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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?

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Your argument is right, but I don't know a way to transfer it to the other eigenvalues.

The analysis of the covering $C\rightarrow C/\varphi$ gives another interpretation of the eigenvalue 1: four points on a $P^1$ give a one-dimensional moduli space (it's $M_{1,4}$) and (most probably - I didn't spell it out) give via the covering of the mentioned type a one-dimensional subfamily of the local universal deformation of $C$ to which the automorphism $\varphi$ deforms. Therefore the corresponding one-dimensional subcurve of the base space of the local universal deformation is fixed pointwise by the induced action of $\varphi$.

I would try to transfer this sort of argument to pin down the remaining eigenvalues. For example the cube of $\varphi$ is the hyperelliptic involution which deforms to every smooth curve of genus 2. Hence the induced action of $\varphi^3$ is the identity, which proves that the eigenvalues of the induced action of $\varphi$ must be third roots of unity (and not primitive sixth roots). The remaining problem would be to prove that these are primitive third roots (i.e. there is no two-dimensional family to which $\varphi$ deforms) and these are two different ones (otherwise the Reid-Tai-sum would be only $\frac{2}{6}+\frac{2}{6}<1$).

It might also be necessary to analyse the induced action of $\varphi$ on the local universal deformation of $C$ in more detail. You'll find information on deformation theory in Harris-Morrison's Moduli of curves (p.102ff). A problem I fought with a lot is the question "Which direction of the action is the right one?": The space of infinitesimal deformations of $C$ is $H^1(C,T_C)$ and isomorphic to the tangent space of the base space of the local universal deformation. Via Riemann-Roch it is also isomorphic to $H^0(C,2K_C)^\ast$. The induced action of $\varphi$ on $H^0(C,2K_C)^\ast$ is dual to the action on $H^0(C,2K_C)$, which - sadly - inverses all eigenvalues. For Reid-Tai it's not important whether you take one or the other, but you have to be extra careful not to use the one in half of the calculations and the other in the second half.

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  • $\begingroup$ Welcome to MathSE! This might be helpful for future questions and answers. $\endgroup$
    – Hirshy
    Jul 29, 2015 at 18:26

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