Let $X$ be a Spherical variety for a reductive group $G$ with a Borel subgroup $B$. A boundary divisor of $X$ is a $G$-invariant divisor and a color of $X$ is a $B$-invariant divisor which is no $G$-invariant. We can also assume that $X$ smooth and the divisor given by the union of the boundary divisors is simple normal crossing. 


Does there exist a formula (in terms of invariants of $G$, $B$ and $X$) for the number of boundary divisors and colors of $X$?



Number of boundary divisors and colors of a Spherical variety
There are relations coming from computing Picard groups in different ways. For simplicity let $G$ be semisimple. Let $x\in X$ be in the open $B$-orbit. Then $Bx=B/B\_x$ and $Gx=G/G\_x$ are open in $X$.


We compute $Piс(Gx)$ in $3$ different ways:


a) Since $G$ is semisimple, we have $Pic(Gx)=Pic^G(Gx)=\Xi(G\_x)$. Here $\Xi$ denotes the character group.


b) There is an exact sequence
$$
1\to k^\times=\mathcal O(Gx)^\times\to\mathcal(Bx)^\times\to\mathbb Z^c\to Pic(Gx)\to Pic(Bx)=0
$$
Here $c$ is the number of colors. The rank of $\mathcal O(Gx)^\times/k^\times$ is by definition the rank ${\rm rk}\,X$ of $X$. It equals ${\rm rk}\,\Xi(B)-{\rm rk}\,\Xi(B\_x)$.


c) Let $b$ be the number of boundary divisors. Then there is a short exact sequence
$$
0\to\mathbb Z^b\to Pic(X)\to Pic(Gx)\to1.
$$


Now a), b), c) imply
$$
b={\rm rk}\,Pic(X)-{\rm rk}\,\Xi(G\_x)
$$
$$
c={\rm rk}\, X+{\rm rk}\,\Xi(G\_x)=
{\rm rk}\,\Xi(B)-{\rm rk}\,\Xi(B\_x)+{\rm rk}\,\Xi(G\_x).
$$


