I'm learning the basics about stable curves. Suppose we have a connected complex projective curve $C$, at worst nodal and of genus $g\geq 2$. Then I want to prove that the following are equivalent:
- $|\textrm{Aut}\,C|<\infty$.
- For every smooth rational component $E\subset C$, $|E\cap\overline{C-E}|\geq 3$.
- The dualizing sheaf $\omega_C$ is ample.
My attempts: for $1\Rightarrow 2$, I notice the following facts (but I might be wrong!): first, an automorphism must send a component onto itself. Second, it has to permute the nodes. Third, we have $E\cap\overline{C-E}=\{\textrm{Nodes of } C\, \textrm{inside }E\}.$ Hence if we have strictly less than $3$ nodes on a component $E\cong\mathbb P^1$, then because Aut $\mathbb P^1$ is 3-dimensional, we have infinitely many choices of automorphisms of $E$, which I can then extend to automorphisms of $C$. Contradiction.
For $2\Rightarrow 1$, let $\phi\in\textrm{Aut}\,C$. By the condition $|E\cap\overline{C-E}|\geq 3$ I can choose $3$ points on every smooth rational component, and their images (permutations) determine a unique automorphism of $E$. Then, I would like to conclude using the fact that there are finitely many components and finitely many permutations. But I wonder if I can really conclude, because my argument does not involve singular components. So how to fix the reasoning?
Furthermore, I have no idea how to deal with the condition on $\omega_C$. Indeed I was able to show just one thing: namely, if $\omega_C^m$ gives me an embedding of $C$ in $\mathbb P^r$, then necessarily $m\geq 3$, and $r=(2m-1)(g-1)-1$. So, a "$3$" magically appeared, but I don't know how to use it!
Last (aside) question. I'll be honest: I don't know why $\omega_C$ is invertible for such a curve $C$. Is there an easy way to prove it?
Thanks for any help.