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?



Local splitting of the tangent bundle with interesting properties
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.


