Background: a symmetric variety is a homogeneous space $G/H$ associated to an involution $\theta$ of a semisimple algebraic group $G$ and $\{g | \theta(g) = g\} = G^\theta \subset H \subset N_G(G^\theta) = \{h| hG^\theta h^{-1} = G^\theta \}$.

A spherical variety is a homogeneous space $G/H$ which contains an open orbit under the action of a Borel subgroup $B$. It is known that symmetric varieties give examples of spherical varieties.

Let $G = SL_n(\mathbb{C})$ and $H = S(GL_k \times GL_{n-k}) = \{(g,g') | \det g \det g' = 1 \}$. It is known that $G/H$ is spherical. $G/H$ is a non compact affine variety. In fact if I write $Gr_{k,n}(\mathbb{C}) = G/P$ then there is a Levi decomposition $P = H U$ so that $G/H$ is a $U$-fiber bundle over $Gr_{k,n}(\mathbb{C})$.

QUESTION Is $G/H$ symmetric? If so, what is the associated involution?

The answer to the first question seems to be yes (at least for small values of $k$) as this particular homogeneous spaces makes numerous appearances in http://arxiv.org/pdf/1012.4171.pdf (see for example figures 3,4,5 starting on page 15)

In fact the cited paper even gives a name to the involution (AIII) referencing a list of the possible involutions of a simple algebraic group. My confusion is that this involution seems to be a composition of usual transposition (swap across the diagonal) with a swap along the anti-diagonal and this does not seem to exhibit $G/H$ as a symmetric space.

Further, I thought the rank of a symmetric variety was supposed to agree with its rank as a spherical variety. In my particular example, the rank of $G/H$ seems to be 1 as a spherical variety while the paper seems to say that the rank of $G/H$ as a symmetric variety is in general greater than 1. I'm likely missing a basic point and if it can be pointed out to me I would greatly appreciate it.

UPDATE: José Figueroa-O'Farrill answered the question over the real numbers and pointed out that the rank is $\min\{k,n-k\}$ and the involution can be taken to be conjugating by the matrix $\left(\begin{array}{cc} -I_k & \\ & I_{n-k} \end{array}\right)$. The negative of this also works as Lev Soukhanov points out below.

The same involution seems to work over $\mathbb{C}$ but I'm not certain the statement about ranks still holds.


1 Answer 1


It seems that you just take the involution which is the conjugation with the diagonal matrix with k-dimensional subspace with eigenvalue 1 and n-k dimensional subspace with eigenvalue -1.

Dunno about ranks at all.


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