5
$\begingroup$

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.

$\endgroup$
2
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.