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Let $G$ be a complex connected reductive Lie group, $\theta$ an involution, and $K = G^\theta$ the fixed-point subgroup. Then $K$ has finitely many orbits on $G/B$, one of which is open and (quite often) several of which are closed.

We can pick a $\theta$-stable Borel $B$ and maximal torus $T$, and using Springer's map $$ \varphi : K\backslash G \to G, \qquad Kg \mapsto \theta(g^{-1}) g $$ we get a map of the same name $$ \varphi: K\backslash G/B \to B\backslash G/B \qquad \cong W := N(T)/T. $$ This map isn't usually injective. For example, if $G = SL(2)$ and $K = SO(2)$, then the LHS has three elements (North and South poles and the rest) while the RHS has only two.

If we fix a closed orbit $v \in K \backslash G/B$, and restrict $\varphi$ to the order ideal {$u \in K \backslash G/B : u \geq v $}, does it become injective?

Motivation: the poset $K \backslash G/B$ is somewhat mysterious, whereas the poset of "twisted involutions" {$w \in W : (\theta\cdot w)^2 = 1$} is at least less so. I need to sum over chains in this order ideal and would rather do it inside $W$.

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up vote 2 down vote accepted

Malheureusement, this is not true, not even for the weak order. This can be seen for example when $G = GL(4)$ and $K = GL(2) \times GL(2)$. Then $K \backslash G / B$ is parameterized by involutions with signs attached to fixed points and the map $\varphi$ simply forgets the markings on the fixed points.

For example, the closed orbit associated to $(1^+)(2^-)(3^-)(4^+)$ lies below both $(14)(2^-)(3^+)$ and $(14)(2^+)(3^-)$. Figures with weak order for a plethora of examples of symmetric subgroups appear in a preprint by Ben Wyser.

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Helas! $\ \ \ \ $ –  Allen Knutson Mar 22 '12 at 1:17
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