Some modular forms related to cubic surfaces

August 1999

E. Freitag Mathematisches Institut Universität Heidelberg Im Neuenheimer Feld 288 69120 Heidelberg E-mail: freitag@mathi.uni-heidelberg.de home-page: http://www.rzuser.uni-heidelberg.de/∼t91/

1. Introduction

The moduli space of marked cubic surfaces can be described as a ball quotient B_{4}/Γ, where Γ ⊂ U(1, 4) is a certain arithmetic subgroup of the unitary group of signature (1, 4) [ACT1,2]. The unitary group can be considered as subgroup of O(2, 8). It is natural to use the lifting constructions of R. Borcherds for the group O(2, n) to construct projective models for the moduli space of marked cubic surfaces. In [AF] such a model has been constructed by means of the singular Borcherds lift. In this paper we will investigate modular forms on Γ in a more systematic way and we will obtain several interesting spaces of modular forms. All of them have meaning for the moduli space of cubics, but we will not discuss applications to the moduli space of cubics in this paper.

I would like to thank D. Allcock, R. Borcherds and B. v. Geemen for helpful discussions.

2. Elliptic modular forms

Borcherds theory can be considered as a kind of generalized Howe duality for the dual pair SL(2), O(2, n), which means that modular forms on O(2, n) are constructed as liftings from elliptic modular forms. The elliptic modular forms in the setting of Borcherds belong to the full elliptic modular group but they are vector valued. This means that one starts with a finite dimensional representation of SL(2, ℤ) on a complex vector space*)

ϱ : SL(2, ℤ) ⟶ GL(V).

*) Borcherds considers more generally the metaplectic covering of SL(2, ℤ). This is

We assume that ϱ factors through SL(2, ℤ/N ℤ) for a suitable natural number N. An (entire) elliptic modular form of weight k ∈ ℤ with respect to ϱ is a holomorphic V -valued function f(τ) on the upper half plane which satisfies

f(Mτ) = ϱ(M)(cτ + d)^{k}f(τ) for all M ∈ SL(2, ℤ)

and which is bounded for Im τ ≥ 1. Let V_{0} ⊂ V be the subspace on which ϱ(−E) (E denotes the unit matrix) acts by multiplication with (−1)^{k}. It is quite clear that the values of f are contained in V_{0} and that V_{0} is invariant under SL(2, ℤ). The dimension of V_{0} is denoted by d.

Let A be a complex matrix of finite order. The eigenvalues are roots of unity which we can write in the form

λ = exp(2πiα) with 0 ≤ α < 1.

We need the invariant

α(A) = ∑_{λ} α,

where λ runs through all eigenvalues (counted with multiplicity). From [Bo2] we take the following formula which can be proved by means of the Riemann-Roch theorem:

2.1 Proposition. Assume k > 2. The dimension of the space of modular forms of weight k and with respect to ϱ is

d + dk/12 − α(e^{πik/2}ϱ(S)) − α((e^{πik/3}ϱ(ST))^{−1}) − α(T)

(We use the notations S = \binom{0 −1}{1 0} and T = \binom{1 1}{0 1}. ρ is considered as representation on V_{0}.)

Special examples of modular forms can be constructed by means of Eisenstein series. Assume that a ∈ V_{0} is a vector with the property

ϱ ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) a = a.

Then (cτ + d)^{−k}ϱ(M)^{−1}a remains unchanged if one replaces M by P M, where P ∈ SL(2, ℤ)_{∞} lies in the stabilizer of ∞ (c = 0). The Eisenstein series is

E_{a}(τ) = ∑_{M∈SL(2,ℤ)_{∞}\ SL(2,ℤ)}(cτ + d)^{−k}ϱ(M)^{−1}a.

not necessary in this paper, because only elliptic modular forms of integral weight will occur.

§2. Elliptic modular forms 3

For k > 2 it defines a modular form of type (k, ϱ) and it has the further property

E_{a}(i∞) := lim_{Im τ→∞} E_{a}(τ) = a.

The Eisenstein series can be also defined in the case k = 2. Its Fourier expansion is formally the same as in the case k > 2 up to an additional non-holomorphic term in its Fourier expansion, namely

C(y) ∑_{M} ϱ(M)a.

Here M runs through SL(2, ℤ/N ℤ). The sum over ϱ(M) is zero if ϱ contains no one-dimensional trivial representation.

2.2 Remark. Assume k > 2. The space of Eisenstein series of weight k and with respect to ϱ is isomorphic to the space of all a ∈ V with

ϱ ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) a = a  and  ϱ(−E)a = (−1)^{k}a .

This statement remains also true in the case k = 2 if ϱ contains no trivial one dimensional representation.

We will need Fourier coefficients of Eisenstein series in the case k > 2. Because ϱ is trivial on the principal congruence group of level N, our Eisenstein series are linear combinations of the standard ones:

G_{k}(τ; c, d, N) := \begin{matrix}∑ & ′ \\ {m≡c \atop n≡d} mod N \end{matrix} \frac{1}{(mτ + n)^{k}}

The Fourier expansion of the Eisenstein series is ([He] No 24, §1)

\begin{matrix}G_{k}(τ; c, d, N) = δ(\frac{c}{N} ) \begin{matrix}∑ & ′ \\ m≡c(N) \end{matrix} \frac{1}{m^{k}} + \\ \frac{(−2πi)^{k}}{N^{k}(k − 1)!} ∑_{{mn>0 \atop m≡c(N)}} n^{k−1}sgn(n)ζ^{dn}q^{\frac{mn}{N}} , \end{matrix}

where

δ(x) = \{ \begin{matrix}1 & for x integral \\ 0 & else \end{matrix} , q^{1/N} = e^{2πiτ/N}.

The series

C(N, k, d) := \begin{matrix}∑ & ′ \\ m≡d(N) \end{matrix} m^{−k}

can be computed. If d ≡ 0 mod N its value is

2N^{−k}ζ(k), if k is even
0 , if k is odd

and

ζ(k) = \frac{(−1)^{k/2−1}2^{k−1}B_{k}}{k!} π^{k} (k even),

where B_{k} denotes the k-th Bernoulli number. If d is not divisible by N we use the formula

∑_{m≡d(N)} \frac{1}{m^{k}} = ζ(f, s) := ∑_{n=1}^{∞} \frac{f(n)}{n^{s}}

where

f(x) = \{ \begin{matrix}1 & for x ≡ d mod N \\ (−1)^{k} & for x ≡ −d mod N \\ 0 & else \end{matrix}

We use the functional equation ([La], XIV, theorem 2.1), which gets the form (use f(−n) = (−1)^{k}f(n))

ζ(f, s) = \frac{1}{2πi}(\frac{2π}{N} )^{s}E(s)ζ(1 − s, f̂),

where

f̂(x) = ∑_{m=1}^{N} f(m)e^{−2πimx/N}

and

E(s) = Γ(1 − s)((−1)^{k}e^{πis/2} − e^{−πis/2}).

For integral s = k one computes by means of Γ(s)Γ(1 − s) = π/ sin πs

E(k) = \frac{π}{(k − 1)!} \{ \begin{matrix}i & if k ≡ 0(4) \\ −1 & if k ≡ 1(4) \\ −i & if k ≡ 2(4) \\ 1 & if k ≡ 3(4) . \end{matrix}

We express ζ(f, s) by means of the Hurwitz Zeta function

ζ(f, s) = N^{−s} ∑_{m=1}^{N} f(m)ζ(s, u) (ζ(s, u) = ∑_{n=0}^{∞} \frac{1}{(n + u)^{s}} )

and use the value at negative integers ([Za], Teil I, §7, Aufgabe 3)

ζ(−n, a) = −\frac{1}{n + 1}B_{n+1}(a)

to obtain

∑_{m≡d(N)} \frac{1}{m^{k}} =

\frac{(2π)^{k}}{k! · N} ∑_{m=1}^{N} B_{k}(\frac{m}{N} )\frac{e^{−2πidm/N} + (−1)^{k}e^{2πidm/N}}{2} \{ \begin{matrix}−1 & if k ≡ 0(4) \\ −i & if k ≡ 1(4) \\ 1 & if k ≡ 2(4) \\ i & if k ≡ 3(4) . \end{matrix}

Here B_{k}(x) denotes the Bernoulli polynomial which is defined by

∑_{n=0}^{∞} B_{n}(x)\frac{t^{n}}{n!} = \frac{te^{xt}}{e^{t} − 1}.

(The Bernoulli numbers are B_{n} = B_{n}(0)).

As an example we consider a four dimensional representation ϱ. It is defined by means of the generators

T = ( \begin{matrix}1 & 1 \\ 0 & 1 \end{matrix} ) , S = ( \begin{matrix}0 & −1 \\ 1 & 0 \end{matrix} )

as

ϱ(T) = ( \begin{matrix}1 \\ & 1 \\ & & ζ^{2} \\ & & & ζ \end{matrix} ) , ϱ(S) = \frac{i}{\sqrt{3}^{5}} ( \begin{matrix}1 & 1 & 1 & 1 \\ 80 & −1 & 8 & −10 \\ 90 & 9 & −9 & 0 \\ 72 & −9 & 0 & 9 \end{matrix} ) ,

where ζ = −1/2 + i\sqrt{3}/2 denotes a third root of unity. This defines actually a unitary representation. Using the Riemann-Roch formula above and the given formulae for the Fourier coefficients of Eisenstein series one easily verifies:

2.3 Proposition. The space of elliptic modular forms of weight 5 with respect to the four dimensional representation ϱ has dimension 2. There are no cusp forms in this space. It is generated by the following two Eisenstein series f and g.

\begin{aligned}f_{1} & = \frac{i\sqrt{3}}{18}E_{1} − \frac{1}{18}(E_{2} + E_{3} + E_{4}) \\ f_{2} & = −\frac{5i\sqrt{3}}{9}E_{1} + \frac{5}{9}(E_{2} + E_{3} + E_{4}) \\ f_{3} & = 0 \\ f_{4} & = E_{2} + ζ^{2}E_{3} + ζE_{4} \end{aligned} \begin{aligned}g_{1} & = \frac{i\sqrt{3}}{18}E_{1} + \frac{1}{18}(E_{2} + E_{3} + E_{4}) \\ g_{2} & = \frac{4i\sqrt{3}}{9}E_{1} + \frac{4}{9}(E_{2} + E_{3} + E_{4}) \\ g_{3} & = E_{2} + ζE_{3} + ζ^{2}E_{4} \\ g_{4} & = 0 \end{aligned} ,

Here we write E_{1}, E_{2}, E_{3}, E_{4} for the 4 Eisenstein series of weight k = 5 and level N = 3 and for the values (c, d) = (0, 1), (1, 0), (1, 1), (1, 2).

The proof is straightforward computation. We only mention, that the eigenvalues of the three matrices, which occur in the Riemann-Roch formula are determined by their traces because 1, ζ, ζ^{2} are linearly independent over ℚ.

We will need later the Eisenstein series with the initial values (−1/2, 0, 0, 0). Using the explicit formulae for the Fourier coefficients one obtains:

2.4 Lemma. The unique element in the space of modular forms 2.3 with constant Fourier coefficient (−1/2, 0, 0, 0) is

h_{1} = −1/2 + 12 q + 225 q^{2} + 1092 q^{3} + 2892 q^{4} + 9360 q^{5} + 16425 q^{6} + ⋯ h_{2} = 1080 q + 16200 q^{2} + 87480 q^{3} + 260280 q^{4} + 673920 q^{5} + 1312200 q^{6} + ⋯ h_{3} = 225 q^{2/3} + 9360 q^{5/3} + 57825 q^{8/3} + 219600 q^{11/3} +

540450 q^{14/3} + 1252800 q^{17/3} + 2255760 q^{20/3} + ⋯ h_{4} = 12 q^{1/3} + 2892 q^{4/3} + 28824 q^{7/3} + 112320 q^{10/3} + 342744 q^{13/3} +

740172 q^{16/3} + 1563864 q^{19/3} + ⋯

3. A Weil representation

The inputs of Borcherds liftings are elliptic modular forms with respect to certain Weil representations of SL(2, ℤ).

Let M be a lattice equipped with a quadratic form q. We assume that q is integral on M. This means that M is an even lattice if we equip it with the bilinear form

(x, y) := q(x + y) − q(x) − q(y) ((x, x) = 2q(x)),

which is assumed to be non-degenerated. The dual lattice is

M′ = { x ∈ M ⊗_{ℤ} ℝ, (x, y) ∈ ℤ ∀y ∈ M }.

We recall the Weil representation, which can be defined by means of the generators

3.1 Proposition. Let (M, q) be an even integral lattice of signature (m, n). We assume that m−n is even. Let N be a natural number such that q(α)·N is integral for all α ∈ M′. There exists a representation ϱ = ϱ_{M} of SL(2, ℤ/N ℤ) on the space ℂ[M′/M] which is given on generators by

ϱ(T)e_{α} = exp(2πiq(α))e_{α},
ϱ(S)e_{α} = \begin{matrix}i & (m−n)/2 \\ √ & \overline{#(M′/M)} \end{matrix} ∑_{β∈M′/M} exp(−2πi(α, β))e_{β}.

(The standard generators of the group ring ℂ[M′/M] are denoted by e_{α}.)

§3. A Weil representation 7

Sometimes it is necessary to have an explicit formula for the Weil representation. From [Sh] we take 3.2 Proposition. Let be σ = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL(2, ℤ). The Weil representation

ϱ(σ) acts by the matrix

i^{(sgn d−1)(m−n)/2}δ_{α,aβ} exp 2πiabq(α)

if c = 0 and otherwise by the matrix

\begin{matrix}i & (sgn c)(n−m)/2 \\ √ & \overline{#(M′/M)} \end{matrix} \frac{1}{|c|^{(m+n)/2}}

∑_{γ∈M/cM} exp 2πi [\frac{a(α + γ, α + γ) − 2(β, α + γ) + d(β, β)}{2c}] .

We consider a special lattice, which can be defined by means of the ring of Eisenstein integers

E := ℤ[ζ] (ζ = −1/2 + i\sqrt{3}/2).

The quadratic form x̄x equips E with a structure as even lattice. The associated bilinear form is 2 Re x̄y. (This is the root lattice A_{2}.) The dual lattice is

E′ := \frac{1}{\sqrt{−3}}E.

The discriminant group

E′/E ≅ 𝔽_{3} = ℤ/3ℤ

has order three. We have to consider the lattice of signature (2, 8)

M := E − E − E − E − E

whose underlying group is E^{5} which is equipped with

q(x) = x̄_{1}x_{1} − x̄_{2}x_{2} − ⋯ − x̄_{5}x_{5}.

The discriminant group M′/M has order 3^{5}, hence (M′/M)/±1 has order 122.

Decomposition of the Weil representation

We describe the irreducible representations of the group SL(2, 𝔽_{3}), 𝔽_{3} = ℤ/3ℤ. It has seven conjugacy classes represented by

±E S, ±ST^{2}, ±ST.

We write down the full character table:

E −E S ST^{2} −ST^{2}  ST  −ST
ϱ_{1} 1 1 1  1  1  1  1
ϱ_{2} 3 3 −1  0  0  0  0
ϱ_{3} 1 1 1 −1 − ζ −1 − ζ  ζ  ζ
ϱ_{4} 1 1 1  ζ  ζ  −1 − ζ −1 − ζ
ϱ_{5} 2 −2 0  −ζ  ζ  −1 − ζ 1 + ζ
ϱ_{6} 2 −2 0  −1  1  1  −1
ϱ_{7} 2 −2 0 1 + ζ −1 − ζ  ζ  −ζ
1 1 6  4  4  4  4

The last line gives the order of the corresponding conjugacy class.

We decompose the Weil representation ϱ of SL(2, 𝔽_{3}) on ℂ[M′/M] into irreducible components. We recall that ϱ(T) is the diagonal matrix with entries exp(2πiq(α)) and that ϱ(S) = \frac{i}{\sqrt{3^{5}}} exp(−2πi(α, β)). It is no problem to compute the character of this representation. The result is

3.3 Proposition. The representations ϱ_{1} …, ϱ_{7} occur in the Weil representation with multiplicities

10 30 6 15 16 20 25 1 3 1 1 2 2 2

The second line gives the dimensions of the representations. The decomposition of the dual representation is obtained by replacing ϱ_{3} ↔ ϱ_{4} and ϱ_{5} ↔ ϱ_{7}.

There is a natural isomorphism

M′/M ⟶^{≅} 𝔽_{3}^{5}, a ⟼ \sqrt{−3} a mod \sqrt{−3}.

The quadratic form on M′ induces a 𝔽_{3}-valued bilinear form x_{1}y_{1} − x_{2}y_{2} − ⋯ − x_{5}y_{5} on 𝔽_{3}^{5}. The orthogonal group O(5, 3) is the subgroup of GL(5, 𝔽_{3}) which preserves this quadratic form. (All non singular quadratic forms are equivalent because 5 is odd). This group acts on ℂ[M′/M]. The action of O(5, 3) commutes with the action of SL(2, 𝔽_{3}). Therefore SL(2, 𝔽_{3}) × O(5, 3) acts on the isotypic subspaces of dimension (s. 3.3)

10 90 6 15 32 40 50.

Recall that finite dimensional irreducible representations of a direct product G × H of two groups are tensor products of irreducible representations of the factors G, H.

The group O(5, 3) is of order 103 680. It contains a copy of the Weyl group W (E_{6}) of the lattice E_{6} as subgroup of index 2. O(5, 3) is generated by this Weyl group and the negative identity −E. The Weyl group contains the simple group G_{25 920} of order 25 920 as subgroup of index 2. The group W (E_{6}) and its character table can be found in the ATLAS of finite groups. Fortunately the group O(5, 3) is also contained in the shared package ”matrix“ of the GAP–3– system and can be called by the name O(0, 5, 3) after loading the package by the command RequirePackage(“matrix“). Its 50 conjugacy classes and the 50 characters of its irreducible representations can be obtained by the commands ConjugacyClasses(O(0, 5, 3)) and Irr(O(0, 5, 3)). It is no problem to compute the trace of an element g ∈ O(5, 3) acting on ℂ[M′/M] because this is merely the number of the elements α ∈ M′/M such that gα = α.

We formulate the result of the calculation and distinguish the two subspaces of dimension 122 resp. 121 where −E ∈ O(5, 3) acts by multiplication with 1 resp. −1. The representations ϱ_{1} …, ϱ_{4} belong to the second type, the rest ϱ_{5}, …, ϱ_{7} to the first type.

3.4 Proposition. Case 1: −E ∈ O(5, 3) acts by multiplication with −1. The representation of SL(2, 𝔽_{3}) × O(5, 3) on the isotypic components of the representations ϱ_{1}, …, ϱ_{4} is irreducible. Case 2: −E acts trivial. The representation of SL(2, 𝔽_{3}) × O(5, 3) on the isotypic component of ϱ_{5} (dimension 32) is the sum of two irreducible representations of dimension 2 and 30. The representation on the isotypic component of ϱ_{6} (dimension 40) is irreducible. Finally the representation on the isotypic component of ϱ_{7} (dimension 50) is the sum of two irreducible representations of dimension 2 and 48.

We don’t write down here the precise characters of the representations, but we want to mention that in the two 2-dimensional representations in case 2 the group O(5, 3) acts trivial. We also mention that in case 1 the group O(5, 3) acts irreducibly on the spaces of dimension 6, 10, 15, because SL(2, 𝔽_{3}) acts by a one-dimensional representation. The 10-dimensional case of is very important:

3.5 Lemma. The space ℂ[M′/M]^{SL(2,ℤ)} is ten dimensional. The group O(5, 3) acts irreducibly on this space. A concrete element (C_{α})_{α∈M′/M} is given by

1. C_{α} = 1 for the vectors α in the list (0, 0, 1, −1, −1), (0, 1, 0, −1, −1), (1, 0, −1, 0, 0) (1, 0, 0, 0, 1), (1, 0, 0, 1, 0), (1, 1, 0, 0, 0), (1, 1, 1, 1, 1)

2. and C_{α} = −1 for the negatives of these vectors.

3. C_{α} = −1 for the vectors α in the list (0, 1, −1, 0, 1), (0, 1, −1, 1, 0), (0, 1, 0, −1, 1), (0, 1, 0, 1, −1), (1, −1, −1, −1, −1), (1, −1, 1, −1, −1), (1, −1, 1, 1, 1), (1, 1, 1, −1, 1), (1, 1, 1, 1, −1)

4. and C_{α} = 1 for the negatives of these vectors.

5. C_{α} = 0 for the rest.

4. Borcherds additive lift

In his paper [Bo1] Borcherds gave two constructions for automorphic forms on O(2, n), the multiplicative lift (Borcherds products) and also an additive lift, which generalizes correspondences of Shimura-Doi-Naganuma-Maass-Oda-Gritsenko and others ([Bo1], §14). The inputs of these lifts are elliptic modular forms which are allowed to have poles at cusps. We will use the additive lift here only in the case where the inputs are entire modular forms, i.e. no poles at cusps are allowed. Even in this special case the setting of Borcherds reaches beyond the constructions known before.

We recall some basic facts about automorphic forms on O(2, n). Let H_{n} denote the hermitean symmetric space associated to O(2, n). It can be realized as an open subset of the zero quadric defined by (z, z) = 0 in the projective space P (V (ℂ)), where

V = M ⊗_{ℤ} ℝ, V (ℂ) = V ⊗_{ℝ} ℂ

and where we extended (·, ·) to a ℂ-bilinear form on V (ℂ). The set H_{n} is one of the two connected components of the open subset of the zero quadric defined by (z, z̄) > 0. A subgroup O′(V) of index two of the orthogonal group O(V) acts on H_{n} as group of biholomorphic transformations. For later use we notice

4.1 Lemma. The map

ℂz ⟼ ℝx + ℝy

defines a bijection from H_{n} to the Grassmanian of all two-dimensional positive definite subspaces of V.

Let H̃_{n} denote the inverse image of H_{n} in V (ℂ). A meromorphic automorphic form of weight k ∈ ℤ with respect to a subgroup of finite index

Γ ⊂ O′(M) = O(M) ∩ O′(V)

and with respect to a character v on Γ is a meromorphic function f on H̃ with the properties

a) f(γz) = v(γ)f(z) for all γ ∈ Γ.

b) f(tz) = t^{−k}f(z) for all t ∈ ℂ^{•}. We are especially interested in entire modular forms, which have no poles (including the cusps). We denote the space of entire modular forms by [Γ, k, v] or simply by [Γ, r] if the character v is trivial.

We have to explain part of Borcherds theorem 14.3 in [Bo1].

§4. Borcherds additive lift 11

We start with an even lattice (M, q) of signature (2, n) (n even). We will produce an automorphic form on O(2, n) of a certain (here integral) weight r > 0. To be more precise this will be a holomorphic automorphic form F : H_{n} → ℂ with respect to the ”discriminant kernel“

Γ = kernel(O′(M) ⟶ Aut(M′/M))

and with trivial multiplier system on Γ. The form F is a theta lift of an elliptic modular f form of weight r + 1 − n/2 with respect to the Weil representation. More precisely, f = (f_{α})_{α∈M′/M} is a vector valued modular form as in §2 of the transformation type

f_{α}(τ + 1) = e^{2πiq(α)}f_{α}(τ), 1.

f_{α}(−\frac{1}{τ} ) = τ^{r+1−n/2} \begin{matrix}i & 1−n/2 \\ √ & \overline{#M^{∗}/M} \end{matrix} ∑_{β∈M^{∗}/M} e^{−2πi⟨α,β⟩}f_{β}(τ), 2.

3. f is holomorphic at the cusp infinity.

Borcherds additive lift f ↦ F is defined by a theta correspondence and in his theorem 14.3 he gives explicit formulae for the Fourier coefficients of F at all the cusps of Γ in terms of the Fourier coefficients of f. We mentioned already that special cases of this lift have been constructed by various authors using different methods. Because of condition 3. (which is not really necessary) we are forced to assume r ≥ n/2 − 1. The border case r = n/2 − 1 is the socalled singular weight. One can show, that beyond this weight, non-vanishing holomorphic automorphic forms do not exist. We see that singular forms (i.e. forms of weight n/2−1 are obtained as additive lifts of modular forms of weight 0, i.e. of constant modular forms! It is natural to ask, whether all singular forms can be obtained in this way.

We return to our case of interest, M = E − E^{4}. We proved (3.3) that the multiplicity of the trivial representation in the Weil representation of M is 10. This gives us:

4.2 Proposition. The space of modular forms of weight 0 with respect to the Weil representation of M = E − E^{4} has dimension 10. The group O(5, 3) acts irreducibly on this space. Borcherds additive lift is an O(5, 3)-equivariant linear map of this 10-dimensional space into the space of automorphic forms of weight three on the introduced group Γ ⊂ O(M) (with trivial multiplier system),

ℂ[M′/M]^{SL(2,ℤ)} ⟶ [Γ, 3].

Another example comes from the unique six dimensional subspace of ℂ[M′/M], which is irreducible under O(5, 3) and which is the sum of six one dimensional representations ϱ_{3} of SL(2, 𝔽_{3}) (s. 3.3 and 3.4). A modular form with respect to the character ϱ_{3} is automatically a cusp form. It has period 3. Hence it is divisible by η^{8}, where η denotes the Dedekind η-function. We consider the weight k = 12. The space of elliptic modular forms of weight 8 is generated by the Eisenstein series (of level one) G_{8}. We obtain that η^{8}G_{8} spans the space of modular forms of weight 12 on SL(2, ℤ) with character ϱ_{3}.

4.3 Proposition. The space of modular forms of weight 12 and with respect to ϱ_{3}×⋯×ϱ_{3} (six copies) has dimension six. By means of the Weil representation of M one obtains an irreducible action of O(5, 3) on this space. Borcherds additive lift defines an O(5, 3)-equivariant map of this space into the space of automorphic forms of weight 15 on Γ (and with trivial multiplier system).

In both cases 4.2 and 4.3 it is clear that the additive lift either is zero or injective. We will show that it is injective at least in the first case.

Fourier expansions of automorphic forms

We consider an even lattice M of signature (2, n), which contains a non-zero isotropic vector. The parabolic subgroups of O(V) which correspond to the zero dimensional boundary components are the stabilizers of isotropic vectors. We want to describe the Fourier expansion of an automorphic form at a cusp. We follow the presentation given in [Bo1]. We select a non zero isotropic vector α ∈ M(ℚ), (q(α) = 0). Let L(ℚ) be the orthogonal complement of α in M(ℚ). The bilinear form on L(ℚ) is degenerated but the induced bilinear form on

K(ℚ) := L(ℚ)/ℚα = (M(ℚ) ∩ α^{⊥})/ℚα

is non-degenerated. More precisely, K(ℚ) is a hyperbolic (i.e. of signature (1, n − 1)) space.

We choose a vector α′ ∈ M(ℚ) such that (α, α′) ≠ 0. The projection

L(ℚ) ∩ α^{′⊥} → K(ℚ)

is an isomorphism. We obtain an embedding K(ℚ) ↪ L(ℚ) (depending on (α, α′)). We use this embedding to identify K(ℚ) with a subspace of M(ℚ). Then we have

K(ℚ) = M(ℚ) ∩ α^{⊥} ∩ α^{′⊥}

(similarly after replacing ℚ by ℝ or ℂ). We obtain an isomorphism

K(ℂ) × ℂ × ℂ ⟶ V (ℂ), (Z, m, n) ⟼ Z + mα′ + nα.

The quadratic form in this model is

q(Z, m, n) = q(Z) + mn(α, α′) + m^{2}q(α′).

Let z = (Z, m, n) be an element which represents a point in H_{n}. Then m is different from zero and we can normalize m = 1. From (z, z) = 0 we obtain after the normalization z = (Z, 1, −(q(Z) + q(α′)/(α, α′)). The assignment z ↦ Z gives an embedding

H_{n} ⟶ K(ℂ).

The image is the tube domain of all Z = X + iY, where Y varies in one of the two connected components of elements with positive norm ((Y, Y) > 0). We call this connected component (which depends on the choice of H_{n}) the positive cone. This is an open subset of K(ℝ) = K(ℚ) ⊗_{ℚ} ℝ.

4.4 Lemma. The linear transformations of V of the form

(X, m, n) ⟼ (X + mS, m, n − ((X, S) + mq(S))/(α, α′)), S ∈ V,

are orthogonal transformations in the stabilizer of z. They act on the tube domain as translations, Z ↦ Z + S.

Let f be an automorphic form with respect to a subgroup of finite Index Γ ⊂ O′(M). We realize f as function on our tube domain by defining

f_{α,α′}(Z) := f(Z, 1, −(q(Z) + q(α′))/(α, α′)).

Then f is periodic with respect to some lattice K ⊂ K(ℚ). Hence it admits a Fourier expansion

f_{α,α′}(Z) = ∑_{T ∈K′} a(T) exp 2πi(T, Z).

The Koecher principle states that up to some low dimensional exceptions a(T) can be different from 0 only if either T = 0 or (T, Y) > 0 for all Y in the positive cone. A different way to express this is that T is contained in the closure of the dual cone.

A good choice for the vectors α and α′ is as follows: We take α as a primitive isotropic vector α ∈ M, i.e.

α ∈ M; q(α) = 0; ℚα ∩ M = ℤα.

Because α is primitive, there exists a vector α′ ∈ M′ such that (α, α′) = 1. We choose α′ in this way. Let L be the orthogonal complement of α in M. This is a lattice in L(ℚ) and

K := L/ℤα = (M ∩ α^{⊥})/ℤα

is a lattice in K(ℚ). We can consider K as a subgroup of M(ℚ) by means of the embedding of K(ℚ) into M(ℚ) (which has been constructed by means of α′). But K is not necessarily contained in M. If f is an automorphic form with trivial multiplier system on the discriminant kernel Γ ⊂ O′(M), then the elements of K obviously are periods of f_{α,α′}.

We recall Borcherds formula for the Fourier coefficients of the additive lifts (using the special choices for α, α′). We denote now by N the natural number with the property

(α, M) = N ℤ,

i.e. the biggest number such that α/N ∈ M′ (and not the level of ϱ).

4.5 Theorem (Borcherds, 14.3 in [Bo1]). Let f = (f_{α})_{α∈M′/M} be an elliptic modular form of weight r + 1 − n/2 with respect to the Weil representation and let

f_{α}(τ) = ∑_{n∈ℚ} c_{α}(n) exp 2πinτ

be the Fourier expansion. Then the Fourier coefficients of Borcherds additive lift F (Z) = ∑ a(T) exp 2πi(T, Z) at the cusp defined by (α, α′) are

a(0) = −\frac{N^{r−1}}{2r} ∑_{ν∈ℤ/N ℤ} c_{να/N}(0) ∑_{ϵ=1}^{N} exp(2πiνϵ/N)B_{r}(ϵ/N),
a(T) = ∑_{{T =nH \atop n∈ℕ, H∈K′}} n^{r−1} ∑_{{δ∈M′/M \atop δ over H}} exp(2πi(δ, α′))c_{δ}(q(H))  (T ≠ 0).

Here ”δ over H“ means that there exists a natural number n such that

δ = H + \frac{nα}{N} mod M  and  q(δ) ≡ q(H) mod 1.

(If the element α ∈ M′/M is represented by a ∈ M′, we use the notation c_{a}(n) = c_{α}(n). Notice that the element να/N occurring in 4.5 is contained in M′.) The formula simplifies considerably if N = 1. In this case we may choose α′ ∈ M and M splits as an orthogonal sum of a hyperbolic plane and K ⊂ M(ℚ),

M = (ℤα′ + ℤα) ⊥ K.

The dual K′ of K is a subgroup of M′, more precisely K′ = K(ℚ) ∩ M′. One obtains

4.6 Corollary of 4.5. Assume α′ ∈ M. Then

\begin{aligned}a(0) & = −\frac{1}{2r} c_{0}(0)B_{r}(1), \\ a(T) & = ∑_{T =nH} n^{r−1}c_{H}(q(H)) (T ≠ 0). \end{aligned}

If especially T ∈ K′ is a primitive nonzero vector, then simply a(T) = c_{T} (q(T)). If T = nH with a non-zero isotropic H, primitive in K′, then

a(T) = n^{r−1} ∑_{d|n} d^{1−r}c_{dH}(0).

As an example we prove that the additive lift described in 4.2, which produces forms of the singular weight 3, is injective. As primitive isotropic vector of the lattice E − E^{4} with bilinear form

(x, y) = 2 Re(x̄_{1}y_{1}) − 2 Re(x̄_{2}y_{2}) − ⋯ − 2 Re(x̄_{5}y_{5})

we take α = (1, 1, 0, …, 0) and as corresponding vector α′ = (−ζ, 0, 0, 0, 0)

(recall ζ = −1/2 + \sqrt{−3}/2). We obtain N = 1.  ⊔ ⊓

§5. Borcherds multiplicative lift 15

4.7 Proposition. The additive lift of the ten dimensional space of modular forms of weight 0 (s. 4.2) is injective. Hence we obtain a ten dimensional space of automorphic forms of weight three such that the group O′(M) acts through the factor group O(5, 3) irreducibly.

5. Borcherds multiplicative lift

Let α ∈ M′ be an element of the dual lattice and n < 0 a negative integer. The Heegner divisor H(α, n) ⊂ H_{n} is the union of all

v^{⊥} ∩ H_{n} (v^{⊥} orthogonal complement of v in P (V (ℂ)),

where v runs through all elements α ∈ M′ with

v ≡ α mod M and q(α) = n.

We consider H(α, n) as a divisor on H_{n} by attaching multiplicity 1 to all components. We have H(α, n) = H(−α, n), more precisely, this divisor depends only on the image of α in (M′/M)/ ± 1. Borcherds products are meromorphic automorphic forms whose zeros and poles lie on Heegner divisors.

Borcherds introduced in [B2] the following ”space of obstructions“ which regulates the existence of these products. It consists of all elliptic modular forms of weight (2+n)/2 with respect to the dual ϱ^{∗} of the Weil representation. The precise transformation law of such a form (f_{α})_{α∈M′/M} is

f_{α}(τ + 1) = e^{−2πiq(α)}f_{α}(τ) 1.

f_{α}(−\frac{1}{τ} ) = −\sqrt{\frac{τ}{i}}^{2+n} \frac{1}{\sqrt{#M^{∗}/M}} ∑_{β∈M^{∗}/M} e^{2πi⟨α,β⟩}f_{β}(τ). 2.

3. f is holomorphic at the cusp infinity.

The weight of this modular form is k = (2 + n)/2.

A fundamental result of Borcherds states [Bo2]:

5.1 Proposition. A finite linear combination

∑_{α∈(M′/M)/±1, n<0} C(α, n)H(α, n) (C(α, n) ∈ ℤ)

is the divisor of a meromorphic automorphic form of weight k if there exist numbers (D_{α})_{α∈M′/M} with D_{0} = 2k and such that for every f in the the space of obstructions,

f_{α}(τ) = ∑_{n∈ℚ} a_{α}(n) exp(2πinτ),

the relation

∑_{α∈M′/M} D_{α}a_{α}(0) + ∑_{n<0, α∈M′/M} a_{α}(−n)C(α, n) = 0

holds.

Using the vector valued Eisenstein series, we can reformulate this theorem in the following convenient form:

5.2 Thorem. Assume n > 2. A finite linear combination

∑_{α∈(L′/L)/±1, n<0} C(α, n)H(α, n) (C(α, n) ∈ ℤ)

is the divisor of a meromorphic automorphic form of weight k if for every cusp form f in the the space of obstructions,

f_{α}(τ) = ∑_{n∈ℚ} a_{α}(n) exp(2πinτ),

the relation

∑_{n<0, α∈L′/L} a_{α}(−n)C(α, n) = 0

holds. The weight of this modular form is

k = ∑_{n∈ℤ, α∈L′/L} b_{α}(n)C(α, −n),

where b_{α}(n) denotes the Fourier coefficients of the Eisenstein series with the constant term

b_{α}(0) = \{ \begin{matrix}−1/2 & if α = 0, \\ 0 & else. \end{matrix}

Corollary. An individual Heegner divisor H(α, n) is the divisor of an automorphic form if and only if a_{α}(n) = 0 for every cusp form in the the space of obstructions.

The type of an element α ∈ M′/M is defined as 00 if α is the zero element and as n ∈ {0, 1, 2} if α is different from zero and q(α) ≡ n/3 mod 1. So we have 4 types 00, 0, 1, 2. One can show that the orbits of M′/M under O(M) are given by the 4 types. The same is true for (M′/M)/ ± 1.

5.3 Remark. The set (M′/M)/±1 contains 1 element of type 00, 40 elements of type 0, 45 elements of type 1 and 36 elements of type 2.

We want to investigate divisors ∑_{α,n<0} C(α, n)H(α, n), where C(α, n) depends only on the type of α. Let h be an element in the space of obstructions. We need for this purpose only the sums ∑ h_{α}, where α runs through a given type. We use the obvious notation h_{00}, h_{0}, h_{1}, h_{2} for these sums. After some calculation one obtains the equations

\begin{aligned}h_{00}(τ + 1) & = h_{00}(τ) \\ h_{0}(τ + 1) & = h_{0}(τ) \\ h_{1}(τ + 1) & = ζ^{2}h_{1}(τ) \\ h_{2}(τ + 1) & = ζh_{2}(τ) \end{aligned} \begin{aligned}h_{00}(−1/τ) & = iτ^{5}/3^{5/2}(h_{00}(τ) + h_{0}(τ) + h_{1}(τ) + h_{2}(τ)) \\ h_{0}(−1/τ) & = iτ^{5}/3^{5/2}(80h_{00}(τ) − h_{0}(τ) + 8h_{1}(τ) − 10h_{2}(τ)) \\ h_{1}(−1/τ) & = iτ^{5}/3^{5/2}(90h_{00}(τ) + 9h_{0}(τ) − 9h_{1}(τ)) \\ h_{2}(−1/τ) & = iτ^{5}/3^{5/2}(72h_{00}(τ) − 9h_{0}(τ) + 9h_{2}(τ)) \end{aligned}

This is precisely the representation which we considered in 2.3:

5.4 Lemma. The space of obstructions h_{00}, h_{0}, h_{1}, h_{2} has dimension two, the same as the dimension of the space of Eisenstein series. Corollary. Every finite combination

∑_{α,n<0} C(α, n)H(α, n) (C(α, n) ∈ ℤ),

where C(α, n) depends only on the type of α is the divisor of a meromorphic automorphic form.

The Fourier coefficients in 2.4 are weights of automorphic forms, for example:

5.5 Proposition. There is a holomorphic automorphic form*) of weight 12 resp. 225 resp. 1080, whose zeros are of order 1 orthogonal to all vectors α ∈ M′ with q(α) = −1/3, resp. q(α) = −2/3 resp. q(α) = −1 and α ∉ M.

The form of weight 12 is due to Borcherds. He proved in an unpublished paper that this form is the infinite product in the sense of his theory [Bo1] attached to the modular form (with singularity at the cusp) with the components

24η(3τ)^{3}η(τ)^{−9}, −3η(3τ)^{3}η(τ)^{−9}, 0, η(τ/3)^{3}η(τ)^{−9}

(corresponding to the types 00, 0, 1, 2). Another interesting example is the coefficient 1080 in the second line. The corresponding automorphic form of this weight vanishes of order 1 on the orthogonal complements α ∈ M′ − M, q(α) = −1.

*) with respect to the full modular group O(M) but possibly with non-trivial multipliers

Each of the 36 vectors of type 2 has a representative α with q(α) = −1/3. This means that the zero divisor of the above form of weight 12 breaks into 36 pieces. It is natural to ask whether this corresponds to a factorization of the modular form as a product of 36. Each factor would have weight 1/3. But this is not possible because the smallest weight of a non-vanishing modular holomorphic automorphic form is the singular weight n/2 − 1 = 3. This shows that in the space of obstructions must occur cusp forms. We have to investigate them. The space of Eisenstein series has dimension 41 because this is the number of elements of α ∈ (M′/N)/ ± 1 such that q(α) ≡ 0 mod 1.

5.6 Lemma. The space of obstructions for the lattice M = E − E^{4} has dimension 61. The space of cusp forms has dimension 20.

The dimension can be computed again by means of the Riemann-Roch formula. On has to compute the traces of the relevant three 122 × 122-matrices. The three α-values in 2.1 are 61/2, 118/3, 42. A different proof comes from the following consideration. The Dedekind η-function

η(τ) = q^{1/24} ∏_{n=1}^{∞}(1 − q^{n}) (q = e^{2πiτ} )

is a modular form of weight 1/2 on the full elliptic modular group. Its eighth power has trivial multiplier system on the principal congruence subgroup of level 3. The vanishing order at the cusp ∞ is 1 if one takes q^{1/3} as uniformzing parameter. The similar statement is true for all other cusps. Hence every cusp form of weight 5 and of level 3 is the product of η^{8} and a modular form of weight 1. There is no nonzero cusp from of weight 1 (because a cusp form of weight one can be divided by η^{8}). Therefore a modular form of weight 1 must be an Eisenstein series and we obtain from the theory of Eisenstein series of weight 1:

5.7 Proposition. Every cusp form in the space of obstructions is of the form η^{8}(τ)E_{a}(τ) where a ∈ ℂ[M′/M] is an element supported on vectors α of type 1 (i.e. q(α) ≡ 1/3 mod 1) and E_{a} is the Eisenstein series of weight 1 (defined by analytic continuation)

E_{a}(τ) = ∑_{M∈SL(2,ℤ)_{∞}\ SL(2,ℤ)}(cτ + d)^{−1}|cτ + d|^{−s}ϱ_{0}(M)^{−1}a _{s=0},

where

ϱ_{0}(M) := χ(M)^{−1}ϱ(M), η^{8}(Mτ) = χ(M)(cτ + d)^{4}η(τ).

Due to a result of Hecke ([He], No24, §2) these Eisenstein series always are entire modular forms. But the constant Fourier coefficients behave complicated. Therefore we use a different method to work out the space of obstructions: The components of the Eisenstein series belong to the principal congruence subgroup of level 3. This group has 4 cusps. But as Hecke proved ([He], No24, §2) there are only two linear independent Eisenstein series. One Eisenstein series of weight 1 is ([La], Chap.XV, H6).

G := −\frac{1}{6} − \frac{q}{1 − q} + ∑_{ν=1}^{∞} [\frac{q^{3ν−1}}{1 − q^{3ν−1}} − \frac{q^{3ν+1}}{1 − q^{3ν+1}} ] .

This form belongs to the group Γ_{1}[3] in Heckes notation. This group is defined by a ≡ d ≡ 1 and c ≡ 0 mod 3. If we replace q by q^{1/3} := exp(2πiτ/3), we obtain

H(τ) := G(τ/3),

which belongs to Γ^{1}[3]. From Heckes dimension formula we obtain that both must generate the space of Eisenstein series (hence all modular forms) of weight 1 and level 3. This has a very remarkable consequence (compare 5.13):

5.8 Lemma. Let

f = ∑_{m∈ \frac{1}{3} ℤ} a_{m}e^{2πimτ}

be a modular form of weight 1 (and trivial character) with respect to the principal congruence subgroup of level three. Then

m ≡ 2/3 mod 1 ⟹ a_{m} = 0.

The form η^{8} has only non zero coefficients for m ≡ 1/3 mod 1 and we obtain

5.9 Lemma. In the space of obstructions all Fourier coefficients a_{m} with respect to integral m vanish.

This implies:

5.10 Proposition. A Heegner divisor H(α, m) is the divisor of an automorphic form if q(α) is integral.

We can apply this result to one of our examples above: There are 40 nonzero elements in α ∈ M′/M with q(α) ≡ 0 mod 1. All can be represented by elements α ∈ M′ with q(α) = −1. Therefore the form of weight 1080 in 5.5 is the product of 40 forms all of the same weight 27.

5.11 Proposition. Let α ∈ M′/M be an element which is different from zero but with q(α) ≡ 0 mod 1. The Heegner divisor H(α, −1) is the divisor of an automorphic form of weight 27.

We want to determine the 20-dimensional space of cusp forms. For this we need the precise transformation formula of G and H under the modular group. We use the Petersson notation

f|M(τ) := (cτ + d)^{−k}f(Mτ) (M ∈ GL_{+}(2, ℝ)).

Firstly we notice that G is the only form in the space generated by both which has period 1, i.e. that it is a modular form on Γ_{1}[3]. The Fricke involution

S_{3} := ( \begin{matrix}0 & −1 \\ 3 & 0 \end{matrix} )

normalizes this group. Therefore

G|S_{3} = C · G

with a certain constant C, which we have to determine. From the equation

(3τ)^{−1}G(−\frac{1}{3τ} ) = CG(τ)

with the special value τ = i/\sqrt{3} we get

C = −\frac{i}{\sqrt{3}}

after verifying by a numerical calculation that G(i/\sqrt{3}) is different from zero. Next we use

B_{3} := ( \begin{matrix}3 & 0 \\ 0 & 1 \end{matrix} ) .

We have G|B_{3}^{−1} = H. From the equality

S = S_{3}B_{3}^{−1}

we obtain

G|S = −\frac{i}{\sqrt{3}}H and H|S = −i\sqrt{3}G.

The transformation under the translation T is easy,

G|T = G and H|T = (1 − ζ)G + ζH.

5.12 Lemma. The space of modular forms of weight 1 and level 3 has a basis G, H which transforms as

\begin{matrix}G(τ + 1) = G(τ) & G(−\frac{1}{τ} ) = −\frac{i}{\sqrt{3}} τ H(τ) \\ H(τ + 1) = (1 − ζ)G(τ) + ζH(τ) & H(−\frac{1}{τ} ) = −i\sqrt{3} τ G(τ). \end{matrix}

Both forms have the same value at ∞.

We need the eigenforms under T. The eigenvalues are 1 and ζ. Corresponding eigenforms are G and F := G − H.

§6. A unitary group 21

5.13 Lemma. The space of modular forms of weight 1 and level 3 has a basis F = G − H, G which transforms as

\begin{matrix}F (τ + 1) = ζF (τ) & F (−\frac{1}{τ} ) = \frac{i}{\sqrt{3}} τ F (τ) + \frac{2i}{\sqrt{3}} τ G(τ) \\ G(τ + 1) = G(τ) & G(−\frac{1}{τ} ) = \frac{i}{\sqrt{3}} τ (F (τ) − G(τ)). \end{matrix}

The value of F at ∞ is 0.

The elements in the space of cuspidal obstructions (f_{α})_{α∈M′/M} are of the form

f_{α}(τ) = \{ \begin{matrix}C_{α}η(τ)^{8}F (τ) & if q(α) ≡ 1/3 mod 1 \\ C_{α}η(τ)^{8}G(τ) & if q(α) ≡ 2/3 mod 1 \\ 0 & if q(α) ≡ 0 mod 1. \end{matrix}

with certain constants C_{α} = C_{−α}. We define C_{α} = 0 if q(α) ≡ 0 mod 1. Using the definition of the space of obstructions and the formula η(−1/τ)^{8} = τ^{4}η(τ)^{8} we obtain the following equations for the constants C_{α} = C_{−α}.

\begin{matrix}C_{α} = & 3^{−2} ∑_{q(β)=1/3} e^{2πi(α,β)}C_{β}, & q(α) = 1/3, \\ 2C_{α} = & 3^{−2} ∑_{q(β)=2/3} e^{2πi(α,β)}C_{β}, & q(α) = 1/3, \\ C_{α} = & 3^{−2} ∑_{q(β)=1/3} e^{2πi(α,β)}C_{β}, & q(α) = 2/3, \\ −C_{α} = & 3^{−2} ∑_{q(β)=2/3} e^{2πi(α,β)}C_{β}, & q(α) = 2/3 \end{matrix}

and the additional equation

∑_{q(β)=1/3} e^{2πi(α,β)}C_{β} = ∑_{q(β)=2/3} e^{2πi(α,β)}C_{β} = 0 if q(α) = 0.

It is easy to solve this system of equations with the help of a computer and to verify that the dimension of the space of solutions is twenty. Because the Fourier coefficients of F and G are known it is now possible to decide which linear combinations of Heegner divisors are Borcherds products. One consequence is:

5.14 Proposition. Let α ∈ M′ be a vector with q(α) = −a/3 where a ∈ {0, 1}. The divisor H(α, −a/3) is not the divisor of a Borcherds product.

It is expected*) that every automorphic form on a lattice of signature (2, n), n ≥ 2, whose divisor is a linear combination of Heegner divisors, is a Borcherds product. This means that in 5.14 one can replace “Borcherds product” by “automorphic form”. Computing traces and comparing them with character tables of G_{25 920} one obtains:

5.15 Proposition. The representation of the orthogonal group G_{25 920} on the twenty dimensional space of cuspidal obstructions is irreducible.

*) This has been proved meanwhile by Bruinier in many cases, [Br1,2].

6. A unitary group

Let V be a complex vector space. We assume that a hermitean form ⟨a, b⟩ = \overline{⟨b, a⟩} on V is given (ℂ-linear in b and antilinear in a). We assume that the signature of ⟨a, b⟩ is (1, n). The domain

B_{n} := { a ∈ P (V); ⟨a, a⟩ > 0 }

in the projective space of V is a realization of the n-ball.

Let 𝔬 be an order in an imaginary number field 𝔽. We consider a lattice M ⊂ V such that E acts on M by multiplication and such that ⟨a, b⟩ is contained in 𝔽 for all a, b ∈ M. The unitary group U(V) consists of all ℂ-linear automorphisms of V which preserve the hermitean form. This group acts on B_{n}. We have to consider the integral unitary group

U(M) := { g ∈ U(V); g(M) = M }.

It is clear how to define the notion of an automorphic form with respect to a subgroup of finite index Γ ⊂ U(M) (compare §2). It consists of meromorphic functions f on the inverse image B̃_{n} of B_{n} in V − {0} with the properties a) f(γz) = v(γ)f(z) for all γ ∈ Γ.

b) f(tz) = t^{−k}f(z) for all t ∈ ℂ^{•}. Here v is a character on Γ and k is an integer, the so-called weight of f. As in the orthogonal case we use the notation [Γ, k, v] for the space of all entire automorphic forms and write simply [Γ, k], when the multipliers are trivial. “Entire” means holomorphic in B_{n} and regular at ∞. We have to explain what this means:

We choose a non-zero isotropic vector α ∈ M ⊗ ℚ and a vector α′ such that ⟨α, α′⟩ ≠ 0. We consider

f_{α,α′}(τ, w) := f(τα + 2i⟨α′, α⟩α′ + w)

where w is orthogonal to α and α′. The positivity condition means that τ can vary in an upper half plane Im τ > C, where C depends on w. It is easy to show that the translations (τ, w) ↦ (τ + ν, w) are induced by transformations from U(V), if ν is real and even more by transformations from U(M) if ν is the multiple of a certain non-zero element. We assume that f is holomorphic and obtain a Fourier expansion

f(τ, w) = ∑ a_{n}(w)e^{2πinτ} .

Regularity at ∞ means that a_{n}(w) vanishes for n < 0 (for all choices of α, α′). Then we can define

f_{α}(i∞) = lim_{Im τ→∞} f_{α,α′} = a_{0}.

It is easy to prove that a_{0} is a constant*), i.e. independent of w.

6.1 Lemma. The value f_{α}(i∞) = lim_{Im τ→∞} f_{α,α′}(τ, w) does not depend on the choice of α′. Moreover it depends only on the Γ-orbit of α, if the multipliers of f are trivial. Finally f_{Cα}(i∞) = C^{k}f_{α}(i∞).

Proof. It is clear that f_{α,α′} depends only on ℂα′. Hence we may restrict to the case ⟨α, α′⟩ = 1. We consider another α′′ with ⟨α, α′′⟩ = 1 and write

τα/2i + α′ + w = C(τ̃α/2i + α′′ + w̃) (⟨w̃, α⟩ = ⟨w̃, α′′⟩ = 0).

Taking scalar products with α, α′ we obtain C = 1 and that τ̃ = τ + A, where A is independent of τ. This proves the first part. The second part is clear.

⊔ ⊓

We have to consider V also as a real vector space equipped with the bilinear form

(a, b) := 2 Re⟨a, b⟩ = ⟨a, b⟩ + ⟨b, a⟩

of signature (2, n). Every unitary transformation is also an orthogonal transformation, which means that we have a natural inclusion

U(1, n) ⊂ O(2, 2n).

There is a corresponding inclusion of the Hermitean domains B_{n} ↪ H_{2n}. If we identify B_{n} with the set of positive definite one-dimensional complex subspaces of (V, ⟨·, ·⟩) and H_{2n} with the set of positive definite two-dimensional real subspaces of (V, (·, ·)) (s. 4.1), the embedding is the tautological one. We have to describe this embedding with some care in the coordinates, we used above. The problem is that H_{2n} sits inside V ⊗_{ℝ} ℂ and that V is already a complex vector space.

To avoid confusion we introduce a second field of complex numbers

ℂ′ := ℝ + jℂ, j^{2} = −1.

and realize the complexification of V concretely as vector space over ℂ′.

Ṽ = V (ℂ′) := V ⊗_{ℝ} ℂ′ = V + jV.

We extend the multiplication by i as a ℂ′-linear map on Ṽ. This means that Ṽ is a vector space over ℂ and over ℂ′ and both actions commute.

*) One shows that a_{0}(w) is invariant under a certain lattice, whereas the higher coefficients are theta functions for this lattice, compare for example [Sh].

6.2 Remark. The map

\begin{matrix}φ : V ⟶ V (ℂ′), \\ φ(v) := iv + jv = (iv) ⊗ 1 + j(v ⊗ 1) = (iv) ⊗ 1 + v ⊗ j, \end{matrix}

has the property

φ(iv) = jφ(v) = iφ(v).

This is clear because iv goes to −v + jiv which is the same as j(iv + jv).  ⊔ ⊓

6.3 Remark. The image of φ : V → Ṽ is defined by the equation ia = ja. The image of a ℂ-subspace of V is a ℂ′-subspace (not only a ℂ-subspace) of Ṽ. Especially a map P (V) → P (Ṽ) is induced, where P (Ṽ) is the projective space associated to the ℂ′-vector space Ṽ.

Recall that we have to extend (a, b) to a symmetric ℂ′-bilinear form on V (ℂ′), which also is denoted by (·, ·).

The image of B_{n} under the map φ : V → Ṽ (6.3) is contained in the set defined by (z, z) = 0 and (z, z̄) > 0. We may assume that it is contained in the connected component H_{2n}. We obtain a commutative diagram

B_{n}  ⟶^{φ}  H_{2n}
∩  ∩
P (V) ⟶^{φ} P (V (ℂ′)) .

If ℂz ∈ B_{n} is a given point then its image ℂ′(iz + jz) corresponds to the two-dimensional definite subspace ℝiz + ℝz in the Grassmanian model (4.1). This is precisely ℂz itself, i.e. the embedding we constructed is the mentioned tautological one. This embedding is equivariant with respect to U(M), which is a subgroup of O′(M).

Recall that we have to extend the action of O′(M) to the complexification of V, which now means the ℂ′-linear extension to Ṽ = V ⊗_{ℝ} ℂ′. Especially the subgroup U(M) acts on Ṽ as a group of ℂ′-linear transformations. The embedding (s. 6.2) φ : V ↪ Ṽ is compatible with the action of U(M).

We consider the isomorphism

σ : ℂ ⟶ ℂ′, x + iy ⟼ x + jy.

We define the restriction of a ℂ′-valued function F : Ṽ → ℂ′ with respect to φ as the ℂ-valued function f : V → ℂ making the diagram

Ṽ ⟶^{F} ℂ′
φ ↑  ↑ σ
V ⟶^{f} ℂ

commutative. We use the notation f := F |V. It is easy to check that f is ℂ-linear if F is ℂ′-linear. More general we can define in the same way the restriction of a ℂ′-valued function F on an open subset W ⊂ Ṽ to obtain a ℂ-valued function f on the open subset φ^{−1}(W) ⊂ V. If F is holomorphic (with respect to the field ℂ′) then f is holomorphic with respect to ℂ. This follows from the stated result about linearity.

Let Γ ⊂ O′(M) be a subgroup of finite index of the integral orthogonal group. The restriction of an automorphic form on H_{2n} with respect to Γ to the ball B_{n} is an automorphic form with respect to Γ ∩ U(M) of the same weight. We recall that the restriction of a ℂ′-valued function F : H_{2n} → ℂ′ is the function f : B_{n} → ℂ making the diagram

H_{2n} ⟶^{F} ℂ′
φ ↑  ↑ σ
B_{n} ⟶^{f} ℂ

commutative.

We discuss the case of our special lattice M = E − E^{4}. The imaginary quadratic field is 𝔽 = ℚ(ζ), ζ = −1/2 + \sqrt{−3}/2. We denote from now on by Γ the kernel of

U(M) ⟶ Aut(M/\sqrt{−3}M) = O(5, 3).

This is the intersection of the kernel of O′(M) → Aut(M′/M) with U(1, 4). Therefore we can restrict the automorphic forms which we constructed on H_{8} to get automorphic forms on B_{4} with respect to Γ.

We defined f_{α}(i∞) for an automorphic form f ∈ [Γ, k] and an isotropic vector α. We assume now that α is a primitive element from M. We assume that k is divisible by 3. Then f_{α}(i∞) remains unchanged if on replaces α by ζα, where ζ is a third root of unity. It depends also only on the Γ-orbit of α.

6.4 Proposition. Two primitive isotropic elements a, b ∈ M have the same image in 𝔽_{3}^{5} if and only if there exists a third root of unity ζ such that b and ζa are in the same Γ-orbit.

The images of the primitive isotropic elements in 𝔽_{3}^{5} are precisely the nonzero primitive elements of 𝔽_{3}^{5}. We know that there total number is 80. If f ∈ [Γ, k] is an automorphic form with weight divisible by 3, then we can define φ(a) = f_{α}(i∞) for a non-zero isotropic element a ∈ 𝔽_{3}^{5} choosing a primitive representative α ∈ M. We extend φ by 0 to a function φ : 𝔽_{3}^{5} → ℂ. Recall that we identified 𝔽_{3}^{5} with M′/M. Therefore we can consider φ as element of ℂ[M′/M].

We constructed a certain ten dimensional space of modular forms on a O(2, 8) by lifting the space of constant elliptic modular forms ℂ[M′/M]^{SL(2,ℤ)}. By restriction we obtain a map

ℂ[M′/M]^{SL(2,ℤ)} ⟶ [Γ, 3].

We can combine this map with

[Γ, 3] ⟶ ℂ[M′/M], f ⟼ φ.

6.5 Theorem. The composition of the maps

ℂ[M′/M]^{SL(2,ℤ)} ⟶^{additive lift} [Γ, 3] ⟶^{boundary values} ℂ[M′/M]

is up to a non-zero constant factor the natural inclusion.

Proof. The map is O(5, 3)-equivariant. From the decomposition of the Weil representation we know that the map is ether zero or that the image is ℂ[M′/M]^{SL(2,ℤ)}. In the latter case the map is multiplication by a non-zero constant (Schur’s lemmma). In the rest of the section we prove that the map is not zero: Firstly we consider an automorphic form F on H_{8}. We want to compute the restriction f := F |B_{n} in terms of a Fourier expansion of F in an arbitrarily chosen “cusp”:

We use the cusp defined by the two vectors

β = (1, 1, 0, 0, 0), β′ = (−ζ, 0, 0, 0, 0).

The vector β is primitive isotropic and β′ is a vector in M′ (actually already in M) with the property (β, β′) = 2 Re⟨β, β′⟩ = 1. We proceed as in section 4 (where we used the letter α instead of β).

As explained in section 4, we have to determine the orthogonal complement of β and β′ with respect to the (real) bilinear form (·, ·). This complement contains the orthogonal complement of β and β′ with respect to the hermitean form ⟨·, ·⟩. We determine first this complement and denote it by W, An easy calculation gives

W := { a = (a_{1}, a_{2}, a_{3}, a_{4}, a_{5}) ∈ ℂ^{5} = E^{5} ⊗_{ℤ} ℝ; a_{1} = a_{2} = 0 }.

Now we determine the orthogonal complement of β and β′ with respect to the real form (·, ·). Obviously this is the sum of W and the orthogonal complement (with respect to (·, ·)) of ℝβ + ℝβ′ in ℂβ + ℂβ′. This is easy to compute:

6.6 Lemma. The orthogonal complement of ℝβ + ℝβ′ in ℂβ + ℂβ′ with respect to (·, ·) is generated (as ℝ-vector space) by

\begin{aligned}α & = (2 + ζ, 2 + ζ, 0, 0, 0) = (2 + ζ)β, \\ α′ & = (2 + ζ, 1 − ζ, 0, 0, 0) = (ζ − 1)β′ + (1 − ζ)β. \end{aligned}

One has

\begin{matrix}⟨α, α⟩ = 0, ⟨α′, α⟩ = −3ζ, ⟨α′, α′⟩ = 6, \\ (α, α) = 0, (α′, α) = 3, (α′, α′) = 12. \end{matrix}

The following computation will show that the Fourier expansion of f_{α,α′} can be expressed in terms of the Fourier expansion of F_{β,β′}. (Recall again that F is an automorphic form on H_{8} and f its restriction to B_{4}.)

We have to consider

f_{α,α}(τ) = f(τα + 2i⟨α′, α⟩α′ + w).

So we have to compute the image of τα + 2i⟨α′, α⟩α′ + w in H_{8} (and express it it terms of the coordinates coming form β and β′.) For simplicity we restrict at the moment to purely imaginary τ = iy, which means no loss of generality. The image of z = iyα − 6ζiα′ + w in H_{8} has been defined as

z(i + j) = −yα + 6ζα′ + jyiα − 6jiζα′ + (i + j)w.

The elements iα, ζα′, iζα′ are contained in ℂβ + ℂβ′. We have to express them by means of the ℝ-basis α, α′, β, β′:

\begin{aligned}iα & = \sqrt{3}α − 2\sqrt{3}β, \\ ζα′ & = −2α′ + 3β − 3β′, \\ iζα′ & = −\sqrt{3}β + \sqrt{3}β′. \end{aligned}

We obtain

z(i + j) = −yα − 12α′ + 18β − 18β′ + jy\sqrt{3}α
− 2jy\sqrt{3}β + 6j\sqrt{3}β − 6j\sqrt{3}β′ + (i + j)w

= −\frac{1}{18 + 6j\sqrt{3}}(β + Cβ′ + Z)

with an uninteresting constant C. The point Z represents z(i + j) in the tube domain coordinates of H_{8} with respect to (β, β′). A short calculation gives

6.7 Lemma. The image of iyα + 2i⟨α′, α⟩α′ + w ∈ B_{4} in H_{8} expressed in the tube domain model of H_{8} with respect to the pair (β, β′) is

Z = \frac{−jy\sqrt{3}}{18}α + (\frac{1}{2} − \frac{j\sqrt{3}}{6})α′ − \frac{i + j}{18 + 6j\sqrt{3}}w

Recall that the positive cone of the tube domain model of H_{8} sits inside the orthogonal complement of β and β′ with respect to the real form (·, ·). If an element of the form yα + y′α′ for real y, y′ is contained in this cone, its norm with respect to (·, ·) must be positive, i.e. yy′ > 0. This domain has two connected components y > 0 and y < 0. Our convention of the choice of the connected component was such that B_{4} is contained in H_{8}. from 6.7 we see:

6.8 Lemma. An element yα + y′α′, y, y′ ∈ ℝ, is contained in the positive cone (with respect to the pair (β, β′)) if y < 0 and y′ < 0.

(If one prefers positive y-s, one should replace the α-s by their negatives.) Now we can describe the restriction of the automorphic form F in terms of its Fourier series

F (Z) = ∑_{T ∈K′} a(T) exp 2πj(T, Z).

(Recall that the complex structure of H_{8} comes from the field ℂ′ = ℝ[j], therefore we have to take the j in the exponent.) Because (β, β′) = 1 the periodicity-lattice K (s. §4) is just the orthogonal complement (with respect to the real form (·, ·)) of β, β′ in M. We write T in the form T = −t_{0}α − t_{2}α′ + 𝔱 with non-negative integers t_{0}, t_{2} and 𝔱 in the definite sublattice of K which is orthogonal to α, α′. Now we specialize. A direct calculation gives

j(Z, T) = −(\frac{\sqrt{3}}{6}y+2\sqrt{3}+6j)t_{2}−(\frac{\sqrt{3}}{2} + \frac{3j}{2})t_{0}+ \frac{1}{9 + 3j\sqrt{3}}(Re⟨𝔱, w⟩+j Im⟨𝔱, w⟩).

This expression is contained in the field ℂ′. We recall that restriction to B_{4} means to replace this by its image in ℂ and finally we will replace iy by the complex variable τ.

6.9 Lemma. The restriction of

F (Z) = ∑_{T ∈K′} a(T) exp 2πj(T, Z)

to B_{4} is given by

f(τ, w) = ∑_{T} exp 2πi\{(\frac{\sqrt{3}}{6}τ − 2\sqrt{3}i + 6)t_{2} + (\frac{\sqrt{3}i}{2} − \frac{3}{2})t_{0} − \frac{i}{9 + 3i\sqrt{3}}⟨𝔱, w⟩\}

We can derive from 6.9 the boundary value y → ∞. Only Fourier coefficients a(T) with t_{2} = 0 survive. From (T, T) ≥ 0 follows 𝔱 = 0. We obtain T = −t_{0}α. The rational number t_{0} is non-negative by the choice of the positive cone. The condition T ∈ K′ means t_{0} = n/3 with an integer n. We obtain for the limit value the following formula

6.10 Lemma. Let F be an automorphic form on H_{4} and f its restriction to B_{4}. We choose α, α′, β, β′ as above. Then

f_{α}(i∞) = ∑_{n=0}^{∞}(−1)^{n}a(−nα/3)e^{−πn\sqrt{3}/3}.

§7. The ten-dimensional space of automorphic forms on the 4-ball 29

We can apply this formula in the special case of an additive lift of an elliptic modular form We consider the rather special case where this modular form is constant, i.e. contained in the 10-dimensional space ℂ[M′/M]^{SL(2,ℤ)}. Our vector α/3 is contained in M′ but not in M. There exists a form c ∈ ℂ[M′/M]^{SL(2,ℤ)} in our space with c_{−α/3} = 1. Then c_{α/3} = −1 or more generally

c_{−nα/3} = \{ \begin{matrix}0 & if n ≡ 0 mod 3, \\ 1 & if n ≡ 1 mod 3, \\ −1 & if n ≡ −1 mod 3. \end{matrix}

From 4.6 we obtain

a(−nα/3) = n^{2}( ∑_{{d|n \atop d≡1 mod 3}} d^{−2} − ∑_{{d|n \atop d≡−1 mod 3}} d^{−2}).

A numerical calculation shows that f_{α}(i∞) is different from zero. This shows

that the constant factor in 6.5 must be different from 0.  ⊔ ⊓

7. The ten-dimensional space of automorphic forms on the 4-ball

We investigate the ten-dimensional space of automorphic forms, which we constructed in the previous section, in more detail. We will exhibit a certain system of automorphic forms with known zero-divisors. This system will play a fundamental role in the investigation of the moduli space of cubic surfaces [AF].

Recall that we consider the lattice M = E^{5} with the hermitean form ā_{0}b_{0} − ā_{1}b_{1} − ⋯ − ā_{4}b_{4}. Most of the results, which we need for this lattice, are due to Allcock and are explained in more detail in [AF]. Recall that we have M/\sqrt{−3}M ≅ 𝔽_{3}^{5} and hence a natural restriction homomorphism U(M) → O(5, 3).

7.1 Proposition. The homomorphism U(M) → O(5, 3) is surjective. Let Γ denote its kernel. By restriction of the automorphic forms of weight three described in 4.2, one obtains a ten dimensional space of modular forms of weight three on B_{4} with respect to Γ and with trivial multiplier system. The group U(M) acts on this space by means of an irreducible representation of O(5, 3). This representation breaks into the direct sum of the two different five dimensional representations of the simple subgroup of order 25 920.

For the investigation of this ten dimensional space we need some information about the lattice M. A non-zero vector a ∈ M is called primitive if it cannot be divided in M by a non-unit of E.

A vector a ∈ M is called

1. isotropic  if ⟨a, a⟩ = 0
2. a short root if ⟨a, a⟩ = −1
3. a long root if ⟨a, a⟩ = −2.

If a, b ∈ M are in the same Γ-orbit, their images with respect to the natural projection M → 𝔽_{3}^{5} agree. In some special cases a converse result is true.

7.2 Propopsition. Let a, b be two primitive isotropic vectors, resp. two short roots resp. two long roots. The following 4 conditions are equivalent:

1. The lines ℂa, ℂb are in the same Γ-orbit.

2. There exists a sixth root of unity ζ such that b and ζa are in the same Γ-orbit.

3. The images of a and b in 𝔽_{3}^{5} generate the same line.

4. The images of a and b in 𝔽_{3}^{5} coincide up to the sign. Corollary. The number of Γ-orbits of lines ℂa, where a is a primitive isotropic vector, resp. a short root, resp. a long roots is 40, resp. 36, resp. 45.

The total number of elements of 𝔽_{3}^{5} is

3^{5} = 243 = 1 + 2(40 + 36 + 45).

If a is a short root and ζ a unit of E, i.e. a sixth root of unity, then the map

v ⟼ v − (1 − ζ)\frac{⟨a, v⟩}{⟨a, a⟩}a

is an automorphism of M. In the special case ζ = ±1 this remains also true for long roots a. This automorphism fixes the orthogonal complement of a and maps a to ζa. We call this automorphism a reflection if ζ ≠ 1. The order of a reflection is two, three or six. We call reflections of this order sometimes biflections, triflections or hexflections.

Third roots of unity are congruent 1 mod \sqrt{−3}. Therefore triflections belong to our congruence group Γ. They generate Γ modulo its center which consists of the six matrices ζE, where ζ is a sixth root of unity.

We collect some results about O(5, 3). This group contains the Weyl group W (E_{6}) of the root lattice E_{6} as subgroup of index two. It can be defined in this connection as the kernel of the spinor norm. We recall the definition of the spinor norm: If a ∈ 𝔽_{3}^{5} is an anisotropic vector, ⟨a, a⟩ ≠ 0, then the reflection

x ⟼ x − 2\frac{⟨x, a⟩}{⟨a, a⟩}a

is defined. The spinor norm of this reflection is

⟨a, a⟩ ∈ 𝔽_{3}^{∗}/(𝔽_{3}^{∗})^{2}.

This extends to a unique homomorphism

O(5, 3) ⟶ 𝔽_{3}^{∗}/(𝔽_{3}^{∗})^{2} ≅ {±1},

whose kernel is W (E_{6}). So we have to differ between two types of elements of 𝔽_{3}^{5}. An element a is called short, if ⟨a, a⟩ = −1 and long, if ⟨a, a⟩ = 1. This means that the short (long) elements are precisely the images of the short (long) roots of M.

The short reflections and the negatives of the long reflections are contained in W (E_{6}). It is known that this group contains the simple group of order 25 920 as subgroup of index two. We denote this simple group by G. This is a normal subgroup of O(5, 3) and the factor group is ℤ/2ℤ ×ℤ/2ℤ. The three subgroups of index two containing G are SO(5, 3), W (E_{6}) and ⟨G, x ↦ −x⟩. It is known that O(5, 3) is generated by the long reflections (their number is 45) and that W (E_{6}) is generated by the short reflections (their number is 36). It is known that W (E_{6}) is generated by short reflections.

Let a ∈ L be a root. We consider the orthogonal complement of a in V. The intersection with B_{n} is not empty because a has negative norm. We call the intersection with B_{n}^{∗} a mirror in B_{n}^{∗}. The mirror is called short (long), if a has this property. The image of a short (long) mirror in X_{Γ} is called a short (long) mirror in X_{Γ}. A mirror in X_{Γ} is an irreducible algebraic subvariety of codimension one. The mirrors in X_{Γ} correspond to the non isotropic pairs of vectors ±a of the quadratic space 𝔽_{3}^{5} (with the bilinear form x_{0}y_{0} − x_{1}y_{1} − ⋯ − x_{4}y_{4}). There are 36 short mirrors in X_{Γ} which correspond to the vectors with norm −1 and 45 long mirrors, which correspond to the vectors with norm −2 = 1 ∈ 𝔽_{3}.

We want to construct modular forms F with the property F (γz) = −F (z), where γ is reflection along a long mirror. If F is the additive lift of a constant elliptic modular form c ∈ ℂ[M′/M]^{SL(2,ℤ)}, this is a consequence of

c_{γα} = −c_{α}.

7.3 Proposition. Let γ ∈ O(5, 3) by the reflection along a long element a ∈ 𝔽_{3}^{5}. The space of constant modular forms (c_{α}) ∈ ℂ[M′/M]^{SL(2,ℤ)} with the property c_{γα} = −c_{α} has dimension two. The additive lifts of these forms vanish on the long mirror belonging to a.

The proof can be given by direct calculation. It is sufficient to take one of the 45 reflections. We take this one which belongs to the long root (0, 1, 1, 0, 0). We give an explicit basis. The first element of the basis is the additive lift of the constant modular form (c_{α}), which is 1 at

(0, 0, 1, −1, −1), (0, 0, 1, 1, 1), (0, 1, 0, −1, −1), (0, 1, 0, 1, 1), (0, 1, 1, −1, 0),
(0, 1, 1, 0, −1), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (1, −1, −1, −1, 1), (1, −1, −1, 1, −1),
(1, −1, 0, 0, 0), (1, 0, −1, 0, 0), (−1, 0, −1, 0, 0), (−1, −1, 0, 0, 0),
(−1, −1, −1, 1, −1),  (−1, −1, −1, −1, 1)

and −1 at their negatives and 0 elsewhere. The second element of the basis is the additive lift of the constant modular form, which is 1 at (0, 0, 1, −1, 1), (0, 0, 1, 1, −1), (0, 1, 0, −1, 1), (0, 1, 0, 1, −1), (0, 1, 1, −1, 0), (0, 1, 1, 0, −1), (0, 1, 1, 0, 1), (0, 1, 1, 1, 0), (1, −1, −1, −1, −1), (1, −1, −1, 1, 1), (1, −1, 0, 0, 0),(1, 0, −1, 0, 0),(−1, 0, −1, 0, 0),(−1, −1, 0, 0, 0),(−1, −1, −1, 1, 1), (−1, −1, −1, −1, −1)

and −1 at their negatives and 0 elsewhere.  ⊔ ⊓

The configuration of a cross is basic for the structure of the moduli space of marked cubic surfaces [AF]:

7.4 Definition. A cross in 𝔽_{3}^{5} is a set of 5 pair-wise orthogonal pairs ±a, one with norm 1 (called long) and 4 with norm −1 (called short).

We also use the notion of a cross in X_{Γ}.

7.5 Definition. A cross in X_{Γ} is a set of 5 pair-wise orthogonal mirrors, one long and 4 short.

7.6 Lemma. Let L be a long mirror in X_{Γ}. There are 12 short mirrors which are orthogonal to a. These 12 mirrors decompose into three sets of 4 mirrors which are pair-wise orthogonal. This decomposition is unique. The stabilizer of L in O(5, 3) permutes the three sets. Corollary. There are 135 crosses, three belonging to each of the 45 long mirrors.

In case of the long mirror which belongs to ±a, a = (0, 1, 1, 0, 0) ∈ 𝔽_{3}^{5}, the decomposition is given by

\begin{matrix}\{±(0, 1, −1, −1, 1), & ±(0, 1, −1, 1, −1), & ±(1, 0, 0, −1, −1), & ±(1, 0, 0, 1, 1)\}; \\ \{±(0, 1, −1, −1, −1), & ±(0, 1, −1, 1, 1), & ±(1, 0, 0, −1, 1), & ±(1, 0, 0, 1, −1)\}; \\ \{±(0, 0, 0, 0, 1), & ±(0, 0, 0, 1, 0), & ±(1, −1, 1, 0, 0), & ±(1, 1, −1, 0, 0)\}. \end{matrix}

7.7 Lemma. Let f be a cross in 𝔽_{3}^{5}. The space of all c ∈ ℂ[M′/M]^{SL(2,ℤ)} changing there sign under the 5 reflections belonging to the cross is onedimensional.

Proof. This can be read of from the basis given during the proof of 7.3. The two basis elements, which we constructed for the proof of 7.3 and their sum have the property that they change their sign under hexflections with respect to four orbits of short roots: The four short roots for the first basis element are

(0, 1, −1, −1, 1), (0, 1, −1, 1, −1), (1, 0, 0, −1, −1), (1, 0, 0, 1, 1)

and for the second basis element

(0, 1, −1, −1, −1), (0, 1, −1, 1, 1), (1, 0, 0, −1, 1), (1, 0, 0, 1, −1)

and for the sum of both

(0, 0, 0, 0, 1), (0, 0, 0, 1, 0), (1, −1, 1, 0, 0), (1, 1, −1, 0, 0)  ⊔ ⊓

We recall the notion of the divisor of a modular form. Let Y ⊂ X_{Γ} be an irreducible subvariety of codimension one. We denote by e_{Y} the ramification order with respect to the natural projection π : B_{4} → X_{Γ} (counted as 1 if π is unramified along Y). If Y is a short mirror this ramification order is three, because the triflections are contained in Γ. For long mirrors it is one. If F ∈ [Γ, k], F ≠ 0, is a modular form on Γ with trivial multipliers, the vanishing order of F along π^{−1}(Y) is divisible by e_{Y}. We call the quotient of this vanishing order by e_{Y} the order of F along Y and denote it by n_{Y} (F). The divisor of F is the finite sum

(F) := ∑_{Y ⊂X_{Γ}} n_{Y} (F)Y.

We consider a cross in X_{Γ} as a divisor with multiplicity one at all 5 mirrors of the cross.

7.8 Theorem. Let F ℂ be the one dimensional space of modular forms belonging to a cross. The divisor of F is exactly this cross.

Proof. We consider the product P of all the 135 modular forms. We compute its divisor. During this proof we will count the multiplicities in B_{4} and not in the quotient. (Borcherds products can have a non-trivial character. Therefore it is better to count multiplicities in B_{4} and not in the quotient.) The form P vanishes along all long mirrors and the vanishing order is at least three. It vanishes also along all short mirrors. But the vanishing order is at least 3 · 4 × 135/36 = 45. Now we claim that that there exists an automorphic form Q of weight 3 · 135 (=weight of P) whose zero divisor is precisely the union of the long mirrors with multiplicity 3 and of the short mirrors with multiplicity 45. The proof of 7.8 then will be complete because P/Q is an automorphic form of weight 0 and hence constant. We will construct Q as a Borcherds product, namely Q = χ_{4}^{45} · χ_{75}^{3}, where χ_{4} and χ_{75} have the following meaning:

7.9 Theorem. The union of all short (resp. long) mirrors is the zero set of an automorphic form χ_{4} of weight 4 (resp. χ_{75} of weight 75). The multiplicities counted in B_{4} are one.

We want to mention that the form of weight 4 has been constructed by Borcherds in an unpublished paper (compare 5.5).

Theorem 7.9 is a consequence of 5.5. One has to use the natural embedding U(1, 4) ↪ O(2, 8) and the holomorphic embedding B_{4} ↪ H_{8}. The short (long) mirrors in B_{4} will be the intersections of the divisors described in 5.5. There are always three divisors in H_{8}, which have the same intersection with B_{4}.

Therefore the vanishing order of the pull back forms on B_{4} are three. We can take cube roots and get weights 12/3 = 4 and 225/3 = 75 as claimed in 7.9.

This completes the proof of 7.9.  ⊔ ⊓

Each long mirror is contained in three crosses. We consider the 45 products F GH of automorphic forms F, G, H belonging to these crosses. They are automorphic forms of weight 9. The zero set consists of one orbit of long mirrors and the 12 orbits of orthogonal short mirrors. The vanishing orders at all components counted in B_{4} is three. We can therefore take a holomorphic cube root, F GH = A^{3}. The A-s are automorphic forms on the congruence group Γ. But the multipliers cannot be trivial.

7.10 Theorem. For every long mirror in B_{4}/Γ there exists an automorpic form A of weight 3 on Γ, whose zero set consists of this long mirror and the 12 orthogonal short mirrors. The multiplicities counted in B_{4} are 1. The multipliers are non-trivial: Let γ ∈ Γ be a triflection with respect to a short root b ∈ M.

1. If A does not vanish on the mirror of b then A(γz) = A(z).

2. If A vanishes on the mirror of c, then A(γz) = ζA(γz), where ζ is the reflection factor of γ (which is a third root of unity).

Proof. The first statement is clear, we prove the second: The function A(γz) agrees with A upto a constant factor. This factor is a third root of unity and hence ζ^{±1}. The minus sign can be excluded because A is holomorphic.

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