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I would like to know all finite subgroups of $\operatorname{Aut}(\mathbb{P}^1)$.

I am aware that any automorphism of $\mathbb{P}^1$ is given by a Möbius transformation $$ z\mapsto\frac{az+b}{cz+d} $$ and thus there is an identification $$ \operatorname{Aut}(\mathbb{P}^1)\cong \operatorname{PSL}(2,\mathbb{C})\cong \operatorname{SO}(3, \mathbb{C}). $$ I thought this solved the question, but what I know is the classification of finite subgroups of real orthogonal group $\operatorname{SO}(3, \mathbb{R})$.

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I think the third groupe of your line of isomorphism is in fact $SO(3,\mathbb{R})$, not $SO(3,\mathbb{C})$ and hence your problem is solved (you can check that $SO(3,\mathbb{C})$ doesn't have the good dimension as a real manifold if you're not covinced...). – Simon Henry Oct 13 '12 at 11:22

1 Answer 1

So pointed out by Simon Henry, $\operatorname{PSL}(2,\mathbb{C})\cong \operatorname{SO}(3, \mathbb{R})$, not $\operatorname{SO}(3, \mathbb{C})$. Recall that $SO(3, \mathbb{R})$ is the set of orientation-preserving isometries of $\mathbb{R}^3$, and hence $S^2$ (which is homeomorphic to $\mathbb{P}^1$).

The finite subgroups of $SO(3, \mathbb{R})$ have been classified, see the groupprops wikipage for example. I will briefly outline them below.

Note that $SO(2, \mathbb{R})$ naturally includes in $SO(3, \mathbb{R})$. More precisely, the map

\begin{align*} SO(2, \mathbb{R}) &\to SO(3, \mathbb{R})\\ A &\mapsto \begin{bmatrix} A & 0\\ 0 & 1\end{bmatrix} \end{align*}

gives an isomorphism between $SO(2, \mathbb{R})$ and a subgroup of $SO(3, \mathbb{R})$. Therefore, any subgroup of $SO(2, \mathbb{R})$ can be realised as a subgroup of $SO(3, \mathbb{R})$; in particular, every finite subgroup of $SO(2, \mathbb{R})$ will give a finite subgroup $SO(3, \mathbb{R})$. Note that the elements of $SO(2, \mathbb{R})$ are simply the rotation matrices and so we have $SO(2, \mathbb{R}) \cong S^1$. The only finite subgroups of $S^1$ are finite cyclic subgroups, so $SO(3, \mathbb{R})$ has such subgroups.

Another infinite family of finite subgroups of $SO(3, \mathbb{R})$ are the dihedral groups $D_{2n}$, which are the symmetries of a regular $n$-gon in a plane. Although these symmetries leave the plane fixed, they do not come from elements of $SO(2, \mathbb{R})$ as above because reflections swap the sides of the plane.

The remaining finite subgroups of $SO(3, \mathbb{R})$ rely on the classification of platonic solids. There is a subgroup isomorphic to $A_4$ given by the symmetries of a tetrahedron, a subgroup isomorphic to $S_4$ given by the symmetries of a cube or octahedron, and a subgroup isomorphic to $A_5$ given by the symmetries of an icosahedron or dodecahedron. Note, the platonic solids which appear in pairs are dual polyhedra (a tetrahedron is self-dual).

In addition to all of the finite subgroups discussed above, we must also consider their subgroups. In doing so, we see that all of them have already been mentioned${}^1$. Therefore, a finite subgroup of $SO(3, \mathbb{R})$ is isomorphic to one of the following: $\mathbb{Z}/n\mathbb{Z}$, $D_{2n}$, $S_3$, $A_4$, $S_4$, $A_5$.

${}^1$ I initially thought $S_3$ (a subgroup of both $S_4$ and $A_5$) was missing, but as pointed out by pjs36, $S_3 \cong D_6$.

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An emphatic +1! But isn't $S_3 \cong D_6$? – pjs36 May 25 at 6:14
@pjs36: Yes it is! I totally overlooked this. I will edit accordingly, thanks. – Michael Albanese May 25 at 6:15

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