TITLE: Local splitting of the tangent bundle with interesting properties
QUESTION [5 upvotes]: Let $(M,g)$ be a Riemannian manifold and let $U\subset M$ be an open subset. Suppose that the tangent bundle over $U$ splits into two orthogonal distributions $TU=\mathcal{E}\oplus \mathcal{F}$.
Is it possible that the two $C^{\infty}(U)$-bilinear maps 
\begin{align*}I:\mathcal{E}\times\mathcal{E}&\to \mathcal{F}\\
&(X,Y)\mapsto pr_{\mathcal{F}}(\nabla_X Y)
\end{align*}
and
\begin{align*}I:\mathcal{F}\times\mathcal{F}&\to \mathcal{E}\\
&(X,Y)\mapsto pr_{\mathcal{E}}(\nabla_X Y)
\end{align*}
are both antisymmetric in $X$ and $Y$ without vanishing?
If $\mathcal{E}$ and $\mathcal{F}$ both were integrable, both maps would be symmetric. So is this in some sense the most non-integrable way, distributions can be?

REPLY [5 votes]: Yes, this can happen.  A little experimentation with the structure equations shows that there is a $3$-parameter family of homogeneous examples in dimension $4$:  Let $c_1,\ldots,c_4$ be nonzero constants satisfying $c_1c_2=c_3c_4$, and consider the simply-connected $4$-dimensional Lie group $G$ that has a basis of left-invariant $1$-forms $\omega_1,\ldots,\omega_4$ that satisfy the structure equations
\begin{aligned}
d\omega_1 &= 2c_1\ \omega_2\wedge\omega_3 + 2c_3\ \omega_4\wedge\omega_3\ ,\\\\
d\omega_2 &= 2c_1\ \omega_3\wedge\omega_1 \ ,\\\\
d\omega_3 &= 2c_2\ \omega_4\wedge\omega_1 + 2c_4\ \omega_2\wedge\omega_1\ ,\\\\
d\omega_4 &= 2c_2\ \omega_1\wedge\omega_3 \ .
\end{aligned}
Now endow $G$ with the Riemannian metric $g$ for which the $\omega_i$ define an orthonormal coframing, let $e_1,\ldots,e_4$ be the dual ($g$-orthonormal) vector fields, and let $\mathcal{E}$ be the $2$-plane field spanned by $e_1$ and $e_2$ while $\mathcal{F}$ is the $2$-plane field defined by $e_3$ and $e_4$.  
One easily checks that this is an example of the desired type:  If $\nabla$ is the Levi-Civita connection of this metric, then
$$
\nabla_{e_1}e_1\equiv0\ ,\quad\nabla_{e_1}e_2\equiv c_4e_3, \quad
\nabla_{e_2}e_1\equiv-c_4e_3\ ,\quad\nabla_{e_2}e_2\equiv 0 \mod \mathcal{E}
$$
and
$$
\nabla_{e_3}e_3\equiv0\ ,\quad\nabla_{e_3}e_4\equiv c_3e_1, \quad
\nabla_{e_4}e_3\equiv-c_3e_1\ ,\quad\nabla_{e_4}e_4\equiv 0 \mod \mathcal{F},
$$
as desired.
There are non-homgeneous examples in dimension $4$ as well.  A little more work with the structure equations shows that there exists a $4$-parameter family of examples of cohomogeneity $2$.  (I don't know how many of these are complete.)  If there is interest, I can give the structure equations of these examples as well.