# The group of automorphisms a pair $(G,X)$ where $X$ is a spherical homogeneous space of $G$

I wish to "compute" $${{\rm Aut}}(G,X)$$ for a spherical homogeneous space $$X$$ of $$G$$ in terms of the spherical datum of $$X$$.

First let $$G$$ be any algebraic group over $$\mathbb C$$, and let $$X$$ be any left $$G$$-variety. We define $${{\rm Aut}}(G,X):=\{(\sigma,\mu)\in{{\rm Aut}}(G)\times{{\rm Aut}}(X)\ |\ \mu(g\cdot x)=\sigma(g)\cdot\mu(x)\text{ for } g\in G,\ x\in X\}.$$ We have a canonical homomorphism $$\pi\colon{{\rm Aut}}(G,X)\to {{\rm Aut}}(G),\quad (\sigma,\mu)\mapsto \sigma.$$ Write $${{\rm Aut}}_G(X):=\ker\,\pi=\{\mu\in{{\rm Aut}}(X)\ |\ \mu(g\cdot x)=g\cdot\mu(x) \text{ for all } g\in G,\ x\in X\},$$ then we have an exact sequence $$1\to{{\rm Aut}}_G(X)\to{{\rm Aut}}(G,X)\to {{\rm Aut}}(G).$$

Now assume that $$G$$ is a connected semisimple group and that $$X$$ is a spherical homogeneous space of $$G$$, $$X=G/H$$. Let $$N(H)$$ denote the normalizer of $$H$$ in $$G$$. Then $${{\rm Aut}}_G(X)=N(H)/H$$ (because $$X=G/H$$) and $$N(H)/H$$ is a group of multiplicative type (because $$X$$ is spherical). This group $$N(H)/H$$ can be computed in terms of the spherical datum of $$X$$, see Friedrich Knop's answer.

Set $${{\rm Aut}}_X(G)={{\rm im}}\, \pi\subset {{\rm Aut}}(G)$$. We compute $${{\rm Aut}}_X(G)$$. Consider the homomorphism $$i\colon G\to {{\rm Aut}}(G,X)$$ sending $$a\in G$$ to the "inner automorphism" $$i_a\in {{\rm Aut}}(G,X)$$, where $$i_a(g,x)=(aga^{-1}, a\cdot x).$$ Clearly $$\pi(i_a)={\rm inn}(a)\in {{\rm Aut}}(G)$$, hence $${{\rm Aut}}_X(G)\supset {{\rm Inn}}(G)$$. Write $${{\rm Out}}(G)={{\rm Aut}}(G)/{{\rm Inn}}(G),\quad {{\rm Out}}_X(G)={{\rm Aut}}_X(G)/{{\rm Inn}}(G).$$ Using Losev's uniqueness theorem and ideas of Akhiezer and Cupit-Foutou, one can describe the subgroup $${{\rm Out}}_X(G)\subset {{\rm Out}}(G)$$ as follows: the group $${{\rm Out}}(G)$$ acts on the Dynkin diagram and on the based root datum of the semisimple group $$G$$, in particular, on the character group $${{\mathcal{X}}}(B)$$ of a Borel subgroup $$B$$ of $$G$$; then $${{\rm Out}}_X(G)$$ is the subgroup of $${{\rm Out}}(G)$$ consisting of the elements that preserve the spherical datum of $$X$$ when acting on $${{\mathcal{X}}}(B)$$.

A pinning $$p$$ of $$G$$ defined a splitting $$s_p\colon {{\rm Out}}(G)\to {{\rm Aut}}(G)$$, and thus a splitting $${{\rm Out}}_X(G)\to {{\rm Aut}}_X(G)$$. Thus we know $${{\rm Aut}}_X(G)$$, it is a semidirect product $${{\rm Inn}}(G)\rtimes{{\rm Out}}_X(G)$$. We know also $${{\rm Aut}}_G(X)=N(H)/H$$.

I wish to understand the extension $$1\to {{\rm Aut}}_G(X)\to {{\rm Aut}}(G,X)\to {{\rm Aut}}_X(G)\to 1.$$ Using the splitting $$s_p\colon {{\rm Out}}_X(G)\to {{\rm Aut}}_X(G)$$, we obtain an extension $$1\to {{\rm Aut}}_G(X)\to\mathfrak{G}\to{{\rm Out}}_X(G)\to 1.$$ This last extension defines a cohomology class $$\eta_X\in H^2({{\rm Out}}_X(G), {{\rm Aut}}_G(X)).$$

Question 1. Are there examples for which this extension does not split, that is, $$\eta_X\neq 0$$ ?

Question 2. How can one compute $$\eta_X$$ in terms of the spherical datum of $$X$$ ?

EDIT (based on a comment of YCor). Set $${\rm Inn}(G,H)={\rm im}\,i\subset{\rm Aut}(G,X)$$, which is a normal subgroup of $${\rm Aut}(G,X)$$. We set $${\rm Out}(G,X)={\rm Aut}(G,X)/{\rm Inn}(G,X)$$. The epimorphism $$\pi\colon {\rm Aut}(G,X)\to {\rm Aut}_X(G)$$ induces an epimorphism $$\bar\pi\colon {\rm Out}(G,X)\to{\rm Out}_X(G)$$ whose kernel seems to be $$N(H)/(Z(G)H)$$. The following question seems to be easier than Question 2.

Question 3. How can one describe the short exact sequence $$1\to N(H)/(Z(G)H)\to {\rm Out}(G,X)\to {\rm Out}_X(G)\to 1$$ in terms of the spherical datum of $$X$$ ?

• The kernel of $\pi$ is rather those $\mu$ such that $\mu(g.x)=g\cdot\mu(x)$ for all $g,x$, isn't it? (The unit element of $Aut(G)$ is not the constant 1, it is the identity map.) It seems to match the sequel of your post. – YCor Apr 15 '17 at 11:15
• You have defined a subgroup $\mathrm{Inn}(G,X)$ (the set of all $i_a$ when $a$ ranges over $G$); is by definition $\mathfrak{S}$ the quotient of $\mathrm{Out}(G,X)=\mathrm{Aut}(G,X)/\mathrm{Inn}(G,X)$? If so I don't see why you need any splitting to define the extension. Also ther kernel of $G\to\mathrm{Inn}(G,X)$ consists in the intersection $Z_X(G)$ of the kernel of the $G$-action on $X$ with the center of $G$. This yields an extension $1\to\mathrm{Aut}_G(X)/Z_X(G)\to\mathrm{Out}(G,X)\to\mathrm{Out}_X(G)\to 1$. – YCor Apr 15 '17 at 11:42
• @YCor: I agree with the first comment; I have edited my post. – Mikhail Borovoi Apr 15 '17 at 14:38
• @YCor: Concerning your second comment, I do not agree with the assertion that $Z_X(G)\subset {\rm Aut}_G(X)$ (this assertion is implicit in the formula ${\rm Aut}_G(X)/Z_X(G)\$). See my edit. – Mikhail Borovoi Apr 15 '17 at 17:07
• You're right, I wasn't careful. – YCor Apr 15 '17 at 19:08