Lifting an automorphism of a curve to an automorphism of its Jacobian

Question

Let $C: Y^3Z = f(X,Z)$, with $f(X,Z)\in K[X,Z]$, a degree 4 homogeneous polynomial, and $K$ a field.

The curve $C$ has an order $3$ automorphism, given by sending $(x,y,z)$ to $(x,\omega y, z)$, where $\omega^3 = 1$ is a non-trivial cube root of unity.

I imagine that the lift of $[\omega]$ to the Jacobian has a nice definition, given by multiplication by some scalar. How can I identify what this scalar is?

I am basing this intuition on the fact that for an hyperelliptic curve, the hyperelliptic involution is lifted onto multiplication by $-1$ on the Jacobian.

Thanks in advance!

Answer 1

By taking $Z=1$, the defining equation of the quartic curve is $y^3-f(x)$, where I denote $f(x)=f(x,1)$. Then use Poincare residue, the space of holomorphic 1-forms $H^{1,0}$ has a basis $$\eta_1=\frac{dx}{y^2}, \eta_2=\frac{xdx}{y^2}, \eta_3=\frac{dx}{y}.$$

Since the action is by multiplying 3rd root of unity $\omega$ on $y$, its action on $\eta_1$ and $\eta_2$ is by multiplying $\omega$, and its action on $\eta_3$ is by multiplying $\omega^2$.

Now, the Jacobian is $H^{0,1}/\text{lattice}$. Take conjugation on $\eta_1,\eta_2,\eta_3$ gives you action on the Jacobian.