complement of Lagrangian Grassmannian The Lagrangian Grassmannian $LG(2,4)$ is a 3 dimensional sub-manifold of the original 4 dimensional Grassmannian manifold $Gr(2,4)$. Is there a nice description of the complement $Gr(2,4) \setminus LG(2,4)$ ? 
 A: The Grassmannian $Gr(2,4)$ is a smooth quadric in $\mathbb{P}^5$, and $LGr(2,4)$ is its smooth hyperplane section. Therefore the complement is a smooth affine quadric. Over $\mathbb{C}$ it can be described in the affine space $\mathbb{A}^5$ as the hypersurface
$$
x_1^2 + x_2^2 + x_3^2 + x_4^2 + x_5^2 = 1.
$$
EDIT. There is also a similar construction for $Gr(3,6) \setminus LGr(3,6)$. If $U \subset V := \mathbb{C}^6$ is a 3-subspace, which is not isotropic, then the restriction of the symplectic form to $U$ has rank 2, hence has a 1-dimensional kernel space. This gives a regular map
$$
Gr(3,6) \setminus LGr(3,6) \to \mathbb{P}(V).
$$
Its fiber over point $v \in V$ is easily seen to be $Gr(2,\bar{V}) \setminus LGr(2,\bar{V})$, where $\bar{V} := v^\perp/v$ with its induced symplectic form. Thus, the map is a fibration with fibers being 4-dimensional affine quadrics. Note, however, that the ambient affine space fibration for it is non-trivial!
A: In this low dimensional case everything are explicit at least for the oriented case: The oriented Grassmanian $G(2, 4)$ is isomorphic to $\mathbb S^2 \times \mathbb S^2$. This is well-known and is explained quite clearly in the answer here. In particular, if we choose the orthornormal basis $\{e_1, e_2, e_3, e_4\}$ so that the symplectic form is 
$$ \omega = e^1 \wedge e^2 + e^3 \wedge e^4, $$
then the Lagrangian plane $L$ are precisely those without the $e_1 \wedge e_2 + e_3\wedge e_4-$ component. Thus 
$$ Gr_+(2, 4) \cong \mathbb S^2 \times \mathbb S^2, \ \ \ \ LG_+(2, 4) = \mathbb S^1 \times \mathbb S^2.$$
So $Gr_+(2, 4)\setminus LG_+(2, 4)$ is diffeomorphic to two disjoint copies of $D \times \mathbb S^2$. 
A: The decomposition of $Gr(2,4)$ is determined by the restriction of the the skew symmetric bilinear form. The Lagrange Grassmannian consists of those planes for which this restriction is zero, the complement of those planes for which the restriction is non-degenerate. (There is no other possibility since a skew symmetric form has to have even rank.)
