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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:

  1. $|\textrm{Aut}\,C|<\infty$.
  2. For every smooth rational component $E\subset C$, $|E\cap\overline{C-E}|\geq 3$.
  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.

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up vote 3 down vote accepted

For (1) implies (2): it is not true that an automorphism must preserve each irreducible component. There is a normal subgroup $H$ of $\mathrm{Aut}(C)$ consisting precisely in those automorphisms which leave each component globablly invariant and which fix individually each intersection point. The quotient is a subgroup of the finite group of permutations of the components and of the intersection points.

If (2) is not satisfied, let $L$ be a smooth rational component meeting the other components at one or two points. Then there are infinitely many automorphisms of $L$ fixing these intersections points. Each of this automorphism extends (by identity on the other components) to an automorphism of $C$, hence $\mathrm{Aut}(C)$ is infinite.

If (2) is satisified, for each element $\sigma\in H$, the restriction of $\sigma$ to any irreducible component belongs to a finite group (this uses the fact that the groups of automorphisms of $\mathbb P^1$ minus three points, of elliptics curve minus one point, and of curves of genus $\ge 2$ are all finite). So $H$, hence $\mathrm{Aut}(C)$, are finite.

As for $\omega_{C}$: it is invertible because $C$ is locally complete interseciton. The proof of the ampleness when $C$ is stable can be found in "Algebraic geometry and arithmetic curves", 10.3.13. The same computations can be used to show the inverse.

It is not true that when $\omega_C^{\otimes n}$ is very ample then $n\ge 3$. Already for smooth $C$, $\omega_C$ can be very ample (enough to take non-hyperelliptic curves).

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Thanks @QiL! I have only one doubt left, in your $2\Rightarrow 1$: what kind of irreducible component can such a curve have? I would say: $\mathbb P^1$, elliptic curves (but why do you remove just one point from them?), smooth curves of genus $\geq 2$ (the nonsingular ones), or nodal curves of any genus (the singular ones). Does your argument include all the possibilities? – Brenin Dec 2 '12 at 9:48
@atricolf: I just forgot singular components. But the group of automorphisms of such a component fixing a point (an intersection point) is finite. It is enough to see what happen at the normalization. – user18119 Dec 2 '12 at 9:54
Thank you very much! – Brenin Dec 2 '12 at 10:03

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