In symplectic geometry, the Grassmannian of isotropic planes for a symplectic vector space is a well known and well studied object; for example, one can realize it as a homogeneous space with a known stabilizer subgroup of the symplectic group, and one can also realize a Schubert cell decomposition.

However, one can relax the symplectic condition in two ways: by relaxing its degeneracy condition, and by allowing it to take vector values (i.e., allowing for general tensors). A good example generalizing both of these situations can be found from the study of distributions: for any given distribution $D$ on a tangent bundle $TM$ of some smooth manifold $M$, there is a natural (O'Neill?) torsion tensor $T \in \Gamma(\wedge^2D^* \otimes TM/D)%$ defined by:

\begin{equation} T(X,Y) = [X,Y] \text{ mod }D \end{equation}

Clearly, $T$ is a skew symmetric bilinear map. However, its degeneracy could vary depending on the distribution. On one hand, $T = 0$ identically gives that $D$ is integrable. On the other, it being nondegenerate gives that $D$ is maximally non-integrable and is a contact structure. One can still make sense of planes isotropic with respect to $T$, as the planes on which $T$ vanish identically.

Is there anything known about the isotropic Grassmannian of planes with respect to the above tensor (e.g., is it homogeneous with respect to a nice known Lie group? Can one describe its Schubert cell decomposition?)? With respect to more general settings/tensors as above?

I imagine in the case of the above particular tensor, one treats each component of the vector as a (possibly degenerate) symplectic form but I have not seen anything about it in the literature, so I'm hesitant to declare everything should be *exactly* the same as for the standard Grassmannian of isotropic planes with respect to an ordinary symplectic form.