Automorphisms of plane curves are linear

Question

I am looking for a proof or a reference for the following fact:

Every automorphism of any non-singular algebraic curve in $\mathbb{P}^2(\mathbb{C})$ of genus $g\geq 2$ is linear, i.e., can be viewed as an element of $\mbox{PGL}_3(\mathbb{C})$.

I read the proof for higher dimensional hypersurfaces (Matsumura-Monsky) but they use the fact that the Picard group of the variety is torsion-free (and in this case is false, since $\mbox{Pic}^0(C)$ is an abelian variety of positive dimension).

Thanks!

Answer 1

Let $C\subset\mathbb{P}^3$ be a smooth curve of degree $d$ and genus $g = \frac{(d-1)(d-2)}{2}\geq 4$. Let us consider the divisor $D = \mathcal{O}_C(1)$.

• Consider the exact sequence $$0\mapsto \mathcal{I}_C\rightarrow \mathcal{O}_{\mathbb{P}^2}\rightarrow\mathcal{O}_C\mapsto 0,$$ that is $$0\mapsto \mathcal{O}_{\mathbb{P}^2}(-d)\rightarrow \mathcal{O}_{\mathbb{P}^2}\rightarrow\mathcal{O}_C\mapsto 0.$$ Twisting by $\mathcal{O}_{\mathbb{P}^2}(1)$ we get $$0\mapsto \mathcal{O}_{\mathbb{P}^2}(-d+1)\rightarrow \mathcal{O}_{\mathbb{P}^2}(1)\rightarrow\mathcal{O}_C(1)\mapsto 0.$$ Finally, taking cohomology we have $H^{0}(C,\mathcal{O}_C(1))\cong H^{0}(\mathbb{P}^2,\mathcal{O}_{\mathbb{P}^2}(1))$. Therefore the linear system cut out on $C$ be the lines in $\mathbb{P}^2$ is complete, and $h^0(C,\mathcal{O}_C(1)) = 3$.
• Let $A$ be another effective divisor on $C$ such that $deg(A) = d$ and $h^{0}(C,A)=3$. Then $deg(D-A) = 0$ and $h^{0}(C,D-A)\neq 0$. Therefore $D-A$ is linearly equivalent to $\mathcal{O}_C$, and $A$ is linearly equivalent to $D$.
• We conclude that $D = \mathcal{O}_C(1)$ is the unique effective divisor of degree $d$ and with $h^{0}(C,D) = 3$.
• Let $\phi:C\rightarrow C$ be an automorphism. Then $\phi^{*}\mathcal{O}_C(1)\cong \mathcal{O_C}(1)$. Therefore $\phi$ acts on the sections of $\mathcal{O}_{\mathbb{P}^2}(1)$. Since $\mathbb{P}^2 = \mathbb{P}(H^{0}(\mathbb{P}^2,\mathcal{O}_{\mathbb{P}^2}(1))) = \mathbb{P}(H^{0}(C,\mathcal{O}_{C}(1)))$ we conclude that $\phi$ in indeed the restriction to $C$ of a linear automorphism of $\mathbb{P}^2$.

Just a couple of remarks:

• This is true for curves of degree $d=1,2$ as well. For a line it is trivial. If $C$ is a smooth conic an automorphism of $\mathbb{P}^2$ has to map five points of $C$ to other five points of $C$ in order to stabilize $C$. This group has dimension $dim(Aut(\mathbb{P}^2))-5 = 3$. On the other and $Aut(C) = PGL(2)$ has dimension $3$ as well. This is clearly false for smooth plane cubic. Indeed one needs nine points to stabilize a cubic. But $C\subseteq Aut(C)$.
• If $X\subset\mathbb{P}^n$, with $n\geq 3$, is a smooth hypersurface of degree $d$ and $d\neq n+1$ (i.e. $\omega_X$ is not trivial) then any automorphism of $X$ is induced by an automorphism of $\mathbb{P}^n$. This is essentialy due to the fact that such a hypersurface is sub-canonical and the Picard group is torsion free. This fact is indeed true for any Calabi-Yau hypersurface with $(n,d)\neq (3,4)$.