The group of $G$-automorphisms of a spherical variety from the spherical datum?

Question

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.

Question. How can one compute ${\rm Aut}_G(X)$ from the spherical datum of $X$?

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 $X$.

Answer 1

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.