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$\DeclareMathOperator{\AGL}{\operatorname{AGL}}\DeclareMathOperator{\PGL}{\operatorname{PGL}}$What is the automorphism group of $\mathbb P^1$ minus $n$ points (let's say over an algebraically closed field of characteristic $0$ if it matters). I want to consider the removed points without order. I can do small cases by hand but it seems hard in general and it seems to depend on which points are removed.

Here's what I have thought about so far:

  • $n = 0$: The automorphism group of $\mathbb P^1$ is $\PGL_2(k)$
  • $n = 1$: The automorphism group of $\mathbb A^1$ is $\AGL(1)$.
  • $n = 2$: The automorphism group of $\mathbb G_m$ is $\mathbb Z/2 \ltimes k^\times$.
  • $n = 3$: Since $\PGL_2$ acts three transitively, it doesn't matter which points we remove. Any automorphism of $\mathbb P^1 - \{0,1,\infty\}$ will extend to an automorphism of $\mathbb P^1$ fixing $\{0,1,\infty\}$ as a set and is determined by what it does to this set. We get all of $S_3$ in this case.
  • $n = 4$: Any automorphism has to preserve the cross ratio and every permutation that does so is obtainable. So we get the Klein $4$ group - $\mathbb Z/2 \times \mathbb Z/2$ in the generic case.

I don't know what happens for $n \geq 5$. We can get non trivial automorphisms for large $n$ by doing the following: Pick a finite subgroup of $\PGL_2(k)$ (these are classified) and pick any finite subset of $\mathbb P^1$ and remove the entire orbit of this set by the finite subgroup.

Also generically, I believe there are no automorphisms for $n\geq 5$ by the following argument: We require all $4$ element subsets to have distinct cross ratios and since automorphisms have to preserve the cross ratio, this means that all $4$ element subsets are preserved. But this implies that the automorphism is trivial.

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    $\begingroup$ The $n=2$ case should be $\mathbb{Z}/2 \ltimes k^\times$, no? $\endgroup$ Commented Oct 16, 2020 at 1:16
  • $\begingroup$ Yes, you are absolutely correct! I guess you mean the $\mathbb Z/2$ acts by inversion on $k^\times$? $\endgroup$
    – Asvin
    Commented Oct 16, 2020 at 1:28
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    $\begingroup$ For $n \geq 5$ there are no generic automorphisms, and we have automorphisms in a special case using the construction you said. What more do you want to say? $\endgroup$
    – Will Sawin
    Commented Oct 16, 2020 at 1:45
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    $\begingroup$ Every configuration with any automorphisms has an automorphism of order $p$ for some prime $p$. This can happen only if $p$ divides $n$, $n-1$, or $n-2$, in which case the $n$ points consist of $0,1,$ or $2$ fixed points, respectively, together with some number of orbits of size $p$. This writes the exceptional cases as a finite union of explicit subvarieties. $\endgroup$
    – Will Sawin
    Commented Oct 16, 2020 at 3:04
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    $\begingroup$ In case $n=4$, $k = \mathbb{C}$, consider the set consisting of $0$ and the three third roots of unity. The automorphism group of this set contains an element of order $3$, so it cannot be the Klein four group. $\endgroup$ Commented Oct 16, 2020 at 10:18

1 Answer 1

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For $n \geq 5$, we can describe the locus of configurations that have nontrivial automorphisms. To do this, note that if there is any nontrivial automorphism, there is an automorphism of order $p$ for some prime $p$. Such an automorphism acts on $\mathbb P^1$ with two fixed points and the remaining points orbits of size $p$.

So the automorphism restricts to $\mathbb P^1$ minus $n$ points if and only if $n=ap+b$ for some $a \in \mathbb N$ and $b \in \{0,1,2\}$, and those $n$ points consist of $b$ of the fixed points as well as $a$ orbits of size $p$.

The space of such configurations has dimension $a$ for any given automorphism, hence $a+2$ in total, which is at most $\frac{n}{p}+ 2 \leq \frac{n}{2} + 2 <n$.

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