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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?

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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.

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  • $\begingroup$ 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 '16 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 '16 at 17:42
  • $\begingroup$ The invariance of both the quadric and the cubic did not follow from the uniqueness ? $\endgroup$ – Holonomia Jan 29 '16 at 17:56
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    $\begingroup$ 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 '16 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 '16 at 18:49

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