To fix the ideas all curves are supposed to be defined over $\mathbb{C}$. Let $C$ be a rational connected projective curve. Note that we don't assume the curve to be smooth. Let $Aut(C)$ be the group of automorphisms of $C$ in the category of algebraic varieties.

Q1 When is $Aut(C)$ infinite?

For example, if $C\simeq \mathbb{P}^1$ then $Aut(C)\simeq PGL_2(\mathbb{C})$ which is infinite.

Q2 When is $Aut(C)$ finite?

Q3 Do we know all possible groups which may occur in Q1 and Q2?


Let $D$ be the normalization of $C$. The map $D \to C$ comes from a canonical construction, so automorphisms of $C$ lift uniquely to automorphisms of $D$.

So $Aut(C) \subseteq PGL_2$.

Next note that $Aut(C)$ is contained in the group of automorphism of $D$ that fix the inverse image of the singular points of $C$. So if this set has size at least $3$, the group is finite.

There are not too many subgroups of $PGL_2$. For instance, the finite ones are all either $C_n$, $D_n$, $A_4$, $S_4$, or $A_5$. To construct a curve whose automorphism group is one of these, any easy method is to take $\mathbb P^1$, view it as a sphere, and glue all the vertices of the appropriate regular polyhedron together.

The infinite ones must have connected component of the identity $\mathbb G_m$, $\mathbb G_a$, or $\mathbb G_a \ltimes \mathbb G_m$. I believe the only infinite ones are these, $\mathbb G_m \ltimes C_2$, and $\mathbb G_a \ltimes C_n$. I think it is not too hard to construct examples of all of these.

This leaves the question of when a rational curve with one or two singular points that lift to one or two points in the normalization has an infinite automorphism group. The answer will come down to local obstructions at the singular point(s) - does scaling induce a symmetry of the singularity? Does the operation $x \to \frac{1}{\frac{1}{x}+t}$ induce a symmetry? I don't know a nice way of checking which singularities have this symmetry.

| cite | improve this answer | |
  • $\begingroup$ How do you get $\mathbb{G}_a$ or $\mathbb{G}_a \rtimes C_n$ ? $\endgroup$ – Jérémy Blanc Sep 20 '13 at 0:36
  • $\begingroup$ For $\mathbb G_a$, $\operatorname {Spec} \mathbb C[x^2-x^3,x^4,x^5]$ should do. I'm less sure about the second one now, but maybe $\operatorname{Spec} \mathbb C[x^{n-1}+x^{2n-1},x^{2n}$ and all higher powers of $x$ works, or something like that. (with a point at $\infty$ glued on.) $\endgroup$ – Will Sawin Sep 20 '13 at 0:47
  • $\begingroup$ Thanks Will for the nice answer. I'm missing something quite elementary here. So let $S$ be the set of singular points of $C$ and assume that $|S|\geq 3$. Then I agree that as a SET, $S$ is fixed by $Aut(C)$ and therefore every element of $Aut(C)$ has finite order. But I don't quite see why this forces $Aut(C)$ to be finite? $\endgroup$ – Hugo Chapdelaine Sep 20 '13 at 12:25
  • $\begingroup$ @HugoChapdelaine: I think Will suggested $S$ to be the preimage of singular points of $C$ in $D$, so if $|S|\geq 3$, then $S$ is fixed under the unique lifting of $Aut(C)$ in $PGL_2$. Since every element in $PGL_2$ fixed $\geq$ 3 points in $D\cong \mathbb{P}^1$ must be the identity map, so $Aut(C)$ can be viewed as a subgroup of the permutation group on $S$. As a result, $Aut(C)$ is finite. $\endgroup$ – Yuchen Liu Sep 20 '13 at 14:09
  • $\begingroup$ That is exactly what I meant. Sorry for not spelling it out. $\endgroup$ – Will Sawin Sep 20 '13 at 15:13

Your Answer

