Computational Aspects of Equivariant Hilbert Series of Canonical Rings for Algebraic Curves

Abstract

We study computational aspects of the problem of decomposing finite group actions on graded modules arising in arithmetic geometry, in the context of ordinary representation theory. We provide an algorithm to compute the equivariant Hilbert series of automorphisms acting on canonical rings of projective curves, using the formulas of Chevalley and Weil. Further, we apply our results on Fermat curves, determine explicitly the respective equivariant Hilbert series and extend the computation to the short exact sequence that arises from Petri’s Theorem. Finally, we implement the above computations in Sage.

This research is co-financed by Greece and the European Union (European Social Fund- ESF) through the Operational Programme $\ll$Human Resources Development, Education and Lifelong Learning 2014-2020$\gg$ in the context of the project “On the canonical ideal of algebraic curves” (MIS 5047968).

1 Introduction

1.1 Equivariant Hilbert Series

One of the most fundamental problems in representation theory of finite groups is that of decomposing representations into direct sums of indecomposables. Namely, given a finite group G acting on a vector space V over an arbitrary field k, the problem amounts to determining, for each indecomposable representation $W\in \mathrm{Ind}(G)$ over k, a natural number $n_{W,V}$ such that $V=\bigoplus _{W\in \mathrm{Ind}(G)}n_{W,V}W$. In the context of modular representation theory, that is, if the characteristic of the ground field is positive and divides the order of G, there are several complications that make the general case of this problem practically impossible; however, if $\mathrm{char}(k)=0$ or $\mathrm{char}(k)=p\not \mid |G|$, every indecomposable representation is irreducible, and there is a direct approach using character theory

$$ n_{W,V}=\langle \chi _V,\chi _W\rangle :=\frac{1}{|G|}\sum _{g\in G}\chi _V(g)\overline{\chi _W(g)}, $$

where $\chi _V$ denotes the character of the representation $\rho :G\rightarrow \mathrm{GL}(V)$.

The above technique can be also used when one generalizes the objects acted on, from vector spaces V over k to modules M over some k-algebra R. Historically, a case of particular interest is that of finite groups acting as automorphisms on polynomial rings: the study of their G-structure is essentially the main motivation behind the development of invariant theory, a subject whose origins date back to Hilbert’s fourteenth problem. The next level of abstraction dictates to consider, instead of a polynomial ring, an arbitrary graded, Noetherian k-algebra $R=\bigoplus _{d=0}^\infty R_d$ acted upon by a finite group G. Since each graded piece $R_d$ is a vector space over k, one can apply the techniques of the first paragraph to obtain for each $d\in \mathbb {N}$ and each $W\in \mathrm{Irr}(G)$, natural numbers $n_{W,d}$ such that $R_d=\bigoplus _{W\in \mathrm{Irr}(G)}n_{W,d}W$. The decomposition of R is then given by

$$ R =\bigoplus _{d=0}^\infty R_d =\bigoplus _{d=0}^\infty \bigoplus _{W\in \mathrm{Irr}(G)}n_{W,d}W =\bigoplus _{W\in \mathrm{Irr}(G)}\sum _{d=0}^{\infty }n_{W,d}W. $$

One obtains for each $W \in \mathrm{Irr}(G)$ a generating function for the sequence $\{n_{W,d}\}_{d=0}^\infty$

$$ H_{R,W}(T)=\sum _{d=0}^{\infty }n_{W,d}T^d. $$

By studying the convergence of $H_{R,W}(T)$, the infinite information of the action of G on R, which is infinite dimensional over k, can be packaged in a finite sequence $\left\{ H_{R,W}(T)\;|\; W\in \mathrm{Irr}(G)\right\}$ which is called the equivariant Hilbert series of the pair (R, G). The best understood case is, again, that of polynomial rings: if $R=\mathrm{Sym}\left( V\right)$ is the symmetric algebra of a finite dimensional k-vector space V, Molien’s theorem [15, Theorem 2.1] says that

$$\begin{aligned} H_{R,W}(T)=\frac{\dim W}{|G|}\sum _{g\in G}\frac{\overline{\chi _W(g)}}{\mathrm{det}\left( \mathrm{Id}_V-gT\right) }. \end{aligned}$$

(1)

Of course, hoping to obtain an analogous formula for arbitrary graded, Noetherian k-algebras R, is unrealistic, unless one has some concrete information on the action of G on R. Since graded, Noetherian k-algebras arise as homogeneous coordinate rings of projective varieties, this can be achieved by switching the viewpoint towards algebraic geometry.

1.2 Petri’s Theorem

From now on we assume that k is algebraically closed. Let X be a smooth, projective curve of genus g over k. Recall that X does not come a priori with a fixed embedding into projective space; however, it is well known that explicit projective embeddings can be constructed using (very ample) line bundles on X. Of all possible projective embeddings of X, there is one that stands out as canonical: that determined by the cotangent bundle $\varOmega _X$, referred to also as the sheaf of holomorphic differentials on X. It is given by

$$ X\rightarrow \mathbb {P}\left( H^0(X,\varOmega _X)\right) \cong \mathbb {P}^{g-1}_k,\quad P\mapsto [\omega _1(P):\cdots :\omega _g(P)], $$

where $\{\omega _1,\ldots ,\omega _g\}$ denotes a k-basis for the global sections $H^0(X,\varOmega _X)$.

To see that this construction gives an embedding, we rephrase the above in the algebraic language. Recall that the homogeneous coordinate ring of the projectivization $\mathbb {P}\left( H^0(X,\varOmega _X)\right)$ is the symmetric algebra $\mathrm{Sym}\left( H^0(X,\varOmega _X)\right)$, which may be identified with a polynomial ring in g variables. The canonical embedding is then determined by the so-called canonical map, as ensured by the following classic theorem [14] due to Max Noether, Federigo Enriques and Karl Petri.

Theorem 1

If X is not hyperelliptic and has genus $g\ge 4$, the canonical map

$$ \phi :S:=\mathrm{Sym}\left( H^0(X,\varOmega _X)\right) \rightarrow S_X:=\bigoplus _{m=0}^\infty H^0(X,\varOmega _X^{\otimes m}), $$

is surjective. Its kernel $I_X$, the canonical ideal, is generated in degrees 2 and 3.

Quoting from [3, Section 2, §3], the canonical ring $S_X$ “is the homogeneous coordinate ring of the canonically embedded curve X”. Any action of a finite group G on X induces an action on $S_X$, and thus, we may seek a formula for its equivariant Hilbert series. Assuming that $\mathrm{char}(k)=p\not \mid |G|$, we may use Molien’s formula to compute the respective series for S and thus obtain the equivariant Hilbert series for the canonical ideal $I_X$. It is worth noting that these calculations are the starting point in computing the action of G on the minimal graded resolution of $S_X$ as an S-module. The latter is well-studied in the non-equivariant case mainly due its connection to Green’s syzygy conjecture [5]; we hope that this work will shed some light to possible generalizations in the equivariant case.

The main results of this paper are:

• 1. General formulas (Theorem 3) and an algorithm (Algorithm 2) that gives the equivariant Hilbert series of $S_X$ for arbitrary curves X.

• 2. Explicit formulas (Theorem 4) for the equivariant Hilbert series of $S_X$ when X is a Fermat curve.

• 3. A Sage [16] program¹$^,$² that computes, when X is a Fermat curve:

  • (a) $\{H_{S_X,V}(T):V\in \mathrm{Irr}(G)\}$, by implementing the formulas of Theorem 4.

  • (b) $\{H_{S,V}(T):V\in \mathrm{Irr}(G)\}$, by implementing Molien’s formula.

  • (c) $\{H_{I_X,V}(T):V\in \mathrm{Irr}(G)\}$, subtracting the two above results.

We remark that similar results were obtained in our preprint [1] using different techniques. We have verified computationally that the two approaches lead to the same answers; a concrete theoretical proof involves complicated calculations, however we can indicatively provide the reader with one, i.e., for one of the irreducible representations, upon request.

2 Equivariant Hilbert Series of Canonical Rings

Let X be a smooth, projective curve of genus g over an algebraically closed field k of arbitrary characteristic $p\ge 0$. Let G be a finite subgroup of its automorphism group $\mathrm{Aut}_k(X)$ of order |G| not divisible by p. For $m\ge 1$, we denote by $\varOmega _X^{\otimes m}$ the sheaf of holomorphic m-differentials on X and by $W_m$ the k-vector space $H^0(X,\varOmega _X^{\otimes m})$ of its global sections. By the Riemann-Roch Theorem [6, IV.1.3],

$$ \dim _k W_m= {\left\{ \begin{array}{ll} 1&{}\text {, if }m=0\\ g&{}\text {, if }m=1\\ (2m-1)(g-1)&{}\text {, if }m\ge 2 \end{array}\right. } $$

it is further well known that the action of G on X induces an action on $W_m$ for all $m\ge 1$. Let $\mathrm{Irr}(G)$ denote the group of irreducible representations of G over k; the isomorphism class of each $W_m$, viewed as a kG-module, is uniquely determined by a collection of integers $\{N_{V,m} \}_{V\in \mathrm{Irr}(G)}$ such that $W_m=\bigoplus _{V\in \mathrm{Irr}(G)}N_{V,m}V$. The classic approach to computing the integers $N_{V,m}$ goes as follows.

The downside of the above algorithm comes from input (3), in that there does not exist a general method to compute explicit bases for the k-vector spaces $W_m=H^0(X,\varOmega _X^{\otimes m})$. Even in the few cases in which bases are known, one of which is that of Fermat curves that we will study in Sect. 3, the sums in step (2) can in practice become rather difficult to compute, see for example our proof of [1, Theorem 20]. An alternative approach, exploited with great success by many authors, see for example [2, 4] and [8], is to express the multiplicities $N_{V,m}$ in terms of the ramification data of the action of G on X. The resulting formulas are much easier to use, both in terms of the input required and in terms of computational complexity; however, as is usually the case in such situations, they require some familiarity with technical aspects of arithmetic geometry, which we briefly recall here. For more details the reader may refer to [6, Chapter IV], [12, Chapters 4 & 10], or [13].

From now on, we assume that the characteristic of k is either 0 or does not divide |G|. Let $Y=X/G$ be the quotient of X by the action of G. The quotient map $\pi :X\rightarrow Y$ is a non-constant, regular morphism of curves of degree |G|, so that the number of points in a generic fiber $\pi ^{-1}\left( Q\right) , Q\in Y$ is equal to |G|. There exists a finite set of points $Q\in Y$ for which the fiber $\pi ^{-1}\left( Q\right)$ has cardinality strictly less than |G|, called the branch locus of $\pi$ and denoted by $\mathscr {B}$. The ramification locus of $\pi$ is $\mathscr {R}=\pi ^{-1}\left( \mathscr {B}\right) \subseteq X$. By [7, Theorem 11.49], the decomposition group of a point $P\in X$ is the cyclic group $G_P=\{\sigma \in G: \sigma (P)=P \}$ and its order is called the ramification index of $P\in X$. Since the ramification index is the same for all points in the orbit of $P\in X$, we denote it by $e_Q$, where $Q=\pi (P)\in Y$. The cyclic group $G_P$ has $e_Q$-many, distinct, one-dimensional irreducible representations, determined by their characters. Fix $\zeta _{e_Q}$ to be a primitive $e_Q$-th root of unity; the irreducible characters of $G_P$ are all of the of the form $\chi _P^d$, $1\le d\le e_Q$, where $\chi _P$, is the fundamental character at the point P, that is, the character obtained by letting $G_P$ act on a local uniformizer $u_P$ at P considered modulo $u_P^2$. The monodromy element $\sigma _P$ is a generator of $G_P$ such that $\sigma _P(u_P)=\zeta _{e_Q} u_P$. For each irreducible representation V of G, we denote by $n_{d,Q,V}$ the multiplicity of the irreducible character $\chi _P^d$ in the decomposition of the restricted representation $\mathrm{Res}_{G_P}^G\left( V\right)$, i.e., $n_{d,Q,V}=\langle \chi _P^d,\mathrm{Res}_{G_P}^G\left( \chi _V\right) \rangle$. We summarize the above in the table below.

Table 1. Notation for the ramification data of the action of G on X

The following result gives an explicit formula for the multiplicities $N_{V,m}$.

Theorem 2

(Chevalley-Weil [2]). For each $V\in \mathrm{Irr}(G)$, we have that

$$\begin{aligned}&N_{V,m}=E_{V,m}+ (2m-1)(g_Y-1) \dim V \\&\qquad \qquad \qquad \;\;\; + \sum _{Q\in \mathscr {B}} \sum _{d=0}^{e_Q-1} \left( (m-1)\left( 1- \frac{1}{e_Q} \right) + \left\langle \frac{m-1-d}{e_Q} \right\rangle \right) n_{d,Q,V}, \end{aligned}$$

where $\mathscr {B},e_Q$ and $n_{d,Q,V}$ are given in Table 1, $g_Y$ is the genus of Y,

$$E_{V,m}= {\left\{ \begin{array}{ll} N_{V^*,1} &{}\text {, if }m=0\;(V^*\text { denotes the dual of } V)\\ 1&{}\text {, if }m=1\text { and }V \text { is the trivial representation}\\ 0&{}\text {, otherwise.} \end{array}\right. } $$

and $\langle x\rangle =x-\left\lfloor x\right\rfloor$ denotes the fractional part of x.

Remark 1

For a proof of the above, see [4, Th. 3.8 & Rem. 3.9] The authors compute the multiplicity of V in the equivariant Euler characteristic $[H^0(X,\varOmega _X^{\otimes m})]-[H^1(X,\varOmega _X^{\otimes m})]$. The formula for $E_{V,m}$, which is the multiplicity in $H^1(X,\varOmega _X^{\otimes m})$, follows from the Riemann-Roch theorem combined with Serre’s duality. It is worth mentioning that the above result was generalized in [8] to the weakly ramified case.

Theorem 3

The equivariant Hilbert series of $S_X=\bigoplus _m H^0(X,\varOmega _X^{\otimes m})$

$$ H_{S_X,V}(T)=\sum _{m=0}^\infty N_{V,m}T^m $$

of an irreducible representation V of G is given by the rational function

$$\begin{aligned}&H_{S_X,V}(T)=N_{V^*,1}+\delta _{V}T+ \frac{3T-1}{(1-T)^2}(g_Y-1) \dim V + \frac{T}{(1-T)^2}\dim V |\mathscr {B}|\\&\qquad \qquad \qquad \qquad \qquad \qquad \quad \;\, - \frac{1}{1-T}\sum _Q \frac{f'_{Q,V}(1)}{e_Q} - \frac{T}{1-T} \sum _{Q} \frac{f_{Q,V}(T)}{1-T^{e_Q}} , \end{aligned}$$

where $\delta _V=1$ for $V=V_\mathrm{triv}$ and 0 otherwise,

$$\begin{aligned} f_{Q,V}(T)=\sum _{d=0}^{e_Q-1}n_{d,Q,V}T^d \end{aligned}$$

and $|\mathscr {B}|$ denotes the cardinality of the branch locus of the cover $X\rightarrow X/G$.

Our computations will be in two steps. Write

$$ H_{S_X,V}(T)=N_{V^*,1}+\delta _VT+F_V(T)+G_V(T), $$

where

$$\begin{aligned} F_V(T)&= \sum _{m=0}^\infty \left( \!(2m-1)(g_Y-1) \dim V\!+\!\sum _Q\!\sum _{d=0}^{e_Q-1} (m-1)\left( 1- \frac{1}{e_Q} \right) n_{d,Q,V} \!\right) T^m \nonumber \\ G_V(T)&=\sum _{m=0}^{\infty } \sum _Q \sum _{d=0}^{e_Q-1} n_{d,Q,V}\left\langle \frac{m-d-1}{e_Q} \right\rangle T^m. \end{aligned}$$

(2)

Lemma 1

$$ F_V(T) = \frac{3T-1}{(1-T)^2}(g_Y-1) \dim V+\frac{2T-1}{(1-T)^2} \dim V \sum _Q \left( 1- \frac{1}{e_Q} \right) . $$

Proof

The result follows from the well-known formulas

$$ \sum _{m=0}^\infty (2m-1)T^m=\frac{3T-1}{(1-T)^2},\;\text { and } \sum _{m=0}^\infty (m-1)T^m=\frac{2T-1}{(1-T)^2}, $$

as well as the fact that $\displaystyle \sum _{d=0}^{e_Q-1} n_{d,Q,V}=\dim V$.

To compute $G_V(T)$, we first prove the following auxiliary lemma.

Lemma 2

For $A\in \mathbb {Z}$ and $1<e\in \mathbb {N}$, we have that

$$\begin{aligned} \sum _{m=0}^{\infty } \left\langle \frac{m+A}{e} \right\rangle T^m=\frac{T}{e(1-T)^2} + \frac{v_A}{e(1-T)} - \frac{T^{e-v_A}}{(1-T^{e})(1-T)}, \end{aligned}$$

where $v_A$ is the remainder of the division of A by e.

Proof

Recall that $\langle x \rangle = x- \left\lfloor x\right\rfloor$. Write $m=\pi e+v$ and $A=\pi _Ae+v_A$, for $0 \le v,v_A < e$ and $\pi ,\pi _A \in \mathbb {Z}$. Then

$$ \left\langle \frac{m+A}{e} \right\rangle = \frac{v+v_A}{e} - \left\lfloor \frac{v+v_A}{e} \right\rfloor , $$

and thus

$$\begin{aligned} \sum _{m=0}^{\infty } \left\langle \frac{m+A}{e} \right\rangle T^m= & {} \sum _{v=0}^{e-1} \sum _{\pi =0}^{\infty } \left( \frac{v+v_A}{e}- \left\lfloor \frac{v+v_A}{e} \right\rfloor \right) \left( T^e\right) ^\pi T^v\\= & {} \frac{1}{1-T^e} \sum _{v=0}^{e-1} \left( \frac{v+v_A}{e}- \left\lfloor \frac{v+v_A}{e} \right\rfloor \right) T^v. \end{aligned}$$

Next, we remark that since

$$ \left\lfloor \frac{v+v_A}{e} \right\rfloor = {\left\{ \begin{array}{ll} 0 &{} \text { if } 0\le v+v_A \le e-1 \\ 1 &{} \text { if } e\le v+v_A < 2e \end{array}\right. }, $$

we have that

$$\begin{aligned} \sum _{v=0}^{e-1} \left( \frac{v+v_A}{e}- \left\lfloor \frac{v+v_A}{e} \right\rfloor \right) T^v= & {} \sum _{v=0}^{e-1} \frac{v+v_A}{e} T^v - \sum _{v=e-v_A}^{e-1} T^v\\= & {} \frac{1}{e} \sum _{v=0}^{e-1} v T^v + \frac{v_A}{e} \sum _{v=0}^{e-1} T^v - T^{e-v_A}\sum _{v=0}^{v_A-1} T^v. \end{aligned}$$

Each of the three sums is given by

$$\begin{aligned} \frac{1}{e} \sum _{v=0}^{e-1} v T^v&= \frac{eT^{e+1}-T^{e+1}-eT^e+T}{e(1-T)^2} =-\frac{T^e}{(1-T)}+\frac{T(1-T^e)}{e(1-T)^2}, \nonumber \\ \frac{v_A}{e} \sum _{v=0}^{e-1} T^v&= \frac{v_A(1-T^e)}{e(1-T)}, \nonumber \\ T^{e-v_A}\sum _{v=0}^{v_A-1} T^v&= T^{e-v_A} \frac{1-T^{v_A}}{1-T} = \frac{T^{e-v_A}}{1-T} - \frac{T^e}{1-T}. \end{aligned}$$

Observe that the first term of the first sum cancels out with the second term of the third sum, and thus

$$\begin{aligned} \sum _{m=0}^{\infty } \left\langle \frac{m+A}{e} \right\rangle T^m&= \frac{1}{1-T^e} \left( \frac{T(1-T^e)}{e(1-T)^2} + \frac{v_A(1-T^e)}{e(1-T)} - \frac{T^{e-v_A}}{1-T} \right) \\&= \frac{T}{e(1-T)^2} + \frac{v_A}{e(1-T)} - \frac{T^{e-v_A}}{(1-T^e)(1-T)}. \end{aligned}$$

Corollary 1

Let $G_V(T)$ be as in Eq. (2) and $f_{Q,V}(T)=\sum _{d=0}^{e_Q-1}n_{d,Q,V}T^d$. Then

$$\begin{aligned}&G_V(T) = \frac{\dim V\cdot T}{(1-T)^2}\sum _{Q}\frac{1}{e_Q} + \frac{\dim V}{1-T} \sum _{Q} \left( 1-\frac{1}{e_Q}\right) \\&\qquad \qquad \qquad \qquad \qquad \quad \;\; -\frac{1}{1-T}\sum _Q \frac{f'_{Q,V}(1)}{e_Q} - \frac{T}{1-T} \sum _{Q} \left( \frac{f_{Q,V}(T)}{1-T^{e_Q}} \right) . \end{aligned}$$

Proof

Observe that if $0\le d\le e_{Q}-1$, the remainder of the division of $A=-d-1$ by $e_{Q}$ is $v_A=e_Q-d-1$. Thus, applying Lemma 2 for $A=-d-1$ we obtain

$$\begin{aligned}&\sum _{m=0}^{\infty } \sum _{Q} \sum _{d=0}^{e_Q-1} n_{d,Q,V} \left\langle \frac{m-d-1}{e_Q} \right\rangle T^m = \sum _{Q} \sum _{d=0}^{e_Q-1} n_{d,Q,V} \sum _{m=0}^{\infty } \left\langle \frac{m-d-1}{e_Q} \right\rangle T^m \\&= \sum _{Q} \sum _{d=0}^{e_Q-1} n_{d,Q,V} \left( \frac{T}{e_Q(1-T)^2} + \frac{e_Q-d-1}{e_Q(1-T)} - \frac{T^{d+1}}{(1-T^{e_Q})(1-T)} \right) \\&= \sum _{Q} \sum _{d=0}^{e_Q-1} \frac{n_{d,Q,V}T}{e_Q(1-T)^2} + \left( 1-\frac{1}{e_Q}\right) \frac{n_{d,Q,V}}{1-T} - \frac{n_{d,Q,V}d}{e_Q(1-T)} - \frac{n_{d,Q,V}T^{d+1}}{(1-T^{e_Q})(1-T)} \\&= \sum _{Q} \frac{f_{Q,V}(1)T}{e_Q(1-T)^2} + \left( 1-\frac{1}{e_Q}\right) \frac{f_{Q,V}(1)}{1-T} - \frac{f'_{Q,V}(1)}{e_Q(1-T)} - \frac{f_{Q,V}(T)T}{(1-T^{e_Q})(1-T)}. \end{aligned}$$

Using again the fact that $f_{Q,V}(1)=\displaystyle \sum _{d=0}^{e_Q-1} n_{d,Q,V}=\dim V$ gives the desired result.

Proof

(of Theorem 3). Let $F_V(T)$ and $G_V(T)$ be as in Eq. (2). By Lemma 1 and Corollary 1 we have that

$$\begin{aligned} F_V(T)= & {} \frac{3T-1}{(1-T)^2}(g_Y-1) \dim V+\frac{2T-1}{(1-T)^2} \dim V \sum _Q \left( 1- \frac{1}{e_Q} \right) , \\ G_V(T)= & {} \frac{\dim V\cdot T}{(1-T)^2}\sum _{Q}\frac{1}{e_Q} + \frac{\dim V}{1-T} \sum _{Q} \left( 1-\frac{1}{e_Q}\right) \\&- \frac{1}{1-T}\sum _Q \frac{f'_{Q,V}(1)}{e_Q} - \frac{T}{1-T} \sum _{Q} \left( \frac{f_{Q,V}(T)}{1-T^{e_Q}} \right) . \end{aligned}$$

Adding the second term of $F_V(T)$ to the second term of $G_V(T)$ gives

$$\begin{aligned}&\frac{(2T-1)\dim V}{(1-T)^2} \sum _{Q} \left( 1-\frac{1}{e_Q}\right) + \frac{\dim V}{1-T} \sum _{Q} \left( 1-\frac{1}{e_Q}\right) \\&\qquad \qquad = \frac{\dim V\cdot T}{(1-T)^2} \sum _{Q} \left( 1-\frac{1}{e_Q}\right) = \frac{\dim V\cdot T}{(1-T)^2}\#\mathscr {B}-\frac{\dim V\cdot T}{(1-T)^2}\sum _{Q}\frac{1}{e_Q}, \end{aligned}$$

and the last term above cancels out with the first term of $G_V(T)$. Thus

$$\begin{aligned}&F_V(T)+G_V(T)= \frac{3T-1}{(1-T)^2}(g_Y-1) \dim V + \frac{T}{(1-T)^2}\dim V |\mathscr {B}|\\&\qquad \qquad \qquad \qquad \qquad \qquad \quad \;\;\, - \frac{1}{1-T}\sum _Q \frac{f'_{Q,V}(1)}{e_Q} - \frac{T}{1-T} \sum _{Q} \frac{f_{Q,V}(T)}{1-T^{e_Q}} . \end{aligned}$$

As a corollary we obtain the below algorithm.

There are two advantages of Algorithm 2 over Algorithm 1. Firstly it can be used in the cases in which explicit k-bases for polydifferentials are not known; secondly the inner products of step 3(b) are taken over the decomposition groups $G_P$ which are strictly smaller than the full automorphism group G. On the other hand, its disadvantages are that one needs to compute the ramification data of the cover $\pi :X\rightarrow Y$, a problem which is wide open in its full generality, and that computing the multiplicities $n_{d,Q,V}$ is not always a straightforward task. We shall demonstrate how this is done in the next section by applying our results to Fermat curves.

3 The Case of Fermat Curves

Let $F_n$ be a Fermat curve with affine model $x^n+y^n+1=0$, defined over an algebraically closed field k of characteristic $p\ge 0$. We assume that $n\ge 4,\;p>3$ and $n-1$ is not a power of p. To describe the automorphism group $G=\mathrm{Aut}_k(X)$, we write

$$\begin{aligned} A:=\mathbb {Z}/n\mathbb {Z}\times \mathbb {Z}/n\mathbb {Z}= & {} \{ \sigma _{\alpha ,\beta }:0\le \alpha ,\beta \le n-1 \}\\ S_3= & {} \langle s,t: s^3=t^2=1,\;tst=s^{-1}\rangle =\{1,s,s^2,t,st,ts \}. \end{aligned}$$

and note that $S_3$ acts on A by conjugation as:

$$ { \begin{array}{|c|c|c|c|c|c|} \hline h\in S_3 &{} s &{} s^2 &{} t &{} ts &{}st \\ \hline \;h^{-1} \sigma _{\alpha ,\beta } h \;\;\; &{} \;\sigma _{\beta -\alpha ,-\alpha } \;\;\; &{}\; \sigma _{-\beta ,\alpha -\beta }\;\;\; &{}\; \sigma _{-\alpha ,\beta -\alpha }\;\;\; &{}\; \sigma _{\beta ,\alpha }\;\;\; &{} \;\sigma _{\alpha -\beta ,-\beta } \;\;\;\\ \hline \end{array} } $$

Remark 2

An automorphism $\sigma : F_n \rightarrow F_n$ acts on functions $f\in k(F_n)$ by $\sigma (f)=f\circ \sigma ^{-1}$. The group acts on the left on points, so $(\sigma _1 \sigma _2) P =\sigma _1 (\sigma _2 P)$, and the action on functions satisfies $(\sigma _1 \sigma _2 f)=f \circ (\sigma _1 \sigma _2)^{-1}= f \circ \sigma _2^{-1} \circ \sigma _1^{-1}= \sigma _1 (\sigma _2 f)$.

In [11] and [18] the authors prove that $F_n$ has genus $g=\frac{(n-1)(n-2)}{2}$, automorphism group $G=A\rtimes S_3$ and that the action of G on the function field $k\left( F_n\right)$, i.e., the field k(x, y) subject to the equation $x^n+y^n+1=0$, is given by

$$ { \begin{array}{|c|c|c|c|c|c|c|} \hline g\in G &{} \sigma _{\alpha ,\beta }&{} s &{} s^2 &{} t &{} ts &{}st \\ \hline \;g(x,y)\;\;\; &{}\;\left( \zeta _n^\alpha x,\zeta _n^\beta y\right) \;\;\; &{} \;\left( \frac{y}{x},\frac{1}{x}\right) \;\;\; &{}\; \left( \frac{1}{y},\frac{x}{y}\right) \;\;\; &{}\; \left( \frac{1}{x},\frac{y}{x}\right) \;\;\; &{}\; \left( y,x\right) \;\;\; &{} \;\left( \frac{x}{y},\frac{1}{y}\right) \;\;\;\\ \hline \end{array} } $$

where $\zeta _n$ is a fixed primitive n-th root of unity. The above gives us the second required input item for Algorithm 2. Regarding the first, we use the character table of G that was computed in [1, Proposition 3]. Recall that $S_3$ has three irreducible representations: the trivial representation, the sign representation and the standard representation, denoted by $\rho _\mathrm{triv},\rho _\mathrm{sgn}$ and $\rho _\mathrm{stan}$ respectively.

Proposition 1

The irreducible representations of G are given in the table below,

$$ { \begin{array}{|c|c|c|} \hline {\textbf {Rep.}} &{} {\textbf { Degree }} &{} {\textbf { Character }} \chi (\sigma _{\alpha ,\beta }x), \text { where } x\in S_3 \\ \hline \theta _{\frac{\nu n}{3},\frac{\nu n}{3},\rho } &{} 1 &{} \zeta ^{\frac{\nu n}{3}(\alpha +\beta )}\chi _\rho (x) \\ \hline \theta _{\frac{\nu n}{3},\frac{\nu n}{3},\rho _\mathrm {stan}} &{} 2 &{}\zeta ^{\frac{\nu n}{3}(\alpha +\beta )}\chi _\mathrm {stan}(x) \\ \hline \theta _{\kappa ,\kappa ,\rho } &{} 3 &{} {\left\{ \begin{array}{ll} \zeta ^{\kappa (\alpha +\beta )} + \zeta ^{\kappa (\alpha -2 \beta )} + \zeta ^{\kappa (\beta -2\alpha ) } &{}, \text { if } x =1 \\ \zeta ^{\kappa (\alpha +\beta )} \chi _{\rho }(x) &{}, \text { if } x=ts \\ \zeta ^{\kappa (\alpha -2 \beta )} \chi _{\rho }(x) &{}, \text { if } x=t \\ \zeta ^{\kappa (\beta -2 \alpha )} \chi _{\rho }(x) &{}, \text { if } x=st \\ 0 &{}, \text { if } x=s,s^2 \end{array}\right. } \\ \hline \theta _{\kappa ,\lambda , \rho _\mathrm {triv} } &{} 6 &{} {\left\{ \begin{array}{ll} \left( \begin{array}{l} \zeta ^{\kappa \alpha + \lambda \beta } + \zeta ^{-(\kappa +\lambda ) \alpha +\kappa \beta }+ \zeta ^{\lambda \alpha -(\kappa +\lambda ) \beta }+ \\ \zeta ^{\lambda \alpha + \kappa \beta }+ \zeta ^{-(\kappa +\lambda )\alpha +\lambda \beta }+ \zeta ^{\kappa \alpha - (\kappa +\lambda ) \beta } \end{array}\right) &{}, \!\text { if } x = 1 \\ 0 &{}, \!\text { if } x\ne 1 \end{array}\right. } \\ \hline \end{array} } $$

where $\nu \in \{0,1,2\},\;\rho \in \{\rho _\mathrm {triv},\rho _\mathrm {sgn}\},\;\kappa ,\lambda \in \mathbb {Z}/n\mathbb {Z},\;\kappa ,\lambda \ne \frac{\nu n}{3},\kappa \ne \lambda , \kappa \ne -2\lambda , \lambda \ne -2\kappa$ and the representations corresponding to $\kappa ,\lambda \in \{\frac{n}{3},\frac{2n}{3}\}$ appear only when $3\mid n$.

In what follows, we fix all primitive roots of unity to be compatible with the chosen $\zeta _n$, in the sense that if $n\mid i$, then $\zeta _i$ must satisfy $\zeta _i^{i/n}=\zeta _n$, whereas if $i \mid n$ then $\zeta _i=\zeta _n^{n/i}.$

Proposition 2

The quotient $F_n/G$ is isomorphic to $\mathbb {P}^1_{k}$, the branch locus of $F_n \rightarrow \mathbb {P}^1_{k}$ consists of three points $P_{\infty },P_{1}, P_{0}$. The points $Q_\infty =(\zeta _{2n},0), Q_1=(1,\root n \of {-2}),Q_0=(\zeta _{6n}^4,\zeta _{6n}^2)$ of $F_n$ lie above each of the three mentioned points, and their isotropy groups and monodromy generators are given in the following table.

$$ { \begin{array}{|c|c|c|} \hline \text {point} &{} \text {group} &{} \text {monodromy} \\ \hline Q_\infty =(\zeta _{2n},0) &{} \mathbb {Z}/2n \mathbb {Z} &{} \sigma _{1,1}t \\ \hline Q_1=(1,\root n \of {-2}) &{} \mathbb {Z}/2 \mathbb {Z} &{} t \\ \hline Q_0=(\zeta _{6n}^4,\zeta _{6n}^2) &{} \mathbb {Z}/3 \mathbb {Z} &{} \sigma _{-1,-1}s^2\\ \hline \end{array} } $$

Proof

The proof can be found at the appendix.

The above implies that, in the notation of Theorem 3, we have $g_Y=0$ and $|\mathscr {B}|=3$. Thus the third and fourth term of $H_{S_X,V}$ simplify as follows

$$ \frac{3T-1}{(1-T)^2}(g_Y-1) \dim V + \frac{T}{(1-T)^2}\dim V |\mathscr {B}| = \frac{\dim V}{(1-T)^2}. $$

Theorem 4

With the above notation, we have that

$$ H_{S_X,V}(T)= N_{V^*,1}+\delta _VT+\frac{\dim V}{(1-T)^2} - \frac{1}{1-T}\sum _Q \frac{f'_{Q,V}(1)}{e_Q} - \frac{T}{1-T} \sum _{Q} \frac{f_{Q,V}(T)}{1-T^{e_Q}} , $$

where $\delta _V=1$ for $V=V_\mathrm{triv}$ and 0 otherwise, the polynomials $f_{Q,V}(T)$ are given in the table below

$$ { \!\!\!\! \begin{array}{|c|c|c|c|} \hline V\in \mathrm{Irr(G)} &{} f_{Q_\infty ,V}(T) &{} f_{Q_0,V}(T) &{} f_{Q_1,V}(T) \\ \hline \theta _{0,0,\rho _\mathrm{triv}} &{} 1 &{} 1 &{} 1 \\ \hline \theta _{0,0,\rho _\mathrm{sgn}} &{} T^{n} &{} 1 &{} T \\ \hline \theta _{\frac{ n}{3},\frac{ n}{3},\rho _\mathrm{triv}} &{} T^{\frac{4 n}{3}} &{} T &{} 1 \\ \hline \theta _{\frac{ n}{3},\frac{ n}{3},\rho _\mathrm{sgn}} &{} T^{\frac{n}{3}} &{} T &{} T \\ \hline \theta _{\frac{2 n}{3},\frac{2 n}{3},\rho _\mathrm{triv}} &{} T^{\frac{2 n}{3}} &{} T^2 &{} 1 \\ \hline \theta _{\frac{2\nu n}{3},\frac{2\nu n}{3},\rho _\mathrm{sgn}} &{} T^{\frac{5n}{3}} &{} T^2 &{} T \\ \hline \theta _{\frac{\nu n}{3},\frac{\nu n}{3},\rho _\mathrm {stan}} &{} T^{\frac{\nu n}{3}}+T^{n+\frac{\nu n}{3}} &{} 1+T+T^2-T^\nu &{} 1+T \\ \hline \theta _{\kappa ,\kappa ,\rho _{\mathrm {triv}}} &{} T^{\kappa }+T^{n+\kappa }+T^{[-2\kappa ]_n} &{} 1+T+T^2 &{} 2+T \\ \hline \theta _{\kappa ,\kappa ,\rho _{\mathrm {sgn}}} &{} T^{\kappa }+T^{n+\kappa }+T^{[-2\kappa ]_n+n} &{} 1+T+T^2 &{} 1+2T \\ \hline \theta _{\kappa ,\lambda , \rho _\mathrm {triv} } &{} \begin{array}{l} T^\kappa + T^\lambda + T^{n+\kappa }+ T^{n+\lambda }\\ +T^{[-(\kappa +\lambda )]_n}+ T^{[-(\kappa +\lambda )]_n+n} \end{array} &{} 2(1+T+T^2) &{} 3+3T \\ \hline \end{array} } $$

and $[x]_n$ denotes the smallest non-negative remainder of the division of x by n.

The proof of the above will be given separately for each of $f_{Q_\infty ,V},\;f_{Q_0,V},\;f_{Q_1,V}$, by considering all irreducible representations of Proposition 1. To do so, one needs to calculate first the multiplicities $n_{d,Q,V},\;Q\in \{Q_\infty ,Q_0,Q_1\}$ as follows:

• 1. For each Q, write $G_Q=\{ \sigma _Q^i:0\le i\le e_Q\}$ where $\sigma _Q$ is the local monodromy and $e_Q$ is the ramification index, both taken from Proposition 2.

• 2. For each $0\le i\le e_Q$, find $\sigma _{\alpha _i,\beta _i}\in A$ and $x_i\in S_3$ such that $\sigma _Q^i=\sigma _{\alpha _i,\beta _i}x_i$.

• 3. Fix a primitive root of unity $\zeta _{e_Q}$ compatible with $\zeta _n$ as discussed above.

• 4. Compute $n_{d,Q ,V} =\langle \mathrm{Res}^G_{G_{Q}}\left( \chi _V\right) ,\chi _{Q}^d \rangle =\sum _{i=0}^{e_Q-1}\chi _V\left( \sigma _P^i\right) \zeta _{e_Q}^{-id}$.

3.1 The Polynomials $f_{Q_\infty ,V}(T)$

By Proposition 2, $Q_\infty =(\zeta _{2n},0)$ and $G_{Q_\infty }$ is generated by the monodromy element $\sigma _{1,1}t$. Since $\left( \sigma _{1,1}t\right) ^2=\sigma _{0,1}$, we have that

$$ (\sigma _{1,1}t)^{2k}=\sigma _{0,k} \text { and } (\sigma _{1,1}t)^{2k+1}=\sigma _{1,k+1}t, \quad \text { for } 0\le k\le n-1, $$

and thus, for $0\le d\le 2n-1$, we have that

$$\begin{aligned} n_{d,Q_{\infty },V}= & {} \langle \mathrm{Res}^G_{G_{Q_\infty }}\left( \chi _V\right) ,\chi _{Q_\infty }^d \rangle \\= & {} \displaystyle \frac{1}{2n}\sum _{\ell =0}^{n-1} \zeta _{2n}^{2\ell (-d)} \chi _V(\sigma _{0,\ell }) + \frac{1}{2n}\sum _{\ell =0}^{n-1} \zeta _{2n}^{(2\ell +1)(-d)} \chi _V(\sigma _{1,\ell +1}t). \end{aligned}$$

When $V=\theta _{\frac{\nu n}{3},\frac{\nu n}{3},\rho }$, $\rho \in \{\rho _{\mathrm {triv}},\rho _{\mathrm {sgn}} \}$, Proposition 1 gives

$$ \chi _V(\sigma _{0,\ell })=\zeta _n^{\frac{\nu n\ell }{3}}=\zeta _{2n}^{\frac{2\nu n\ell }{3}} \text { and } \chi _V(\sigma _{1,\ell +1}t)=\zeta _n^{\frac{\nu n(\ell +2)}{3}}\chi _{\rho }(t)=\zeta _{2n}^{\frac{2\nu n(\ell +2)}{3}}\chi _{\rho }(t). $$

Thus, we compute

$$\begin{aligned} n_{d,Q_{\infty },V}= & {} \frac{1}{2n}\sum _{\ell =0}^{n-1} \zeta _{2n}^{2\ell (-d)} \zeta _{2n}^{\frac{ 2\nu n \ell }{3}} + \frac{1}{2n}\sum _{\ell =0}^{n-1} \zeta _{2n}^{(2\ell +1)(-d)} \zeta _{2n}^{\frac{2\nu n(\ell +2)}{3}}\chi _{\rho }(t) \\= & {} \frac{1}{2n}\sum _{\ell =0}^{n-1} \zeta _{2n}^{2\ell (\frac{ \nu n}{3}-d)} + \frac{1}{2n}\sum _{\ell =0}^{n-1} \zeta _{2n}^{2\ell (\frac{ \nu n}{3}-d)+\left( \frac{ 4\nu n}{3}-d\right) } \chi _{\rho }(t)\\= & {} {\left\{ \begin{array}{ll} \frac{1}{2}+\frac{1}{2} \zeta _{2n}^{\frac{ 4\nu n}{3}-d}\chi _{\rho }(t) &{} \text {, if } n\mid \frac{\nu n}{3}-d\\ 0 &{} \text {, otherwise. } \end{array}\right. } \end{aligned}$$

The only two values of d such that $0 \le d \le 2n-1$ and $n\mid \frac{\nu n}{3}-d$ are $d=\frac{\nu n}{3}$ and $d=n+\frac{\nu n}{3}$. So

$$ f_{Q_{\infty },V}(T)= \left( \frac{1}{2}+\frac{1}{2}\zeta _{2n}^{\nu n}\chi _{\rho }(t) \right) T^{\frac{\nu n}{3}} + \left( \frac{1}{2}+\frac{1}{2}\zeta _{2n}^{(\nu -1)n}\chi _{\rho }(t) \right) T^{n+\frac{\nu n}{3}}. $$

Notice that for $\nu \in \{0,1,2\}$ we always have that $\{\zeta _{2n}^{\nu n},\zeta _{2n}^{(\nu -1)n}\}=\{-1,1\}.$

When $V=\theta _{\frac{\nu n}{3},\frac{\nu n}{3},\rho _{\mathrm {stan}}}$, Proposition 1 gives

$$ \chi _V(\sigma _{0,\ell })=2\zeta _{2n}^{\frac{2\nu n\ell }{3}} \text { and } \chi _V(\sigma _{1,\ell +1}t)=0. $$

We compute as above

$$ n_{d,Q_\infty ,V}= \frac{1}{n}\sum _{\ell =0}^{n-1} \zeta _{2n}^{2\ell \left( \frac{\nu n}{ 3}-d \right) } + 0 = {\left\{ \begin{array}{ll} 1 &{} \text {, if } n\mid \frac{\nu n}{3}-d\\ 0 &{} \text {, otherwise } \end{array}\right. }, $$

and thus

$$ f_{Q_\infty ,V}(T)=T^{\frac{\nu n}{3}}+T^{n+\frac{\nu n}{3}}. $$

When $V=\theta _{\kappa ,\kappa ,\rho }$, $\rho \in \{\rho _{\mathrm {triv}},\rho _{\mathrm {sgn}}\}$, Proposition 1 gives

$$\begin{aligned} \chi _V(\sigma _{0,\ell })= & {} 2\zeta _n^{\kappa \ell }+\zeta _n^{-2\kappa \ell } =2\zeta _{2n}^{2\kappa \ell }+\zeta _{2n}^{-4\kappa \ell } \text { and }\\ \chi _V(\sigma _{1,\ell +1}t)= & {} \zeta _n^{-\kappa (2\ell +1)}\chi _{\rho }(t) =\zeta _{2n}^{-2\kappa (2\ell +1)}\chi _{\rho }(t). \end{aligned}$$

We then have

$$\begin{aligned} n_{d,Q_\infty ,V}= & {} \frac{1}{n} \sum _{\ell =0}^{n-1} \zeta _{2n}^{2\ell (\kappa -d)} + \frac{1}{2n}\sum _{\ell =0}^{n-1} \zeta _{2n}^{-2 \ell (2 \kappa +d)} + \frac{1}{2n} \sum _{\ell =0}^{n-1} \zeta _{2n}^{-2\ell (2 \kappa +d) }\zeta _{2n}^{-(2 \kappa +d)} \chi _{\rho }(t) \\= & {} {\left\{ \begin{array}{ll} 1 &{} \text {, if } n \mid \kappa -d \\ \frac{1}{2}+\frac{1}{2} \zeta _{2n}^{-(2 \kappa +d)} \chi _{\rho }(t) &{} \text {, if } n\mid 2 \kappa +d \\ 0 &{} \text { otherwise.} \end{array}\right. } \end{aligned}$$

The first case gives $d=\kappa$ or $d=n+\kappa$, while the second gives $d=[-2k]_n$ or $d=n+[-2k]_n$ so

$$ f_{Q_\infty ,V}(T)=T^{\kappa }+T^{n+\kappa }+ \left( \frac{1}{2}+\frac{1}{2}\chi _\rho (t) \right) T^{[-2\kappa ]_n} + \left( \frac{1}{2}-\frac{1}{2}\chi _\rho (t) \right) T^{[-2 \kappa ]_n+n}. $$

Finally, for $V=\theta _{\kappa ,\lambda ,\rho _{\mathrm {triv}}}$ we have that

$$ \chi _V(\sigma _{0,\ell }) = 2 \zeta _{2n}^{2 \lambda \ell } + 2\zeta _{2n}^{2\kappa \ell } + 2\zeta _{2n}^{-2(\kappa + \lambda )\ell } \text { and } \chi _V(\sigma _{1,\ell +1}t)=0. $$

Thus

$$\begin{aligned} n_{d,Q_\infty ,V}= & {} \frac{1}{n}\sum _{\ell =0}^{n-1} \left( \zeta _{2n}^{2\ell (\lambda -d)} + \zeta _{2n}^{2\ell (\kappa -d)} + \zeta _{2n}^{-2\ell (\kappa + \lambda +d)} \right) \\= & {} {\left\{ \begin{array}{ll} 1 &{} \text {, if } n\mid \lambda -d \text { or }n\mid \kappa -d \text { or } n\mid \kappa + \lambda +d \\ 0 &{} \text {, otherwise} \end{array}\right. } \end{aligned}$$

and so

$$ f_{Q_\infty ,V}(T)= T^\kappa + T^\lambda + T^{n+\kappa }+ T^{n+\lambda }+ T^{[-(\kappa +\lambda )]_n}+ T^{[-(\kappa +\lambda )]_n+n}. $$

3.2 The Polynomials $f_{Q_0,V}(T)$

By Proposition 2, $Q_0=(\zeta _{6n}^4,\zeta _{6n}^2)$ and $G_{Q_0}$ is generated by the monodromy element $\sigma _{-1,-1}s^2$. For $d=0,1,2$ we have

$$\begin{aligned} n_{d,Q_0,V}= & {} \langle \mathrm{Res}^G_{G_{Q_0}}\left( \chi _V\right) ,\chi _{Q_0}^d \rangle \\= & {} \frac{1}{3} \left( \chi _V(1)+ \zeta _3^{-d} \chi _V(\sigma _{-1,-1}s) + \zeta _3^{2(-d)} \chi _V\big ( (\sigma _{-1,-1}s^2 )^2 \big ) \right) . \end{aligned}$$

For both $V=\theta _{\frac{\nu n}{3},\frac{\nu n}{3},\rho }$, $\rho \in \{\rho _{\mathrm {triv}},\rho _{\mathrm {sgn}}\}$ we have

$$\begin{aligned} n_{d,Q_0,V}= & {} \frac{1}{3} \left( 1 + \zeta _3^{-d}\zeta _n^{- \frac{2\nu n}{3}}+ \zeta _3^{2(-d)}\zeta _n^{- \frac{4\nu n}{3}}\right) = \frac{1}{3}\left( 1 + \zeta _3^{-d-2\nu }+ \zeta _3^{-2d-4\nu }\right) \\= & {} {\left\{ \begin{array}{ll} 1 &{} \text {, if } d \equiv {-2\nu }\;\mathrm{mod}\; 3\;{\equiv \nu }\;\mathrm{mod}\; 3\\ 0 &{} \text {, otherwise.} \end{array}\right. } \end{aligned}$$

and so $f_{Q_0,V}(T)=T^{[{\nu }]_3}=T^\nu$.

When $V=\theta _{\frac{\nu n}{3},\frac{\nu n}{3},\rho _{\mathrm {stan}}}$ we have

$$\begin{aligned} n_{d,Q_0,V}= & {} \frac{1}{3} \left( 2 - \zeta _3^{-d{-2\nu }} -\zeta _3^{-2 d{-4\nu }} \right) = \frac{1}{3} \left( 3 - 1 - \zeta _3^{-d-2\nu }- \zeta _3^{-2d -4\nu } \right) \\= & {} {\left\{ \begin{array}{ll} 1 &{} \text {, if } d\not \equiv -2\nu \;\mathrm{mod}\; 3\\ 0 &{} \text {, otherwise} \end{array}\right. } \end{aligned}$$

and so $f_{Q_0,V}(T)=1+T+T^2-T^{[-2\nu ]_3}=1+T+T^2-T^{\nu }$.

When $V=\theta _{\kappa ,\kappa ,\rho }$, $\rho \in \{\rho _{\mathrm {triv}},\rho _{\mathrm {sgn}}\}$, we have $n_{d,Q_0,V}=1$ for $d\in \{0,1,2\}$ and so $f_{Q_0,V}=1+T+T^2$.

When $V=\theta _{\kappa ,\lambda ,\rho _{\mathrm {triv}}}$, $n_{d,Q,V}=2$ for $d\in \{0,1,2\}$ and so $f_{Q_0,V}=2(1+T+T^2)$.

3.3 The Polynomials $f_{Q_1,V}(T)$

By Proposition 2, $Q_1=(1,\root n \of {-2})$ and $G_{Q_1}$ is generated by the monodromy element t. For $d\in \{0,1\}$ we have

$$\begin{aligned} n_{d,Q_1,V}= & {} \langle { \mathrm{Res}^G_{G_{Q_1}}\left( \chi _V\right) ,\chi _{Q_1}^d } \rangle \\= & {} \frac{1}{2} \left( \chi _V(1) + (-1)^{-d} \chi _V(t) \right) = \frac{1}{2} \left( \dim V + (-1)^{d} \chi _V(t) \right) \end{aligned}$$

When $V=\theta _{\frac{\nu n}{3},\frac{\nu n}{3},\rho }$, $\rho \in \{\rho _{\mathrm {triv}},\rho _{\mathrm {sgn}},\rho _\mathrm{stan}\}$, we get

$$ f_{Q_1,V}(T)= {\left\{ \begin{array}{ll} 1 &{} \text {, if } \rho =\rho _{\mathrm {triv}} \\ T &{} \text {, if }\rho =\rho _{\mathrm {sgn}} \\ 1+T &{} \text {, if }\rho =\rho _{\mathrm {stan}}. \end{array}\right. } $$

When $V=\theta _{\kappa ,\kappa ,\rho }$, $\rho \in \{\rho _{\mathrm {triv}},\rho _{\mathrm {sgn}}\}$, we get

$$ f_{Q_1,V}(T)= {\left\{ \begin{array}{ll} 2+T &{} \text {, if } \rho =\rho _{\mathrm {triv}} \\ 1+2T &{} \text {, if }\rho =\rho _{\mathrm {sgn}}. \end{array}\right. } $$

Finally, when $V=\theta _{\kappa ,\lambda ,\rho _{\mathrm {triv}}},\; f_{Q_1,V}(T)=3+3T$.

4 Implementation and Examples

Let $F_n$ be a Fermat curve over k, with the assumptions on n and k as in the previous section. By Petri’s Theorem 1, there exists a short exact sequence

$$\begin{aligned} 0\rightarrow I_X:=\ker \phi \hookrightarrow S:=\mathrm{Sym}\left( H^0(X,\varOmega _X)\right) \overset{\phi }{\twoheadrightarrow }S_X:=\bigoplus _{m=0}^\infty H^0(X,\varOmega _X^{\otimes m})\rightarrow 0, \end{aligned}$$

which is split over kG, since $\mathrm{char}(k)\not \mid |G|$.

In this section, we present our Sage [16] program³$^,$⁴, which, as mentioned in the introduction, computes for each $V\in \mathrm{Irr}(G)$:

• 1. $H_{S_X,V}(T)$, by implementing the formulas of Theorem 4.

• 2. $H_{S,V}(T)$, by implementing Molien’s formula (see Eq. 1).

• 3. $H_{I_X,V}(T)=H_{S,V}(T)-H_{S_X,V}(T)$, by subtracting the two above results.

The computation of $H_{S_X,V}(T)$ follows by Theorem 4. First we implement the difference $H_{S_X,V}(T)-N_{V^*,1}-\delta _V T$ for each $V\in \mathrm{Irr(G)}$. Then we read each $N_{V^*1}$ from the implementation of $H_{S_X,V^*}(T)-N_{V,1}-\delta _{V^*} T$ and add T if V is trivial. We remark that implementing the algorithm for $n=6$ we retrieve same results as in [1, Table 1, pg. 18], where we computed the kG-structure of $H^0(X,\varOmega _X^{\otimes m})$ using an alternative approach. For example, the series for $V=\theta _{0,1,\mathrm {triv}}$ is

$$ H_{S_X,V}(T)=\frac{T^{3}}{T^{5} - 2 \, T^{4} + T^{3} + T^{2} - 2 \, T + 1}. $$

To implement Molien’s formula, it is required to input the character table of G and the representation $G\rightarrow \mathrm{GL}\left( H^0(X,\varOmega _X) \right)$. The former is taken directly from Proposition 1, while the latter was implemented using the action of G on a basis $\{\omega _{i,j}\}$ of $H^0(X,\varOmega _X)$ which we computed in [1, Prop. 6]:

$$ \small { \begin{array}{|c|c|c|c|c|c|} \hline \sigma _{\alpha ,\beta }(\omega _{i,j}) &{} s(\omega _{i,j}) &{} t (\omega _{i,j})&{} ts (\omega _{i,j})&{} st (\omega _{i,j})&{} s^2(\omega _{i,j}) \\ \hline \zeta ^{\alpha (i+1)+\beta (j+1)}\omega _{i,j}&{}\omega _{n-3-(i+j),i} &{} -\omega _{n-3-(i+j),j} &{} -\omega _{j,i} &{}-\omega _{i,n-3-(i+j)} &{}\omega _{j,n-3-(i+j)} \\ \hline \end{array} } $$

The output is much more complicated than $H_{S_X,V}(T)$: for instance when $n=6$ and $V=\theta _{0,1,\mathrm {triv}}$ we obtain a rational function with numerator of degree 18 and denominator of degree 30.

The final step is to compute the equivariant Hilbert series of $I_X$ using Petri’s Theorem 1. For $n=6$ and $V=\theta _{0,1,\mathrm {triv}},\;H_{I_X,V}(T)=H_{S,V}(T)-H_{S_X,V}(T)$ has power series expansion

$$\begin{aligned} 8\,T^{3} + 20\,T^{4} + 49\,T^{5} + 130\,T^{6} + 319\,T^{7} + 667\,T^{8} + 1363\, T^{9} + 2557\, T^{10} + \text {higher order terms}. \end{aligned}$$

The interpretation is that the representation $\theta _{0,1,\mathrm {triv}}$ appears, for example, 2557 times in the decomposition of the degree 10 graded piece of $I_{X}$ into irreducible summands.

5 Appendix - The Ramification Data of Fermat Curves

In this section we give the details for the proof of Proposition 2. We shall work over $k=\mathbb {C}$ for simplicity, even though the arguments are valid over any algebraically closed field of characteristic prime to the order of G. Recall that all roots of unity are fixed, as per the discussion preceding Proposition 2.

The Fermat curve can be seen as double Kummer cover of the projective line $\mathbb {P}^1$. We will work with Galois extensions of the corresponding function fields and in this way we have the Kummer extension of function fields $\mathbb {C}(F_n)/\mathbb {C}(x)$, where $\mathbb {C}(F_n)$ is the extension obtained by the rational function field $\mathbb {C}(x)$ by adjoining the quantity $y=(-1-x^{n})^{\frac{1}{n}}$. Then we can consider the cyclic extension of function fields $\mathbb {C}(x)/\mathbb {C}(x^n)$. The ramification in such extensions is well known, see for example [9, 10], namely there is ramification in the cover $\mathbb {C}\left( F_n\right) /\mathbb {C}\left( F_n\right) ^{A}$ over the points $x^n=-1,x^n=0, x^n=\infty$, where $A=\mathbb {Z}/n\mathbb {Z}\times \mathbb {Z}/n\mathbb {Z}$.

Since $G=\mathrm {Aut}(F_n)=A \rtimes S_3$, the Galois extension $\mathbb {C}(F_n)/\mathbb {C}(F_n)^G$ corresponding to the cover $F_n \rightarrow F_n/G$ has the intermediate subfield $\mathbb {C}(F_n)^A=\mathbb {C}(x^n)$, and $\mathbb {C}(F_n)^A/\mathbb {C}(F_n)^G$ is Galois with Galois group the symmetric group $S_3$. Moreover, the extension $\mathbb {C}(F_n)^A/\mathbb {C}(F_n)^G$ corresponds to a ramified cover $\mathbb {P}^1 \rightarrow \mathbb {P}^1$ ramified over three points. Such covers can be explained in terms of the j invariant, see [19]. Indeed, if we set $X=-x^n$ then the group $S_3$ can be realized by the six Möbius automorphisms:

$$ X\mapsto \left\{ X, \frac{1}{X}, 1-X, \frac{1}{1-X}, \frac{X}{1-X}, \frac{X-1}{X} \right\} . $$

The fixed points of these maps are given in the following table:

transform	order	equation	fixed points
$\frac{1}{X}$	2	$X^2-1=0$	$1,-1$
$1-X$	2	$2X-1$	$\frac{1}{2}, \infty$
$\frac{X}{X-1}$	2	$X^2-2X=0$	0, 2
$\frac{1}{1-X}, \frac{X-1}{X}$	3	$X^2-X+1=0$	$\zeta _6,\frac{1}{\zeta _6}$

and the function

$$ j(X)=\frac{4}{27} \frac{(X^2-X+1)^3}{X^2(X-1)^2} $$

is a generator of the fixed field $\mathbb {C}(X)^{S_3}=\mathbb {C}(F_n)^G=\mathbb {C}(j)$. The fixed points of the $S_3$-cover $\mathbb {P}^1 \rightarrow \mathbb {P}^1$ are $P_{(j=0)},P_{(j=1)},P_{(j=\infty )}$. The map j maps the fixed points as follows:

$$ \begin{array}{ccc} X&{} &{} j(X) \\ \hline 0,1,\infty &{} \longmapsto &{} \infty \\ -1,2,\frac{1}{2} &{} \longmapsto &{} 1 \\ \zeta _6,\frac{1}{\zeta _6} &{} \longmapsto &{} 0 \end{array} $$

In Fig. 1 we display the ramification diagram above the point $P_{(j=\infty )}$ and in Fig. 2 the respective diagram above the points $P_{(j=1)}$ and $P_{(j=0)}$. Note that in the first row we denote by $P_{i,i'}$ the i-th ramification point above $P_{(X=i')}$, for $i'\in \{0,1,\infty \}$, the labels in the vertical lines of the first column indicate the Galois groups, whereas in all other columns they indicate ramification indices.

Fig. 1.

Ramification diagram for $P_{(j=\infty )}$

Fig. 2.

Ramification diagram for $P_{(j=1)}$ and $P_{(j=\infty )}$

Each of the points $P_{(X=-1)},P_{(X=2)}, P_{(X=\frac{1}{2})},P_{(X=\zeta _6)},P_{(X=\frac{1}{\zeta _6})}$ has $n^2$ points in the Fermat curve. For instance the point $X=-x^n=\zeta _6$ is lifted to the points (x, y) where $x=(-\zeta _6)^{1/n}=\zeta _{n}^{\ell } \zeta _{2n} \zeta _{6 n}=\zeta _{6n}^{6\ell +4}$, for $0\le \ell \le n-1$, and similarly, $y^n=-1-x^n=-1+\zeta _6=\zeta _6^2$, since $\zeta _6^2-\zeta _6+1=0$. Therefore, for $0\le k\le n-1$, $y=\zeta _n^k \zeta _{6n}^2=\zeta _{6n}^{6k+2}$. This means that the set of points $\{(\zeta _{6n}^{6 \ell +4},\zeta _{6n}^{6k+2}): 0\le k,\ell < n\}$ are the $n^2$ points above the point $P_{(X=\zeta _6)}$.

We will now select an arbitrary point above each $P_{(j=\infty )}, P_{(j=1)}, P_{(j=0)}$ and for each such point we will find the cyclic subgroup and the monodromy element. Recall that by Remark 2, automorphisms $\sigma \in G$ act on functions $f\in \mathbb {C}\left( F_n\right)$ by $\sigma (f)=f\circ \sigma ^{-1}$.

Consider the point $Q_\infty =(\zeta _{2n},0)$ above $P_{(j=\infty )}$. The isotropy subgroup is a cyclic group of order 2n. For example we can verify that it is fixed by the element $\sigma _Q=\sigma _{1,1}t$. Further, since $\left( \sigma _{1,1}t\right) ^2=\sigma _{0,1}$, we have that

$$ (\sigma _{1,1}t)^{2k}=\sigma _{0,k} \text { and } (\sigma _{1,1}t)^{2k+1}=\sigma _{1,k+1}t, \quad \text { for } 0\le k\le n-1 $$

A local uniformizer at $Q_\infty$ is y, which is acted on by $(\sigma _{1,1}t)^2=\sigma _{0,1}$ by $y\mapsto \zeta _n y$. Hence, the monodromy element at $Q_\infty$ is $\sigma _{1,1}t$.

Consider the point $Q_1$ above $P_{(j=1)}$ given by affine coordinates $(1,\root n \of {-2})$, which is fixed by the automorphism t acting on functions as $t(x)=1/x,t(y)=y/x$. Since the decomposition group at $Q_1$ is a cyclic group of order 2 the monodromy at $Q_1$ is the element t.

A point $Q_0=(x_0,y_0)$ above $P_{(j=0)}$ is given by $X=\zeta _6$, that is, $x_0^n=-\zeta _6$, therefore $x_0=(- \zeta _6)^{1/n}=\zeta _{n}^{\ell } \zeta _{2n} \zeta _{6 n}=\zeta _{6n}^{6\ell +4},$ for $0\le \ell \le n-1$. Similarly $y_0^n=-1-x^n=-1+\zeta _6=\zeta _6^2$, since $\zeta _6^2-\zeta _6+1=0$. Therefore, $y_0=\zeta _n^k \zeta _{6n}^2=\zeta _{6n}^{6k+2}$, for $0\le k\le n-1$.

Let s be the automorphism acting on functions by $s(x)=y/x$, $s(y)=1/x$, so that $s^2(x)=1/y,s^2(y)=x/y$. Observe that the point with coordinates $(x_0,y_0)=\left( \zeta _{6n}^{4}, \zeta _{6n}^{2} \right)$ is sent by $\sigma _{1,1}s$ to $(x_0,y_0)$. Indeed,

$$ \left( \zeta _{6n}^{4},\zeta _{6n}^{2} \right) {\mathop {\longmapsto }\limits ^{s}} \left( \zeta _{6n}^{2-4},\zeta _{6n}^{-4} \right) {\mathop {\longmapsto }\limits ^{\sigma _{1,1}}} \left( \zeta _{6n}^{4},\zeta _{6n}^{2} \right) . $$

The function $x-x_0=x- \zeta _{6n}^4$ is a local uniformizer at $(x_0,y_0)$. By Remark 2, $\sigma _{1,1}s$ acts on functions as $\sigma _{-1,-1}s^2$ and thus

$$\begin{aligned} \sigma _{-1,-1}s^2(x-\zeta _{6n}^4)&=\sigma _{-1,-1}\left( \frac{1}{y}-\zeta _{6n}^4 \right) = \frac{\zeta _{6n}^{6}}{y}-\zeta _{6n}^4= -\zeta _{6n}^4 \frac{y- \zeta _{6n}^2}{y} \nonumber \\&=-\zeta _{6n}^4 \frac{y- \zeta _{6n}^2}{\zeta _{6n}^2 + (y-\zeta _{6n}^2)} \nonumber \\&= -\zeta _{6n}^2 (y-\zeta _{6n}^2 ) \left( 1 + \sum _{\nu =1}^\infty \frac{-1}{\zeta _{6n}^2} (y-\zeta _{6n}^2)^\nu \right) . \end{aligned}$$

(3)

On the other hand Taylor expansion of the Fermat equation at $(x_0,y_0)$ gives

$$ 0=x^n+y^n+1 =x_0^n+y_0^n +1 + nx_0^{n-1}(x-x_0) + ny_0^{n-1}(y-y_0))+\text {higher order terms}, $$

that is

$$ \zeta _{6n}^{4(n-1)}(x-\zeta _{6n}^4)+ \zeta _{6n}^{2(n-1)}(y-\zeta _{6n}^2) \!\!\!\mod \mathfrak {m}_{(x_0,y_0)}^2 $$

and this combined with Eq. (3) gives

$$ \sigma _{-1,-1}s^2(x-\zeta _{6n}^4)=\zeta _{6n}^2 \zeta _{6n}^{2(n-1)} (x-\zeta _{6n}^4)= \zeta _{3} (x-\zeta _{6n}^4), $$

i.e., $\sigma _{-1,-1}s^2$ is indeed the monodromy at the point $Q_0=(x_0,y_0)$.