Let $M$ be a (smooth) manifold. Given tangent vectors $X_q,Y_q \in T_qM$ ($q \in M$), there exist (locally about $q$) vector fields $X,Y \in \Gamma(TM)$ extending $X_q$, $Y_q$ (i.e., $X(q) = X_q$ and $Y(q) = Y_q$), and such that $[X,Y] = 0$. My question is whether this result will extend to the following situation.
Suppose that $\mathcal{D}$ is a nonintegrable subbundle of $M$, and $\mathcal{E}$ is a complement to $\mathcal{D}$ (so that $TM = \mathcal{D} \oplus \mathcal{E}$). Let $\pi : TM \to \mathcal{D}$ be the corresponding projection.
Given $X_q,Y_q \in \mathcal{D}_q$, do there exist (local) $X,Y \in \Gamma(\mathcal{D})$ extending $X_q$ and $Y_q$ and such that $\pi([X,Y]) = 0$?
I can think of two situations when this will be true:
- $M$ is a Lie group $G$, and $\mathcal{D}$, $\mathcal{E}$ are left invariant. (Take $X$ to be the left-invariant vector field corresponding to $TL^{-1}_q\cdot X_q$, and $Y$ to be the right-invariant vector field corresponding to $TL^{-1}_q\cdot Y_q$, where $L_q$ is left translation; then $X,Y \in \Gamma(\mathcal{D})$ extend $X_q,Y_q$, and $[X,Y] = 0$ since left- and right-invariant vector fields commute; hence $\pi([X,Y]) = 0$.)
- $\mathcal{E}$ is integrable. (Let $X,Y \in \Gamma(TM)$ be commuting extensions of $X_q$ and $Y_q$; then $\pi(X),\pi(Y) \in \Gamma(\mathcal{D})$ extend $X_q,Y_q$, and $\pi([\pi(X),\pi(Y)]) = 0$, since the $\pi([\mathcal{D},\mathcal{E}])$ and $\pi([\mathcal{E},\mathcal{E}])$ terms vanish.)
This seems like a fairly straightforward question, but I haven't had any luck in the general case (either in finding a counterexample, or proving it).