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Integrality of the Fourier coefficients of the j-invariant

Statement

With $q = e^{2 π i τ}$ , the $j$ -invariant has a Laurent expansion convergent for $0 < ∣ q ∣ < 1$ , $j (τ) = q^{- 1} + 744 + 196884 q + 21493760 q^{2} + \sum_{n \geq 3} c (n) q^{n}, c (n) \in Z .$ Equivalently $q j$ is holomorphic in the unit disc with integer Taylor coefficients and constant term $1$ .

Facts & Assumptions

Given: $Δ = q P (q)$ with $P$ holomorphic and zero-free on $∣ q ∣ < 1$ and $P (q) = \prod_{n \geq 1} (1 - q^{n})^{24}$ (The Jacobi product formula for the discriminant), and $j = E_{4}^{3} / Δ$ (The modular discriminant and the j-invariant).

[F1]

$P$ has integer Taylor coefficients and $P (0) = 1$ (The Jacobi product formula for the discriminant, The binomial theorem over the complex field).

[F2]

$E_{4} = 1 + 240 \sum_{n \geq 1} σ_{3} (n) q^{n}$ , with $σ_{3} (n) \in Z$ and $σ_{3} (1) = 1$ , $σ_{3} (2) = 9$ , $σ_{3} (3) = 28$ (Eisenstein series are modular forms; their Fourier coefficients, The divisor power sums $σ_{k}$ ).

[F3]

A holomorphic function on a disc has a convergent Taylor expansion with the coefficients given by the derivatives at the centre (A holomorphic function equals its Taylor series throughout the largest centred disc in its domain); locally uniform convergence passes to all derivatives (Locally uniform limits of holomorphic functions are holomorphic and their derivatives converge locally uniformly); the coefficient sequence of a product of two absolutely convergent complex series is the Cauchy product, absolutely convergent (The Cauchy product of two absolutely convergent complex series converges absolutely to the product of their sums, The binomial theorem over the complex field).

Proof

1.1F1F2F3givenalgebra

Since $P$ is zero-free on the unit disc and $Δ = q P$ , the function $q j = E_{4} (q)^{3} / P (q)$ is holomorphic on $∣ q ∣ < 1$ . The powers $E_{4}^{3}$ have integer Taylor coefficients: by [F2] the expansion of $E_{4}$ has integer coefficients, and the coefficients of the cube are finite sums of products of integers, which by [F3] are exactly the Cauchy-product coefficients.

2.1F1F2F3step 1.1givenalgebra

Write $P (q) = 1 + \sum_{n \geq 1} p_{n} q^{n}$ with $p_{n} \in Z$ by [F1], and let $P (q)^{- 1} = \sum_{n \geq 0} b_{n} q^{n}$ be its Taylor expansion at $0$ , which exists and converges on $∣ q ∣ < 1$ because $P$ is holomorphic and zero-free there [F3]. Comparing coefficients in $P \cdot P^{- 1} = 1$ gives $b_{0} = 1$ and $b_{n} = - \sum_{r = 1}^{n} p_{r} b_{n - r}$ for $n \geq 1$ ; by induction on $n$ every $b_{n}$ is an integer. Hence $q j = E_{4}^{3} \cdot P^{- 1}$ has integer Taylor coefficients by the Cauchy product [F3], and its constant term is $1 \cdot 1 = 1$ . Since $j = q^{- 1} (q j)$ , the Laurent expansion of $j$ on $0 < ∣ q ∣ < 1$ has the stated integer coefficients.

3.1F2F3step 2.1givenalgebra∎

The displayed coefficients are obtained by computing finitely many terms. From [F2] and $σ_{3} (1) = 1$ , $σ_{3} (2) = 9$ , $σ_{3} (3) = 28$ : $E_{4} = 1 + 240 q + 2160 q^{2} + 6720 q^{3} + O (q^{4})$ , so $E_{4}^{3} = 1 + 720 q + 179280 q^{2} + 16954560 q^{3} + O (q^{4})$ ; from the product formula $Δ = q (1 - 24 q + 252 q^{2} - 1472 q^{3} + O (q^{4}))$ , so $P = 1 - 24 q + 252 q^{2} - 1472 q^{3} + O (q^{4})$ and its inverse begins $P^{- 1} = 1 + 24 q + 324 q^{2} + 3200 q^{3} + O (q^{4})$ (coefficient comparison, as in 2.1). Multiplying, $q j = E_{4}^{3} P^{- 1} = 1 + 744 q + 196884 q^{2} + 21493760 q^{3} + O (q^{4})$ , hence $j = q^{- 1} + 744 + 196884 q + 21493760 q^{2} + \sum_{n \geq 3} c (n) q^{n}$ with the stated initial coefficients.

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Sources

J. S. Milne, Modular Functions and Modular Forms (v1.31, 2017) (standard reference, not scraped)
C. T. McMullen, Advanced Complex Analysis, Math 213a course notes (Harvard, 2010) (standard reference, not scraped)
D. Zagier, Elliptic Modular Forms and Their Applications, in The 1-2-3 of Modular Forms (Universitext, Springer, 2008) (standard reference, not scraped)