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 $\pi_0(H)=H/H^0$.

Question 1. Is it true that the finite group $\pi_0(H)$ is always abelian?

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)?

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.)?

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.


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.

  • $\begingroup$ Thank you, Friedrich! Q2. Which paper of Losev do you mean? Q3. What do you mean by parabolic induction here? $\endgroup$ – Mikhail Borovoi Jul 16 '16 at 10:43
  • $\begingroup$ Losev's paper is called "Uniqueness properties..." (arxiv.org/abs/math/0703543). Parabolic induction means the following: Let $P=LN$ be a parabolic in $G$ and $J$ a spherical subgroup of $L$. Then the parabolically induced spherical subgroup is $H=JN$. $\endgroup$ – Friedrich Knop Jul 16 '16 at 12:15

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