# Unitary orbits on the Grassmann manifold of 2-planes in complex affine space

MathOverflow Asked by Norman Goldstein on November 26, 2020

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.

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 $$frak{a}subsetfrak{m}$$ where $$frak{g} = frak{k}oplusfrak{m}$$, 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 $$frak{a}simeqmathbb{R}^2$$. One vertex is the fixed point $$mathbb{C}^2 = eKin G/K$$, 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 $$costheta_1,e_1+sintheta_1,e_3quadtext{and}quad costheta_2,e_2+sintheta_2,e_4$$ where $$0le theta_1letheta_2letfrac12pi$$, and the values $$(theta_1,theta_2)$$ in this triangle distinguish the orbits.

Answered by Robert Bryant on November 26, 2020