An automorphic form related to cubic surfaces. First draft. 1997.

R. E. Borcherds.

(Remark added 2000: See the paper “Cubic Surfaces and Borcherds Products” by Allcock and Freitag, math.AG/0002066)

In the paper [A-C-T] the authors showed that the moduli space of cubic surfaces was (CH^{4}\H)/G, where H is the union of the reflection hyperplanes of the reflection group G. The purpose of this note is to construct an automorphic form, called the discriminant, on complex hyperbolic space CH^{4} whose zeros are exactly the reflection hyperplanes of G, each with multiplicity 1.

The complex hyperbolic space of E^{1,4} embeds naturally in the Grassmannian of 2 dimensional positive definite subspaces of the underlying even integral lattice M of E^{1,4}. We will construct the automorphic form on complex hyperbolic space by constructing an automorphic form Ψ on the Grassmannian and restricting it to complex hyperbolic space. The construction of Ψ is similar to that of example 13.7 of [B]. The lattice M is isomorphic to A_{2} ⊕ A_{2}^{4}(−1), so it has dimension 10 and determinant 3^{5}. We let the elements e_{γ} for γ ∈ M′/M stand for the obvious basis of the group ring C[M′/M]. We will construct a modular form of weight −3 and type ρ_{M} for SL_{2}(Z) which is holomorphic on the upper half plane and meromorphic at cusps, by which we mean a holomorphic vector valued function f = ∑_{γ∈M′/M} e_{γ}f_{γ}(τ) such that

f_{γ}(τ + 1) = e^{πiγ^{2}}f_{γ}(τ), f_{γ}(−1/τ) = −i3^{−5/2}τ^{−3} ∑_{δ∈M′/M} e^{−2πi(δ,γ)}f_{δ}(τ)

Recall that if γ, δ ∈ M′/M then γ^{2} is well defined mod 2 and (γ, δ) is well defined mod 1. There are 4 orbits of vectors γ ∈ M′/M under Aut(M), which we name as follows: a nonzero vector γ with γ^{2}/2 ≡ n/3 mod 1 will be called a vector of type n (n = 0, 1, 2), and the zero vector will be called a vector of type 00.

We will construct a modular form f such that the components f_{γ} of f are given by functions f_{00}, f_{0}, f_{1}, or f_{2} depending only on the type of γ. We now work out the conditions that these functions have to satisfy for f to transform correctly. In order to check that f is a modular form under τ ↦ −1/τ we need to know, given some fixed vector u, how many vectors v there are with given type and given inner product with u. These numbers are given in the following table.

type of u 00 00 00 00 0 0 0 0 1 1 1 1 2 2 2 2
type of v 00 0 1 2 00 0 1 2 00 0 1 2 00 0 1 2
(u, v) ≡ 0 1 80 90 72 1 26 36 18 1 32 24 24 1 20 30 30
(u, v) ≡ 1/3 0 0 0 0 0 27 27 27 0 24 33 24 0 30 30 21
(u, v) ≡ 2/3 0 0 0 0 0 27 27 27 0 24 33 24 0 30 30 21

Using this table we see that f_{00}, f_{0}, f_{1}, and f_{2} have to satisfy the equations

f_{00}(τ + 1) = f_{00}(τ), f_{00}(−1/τ) = −i3^{−5/2}τ^{−3}(f_{00}(τ) + 80f_{0}(τ) + 90f_{1}(τ) + 72f_{2}(τ))

f_{0}(τ + 1) = f_{0}(τ), f_{0}(−1/τ) = −i3^{−5/2}τ^{−3}(f_{00}(τ) − f_{0}(τ) + 9f_{1}(τ) − 9f_{2}(τ))

f_{1}(τ + 1) = e^{2πi/3}f_{1}(τ), f_{1}(−1/τ) = −i3^{−5/2}τ^{−3}(f_{00}(τ) + 8f_{0}(τ) − 9f_{1}(τ))

f_{2}(τ + 1) = e^{4πi/3}f_{2}(τ), f_{2}(−1/τ) = −i3^{−5/2}τ^{−3}(f_{00}(τ) − 10f_{0}(τ) + 9f_{2}(τ))

One solution of these equations is given as follows.

\begin{aligned}f_{00}(τ) & = 24η(3τ)^{3}η(τ)^{−9} = 24(1 + 9q + 54q^{2} + O(q^{3})) \\ f_{0}(τ) & = −3η(3τ)^{3}η(τ)^{−9} = −3 + O(q) \\ f_{1}(τ) & = 0 \\ f_{2}(τ) & = η(τ/3)^{3}η(τ)^{−9} + 3η(3τ)^{3}η(τ)^{−9} = q^{−1/3} + 14q^{2/3} + 92q^{5/3} + O(q^{8/3}) \end{aligned}

Most of the transformations follow formally from the functional equations η(τ + 1) = e^{2πi/24}η(τ) and η(−1/τ) = \sqrt{τ/i}η(τ) of η. The only one which takes slightly more work is the transformation of f_{2} under τ ↦ τ + 1, and this follows from the identity η(τ)^{3} = ∑_{n∈Z}(4n + 1)q^{(4n+1)^{2}/8} and its consequence

η(τ/3)^{3}η(τ)^{−9} + η((τ + 1)/3)^{3}η(τ + 1)^{−9} + η((τ + 2)/3)^{3}η(τ + 2)^{−9} = −9η(3τ)^{3}η(τ)^{−9}.

By theorem 13.3 of [B] there is an automorphic form Ψ on the symmetric space of M with the following properties. It has weight 12 = (coefficient of q^{0} in f_{00})/2. The zeros of Ψ correspond to the negative powers of q in f, so are zeros of order 1 orthogonal to all the norm −2/3 vectors of M′. Ψ is holomorphic on the symmetric space, and therefore holomorphic at cusps as well by the Koecher boundedness principle. Ψ is an automorphic form for some one dimensional representation of Aut(M).

By restricting Ψ to complex hyperbolic space we get an automorphic form which has zeros of order 3 along the reflection hyperplanes (because f has zeros of order 1, but every reflection hyperplane is the restriction of 3 hyperplanes of the symmetric space of M). So by taking the cube root of the restriction of Ψ we get an automorphic form for a one dimensional character of G whose zeros are exactly the reflection hyperplanes with multiplicity 1.

References.

[A-C-T] D. J. Allcock, J. Carlson, D. Toledo, A Complex Hyperbolic Structure for Moduli of Cubic Surfaces, alg-geom/9709016

[B] R. E. Borcherds, Automorphic forms with singularities on Grassmannians, alggeom/9609022, to appear in Invent. Math.