Monodromy of Fermat Surfaces and Modular Symbols for Fermat Curves

MONODROMY OF FERMAT SURFACES AND MODULAR SYMBOLS FOR FERMAT CURVES

OZLEM EJDER

Abstract. In this paper, we begin with a parametrized family of Fermat curves. Using the modular description of Fermat curves and the modular symbols, we ﬁrst compute a set of generators for the ﬁrst integral homology group and then compute the monodromy of this family.

1. Introduction This paper is concerned with a geometric problem of computing the monodromy. The monodromy describes how certain objects behave when it runs around the singularity. As an example, consider the map → taking z 7→ zn which is a branched covering of 1 and C C AC the monodromy interchanges the sheets of the covering. Instead of looking at one curve, we may also take a family curves parametrized by another curve, which is an example of a fibration. We call the members of this family as fibers, the monodromy in this case is often described in terms of homology or cohomology of a smooth fiber. The first non-trivial example is a family of elliptic curves which may also be called an elliptic surface. In [Kod64, Kod66], Kodaira classifies all singularity types in elliptic fibrations and com- putes the local monodromy around each singularity type. Since the first homology group of an elliptic curve is free abelian group of rank 2, the local monodromy is given as a matrix in SL2(Z). Although the local monodromy of elliptic surfaces are completely known, we do not have a general result on families of higher genus. For the purposes of this thesis, we will work with a surface which fibers over an affine curve. Our main object in § 4 is a family of Fermat curves of fixed degree n, i.e., xn + yn = zn. In [Bel79], Belyi shows that for any smooth projective curve C, there exist a finite index subgroup G of SL2(Z) such that the quotient of the upper half plane by the action of G is isomorphic to C. For example, Fermat’s cubic x3 + y3 = z3 is an elliptic curve with j invariant 0 and it is isomorphic to X0(36); hence G is Γ0(36). Similarly, Fermat’s quartic 4 4 4 x + y = z is isomorphic to the modular curve X0(64). This leads us to the question of whether there are subgroups of SL2(Z) that are known for Fermat curve of each degree n. In [Roh77], Rohrlich gives a modular description of Fermat curves in terms of certain subgroups Φ(n) of Γ(2). The homology of modular curves can be easily described by a subset of modular symbols. A modular symbol is a pair of cusps i.e., {α, β} for some α, β in P1(Q) under some equivalence relation. One should see this symbol as a path from α to β. Manin [Man72] describes a set of generators for the group of modular symbols for a finite index subgroup G of SL2(Z); moreover he gives all the relations between these generators, so called the Manin symbols. 1 2 OZLEM EJDER

In Proposition 3.1, we compute the Manin symbols for the group Φ(n) and find a set of generators for the first integral homology group in Proposition 3.2. Moreover, the group of modular symbols computed in Proposition 3.1 is the homology group of a Fermat curve of degree n relative to the set of cusps (or the points at infinity), hence Proposition 3.1 shows that the relative homology group has rank n2 + 1. In § 4, we describe a family of Fermat curves, its singularities, and the fundamental group of the parameter space. We show that the monodromy action on the fibers of this family is the automorphism given by (x : y : z) 7→ (ζnx : ζny : z) and determine its action on the generators of the homology given in Proposition 3.2.

2. Background 2.1. Modular Interpretation of Fermat curves. Let H¯ denote the extended upper half 1 ¯ plane H ∪ P (Q). The group SL2(Z) acts on H by fractional linear transformations. Let a b M = be in SL ( ). Then c d 2 Z az + b M.z := cz + d ¯ ¯ The quotient H/G of H by the action of a ﬁnite index subgroup G of SL2(Z) is a projective curve and it is denoted as X(G). In the following, we describe Φ(n); subgroups of SL2(Z) such that the Fermat curve xn + yn = zn

is isomorphic to X(Φ(n)) for every n ∈ Z≥1. We begin with defining a principle congruence subgroup of SL2(Z). a b Γ(2) = ∈ SL ( ): a, d ≡ 1 mod 2 and b, c ≡ 0 mod 2. c d 2 Z It is known that Γ(2) = ±hA, Bi with 1 2 1 0 A := and B := . 0 1 2 1 Let Φ(n) = hAn,Bn, Γ(2)0i be the subgroup of Γ(2) generated by An,Bn, and the commutator subgroup Γ(2)0 of Γ(2). Then H¯ → X(Γ(2)) induces a map X(Φ(n)) −→ X(Γ(2)). The modular curve X(Φ(n)) is isomorphic to the Fermat curve C(n): xn + yn = zn and the modular curve X(Γ(2)) is a genus 0 curve (See [Roh77], [Lon08]). Moreover, the map X(Φ(n)) −→ X(Γ(2)) corresponds to the projection map 1 g : C(n) −→ P (x : y : z) 7→ (xn : zn) The morphism g has degree n2 with ramification at the points 0, 1 and ∞ which are cusps of X(Γ(2)). At each of these points, the ramification degree is n and hence we can conclude that the modular curve X(Φ(n)) has 3n cusps and there are exactly n of them lying above each of the points 0, 1, and ∞. Thus the cuspidal points on C(n) are the points [x : y : z] with xyz = 0 which are known as the points at infinity. MONODROMY OF FERMAT SURFACES AND MODULAR SYMBOLS FOR FERMAT CURVES 3

2.2. Cusps of X(Φ(n)). Deﬁne ai, bi, and ci for 1 ≤ i ≤ n as: i i i ai = (0 : ζ : 1), bi = (ζ : 0 : 1), ci = (ζ : 1 : 0) where = eπ/n and ζ is a primitive n’th root of unity. Using the map g described above, we see that the points ai, bi and ci correspond to the cusps above 0, 1 and ∞ respectively. The j point ci corresponds to the equivalence class π(B .∞) for some j since the Galois group of the morphism g is Γ(2)/Φ(n). Lemma 2.1. Let π denote the quotient map H¯ → H¯/Φ(n). Then π(Ak.1) = π(Bk.1). Proof. We will do induction on k. If k = 1, then π(A.1) = π(B.1) since BA−1.1 = 1 and A.1 = (ABA−1B−1)B.1. Let γ1, γ2 be in Γ(2), then we have

π(γ1γ2.z) = π(γ2γ1.z) since Φ(n) contains the commutator subgroup of Γ(2). Assume π(Ak−1.1) = π(Bk−1.1). Remember that π(z) = π(z0) if and only if there exist a γ ∈ Φ(n) such that z0 = γz. Then

k k−1 k−1 k−1 k A .1 = A.(A .1) = A(γ1B ).1 = γ2γ1.B A.1 = γB .1. where γ1, γ2, γ in Φ(n). 2.3. Automorphisms of X(Φ(n)). It is proven in [Tze95] that the automorphism group of the Fermat curve xn + yn = zn is generated by two automorphisms of order n which are given by

(x : y : z) 7→ (ζnx : y : z),

(x : y : z) 7→ (x : ζny : z) and the permutation group S3 on three letters. On the other side, the normalizer of Φ(n) in k SL2(Z) acts on X(Φ(n)). Notice that A is in the normalizer of Φ(n) and it stabilizes B .∞ k k k (or the points ci) since π(AB .∞) = π(B A.∞) = π(B ∞). The only automorphism of order n of X(Φ(n)) ﬁxing ck for all k is given by δ :(x : y : z) 7→ (ζx : ζy : z). Hence the automorphism δ corresponds to some power of A.

2.4. Modular Symbols. Let M2 be the free abelian group on the set of symbols {α, β} with α, β ∈ P1(Q) modulo the 3-term relations {α, β} + {β, γ} + {γ, α} = 0 and modulo any torsion. Deﬁne a left action of SL2(Z) on M2 by letting g ∈ SL2(Z) act by g{α, β} = {gα, gβ} 4 OZLEM EJDER

a b where the action of g = on α and β is as it is defined in § 2.1. Let G be a finite index c d subgroup of SL2(Z). Then we can define M2(G), the group of modular symbols for G as the quotient of M2 by the submodule generated by the set {x − g.x : x ∈ M2, g ∈ G} and modulo any torsion. Suppose g ∈ SL2(Z). Then the Manin symbol associated to g is g{0, ∞}. Notice here 0 0 that if Gg = Gg , then g{0, ∞} = g {0, ∞} in M2(G). Hence, there is a well-defined Manin symbol associated to each right coset of G in SL2(Z). We will use the notation [x] for the Manin symbol of Gx. (For more details, one can read [Ste07].) Define a right action of G on the set of Manin symbols as follows: {0, ∞}.(hg) = h (g{0, ∞}) .

In other words, the group SL2(Z) acts on Manin symbols as h.[g] := [gh] for any g, h ∈ SL2(Z). Manin [Man72] showed that every modular symbol can be written as a Z-linear combina- tion of Manin symbols and determined the relations between these generators. Let 0 1 0 −1 −1 −0 σ = , τ = ,J = . −1 0 1 −1 0 −1 Theorem 2.2 ( [Man72]). If x is a Manin symbol, then x + xσ = 0 (1) x + xτ + xτ 2 = 0 (2) x − xJ = 0. (3) Moreover, these are all the only relations between Manin symbols.

Let G be a ﬁnite index subgroup of SL2(Z). Then the following result of Manin describes the integral homology of a modular curve using modular symbols.

Theorem 2.3 ( [Man72]). The representation of any class h ∈ H1(X(G), Z) as a sum P {α , β } of Manin symbols can be chosen so that P n (φ(β ) − φ(α )) = 0 as a zero nk k k k k k k dimensional cycle on X(G).

3. Modular Symbols for Fermat Curves

We would like to compute M2(Φ(n)). Hence, we will begin with ﬁnding coset representatives of Φ(n) in SL2(Z). We already mentioned that the cosets of Φ(n) in Γ(2) are given by AiBj : 0 ≤ i, j ≤ (n − 1).

Lemma 3.1. A complete set of right coset representatives of Γ(2) in SL2(Z) consists of the following matrices: 1 1 0 −1 0 1 1 1 1 0 1 0 , , , , , . −1 0 1 −1 −1 0 0 1 1 1 0 1

Proof. We ﬁrst label the matrices with the given order above as αi’s for i = 1 ... 6. Then −1 one can show that αiαj 6∈ Γ(2) for any i 6= j. Since the index of Γ(2) in SL2(Z) is 6, we are done. MONODROMY OF FERMAT SURFACES AND MODULAR SYMBOLS FOR FERMAT CURVES 5

Notice that τ = α2 and σ = α3. By Lemma 3.1, the set of right cosets of Φ(n) in SL2(Z) is given by i j {A B αk : 0 ≤ i, j ≤ (n − 1) and 1 ≤ k ≤ 6.} In the following computations, the powers of A and B are considered in Zn. We will begin with computing the relation 1 in Theorem 2.2. 3.1. σ-relations. We ﬁrst compute that −1 α1σ = A Jα4, α2σ = α5, α3σ = α6J,

α4σ = Aα1, α5σ = Jα2, α6σ = σ. Now there is a unique Manin symbol for each right coset of Φ(n) in SL2(Z). Thus we have

i j i−1 j i j i j i j i j [A B α1σ] = [A B α4], [A B α2σ] = [A B α5], [A B α3σ] = [A B ], i j i+1 j i j i j i j i j [A B α4σ] = [A B α1], [A B α5σ] = [A B α2], [A B α6σ] = [A B α3]. Notice that the right cosets of Φ(n) in Γ(2) commute since Φ(n) contains the commutator subgroup of Γ(2). Now using Equation 1, we have the following equalities:

i j i−1 j [A B α1] + [A B α4] = 0 (4) i j i j [A B α2] + [A B α5] = 0 (5) i j i j [A B α3] + [A B ] = 0. (6)

Therefore M2(Φ(n)) is generated by i j {[A B αk] : 0 ≤ i, j ≤ (n − 1) and k = 1, 2, and 6.}

3.2. τ-relations. Similarly we compute αkτ for k = 1 ... 6 as the folowing:

−1 −1 α1τ = A , α2τ = Aα1, α3τ = A α4, −1 α4τ = AB α5J, α5τ = Bα3J, α6τ = τ.

And the Manin symbol of αkτ is: i j i−1 j i j i+1 j i j i−1 j [A B α1τ] = [A B ], [A B α2τ] = [A B α1], [A B α3τ] = [A B α4], i j i+1 j−1 i j i j+1 i j i j [A B α4τ] = [A B α5], [A B α5τ] = [A B α3], [A B α6τ] = [A B α2]. Hence by Equation 2 we obtain:

i j i j i+1 j [A B ] + [A B α2] + [A B α1] = 0 (7) i j i−1 j i j−1 [A B α3] + [A B α4] + [A B α5] = 0. (8) Now using σ-relations, Equation 8 becomes i j i j i j−1 [A B ] + [A B α1] + [A B α2] = 0 (9) Hence by Equation 7, we conclude that the set of Manin symbols i j i j {[A B ] and [A B α2] : 0 ≤ i, j ≤ (n − 1).} generate M2(Φ(n)). Therefore using Equation 9 and Equation 7, we obtain the following relation: [Ai+1Bj] + [Ai+1Bj−1τ] = [AiBj] + [AiBjτ] (10) 6 OZLEM EJDER

for 0 ≤ i ≤ (n − 1) and 0 ≤ j ≤ n − 1. Remember that α2 = τ. Proposition 3.1. The group of modular symbols for the Fermat group Φ(n) is free of rank n2 + 1 and it is generated by {[AiBjτ] : 1 ≤ i ≤ (n − 1), 0 ≤ j ≤ (n − 1)}

{[An−1Bj] : 0 ≤ j ≤ (n − 1)} and [Bn−1τ].

i j i j Proof. We will use xi,j := [A B ] and yi,j := [A B τ] to ease the notation. We put Lexico- graphical order on the set {xi,j, yi,j : i = 1, . . . , n, j = 0, . . . n − 1} as follows: For any i, j, k, l,

xi,j ≤ yk,l and

xi,j ≤ xk,l or yi,j ≤ yk,l if and only if i < k or i = k and j ≤ l.

Also note that the index set for i, j is Zn. Hence we see that if i = n − 1 and j = 0, then xi+1,j−1 = x0,n−1. We already showed that xi,j and yi,j generate the group of modular symbols for the Fermat group Φ(n). Furthermore we know from Equation 10 that they satisfy the relation

xi,j + yi,j = xi+1,j + yi+1,j−1. (11) Equation 11 gives us a system of linear equations. Let R be the (n2 × 2n2) matrix obtained by this system of equations with respect to the ordered basis

{xi,j, yi,j : i = 1, . . . , n − 1, j = 0, . . . n − 1}. Let Ri,j denote the ij’th row of R, i.e., the row corresponding to the equation:

xi,j + yi,j − xi+1,j − yi+1,j−1 = 0. Notice that for i ≤ n − 2 the first non-zero entry of each Ri,j is 1 and it is the coefficient of P i,j n−1,j xi,j. Assume i = n − 1. For j fixed, we add i6=(n−1) R to R and denote the new row n−1,j 2 n−1,j we obtain by R0 . Notice that the first n entry of R0 is 0 since n−1 X (xi,j − xi+1,j) = 0. i=0 n−1,j The equation corresponding to R0 gives us that n−1 n−1 X X xi,j + yi,j − xi+1,j − yi+1,j−1 = (yi,j − yi+1,j−1) = 0 i=0 i=0 n−1,j If j ≥ 1, then the first non-zero term of R0 is the coefficient of y0,j−1. Similarly if j = 0, then it is the coefficient of y0,0. Now let

n−1,j X n−1,k R1 = R0 k≤j MONODROMY OF FERMAT SURFACES AND MODULAR SYMBOLS FOR FERMAT CURVES 7

n−1,j n−1,j n−1,j and replace R0 by R0 . Assume j 6= n − 1, then the ﬁrst non-zero entry of R1 is 1 n−1,n−1 and it is the coeﬃcient of y0,j. If j = n − 1, then R1 is the zero row since ! X X yi,k − yi+1,k−1 = 0. k i To summarize, we obtain a matrix of the form:  1   ..   .   .   0 ..     1     1     .   0 ..     1  0 0 Hence

{yi,j : 1 ≤ i ≤ (n − 1), 0 ≤ j ≤ (n − 1)}

{xn−1,j : 0 ≤ j ≤ (n − 1)} and

[y0,n−1] are the free variables and they generate the group of modular symbols for Φ(n). The proof of the proposition gives us that if j 6= n − 1, then

n−1 ! n−1 X X X (yi,k − yi+1,k−1) = yi,n−1 − yi,j = 0. (12) k≤j i=0 i=0 We will need Equation 12 in Section § 5.

3.3. Homology of Fermat Curves. In this section, we will describe the ﬁrst integral homology group of Fermat curves in terms of modular symbols.

Proposition 3.2. The homology group H1(X(Φ(n)), Z) of X(Φ(n)) is generated by i i+1 n−1 i+1 j i j+1 γi,j := [A τ] − [A B τ] + [A B τ] − [A B τ] for 1 ≤ i ≤ (n − 2) and 0 ≤ j ≤ (n − 2). Proof. We ﬁrst compute that τ{0, ∞} = {1, 0}. Then Ai{1, 0} − AiBj{1, 0} = {Ai.1,AiBj.1} and Ai+1Bj−1{1, 0} − Ai+1Bn−1{1, 0} = {Ai+1Bj−1.1,Ai+1Bn−1.1}. 8 OZLEM EJDER

We proved in §§ 2.2 that φ(Ai(1)) = φ(Bi(1)). Hence,

φ(Ai+1Bn−1.1) = φ(Ai+1gAn−1.1) for some g ∈ Φ(n). = φ(g0Ai+1An−1.1) for some g0 ∈ Φ(n). = φ(Ai.1). One can similarly show that φ(Ai+1Bj−1.1) = φ(AiBj.1). Therefore we obtain that φ(Ai.1) − φ(AiBj.1) + φ(Ai+1Bj−1.1) − φ(Ai+1Bn−1.1) = 0. Hence by Theorem 2.3,

i i+1 n−1 i+1 j−1 i j γi,j = [A τ] − [A B τ] + [A B τ] − [A B τ]

is in H1(X(Φ(n)), Z) for every i, j such that 1 ≤ i ≤ n − 2 and 0 ≤ j ≤ n − 2. We will show i j that γi,j’s are Z-linearly independent. To ease the notation, we will use yi,j for [A B τ] as in the proof of Proposition 3.1. Hence γi,j denotes the cycle

yi,0 − yi+1,n−1 + yi+1,j − yi,j+1 and suppose that X ni,jγi,j = 0. (13) i,j

We will prove by induction that ni,j = 0 for all i = 1, . . . , n − 2 and for all j = 0, . . . , n − 2. We know by Proposition 3.1 that yi,j’s are Z-linearly independent. If j is ﬁxed, then the coeﬃcient of y1,j+1 in the sum 13 is −n1,j. Hence

n1,j = 0 for all 0 ≤ j ≤ n − 3. P Similarly the coeﬃcient of y1,0 is j n1,j and thus

n1,n−2 = 0.

This proves the base step of the induction. Assume ni−1,j = 0 for all j = 0, . . . , n − 2. If j ≥ 1, then the coeﬃcient of yi,j is −ni,j−1 + ni−1,j and using the induction hypothesis, we obtain

ni,j−1 = 0 for all j = 1, . . . , n − 2. P For j = 0, we look at the coeﬃcient of yi, 0 which equals to ni−1,0 + j ni,j. Since we showed ni,j = 0 for every j ≤ n − 3 we see that ni,n−2 must also equal to 0. Hence the set

{γi,j : 1 ≤ i ≤ n − 2, 0 ≤ j ≤ n − 2} is linearly independent and they form a basis for the homology group since the genus of the Fermat curve is 1/2(n − 1)(n − 2) which implies that the rank of the ﬁrst homology group is (n − 1)(n − 2). MONODROMY OF FERMAT SURFACES AND MODULAR SYMBOLS FOR FERMAT CURVES 9

4. Fermat Surfaces 4.1. Definition of Monodromy. Let X and Y be algebraic varieties over C. Suppose that f : X → Y is a fibration. Let y0 be in Y and let X0 be the fiber of f at y0. For γ in π1(Y, y0), define H : X0 × [0, 1] → Y as H(x, t) := γ(t)

for all x ∈ X0. Then by the homotopy lifting property of f, there exists a lift ˜ H : X0 × [0, 1] → X as given in the following diagram (See [DK01] for more details.).

X0 × {0} X H˜ f

H X0 × [0, 1] Y

The commutativity of the above diagram induces a morphism X0 → X0 (at t = 1) which induces a homomorphism on the homology groups. Hence we obtain an action of the fundamental group π1(Y, y0) of Y on H1(X0, Z): ρ : π1(Y, y0) −→ Aut(H1(X0, Z))

which we call the monodromy action. It is worth noting here that the induced map X0 → X0 is a self homotopy equivalence and it is unique up to a homotopy. 4.2. Description of the Fibration. Let X denote the Fermat surface given by the equation xn + yn + zn = wn. in 3 . Consider the rational map f : X → 1 given by PC P f :(x : y : z : w) 7→ (z : w). We first blow up the variety X successively at the points of the set 3 n n B = {(x : y : 0 : 0) ∈ P : x + y = 0}. We obtain a variety X¯, a proper morphism f¯ : X¯ → P1 and a birational map ϕ : X¯ → X.A generic fiber of f is given by the equation xn + yn = 1. The fibers of f¯ are projective curves since f¯ is proper. Then a generic fiber of f¯ is isomorphic to the curve xn + yn = zn. The singular fibers are given by xn + yn = 0 when zn = wn. Let U denote the variety f¯−1(P1 − S) with k 1 S = {[ζn : 1] ∈ P : k = 0, . . . , n − 1.}. 1 We obtain a fibration U → V where V = P − S by Ehremann’s theorem [Ehr95]. Let y0 be the point [0 : 1] ∈ P1. To describe the monodromy action, we first formulate the generators of the fundamental group π1(V, y0) in the following way: n If n is even, then take leaves of the polar rose r = 2cos( 2 θ) and otherwise take leaves of r = 2cos(nθ) (in the complex plane given by w = 1). When n = 3, one can visualize these generators as in the following figure. 10 OZLEM EJDER

90 120 60

σ2 150 30

σ1 0 0.5 1 1.5 2 180 0

210 σ3 330

240 300 270

Figure 1. Generators of π1(V, y0) in w = 1 plane for n = 3.

k Let σk be the loop around the singular point [ζn : 1] and let ρ denote the monodromy −2πki n n representation. First we notice that σke n is σ1 and hence σk = σ1 . n n Consider γ(t) = 1 − σk(t) = 1 − σ1(t) . The function γ has the same start and the end point, i.e., it is a loop. Since the lines at θ = ±π/2n (respectively θ = ±π/n) are tangent to σ1 when n is odd (respectively when n is even), γ(t) will be tangent to the lines θ = ±π/2 or ±π. Hence we can find continuous functions r(t) and g(t) such that n 2πig(t) 1 − σk(t) = r(t)e . and g(0) = 0 and g(1) = 1. Let (u : v : w) be a point on the fiber xn + yn = zn. Note that σk(t) lies in the complex plane w = 1. ¯ We would like to find a liftσ ¯k for σk such that f(¯σ) = σ. Since σ(t) lies in the plane w = 1,σ ¯(t) also should be in w = 1. We will assume that w 6= 0. Since X¯ and X are isomorphic on w 6= 0, we may assume that U is given by the equation xn + yn = wn(1 − un) for un 6= 1. Remember that we pick the base point y0 as u = 0. Let (x : y : w : 0) be a point in the fiber of u = 0. We observe the followings: (1) The function r(t) only takes positive real values and there exist a function whose n’th power is r(t) such that r1/n(t) is also positive real for every t. (2) Define σ˜ : I → U σ˜(t) = (λ(t)x, λ(t)y : w : σ(t)) 1/n 2πig(t) where λ(t) = r(t) e n . (3) The pathσ ¯(t) takes values in U, i.e., (t)wn(1 − σ(t)n) = (1 − σ(t)n)(xn + yn).

(4) The pathσ ¯ begins atσ ˜(0) = (x : y : w) and ends atσ ˜(1) = (ζnx : ζny : w). Henceσ ˜ induces an action on the ﬁber xn + yn = wn as

(x : y : w) 7→ (ζnx, ζny : w). In particular, this action only depends on the choice of the n’th root of unity.

5. Calculation of the Monodromy In §§ 4, we showed that the action of each generator of the fundamental group on a generic ﬁber is identical. We will compute the monodromy by describing the action of A on the MONODROMY OF FERMAT SURFACES AND MODULAR SYMBOLS FOR FERMAT CURVES 11 modular symbols γi,j using our results from §§ 2.3. Remember that C(n) denotes the Fermat curve xn + yn = zn.

Theorem 5.1. Let σk be the generators of π1(V, y0) as given in Section § 4. Then for any k the action of σk on the homology group H1(C(n), Z) is given by the following:

γi,j 7→ γi+1,j for 1 ≤ i ≤ n − 3, 0 ≤ j ≤ n − 2. and if i = n − 2, then

n−2 X γn−2,0 7→ − γk,n−1−k k=1 and

j−1 n−l−2 ! n−2 X X X γn−2,j 7→ γs,l − γs,n−2−s+j−l − γi,n−i+j−1 l=0 s=j−l i=j+1 for 1 ≤ j ≤ n − 2.

Proof. We recall from equation (12) that

n−1 X (yi,n−1 − yi,j) = 0 i=0

for any 0 ≤ j ≤ n − 2. If i < n − 2, then it is clear that Aγi,j = γi+1,j. Assume i = n − 2, then

Aγn−2,j = A(yn−2,0 − yn−1,n−1 + yn−1,j − yn−2,j+1)

= yn−1,0 − y0,n−1 + y0,j − yn−1,j+1 n−1 X = yn−1,0 − y0,n−1 + y0,j − yn−1,j+1 + (yi,n−1 − yi,j). i=0 n−1 X = yn−1,0 − yn−1,j+1 + (yi,n−1 − yi,j) i=1 12 OZLEM EJDER

We will separate the cases j = 0 and j ≥ 1. Assume j = 0, then

n−1 X Aγn−2,0 = yn−1,0 − yn−1,1 + (yi,n−1 − yi,0) i=1 n−1 n−2 X X = −yn−1,1 + y1,n−1 + yi,n−1 − yi,0 i=2 i=1 n−2 n−2 X X = (yi,n−i − yi+1,n−i−1) + (yi+1,n−1 − yi,0) i=1 i=1 n−2 X = (yi+1,n−1 − yi,0 + yi,n−i − yi+1,n−i−1) i=1 n−2 X = − γi,n−i−1. i=1

Assume j ≥ 1. The proof is similar to the case j = 0 but it requires more work.

n−1 X Aγn−2,j = yn−1,0 − yn−1,j+1 + (yi,n−1 − yi,j) i=1 n−1 n−1 X X = yn−1,0 − yn−1,j+1 − yi,n−1 + yn−i,j i=1 i=1 n−1 n−1 X X = yn−1,0 − yn−1,j+1 + yj+1,n−1 − yn−1−j,j + yi,n−1 − yn−i,j i=1 i=1 i6=j+1 i6=j+1 n−1 ! j ! X X = (yj+1,n−1 − yn−1,j+1) + yi,n−1 − yi−1−j,j + yi,n−1 − yn+i−1−j,j i=j+2 i=1

+ yn−1,0 − yn−1−j,j. since

n−1 n−1 j j X X X X yn−i,j = yi−1−j,j and yn−i,j = yn+i−1−j,j. i=j+2 i=j+2 i=1 i=1

We ﬁrst compute that

n−1 n−1 n−1 X X X (yi,n−1 − yi−1−j,j) = (yi,n−1 − yi−1,0) + (yi−1,0 − yi−1−j,j). i=j+2 i=j+2 i=j+2 MONODROMY OF FERMAT SURFACES AND MODULAR SYMBOLS FOR FERMAT CURVES 13

We can easily calculate that

n−1 ! X yi,n−1 − yi−1,0 + (yj+1,n−1 − yn−1,j+1) i=j+2 n−1 n−2 X X = (yi,n−1 − yi−1,0) + (yi,n+j−i − yi+1,n+j−i−1) i=j+2 i=j+1 n−2 X = (yi+1,n−1 − yi,0 + yi,n+j−i − yi+1,n+j−i−1) i=j+1 n−2 X = − γi,n−i+j−1. i=j+1 Now we will show that

n−1 ! j X X yi−1,0 − yi−1−j,j + yn−1,0 − yn−1−j,j + yi,n−1 − yn+i−1−j,j (14) i=j+2 i=1 j−1 n−l−2 ! X X = γs,l − γs,n−2−s+j−l . (15) l=0 s=j−l

Left hand side of equation (14) becomes

n−1 j X X yi,0 − yi−j,j + yi,n−1 − yn+i−1−j,j. i=j+1 i=1 We can rearrange the ﬁrst sum as the following:

n−1 n−1 i ! X X X yi,0 − yi−j,j = yl,i−l − yl−1,i−l+1 i=j+1 i=j+1 l=i−j+1 j−1 n−2−j ! X X = y1+j+k−l,l − yj+k−l,l+1 l=0 k=0 j−1 n−l−2 ! X X = y1+s,l − ys,l+1 . l=0 s=j−l We may also rearrange the second sum as:

j j n−2+i−j ! X X X yi,n−1 − yn+i−1−j,j = yk,n−1+i−k − yk+1,n−2+i−k i=1 i=1 k=i j−1 n−l−2 ! X X = ys,n−1−s+j−l − ys+1,n−2−s+j−l . l=0 s=j−l 14 OZLEM EJDER

Combining these two sums, j−1 n−l−2 ! j−1 n−l−2 ! X X X X = y1+s,l − ys,l+1 + ys,n−1−s+j−l − ys+1,n−2−s+j−l l=0 s=j−l l=0 s=j−l j−1 n−l−2 ! X X = y1+s,l − ys,l+1 + ys,n−1−s+j−l − ys+1,n−2−s+j−l l=0 s=j−l j−1 n−l−2 ! X X = γs,l − γs,n−2−s+j−l . l=0 s=j−l d Corollary 5.1.1. The action of A for d ≥ 1 on the homology group H1(X(Φ(n)), Z) is given by the following:

γi,j 7→ γi+d,j for 1 ≤ i ≤ n − 2 and i 6= n − 1 − d, 0 ≤ j ≤ n − 2. and if i = n − 1 − k, then n−2 X γn−1−d,0 7→ − γk,n−1−k k=1 and j−1 n−l−2 ! n−2 X X X γn−1−d,j 7→ γs,l − γs,n−2−s+j−l − γk,n−2−k l=0 s=j−l k=1+j for 1 ≤ j ≤ n − 2. Remark. Let n = 3. Then the monodromy of each generator is given by 0 1 −1 −1 which coincides with monodromy of an elliptic surface with a singularity of three lines inter- secting at a single point given in [Kod64], [Kod66], [Mir89].

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Department of Mathematics, University of Southern California, Los Angeles, California 90089 E-mail address: [email protected]