# What is the automorphism group of the projective line minus $n$ points?

$$\DeclareMathOperator{\AGL}{\operatorname{AGL}}\DeclareMathOperator{\PGL}{\operatorname{PGL}}$$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.

• The $n=2$ case should be $\mathbb{Z}/2 \ltimes k^\times$, no? Commented Oct 16, 2020 at 1:16
• Yes, you are absolutely correct! I guess you mean the $\mathbb Z/2$ acts by inversion on $k^\times$? Commented Oct 16, 2020 at 1:28
• 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? Commented Oct 16, 2020 at 1:45
• 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. Commented Oct 16, 2020 at 3:04
• 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. Commented Oct 16, 2020 at 10:18

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 .