7
$\begingroup$

Let $V_2$ and $V_3$ be the two hypersurfaces of $\mathbb P^3$ defined by \begin{equation*} V_2:={x_2x_3 + r(x_0, \, x_1)=0}, \quad V_3:={x_2^3+x_3^3+s(x_0, \, x_1)=0}, \end{equation*} where $r, \, s \in \mathbb{C}[x_0, \, x_1]$ are general homogeneous forms of degree $2$ and $3$, respectively.

Then $C_4:=V_2 \cap V_3$ is a smooth, canonical curve of genus $4$. Denoting by $\xi$ a primitive third root of unity, then $C_4$ admits a free action of the cyclic group $\langle \xi \rangle \cong \mathbb Z/3 \mathbb Z$, defined by \begin{equation*} \xi \cdot [x_0: x_1:x_2:x_3] = [x_0: x_1: \xi x_2: \xi^2 x_3] \end{equation*} and the quotient $C_2 := C_4/ \langle \xi \rangle$ is a smooth curve of genus $2$.

A naive count of parameters shows that the number of moduli on which this construction depends is $$\dim |\mathcal{O}_{\mathbb{P}^1}(2)| + \dim |\mathcal{O}_{\mathbb{P}^1}(3)| - \dim G= 7 - \dim G,$$ where $G \subset \textrm{Aut}(\mathbb{P}^4)$ is the subgroup of projectivities sending both $V_2$ and $V_3$ to hypersurfaces of the same form. If my computations are correct, $G$ has dimension $4$ so the construction actually depends on $3$ parameters.

On the other hand, $3$ is also the dimension of the moduli space $\mathcal{M}_2$. This shows that the general curve $C_2$ of genus $2$ can be constructed in this way.

Question. Is it possible to obtain all smooth curves of genus $2$ by means of this construction?

Improve this question
$\endgroup$

1 Answer 1

Reset to default
14
$\begingroup$

Yes. Start from a genus 2 curve $C_2$, and choose a point of order 3 in $JC_2$, giving rise to an étale $\mathbb{Z}/3$-covering $C_4\rightarrow C_2$. Then $C_4$ cannot be hyperelliptic: a $g^1_2$ on $C_4$ would be stable under the covering automorphism $\sigma $, hence descend to $C_2$, which is impossible for degree reasons. $\sigma $ acts on $H^0(C_4,K_{C_4})$ with eigenvalues $(1,1,\xi ,\xi ^2)$: the invariant part comes from $C_2$, and the trace must be 1 by the holomorphic Lefschetz formula. Writing down the invariant quadric and cubic(s) containing $C_4$ leads to your formulas.

Improve this answer
$\endgroup$
11
  • $\begingroup$ Thank you for the answer, it is really helpful. I have a couple of questions. (1) Your computations show that the curve $C_4$ embeds in $\mathbb P^4$ and must be invariant by the action $$[x_0: x_1 : x_2 : x_3] \mapsto [x_0: x_1 : \xi x_2 : \xi^2 x_3].$$ On the other hand, it is a canonical curve, so it is intersection of a (unique) quadric and a cubic. But why both the quadric and the cubic must be invariant, too? $\endgroup$
    – Francesco Polizzi
    Jan 29, 2016 at 17:37
  • $\begingroup$ (2) How can I detect the space of invariant cubics through the canonical curve? Using Koszul resolution I can compute $$h^0(\mathscr{I}_{C_4})=5,$$ and it seems to me that there are $8$ invariant monomials, namely $$x_0^3, \, x_0^2x_1, \, x_0x_1^2, \, x_1^3, \, x_0x_2x_3, \, x_1x_2x_3, x_2^3, \, x_3^3.$$ Why you say Writing down the invariant quadric and cubic(s) containing $C_4$ leads to your formulas? How can I check that there exist an invariant cubic containing $C_4$ and whose equation does not contain the invariant monomials $x_0x_2x_3, \, x_1x_2x_3$? $\endgroup$
    – Francesco Polizzi
    Jan 29, 2016 at 17:42
  • $\begingroup$ The invariance of both the quadric and the cubic did not follow from the uniqueness ? $\endgroup$
    – Holonomia
    Jan 29, 2016 at 17:56
  • 1
    $\begingroup$ The quadric is in the kernel of $\mathsf{S}^2H^0(C_4,K)\rightarrow H^0(C_4,2K)$. The invariant subspace in $\mathsf{S}^2$ has dimension 4; in $H^0(C_4,2K)$, it is $H^0(C_2,2K)$, which has dimension 3. Similarly, the kernel of $\mathsf{S}^3H^0(C_4,K)\rightarrow H^0(C_4,3K)$ has dimension 5, with eigenvalues $(1,1,1,\xi ,\xi ^2)$. The eigenvalues $(1,1,\xi ,\xi ^2)$ correspond to the multiples of the quadric, it remains one invariant cubic. $\endgroup$
    – abx
    Jan 29, 2016 at 18:08
  • $\begingroup$ @Holonomia: the quadric through $C_4$ is unique (so necessarily invariant), but the cubic is not: there is a $5$-dimensional vector space of them. $\endgroup$
    – Francesco Polizzi
    Jan 29, 2016 at 18:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.