Take the generalized flag variety $SO(2n,\mathbb{C})/B$, considered as the moduli of isotropic flags (according to the form $\langle e\_i, e\_{2n+1-j}\rangle=\delta\_{ij}$)
$$F\_1\subset F\_2\subset\cdots F\_n\subset\mathbb{C}^{2n}$$
with $\dim(F\_i)=i$ and $\dim(F\_n\cap\mathrm{Span}(e\_1,\ldots,e\_n))$ even.


Let $P$ be the maximal parabolic subgroup of $SO(2n)$ omitting one of the two nodes at the forked end. Then $G/P$ is the moduli of isotropic subspaces of dimension $n$ in $\mathbb{C}^{2n}$.


For one of those end nodes, the description of the projection $G/B\rightarrow G/P$ is obvious; one just sends the point corresponding to the full isotropic flag to the point corresponding to the maximal subspace $F\_n$.


In explicit linear algebra terms, what is the projection $G/B\rightarrow G/P$ for the maximal parabolic corresponding to the other end node on the forked end? In other words, what is the other way to get a maximal isotropic subspace out of a full isotropic flag?



The weird projection from SO(2n)/B to maximal isotropic grassmannian
The other projection goes to the other Lagrangian subspace containing $F\_{n-1}$. If you look at $F\_{n-1}^\perp/F\_{n-1}$, this is a 2-dimensional space with symmetric non-degenerate bilinear form. Thus, it has exactly 2 isotropic lines (there are coordinates where the matrix of the form is $\begin{bmatrix}0 & 1\\ 1&0\end{bmatrix}$, in which case the coordinate lines are the only isotropic ones), whose preimages are the spaces $F\_n$ and $F\_n'$ canonically. 


