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 obtainallsmooth curves of genus $2$ by means of this construction?