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Level-One Modular Forms and the j-Invariant

1 · Prerequisites

2 · Summary

This page builds the classical theory of level-one modular forms from the action of the modular group. The group $P S L_{2} (Z)$ acts on the upper half-plane by Möbius transformations, and the reduction argument produces the standard fundamental domain $D$ with its boundary identifications and its two elliptic classes, represented by $i$ , $ω$ and $ω + 1$ in its closure; the quotient chart lemma then gives $Γ \ H$ a Riemann surface structure, with the $ν$ -th-power chart at an elliptic point, and the cusp chart and compactness lemma adjoin the orbit of $\infty$ so that $X (1) = P S L_{2} (Z) \ H^{*}$ is a compact Riemann surface. On this curve a modular form of weight $k$ is defined as a holomorphic function on $H$ transforming by the factor $(c τ + d)^{k}$ and bounded at the cusp; equivalently its $q$ -expansion has no negative powers, and the $q$ -expansion principle records how the growth condition descends to the quotient.

The supply of forms comes from the lattice sums. The Eisenstein series $G_{k} (τ) = \sum^{'} (m τ + n)^{- k}$ converge absolutely and normally for even $k \geq 4$ , uniformly on compact subsets of the half-plane, and the Lipschitz formula turns the sum into $2 ζ (k)$ plus an explicit power series in $q = e^{2 π i τ}$ , so that the normalised series $E_{k} = G_{k} / (2 ζ (k))$ has $q$ -expansion beginning $1 + O (q)$ with coefficients $- \frac{2 k}{B_{k}} σ_{k - 1} (n)$ . The transformation law $G_{k} (γ τ) = (c τ + d)^{k} G_{k} (τ)$ then makes $E_{4}$ and $E_{6}$ modular forms of weights four and six, while the weight-two series is treated separately: its regularisation satisfies a transformation law with a non-vanishing correction term, so it is quasimodular rather than modular, and it is used in the discriminant computation.

The analytic heart of the page is the valence formula. Subtracting the contributions of the cusp and of the elliptic points from the boundary term of the argument principle gives $\sum_{P} {ord}_{P} (f) / ν_{P} + {ord}_{\infty} (f) = k / 12$ for every nonzero weight- $k$ form, and the boundary arc computation is what produces the $k / 12$ . The valence formula yields the location of the zeros of $E_{4}$ and $E_{6}$ , the dimension formula $\dim M_{k} = ⌊ k / 12 ⌋ + 1$ for even $k \geq 0$ with $k \equiv̸ 2 (m o d 12)$ and $\dim M_{k} = ⌊ k / 12 ⌋$ for $k \equiv 2 (m o d 12)$ , with $\dim S_{k} = \dim M_{k} - 1$ for even $k \geq 4$ , and the fact that $Δ = (E_{4}^{3} - E_{6}^{2}) / 1728$ is a cusp form of weight twelve with no zeros on $H$ . The logarithmic derivative of the infinite product and the $E_{2}$ transformation law give the product formula $Δ = q \prod_{n \geq 1} (1 - q^{n})^{24}$ and the integrality of its Fourier coefficients.

The final block assembles the graded ring and the invariant. The monomials $E_{4}^{a} E_{6}^{b}$ with $4 a + 6 b = k$ form a basis of $M_{k}$ , so $M_{*} = C [E_{4}, E_{6}]$ with $E_{4}$ and $E_{6}$ algebraically independent; the function $j = E_{4}^{3} / Δ$ is holomorphic on $H$ , invariant under the modular group, and has a simple pole at the cusp, and the $q$ -expansion $j = q^{- 1} + 744 + 196884 q + \dots$ has integral coefficients. The resulting map $\overset{ˉ}{j} : X (1) \to \hat{C}$ is a biholomorphism, and through it the $j$ -invariant classifies complex tori up to biholomorphism, with the square and hexagonal tori realising the values $1728$ and $0$ .

3 · Logical flowchart

4 · Definitions, theorems and proofs

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The modular group and its action on the upper half-plane

Definition

Let

$S L_{2} (Z) = {γ = (\begin{array}{cc} a & b \\ c & d \end{array}) : a, b, c, d \in Z, a d - b c = 1},$

a subgroup of $G L_{2} (R) \subseteq G L_{2} (C)$ (Invertible matrices and the general linear group ${GL}_{n} (F)$ , ${GL}_{n} (F)$ is a group under matrix multiplication, including the trivial group ${GL}_{0} (F)$ , The integers as equivalence classes of pairs of naturals, For same-sized finite square matrices over a commutative ring, $\det (A B) = \det (A) \det (B)$ ). Its centre is ${\pm I}$ : the inclusion ${\pm I} \subseteq Z (S L_{2} (Z))$ is immediate, ${\pm I}$ is normal (The center of a group is a normal subgroup, Normal subgroup: invariance under conjugation), and conversely every $2 \times 2$ matrix commuting with all of $S L_{2} (Z)$ is scalar — in particular a matrix $A = (\begin{matrix} a & b \\ c & d \end{matrix})$ commuting with both $U = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ and $L = (\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix})$ satisfies $c = 0$ and $a = d$ from $A U = U A$ , and then $b = 0$ from $A L = L A$ , by comparing entries of the two products, so $A = a I$ is scalar; a scalar matrix $λ I$ of determinant $1$ has $λ^{2} = 1$ , so $λ = \pm 1$ (Rectangular matrix multiplication and the identity matrix $I_{n}$ , including zero-sized shapes). The modular group is the quotient

$P S L_{2} (Z) : = S L_{2} (Z) / {\pm I},$

a group by For $N ⊴ G$ , the cosets form a group with identity $N$ and inverse $(g N)^{- 1} = g^{- 1} N$ , The quotient group $G / N$ and coset product $(g N) (h N) = g h N$ , The canonical projection $π : G \to G / N$ , $π (g) = g N$ , is a surjective group homomorphism.

For $γ \in S L_{2} (R)$ and $τ \in H$ set $γ \cdot τ : = \frac{a τ + b}{c τ + d}$ , the Möbius transformation attached to $γ$ (Möbius transformations of the Riemann sphere, The unit disc, the upper half-plane, and Blaschke factors). The denominator does not vanish: $c τ + d = 0$ would give $τ = - d / c \in R$ when $c \neq 0$ , and when $c = 0$ invertibility gives $d \neq 0$ . Writing $c τ + d = (c Re τ + d) + i c Im τ$ and using $\overline{c τ + d} = c \overset{ˉ}{τ} + d$ , a direct computation gives

$Im (γ \cdot τ) = \frac{Im τ}{∣ c τ + d ∣^{2}} > 0,$

so $H$ is stable under each $γ$ , and $γ \mapsto γ \cdot$ is the restriction of the matrix-to-Möbius map, a group homomorphism (Möbius transformations form a group and identify with the projective linear quotient of GL_2(C), Actions of $G$ on $X$ correspond exactly to homomorphisms $G \to Sym (X)$ ); hence $γ \mapsto (τ \mapsto γ \cdot τ)$ is a left action of $S L_{2} (R)$ on $H$ by biholomorphisms (Left group actions, transitive actions, and faithful actions, Every Möbius transformation is a biholomorphism of the Riemann sphere, Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive, Real and imaginary parts, complex conjugation, and modulus)).

The kernel of the restricted action of $S L_{2} (Z)$ is exactly ${\pm I}$ : $\pm I$ act trivially, while a matrix acting trivially fixes the three distinct points $i$ , $i + 1$ , $2 i \in H$ , so it is the identity Möbius transformation and hence scalar — knowing that a Möbius transformation is determined by its values at three distinct points and that the kernel of the matrix-to-Möbius map is the scalar subgroup (A unique Möbius transformation carries any ordered triple of distinct sphere points to any other, Möbius transformations form a group and identify with the projective linear quotient of GL_2(C)) — and a scalar in $S L_{2} (Z)$ is $\pm I$ . The action therefore descends to a faithful action of $P S L_{2} (Z)$ , the quotient by the normal subgroup ${\pm I}$ (First isomorphism theorem for groups: $G / \ker f ≅ im f$ ).

The elements $S = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ and $T = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ act by $S \cdot τ = - 1 / τ$ and $T \cdot τ = τ + 1$ . In $S L_{2} (Z)$ one computes $S^{2} = - I$ and $(S T)^{3} = - I$ : $S T = (\begin{matrix} 0 & - 1 \\ 1 & 1 \end{matrix})$ has $(S T)^{2} - S T + I = 0$ by direct multiplication, whence $(S T)^{3} = (S T) (S T - I) = - I$ . Hence $S$ and $S T$ have order $2$ and $3$ in $P S L_{2} (Z)$ , so $S^{2} = 1$ and $(S T)^{3} = 1$ there.

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Reduction of orbits to the standard domain

Statement

Let $Γ^{'} = ⟨ S, T ⟩ \leq P S L_{2} (Z)$ and $\overline{D} = {τ \in H : ∣ ℜ τ ∣ \leq 1 / 2, ∣ τ ∣ \geq 1}$ . (a) Every $τ \in H$ is $Γ^{'}$ -equivalent to a point of $\overline{D}$ : among the points of the orbit $Γ^{'} \cdot τ$ some point $τ_{0}$ has maximal imaginary part; after applying a power of $T$ one has $∣ ℜ τ_{0} ∣ \leq 1 / 2$ , and then necessarily $∣ τ_{0} ∣ \geq 1$ . (b) For fixed $τ \in H$ and $N > 0$ there are only finitely many pairs $(c, d) \in Z^{2}$ with $∣ c τ + d ∣ \leq N$ .

Facts & Assumptions

Given: $τ = x + i y \in H$ , so $y > 0$ , and the action of $Γ^{'}$ with $ℑ (γ \cdot τ) = ℑ τ / ∣ c τ + d ∣^{2}$ for the bottom row $(c, d)$ of $γ$ ; $T \cdot τ = τ + 1$ , $S \cdot τ = - 1 / τ$ (The modular group and its action on the upper half-plane).

[F1]

For $z \in C$ , $∣ Re z ∣ \leq ∣ z ∣$ , $∣ Im z ∣ \leq ∣ z ∣$ , and $∣ z ∣^{2} = z \overset{ˉ}{z}$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive, Real and imaginary parts, complex conjugation, and modulus).

[F2]

A nonempty subset of $Z$ that is bounded above has a greatest element and one that is bounded below has a least element; in particular the integers in a bounded interval form a finite set (A nonempty set of integers bounded above has a greatest element, and a nonempty set of integers bounded below has a least element). Every real has an integer part $⌊ u ⌋$ with $⌊ u ⌋ \leq u < ⌊ u ⌋ + 1$ (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

[F3]

If $0 < r < 1$ then $1 / r^{2} > 1$ (Basic properties of the absolute value).

Proof

1.1F1F2givenalgebra

Fix $N > 0$ and suppose $∣ c τ + d ∣ \leq N$ . Since $Im (c τ + d) = c y$ and $Re (c τ + d) = c x + d$ , [F1] gives $∣ c ∣ y \leq N$ , that is $∣ c ∣ \leq N / y$ ; by [F2] only finitely many integers $c$ satisfy this. For each such $c$ , [F1] gives $∣ c x + d ∣ \leq N$ , so $- N - c x \leq d \leq N - c x$ , and [F2] leaves only finitely many integers $d$ . Hence only finitely many pairs $(c, d)$ satisfy $∣ c τ + d ∣ \leq N$ .

2.1F1F2F3step 1.1givenalgebra∎

The set $V : = {∣ c τ + d ∣ : γ \in Γ^{'} with bottom row (c, d)}$ is nonempty (the identity has value $∣ 0 \cdot τ + 1 ∣ = 1$ ) and every element is positive. Applying 1.1 with $N = 1$ shows that the elements of $V$ are among the finitely many numbers $∣ c τ + d ∣$ attached to pairs with $∣ c τ + d ∣ \leq 1$ , together with values $> 1$ ; hence $V$ has a least element $m > 0$ , realized by some $γ_{0} \in Γ^{'}$ . Since $ℑ (γ \cdot τ) = y / ∣ c τ + d ∣^{2}$ , the point $τ_{0} : = γ_{0} \cdot τ$ of the orbit has maximal imaginary part $y / m^{2}$ : for every $γ \in Γ^{'}$ with bottom row $(c, d)$ one has $∣ c τ + d ∣ \geq m$ , so $ℑ (γ \cdot τ) = y / ∣ c τ + d ∣^{2} \leq y / m^{2} = ℑ τ_{0}$ . Choose $n \in Z$ with $∣ ℜ τ_{0} - n ∣ \leq 1 / 2$ , possible by taking $n = ⌊ ℜ τ_{0} + 1 / 2 ⌋$ [F2]; then $τ_{1} : = T^{- n} \cdot τ_{0}$ satisfies $ℑ τ_{1} = ℑ τ_{0}$ (translation does not change the imaginary part) and $∣ ℜ τ_{1} ∣ = ∣ ℜ τ_{0} - n ∣ \leq 1 / 2$ . If $∣ τ_{1} ∣ < 1$ , then $S τ_{1} \in Γ^{'} \cdot τ$ and $ℑ (S τ_{1}) = ℑ τ_{1} / ∣ τ_{1} ∣^{2} > ℑ τ_{1}$ by [F1] and [F3], contradicting the maximality of $ℑ τ_{0} = ℑ τ_{1}$ . Hence $∣ τ_{1} ∣ \geq 1$ , and $τ_{1} \in \overline{D}$ is the required $Γ^{'}$ -equivalent point.

TheoremStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

The standard fundamental domain, boundary identifications and elliptic stabilisers

Statement

Let $D = {τ \in H : ∣ ℜ τ ∣ < 1 / 2, ∣ τ ∣ > 1}$ and let $\overline{D}$ be its closure in $H$ . (a) $D$ is a fundamental domain for $P S L_{2} (Z)$ : every orbit meets $\overline{D}$ , no two points of $D$ are equivalent, and two distinct points $z, z^{'} \in \overline{D}$ are equivalent if and only if either $z^{'} = z \pm 1$ with $ℜ z = \mp 1 / 2$ , or $z^{'} = - 1 / z$ with $∣ z ∣ = 1$ . (b) The only points of $\overline{D}$ with nontrivial stabiliser are $i$ , $ω : = e^{2 π i / 3}$ and $ω + 1 = e^{i π / 3}$ : $Stab (i) = ⟨ S ⟩$ has order $2$ , $Stab (ω) = ⟨ S T ⟩$ has order $3$ , and $Stab (ω + 1) = ⟨ T S ⟩$ has order $3$ in $P S L_{2} (Z)$ . (c) $P S L_{2} (Z) = ⟨ S, T ⟩$ .

Facts & Assumptions

Given: $Γ : = P S L_{2} (Z)$ acting on $H$ , its subgroup $Γ^{'} : = ⟨ S, T ⟩$ , the set $D$ and its closure $\overline{D}$ ; $S \cdot z = - 1 / z$ , $T \cdot z = z + 1$ , and $S^{2} = (S T)^{3} = 1$ in $Γ$ (The modular group and its action on the upper half-plane, The subgroup $⟨ S ⟩$ generated by a subset, the cyclic subgroup $⟨ g ⟩$ , and cyclic groups, The orbit $G \cdot x$ and stabilizer $G_{x}$ of a point in a group action, The stabilizer $G_{x}$ is a subgroup of $G$ , A free group action has no nonidentity element fixing a point).

[F1]

Every $Γ^{'}$ -orbit meets $\overline{D}$ (Reduction of orbits to the standard domain), and for $γ = (\begin{matrix} a & b \\ c & d \end{matrix}) \in S L_{2} (Z)$ one has $ℑ (γ \cdot z) = ℑ z / ∣ c z + d ∣^{2}$ , so $ℑ (γ \cdot z) \geq ℑ z$ is equivalent to $∣ c z + d ∣ \leq 1$ (The modular group and its action on the upper half-plane).

[F2]

Every $z \in \overline{D}$ satisfies $∣ ℜ z ∣ \leq 1 / 2$ and $∣ z ∣ \geq 1$ , hence $Im z \geq \sqrt{3} / 2$ (The modular group and its action on the upper half-plane, Reduction of orbits to the standard domain).

[F3]

Orbits partition the set and stabilisers are subgroups of the acting group; two points are equivalent exactly when they lie in the same orbit (The orbits of a group action are the equivalence classes of $x \sim y$ iff $y = g \cdot x$ for some $g$ , and hence partition the acted-on set, The orbit $G \cdot x$ and stabilizer $G_{x}$ of a point in a group action, The stabilizer $G_{x}$ is a subgroup of $G$ ).

Proof

1.1F1F2givenalgebra

Every $Γ$ -orbit meets $\overline{D}$ , because $Γ^{'} \leq Γ$ and every $Γ^{'}$ -orbit does [F1]. Suppose $z, z^{'} \in \overline{D}$ satisfy $z^{'} = γ \cdot z$ with $γ = (\begin{matrix} a & b \\ c & d \end{matrix}) \in S L_{2} (Z)$ and $Im z^{'} \geq Im z$ ; replacing $γ$ by $- γ$ , which gives the same element of $Γ$ , we may assume $c \geq 0$ . Then $∣ c z + d ∣ \leq 1$ by [F1], while $Im z \geq \sqrt{3} / 2$ by [F2]; hence $∣ c ∣ \frac{\sqrt{3}}{2} \leq ∣ c ∣ Im z = ∣ Im (c z + d) ∣ \leq ∣ c z + d ∣ \leq 1$ , so $∣ c ∣ \leq 2 / \sqrt{3} < 2$ and therefore $c \in {0, 1}$ .

2.1F1F2step 1.1givencasesalgebra

Case $c = 0$ . Then $d = \pm 1$ and $γ = \pm (\begin{matrix} 1 & m \\ 0 & 1 \end{matrix})$ , so $z^{'} = z + m$ with $m \in Z$ . Since both points lie in $\overline{D}$ , $∣ m ∣ = ∣ ℜ z^{'} - ℜ z ∣ \leq 1$ . If $m = 0$ then $z^{'} = z$ . If $m = 1$ then $ℜ z^{'} = ℜ z + 1 \leq 1 / 2$ and $ℜ z \geq - 1 / 2$ force $ℜ z = - 1 / 2$ ; conversely for $z \in \overline{D}$ with $ℜ z = - 1 / 2$ the point $z^{'} = z + 1$ lies in $\overline{D}$ , since $1 / 2 \geq ℜ z^{'}$ and $∣ z^{'} ∣ = ∣ z - (- 1) ∣ = ∣ ∣ z ∣ ∣ = ∣ z ∣$ holds because $∣ z^{'} ∣^{2} = ∣ z ∣^{2} + 2 Re z + 1 = ∣ z ∣^{2}$ . The case $m = - 1$ is the mirror image with $ℜ z = 1 / 2$ .

2.2F1F2step 1.1givencasesalgebra

Case $c = 1$ . First suppose $d = 0$ . Then $b = a d - 1 = - 1$ and $∣ z ∣ = ∣ c z + d ∣ \leq 1 \leq ∣ z ∣$ shows $∣ z ∣ = 1$ ; the map is $z^{'} = a - 1 / z = a - \overset{ˉ}{z}$ with $a \in Z$ , and $∣ ℜ z^{'} ∣ \leq 1 / 2$ , $∣ ℜ z ∣ \leq 1 / 2$ give $∣ a ∣ \leq 1$ . For $a = 0$ this is $z^{'} = - 1 / z$ with $∣ z ∣ = 1$ ; for $a = - 1$ the condition $ℜ z^{'} = - 1 - ℜ z \geq - 1 / 2$ forces $ℜ z = - 1 / 2$ , so $z = ω$ and $z^{'} = - 1 - \overset{ˉ}{z} = z$ ; for $a = 1$ we get $ℜ z^{'} = 1 - ℜ z \geq 1 / 2$ , forcing $ℜ z^{'} = 1 / 2$ and $ℜ z = 1 / 2$ , after which $∣ z^{'} ∣^{2} = 1 - 2 ℜ z + ∣ z ∣^{2} = ∣ z ∣^{2} = 1$ and $z^{'} = 1 - \overset{ˉ}{z} = 1 - (\frac{1}{2} - i y) = \frac{1}{2} + i y = z$ . So for $d = 0$ the only new identification is $z^{'} = - 1 / z$ when $∣ z ∣ = 1$ . Now suppose $d \geq 1$ . Since $∣ c z + d ∣^{2} - ∣ z ∣^{2} = 2 d Re z + d^{2} \leq 1 - 1 = 0$ , we get $ℜ z \leq - d / 2 \leq - 1 / 2$ , so $d = 1$ and $ℜ z = - 1 / 2$ ; then $∣ z + 1 ∣^{2} = ∣ z ∣^{2}$ and $1 \leq ∣ z ∣$ , $∣ z + 1 ∣ \leq 1$ force $∣ z + 1 ∣ = 1 = ∣ z ∣$ , whence $z = ω$ . Here $b = a - 1$ and $z^{'} = a + ω$ (because $1 / (1 + ω) = - ω$ ), so $ℜ z^{'} = a - 1 / 2 \in [- 1 / 2, 1 / 2]$ gives $a \in {0, 1}$ : for $a = 0$ one has $z^{'} = z$ , and for $a = 1$ one has $z^{'} = ω + 1 = z + 1$ . If $d \leq - 1$ , write $d = - k$ , $k \geq 1$ ; then $∣ c z + d ∣^{2} - ∣ z ∣^{2} = - 2 k ℜ z + k^{2} \leq 0$ gives $ℜ z \geq k / 2 \geq 1 / 2$ , so $k = 1$ , $d = - 1$ , $ℜ z = 1 / 2$ , and $∣ z - 1 ∣ = ∣ z ∣ = 1$ forces $z = ω + 1$ ; with $b = - a - 1$ and $ω = ω + 1 - 1$ one computes $z^{'} = a + 1 + ω$ , so $ℜ z^{'} = a + 1 / 2 \in [- 1 / 2, 1 / 2]$ gives $a \in {- 1, 0}$ : for $a = 0$ one has $z^{'} = z$ , and for $a = - 1$ one has $z^{'} = ω = z - 1$ .

3.1F1F3step 2.1step 2.2givenalgebra

Combining 1.1, 2.1 and 2.2: for any two $Γ$ -equivalent points $z, z^{'} \in \overline{D}$ one of them, say the one with larger imaginary part, is of the listed shape; the case analysis gives either $z^{'} = z$ , or $z^{'} = z \pm 1$ with $ℜ z = \mp 1 / 2$ , or $z^{'} = - 1 / z$ with $∣ z ∣ = 1$ . Since the two identifications force $∣ z ∣ = 1$ or $∣ ℜ z ∣ = 1 / 2$ , neither can occur for two distinct points of $D$ , whose points have $∣ ℜ z ∣ < 1 / 2$ and $∣ z ∣ > 1$ ; this proves (a). For (b), a stabiliser element is a case with $z^{'} = z$ in the same analysis: besides the identity, this happens exactly for $z = i$ with $γ = \pm S$ , for $z = ω$ with $γ$ representing $S T$ or $(S T)^{2}$ , and for $z = ω + 1$ with $γ$ representing $T S$ or $(T S)^{2}$ . Since $S^{2} = (S T)^{3} = (T S)^{3} = 1$ in $Γ$ and $S \neq 1$ , $S T \neq 1$ , these give $Stab (i) = ⟨ S ⟩$ of order $2$ and $Stab (ω) = ⟨ S T ⟩$ , $Stab (ω + 1) = ⟨ T S ⟩$ of order $3$ , by [F3]; all other points of $\overline{D}$ have trivial stabiliser.

4.1F3step 3.1givenalgebra∎

For (c) let $γ \in Γ$ and fix $z_{0} : = 2 i \in D$ . By 1.1 there is $γ_{1} \in Γ^{'}$ with $γ_{1} \cdot (γ \cdot z_{0}) \in \overline{D}$ ; set $δ : = γ_{1} γ \in Γ$ . Then $z_{0} \in D \subseteq \overline{D}$ and $δ \cdot z_{0} \in \overline{D}$ are equivalent, so by (a) either $δ \cdot z_{0} = z_{0}$ or one of the two boundary identifications holds; the latter are impossible because $∣ z_{0} ∣ = 2 > 1$ and $∣ ℜ z_{0} ∣ = 0 < 1 / 2$ . Hence $δ$ stabilises $z_{0}$ , and by (b) the only points of $\overline{D}$ with nontrivial stabiliser are $i, ω, ω + 1$ , none of which equals $z_{0}$ ; so $δ = 1$ and $γ = γ_{1}^{- 1} \in Γ^{'}$ . Therefore $Γ = Γ^{'} = ⟨ S, T ⟩$ .

LemmaStatement: Literature-sourcedProof: AI-adaptedjudge pass (gpt-6.1-sol)Open item page →

Local charts and the Riemann surface structure of a modular quotient

Statement

Let $Γ \leq P S L_{2} (Z)$ have finite index, with quotient map $p : H \to Γ \ H$ . (a) For every $τ$ there is an open neighbourhood $U$ of $τ$ such that ${γ \in Γ : γ U \cap U \neq \emptyset} = {Stab}_{Γ} (τ)$ ; stabilisers are finite cyclic, distinct orbits have disjoint invariant neighbourhoods, and $p$ is open with Hausdorff quotient. (b) $Γ \ H$ carries a Riemann surface structure for which $p$ is holomorphic and which is unique with that property: at a point with trivial stabiliser the local inverse of $p$ is a chart; at an elliptic point, in a local coordinate $z$ centred at it in which a generator of the stabiliser acts by $z \mapsto e^{2 π i / ν} z$ , the $ν$ -th power $z^{ν}$ descends to a chart on the quotient.

Facts & Assumptions

Given: A finite-index subgroup $Γ \leq G : = P S L_{2} (Z)$ acting on $H$ by biholomorphisms (The modular group and its action on the upper half-plane, Left group actions, transitive actions, and faithful actions).

[F1]

Every point of $H$ is $G$ -equivalent to a point of $\overline{D} = {τ : ∣ ℜ τ ∣ \leq 1 / 2, ∣ τ ∣ \geq 1}$ , and the only points of $\overline{D}$ with nontrivial $G$ -stabiliser are $i, ω, ω + 1$ , with stabilisers cyclic of orders $2, 3, 3$ ; $G$ acts faithfully (The standard fundamental domain, boundary identifications and elliptic stabilisers, The modular group and its action on the upper half-plane).

[F2]

For fixed $τ$ and $N > 0$ only finitely many pairs $(c, d) \in Z^{2}$ satisfy $∣ c τ + d ∣ \leq N$ (Reduction of orbits to the standard domain).

[F3]

A nonidentity Möbius transformation with two fixed points is conjugate to $z \mapsto λ z$ , $λ \neq 0, 1$ (Nonidentity Möbius transformations are parabolic or conjugate to a dilation, with the projective trace invariant); in a coordinate centred at a fixed point of an element of finite order $ν$ , that element acts as $z \mapsto ζ z$ with $ζ$ a primitive $ν$ -th root of unity, and Möbius transformations are biholomorphisms (Biholomorphic maps between complex domains).

[F4]

If $f$ is holomorphic near $a$ with $f^{'} (a) \neq 0$ , then $f$ is biholomorphic between suitable neighbourhoods of $a$ and $f (a)$ (A nonzero complex derivative gives a local biholomorphism); a nonconstant holomorphic function has the local normal form $ϕ (z)^{m}$ (Local normal form of a nonconstant holomorphic map).

[F5]

The quotient topology makes $p$ continuous and satisfies the universal property: a map out of $Γ \ H$ is continuous exactly when its composite with $p$ is (The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection, For a quotient map $q : X \to Y$ , a map out of $Y$ is continuous iff its composite with $q$ is; a continuous map on $X$ constant on the fibres of $q$ factors uniquely through $q$ ; and a composite of quotient maps is a quotient map, For a map of spaces the following agree: continuity at every point, preimages of open sets open, preimages of closed sets closed, preimages of subbasic open sets open, and $f (\overline{A}) \subseteq \overline{f (A)}$ , Continuity of a map of topological spaces at a point and globally). Riemann surface structures, holomorphic maps and biholomorphisms are defined by atlases and charts (Riemann surfaces and holomorphic atlases, Holomorphic maps and meromorphic functions on Riemann surfaces, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological, A continuous bijection is a homeomorphism iff it is open iff it is closed, and homeomorphy is an equivalence relation on spaces).

[F7]

Holomorphic functions admit convergent Taylor series; injective holomorphic functions have nonzero derivative (A holomorphic function equals its Taylor series throughout the largest centred disc in its domain, An injective holomorphic map has no critical point and is biholomorphic onto its image).

[F6]

A subgroup of a cyclic group is cyclic (Every subgroup of a cyclic group is cyclic; the least positive exponent in a nontrivial subgroup supplies a generator).

Proof

1.1F1F2F6givenalgebra

For compact sets $K, L \subset H$ , only finitely many $γ \in Γ$ satisfy $γ K \cap L \neq \emptyset$ . Write $z \in K$ , $w = γ z \in L$ ; if $y_{K} \leq ℑ z \leq Y_{K}$ , $ℑ w \geq y_{L} > 0$ and $∣ ℜ z ∣ \leq X_{K}$ , the height formula gives $∣ c z + d ∣^{2} \leq Y_{K} / y_{L}$ . Thus $∣ c ∣ \leq \sqrt{Y_{K} / y_{L}} / y_{K}$ and $∣ d ∣ \leq \sqrt{Y_{K} / y_{L}} + ∣ c ∣ X_{K}$ , leaving finitely many integer bottom rows. For any fixed bottom row, all determinant-one top rows are $(a_{0}, b_{0}) + k (c, d)$ , so the corresponding maps are $γ_{0} z + k$ ; the real parts of $γ_{0} (K)$ and $L$ are bounded, leaving finitely many $k$ . Apply this to a closed small disc about $τ$ . The stabiliser is finite cyclic: conjugation into $\overline{D}$ and [F1, F6] identify it with a subgroup of a cyclic group of order $1, 2$ or $3$ . For each of the finitely many non-stabilising maps meeting that disc, disjointness of its image of $τ$ and $τ$ permits shrinking the neighbourhood. Intersect its images under the finite stabiliser to obtain an invariant open $U$ with $γ U \cap U \neq \emptyset$ exactly for stabiliser elements.

2.1F5step 1.1givenalgebra

If $q, q^{'} \in Γ \ H$ are distinct, choose $τ, τ^{'}$ with $p (τ) = q$ , $p (τ^{'}) = q^{'}$ ; the set ${γ \in Γ : γ B (τ, δ) \cap B (τ^{'}, δ) \neq \emptyset}$ is finite for small $δ$ by the same boundedness argument as in 1.1, and no element of it carries $τ$ to $τ^{'}$ because $q \neq q^{'}$ ; shrinking $δ$ therefore gives disjoint open neighbourhoods $U, U^{'}$ of $τ, τ^{'}$ with $γ U \cap U^{'} = \emptyset$ for all $γ \in Γ$ . Then $Γ U$ and $Γ U^{'}$ are disjoint $Γ$ -invariant open sets, so $q, q^{'}$ have disjoint neighbourhoods and the quotient is Hausdorff. The map $p$ is open: for open $V \subseteq H$ , the preimage $p^{- 1} (p (V)) = ⋃_{γ \in Γ} γ V$ is open, so $p (V)$ is open by definition of the quotient topology.

3.1F3F4F5step 2.1givenalgebra

Construction of charts. If ${Stab}_{Γ} (τ) = {1}$ , take $U$ as in 1.1; then $p ∣_{U}$ is injective, $p (U)$ is open by 2.1, and $p ∣_{U}^{- 1}$ is a homeomorphism onto its image by [F5]; declare it a chart, and $p$ is the identity map in these coordinates. If ${Stab}_{Γ} (τ) = ⟨ γ_{0} ⟩$ has order $ν > 1$ , then $ν \in {2, 3}$ by 1.1; by [F3] there is a biholomorphic coordinate $z$ on a disc $Δ$ centred at $τ$ , $z (τ) = 0$ , in which $γ_{0}$ acts by $z \mapsto ζ z$ with $ζ$ a primitive $ν$ -th root of unity. Choose $Δ$ so that ${γ : γ Δ \cap Δ \neq \emptyset} = ⟨ γ_{0} ⟩$ ; then $z_{1}^{ν} = z_{2}^{ν}$ for $z_{1}, z_{2} \in Δ$ exactly when $z_{2} = ζ^{k} z_{1}$ for some $k$ (both are $ν$ -th roots of the same number), so $ψ : = z^{ν}$ induces a bijection $Δ / ⟨ γ_{0} ⟩ \to Δ^{'}$ onto a disc $Δ^{'}$ and this bijection is a homeomorphism by [F5]; since $p (Δ) = Δ / ⟨ γ_{0} ⟩$ , the induced map $p (Δ) \to Δ'$ is a chart on the quotient. In these coordinates $p$ is the holomorphic map $z \mapsto z^{ν}$ , so $p$ is holomorphic for the atlas.

4.1F4F5F7step 3.1algebra∎

Transition maps are holomorphic away from elliptic centres by the local inverse theorem [F4]. At a centre of order $ν$ , a quotient-coordinate function pulled back to the uniformising coordinate has a holomorphic Taylor series $h (z)$ invariant under $z \mapsto ζ z$ ; coefficient comparison gives $h (z) = \sum_{m \geq 0} a_{m ν} z^{m ν} = g (z^{ν})$ , with $g$ holomorphic. Indeed its power series converges for $∣ z^{ν} ∣ < r^{ν}$ whenever $h$ converges for $∣ z ∣ < r$ . This proves transition holomorphy also at the centre. Any other surface structure making $p$ holomorphic has the same property for the pullback of each of its charts, so its charts are holomorphic functions of our quotient charts. These functions are injective, since both charts are homeomorphisms; their derivatives are therefore nonzero and their inverses are holomorphic. The two maximal atlases agree, proving uniqueness. The quotient is second countable: images under the open map $p$ of a countable disc basis of $H$ form a basis; it is connected as the continuous image of $H$ . Thus the atlas defines a Riemann surface with all the topological hypotheses.

LemmaStatement: Literature-sourcedProof: AI-adaptedOpen item page →

The cusp chart and compactness of X(1)

Statement

Let $H^{*} = H \cup Q \cup {\infty}$ be the space obtained by adding the cusps, topologised by the usual topology on $H$ together with, at $γ \cdot \infty$ , the images under $γ \in P S L_{2} (Z)$ of the basic neighbourhoods ${ℑ τ > N} \cup {\infty}$ of $\infty$ . Then the $P S L_{2} (Z)$ -action extends continuously to $H^{*}$ , the cusps form the single orbit $Q \cup {\infty} = P S L_{2} (Z) \cdot \infty$ , and $X (1) = P S L_{2} (Z) \ H^{*}$ is compact Hausdorff with $Y (1) = P S L_{2} (Z) \ H$ as a dense open subset whose complement is the single cusp class. The function $q (τ) = e^{2 π i τ}$ descends to a homeomorphism of a neighbourhood of the cusp class onto an open disc in $C$ and provides the cusp chart, so that $X (1)$ is a compact Riemann surface.

Facts & Assumptions

Given: The action of $G : = P S L_{2} (Z) = ⟨ S, T ⟩$ on $H$ with $S τ = - 1 / τ$ , $T τ = τ + 1$ , its fundamental domain $D$ , and the identification of $G$ with its Möbius transformations on $\hat{C}$ (The modular group and its action on the upper half-plane, The standard fundamental domain, boundary identifications and elliptic stabilisers).

[F1]

Every orbit meets $\overline{D} = {τ : ∣ ℜ τ ∣ \leq 1 / 2, ∣ τ ∣ \geq 1}$ , no two distinct points of $D$ are equivalent, and two distinct $z, z^{'} \in \overline{D}$ are equivalent exactly when $z^{'} = z \pm 1$ with $ℜ z = \mp 1 / 2$ or $z^{'} = - 1 / z$ with $∣ z ∣ = 1$ (The standard fundamental domain, boundary identifications and elliptic stabilisers).

[F2]

For finite-index $Γ \leq G$ the quotient $Γ \ H$ has the properties of Local charts and the Riemann surface structure of a modular quotient; in particular its quotient map is open and the quotient is Hausdorff.

[F3]

Quotient topologies, continuous maps, homeomorphisms, compactness and Hausdorffness are as in The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection, For a quotient map $q : X \to Y$ , a map out of $Y$ is continuous iff its composite with $q$ is; a continuous map on $X$ constant on the fibres of $q$ factors uniquely through $q$ ; and a composite of quotient maps is a quotient map, Homeomorphism, open map, closed map, embedding, and what it means for a property to be topological, A continuous bijection is a homeomorphism iff it is open iff it is closed, and homeomorphy is an equivalence relation on spaces, Hausdorff space: distinct points have disjoint open neighbourhoods; every metrizable space is Hausdorff and the indiscrete topology on two points is not, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, Subspace topology: the traces of the open sets, its closed sets and its bases, the continuity of the inclusion, and the characteristic property of a map into a subspace, For a metric space with its metric topology, compactness in the topological sense is compactness in the metric sense, and the two notions of compact subset coincide, A continuous image of a compact space is compact; a continuous real-valued map on a nonempty compact space attains a maximum and a minimum; and a continuous bijection from a compact space to a Hausdorff space is a homeomorphism, In a Hausdorff space a point and a disjoint compact set, and two disjoint compact sets, have disjoint open neighbourhoods; hence every compact subset is closed, and in a compact Hausdorff space the compact subsets are exactly the closed ones, A continuous image of a connected space is connected, and connectedness is a topological property; complex structures and holomorphic maps are as in Riemann surfaces and holomorphic atlases and Holomorphic maps and meromorphic functions on Riemann surfaces.

[F4]

The $Z$ -action on $H$ by $T^{n}$ is a covering-space action, so its orbit map is a covering, and $e^{2 π i τ} = e^{2 π i τ^{'}}$ exactly when $τ - τ^{'} \in Z$ (The orbit map of a covering-space action is a covering, with the acting group equal to the deck group when the total space is path-connected, Covering-space actions by disjoint translates of neighbourhoods, $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ , The complex exponential is entire and its complex derivative is itself, $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential).

Proof

1.1F2F3givenconstructalgebra

The action extends to $H^{*}$ : for $γ \in G$ and a cusp $γ_{0} \cdot \infty = m / n$ with $m, n$ coprime, Bézout gives $b, d \in Z$ with $m d - n b = 1$ , so $(\begin{matrix} m & b \\ n & d \end{matrix}) \in S L_{2} (Z)$ and $γ γ_{0} \cdot \infty = γ \cdot (m / n)$ ; Möbius maps are homeomorphisms of $\hat{C}$ carrying $Q \cup {\infty}$ to itself and the basic cusp neighbourhoods of $γ_{0} \cdot \infty$ to basic cusp neighbourhoods of $γ γ_{0} \cdot \infty$ , so the extended action is continuous and well defined. Every rational $m / n$ in lowest terms equals $γ \cdot \infty$ for the matrix above with $γ = (\begin{matrix} m & b \\ n & d \end{matrix})$ , and $\infty$ itself is in the orbit, so the cusps form the single orbit $Q \cup {\infty} = G \cdot \infty$ .

1.2F2F3F4givenconstructalgebra

Fix $N > 1$ and put $B_{N} : = {ℑ τ > N} \cup {\infty}$ . If $γ \in S L_{2} (Z)$ has $c \neq 0$ and $z, γ z \in H$ with $ℑ z, ℑ γ z > N$ , then $ℑ (γ z) = ℑ z / ∣ c z + d ∣^{2} \leq ℑ z / (c ℑ z)^{2} = 1 / (c^{2} ℑ z) < 1 / N < N$ , a contradiction; hence every $γ$ mapping a point of $B_{N}$ back into $B_{N}$ has $c = 0$ , i.e. lies in $⟨ T ⟩$ . Consequently the $G$ -orbit of a point of $B_{N}$ meets $B_{N}$ exactly in its $⟨ T ⟩$ -orbit, and $B_{N} / ⟨ T ⟩$ is identified with its image $p (B_{N})$ . By [F4] the map $q (τ) = e^{2 π i τ}$ realises $H / ⟨ T ⟩ ≅ {0 < ∣ q ∣ < 1}$ , so it descends to a homeomorphism of $p (B_{N})$ onto ${∣ q ∣ < e^{- 2 π N}}$ sending the cusp class to $0$ ; the topology at the cusp was defined exactly so that ${ℑ τ > N^{'}}$ corresponds to ${∣ q ∣ < e^{- 2 π N^{'}}}$ , so this is a homeomorphism onto the open disc and provides the cusp chart.

2.1F1F3step 1.1givenalgebra

Put $K : = \overline{D} \cup {\infty}$ with its subspace topology in $H^{*}$ . Given an intrinsic open cover $U$ of $K$ , take $U_{\infty} \in U$ containing $\infty$ ; the subspace topology and the cusp neighbourhood basis give $N > 1$ with $K \cap B_{N} \subseteq U_{\infty}$ . The set $L : = \overline{D} \cap {ℑ τ \leq N}$ is closed and bounded in $C$ , with imaginary part at least $\sqrt{3} / 2$ , hence compact by Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line and the metric/topological compactness agreement in [F3]. Since $H$ has its usual topology inside $H^{*}$ , the topology induced on $L$ from $K$ is its usual subspace topology. Thus ${U \cap L : U \in U}$ is an intrinsic open cover of $L$ and has a finite subcover. The corresponding finitely many members of $U$ , together with $U_{\infty}$ , cover $K$ , proving its compactness. The quotient map $p$ is continuous, so $p (\overline{D} \cup {\infty})$ is compact by [F3]; it equals $X (1)$ because every point of $H$ is $G$ -equivalent to a point of $\overline{D}$ by [F1] and every cusp lies in the orbit of $\infty$ by 1.1. Hence $X (1)$ is compact.

3.1F1F2F3F4step 1.2step 2.1algebra∎

To separate the cusp from an interior point $p (τ)$ , choose a relatively compact open neighbourhood $U$ of $τ$ with $0 < y_{0} \leq ℑ z \leq Y$ on $U$ . For every $γ \in G$ , its height on $U$ is at most $M = \max (Y, 1 / y_{0})$ : if $c = 0$ height is unchanged, while if $c \neq 0$ , $ℑ (γ z) \leq 1 / (c^{2} ℑ z) \leq 1 / y_{0}$ . For $N > \max (1, M)$ the open sets $p (U)$ and $p (B_{N})$ are disjoint. The quotient map on $H^{*}$ is open, since the saturation of each open set is the union of its translates; hence these are open neighbourhoods in $X (1)$ . Two interior points are separated by [F2], so $X (1)$ is Hausdorff. The interior quotient is open and dense, since every cusp neighbourhood meets $H$ ; its complement is the unique cusp class. Its atlas [F2] is compatible with the cusp coordinate: for $N > 1$ stabilisers on $B_{N} \cap H$ are trivial and $q^{'} = 2 π i q \neq 0$ , so the transition to any local lift chart and its inverse are holomorphic. The inherited countable interior basis plus $p (B_{n})$ , $n \geq 2$ , gives second countability; connectedness follows from the connected dense interior. Together with 2.1 this makes $X (1)$ a compact Riemann surface.

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

The compactified level-one modular curve X(1)

Definition

Let $H^{*} = H \cup Q \cup {\infty}$ be the space of The cusp chart and compactness of X(1) with its cusp-neighbourhood topology and the action of $P S L_{2} (Z)$ (The modular group and its action on the upper half-plane, The quotient topology of a surjection, quotient maps, saturated sets, and the quotient of a space by an equivalence relation with its canonical projection). The compactified level-one modular curve is the quotient

$X (1) : = P S L_{2} (Z) \ H^{*},$

with the quotient topology and the structure of a compact Riemann surface whose interior quotient map $H \to X (1)$ is holomorphic and whose cusp coordinate is $q = e^{2 π i τ}$ (The cusp chart and compactness of X(1), Riemann surfaces and holomorphic atlases); its single cusp is the class $[\infty]$ . The open modular curve is the dense open subset

$Y (1) : = P S L_{2} (Z) \ H = X (1) ∖ {[\infty]},$

whose Riemann surface structure is the one supplied by the local chart lemma for the quotient of the upper half-plane (Local charts and the Riemann surface structure of a modular quotient). More generally, for a finite-index subgroup $Γ \leq P S L_{2} (Z)$ we write $X_{Γ} : = Γ \ H^{*}$ and $Y_{Γ} : = Γ \ H$ for the corresponding quotient spaces, with $Y_{Γ}$ carrying the complex structure of Local charts and the Riemann surface structure of a modular quotient.

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The q-expansion principle at the cusp

Statement

Let $f$ be holomorphic on $H$ with $f (τ + 1) = f (τ)$ for all $τ$ , and put $q = e^{2 π i τ}$ . Then there is a unique holomorphic $F$ on the punctured unit disc $D^{*} = {0 < ∣ q ∣ < 1}$ with $f (τ) = F (e^{2 π i τ})$ . Moreover $f$ is bounded on ${ℑ τ > Y_{0}}$ for some $Y_{0}$ if and only if $F$ extends holomorphically to $q = 0$ , and then $F (0) = \lim_{ℑ τ \to \infty} f (τ)$ and $f (τ) - F (0) = O (e^{- 2 π δ ℑ τ})$ for some $δ > 0$ as $ℑ τ \to \infty$ . In particular $f (τ) = \sum_{n \geq 0} a_{n} q^{n}$ with $a_{n} = (1 / 2 π i) \oint F (q) q^{- n - 1} d q$ when the extension exists.

Facts & Assumptions

Given: A holomorphic, $1$ -periodic $f$ on $H$ and $q = e^{2 π i τ}$ ; the unit disc and half-plane are those of The unit disc, the upper half-plane, and Blaschke factors.

[F1]

$∣ e^{2 π i τ} ∣ = e^{- 2 π ℑ τ}$ for every $τ \in C$ ( $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ), and $e^{2 π i τ} = e^{2 π i τ^{'}}$ if and only if $τ - τ^{'} \in Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

[F2]

The principal logarithm $L = Log$ is holomorphic on the slit plane $C ∖ (- \infty, 0]$ and satisfies $e^{L (w)} = w$ there (The principal logarithm is the normalised holomorphic branch on the slit plane, Complex logarithms, the principal logarithm, and principal and multivalued complex powers).

[F3]

A composition of holomorphic functions is holomorphic, by the chain rule (The chain rule for complex derivatives, Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions, A complex function is holomorphic if and only if it is analytic, Complex analytic functions as locally representable by convergent power series).

[F4]

On a punctured disc $0 < ∣ z - a ∣ < R$ , the singularity $a$ is removable if and only if the function is bounded on some punctured neighbourhood of $a$ ; then the extension has value the limit at $a$ (Characterizations of removable singularities).

[F5]

If a holomorphic function vanishes at $a$ , then either it vanishes on a neighbourhood of $a$ , or it has finite order $m \geq 1$ and factors locally as $(z - a)^{m} g$ with $g$ holomorphic and $g (a) \neq 0$ . In either case a function vanishing at $0$ is $q$ times a holomorphic function bounded near $0$ (take the latter function to be zero in the first case) (The order of a zero is the exponent in its local holomorphic factorization).

[F6]

Every holomorphic function on an annulus has a convergent Laurent series (Laurent expansion on an annulus). The coefficients of a convergent Laurent series are the contour integrals $\frac{1}{2 π i} \oint f (ζ) (ζ - a)^{- n - 1} d ζ$ , and they are unique (Laurent coefficients are given by contour integrals and are unique).

Proof

1.1F1F2F3givenconstruct

Let $p (τ) : = e^{2 π i τ}$ . For every $q_{0} \in D^{*}$ there are an open disc $V \subseteq D^{*}$ about $q_{0}$ and a holomorphic $σ : V \to H$ with $p (σ (q)) = q$ on $V$ . Indeed, choose an angle $α$ with $e^{- i α} q_{0} \notin (- \infty, 0]$ and let $S_{α} = {q : e^{- i α} q \notin (- \infty, 0]}$ ; the function $v (q) : = L (e^{- i α} q) + i α$ is holomorphic near $q_{0}$ by [F2], and $e^{v (q)} = e^{- i α} q \cdot e^{i α} = q$ by the addition law. Taking $V$ a small disc contained in $D^{*} \cap S_{α}$ and setting $σ (q) : = - i v (q) / (2 π)$ gives $e^{2 π i σ (q)} = e^{v (q)} = q$ , and $σ$ is holomorphic by [F3]. Here $ℑ σ (q) > 0$ holds automatically: $e^{- 2 π ℑ σ (q)} = ∣ e^{2 π i σ (q)} ∣ = ∣ q ∣ < 1$ by [F1], so $σ$ maps $V$ into $H$ .

2.1F1F3step 1.1givenalgebra

Define $F (q) : = f (σ (q))$ for a local section $σ$ as in 1.1; this is well defined. If $σ, σ^{'}$ are two such sections near $q$ , then $p (σ (q)) = q = p (σ^{'} (q))$ , so $σ (q) - σ^{'} (q) \in Z$ by [F1], whence $f (σ (q)) = f (σ^{'} (q))$ by the periodicity of $f$ . Since local sections exist near every $q \in D^{*}$ , this gives a function $F$ on all of $D^{*}$ , holomorphic because near each point it is the composition $f \circ σ$ of holomorphic functions [F3]. If $\tilde{F}$ is holomorphic on $D^{*}$ with $f (τ) = \tilde{F} (e^{2 π i τ})$ for all $τ$ , then for $q \in D^{*}$ and a local section $σ$ with $p (σ (q)) = q$ we get $\tilde{F} (q) = \tilde{F} (p (σ (q))) = f (σ (q)) = F (q)$ ; hence $\tilde{F} = F$ and $F$ is unique.

3.1F1F4F5step 1.1step 2.1givenalgebra

If $F$ extends holomorphically to $0$ , then $F$ is bounded on $∣ q ∣ < δ$ for some $δ > 0$ ; for $ℑ τ > - \log δ / (2 π)$ one has $∣ q ∣ < δ$ by [F1], so $∣ f (τ) ∣ = ∣ F (q) ∣$ is bounded on that half-plane. Conversely, if $∣ f ∣ \leq M$ on ${ℑ τ > Y_{0}}$ , then for $0 < ∣ q ∣ < e^{- 2 π Y_{0}}$ any local section $σ (q)$ of 1.1 satisfies $ℑ σ (q) = - \log ∣ q ∣ / (2 π) > Y_{0}$ by [F1], so $∣ F (q) ∣ = ∣ f (σ (q)) ∣ \leq M$ ; by [F4] the singularity of $F$ at $0$ is removable, $F$ extends holomorphically, and $F (0) = \lim_{q \to 0} F (q) = \lim_{ℑ τ \to \infty} f (τ)$ (given $ε > 0$ choose $δ$ with $∣ F (q) - F (0) ∣ < ε$ for $∣ q ∣ < δ$ ; then $ℑ τ > - \log δ / (2 π)$ gives $∣ f (τ) - F (0) ∣ < ε$ ). Finally, if the extension exists, either $F - F (0)$ vanishes on a neighbourhood of $0$ , in which case take $G : = 0$ , or it vanishes to finite order $m \geq 1$ and [F5] writes $F (q) - F (0) = q G (q)$ with $G$ holomorphic near $0$ ; in either case $G$ is bounded near $0$ , so $∣ f (τ) - F (0) ∣ = ∣ q ∣ ∣ G (q) ∣ \leq C e^{- 2 π ℑ τ}$ for all large $ℑ τ$ , which is the asserted $O (e^{- 2 π δ ℑ τ})$ with $δ = 1$ .

4.1F4F6step 3.1algebra∎

Assume now that $F$ extends holomorphically to $0$ . On the annulus $0 < ∣ q ∣ < 1$ the holomorphic $F$ has Laurent expansion $\sum_{n \in Z} a_{n} q^{n}$ , and by [F6] $a_{n} = \frac{1}{2 π i} \oint_{∣ q ∣ = ρ} F (q) q^{- n - 1} d q$ for every $0 < ρ < 1$ . Since $F$ is holomorphic at $0$ , all coefficients $a_{n}$ with $n < 0$ vanish (their principal part is zero, [F4] applied to the Laurent expansion), so $F (q) = \sum_{n \geq 0} a_{n} q^{n}$ and $f (τ) = F (e^{2 π i τ}) = \sum_{n \geq 0} a_{n} q^{n}$ converges for $∣ q ∣ < 1$ .

DefinitionDefinition: Literature-sourcedProof: Not applicablejudge pass (gpt-6.1-sol)Open item page →

Level-one modular forms and cusp forms

Definition

Fix an integer $k$ . A modular form of weight $k$ for $P S L_{2} (Z)$ is a holomorphic function $f : H \to C$ (Complex differentiability at a point, the complex derivative, holomorphic functions, and entire functions) such that

$f (γ \cdot τ) = (c τ + d)^{k} f (τ)$

for every $γ = (\begin{matrix} a & b \\ c & d \end{matrix}) \in S L_{2} (Z)$ and every $τ \in H$ (The modular group and its action on the upper half-plane), and such that the associated periodic function $F$ of The q-expansion principle at the cusp is holomorphic at $q = 0$ . Since $T = (\begin{matrix} 1 & 1 \\ 0 & 1 \end{matrix})$ acts by $τ \mapsto τ + 1$ with factor $1$ , the law gives $f (τ + 1) = f (τ)$ , so the q-expansion principle applies and $F$ is the unique holomorphic function on $0 < ∣ q ∣ < 1$ with $f (τ) = F (e^{2 π i τ})$ . The form $f$ is a cusp form if additionally $F (0) = 0$ .

Write $M_{k}$ and $S_{k}$ for the sets of modular and cusp forms of weight $k$ . They are $C$ -subspaces of the space of holomorphic functions on $H$ : sums and scalar multiples of functions satisfying the transformation law satisfy it again, and the q-expansion condition is preserved because $F_{a f + b g} = a F_{f} + b F_{g}$ by uniqueness (Vector space over a field, Meromorphic functions on a plane domain). The transformation law is well posed: the factor $(c τ + d)^{k}$ depends only on $γ$ and, by the cocycle identity for the Möbius action, the conditions for all $γ$ are consistent (Möbius transformations form a group and identify with the projective linear quotient of GL_2(C)). Since $- I \in S L_{2} (Z)$ acts as the identity with factor $(- 1)^{k}$ , one has $f = (- 1)^{k} f$ , so $M_{k} = {0}$ for odd $k$ . The weight condition is compatible with multiplication: $M_{k} M_{ℓ} \subseteq M_{k + ℓ}$ and $S_{k} M_{ℓ} \subseteq S_{k + ℓ}$ , because the products of the factors are $(c τ + d)^{k + ℓ}$ and the q-expansion of a product has constant term $F_{f} (0) F_{g} (0)$ .

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Absolute convergence and holomorphy of the lattice Eisenstein sums

Statement

For every even $k \geq 4$ and every $τ \in H$ the family $((m τ + n)^{- k})_{(m, n) \in Z^{2} ∖ {(0, 0)}}$ is absolutely summable, and the convergence is uniform on compact subsets of $H$ . Consequently $G_{k} (τ) : = \sum_{(m, n)}^{'} (m τ + n)^{- k}$ defines a holomorphic function on $H$ , and it depends only on the lattice $Λ_{τ} = Z + Z τ$ assigned to $τ$ .

Facts & Assumptions

Given: An even integer $k \geq 4$ , the upper half-plane $H$ with $Im > 0$ (The modular group and its action on the upper half-plane), and a compact $K \subseteq H$ with $Im τ \geq y_{0} > 0$ and $∣ Re τ ∣ \leq X$ for all $τ \in K$ .

[F1]

Convergence, absolute convergence and nonnegative comparison of series are as in Series, partial sums, convergence and the sum, divergence, and the tail series, Absolutely convergent and conditionally convergent series, and the general starting index and A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum.

[F2]

If $0 \leq a_{j} \leq b_{j}$ eventually and $\sum b_{j}$ converges then $\sum a_{j}$ converges (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ ); $\sum_{j} j^{- p}$ converges for rational $p > 1$ (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ).

[F3]

For a complex double family with summable absolute values, apply the real double-series theorem separately to its real and imaginary parts. The family may then be summed in any order, in particular iterated or regrouped into shells (Fubini for double series: if $\sum_{i} \sum_{j} ∣ a_{i j} ∣$ converges then both iterated sums and the sum along every bijection $N \to N \times N$ converge to one and the same value, If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges).

[F4]

If holomorphic functions on an open set converge locally uniformly, their limit is holomorphic (Locally uniform limits of holomorphic functions are holomorphic and their derivatives converge locally uniformly).

[F5]

$∣ z + w ∣ \geq ∣ ∣ z ∣ - ∣ w ∣ ∣$ , $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ , and $∣ Re z ∣ \leq ∣ z ∣$ (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive, Real and imaginary parts, complex conjugation, and modulus).

Proof

1.1F5givenalgebra

Since $m τ + n = 0$ with $(m, n) \neq (0, 0)$ would give $τ = - n / m \in R$ when $m \neq 0$ and $n = 0$ when $m = 0$ , no denominator vanishes. Let $c_{K} : = \min (1 / 2, y_{0} / \max (1, 2 X)) > 0$ . For $(m, n) \neq (0, 0)$ and $τ \in K$ we claim $∣ m τ + n ∣ \geq c_{K} \max (∣ m ∣, ∣ n ∣)$ . If $∣ n ∣ \geq 2 X ∣ m ∣$ , then $∣ Re (m τ + n) ∣ = ∣ m Re τ + n ∣ \geq ∣ n ∣ - X ∣ m ∣ \geq ∣ n ∣ / 2$ by [F5]; if moreover $∣ n ∣ \geq ∣ m ∣$ then $∣ m τ + n ∣ \geq ∣ n ∣ / 2 \geq \max (∣ m ∣, ∣ n ∣) / 2$ , while if $∣ n ∣ < ∣ m ∣$ then the imaginary part gives $∣ m τ + n ∣ \geq y_{0} ∣ m ∣ = y_{0} \max (∣ m ∣, ∣ n ∣)$ . If instead $∣ n ∣ < 2 X ∣ m ∣$ , then $\max (∣ m ∣, ∣ n ∣) \leq \max (1, 2 X) ∣ m ∣$ and the imaginary part gives $∣ m τ + n ∣ \geq y_{0} ∣ m ∣ \geq \frac{y_{0}}{\max (1, 2 X)} \max (∣ m ∣, ∣ n ∣)$ . In all cases the claim holds.

2.1F1F2F3step 1.1givenalgebra

Regroup the nonzero pairs by the shell $j : = \max (∣ m ∣, ∣ n ∣) \geq 1$ ; the shell has $(2 j + 1)^{2} - (2 j - 1)^{2} = 8 j$ elements, and by 1.1 each contributes at most $(c_{K} j)^{- k}$ , so shell $j$ contributes at most $8 c_{K}^{- k} j^{1 - k}$ . Since $k \geq 4$ gives $k - 1 > 1$ , the bound $\sum_{j \geq 1} 8 c_{K}^{- k} j^{1 - k} < \infty$ follows from [F2]. The shell partial sums of the family are therefore nonnegative and bounded uniformly in $τ \in K$ by an absolute constant, so they converge by the bounded-partial-sums criterion of [F1]; in the terminology of [F1] the family $((m τ + n)^{- k})_{(m, n) \neq (0, 0)}$ is absolutely summable for each $τ \in K$ , the regrouping being licensed by [F3], and the tail beyond shell $J$ is bounded uniformly on $K$ by $8 c_{K}^{- k} \sum_{j > J} j^{1 - k}$ .

3.1F3F4step 2.1givenalgebra∎

Each term $τ \mapsto (m τ + n)^{- k}$ is holomorphic on $H$ for $(m, n) \neq (0, 0)$ , being a power of the nonvanishing holomorphic function $τ \mapsto m τ + n$ . By 2.1 the partial sums over shells converge uniformly on every compact $K \subseteq H$ (compactness supplies some $y_{0}, X$ ), so [F4] makes the shell-sum limit $G_{k}$ holomorphic on $H$ , and this limit agrees with the (absolutely summable, hence order-independent by [F3]) family sum. Finally $(m, n) \mapsto m τ + n$ is a bijection of $Z^{2}$ onto the lattice $Λ_{τ} = Z + Z τ$ , so the summed family is exactly the family of values $λ^{- k}$ over the nonzero points $λ \in Λ_{τ}$ ; hence $G_{k} (τ)$ depends only on $Λ_{τ}$ and not on the enumeration.

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The divisor power sums $σ_{k}$

Definition

For an integer $k \geq 0$ and a positive integer $n$ let

$σ_{k} (n) : = \sum_{d ∣ n, 1 \leq d \leq n} d^{k},$

the sum of the $k$ -th powers of the positive divisors of $n$ (Divisibility in $Z$ : $d ∣ a$ when $a = d q$ for some integer $q$ , Integer powers $a^{m}$ , Finite sums and finite products, by recursion, The integers as equivalence classes of pairs of naturals). Thus $σ_{0} (n)$ is the number of positive divisors and $σ_{1} (n)$ is their sum. The sum is over the set of positive divisors of $n$ , a finite set: it is nonempty because $1 ∣ n$ , and it is bounded above by $n$ when $n \geq 1$ (If $d ∣ a$ and $a \neq 0$ then $d \neq 0$ and $∣ d ∣ \leq ∣ a ∣$ ; hence the set of divisors of a nonzero integer is bounded above by $∣ a ∣$ ), so the displayed sum is a finite sum of integers. Each $σ_{k} (n)$ is therefore a positive integer: by induction on $k$ , $d^{0} = 1 > 0$ and $d^{k + 1} = d^{k} d > 0$ for every positive integer $d$ , since positive integers are closed under multiplication (Integer powers $a^{m}$ , The principle of mathematical induction, The integers form a totally ordered ring). The divisor $d = 1$ contributes $1^{k} = 1$ , and adding the other positive integer summands preserves positivity by compatibility of the integer order with addition (The integers form a totally ordered ring). These functions are used only to express the Fourier coefficients of the Eisenstein series on this page.

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The level-one Eisenstein series E_k and the weight-two series E_2

Definition

For even $k \geq 4$ put

$G_{k} (τ) = {\sum_{(m, n) \in Z^{2}}}^{'} (m τ + n)^{- k}, τ \in H,$

the prime excluding $(m, n) = (0, 0)$ ; this is an absolutely summable family whose sum is holomorphic in $τ$ (Absolute convergence and holomorphy of the lattice Eisenstein sums). Define the normalised Eisenstein series

$E_{k} : = \frac{G_{k}}{2 ζ (k)}, ζ (k) = \sum_{n \geq 1} n^{- k},$

the Riemann zeta value (The Riemann zeta function on the half-plane $Re s > 1$ ). For $k > 1$ the series for $ζ (k)$ converges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ), so $2 ζ (k) > 0$ and the normalisation is well defined.

The constant term and the vanishing of the higher terms. Splitting the absolutely convergent sum into $m = 0$ and $m \neq 0$ , and bounding the latter with the Lipschitz formula (The Lipschitz formula for the reciprocal-power sums),

$G_{k} (τ) = 2 ζ (k) + 2 \sum_{m \geq 1} \frac{(- 2 π i)^{k}}{(k - 1)!} \sum_{r \geq 1} r^{k - 1} q^{m r}, q = e^{2 π i τ},$

so $E_{k} (τ) = 1 + O (q)$ as $Im τ \to \infty$ : the correction is a power series in $q$ divisible by $q$ , convergent for $∣ q ∣ < 1$ . The same computation yields the Fourier expansion of Eisenstein series are modular forms; their Fourier coefficients.

The weight-two series. Separately define

$E_{2} (τ) : = 1 - 24 \sum_{n \geq 1} σ_{1} (n) q^{n}, q = e^{2 π i τ},$

with $σ_{1} (n)$ the sum of the positive divisors of $n$ (The divisor power sums $σ_{k}$ ). Since $σ_{1} (n) \leq n \cdot n = n^{2}$ and $\sum_{n \geq 1} n^{2} ∣ q ∣^{n}$ converges for $∣ q ∣ < 1$ by the ratio test (Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence), the $M$ -test (Weierstrass M-test for complex-valued function series, A complex power series converges absolutely and uniformly on every closed subdisc strictly inside its disc of convergence) makes $E_{2}$ a holomorphic function of $τ \in H$ . It is a quasimodular comparison object used in the discriminant product proof and in the counterexample item of the companion page; it is not itself a modular form, since its transformation law carries a nonzero correction term. The half-plane conventions are The modular group and its action on the upper half-plane.

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The Lipschitz formula for the reciprocal-power sums

Statement

For every integer $k \geq 2$ and every $τ \in H$ , with $q = e^{2 π i τ}$ , $\sum_{n \in Z} \frac{1}{(τ + n)^{k}} = \frac{(- 2 π i)^{k}}{(k - 1)!} \sum_{r = 1}^{\infty} r^{k - 1} q^{r},$ where the series on the left converges absolutely and the identity is independent of the order of summation.

Facts & Assumptions

Given: An integer $k \geq 2$ , a point $τ \in H$ , and $q = e^{2 π i τ}$ .

[F1]

$π \cot (π z) = \frac{1}{z} + \sum_{n \geq 1} (\frac{1}{z - n} + \frac{1}{z + n})$ , and the series converges locally uniformly on $C ∖ Z$ (The Mittag-Leffler expansion of pi cotangent).

[F2]

If the partial sums of $\sum_{j} g_{j}$ of holomorphic functions converge locally uniformly to $g$ , then $g$ is holomorphic and $g^{(m)} = \sum_{j} g_{j}^{(m)}$ for every $m$ , the derivative series again converging locally uniformly (A locally uniformly convergent series of holomorphic functions may be differentiated term by term, Complex analytic functions as locally representable by convergent power series).

[F3]

$\sin z = \frac{e^{i z} - e^{- i z}}{2 i}$ , $\cos z = \frac{e^{i z} + e^{- i z}}{2}$ , and $\cot z = \cos z / \sin z$ wherever $\sin z \neq 0$ (Complex sine, cosine, hyperbolic sine, and hyperbolic cosine from the complex exponential, Tangent, cotangent, secant, and cosecant on their exact natural domains).

[F4]

$\exp (w + w^{'}) = \exp (w) \exp (w^{'})$ , $\exp (2 w) = \exp (w)^{2}$ , $\exp (- w) = 1 / \exp (w)$ , and $∣ \exp (x + i y) ∣ = e^{x}$ for real $x, y$ ( $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[F5]

$\exp^{'} = \exp$ ; the chain rule and the algebra of derivatives give $\frac{d^{m}}{d w^{m}} e^{a w} = a^{m} e^{a w}$ and $\frac{d^{m}}{d w^{m}} \frac{1}{w + n} = \frac{(- 1)^{m} m!}{(w + n)^{m + 1}}$ on their domains (The complex exponential is entire and its complex derivative is itself, The chain rule for complex derivatives, Linearity, product, reciprocal, and quotient rules for complex derivatives).

[F6]

Convergence in $C$ is convergence in the metric $d (z, w) = ∣ z - w ∣$ (The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane); an absolutely convergent complex series converges, and every rearrangement of it has the same sum (Every absolutely convergent complex series converges, and rearrangements preserve its sum, Complex series, absolute convergence, complex power series, and radius of convergence).

[F7]

Let $k \geq 2$ . The real series $\sum_{n \geq 1} n^{- k}$ converges (For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ ), so $\sum_{n \geq 1} ∣ τ + n ∣^{- k}$ converges by comparison (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ , If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges); for $∣ r ∣ < 1$ the real series $\sum_{m \geq 0} r^{m}$ converges and $r^{m} \to 0$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, For $∣ r ∣ < 1$ the sequence $r^{k}$ is null, and for $∣ r ∣ > 1$ the sequence $∣ r ∣^{k}$ diverges to $+ \infty$ ).

Proof

1.1F1F3F4F6F7givenalgebra

The function $z \mapsto π \cot (π z)$ is holomorphic on the open set $H \subseteq C ∖ Z$ , and by [F1] the partial sums of $z^{- 1} + \sum_{n \geq 1} ((z - n)^{- 1} + (z + n)^{- 1})$ converge locally uniformly on $H$ to it. Fix $w$ with $∣ w ∣ < 1$ : the telescoping identity $(1 - w) \sum_{m = 0}^{N} w^{m} = 1 - w^{N + 1}$ , the convergence $∣ w ∣^{N + 1} \to 0$ of [F7] and the metric description [F6] give $∣ (1 - w) \sum_{m \leq N} w^{m} - 1 ∣ \to 0$ ; dividing by the nonzero $∣ 1 - w ∣$ proves $\sum_{m \geq 0} w^{m} = 1 / (1 - w)$ . Now put $u = e^{i π z}$ and $Q = u^{2} = e^{2 π i z}$ by [F4]; then [F3] and [F4] give $\sin (π z) = (u - u^{- 1}) / (2 i) = (Q - 1) / (2 i u)$ and $\cos (π z) = (u + u^{- 1}) / 2 = (Q + 1) / (2 u)$ , so $π \cot (π z) = π i \frac{Q + 1}{Q - 1} = - π i (1 + 2 \sum_{r \geq 1} Q^{r})$ , because $∣ Q ∣ = e^{- 2 π Im z} < 1$ for $z \in H$ by [F4] and $\frac{1 + Q}{1 - Q} = 1 + 2 \sum_{r \geq 1} Q^{r}$ .

2.1F2F5step 1.1algebra

The series in 1.1 has holomorphic terms and locally uniformly convergent partial sums on $H$ , so [F2] applies and, for $m = k - 1$ , differentiating termwise gives $\frac{d^{k - 1}}{d z^{k - 1}} π \cot (π z) = (- 1)^{k - 1} (k - 1)! \sum_{n \in Z} (z + n)^{- k}$ on $H$ , the displayed sum being understood through the absolutely convergent paired series of [F1] together with the term $z^{- k}$ differentiated from $z^{- 1}$ ; here $\frac{d^{m}}{d z^{m}} (z + n)^{- 1} = (- 1)^{m} m! (z + n)^{- m - 1}$ by [F5]. On the other side the $Q$ -series of 1.1 consists of entire terms with locally uniformly convergent partial sums, so differentiating it $k - 1$ times termwise by [F2] and [F5] gives $- 2 π i (2 π i)^{k - 1} \sum_{r \geq 1} r^{k - 1} Q^{r}$ as functions of $z$ .

3.1F6F7step 2.1givenalgebra∎

Evaluating the two expressions of 2.1 at $z = τ$ , where $Q = e^{2 π i τ} = q$ , gives $(- 1)^{k - 1} (k - 1)! \sum_{n \in Z} (τ + n)^{- k} = - 2 π i (2 π i)^{k - 1} \sum_{r \geq 1} r^{k - 1} q^{r}$ . Dividing by the nonzero real number $(- 1)^{k - 1} (k - 1)!$ yields the stated identity, since $- 2 π i (2 π i)^{k - 1} / (- 1)^{k - 1} = (- 2 π i)^{k}$ . Finally, for $n \geq 2 ∣ τ ∣$ one has $∣ τ \pm n ∣ \geq n - ∣ τ ∣ \geq n / 2$ , so $\sum_{n \geq 1} ∣ τ \pm n ∣^{- k} \leq C_{τ} + 2^{k + 1} \sum_{n \geq 1} n^{- k} < \infty$ by [F7]; hence the family $((τ + n)^{- k})_{n \in Z}$ is absolutely summable and, by [F6], every enumeration of it converges to the same sum, which is the order-independence asserted in the Statement.

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Eisenstein series are modular forms; their Fourier coefficients

Statement

For every even $k \geq 4$ : (a) $E_{k}$ is a modular form of weight $k$ for $P S L_{2} (Z)$ ; (b) writing $B_{k}$ for the $k$ -th Bernoulli number, $E_{k} (τ) = 1 - \frac{2 k}{B_{k}} \sum_{n \geq 1} σ_{k - 1} (n) q^{n}$ , $q = e^{2 π i τ}$ , with $σ_{k - 1} (n) = \sum_{d ∣ n} d^{k - 1}$ ; (c) $E_{k} (\infty) = 1$ , and $G_{k} = 2 ζ (k) E_{k}$ ; (d) in particular $E_{4} = 1 + 240 q + 2160 q^{2} + O (q^{3})$ and $E_{6} = 1 - 504 q - 16632 q^{2} + O (q^{3})$ .

Facts & Assumptions

Given: Even $k \geq 4$ , $G_{k} (τ) = \sum_{(m, n)}^{'} (m τ + n)^{- k}$ and $E_{k} = G_{k} / (2 ζ (k))$ (The level-one Eisenstein series E_k and the weight-two series E_2, The Riemann zeta function on the half-plane $Re s > 1$ , The divisor power sums $σ_{k}$ ).

[F1]

The family $((m τ + n)^{- k})$ is absolutely summable with locally uniform convergence on $H$ , and the action is by $ℑ (γ τ) = ℑ τ / ∣ c τ + d ∣^{2}$ with $γ = (\begin{matrix} a & b \\ c & d \end{matrix})$ (Absolute convergence and holomorphy of the lattice Eisenstein sums, The modular group and its action on the upper half-plane).

[F2]

Lipschitz: $\sum_{n \in Z} (z + n)^{- k} = \frac{(- 2 π i)^{k}}{(k - 1)!} \sum_{r \geq 1} r^{k - 1} e^{2 π i r z}$ for $Im z > 0$ , absolutely convergent (The Lipschitz formula for the reciprocal-power sums).

[F3]

$M_{k}$ is defined by the weight- $k$ transformation law and holomorphy at the cusp, with $E_{k} = 1 + O (q)$ (Level-one modular forms and cusp forms, The level-one Eisenstein series E_k and the weight-two series E_2); $2 ζ (k) > 0$ and $(m, n) \mapsto (m a + n c, m b + n d)$ is a bijection of $Z^{2}$ for $γ \in S L_{2} (Z)$ (The Riemann zeta function on the half-plane $Re s > 1$ , Fubini for double series: if $\sum_{i} \sum_{j} ∣ a_{i j} ∣$ converges then both iterated sums and the sum along every bijection $N \to N \times N$ converge to one and the same value).

[F4]

The even zeta values follow locally without a choice assumption. For $∣ z ∣ < 1$ , the cotangent expansion (The Mittag-Leffler expansion of pi cotangent) gives $π z \cot (π z) = 1 - 2 \sum_{m \geq 1} ζ (2 m) z^{2 m}$ : expand $2 z^{2} / (z^{2} - n^{2})$ geometrically and interchange the sums, whose absolute total on $∣ z ∣ \leq r < 1$ is bounded by $2 r^{2} (1 - r^{2})^{- 1} \sum_{n \geq 1} n^{- 2}$ . Put $t = 2 π i z$ . The exponential definitions of sine and cosine give $π z \cot (π z) = t / (e^{t} - 1) + t / 2$ . Comparing the coefficient of $z^{2 m}$ in the Bernoulli generating series yields $ζ (2 m) = (- 1)^{m + 1} B_{2 m} (2 π)^{2 m} / (2 (2 m)!)$ (The Bernoulli numbers are defined by the generating series $t / (e^{t} - 1)$ , Complex sine, cosine, hyperbolic sine, and hyperbolic cosine from the complex exponential, $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, Fubini for double series: if $\sum_{i} \sum_{j} ∣ a_{i j} ∣$ converges then both iterated sums and the sum along every bijection $N \to N \times N$ converge to one and the same value, Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence).

Proof

1.1F1F3givenalgebra

Let $γ = (\begin{matrix} a & b \\ c & d \end{matrix}) \in S L_{2} (Z)$ . Writing $m γ τ + n = \frac{(m a + n c) τ + (m b + n d)}{c τ + d}$ and using that $(m, n) \mapsto (m a + n c, m b + n d)$ is a bijection [F3], the absolute summability [F1] permits reindexing, so $G_{k} (γ τ) = (c τ + d)^{k} G_{k} (τ)$ . Hence $E_{k} (γ τ) = (c τ + d)^{k} E_{k} (τ)$ . By [F3], $E_{k} = 1 + O (q)$ as $Im τ \to \infty$ , so in particular $E_{k}$ is bounded near the cusp, and its periodic function $F$ extends holomorphically to $q = 0$ by the q-expansion principle; therefore $E_{k} \in M_{k}$ and (a) holds.

1.2F2F3givenalgebra

Split the absolutely summable sum defining $G_{k}$ into $m = 0$ and $m \neq 0$ . The $m = 0$ part is $\sum_{n \neq 0} n^{- k} = 2 ζ (k)$ because $k$ is even. For $m \neq 0$ pair $m$ with $- m$ and $n$ with $- n$ ; by [F3] this reindexing preserves the sum, and each remaining term with $m \geq 1$ is handled by [F2] applied to $z = m τ$ (whose imaginary part is positive): $\sum_{n \in Z} (m τ + n)^{- k} = \frac{(- 2 π i)^{k}}{(k - 1)!} \sum_{r \geq 1} r^{k - 1} q^{m r}$ . Therefore $G_{k} = 2 ζ (k) + 2 \frac{(- 2 π i)^{k}}{(k - 1)!} \sum_{m \geq 1} \sum_{r \geq 1} r^{k - 1} q^{m r} = 2 ζ (k) + 2 \frac{(- 2 π i)^{k}}{(k - 1)!} \sum_{n \geq 1} σ_{k - 1} (n) q^{n}$ , the last regrouping being the absolutely summable family $(r^{k - 1} q^{m r})_{m, r \geq 1}$ grouped by $n = m r$ [F3]: for $∣ q ∣ \leq ρ < 1$ , its absolute sum is at most $(1 - ρ)^{- 1} \sum_{r \geq 1} r^{k - 1} ρ^{r} < \infty$ by the ratio test (Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence). Dividing by $2 ζ (k)$ gives $E_{k} = 1 + \frac{(- 2 π i)^{k}}{(k - 1)! ζ (k)} \sum_{n \geq 1} σ_{k - 1} (n) q^{n}$ .

2.1F4step 1.2givenalgebra∎

For even $k$ write $k = 2 m$ . Then $(- 2 π i)^{k} = 2^{k} π^{k} (- i)^{2 m} = (- 1)^{m} 2^{k} π^{k}$ , and by [F4] $ζ (2 m) = (- 1)^{m + 1} \frac{B_{k} (2 π)^{k}}{2 k!}$ , so $(k - 1)! ζ (k) = (- 1)^{m + 1} \frac{B_{k} (2 π)^{k}}{2 k}$ and the coefficient of $\sum σ_{k - 1} (n) q^{n}$ in 1.2 is $\frac{(- 1)^{m} 2^{k} π^{k} \cdot 2 k}{(- 1)^{m + 1} B_{k} (2 π)^{k}} = - \frac{2 k}{B_{k}}$ , proving (b). The constant term gives $E_{k} (\infty) = 1$ and $G_{k} = 2 ζ (k) E_{k}$ , which is (c) (see [F3]). For (d): $B_{4} = - 1 / 30$ and $B_{6} = 1 / 42$ , so $- \frac{2 \cdot 4}{B_{4}} = 240$ and $- \frac{2 \cdot 6}{B_{6}} = - 504$ ; with $σ_{3} (1) = 1$ , $σ_{3} (2) = 1 + 8 = 9$ and $σ_{5} (1) = 1$ , $σ_{5} (2) = 1 + 32 = 33$ this gives $E_{4} = 1 + 240 q + 2160 q^{2} + O (q^{3})$ and $E_{6} = 1 - 504 q - 16632 q^{2} + O (q^{3})$ , as claimed. The special values used here are the local coefficient computation in [F4].

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The boundary arc contribution in the valence computation

Statement

Let $k$ be even and let $0 \neq f \in M_{k}$ . On the unit-circle arc of the fundamental domain (away from the finitely many zeros and poles of $f$ ) one has the identity $d \log f (S τ) = d \log f (τ) + k d τ / τ$ with $S τ = - 1 / τ$ . Consequently, with the boundary orientation used in the argument-principle computation, the circular arc, traversed clockwise from $ω$ to $ω + 1$ , of the positively oriented boundary of the truncated fundamental domain contributes the net amount $π i k / 6$ to $\oint d \log f$ (hence $k / 12$ after division by $2 π i$ ), while the two vertical boundary sides cancel by $T$ -invariance. If zeros lie on the boundary, these contributions mean limits after deleting paired neighbourhoods under $S$ and $T$ ; the small indentation arcs are counted separately.

Facts & Assumptions

Given: Even $k$ , a nonzero modular form $f \in M_{k}$ (Level-one modular forms and cusp forms), and the standard domain $D$ with $S τ = - 1 / τ$ , $T τ = τ + 1$ , the arc being $∣ τ ∣ = 1$ between $ω = e^{2 π i / 3}$ and $ω + 1 = e^{i π / 3}$ through $i$ (The standard fundamental domain, boundary identifications and elliptic stabilisers).

[F1]

$f (S τ) = τ^{k} f (τ)$ for $S = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ and $f (T τ) = f (τ)$ (Level-one modular forms and cusp forms).

[F2]

Logarithmic derivative and the chain rule: on any region where a holomorphic $g$ has no zeros, $d \log g = g^{'} / g d τ$ ; for $c \neq 0$ , $\frac{d}{d τ} \log f (S τ) = \frac{f^{'} (S τ)}{f (S τ)} \cdot \frac{1}{τ^{2}}$ (The logarithmic derivative of a meromorphic function, The chain rule for complex derivatives, Linearity, product, reciprocal, and quotient rules for complex derivatives).

[F3]

For a closed contour, $\oint d \log f$ is computed by the argument principle; reversal changes the sign, concatenation adds, and $\int_{γ} d τ / τ$ over a contour from $a$ to $b$ equals the increment of a logarithm, in particular $\int_{ω}^{i} d τ / τ = \log i - \log ω$ (The integral of $d z / (z - p)$ along a contour is the increment of a continuous logarithm, Rectifiable complex contours, reversal, concatenation, closedness, and orientation, The contour integral of a constant c is c times the endpoint displacement).

Proof

1.1F1F2givenalgebra

On a neighbourhood avoiding zeros, differentiate $f (S τ) = τ^{k} f (τ)$ and divide by that equality. The chain and product rules [F2] give $\frac{f' (S τ)}{f (S τ)} τ^{- 2} = \frac{f' (τ)}{f (τ)} + k / τ$ , hence $d \log f (S τ) = d \log f (τ) + k d τ / τ$ . No global logarithm of $f$ is required.

2.1F3step 1.1givenalgebra

Let $L$ be the clockwise half-arc $ω \to i$ and $R$ the half-arc $i \to ω + 1$ . Since $S (L)$ is $R$ with reversed orientation, integrating 1.1 gives $- \int_{R} d \log f = \int_{L} d \log f + k \int_{L} d τ / τ$ . Thus $A : = \int_{L} d \log f + \int_{R} d \log f = - k \int_{L} d τ / τ = - k (i π / 2 - 2 π i / 3) = π i k / 6$ , and $A / (2 π i) = k / 12$ . When boundary zeros occur, delete matching subarcs under $S$ ; the same equality holds on the remaining half-arcs, and the omitted integral of $d τ / τ$ tends to zero. In particular it also applies when $i$ or the endpoints are zeros.

3.1F1F3givenalgebra∎

Vertical sides: the right vertical side of the truncated domain is the image under $T$ of the left vertical side, the map being orientation preserving on $H$ ; since $f (T τ) = f (τ)$ the logarithmic derivative is $T$ -invariant, and the two sides are traversed in opposite directions as parts of the boundary, so their contributions to $\oint d \log f$ cancel. Hence the net contribution of the boundary pieces lying on $\partial D$ is the arc contribution $π i k / 6$ .

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The level-one valence formula

Statement

Let $k$ be even and let $f \in M_{k}$ with $f \neq 0$ . Then $\sum_{P \in P S L_{2} (Z) \ H} \frac{1}{ν_{P}} {ord}_{P} (f) + {ord}_{\infty} (f) = \frac{k}{12},$ where $ν_{P} = 2$ if $P$ is the class of $i$ , $ν_{P} = 3$ if $P$ is the class of $ω = e^{2 π i / 3}$ , and $ν_{P} = 1$ otherwise; ${ord}_{\infty} (f)$ is the order of vanishing at $q = 0$ of the $q$ -expansion from The q-expansion principle at the cusp.

Facts & Assumptions

Given: An even integer $k$ , a nonzero $f \in M_{k}$ , its $q$ -expansion $f (τ) = q^{n} g (q)$ with $n = {ord}_{\infty} (f)$ , $g$ holomorphic on $∣ q ∣ < 1$ and $g (0) \neq 0$ (Level-one modular forms and cusp forms, The q-expansion principle at the cusp); the standard domain $D$ with its closure $\overline{D}$ , the elliptic points $i, ω, ω + 1$ and stabiliser orders $ν_{i} = 2$ , $ν_{ω} = ν_{ω + 1} = 3$ (The standard fundamental domain, boundary identifications and elliptic stabilisers).

[F1]

$f$ is holomorphic on $H$ ; its zeros are isolated, and the identity theorem forces $f \equiv̸ 0$ on every connected open subset (Zeros of a nonzero holomorphic function are isolated, Identity theorem for holomorphic functions). Orders of zeros are defined by The order of a zero of a holomorphic function and the logarithmic derivative $f^{'} / f$ has a simple pole of residue $m$ at a zero of order $m$ (The logarithmic derivative has residue equal to local order, The logarithmic derivative of a meromorphic function, Meromorphic functions on a plane domain).

[F2]

Argument principle: for a meromorphic $f$ admissible on a cycle $Γ$ enclosing a region and nonvanishing on $Γ$ , $\frac{1}{2 π i} \int_{Γ} \frac{f^{'}}{f} d z = Z (f, Γ) - P (f, Γ)$ with the weighted counts of Zero and pole counts weighted by multiplicity and winding number (The argument principle for an admissible null-homologous cycle).

[F3]

On the truncated fundamental domain the circular arc contributes $k / 12$ and the vertical sides cancel, in the sense of The boundary arc contribution in the valence computation; the truncation at height $Y$ is legitimate because $f = q^{n} g (q)$ with $g (0) \neq 0$ gives $f^{'} / f \to 2 π i n$ uniformly in $Re τ$ as $Im τ \to \infty$ .

[F4]

Representatives and angles: every class in $P S L_{2} (Z) \ H$ has a representative in $\overline{D}$ , points of $D$ have distinct classes, and the identifications of $\partial D$ are only the $T$ - and $S$ -identifications; at $i$ the domain subtends the angle $π = 2 π / ν_{i}$ , at each of $ω, ω + 1$ the angle $π / 3$ , and each of the two points is a representative of the single class of $ω$ (The standard fundamental domain, boundary identifications and elliptic stabilisers, Local charts and the Riemann surface structure of a modular quotient, The compactified level-one modular curve X(1)).

Proof

1.1F1F2givenalgebra

Since $f \equiv̸ 0$ and $f$ has isolated zeros, and since $q^{n} g (q)$ with $g (0) \neq 0$ has no zeros for small $∣ q ∣$ , there are finitely many zeros of $f$ in $\overline{D} \cap {Im τ \leq Y}$ for each $Y$ , and for large $Y$ all zero classes have a representative there. Choose such a $Y$ and $ε > 0$ small, let $R$ be the region obtained from $\overline{D} \cap {Im τ \leq Y}$ by deleting the open discs of radius $ε$ around each zero of $f$ in that set (together with the strip $Im τ > Y$ ), and let $Γ = \partial R$ with the positive orientation. Then $f$ is holomorphic on a neighbourhood of $\overline{R}$ and has no zeros on $Γ$ , so by [F2] $\frac{1}{2 π i} \oint_{Γ} \frac{f^{'}}{f} d τ = 0$ , and $Γ$ consists of the top horizontal segment, the two vertical sides, the circular arc, cut where zeros occur, and the small circles (or circular arcs) around the zeros.

2.1F2F3step 1.1givenalgebra

The top segment is traversed from right to left; on it $f^{'} / f = 2 π i n + o (1)$ uniformly in $Re τ$ as $Y \to \infty$ by [F3], so its contribution to $\frac{1}{2 π i} \oint$ tends to $- n$ . On the parts of $Γ$ lying on the vertical sides of $\partial D$ the integrand is $T$ -invariant and the two sides are oppositely oriented, so they cancel exactly; the parts lying on the circular arc contribute $k / 12$ in the limit $ε \to 0$ by [F3] (the cuts near zeros are accounted for with the small circles below).

2.2F1F4step 1.1givenalgebra

Consider a class $P \neq \infty$ with ${ord}_{P} (f) = m > 0$ and all its representatives in the truncated domain. Near a zero $τ_{0}$ of order $m$ , $f^{'} / f = m / (τ - τ_{0}) +$ holomorphic by [F1], so over a circular arc of angle $θ$ around $τ_{0}$ inside $R$ the integral equals $i m θ + o (1)$ as $ε \to 0$ , contributing $- θ m / (2 π)$ to $\frac{1}{2 π i} \oint_{Γ}$ (the boundary is traversed clockwise around the deleted disc). By [F4] the total angle of the sectors of $R$ at all representatives of $P$ is: $2 π$ if $P$ is a non-elliptic class (one interior representative, or two boundary representatives each contributing $π$ ), $π = 2 π / ν_{i}$ for the class of $i$ , and $π / 3 + π / 3 = 2 π / 3 = 2 π / ν_{ω}$ for the class of $ω$ (represented by the two points $ω$ and $ω + 1$ ). Hence each class $P \neq \infty$ contributes $- m / ν_{P} = - {ord}_{P} (f) / ν_{P}$ .

3.1F3step 2.1step 2.2givenalgebra∎

Summing 2.1 and 2.2 in the identity of 1.1 and letting $Y \to \infty$ , $ε \to 0$ gives $0 = \frac{k}{12} - n - \sum_{P \neq \infty} \frac{{ord}_{P} (f)}{ν_{P}}$ , that is $\sum_{P \neq \infty} \frac{1}{ν_{P}} {ord}_{P} (f) + {ord}_{\infty} (f) = \frac{k}{12}$ . All sums are finite by 1.1, and the term ${ord}_{\infty} (f) = n$ appears as the negative of the top-segment limit.

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The zeros of E4 and E6 at the elliptic points

Statement

$E_{4}$ has a simple zero at the class of $ω = e^{2 π i / 3}$ and no other zeros; $E_{6}$ has a simple zero at the class of $i$ and no other zeros. In particular $E_{4}$ and $E_{6}$ are nonzero modular forms with $E_{4} (\infty) = E_{6} (\infty) = 1$ , and $E_{4}$ does not vanish at $i$ while $E_{6}$ does not vanish at $ω$ .

Facts & Assumptions

Given: The Eisenstein series $E_{4} \in M_{4}$ , $E_{6} \in M_{6}$ with $E_{4} (\infty) = E_{6} (\infty) = 1$ (Eisenstein series are modular forms; their Fourier coefficients, The level-one Eisenstein series E_k and the weight-two series E_2), and the valence formula (The level-one valence formula).

[F1]

In the valence formula $\sum_{P} {ord}_{P} (f) / ν_{P} + {ord}_{\infty} (f) = k / 12$ , each summand ${ord}_{P} (f) / ν_{P}$ is either $0$ or at least $1 / ν_{P}$ ; the elliptic classes have $ν_{i} = 2$ and $ν_{ω} = ν_{ω + 1} = 3$ , all other classes have $ν_{P} = 1$ (The level-one valence formula, The standard fundamental domain, boundary identifications and elliptic stabilisers).

Proof

1.1F1givenalgebra

For $E_{4}$ , $k / 12 = 1 / 3$ and ${ord}_{\infty} (E_{4}) = 0$ because $E_{4} (\infty) = 1 \neq 0$ , so $\sum_{P} {ord}_{P} (E_{4}) / ν_{P} = 1 / 3$ . Every nonzero summand is at least $1 / 3$ by [F1], with equality exactly for a simple zero at a class with $ν_{P} = 3$ , and the doubles $1 / 2, 2 / 3, \dots$ and the integers $1, 2, \dots$ are all strictly larger than $1 / 3$ . Hence exactly one summand is nonzero, namely a simple zero at a class with $ν_{P} = 3$ , and the only such class is that of $ω$ . So $E_{4}$ vanishes simply at the class of $ω$ and nowhere else; in particular it does not vanish at $i$ .

2.1F1step 1.1givenalgebra∎

For $E_{6}$ , $k / 12 = 1 / 2$ and ${ord}_{\infty} (E_{6}) = 0$ , so $\sum_{P} {ord}_{P} (E_{6}) / ν_{P} = 1 / 2$ . A nonzero summand at the class of $ω$ would be at least $1 / 3$ , leaving a total of at most $1 / 6$ to be supplied by the remaining summands, of which every nonzero one is at least $1 / 2$ (at $i$ ) or $1$ (non-elliptic); this is impossible, so the class of $ω$ is not a zero. The total $1 / 2$ must then be a single summand at $i$ (any non-elliptic zero contributes at least $1$ ), and it equals $1 / 2$ exactly for a simple zero at $i$ . Hence $E_{6}$ vanishes simply at the class of $i$ and nowhere else, in particular not at $ω$ .

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The dimension of the space of level-one modular forms

Statement

For even $k \geq 0$ , $\dim_{C} M_{k} = {\begin{cases} ⌊ k / 12 ⌋ + 1, & k \equiv̸ 2 (m o d 12), \\ ⌊ k / 12 ⌋, & k \equiv 2 (m o d 12), \end{cases}$ and $\dim S_{k} = \dim M_{k} - 1$ for $k \geq 4$ , while $S_{0} = 0$ and $S_{k} = 0$ for $2 \leq k < 12$ . In particular $M_{0} = C$ , $M_{2} = 0$ , $M_{4}, M_{6}, M_{8}, M_{10}$ are one-dimensional, and $M_{12}$ is two-dimensional. Moreover the forms $E_{4}^{a} E_{6}^{b}$ with $4 a + 6 b = k$ , $a, b \geq 0$ , are linearly independent.

Facts & Assumptions

Given: The spaces $M_{k}, S_{k}$ of level-one modular and cusp forms (Level-one modular forms and cusp forms), the valence formula, and the Eisenstein forms $E_{4} \in M_{4}$ , $E_{6} \in M_{6}$ with constant term $1$ at the cusp and zeros only at $ω$ (for $E_{4}$ , simple) and $i$ (for $E_{6}$ , simple) (The level-one valence formula, Eisenstein series are modular forms; their Fourier coefficients, The zeros of E4 and E6 at the elliptic points).

[F1]

Valence: for even $k$ and $0 \neq f \in M_{k}$ , $\sum_{P \neq \infty} {ord}_{P} (f) / ν_{P} = k / 12 - {ord}_{\infty} (f)$ , all terms nonnegative with $ν_{P} \in {1, 2, 3}$ (The level-one valence formula).

[F2]

The constant-term functional $ε_{\infty} : M_{k} \to C$ , $f \mapsto F_{f} (0) = f (\infty)$ , is linear and nonzero for $k \geq 4$ since $ε_{\infty} (E_{k}) = 1$ ; $S_{k} = \ker ε_{\infty}$ by definition, and rank-nullity gives $\dim S_{k} = \dim M_{k} - \dim im ε_{\infty}$ (Level-one modular forms and cusp forms, Eisenstein series are modular forms; their Fourier coefficients, Kernel and image of a linear map, The kernel and image are linear subspaces, and a linear map is injective if and only if its kernel is trivial, Rank and nullity of a linear map with finite-dimensional domain, Rank-nullity: $\dim_{F} V = nullity T + rank T$ , Vector space over a field).

[F3]

At the class of $ω$ , $ord (E_{4}) = 1$ and $ord (E_{6}) = 0$ ; hence ${ord}_{ω} (E_{4}^{a} E_{6}^{b}) = a$ for all $a, b \geq 0$ (The zeros of E4 and E6 at the elliptic points, The level-one valence formula).

Proof

1.1F1givenalgebra

Upper bound. Choose $d : = ⌊ k / 12 ⌋ + 1$ distinct non-elliptic classes of points of $H$ if $k \equiv̸ 2 (m o d 12)$ , and $d : = ⌊ k / 12 ⌋$ such classes if $k \equiv 2 (m o d 12)$ ; such classes exist because non-elliptic classes are infinite. Suppose $0 \neq f \in M_{k}$ vanished at all chosen classes. Their total contribution to the valence sum is $N \geq d$ (each has $ν_{P} = 1$ ). If $k \equiv̸ 2 (m o d 12)$ then $d = ⌊ k / 12 ⌋ + 1 > k / 12 \geq k / 12 - {ord}_{\infty} (f)$ , contradicting [F1]. If $k \equiv 2 (m o d 12)$ , write $k = 12 q + 2$ , so $k / 12 = q + 1 / 6$ and $d = q$ . If ${ord}_{\infty} (f) \geq 1$ then the valence sum is at most $q + 1 / 6 - 1 = q - 5 / 6 < q \leq N$ , a contradiction. If ${ord}_{\infty} (f) = 0$ , let $m = {ord}_{i} (f)$ , $r = {ord}_{ω} (f)$ and $N \geq q$ the contribution of the remaining classes; multiplying the valence identity by $6$ gives $6 N + 3 m + 2 r = 6 q + 1$ , so $m$ is odd and $r \equiv 2 (m o d 3)$ ; hence $m \geq 1$ , $r \geq 2$ and $N + m / 2 + r / 3 \geq q + 1 / 2 + 2 / 3 = q + 7 / 6 > q + 1 / 6 = k / 12$ , again contradicting [F1]. Therefore evaluation at the $d$ chosen classes is injective on $M_{k}$ , so $\dim M_{k} \leq d$ .

2.1F3step 1.1givenalgebra

Lower bound. If $4 a + 6 b = 4 a^{'} + 6 b^{'} = k$ with $a < a^{'}$ then ${ord}_{ω} (E_{4}^{a} E_{6}^{b}) = a < a^{'} = {ord}_{ω} (E_{4}^{a^{'}} E_{6}^{b^{'}})$ by [F3], so in a linear relation $\sum_{a} c_{a} E_{4}^{a} E_{6}^{b (a)} = 0$ the term with least $a$ , if its coefficient were nonzero, would give the sum the finite order $a$ at the class of $ω$ ; since the sum is identically zero its order is infinite, so $c_{a} = 0$ for the least $a$ , and induction gives that all coefficients vanish. Hence the monomials are linearly independent, so $\dim M_{k}$ is at least their number. The pairs $(a, b)$ with $4 a + 6 b = k$ , $a, b \geq 0$ , are indexed by the integers $a \geq 0$ with $a \equiv k (m o d 3)$ and $a \leq k / 4$ (then $b = (k - 4 a) / 6 \geq 0$ is a nonnegative integer). Writing $k = 12 q + s$ with $s \in {0, 2, 4, 6, 8, 10}$ , put $a_{0} \in {0, 1, 2}$ for the least nonnegative residue of $s$ modulo $3$ . The solutions are $a = a_{0} + 3 j$ for $0 \leq j \leq ⌊ (k / 4 - a_{0}) / 3 ⌋$ , so their number is $\max (0, ⌊ (k / 4 - a_{0}) / 3 ⌋ + 1) = q + 1$ for $s = 0, 4, 6, 8, 10$ and $q$ for $s = 2$ ; this is exactly $⌊ k / 12 ⌋ + 1$ for $k \equiv̸ 2 (m o d 12)$ and $⌊ k / 12 ⌋$ for $k \equiv 2 (m o d 12)$ . With 1.1 this proves the dimension formula.

3.1F2step 1.1step 2.1givenalgebra∎

Cusp forms and examples. For $k \geq 4$ , $ε_{\infty}$ is nonzero and surjective onto $C$ , so by [F2] $\dim S_{k} = \dim M_{k} - 1$ ; for $0 \leq k < 12$ the formula gives $\dim M_{k} = 1$ for $k = 0, 4, 6, 8, 10$ and $\dim M_{2} = 0$ , so $S_{k} = 0$ there (for $k = 0, 4, 6, 8, 10$ because the kernel of a nonzero functional on a one-dimensional space is zero, for $k = 2$ because $M_{2} = 0$ , and $S_{0} \subseteq M_{0}$ ). In particular $M_{0} = C$ (the constants lie in $M_{0}$ and it is one-dimensional), $M_{2} = 0$ , $M_{4}, M_{6}, M_{8}, M_{10}$ are one-dimensional, and $M_{12}$ is two-dimensional. All the listed monomial counts are covered by 2.1, which also gives the asserted linear independence.

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The transformation law of the weight-two Eisenstein series E_2

Statement

The weight-two Eisenstein series $E_{2} (τ) = 1 - 24 \sum_{n \geq 1} σ_{1} (n) q^{n}$ satisfies $E_{2} (τ + 1) = E_{2} (τ)$ and, for every $γ = (\begin{matrix} a & b \\ c & d \end{matrix}) \in S L_{2} (Z)$ , $E_{2} ⁣ (\frac{a τ + b}{c τ + d}) = (c τ + d)^{2} E_{2} (τ) - \frac{6 i c}{π} (c τ + d) .$ In particular $E_{2} (- 1 / τ) = τ^{2} E_{2} (τ) - \frac{6 i}{π} τ$ , so $E_{2}$ is not a modular form of weight $2$ .

Facts & Assumptions

Given: The series $E_{2} = 1 - 24 \sum_{n \geq 1} σ_{1} (n) q^{n}$ , $q = e^{2 π i τ}$ , and its half-normalisation $H : = (π^{2} / 6) E_{2}$ , all on $H$ (The level-one Eisenstein series E_k and the weight-two series E_2, The unit disc, the upper half-plane, and Blaschke factors).

[F1]

$π \cot (π z) = \frac{1}{z} + \sum_{n \geq 1} \frac{2 z}{z^{2} - n^{2}}$ with local uniform convergence on $C ∖ Z$ (The Mittag-Leffler expansion of pi cotangent), and $\sin w = w - \frac{w^{3}}{6} + O (w^{5})$ , $\cos w = 1 - \frac{w^{2}}{2} + O (w^{4})$ (The exponential definitions of complex sine, cosine, hyperbolic sine, and hyperbolic cosine equal their entire power series).

[F2]

Lipschitz: for $τ \in H$ , $\sum_{n \in Z} (τ + n)^{- 2} = (- 2 π i)^{2} \sum_{r \geq 1} r q^{r} = - 4 π^{2} \sum_{r \geq 1} r q^{r}$ , absolutely on the left (The Lipschitz formula for the reciprocal-power sums).

[F3]

On every compact $K \subseteq H$ there is $c_{K} > 0$ with $∣ m z + n ∣ \geq c_{K} \max (∣ m ∣, ∣ n ∣)$ for all $(m, n) \neq (0, 0)$ and $z \in K$ This estimate is derived locally: if $ℑ z \geq y_{0} > 0$ , $∣ ℜ z ∣ \leq X$ , then $∣ m ∣ \leq ∣ m z + n ∣ / y_{0}$ and $∣ n ∣ \leq ∣ m z + n ∣ + X ∣ m ∣ \leq (1 + X / y_{0}) ∣ m z + n ∣$ , so one may take $c_{K} = \min (y_{0}, (1 + X / y_{0})^{- 1})$ .

[F4]

The $M$ -test gives uniform convergence of a series dominated by a summable real majorant, which may depend on a parameter (Weierstrass M-test for complex-valued function series); comparison with a convergent $p$ -series and bounded monotone partial sums give convergence of positive series (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ , For rational $p > 0$ , $\sum 1 / k^{p}$ converges iff $p > 1$ , A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum); an absolutely summable double family may be regrouped and reindexed (Fubini for double series: if $\sum_{i} \sum_{j} ∣ a_{i j} ∣$ converges then both iterated sums and the sum along every bijection $N \to N \times N$ converge to one and the same value, If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges).

[F5]

If $G$ is differentiable on $[n, n + 1]$ with integrable $G^{'} = f$ , then $\int_{n}^{n + 1} f = G (n + 1) - G (n)$ ; consequently $∣ f (n) - \int_{n}^{n + 1} f ∣ \leq 2 \sup_{[n, n + 1]} ∣ f^{'} ∣$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ ).

[F6]

If $φ (t, ε)$ and $\partial_{ε} φ (t, ε)$ are continuous on $R \times I$ , one slice is absolutely integrable, and $\partial_{ε} φ$ is dominated on each compact interval of $ε$ by a $t$ -integrable function, then $Φ (ε) = \int_{R} φ (t, ε) d t$ is differentiable with $Φ^{'} (ε) = \int_{R} \partial_{ε} φ (t, ε) d t$ (Differentiation under an improper multiple integral under an integrable derivative bound).

Proof

1.1F1F4givenalgebra

We first record $ζ (2) = π^{2} / 6$ . By [F1], $π \cot (π z) = \frac{1}{z} - 2 z \sum_{n \geq 1} \frac{1}{n^{2} - z^{2}}$ for $z \notin Z$ ; for $∣ z ∣ \leq 1 / 2$ one has $\frac{1}{n^{2} - z^{2}} = \frac{1}{n^{2}} + \frac{z^{2}}{n^{2} (n^{2} - z^{2})}$ with $\frac{∣ z ∣^{2}}{n^{2} ∣ n^{2} - z^{2} ∣} \leq \frac{2 ∣ z ∣^{2}}{n^{4}}$ , so $\sum_{n \geq 1} \frac{2 z}{z^{2} - n^{2}} = - 2 z \sum_{n \geq 1} \frac{1}{n^{2}} + O (z^{3})$ by [F4] and $π \cot (π z) = \frac{1}{z} - 2 z \sum_{n \geq 1} n^{- 2} + O (z^{3})$ . On the other hand [F1] gives $π \cot (π z) = π \frac{\cos π z}{\sin π z} = π \frac{1 - (π z)^{2} / 2 + O (z^{4})}{π z - (π z)^{3} / 6 + O (z^{5})} = \frac{1}{z} - \frac{π^{2}}{3} z + O (z^{3})$ . Comparing coefficients of $z$ and using $\sum n^{- 2} > 0$ gives $\sum_{n \geq 1} n^{- 2} = π^{2} / 6$ .

2.1F2F3F4step 1.1givenalgebra

With $q = e^{2 π i τ}$ and $∣ q ∣ < 1$ , [F2] gives $\sum_{n \in Z} (m τ + n)^{- 2} = - 4 π^{2} \sum_{r \geq 1} r q^{m r}$ for each $m \geq 1$ , so $\sum_{m \geq 1} \sum_{n \in Z} (m τ + n)^{- 2} = - 4 π^{2} \sum_{m, r \geq 1} r q^{m r} = - 4 π^{2} \sum_{n \geq 1} σ_{1} (n) q^{n}$ , the last step regrouping the absolutely summable family $(r q^{m r})_{m, r \geq 1}$ by $n = m r$ [F4]. Hence, by 1.1, $H (τ) = ζ (2) + \sum_{m \geq 1} \sum_{n \in Z} (m τ + n)^{- 2} .$ For $ε > 0$ define the regularised series $H_{ε} (z) : = \frac{1}{2} \sum_{(m, n) \neq (0, 0)}^{'} (m z + n)^{- 2} ∣ m z + n ∣^{- 2 ε}$ ; pairing $(m, n)$ with $(- m, - n)$ and writing $f_{m, ε} (t) : = (m z + t)^{- 2} ∣ m z + t ∣^{- 2 ε}$ , we have $H_{ε} (z) = ζ (2 + 2 ε) + \sum_{m \geq 1} \sum_{n \in Z} f_{m, ε} (n)$ , and this family is absolutely summable for $ε > 0$ : on a compact $K$ with $c = c_{K}$ , the shell of pairs with $\max (∣ m ∣, ∣ n ∣) = j$ has $8 j$ elements each bounded by $(c j)^{- 2 - 2 ε}$ , so the shell sum is $8 c^{- 2 - 2 ε} j^{- 1 - 2 ε}$ , summable: compare with $j^{- 1 - r}$ for a rational $0 < r < 2 ε$ and apply [F4].

3.1F3F4step 2.1givenalgebra

The regularised series satisfies $H_{ε} (γ τ) = (c τ + d)^{2} ∣ c τ + d ∣^{2 ε} H_{ε} (τ)$ for every $γ = (\begin{matrix} a & b \\ c & d \end{matrix}) \in S L_{2} (Z)$ . Indeed $m γ τ + n = \frac{m^{'} τ + n^{'}}{c τ + d}$ with $(m^{'}, n^{'}) = (m a + n c, m b + n d)$ , and $(m, n) \mapsto (m^{'}, n^{'})$ is a bijection of $Z^{2}$ because its matrix $(\begin{matrix} a & c \\ b & d \end{matrix})$ has determinant $a d - b c = 1$ ; since $(m γ τ + n)^{- 2} ∣ m γ τ + n ∣^{- 2 ε} = (c τ + d)^{2} ∣ c τ + d ∣^{2 ε} (m^{'} τ + n^{'})^{- 2} ∣ m^{'} τ + n^{'} ∣^{- 2 ε}$ , the absolutely summable family of 2.1 may be reindexed and regrouped by [F4], giving the displayed law.

4.1F3F4F5step 3.1givenalgebra

Fix $τ \in H$ , $y : = Im τ > 0$ , and $m \geq 1$ . Set $I_{ε} (m τ) : = \int_{R} f_{m, ε} (t) d t$ and $A_{m} (ε) : = \sum_{n \in Z} [f_{m, ε} (n) - \int_{n}^{n + 1} f_{m, ε} (t) d t]$ . The improper integral $I_{ε} (m τ)$ is the sum over $n \in Z$ of $\int_{n}^{n + 1} f_{m, ε}$ , so $\sum_{n} f_{m, ε} (n) = I_{ε} (m τ) + A_{m} (ε)$ , and summing over $m$ gives the exact identity $H_{ε} (τ) = ζ (2 + 2 ε) + \sum_{m \geq 1} A_{m} (ε) + \sum_{m \geq 1} I_{ε} (m τ)$ for $ε > 0$ . The substitution $m τ + t = m y (u + i)$ shows $I_{ε} (m τ) = (m y)^{- 1 - 2 ε} I (ε)$ with $I (ε) : = \int_{R} (u + i)^{- 2} (1 + u^{2})^{- ε} d u$ . Finally $\sum_{m \geq 1} A_{m} (ε)$ converges uniformly for $ε \in [- 1 / 4, 1 / 4]$ : by [F5], $∣ f_{m, ε} (n) - \int_{n}^{n + 1} f_{m, ε} ∣ \leq 2 \sup_{[n, n + 1]} ∣ f_{m, ε}^{'} ∣$ , and differentiating the product gives $∣ f_{m, ε}^{'} (t) ∣ \leq (2 + 2 ∣ ε ∣) ∣ m τ + t ∣^{- 3 - 2 ε}$ ; on $[n, n + 1]$ the quantity $∣ m τ + t ∣$ is comparable to $∣ m τ + n ∣$ (their difference has modulus at most $1$ , and $∣ m τ + t ∣ \geq Im m τ = m y$ ; for $∣ m τ + n ∣ > 2$ this gives comparison, and only finitely many $n$ fail it for a given $m$ , none at all once $m y > 2$ ), so the absolute sum of all row differences is dominated uniformly in $ε$ by a constant times $\sum_{j \geq 1} 8 j j^{- 5 / 2} < \infty$ using [F3] and [F4]. The double series is thus uniformly convergent by the M-test; each row difference is continuous in $ε$ , and regrouping into $A_{m}$ preserves the limit.

5.1F4F5F6step 2.1step 4.1givenalgebra

We let $ε \to 0^{+}$ . First, $ζ (2 + 2 ε) \to ζ (2)$ : for $0 < ε \leq 1 / 4$ and every $N$ , $∣ \sum_{n} n^{- 2 - 2 ε} - \sum_{n} n^{- 2} ∣ \leq \sum_{n \leq N} ∣ n^{- 2 - 2 ε} - n^{- 2} ∣ + 2 \sum_{n > N} n^{- 2}$ , and the inner estimate is as small as desired for $N$ large and then $ε$ small. Second, $\sum_{m \geq 1} A_{m} (ε) \to \sum_{m \geq 1} A_{m} (0)$ because the convergence is uniform on $[- 1 / 4, 1 / 4]$ by 4.1 and each $A_{m}$ is continuous in $ε$ ; and $A_{m} (0) = \sum_{n \in Z} (m τ + n)^{- 2}$ because $I_{0} (m τ) = \int_{R} (m τ + t)^{- 2} d t = 0$ , the primitive being $- (m τ + t)^{- 1}$ . Third, $I (0) = \int_{R} (u + i)^{- 2} d u = 0$ (same primitive) and $I^{'} (0) = - \int_{R} (u + i)^{- 2} \log (1 + u^{2}) d u = - π$ : the differentiation is licensed by [F6] separately on real and imaginary parts, with base slice $ε = 0$ absolutely integrable (its modulus is $(1 + u^{2})^{- 1}$ ), since for $∣ ε ∣ \leq 1 / 4$ the $ε$ -difference quotients of $(u + i)^{- 2} (1 + u^{2})^{- ε}$ are bounded by $(1 + u^{2})^{- 3 / 4} \log (1 + u^{2})$ , which is integrable on $R$ , and a primitive of $- \log (1 + t^{2}) / (t + i)^{2}$ is $\frac{1 + \log (1 + t^{2})}{t + i} - \arctan t$ , whose endpoint difference is $- π$ . Fourth, by [F5] and [F4] the decreasing function $t \mapsto t^{- 1 - 2 ε}$ satisfies $\int_{1}^{\infty} t^{- 1 - 2 ε} d t \leq \sum_{m \geq 1} m^{- 1 - 2 ε} \leq 1 + \int_{1}^{\infty} t^{- 1 - 2 ε} d t$ , that is $\sum_{m \geq 1} m^{- 1 - 2 ε} = \frac{1}{2 ε} + O (1)$ . Hence $\sum_{m \geq 1} I_{ε} (m τ) = y^{- 1 - 2 ε} I (ε) \sum_{m \geq 1} m^{- 1 - 2 ε} \to \frac{1}{y} \cdot (- \frac{π}{2}) = - \frac{π}{2 y}$ as $ε \to 0^{+}$ , and therefore $H_{ε} (τ) \to H (τ) - \frac{π}{2 y}$ , using 2.1, 4.1 and $2 ε \to 0$ .

6.1step 3.1step 5.1givenalgebra∎

Taking $ε \to 0^{+}$ in 3.1 and using 5.1 with $∣ c τ + d ∣^{2 ε} \to 1$ , $H (γ τ) - \frac{π}{2 Im (γ τ)} = (c τ + d)^{2} (H (τ) - \frac{π}{2 y}) .$ Since $Im (γ τ) = y / ∣ c τ + d ∣^{2}$ , this rearranges to $H (γ τ) = (c τ + d)^{2} H (τ) + \frac{π}{2 y} (∣ c τ + d ∣^{2} - (c τ + d)^{2})$ . Writing $w : = c τ + d$ we have $w^{2} - ∣ w ∣^{2} = w (w - \overline{w}) = 2 i c y w$ , so the correction is $- π i c (c τ + d)$ and $H (γ τ) = (c τ + d)^{2} H (τ) - π i c (c τ + d)$ . Multiplying by $6 / π^{2}$ proves the stated transformation law; $E_{2} (τ + 1) = E_{2} (τ)$ is immediate from $E_{2} (τ + 1) = 1 - 24 \sum σ_{1} (n) q^{n} = E_{2} (τ)$ ; and at $γ = S$ , where $c = 1$ and $d = 0$ , the law reads $E_{2} (- 1 / τ) = τ^{2} E_{2} (τ) - \frac{6 i}{π} τ$ , which differs from $τ^{2} E_{2} (τ)$ for $τ \neq 0$ , so $E_{2}$ is not a modular form of weight $2$ .

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The discriminant is a nonvanishing cusp form of weight 12

Statement

Define $Δ : = (E_{4}^{3} - E_{6}^{2}) / 1728$ . Then $Δ$ is a cusp form of weight $12$ whose $q$ -expansion is $Δ (τ) = q - 24 q^{2} + O (q^{3}), q = e^{2 π i τ};$ in particular $Δ \neq 0$ and ${ord}_{\infty} (Δ) = 1$ , and the valence formula gives $Δ (τ) \neq 0$ for every $τ \in H$ . Moreover $E_{4}^{3}$ and $E_{6}^{2}$ are linearly independent in $M_{12}$ , this space being two-dimensional with basis ${E_{4}^{3}, E_{6}^{2}}$ .

Facts & Assumptions

Given: $E_{4} \in M_{4}$ and $E_{6} \in M_{6}$ with $E_{4} = 1 + 240 q + 2160 q^{2} + O (q^{3})$ and $E_{6} = 1 - 504 q - 16632 q^{2} + O (q^{3})$ (Eisenstein series are modular forms; their Fourier coefficients, The level-one Eisenstein series E_k and the weight-two series E_2, The divisor power sums $σ_{k}$ ).

[F1]

$M_{12}$ is a two-dimensional complex vector space. Multiplication adds weights, with $M_{4} M_{4} \subseteq M_{8}$ , $M_{8} M_{4} \subseteq M_{12}$ and $M_{6} M_{6} \subseteq M_{12}$ ; the cusp forms are the kernel of the constant-term functional $f \mapsto f (\infty)$ (The dimension of the space of level-one modular forms, Level-one modular forms and cusp forms).

[F2]

Valence formula: for even $k$ and $0 \neq f \in M_{k}$ , $\sum_{P \neq \infty} {ord}_{P} (f) / ν_{P} = k / 12 - {ord}_{\infty} (f)$ (The level-one valence formula).

Proof

1.1F1givenalgebra

$E_{4}^{3} \in M_{12}$ and $E_{6}^{2} \in M_{12}$ by [F1], hence $Δ = (E_{4}^{3} - E_{6}^{2}) / 1728 \in M_{12}$ . From the displayed expansions, $E_{4}^{3} = 1 + 720 q + (3 \cdot 2160 + 3 \cdot 240^{2}) q^{2} + O (q^{3}) = 1 + 720 q + 179280 q^{2} + O (q^{3})$ and $E_{6}^{2} = 1 - 1008 q + (504^{2} - 2 \cdot 16632) q^{2} + O (q^{3}) = 1 - 1008 q + 220752 q^{2} + O (q^{3})$ . Therefore $Δ = \frac{1}{1728} (1728 q - 41472 q^{2} + O (q^{3})) = q - 24 q^{2} + O (q^{3})$ . In particular $Δ \neq 0$ , $Δ (\infty) = 0$ , so $Δ$ is a cusp form, and ${ord}_{\infty} (Δ) = 1$ .

2.1F1F2step 1.1givenalgebra∎

Applying the valence formula [F2] to $Δ$ of weight $12$ gives $\sum_{P \neq \infty} {ord}_{P} (Δ) / ν_{P} = 12 / 12 - 1 = 0$ ; every summand is a nonnegative multiple of $1 / ν_{P}$ , so all vanish and $Δ$ has no zeros on $H$ . Finally, if $a E_{4}^{3} + b E_{6}^{2} = 0$ then comparing constant terms gives $a + b = 0$ and comparing $q$ -coefficients gives $720 a - 1008 b = 0$ , whence $(720 + 1008) a = 0$ and $a = b = 0$ ; so $E_{4}^{3}, E_{6}^{2}$ are linearly independent in the two-dimensional space $M_{12}$ and therefore form a basis.

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The modular discriminant and the j-invariant

Definition

The modular discriminant is

$Δ (τ) : = \frac{E_{4} (τ)^{3} - E_{6} (τ)^{2}}{1728},$

a cusp form of weight $12$ with $q$ -expansion $q - 24 q^{2} + O (q^{3})$ and without zeros on $H$ (The discriminant is a nonvanishing cusp form of weight 12, Eisenstein series are modular forms; their Fourier coefficients). The $j$ -invariant is the quotient

$j (τ) : = \frac{E_{4} (τ)^{3}}{Δ (τ)}, τ \in H,$

which is holomorphic on $H$ because $Δ$ does not vanish there (Meromorphic functions on a plane domain). Since $E_{4}^{3}$ and $Δ$ both transform with the factor $(c τ + d)^{12}$ under every $γ \in S L_{2} (Z)$ , the quotient satisfies $j (γ \cdot τ) = j (τ)$ (Level-one modular forms and cusp forms).

The $q$ -expansion and the cusp. From $E_{4}^{3} = 1 + 720 q + O (q^{2})$ and $Δ = q (1 - 24 q + O (q^{2}))$ (Eisenstein series are modular forms; their Fourier coefficients, The discriminant is a nonvanishing cusp form of weight 12),

$j (τ) = q^{- 1} \frac{1 + 720 q + O (q^{2})}{1 - 24 q + O (q^{2})} = q^{- 1} + 744 + O (q),$

so $j$ has a simple pole at the cusp: its associated function of $q$ is meromorphic at $q = 0$ with a pole of order one. The special values follow from the zeros of $E_{4}, E_{6}$ (The zeros of E4 and E6 at the elliptic points): $E_{6} (i) = 0$ and $E_{4} (i) \neq 0$ give $j (i) = \frac{E_{4} (i)^{3}}{E_{4} (i)^{3} / 1728} = 1728$ , while $E_{4} (ω) = 0$ gives $j (ω) = 0$ ; the valence formula (The level-one valence formula) is what makes these the only relevant zeros.

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The graded ring of level-one modular forms

Statement

The graded $C$ -algebra $M_{*} = ⨁_{k e v e n} M_{k}$ is freely generated by $E_{4}$ and $E_{6}$ : the substitution $X \mapsto E_{4}$ , $Y \mapsto E_{6}$ defines an isomorphism of graded $C$ -algebras $C [X, Y] \to M_{*}$ , where $\deg X = 4$ and $\deg Y = 6$ . Equivalently, for every even $k \geq 0$ the forms $E_{4}^{a} E_{6}^{b}$ with $4 a + 6 b = k$ , $a, b \geq 0$ , form a basis of $M_{k}$ , and the cusp forms form the principal ideal $S_{*} = Δ M_{*}$ ; explicitly $S_{k} = Δ M_{k - 12}$ for all $k \geq 0$ (with $M_{k - 12} = 0$ for $k < 12$ ).

Facts & Assumptions

Given: The graded spaces $M_{k}, S_{k}$ and the Eisenstein forms $E_{4}, E_{6}$ with $E_{4} (\infty) = E_{6} (\infty) = 1$ , together with $Δ = (E_{4}^{3} - E_{6}^{2}) / 1728 \in S_{12}$ of order one at the cusp and without zeros on $H$ (Level-one modular forms and cusp forms, Eisenstein series are modular forms; their Fourier coefficients, The discriminant is a nonvanishing cusp form of weight 12, The modular discriminant and the j-invariant).

[F1]

For even $k$ the monomials $E_{4}^{a} E_{6}^{b}$ with $4 a + 6 b = k$ are linearly independent in $M_{k}$ , and their number equals $\dim M_{k}$ from the dimension formula; every such monomial has value $1$ at the cusp (The dimension of the space of level-one modular forms, Eisenstein series are modular forms; their Fourier coefficients).

[F2]

A nonzero weight- $k$ form has a well-defined constant term $f (\infty)$ at the cusp, and $S_{k}$ is exactly the kernel of $f \mapsto f (\infty)$ (Level-one modular forms and cusp forms).

[F3]

$Δ$ has vanishing order $1$ at the cusp and no zeros on $H$ ; therefore $f \mapsto f / Δ$ maps $S_{k}$ bijectively onto $M_{k - 12}$ for every even $k$ : it is holomorphic on $H$ , has the correct weight, and at the cusp the order drops by exactly $1$ ; conversely $Δ h \in S_{k}$ for $h \in M_{k - 12}$ (The discriminant is a nonvanishing cusp form of weight 12, The modular discriminant and the j-invariant, The level-one valence formula, The zeros of E4 and E6 at the elliptic points).

[F4]

For every even $k \geq 4$ there is a monomial of weight $k$ (and for $k = 0$ the empty monomial $1$ ); $M_{2} = 0$ and $M_{k}$ is one-dimensional for $k = 0, 4, 6, 8, 10$ (The dimension of the space of level-one modular forms).

Proof

technique · induction on the weight $k$ in steps of $12$

1.1baseF1F2F4givenalgebra

Base cases: for $k = 0$ , $M_{0} = C$ consists of constants and the empty monomial $1$ has weight $0$ ; for $k = 2$ , $M_{2} = 0$ ; for $k = 4, 6, 8, 10$ the space $M_{k}$ is one-dimensional and contains the monomial $E_{4}, E_{6}, E_{4}^{2}, E_{4} E_{6}$ respectively, whose value at the cusp is $1$ , so every $f \in M_{k}$ equals $f (\infty)$ times that monomial. Hence in these degrees the monomials span $M_{k}$ .

1.2ihgiven

Assume, as induction hypothesis, that for some even $k \geq 12$ the weight- $(k - 12)$ monomials span $M_{k - 12}$ ; equivalently every element of $M_{k - 12}$ is a polynomial in $E_{4}, E_{6}$ .

2.1F2F3F4step 1.1step 1.2givenalgebra

Let $f \in M_{k}$ and choose a monomial $P = E_{4}^{a} E_{6}^{b}$ of weight $k$ [F4]; then $P (\infty) = 1$ and $f - f (\infty) P \in M_{k}$ has zero constant term at the cusp, so $f - f (\infty) P \in S_{k}$ [F2]. By [F3] there is $h \in M_{k - 12}$ with $f - f (\infty) P = Δ h$ ; by the induction hypothesis 1.2 the form $h$ is a polynomial in $E_{4}, E_{6}$ , so $f = f (\infty) P + Δ h$ is a polynomial in $E_{4}, E_{6}$ (using $Δ = \frac{1}{1728} (E_{4}^{3} - E_{6}^{2})$ ). Hence the monomials span $M_{k}$ for every even $k \geq 0$ .

3.1F1F3step 2.1discharge-induction∎

The graded algebra map $φ : C [X, Y] \to M_{*}$ with $φ (X) = E_{4}$ , $φ (Y) = E_{6}$ is well defined because $E_{4}^{a} E_{6}^{b} \in M_{4 a + 6 b}$ and multiplication of modular forms adds weights. In degree $k$ the source is spanned by the monomials of weight $k$ , whose images are linearly independent and as numerous as $\dim M_{k}$ [F1]; since they also span $M_{k}$ by the induction result, $φ$ is an isomorphism in each degree and hence an isomorphism of graded algebras. Therefore the monomials form a basis and $E_{4}, E_{6}$ are algebraically independent. Finally [F3] gives $S_{k} = Δ M_{k - 12}$ for every $k$ , that is $S_{*} = Δ M_{*}$ .

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The j-invariant classifies complex tori

Statement

For a full lattice $Λ \subset C$ with oriented basis $(ω_{1}, ω_{2})$ put $τ = ω_{2} / ω_{1} \in H$ and $j (Λ) : = j (τ)$ , with $j$ the modular function of The modular discriminant and the j-invariant; this is well defined because another oriented basis changes $τ$ by an element of $S L_{2} (Z)$ and $j$ is invariant. For full lattices $Λ, Λ^{'}$ the following are equivalent: (i) $Λ^{'} = c Λ$ for some $c \in C^{\times}$ (the lattices are homothetic); (ii) the complex tori $C / Λ$ and $C / Λ^{'}$ are biholomorphic; (iii) $j (Λ) = j (Λ^{'})$ .

Facts & Assumptions

Given: Full lattices $Λ, Λ^{'}$ with oriented bases, the complex tori $T_{Λ}, T_{Λ^{'}}$ and their class maps $π, π^{'}$ (Complex lattice and quotient torus, The quotient $C / Λ$ is a compact Riemann surface), and the modular function $j$ with $j (γ τ) = j (τ)$ for $γ \in S L_{2} (Z)$ (The modular discriminant and the j-invariant, The modular group and its action on the upper half-plane).

[F1]

A change of oriented basis is an element of $S L_{2} (Z)$ acting on $τ = ω_{2} / ω_{1}$ by the associated Möbius transformation; hence $j (Λ)$ is well defined (Complex lattice and quotient torus, The modular discriminant and the j-invariant).

[F2]

For $c \in C^{\times}$ , multiplication by $c$ maps $Λ$ onto $c Λ$ and induces a biholomorphism $T_{Λ} \to T_{c Λ}$ ; the oriented basis $(c ω_{1}, c ω_{2})$ has ratio $τ = ω_{2} / ω_{1}$ unchanged, so $j$ is a homothety invariant (Complex lattice and quotient torus, Biholomorphic maps between complex domains).

[F3]

For a lattice $Λ$ , $π : C \to T_{Λ}$ is a holomorphic covering map and $C$ is simply connected; hence for any continuous $G : C \to T_{Λ}$ and basepoint there is a unique based lift $f : C \to C$ (The quotient $C / Λ$ is a compact Riemann surface, Covering maps, evenly covered neighbourhoods, fibres, sheets, and trivial coverings, Every nonempty convex subset of $R^{n}$ is simply connected, Lifting criterion for maps from path-connected locally path-connected spaces).

[F4]

Every biholomorphic self-map of $C$ is affine $f (z) = a z + b$ , $a \neq 0$ (Every biholomorphic self-map of the complex plane is affine); an injective holomorphic map has nowhere-vanishing derivative (An injective holomorphic map has no critical point and is biholomorphic onto its image, Local normal form of a nonconstant holomorphic map).

[F5]

For $f = E_{4}^{3} - μ Δ$ with finite $μ$ , $f (\infty) = 1$ and the valence sum is $1$ . Moreover $f = Δ (j - μ)$ and $Δ$ is nonzero on $H$ . The invariant holomorphic function $j - μ$ factors in each elliptic chart through $z^{ν}$ , where $ν = 2$ or $3$ (Local charts and the Riemann surface structure of a modular quotient); consequently each zero of $f$ has order a positive multiple of $ν$ at such a point, and weighted contribution at least $1$ . At ordinary points its order is at least $1$ . Thus $f$ has exactly one zero class (The level-one valence formula, The discriminant is a nonvanishing cusp form of weight 12, The modular discriminant and the j-invariant).

[F6]

$Λ_{γ τ} = (c τ + d)^{- 1} Λ_{τ}$ for $γ = (\begin{matrix} a & b \\ c & d \end{matrix}) \in S L_{2} (Z)$ , directly from $Z + Z γ τ = Z + Z \frac{a τ + b}{c τ + d} = (c τ + d)^{- 1} (Z (c τ + d) + Z (a τ + b))$ (Complex lattice and quotient torus, The modular group and its action on the upper half-plane); and $\dim M_{12} = 2$ with $E_{4}^{3}, Δ$ a basis (The graded ring of level-one modular forms).

Proof

1.1F1F2givenalgebra

If $Λ^{'} = c Λ$ then multiplication by $c$ is a biholomorphism $T_{Λ} \to T_{Λ^{'}}$ by [F2], so (i) implies (ii). Writing $Λ = Z ω_{1} + Z ω_{2} = ω_{1} Λ_{τ}$ with $τ = ω_{2} / ω_{1}$ , the lattice $Λ^{'}$ has the same normalised parameter $τ$ , so $j (Λ^{'}) = j (τ) = j (Λ)$ ; hence (i) implies (iii). If $Λ^{''}$ is another basis of $Λ$ , it is related to $(ω_{1}, ω_{2})$ by a matrix in $S L_{2} (Z)$ and the new parameter is $γ τ$ , so $j$ is unchanged by [F1].

2.1F2F3F4step 1.1givenalgebra

Suppose (ii): let $F : T_{Λ} \to T_{Λ^{'}}$ be a biholomorphism. Composing with a translation of $T_{Λ^{'}}$ if necessary we may assume $F ([0]) = [0]$ , since translations are biholomorphisms. Put $G : = F \circ π : C \to T_{Λ^{'}}$ ; as $C$ is simply connected and $π^{'}$ is a holomorphic covering [F3], there is a unique based lift $f : C \to C$ with $π^{'} \circ f = G$ and $f (0) = 0$ . Similarly the inverse $F^{- 1}$ admits a based lift $\tilde{f}$ with $\tilde{f} (0) = 0$ . Both $\tilde{f} \circ f$ and the identity are based lifts of $F^{- 1} \circ F \circ π = π = π \circ id$ , so by uniqueness $\tilde{f} \circ f = id$ , and symmetrically $f \circ \tilde{f} = id$ ; thus $f$ is a biholomorphism of $C$ (it is holomorphic because locally it is a branch of the holomorphic covering $π^{'}$ composed with $G$ ). By [F4] $f (z) = a z + b$ with $a \neq 0$ , and $f (0) = 0$ gives $b = 0$ , so $f (z) = a z$ . For every $λ \in Λ$ , $π^{'} (f (z + λ)) = G (z + λ) = G (z)$ because $π$ is $Λ$ -periodic, so $f (z + λ) - f (z) = a λ \in \ker π^{'} = Λ^{'}$ ; hence $a Λ \subseteq Λ^{'}$ . Applying the same argument to $\tilde{f}$ , whose affine form is $a^{- 1} z$ , gives $a^{- 1} Λ^{'} \subseteq Λ$ , so $a Λ = Λ^{'}$ and $Λ^{'}$ is homothetic to $Λ$ : (ii) implies (i).

3.1F5F6step 2.1givenalgebra∎

Suppose (iii): $j (τ) = j (τ^{'})$ where $Λ = ω_{1} Λ_{τ}$ , $Λ^{'} = ω_{1}^{'} Λ_{τ^{'}}$ ; set $μ = j (τ) = j (τ^{'})$ . The form $f : = E_{4}^{3} - μ Δ \in M_{12}$ has $f (τ) = 0 = f (τ^{'})$ and $f (\infty) = 1$ , so $f \neq 0$ and by [F5] all zeros of $f$ , in particular $τ$ and $τ^{'}$ , lie in one $P S L_{2} (Z)$ -class: there is $γ \in S L_{2} (Z)$ with $τ^{'} = γ τ$ . Then $Λ_{τ^{'}} = Λ_{γ τ} = (c τ + d)^{- 1} Λ_{τ}$ by [F6] is homothetic to $Λ_{τ}$ , hence $Λ^{'}$ is homothetic to $Λ$ and (i) holds. The three implications close the equivalence.

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The j-invariant uniformizes X(1)

Statement

The function $j$ descends to a holomorphic map $\overset{ˉ}{j} : X (1) \to \hat{C}$ of compact Riemann surfaces, with $\overset{ˉ}{j} ([\infty]) = \infty$ and a simple pole at the cusp, and $\overset{ˉ}{j}$ is a biholomorphism. Consequently $X (1)$ is biholomorphic to the Riemann sphere and $j$ is a Hauptmodul. The quotient map $π : H \to Y (1) \subset X (1)$ has local degree $2$ at $i$ and $3$ at $ω$ and $ω + 1$ , representatives of the classes of $i$ and $ω$ ; more generally its local degree is $2$ throughout $P S L_{2} (Z) \cdot i$ , $3$ throughout $P S L_{2} (Z) \cdot ω$ , and $1$ elsewhere. These ramification indices belong to $π$ , while $\overset{ˉ}{j}$ itself is unramified, so the composed map $j = \overset{ˉ}{j} \circ π : H \to \hat{C}$ has local degree $2$ at $i$ and $3$ at $ω$ and $ω + 1$ .

Facts & Assumptions

Given: The compact Riemann surface $X (1) = P S L_{2} (Z) \ H^{*}$ with its cusp $[\infty]$ , the open part $Y (1)$ , and the quotient charts at elliptic points and at the cusp (The compactified level-one modular curve X(1), The cusp chart and compactness of X(1), Local charts and the Riemann surface structure of a modular quotient); the modular function $j$ with $j (γ τ) = j (τ)$ , holomorphic on $H$ , and with $j = q^{- 1} + 744 + O (q)$ at the cusp (The modular discriminant and the j-invariant).

[F1]

$j$ is injective on $P S L_{2} (Z)$ -classes: if $j (τ) = j (τ^{'})$ then the lattices $Λ_{τ}, Λ_{τ^{'}}$ are homothetic, hence $τ^{'} = γ τ$ for some $γ \in S L_{2} (Z)$ and $[τ] = [τ^{'}]$ in $Y (1)$ (The j-invariant classifies complex tori).

[F2]

For $λ \in C$ the form $E_{4}^{3} - λ Δ \in M_{12}$ is nonzero with value $1$ at the cusp, and its valence sum is $12 / 12 = 1 > 0$ , so it has a zero in $H$ ; hence $j$ takes the value $λ$ (The level-one valence formula, The modular discriminant and the j-invariant, The zeros of E4 and E6 at the elliptic points, The graded ring of level-one modular forms).

[F3]

Local normal form: a nonconstant holomorphic map of Riemann surfaces has, near a point, the form $z \mapsto z^{ν}$ in suitable coordinates; an injective holomorphic map of domains has nowhere-vanishing derivative and is biholomorphic onto its image (Local power-map normal form on Riemann surfaces, Ramification index, ramification order and branch value, An injective holomorphic map has no critical point and is biholomorphic onto its image, Biholomorphic maps between complex domains).

[F4]

In the quotient chart of Local charts and the Riemann surface structure of a modular quotient the map $π$ near $i$ is $z \mapsto z^{2}$ and near $ω$ or $ω + 1$ is $z \mapsto z^{3}$ , these being the local degrees $2$ and $3$ ; $\overset{ˉ}{j}$ will be unramified once it is shown biholomorphic, since biholomorphic maps have local degree one (Ramification index, ramification order and branch value).

Proof

1.1F4givenalgebra

On $Y (1)$ the invariance and holomorphy of $j$ produce a holomorphic function $\overset{ˉ}{j} : Y (1) \to C$ : at ordinary points use a local inverse of $π$ ; at a point of stabiliser order $ν$ , the invariant Taylor series in a uniformising coordinate $z$ has only powers $z^{m ν}$ , so it is holomorphic in the quotient coordinate $z^{ν}$ . Near the cusp, in the $q$ -chart of The cusp chart and compactness of X(1), $j$ is $q^{- 1} (1 + 744 q + O (q^{2}))$ , a meromorphic function of $q$ with a simple pole at $q = 0$ ; since the cusp chart identifies the cusp with $q = 0$ , the formula $q \mapsto q^{- 1} (1 + 744 q + \dots)$ defines a holomorphic map of a punctured disc into $\hat{C}$ extending to $0$ with value $\infty$ . Hence $\overset{ˉ}{j}$ extends to a holomorphic map $X (1) \to \hat{C}$ with $\overset{ˉ}{j} ([\infty]) = \infty$ and a simple pole in the $q$ -coordinate.

2.1F1step 1.1givenalgebra

$\overset{ˉ}{j}$ is injective. If $\overset{ˉ}{j} ([τ]) = \overset{ˉ}{j} ([τ^{'}])$ then $j (τ) = j (τ^{'})$ (for $τ, τ^{'} \in H$ ) and [F1] gives $[τ] = [τ^{'}]$ ; on $Y (1)$ this is the claim, and the cusp is the only remaining point, with $\overset{ˉ}{j} ([\infty]) = \infty$ not attained on $Y (1)$ because $j$ is holomorphic on $H$ .

3.1F2step 2.1givenalgebra

$\overset{ˉ}{j}$ is surjective: given $λ \in C$ , [F2] provides $τ \in H$ with $j (τ) = λ$ , so $λ$ is in the image, while $\infty = \overset{ˉ}{j} ([\infty])$ ; hence $\overset{ˉ}{j} (X (1)) = \hat{C}$ .

4.1F3F4step 3.1givenalgebra∎

A bijective holomorphic map of compact Riemann surfaces is a biholomorphism: at every point the local normal form is $z \mapsto z^{ν}$ , and injectivity forces $ν = 1$ , so the derivative is never zero and the map is locally biholomorphic by [F3]; a locally biholomorphic bijection has holomorphic inverse, so $\overset{ˉ}{j}$ is biholomorphic and $X (1) ≅ \hat{C}$ with $j$ a Hauptmodul. The local degrees of $π$ at $i$ and at $ω, ω + 1$ are $2$ and $3$ by [F4], the same indices hold at all their modular translates because the group acts by biholomorphisms and $π \circ γ = π$ ; outside these two orbits the stabilisers are trivial and the local degree is $1$ ; since $\overset{ˉ}{j}$ is unramified, the composition $j = \overset{ˉ}{j} \circ π$ has exactly these local degrees at the corresponding points, in particular $2$ at $i$ and $3$ at $ω$ and $ω + 1$ .

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Jacobi theta triple product and nonvanishing of the theta constant

Statement

For $τ \in H$ , $z \in C$ and $Q = e^{π i τ}$ , $Θ (z ∣ τ) : = \sum_{m \in Z} Q^{m^{2}} e^{2 π i m z} = \prod_{n \geq 1} (1 - Q^{2 n}) (1 + Q^{2 n - 1} e^{2 π i z}) (1 + Q^{2 n - 1} e^{- 2 π i z}) .$ The series and product converge locally uniformly in each variable, uniformly on compact products. For fixed $τ$ , $Θ$ is entire in $z$ with exactly the simple zeros $z = (1 + τ) / 2 + a + b τ$ , $a, b \in Z$ . In particular $θ (τ) : = Θ (0 ∣ τ) = \prod_{n \geq 1} (1 - Q^{2 n}) (1 + Q^{2 n - 1})^{2} \neq 0.$ Here $Q^{2} = q$ ; these cusp parameters must not be confused.

Facts & Assumptions

Given: $τ \in H$ , $z \in C$ and $Q = e^{π i τ}$ , so $∣ Q ∣ = e^{- π Im τ} < 1$ (The unit disc, the upper half-plane, and Blaschke factors).

[F1]

Weierstrass $M$ -test (Weierstrass M-test for complex-valued function series).

[F2]

A locally uniform limit of holomorphic functions is holomorphic (Locally uniform limits of holomorphic functions are holomorphic and their derivatives converge locally uniformly).

[F3]

Normally convergent products of holomorphic functions have holomorphic limits; on a compact set the zeros are exactly those of the finitely many factors that vanish there, and the tail is zero-free (Normally convergent products define holomorphic functions with the expected zeros, Normal convergence of holomorphic products).

[F4]

A bounded entire function is constant (Liouville's theorem: every bounded entire function is constant).

[F5]

If $f$ is holomorphic near $a$ and vanishes there to order one, then $f = (z - a) g$ with $g$ holomorphic and $g (a) \neq 0$ (The order of a zero is the exponent in its local holomorphic factorization).

[F6]

$\ker \exp = 2 π i Z$ , $\exp (w + w^{'}) = \exp w \exp w^{'}$ , $\exp (2 w) = \exp (w)^{2}$ , $\exp w \neq 0$ , and $∣ \exp (x + i y) ∣ = e^{x}$ for real $x, y$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ , $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential).

[F7]

$\exp^{'} = \exp$ , and the chain rule and algebra of derivatives give $\frac{d}{d z} e^{2 π i m z} = 2 π i m e^{2 π i m z}$ (The complex exponential is entire and its complex derivative is itself, The chain rule for complex derivatives, Linearity, product, reciprocal, and quotient rules for complex derivatives).

[F8]

The ratio test detects convergence of series of positive terms (Ratio test: $lim sup ∣ a_{k + 1} / a_{k} ∣ < 1$ gives absolute convergence and hence convergence, and $lim inf ∣ a_{k + 1} / a_{k} ∣ > 1$ gives divergence), and an absolutely convergent complex series converges with every rearrangement having the same sum (Every absolutely convergent complex series converges, and rearrangements preserve its sum, Complex series, absolute convergence, complex power series, and radius of convergence, If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges).

Proof

1.1F1F2F6F7F8givenalgebra

Let $K$ be a compact subset of the product, so $Im τ \geq y_{0} > 0$ and $∣ Im z ∣ \leq M$ on $K$ . Put $r : = e^{- π y_{0}} < 1$ . Then $∣ Q ∣ \leq r$ and $∣ Q^{m^{2}} e^{2 π i m z} ∣ \leq r^{m^{2}} e^{2 π M ∣ m ∣}$ for every $(m, z, τ)$ under consideration; the bounding series converges by [F8], since the ratio of consecutive terms is $r^{2 m + 1} e^{2 π M} \to 0$ . The $M$ -test [F1] therefore gives uniform convergence on $K$ ; each term is entire in $z$ and holomorphic in $τ$ by [F6] and [F7], so [F2] shows that $Θ$ is entire in $z$ for fixed $τ$ , holomorphic in $τ$ , and that the series converges locally uniformly in each variable, uniformly on compact products. Termwise index shifts, licensed by absolute convergence [F8], give $Θ (z + 1 ∣ τ) = Θ (z ∣ τ)$ , since $e^{2 π i m} = 1$ , and, using $Q^{m^{2}} e^{2 π i m τ} = Q^{(m + 1)^{2} - 1}$ , also $Θ (z + τ ∣ τ) = Q^{- 1} e^{- 2 π i z} Θ (z ∣ τ)$ .

1.2F3F6F7givenalgebra

Put $Π (z ∣ τ) : = \prod_{n \geq 1} (1 - Q^{2 n}) (1 + Q^{2 n - 1} e^{2 π i z}) (1 + Q^{2 n - 1} e^{- 2 π i z})$ and $g_{n} (z, τ) : = (1 - Q^{2 n}) (1 + Q^{2 n - 1} e^{2 π i z}) (1 + Q^{2 n - 1} e^{- 2 π i z})$ . For finite grouped products put $Π_{0} = 1$ (the empty product) and $Π_{N} = \prod_{1 \leq n \leq N} g_{n}$ for $N \geq 1$ ; normal convergence concerns this explicit sequence including its empty prefix. On $K$ as above, with $B : = e^{2 π M}$ , the estimate $∣ g_{n} - 1 ∣ \leq (1 + r^{2 n}) (1 + B r^{2 n - 1})^{2} - 1 \leq C_{K} r^{2 n - 1}$ shows $\sum_{n} \sup_{K} ∣ g_{n} - 1 ∣ < \infty$ , so $Π$ is normally convergent and [F3] makes it holomorphic in each variable with the displayed product. The factor $1 - Q^{2 n}$ never vanishes because $∣ Q^{2 n} ∣ < 1$ . By [F6], $1 + Q^{2 n - 1} e^{2 π i z} = 0$ is equivalent to $e^{2 π i z} = - Q^{- (2 n - 1)} = e^{π i (1 - (2 n - 1) τ)}$ , that is to $z \equiv \frac{1}{2} - \frac{2 n - 1}{2} τ (m o d Z)$ , and likewise $1 + Q^{2 n - 1} e^{- 2 π i z} = 0$ is equivalent to $z \equiv - \frac{1}{2} + \frac{2 n - 1}{2} τ (m o d Z)$ . Writing a solution of the first type as $\frac{1 + τ}{2} + a + b τ$ with $a = k$ and $b = - n \leq - 1$ , and one of the second type with $a = k - 1$ and $b = n - 1 \geq 0$ , shows that the union of all solutions over $n \geq 1$ is exactly the coset $\frac{1 + τ}{2} + Z + τ Z$ , each of whose points arises from exactly one $n$ and one factor because the representation $a + b τ$ with $a, b \in Z$ is unique ( $Im τ > 0$ ). At such a solution the derivative of the vanishing factor with respect to $z$ is $\pm 2 π i Q^{2 n - 1} e^{\pm 2 π i z} \neq 0$ by [F6] and [F7], so every zero of $Π$ is simple and is a zero of exactly one factor. Finally, cancelling the shifted factors in the normally convergent product, with the reindexings $n \mapsto n + 1$ in the second product and $n \mapsto n \pm 1$ in the third, gives $Π (z + 1 ∣ τ) = Π (z ∣ τ)$ and $Π (z + τ ∣ τ) = \frac{1 + Q^{- 1} e^{- 2 π i z}}{1 + Q e^{2 π i z}} Π (z ∣ τ) = Q^{- 1} e^{- 2 π i z} Π (z ∣ τ)$ .

2.1F4F5F6F8step 1.1step 1.2givenalgebra

At $z_{0} : = \frac{1 + τ}{2}$ we have $Θ (z_{0} ∣ τ) = \sum_{m \in Z} (- 1)^{m} Q^{m^{2} + m}$ ; the terms with indices $m$ and $- m - 1$ are negatives of each other, and the series is absolutely convergent by [F8], so it sums to $0$ . The shift laws of 1.1 then propagate the vanishing to all points of $z_{0} + Z + τ Z$ , the multiplier $Q^{- 1} e^{- 2 π i z}$ being nonzero by [F6]. By 1.2 that coset is exactly the zero set of $Π$ , all its zeros are simple, and $Θ$ is entire in $z$ by 1.1; hence $F_{τ} : = Θ (\cdot ∣ τ) / Π (\cdot ∣ τ)$ is holomorphic off the coset and, by the Taylor series of its vanishing numerator and the simple-zero factorisation [F5] of its denominator at each zero, extends holomorphically across it to an entire function. The shift laws of 1.1 and 1.2 give $F_{τ} (z + 1) = F_{τ} (z)$ and $F_{τ} (z + τ) = F_{τ} (z)$ off the coset, hence everywhere. Every $z \in C$ is $w + m + n τ$ with $m, n \in Z$ and $w$ in the compact parallelogram ${s + t τ : 0 \leq s, t \leq 1}$ (take $m, n$ the integer parts of the coefficients of $z$ in the basis $1, τ$ ), so $z \mapsto F_{τ} (z)$ is bounded on $C$ by its supremum on that compact set; [F4] gives $F_{τ} \equiv c (τ)$ for a number $c (τ)$ depending only on $τ$ .

3.1F3F6F8step 2.1givenalgebra

Evaluating the constant of 2.1 at $z = \frac{1}{4}$ , where no factor of $Π$ vanishes, gives $c (τ) = Θ (\frac{1}{4} ∣ τ) / Π (\frac{1}{4} ∣ τ)$ . The series identity is $Θ (\frac{1}{4} ∣ τ) = \sum_{m} Q^{m^{2}} i^{m} = \sum_{k} (- 1)^{k} Q^{4 k^{2}} = Θ (\frac{1}{2} ∣ 4 τ)$ : the odd- $m$ terms cancel in the pairs $m \leftrightarrow - m$ , and the even ones $m = 2 k$ contribute $i^{2 k} Q^{4 k^{2}} = (- 1)^{k} Q^{4 k^{2}}$ , all licensed by [F8]. The product identity is $Π (\frac{1}{4} ∣ τ) = \prod_{n} (1 - Q^{2 n}) (1 + Q^{4 n - 2}) = \prod_{k} (1 - Q^{4 k}) (1 - Q^{8 k - 4}) = \prod_{j} (1 - Q^{8 j}) (1 - Q^{8 j - 4})^{2} = Π (\frac{1}{2} ∣ 4 τ)$ , obtained from $\prod_{n} (1 + Q^{4 n - 2}) = \prod_{n} (1 - Q^{8 n - 4}) / \prod_{n} (1 - Q^{4 n - 2})$ and from splitting $\prod_{k} (1 - Q^{4 k}) = \prod_{j} (1 - Q^{8 j}) \prod_{j} (1 - Q^{8 j - 4})$ ; all rearrangements are legitimate by normal convergence [F3]. Hence $c (τ) = c (4 τ)$ , and iteration gives $c (τ) = c (4^{k} τ)$ for every $k \geq 0$ . Writing $Q_{k} : = Q^{4^{k}} \to 0$ , the evaluation of 2.1 at $z = \frac{1}{2}$ gives $c (4^{k} τ) = Θ (\frac{1}{2} ∣ 4^{k} τ) / Π (\frac{1}{2} ∣ 4^{k} τ)$ with $Θ (\frac{1}{2} ∣ 4^{k} τ) = 1 + 2 \sum_{m \geq 1} (- 1)^{m} Q_{k}^{m^{2}}$ and $Π (\frac{1}{2} ∣ 4^{k} τ) = \prod_{n} (1 - Q_{k}^{2 n}) (1 - Q_{k}^{2 n - 1})^{2}$ . Since $∣ Q_{k} ∣ \to 0$ , $∣ Θ (\frac{1}{2} ∣ 4^{k} τ) - 1 ∣ \leq 2 ∣ Q_{k} ∣ / (1 - ∣ Q_{k} ∣) \to 0$ , and with $ρ = ∣ Q_{k} ∣$ , $∣ Π (\frac{1}{2} ∣ 4^{k} τ) - 1 ∣ \leq \exp (ρ (2 + ρ) / (1 - ρ^{2})) - 1 \to 0$ by the elementary product bound $\prod (1 + a_{n}) \leq \exp \sum a_{n}$ . Therefore $c (τ) = \lim_{k} c (4^{k} τ) = 1$ .

4.1F6step 2.1step 3.1givenalgebra∎

By 3.1, $Θ (z ∣ τ) = Π (z ∣ τ)$ for all $z$ , so by 2.1 the zeros of $Θ$ in $z$ are exactly the simple zeros $z = \frac{1 + τ}{2} + a + b τ$ , $a, b \in Z$ . Moreover $Θ (0 ∣ τ) = Π (0 ∣ τ) = \prod_{n} (1 - Q^{2 n}) (1 + Q^{2 n - 1})^{2} \neq 0$ , since $∣ Q^{2 n} ∣ < 1$ and $∣ Q^{2 n - 1} ∣ < 1$ make every factor nonzero. This is the zero-free theta constant $θ (τ)$ , and $Q^{2} = q$ by the addition law [F6].

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Transformation laws of the Jacobi theta function

Statement

Assume countable choice. For $τ \in H$ and $z \in C$ , let $Θ (z ∣ τ)$ and $θ (τ) = Θ (0 ∣ τ)$ be the functions of Jacobi theta triple product and nonvanishing of the theta constant, and set $s (τ) = \exp (\frac{1}{2} Log (τ / i)) .$ The principal logarithm is defined here because $Re (τ / i) > 0$ ; thus $s$ is the holomorphic square root of $τ / i$ that is positive for $τ = i t$ , $t > 0$ . Then $Θ (z ∣ - 1 / τ) = s (τ) e^{π i τ z^{2}} Θ (τ z ∣ τ), Θ (z ∣ τ + 2) = Θ (z ∣ τ) .$ In particular $θ (- 1 / τ) = s (τ) θ (τ), θ (1 - 1 / τ) = s (τ) \sum_{n \in Z} e^{π i (n + 1 / 2)^{2} τ} .$ As $Im τ \to \infty$ , uniformly in $Re τ$ , $θ (1 - 1 / τ) = 2 s (τ) e^{π i τ / 4} (1 + O (e^{- 2 π Im τ})) .$

Facts & Assumptions

Given: Countable choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )), $τ \in H$ and $z \in C$ .

[F1]

The theta series $Θ (z ∣ τ) = \sum_{n \in Z} e^{π i n^{2} τ + 2 π i n z}$ converges absolutely and uniformly on compact products; it is entire in $z$ and holomorphic in $τ$ (Jacobi theta triple product and nonvanishing of the theta constant).

[F2]

Under countable choice, shifted Poisson summation for a Schwartz function $f$ on $R$ gives $\sum_{n} f (a + n) = \sum_{m} \hat{f} (m) e^{2 π i m a}$ , and both sums converge absolutely. For $f_{t} (u) = e^{- π t u^{2}}$ , $t > 0$ , the Fourier transform is $t^{- 1 / 2} e^{- π ξ^{2} / t}$ (Poisson summation for Schwartz functions, Euclidean Gaussian transform with the 2π normalization).

[F3]

Summable uniform majorants give uniformly convergent function series; locally uniform holomorphic limits are holomorphic, and holomorphic functions on a connected domain agreeing on a set with an interior accumulation point agree everywhere (Weierstrass M-test for complex-valued function series, Locally uniform limits of holomorphic functions are holomorphic and their derivatives converge locally uniformly, Identity theorem for holomorphic functions).

[F4]

The principal logarithm is holomorphic off the nonpositive real ray and exponentiates to its argument. The exponential is entire, satisfies its addition law and has kernel $2 π i Z$ ; composition preserves holomorphy (The principal logarithm is the normalised holomorphic branch on the slit plane, Complex logarithms, the principal logarithm, and principal and multivalued complex powers, The complex exponential is entire and its complex derivative is itself, $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ , The chain rule for complex derivatives).

Proof

1.1F1F2F4givenalgebra

For $t > 0$ , every derivative of $f_{t} (u) = e^{- π t u^{2}}$ is a polynomial times this Gaussian. Exponential domination (The exponential dominates every fixed nonnegative integer power at $+ \infty$ ) bounds each such derivative times every power of $u$ , so $f_{t}$ is Schwartz in Schwartz space and its seminorms. Apply [F2] at the real shift $a = x$ to obtain $\sum_{n} e^{- π t (n + x)^{2}} = t^{- 1 / 2} \sum_{m} e^{- π m^{2} / t} e^{2 π i m x}$ . Multiplying by $\sqrt{t}$ and expanding $(n + x)^{2}$ gives $Θ (x ∣ i / t) = \sqrt{t} e^{- π t x^{2}} Θ (i t x ∣ i t)$ for every real $x$ . This is the asserted inversion identity for $τ = i t$ , since $- 1 / (i t) = i / t$ , $s (i t) = \sqrt{t}$ and $e^{π i (i t) x^{2}} = e^{- π t x^{2}}$ . Countable choice is used precisely through the two suppliers in [F2].

2.1F1F3F4step 1.1algebra

Fix real $x$ . Both sides of the claimed inversion identity are holomorphic in $τ \in H$ . For the right-hand theta series, on any compact $K \subset H$ its terms have modulus at most $e^{- π y_{0} n^{2} + C ∣ n ∣}$ , where $y_{0} = \min_{K} Im τ > 0$ and $C$ bounds $2 π ∣ x Im τ ∣$ ; this summable Gaussian majorant proves holomorphy by [F3]. The other factors are holomorphic by [F4], including $s$ , since $τ / i$ lies in the right half-plane. The left side is holomorphic by [F1] and composition with $- 1 / τ$ . By 1.1 the two sides agree on the positive imaginary axis, whose points accumulate within $H$ , so the identity theorem [F3] proves equality for all $τ$ . Now fix such $τ$ . The two sides are entire in $z$ by [F1, F4] and agree for real $z = x$ , so a second identity-theorem application proves the inversion formula for every $z \in C$ .

3.1F1F4step 2.1algebra

Each term of the theta series is unchanged on replacing $τ$ by $τ + 2$ , since $e^{2 π i n^{2}} = 1$ for integral $n$ ; absolute convergence therefore proves the stated period-two identity. Setting $z = 0$ in 2.1 gives $θ (- 1 / τ) = s (τ) θ (τ)$ . Since $n^{2}$ and $n$ have the same parity, [F1, F4] give $θ (1 + u) = Θ (1 / 2 ∣ u)$ for $u \in H$ . Apply 2.1 with $z = 1 / 2$ and complete the square: $e^{π i τ / 4} Θ (τ / 2 ∣ τ) = \sum_{n} e^{π i (n + 1 / 2)^{2} τ}$ . This proves the shifted-constant formula.

4.1F1F4step 3.1algebra∎

Pair $n = j$ with $n = - j - 1$ , $j \geq 0$ , in the absolutely convergent shifted series. Its value is $2 e^{π i τ / 4} (1 + \sum_{j \geq 1} e^{π i τ j (j + 1)})$ . If $y = Im τ$ , then $j (j + 1) \geq 2 j$ gives $∣ \sum_{j \geq 1} e^{π i τ j (j + 1)} ∣ \leq e^{- 2 π y} / (1 - e^{- 2 π y})$ . This proves the uniform asymptotic and its stated error term. The principal square root never vanishes on $H$ ; no alternative square-root sign, boundary point $τ = 0$ , or half-weight multiplier convention is implicit in any formula.

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The Jacobi product formula for the discriminant

Statement

For $τ \in H$ and $q = e^{2 π i τ}$ , $Δ (τ) = q \prod_{n = 1}^{\infty} (1 - q^{n})^{24}, Δ = (E_{4}^{3} - E_{6}^{2}) / 1728.$ The product converges locally uniformly for $∣ q ∣ < 1$ ; its factor after $q$ is holomorphic and zero-free in the unit disc and equals $1$ at $q = 0$ . Consequently $Δ$ has integral Fourier coefficients and a simple zero at the cusp.

Facts & Assumptions

Given: $Δ = (E_{4}^{3} - E_{6}^{2}) / 1728 \in S_{12}$ with $q$ -expansion $q - 24 q^{2} + O (q^{3})$ , and $E_{2} (τ) = 1 - 24 \sum_{n \geq 1} σ_{1} (n) q^{n}$ with its transformation law (The discriminant is a nonvanishing cusp form of weight 12, The level-one Eisenstein series E_k and the weight-two series E_2, The transformation law of the weight-two Eisenstein series E_2).

[F1]

A normally convergent product of holomorphic functions on a domain is holomorphic; on compacta all but finitely many factors are zero-free and the zeros come from the finitely many exceptional factors (Normally convergent products define holomorphic functions with the expected zeros).

[F2]

Locally uniform limits of holomorphic functions have locally uniformly convergent derivatives (Locally uniform limits of holomorphic functions are holomorphic and their derivatives converge locally uniformly).

[F3]

If a holomorphic function on a domain has zero derivative it is constant (A holomorphic function with zero derivative on a domain is constant).

[F4]

$\sum_{r \geq 0} w^{r} = 1 / (1 - w)$ for $∣ w ∣ < 1$ and the same geometric identity for complex $w$ follows from the finite geometric sum and $∣ w ∣^{N + 1} \to 0$ ; absolutely convergent complex double families may be regrouped by applying the real double-series theorem to real and imaginary parts (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges, Fubini for double series: if $\sum_{i} \sum_{j} ∣ a_{i j} ∣$ converges then both iterated sums and the sum along every bijection $N \to N \times N$ converge to one and the same value); finite products of $(1 - q^{n})^{24}$ have integer coefficients by the binomial theorem (The binomial theorem over the complex field).

[F5]

$\dim S_{12} = 1$ and $Δ \neq 0$ has leading coefficient $1$ at $q$ ; $S, T$ generate $S L_{2} (Z)$ (The dimension of the space of level-one modular forms, The discriminant is a nonvanishing cusp form of weight 12, The standard fundamental domain, boundary identifications and elliptic stabilisers).

[F6]

$F \in S_{12}$ means: $F$ holomorphic on $H$ , $F (γ τ) = (c τ + d)^{12} F (τ)$ for all $γ \in S L_{2} (Z)$ , and the associated function of $q$ is holomorphic at $0$ with value $0$ (Level-one modular forms and cusp forms).

Proof

1.1F1F2F4givenalgebra

Put $P_{0} (q) = 1$ for the empty product, $P_{N} (q) = \prod_{1 \leq n \leq N} (1 - q^{n})^{24}$ for $N \geq 1$ , $P (q) : = \lim_{N} P_{N} (q)$ and $F (τ) : = q P (q)$ . For $∣ q ∣ \leq r < 1$ we have $∣ (1 - q^{n})^{24} - 1 ∣ \leq (2^{24} - 1) r^{n}$ , a summable majorant independent of $q$ , so the product is normally convergent on the disc and [F1] makes $P$ holomorphic and zero-free there with $P (0) = 1$ ; hence $F$ is holomorphic and zero-free on $H$ and $F (τ) = q + O (q^{2})$ (as a function of $q$ ). By [F2] the logarithmic derivative of $P$ may be computed from the finite products and the geometric series [F4]: $\frac{1}{2 π i} \frac{F^{'}}{F} = 1 - 24 \sum_{n \geq 1} \frac{n q^{n}}{1 - q^{n}} = 1 - 24 \sum_{n \geq 1} \sum_{r \geq 1} n q^{n r} = 1 - 24 \sum_{m \geq 1} σ_{1} (m) q^{m} = E_{2} (τ)$ , the regrouping being justified by the bound $\sum_{n, r} n ∣ q ∣^{n r} \leq (1 - r)^{- 1} \sum_{n} n r^{n} < \infty$ on $∣ q ∣ \leq r < 1$ [F4]; the last series converges by the ratio test.

2.1F3F5F6step 1.1givenalgebra

For $γ = (\begin{matrix} a & b \\ c & d \end{matrix}) \in S L_{2} (Z)$ put $R_{γ} (τ) : = F (γ τ) / [(c τ + d)^{12} F (τ)]$ ; this is holomorphic and zero-free on $H$ . Its logarithmic derivative, divided by $2 π i$ , equals $\frac{E_{2} (γ τ)}{(c τ + d)^{2}} - \frac{12 c}{2 π i (c τ + d)} - E_{2} (τ)$ by 1.1, and by the $E_{2}$ transformation law this is $[E_{2} (τ) - \frac{6 i c}{π (c τ + d)}] - [- \frac{6 i c}{π (c τ + d)}] - E_{2} (τ) = 0$ (using $\frac{12 c}{2 π i (c τ + d)} = - \frac{6 i c}{π (c τ + d)}$ ). Hence $R_{γ}$ is constant, $C_{γ}$ , by [F3], and $C_{γ δ} = C_{γ} C_{δ}$ because $(c_{γ δ} τ + d_{γ δ}) = (c_{γ} δ τ + d_{γ}) (c_{δ} τ + d_{δ})$ . For $T$ we have $F (τ + 1) = F (τ)$ , so $C_{T} = 1$ ; for $S$ we have $R_{S} (τ) = F (- 1 / τ) / [τ^{12} F (τ)]$ and at $τ = i$ , where $- 1 / i = i$ and $i^{12} = 1$ , this gives $C_{S} = F (i) / F (i) = 1$ . Since $S, T$ generate $S L_{2} (Z)$ [F5] and $(- I)$ acts with factor $(- 1)^{12} = 1$ , multiplicativity gives $C_{γ} = 1$ for every $γ$ ; thus $F (γ τ) = (c τ + d)^{12} F (τ)$ for all $γ$ , and since $F = q + O (q^{2})$ it satisfies the cusp condition. Hence $F \in S_{12}$ by [F6].

3.1F5step 2.1givenalgebra

By [F5], $S_{12}$ is one-dimensional and contains the nonzero $Δ = q - 24 q^{2} + O (q^{3})$ ; so $F = c Δ$ for some $c \in C$ . Comparing $q$ -coefficients, $1 = c \cdot 1$ , so $F = Δ$ , which is the product formula.

4.1F1F2F4step 2.1givenalgebra∎

Integrality of the coefficients: the coefficient of $q^{N}$ in $P$ equals the coefficient of $q^{N}$ in the finite product $\prod_{n \leq N} (1 - q^{n})^{24}$ , because all factors with $n > N$ are congruent to $1$ modulo $q^{N + 1}$ ; the finite product has integer coefficients by [F4], and its coefficients agree with those of the analytic expansion of $P$ because the finite products converge locally uniformly, with all derivatives, to $P$ near $0$ by [F1] and [F2]. Hence every Taylor coefficient of $P$ is an integer, and the coefficients of $Δ = q P$ are integers as well; the simple zero at the cusp is ${ord}_{\infty} (Δ) = 1$ from the expansion $q - 24 q^{2} + \dots$ , consistent with [F5].

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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.

5 · Examples, counterexamples and false statements

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Level-One Modular Forms and the j-Invariant

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

The modular group and its action on the upper half-plane

Definition

Reduction of orbits to the standard domain

Statement

Facts & Assumptions

Proof

The standard fundamental domain, boundary identifications and elliptic stabilisers

Statement

Facts & Assumptions

Proof

Local charts and the Riemann surface structure of a modular quotient

Statement

Facts & Assumptions

Proof

The cusp chart and compactness of X(1)

Statement

Facts & Assumptions

Proof

The compactified level-one modular curve X(1)

Definition

The q-expansion principle at the cusp

Statement

Facts & Assumptions

Proof

Level-one modular forms and cusp forms

Definition

Absolute convergence and holomorphy of the lattice Eisenstein sums

Statement

Facts & Assumptions

Proof

The divisor power sums σkσk​

Definition

The level-one Eisenstein series E_k and the weight-two series E_2

Definition

The Lipschitz formula for the reciprocal-power sums

Statement

Facts & Assumptions

Proof

Eisenstein series are modular forms; their Fourier coefficients

Statement

Facts & Assumptions

Proof

The boundary arc contribution in the valence computation

Statement

Facts & Assumptions

Proof

The level-one valence formula

Statement

Facts & Assumptions

Proof

The zeros of E4 and E6 at the elliptic points

Statement

Facts & Assumptions

Proof

The dimension of the space of level-one modular forms

Statement

Facts & Assumptions

Proof

The transformation law of the weight-two Eisenstein series E_2

Statement

Facts & Assumptions

Proof

The discriminant is a nonvanishing cusp form of weight 12

Statement

Facts & Assumptions

Proof

The modular discriminant and the j-invariant

Definition

The graded ring of level-one modular forms

Statement

Facts & Assumptions

Proof

The j-invariant classifies complex tori

Statement

Facts & Assumptions

The divisor power sums $σ_{k}$