# Automorphisms of plane curves are linear

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!

• For degree $4$, this is because the embedding is canonical. For higher degree, I do not see why it should be true. May 2, 2013 at 19:18
• ACGH Exercise 18 pg 56 May 2, 2013 at 20:20
• Excuse me for the delay in responding, but I was watching the reference and then get the text of Makoto Namba which prove that there exists a unique $g_d^2$ for a plane curve of degree $d\geq4$, In what way this implies the result? Thanks a lot! May 16, 2013 at 19:14
• The statement ("every automorphism... can be viewed as an an element of..."), as written, is senseless. Maybe the correct statement is that the automorphism group has an injective homomorphism into the given group?
– YCor
Dec 26, 2022 at 17:02

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 $(n,d)\neq (3,4)$.
• Just to add, the same approach works for singular curves using Theorem 2.1 from projecteuclid.org/euclid.kjm/1250520873 and Hartshorne's "generalized divisors". Feb 8, 2019 at 16:25
• How did you deduce $h^0(C,D-A) \ne 0$ in the argument above? Any correct argument must at some point use the hypothesis $g \ge 2$, but it's not clear where you are using this. Anyway, the proof outlined in Arbarello, Cornalba, Griffiths, and Harris, Geometry of algebraic curves, volume I, page 56, exercise 18 mentioned by Felipe Voloch works. Dec 9, 2022 at 2:46