# Grassmannians of planes isotropic with respect to general tensors

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.

What you are asking about is very classical in the theory of exterior differential systems. The subspaces of $$D$$ that you are calling `isotropic' are what Élie Cartan called the integral elements of the differential ideal $$\mathcal{I}$$ generated by the sections of $$D^\perp\subset T^*M$$ (the annihilator subbundle of $$D$$). The geometry and topology of these spaces of integral elements can be quite subtle, and a significant part of the basic theory of exterior differential systems is devoted to developing ways to describe these subspaces of the Grassmann bundles, particularly deriving necessary and sufficient conditions that a given integral element $$E\subset D_x\subset T_xM$$ should be a tangent space of an integral manifold, i.e., a submanifold $$N\subset M$$ all of whose tangent spaces are sub-planes of the plane field $$D$$. This leads to the Cartan-Kähler theory and beyond.
• @TobiasShin: I believe that the answer to the question over $\mathbb{C}$ is 'no'. The first example I can think of is the generic case when $M$ has dimension $10$ and $D$ has dimension $5$. In that case, generically, there will be no $3$-dimensional integral elements, and, at each point, the space of $2$-dimensional integral elements will have dimension $1$. In fact, it will be a smooth curve of degree $5$ lying in a projective space of dimension $4$ and it will have genus $1$, so it will be an elliptic curve, which is topologically a (real) $2$-torus. Hence, it will not be simply connected. – Robert Bryant Nov 21 '18 at 21:28
• how do you reconcile the above with the following theorem (in Gunning and Rossi's Analytic Functions of Several Complex Variables): if $X$ is an irreducible complex variety whose universal cover $\tilde{X}$ is an irreducible analytic space, then for any closed analytic subspace $Z$ inside $X$, we have $\pi_1(X-Z)$ surjects onto $\pi_1(Z)$. It seems like in the above situation, your $X$ is the Grassmannian of 2-planes in 5-space, and your $Z$ is of codimension 5. But removing a submanifold of real codimension at least 3 should not change the fundamental group. – Tobias Shin Nov 25 '18 at 21:34
• Ack sorry I think there might be a typo... they might mean surjective onto $\pi_1(X)$ – Tobias Shin Nov 25 '18 at 23:47
• @TobiasShin: I'm not familiar with this result, but you must be misunderstanding it. If what you claim were true, then an elliptic curve could not appear embedded in a $6$-dimensional projective space (which is clearly simply connected, compact, and irreducible), and that is obviously false, since an elliptic curve can appear embedded in a projective space of any dimension at least 2. Until you explain why this result doesn't apply in this simpler and obvious case of a non-simply connected curve in a simply connected compact variety of high dimension, I don't really have anything to do. – Robert Bryant Nov 26 '18 at 0:47