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.

One Answer

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

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