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 $W$.



Chains in $K\backslash G/B$ lying over a closed $K$-orbit
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](http://arxiv.org/abs/1201.4397).


