Given a (smooth) curve $C$ and an automorphism $\phi$ of $C$. In the first part of their paper On the Kodaira dimension of the moduli space of curves Harris and Mumford calculate the eigenvalues of the induced action of $\phi$ on $H^0(2 K_C)$ (to study the singularities of $\mathcal{M}_g$). They don't do those calculations, instead only state the results and I seem to be missing something to understand how this is done in principle.
To be more concrete, the easiest example they consider is (p. 31): $C$ given as $y^2 = (x^3 - 1) (x^3 - a)$ with automorphism $\phi(x,y) = (\zeta_3 x, - y)$, where $\zeta_3$ is a primitive third root of $1$. $C$ has genus $2$, so $H^0(2 K_C)$ has dimension $3$. The automorphism is of order six, so the eigenvalues are given as $\zeta_6^{a_i}$, $i = 1,2,3$ with $\zeta_6$ a primitive sixth root of $1$. They state that the $a_i$ are given as $0,2,4$.
How to calculate those?
The only thing I see is that $a_i = 0$ gives an eigenvalue: Considering $C \rightarrow C/\phi$, we get a six to one cover of $\mathbb{P}^1$ with four branch points of profile $(2,2,3,3)$. Now a quadratic differential on $C$ has eigenvalue one iff it is the lift of a quadratic differential under this map. There are no smooth quadratic differentials on $\mathbb{P}^1$. But if a quadratic differential on $\mathbb{P}^1$ has a zero of order $m$ at $p$, its lift to $C$ has a zero of order $m k + 2(k-1)$ (where negative numbers denote poles) at a point in the fiber over $p$, at which locally the cover is given as $z \rightarrow z^k$. So in this example, a quadratic differential with simple poles at the branch points lifts to a smooth quadratic differential on $C$ (with prescribed zeroes at four of the ten ramification points). Now the space of quadratic differentials on $\mathbb{P}^1$ with poles at at most four points is one dimensional and in this way we get the one-dimensional eigenspace corresponding to the eigenvalue $1$. Is that right? Is it possible to calculate the other eigenvalues in a similar manner?
