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Let $G$ be a semisimple group over $\mathbb C$, and $X=G/H$ be a spherical homogeneous space of $G$. Let $T\subset B\subset G$ be a maximal torus and a Borel subgroup. Let $S=S(G,T,B)$ denote the corresponding set of simple roots.

Let ${\mathcal{P}}(S)$ denote the set of subsets of $S$. Let $M$ denote the weight lattice of $X$, and set $N:={\rm Hom}(M,\mathbb Z)$. Let $\mathcal D$ denote the set of colors of $X$.

We have maps $\rho\colon \mathcal D\to N$ and ${\varsigma}\colon\mathcal D\to{\mathcal{P}}(S)$. Here ${\varsigma}(D)$ for $D\in\mathcal D$ is the set of simple roots $\alpha\in S$ such that the corresponding minimal parabolic subgroup $P_\alpha\supset B$ moves the color $D$. Thus we obtain a map $$ {\varsigma}\times\rho\colon\ \mathcal D\ \longrightarrow\ {\mathcal{P}}(S)\times N.$$ This map need not be injective, but by Proposition 3.2.3 of Losev's paper "Uniqueness property for spherical homogeneous spaces" each of its fibers has $\le 2$ elements.

Now consider the group ${{\rm Aut}}_G(X)=\mathcal N_G(H)/H$, this group acts on $\mathcal D$. One can easily see that ${{\rm Aut}}_G(X)$ acts on the fibers of ${\varsigma}\times\rho$.

Question 1. Is it true that ${{\rm Aut}}_G(X)$ acts transitively on each fiber of ${\varsigma}\times\rho$ ?

Question 2. In particular, is it true that if $\mathcal N_G(H)=H$, then then the map ${\varsigma}\times\rho$ is injective?

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    $\begingroup$ What is a color of a spherical homogeneous space? $\endgroup$
    – LSpice
    Commented Jun 9, 2017 at 12:06
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    $\begingroup$ @LSpice: A homogeneous space $X=G/H$ is called spherical if $B$ has an open orbis in $X$. A color is a $B$-orbit of codimension 1 in $X$. The set of colors $\mathcal D$ is finite. $\endgroup$ Commented Jun 9, 2017 at 12:11

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The answer to both questions is "yes" as was communicated to me by Losev.

The main reason is that already $\varsigma$ is almost injective. More precisely, the sets $\varsigma(D)$ are always non-empty and they are pairwise disjoint except for a very well controlled situation.

Assume $I:=\varsigma(D_1)\cap\varsigma(D_2)\ne\emptyset$. Then $D_1$ and $D_2$ are moved by the same simple root $\alpha$. Luna has shown that this can only happen if $\alpha$ is also a spherical root. He showed further that then all elements of $\rho(D_i)$ are simple roots which are spherical. The Luna axioms easily imply $I=\{\alpha\}$.

So the only way for $\varsigma(D_1)=\varsigma(D_2)$ is when $\varsigma(D_1)=\varsigma(D_2)=\{\alpha\}$. At this point one has to look at $\rho$. If $\rho(D_1)=\rho(D_2)$ then both are equal to $\frac12\alpha^\vee$ (since their sum is $\alpha^\vee$). This implies that it is possible to replace $\alpha$ by $2\alpha$ in the homogeneous spherical datum for $G/H$. This modified datum corresponds to a subgroup $\tilde H$ in which $H$ has index two. So $G/H\to G/\tilde H$ is obtained an automorphism $\phi$ of order $2$. Since $D_1$ and $D_2$ are mapped to the same color in $G/\tilde H$ they are swapped by $\phi$.

As an extra bonus one gets that $\phi$ interchanges only $D_1$ and $D_2$ and leaves all other colors fixed.

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  • $\begingroup$ Thank you, Friedrich! When I ask a question on spherical varieties, I hope that you will answer... $\endgroup$ Commented Jun 9, 2017 at 14:18
  • $\begingroup$ For a detailed version of Friedrich's answer see the appendix by Giuliano Gagliardi to this preprint. $\endgroup$ Commented Oct 9, 2017 at 6:29

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