Unitary orbits on the Grassmann manifold of 2-planes in complex affine space The unitary group, $U(n)$, acts transitively on the Grassmann manifold $X = Gr(2, C^n)$.  The isotropy group is $H = U(2)\times U(n-2)$, i.e. the group elements leaving some $x$ fixed.  What are the dimensions of the orbits of $H$ in $X$?  The brute force calculation gets messy for some of the orbits, so I am wondering if these results are published somewhere.  The generic orbit is codimension 2, and there are the special orbits of $\{x\}$ and $x_\perp = Gr(2,C^{n-2})$, but there are other orbits, too.
 A: The Grassmannian $$\mathrm{Gr}(2,\mathbb{C}^n) = \frac{\mathrm{SU}(n)}{\mathrm{S}\bigl(\mathrm{U}(2){\times}\mathrm{U}(n{-}2)\bigr)}$$ is the compact Hermitian symmetric space of type AIII of rank $r = \min(2,n{-}2)$.  When $n=3$, this is just $\mathbb{CP}^2$, of rank $1$, so assume $n>3$.
There is a standard method to classify the orbits of $K$ acting on a compact symmetric space $G/K$ of rank $r$, and you are asking about the special case $K=\mathrm{S}\bigl(\mathrm{U}(2){\times}\mathrm{U}(n{-}2)\bigr)$ and $G=\mathrm{SU}(n)$ and $r=2$.  The basic result is that the space of orbits naturally forms a convex polytope in an 'abelian' subspace $\frak{a}\subset\frak{m}$ where $\frak{g} = \frak{k}\oplus\frak{m}$, and that abelian subspace has (real) dimension $r$.
In your case, $r=2$, and I think I remember that it turns out that the space of orbits is a triangle in $\frak{a}\simeq\mathbb{R}^2$.  One vertex is the fixed point $\mathbb{C}^2 = eK\in G/K$, one vertex is the set of $2$-planes in the $\mathbb{C}^{n-2}$ perpendicular to $\mathbb{C}^2$, and one vertex is the set of $2$-planes that meet each of $\mathbb{C}^2$ and $\mathbb{C}^{n-2}$ in a complex line.
A good source for the general theory is O. Loos' $2$-volume work Symmetric Spaces.  I would expect this example to be explicitly computed there.
Addendum:  So, after thinking about it, I realized that the answer is this:  Let $e_1,\ldots, e_n$ be a unitary basis of $\mathbb{C}^n$ with $e_1,e_2$ a basis of the fixed $\mathbb{C}^2$.  Then each $2$-plane is in the $K$-orbit of a plane spanned by
$$
\cos\theta_1\,e_1+\sin\theta_1\,e_3\quad\text{and}\quad
\cos\theta_2\,e_2+\sin\theta_2\,e_4
$$
where $0\le \theta_1\le\theta_2\le\tfrac12\pi$, and the values $(\theta_1,\theta_2)$ in this triangle distinguish the orbits.
