Let $G$ be a semisimple group over $\mathbb{C}$ and let $X=G/H$ be a homogeneous spherical variety.
By Losev's theorem, the spherical $G$-variety $X$ is uniquely determined by its spherical datum, see below.
In particular, the group ${\rm Aut}\_G(X)=\mathcal{N}\_G(H)/H$ is uniquely determined by the spherical datum.



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> **Question.** How can one compute ${\rm Aut}\_G(X)$ from the spherical datum of $X$?
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We specify a version of the spherical datum of $X$.
Let $B\subset G$ be a Borel subgroup, and let $T\subset B\subset G$ be a maximal torus.
Let $S=S(G,T,B)$ denote the corresponding system of simple roots.
Let ${\mathcal{P}}(S)$ denote the set of subsets of $S$.


Let ${\mathcal{X}}(B)$ denote the character group of $B$, and let $M\subset {\mathcal{X}}(B)$ denote the weight lattice of $X$.
Set $N={\rm Hom}(M,{\mathbb{Z}})$, $N\_{\mathbb{Q}}={\rm Hom}(M,{\mathbb{Q}})$.
Let $\mathcal{V}\subset N\_{\mathbb{Q}}$ denote the valuation cone of $X$, and let $\Sigma\subset M$ denote the corresponding set of spherical roots.


Let ${\mathcal{D}}$ denote the set of colors of $X$ (the set of $B$-invariant prime divisors of $X$). We have two maps:
$$\rho\colon {\mathcal{D}}\to N,\qquad \varsigma\colon {\mathcal{D}}\to {\mathcal{P}}(S). $$
Here, for $D\in{\mathcal{D}}$,
$$ \varsigma(D)=\{\alpha\in S\ | \ P\_\alpha\cdot D\neq D\}, $$
where $P\_\alpha\supset B$ denotes the parabolic subgroup corresponding to $\alpha\in S$.


I wish to express ${\rm Aut}\_G(X)$ in terms of the spherical datum $(M,\Sigma,\mathcal D,\rho,\varsigma)$ of $X$.



The group of $G$-automorphisms of a spherical variety from the spherical datum?
This question has been answered by Losev in Theorem 2 of his paper "Uniqueness property for spherical homogeneous spaces".


The answer is roughly as follows: The space $\overline X=G/\mathcal N\_G(H)$ is also spherical with weight lattice $\overline M$ and spherical roots $\overline\Sigma$. It is well known that $\mathrm{Aut}^GX=\mathcal N\_G(H)/H$ is of multiplicative type. Its character group is $M/\overline M$, so it suffices to determine the subgroup $\overline M\subseteq M$.


First of all $\overline M=\langle\overline\Sigma\rangle\_{\mathbb Z}$ by a theorem of myself. So it suffices to compute $\overline\Sigma$.


For this, let $\Sigma\_2\subseteq\Sigma$ be the set of $\sigma\in\Sigma$ such that 


* $2\sigma$ is a spherical root of $G$ which is compatible with $S^{(p)}=S\setminus \bigcup\_{D\in\mathcal D}\varsigma(D)$


and


* in case $\sigma$ is also a simple root of $G$, then $\rho$ maps the two colors associated to $\sigma$ to the same element of $N$.


Put $\Sigma\_1=\Sigma\setminus\Sigma\_2$. Then Losev shows that $\overline\Sigma=\Sigma\_1\cup2\Sigma\_2$.


In other words, $\overline\Sigma$ is obtained from $\Sigma$ by doubling all elements which can be doubled without violating Luna's axioms.


