Let $G$ be *simply connected, simple* algebraic group over $\mathbb{C}$.
Let $H\subset G$ be a *self-normalizing* spherical subgroup of $G$,
not necessarily connected or reductive.
Here "self-normalizing" means that $\mathcal{N}\_G(H)=H$.


I am interested in a certain quotient $H^{\mathrm{mt}}$ of $H$.
Namely, let $H^0$ denote the identity component of $H$, and let $H^{\mathrm{u}}$ denote the unipotent radical of $H^0$.
Set $H^{\mathrm{red}}=H^0/H^{\mathrm{u}}$, which is a connected reductive group.
Let $H^{\mathrm{ss}}$ denote the commutator subgroup of $G^{\mathrm{red}}$, which is a connected semisimple group.
Set $H^{\mathrm{tor}}=H^{\mathrm{red}}/H^{\mathrm{ss}}$, which is a torus.
Then $H^{\mathrm{u}}$ is a normal subgroup in $H$, and $H^{\mathrm{ss}}$ is a normal subgroup in $H/H^{\mathrm{u}}$.
We set
$$ H^{\mathrm{mt}}=(H/H^{\mathrm{u}})/H^{\mathrm{ss}}.$$
Then we have a short exact sequence
$$ 1\to H^{\mathrm{tor}}\to H^{\mathrm{mt}}\to\pi\_0(H)\to 1,$$
where $\pi\_0(H)=H/H^0$.



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> **Question 1.** Is it true that the finite group $\pi\_0(H)$ is always abelian?
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> **Question 2.** When $\pi\_0(H)$ is abelian, is it true that it acts on $H^{\mathrm{tor}}$ trivially and that the group $H^{\mathrm{mt}}$ is abelian (hence of multiplicative type)?
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> **Question 3.** What are nontrivial examples of $H$ and $H^{\mathrm{mt}}$ (say, when $\pi\_0(H)$ is not cyclic, or when $\mathrm{dim}(H^{\mathrm{tor}})>1$, but $H$ is not parabolic, etc.)?
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Motivation: I would like to compute the Tate-Shafarevich kernel $Ш^2(k,H)$ (whatever this means) over a number field $k$, and then $H^{\mathrm{u}}$ and $H^{\mathrm{ss}}$ play no role, so I consider the quotient $H^{\mathrm{mt}}$ of $H$.


Any examples and/or references will be appreciated.



A quotient group of a self-normalizing spherical subgroup
Q1: Yes. More generally, for any spherical subgroup $J\subseteq G$, the quotient $N\_G(J)/J$ is multiplicative (see the Brion-Pauer paper). Apply this to $J=H^0$. Then $\pi\_0(H)\subseteq N\_G(H^0)/H^0$ shows that $\pi\_0(H)$ is multiplicative.


Q2: No. Take $H=N\_G(T)$ where $G=SL(2)$ and $T$ is a maximal torus. Then $H^{\rm mt}=H$ is non-abelian. Losev has shown that the image of $\pi\_0(T)$ in $Aut(H^{\rm tor})$ is an elementary abelian $2$-group.


Q3: Depends on your notion of triviality. There are zillions of solvable spherical subgroups containing a maximal torus of $G$, e.g. $H=T\ltimes (U,U)$. Then $H^{\rm mt}=T$. As for non-cyclic $\pi\_0$: Take a direct product of $N\_G(T)\subseteq SL(2)$. Then the group is not simple. But you can take a parabolic induction in a larger simple group.


