Let $G$ be a complex semisimple Lie group, $B$ be a Borel subgroup of $G$. Denote by $X$ the quotient $G/B$. It is a complex projective variety. Let $L$ be a $G$-equivariant line bundle on $X$ such that the tensor product $L\otimes K_X$ is very ample. Here $K_X$ is the canonical line bundle on $X$. In particularly this means that $L$ is very ample itself. Fix a nonzero section $s\in \Gamma(X,L)$ such that its zero locus $Z(s):=\{x\in X\,|\, s(x)=0\}$ is a smooth divisor of $X$. The group $G$ acts on the projectivization $\mathbb{P}(\Gamma(X,L))$ and we can consider the stabilizer group of the image of the fixed section $\mathrm{Stab}([s])$.

On the other hand, we can consider the automorphism group of its zero locus $\mathrm{Aut}(Z(s))$ (as an algebraic variety). By the adjunction formula the canonical line bundle on $Z(s)$ is ample, so $Z(s)$ is a variety of general type. In particularly the group $\mathrm{Aut}(Z(s))$ is finite. I would like to estimate the order of this group. For this I need to know how groups $\mathrm{Stab}([s])$ and $\mathrm{Aut}(Z(s))$ are related.

More specific, there is a natural map $r\colon \mathrm{Stab}([s])\to \mathrm{Aut}(Z(s))$ which is defined by the following rule. Consider an element $g\in \mathrm{Stab}([s])$ as the automorphism of $X$. Then $g$ preserves the zero locus $Z(s)$ and define $r(g)$ as the restriction of $g$ to $Z(s)$.

I would like to know two things on this map. First, the description of the kernel. And second, under what conditions on $L$, $s$ and $G$ the map $r$ is surjective? I hope these questions appeared somewhere in literature earlier.


1 Answer 1


A very partial answer:

  1. The kernel of $r$ is precisely the center of $G$: let $V=H^0(L\otimes K_X)$. Then $X$ embeds into $\mathbf P(V)$ and $G/\mathrm{center}\to\mathrm{Aut}(\mathbf P(V))$ is injective. Moreover, $Z(s)\subset X$ is not a hyperplane section, so it contains a basis.

  2. $r$ is in general not surjective for trivial reasons: Let $G=PGL(2)$ and $X=\mathbb P^1$. Then $Z(s)$ is just a finite set. Its automorphism group is the symmetric group $S_n$ which does not "fit" into $G$ for $n>\!\!>0$.

There is a good chance that $r$ is locally surjective, i.e., that its image contains the connected component $\mathrm{Aut}(Z(s))^0$.

  • $\begingroup$ Thank you. I understand your argument that the kernel is the center of $G$. But could you explain in more detail what you mean by locally surjective map? The target of the map $r$ is already is $\mathrm{Aut}(Z(s))$. $\endgroup$ Aug 19, 2016 at 7:27
  • $\begingroup$ @Nikolay Konovalov: Sorry, that was a typo. I meant the image contains the connected component of Aut. Fixed. $\endgroup$ Aug 19, 2016 at 10:21
  • $\begingroup$ @Fredrich Knop: OK, I see. Maybe at first I should prove this statement (for an arbitrary $L$). But I am mostly interested in the case when $Z(s)$ is a variety of general type. In this case the group $\mathrm{Aut}(Z(s))$ is finite and the map $r$ is locally surjective by trivial reasons. $\endgroup$ Aug 19, 2016 at 12:41

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