3
$\begingroup$

Let $G$ be a finite subgroup of $\mathrm{Aut}(\mathbb P^{1}(k))=\mathrm{PGL}(2,k)$, where $k$ is an algebrically closed field of characteristic $0$. Suppose that there is a common fixed point for all the elements of $G$. Is $G$ cyclic?

In other words, is it true that $ \exists [p] \in \mathbb P^{1}(k): \, \phi([p])=[p] \, \quad \forall \phi \in G \Rightarrow G \quad \text{is cyclic?} $

I am really puzzled and I do not know what to say. What do you think?

I started considering $k = \mathbb C$ but I couldn't conclude anything. Can you help me, please?

Thanks.

1 Answers 1

1

Here is a proof:

For $\mathbb{C}$, I identify $P^1 \mathbb{C}$ with the complex Riemann-sphere, then $Aut =PSL_2(\mathbb{C})$ are the Moebius transformation. A stabilizer of $0$ are the matrices, which are lower triangular. Every stabilizer at every other point will be conjugate to this, since the action is transitive. Now every finite group is actually a finite subgroup of the diagonal matrices with entries having absolute value $| z |=1$, so it will be cyclic.

For a general field $F$, you will make the same observation that it will be equivalent to an upper triangular matrix, so it will be a subgroup of $n$-th roots of unity for some $n$, so they will be cyclic as a subgroup of a cyclic one.

What I am using is the Levi-decomposition of a parabolic subgroup.

  • 0
    Sorry, for the original mess. I was answering to quickly. In particular the stabilizer of a transitive group action of non compact group on a compact space, can never be compact. I must have been thinking about Moebius transformation on the upper halfplane by matrices of $PSL_2(\mathbb{R})$2012-04-15