I am a bit confused by what Harris and Morrison write about the finiteness condition for stable curves in Moduli of Curves.
First the defintions:
Definition (2.12) A stable curve is a complete connected curve that
has only nodes as singularities and has only finitely many automorphisms.
Directly after this he writes
In view of the connectedness of C, its automorphism group can fail
to be finite only if C contains rational components.
But what about an elliptic curve $E$? This is a complete, connected, smooth curve, and it's not rational. But according to Lemma IV 4.2 in Hartshorne's Algebraic Geometry, given two points $P, Q \in E$, there always exists an automorphism $\sigma$ with $\sigma(P) = Q$, and $\sigma^2 = \text{id}_E$. So if $E$ has infinitely many points (e.g. $E$ is an elliptic curve over $\mathbb{C}$), then $E$ has infinitely many automorphisms as well.
Have I overlooked something?
Best Answer
You are right! A genus 1 curve has infinitely many automorphisms, and is not rational.
Of course, Harris and Morrison are aware of this, so what is going on?
My diagnosis: this is a case of terrible writing. I think that in the passages you quote, the authors only mean to discuss curves of genus $g \geq 2$. This is highly unclear in context, in particular given the example preceding the definition. However, if you flip forward to Chapter 4 and the discussion of the construction of $\overline{M_g}$ using GIT, you will see that the context is restricted to $g \geq 2$.
Finally, note that for curves of genus $g \geq 2$, your second quoted sentence is correct. Any component $C$ of genus 1 has to meet another component of the curve, so (up to finite index) the automorphism group acts on $C$ as a group of automorphisms fixing a point, hence a finite group.