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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.

gmvh
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1 Answers1

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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.

Robert Bryant
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  • Related: https://mathoverflow.net/questions/91583/the-rank-of-a-symmetric-space#comment235668_91602 – Vít Tuček Nov 18 '20 at 16:09
  • Thanks for the intro and pointers! I'll see if this leads to the dimensions of the orbits. – Norman Goldstein Nov 18 '20 at 16:13
  • @NormanGoldstein: It does. There are 7 stabilizer types, one for each of the 3 vertices of the triangle, one for each of the interiors of the three sides, and one for the interior points of the triangle. – Robert Bryant Nov 18 '20 at 17:02