One can even define a *holomorphic* (i.e. integrable almost complex) structure on this Grassmanian. To give the definition it is easier to consider instead $Gr_2^{-}(E^{n+2}_{2})$ which is obviously the same object.

**Definition.** We will identify the Grassmanian of two-planes with a part of the quadric in $\mathbb CP^{n+1}$. Namely, let $\mathbb C^{n+2}$ be the complexification of $\mathbb R^{n+2}$: $\mathbb C^{n+2}=\mathbb R^{n+2}\oplus i \mathbb R^{n+2}$. Let $x_1,...,x_{n+2}$ be the coordinates in $\mathbb R^{n+2}$, so that the quadratic form corresponding to the metric is $x_1^2+...+x_{n}^2-x_{n+1}^2-x_{n+2}^2$. Let $z_1,...,z_n$ be the complexified coordinates, and let $Q(z)=z_1^2+...+z_{n}^2-z_{n+1}^2-z_{n+2}^2$ be the complexified quadratic form.

Now, consider the quadric cone $Q(z)=0$, in $\mathbb C^{n+2}$ ($z=x+iy$, $x,y\in \mathbb R^{n+2}$).  
Notice that $Gr_2^{-}(E^{n+2}_{2})$ is naturally is an  open subset of the projectivisation of the cone $Q(z)=0$. Indeed, for any pair of orthogonal unite vectors  $x\perp y$ in $\mathbb R^2\subset \mathbb R^{n+2}$ we can associate a complex vector $z=x+iy$, $z\in \mathbb C^{n+2}$, satisfying $Q(z)=0$. The line $\lambda z\subset \mathbb C^{n+2}$  is independent on the choice of orthogonal unite vectors $x,y$ in $\mathbb R^2$.

It is easy to see that the complex structure constructed this way is invariant with respect to the action of isometries of $\mathbb R^{n+2}$ on $Gr_2^{-}(E^{n+2}_{2})$.