappa|_H = 0$, and the right hand simplifies since $\alpha|_H = 0$. Lastly \begin{align} d\alpha|_H \wedge \rho^{i} = 0\ \ \forall \ i \end{align} but non-degeneracy of $d\alpha|_H$ in the definition of a contact manifold (equivalent to the maximal non-integrability condition) assures us that $\rho^i=0$ if $2n+1 \ge 5$. Then the component of $\kappa(s)$ in $L \otimes T^*M|_H \otimes E$ vanishes which was all that was left to show $\kappa = 0$.
Interestingly this doesn't work for contact manifolds of dimension $3$. Also this method of proof really does rely on the maximal non-integrability of the contact distribution.