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Level-One Modular Forms and the j-Invariant — Examples

1 · Prerequisites

2 · Summary

The examples make the constructions of the companion page explicit. The standard fundamental domain is shown to tessellate the upper half-plane: the tiles $γ \overline{D}$ cover $H$ with disjoint interiors and meet in a common edge, half-edge or vertex, the edge identifications being $τ \sim τ + 1$ on the vertical sides and $τ \sim - 1 / τ$ on the circular arc. The modular group has exactly two elliptic classes, those of $i$ and of $ω = e^{2 π i / 3}$ , with stabilisers of orders two and three; the quotient map has local degrees two and three there, and the values $j (i) = 1728$ and $j (ω) = 0$ are computed from the zeros of $E_{6}$ and $E_{4}$ . On the torus side the square and hexagonal lattices $Z [i]$ and $Z [ω]$ have these same $j$ -invariants, the level sets of $1728$ and $0$ are exactly their homothety classes, and multiplication by $i$ and by $ω$ realises automorphisms of the corresponding tori of orders four and three.

The arithmetic examples read coefficients off the $q$ -expansions: $E_{4} = 1 + 240 q + 2160 q^{2} + \dots$ , $E_{6} = 1 - 504 q - 16632 q^{2} + \dots$ , $Δ = q - 24 q^{2} + 252 q^{3} - \dots$ and $j = q^{- 1} + 744 + 196884 q + \dots$ , the last from the division of $E_{4}^{3}$ by $Δ$ . For odd weight the transformation law applied to $- I$ reads $f = (- 1)^{k} f = - f$ , so the only odd-weight form is the zero form. The final entry records a false statement, that the weight-two Eisenstein series is a modular form; its transformation law carries a correction term, and the false statement is kept as a flagged non-result rather than a theorem.

The remaining figures develop the level-two theory of the modular lambda function as a worked counterpart of the level-one picture. The principal congruence subgroup $Γ (2) \leq S L_{2} (Z)$ has projective image $\overset{ˉ}{Γ} (2) = Γ (2) / {\pm I}$ of index six in the modular group. $P S L_{2} (Z)$ acts on $λ$ through six fractional-linear substitutions, whose values may coincide; $\overset{ˉ}{Γ} (2)$ is torsion-free and acts freely, the fibres of $λ$ are exactly its orbits, and the Legendre normal form identifies the values of $j$ with $J (λ) = 256 (λ^{2} - λ + 1)^{3} / λ^{2} (λ - 1)^{2}$ . The lambda function is then shown to induce a biholomorphism from $Y (2) = H / Γ (2)$ onto the twice-punctured plane and a biholomorphism from the interior of the standard ideal quadrilateral onto the plane slit along the two closed real rays, so the abstract quotient of the companion page acquires an explicit coordinate.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

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The standard fundamental domain tessellates the upper half-plane

Example

The closed tiles $γ \cdot \overline{D}$ , $γ \in P S L_{2} (Z)$ , have union $H$ , pairwise disjoint interiors, and any two distinct tiles have empty intersection or meet in a common edge, a half-edge or a vertex; the full edge identifications are $τ \sim τ + 1$ on the vertical sides and $τ \sim - 1 / τ$ on the circular arc. The tiling is $P S L_{2} (Z)$ -invariant and locally finite.

Facts & Assumptions

Given: $D = {τ \in H : ∣ ℜ τ ∣ < 1 / 2, ∣ τ ∣ > 1}$ , its closure $\overline{D}$ , and the action of $G = P S L_{2} (Z)$ (The modular group and its action on the upper half-plane).

[F1]

Every orbit meets $\overline{D}$ ; no two distinct points of $D$ are equivalent; two distinct points $z, z^{'} \in \overline{D}$ are equivalent if and only if $z^{'} = z \pm 1$ with $ℜ z = \mp 1 / 2$ , or $z^{'} = - 1 / z$ with $∣ z ∣ = 1$ ; the points of $\overline{D}$ with nontrivial stabiliser are only $i, ω, ω + 1$ (The standard fundamental domain, boundary identifications and elliptic stabilisers, The orbit $G \cdot x$ and stabilizer $G_{x}$ of a point in a group action).

[F2]

Each $τ \in H$ has a neighbourhood meeting only the finitely many stabiliser translates of $τ$ ; equivalently the action is properly discontinuous and the quotient map is open (Local charts and the Riemann surface structure of a modular quotient).

Verification

1.1F1givenalgebra

Every point of $H$ lies in some tile, because its orbit meets $\overline{D}$ [F1]; thus $⋃_{γ} γ \overline{D} = H$ . If two tiles have a common interior point, then $γ z = γ^{'} z^{'}$ with $z, z^{'} \in D$ , so $z, z^{'}$ are equivalent points of $D$ ; by [F1] they are equal and $γ^{- 1} γ^{'}$ stabilises $z \in D$ , which by [F1] has trivial stabiliser, so $γ = γ^{'}$ . Hence distinct tiles have disjoint interiors.

1.2F1givenalgebra

$T$ identifies the two vertical sides, and $S$ identifies the two halves of the circular side, fixing $i$ . To check incidence, translate one of two meeting tiles to $\overline{D}$ . At a boundary point other than $i, ω, ω + 1$ the stabiliser is trivial; [F1] then forces the other tile to be $T \overline{D}$ , $T^{- 1} \overline{D}$ , or $S \overline{D}$ , according to the side containing that point. Direct substitution shows that these share respectively a full vertical side or the full circular side. Any other tile can meet $\overline{D}$ only at the three exceptional points. Such an intersection has at most one point: each tile is an intersection of three half-planes bounded by vertical lines or circles orthogonal to the real axis, hence is convex along those real-orthogonal circular or vertical geodesics. Indeed, a real Möbius map sending a given geodesic to the imaginary axis carries each bounding half-plane to one whose intersection with that axis is an interval. Two distinct common points would therefore give a common segment, including a nonexceptional point, which is the already listed side case. Thus every nonempty intersection is a side or a vertex, as asserted.

2.1F1F2step 1.2givenalgebra∎

The tiles are invariant by construction. For local finiteness let $K \subset H$ be compact and put $a : = \min_{K} Im > 0$ . If $w = γ z \in K$ with $z \in \overline{D}$ and $c \neq 0$ , then $a \leq Im w \leq 1 / (c^{2} Im z)$ , so $Im z \leq 1 / a$ . Therefore such a tile meets $K$ through the compact set $L : = \overline{D} \cap {Im z \leq 1 / a}$ ; the compact-set finiteness proved in Local charts and the Riemann surface structure of a modular quotient, step 1.1, leaves only finitely many $γ$ with $γ L \cap K \neq \emptyset$ . For $c = 0$ the maps are translations $T^{n}$ , and the real-part bounds on $K$ and $∣ Re z ∣ \leq 1 / 2$ leave only finitely many $n$ . Thus every compact $K$ meets finitely many tiles; a compact disc neighbourhood at each point proves local finiteness.

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The elliptic points of the modular group and their images under j

Example

In $P S L_{2} (Z) \ H$ there are exactly two elliptic classes: the class of $i$ , with stabiliser of order $2$ generated by $S$ , and the class of $ω = e^{2 π i / 3}$ , with stabiliser of order $3$ generated by $S T$ ; every other stabiliser is trivial. The corresponding orbifold points have orders $2$ and $3$ ; the quotient map $H \to X (1)$ has local degrees $2$ at $i$ and $3$ at $ω$ , and $j (i) = 1728$ , $j (ω) = 0$ .

Facts & Assumptions

Given: The action of $G = P S L_{2} (Z)$ on $H$ with $S \cdot z = - 1 / z$ , $T \cdot z = z + 1$ and the closure $\overline{D} = {τ : ∣ ℜ τ ∣ \leq 1 / 2, ∣ τ ∣ \geq 1}$ of the standard fundamental domain (The standard fundamental domain, boundary identifications and elliptic stabilisers); the quotient map $π : H \to Y (1) \subset X (1)$ with its local charts at the elliptic points, where it is $z \mapsto z^{ν}$ in a centred coordinate (Local charts and the Riemann surface structure of a modular quotient, The j-invariant uniformizes X(1)); the modular function $j = E_{4}^{3} / Δ$ with $Δ = (E_{4}^{3} - E_{6}^{2}) / 1728$ nonvanishing on $H$ (The modular discriminant and the j-invariant).

[F1]

Every $G$ -orbit meets $\overline{D}$ ; the only points of $\overline{D}$ with nontrivial $G$ -stabiliser are $i$ , $ω$ and $ω + 1$ , with $Stab (i) = ⟨ S ⟩$ of order $2$ , $Stab (ω) = ⟨ S T ⟩$ of order $3$ and $Stab (ω + 1) = ⟨ T S ⟩$ of order $3$ , while every other point of $\overline{D}$ has trivial stabiliser; $T \cdot ω = ω + 1$ (The standard fundamental domain, boundary identifications and elliptic stabilisers, Local charts and the Riemann surface structure of a modular quotient).

[F2]

In the quotient chart at an elliptic point a generator of the stabiliser acts by $z \mapsto e^{2 π i / ν} z$ and $π$ becomes the map $z \mapsto z^{ν}$ ; for $G$ this is $ν = 2$ at $i$ and $ν = 3$ at $ω$ and $ω + 1$ , so $π$ has local degree $2$ at $i$ and $3$ at $ω$ and $ω + 1$ , representatives of the two elliptic classes; the same local degrees hold at all their modular translates (Local charts and the Riemann surface structure of a modular quotient, The j-invariant uniformizes X(1)).

[F3]

$E_{4}$ has a simple zero at the class of $ω$ and no other zeros, and $E_{6}$ has a simple zero at the class of $i$ and no other zeros; in particular $E_{6} (i) = 0$ , $E_{4} (i) \neq 0$ , $E_{4} (ω) = 0$ and $E_{6} (ω) \neq 0$ (The zeros of E4 and E6 at the elliptic points).

[F4]

$Δ = (E_{4}^{3} - E_{6}^{2}) / 1728$ has no zeros on $H$ and $j = E_{4}^{3} / Δ$ is a holomorphic $G$ -invariant function with $j (i) = 1728$ and $j (ω) = 0$ (The modular discriminant and the j-invariant).

Verification

1.1F1givenalgebra

Exactly two elliptic classes. Let $τ \in H$ have nontrivial stabiliser in $G$ . By [F1] there is $γ \in G$ with $γ \cdot τ \in \overline{D}$ , and $Stab (γ \cdot τ) = γ Stab (τ) γ^{- 1}$ is then nontrivial, so $γ \cdot τ \in {i, ω, ω + 1}$ by [F1]. Since $T \cdot ω = ω + 1$ , every point with nontrivial stabiliser is $G$ -equivalent to $i$ or to $ω$ . The classes of $i$ and $ω$ are distinct: if $ω = γ \cdot i$ , then conjugation would give $Stab (ω) = γ Stab (i) γ^{- 1}$ , a group of order $2$ , whereas $Stab (ω) = ⟨ S T ⟩$ has order $3$ by [F1]. So there are exactly two elliptic classes, the classes of $i$ and $ω$ , and their stabilisers are of orders $2$ and $3$ generated by $S$ and $S T$ ; every point outside these two classes has trivial stabiliser, since a point with nontrivial stabiliser is equivalent to $i$ or $ω$ and all other points of $\overline{D}$ have trivial stabiliser [F1].

2.1F2step 1.1givenalgebra

By [F2], the quotient map has local degree $2$ at $i$ and $3$ at $ω$ and $ω + 1$ . For every $γ \in G$ , $π \circ γ = π$ and $γ$ is a biholomorphism, so the local degree is unchanged at $γ i$ or $γ ω$ . Thus its ramification locus is exactly $G \cdot i \cup G \cdot ω$ , with local degrees $2$ and $3$ respectively; outside these orbits the stabiliser is trivial by 1.1 and the quotient chart is a local inverse of $π$ , giving degree $1$ .

3.1F3F4givenalgebra∎

Special values. At $τ = i$ , [F3] gives $E_{6} (i) = 0$ and $E_{4} (i) \neq 0$ , so $Δ (i) = (E_{4} (i)^{3} - 0) / 1728 = E_{4} (i)^{3} / 1728 \neq 0$ and $j (i) = E_{4} (i)^{3} / Δ (i) = 1728$ . At $τ = ω$ , [F3] gives $E_{4} (ω) = 0$ , so $j (ω) = 0 / Δ (ω) = 0$ , the denominator being nonzero by [F4]; the same two values are recorded in [F4].

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The first Fourier coefficients of E4, E6, Delta and j

Example

With $q = e^{2 π i τ}$ , $E_{4} = 1 + 240 q + 2160 q^{2} + 6720 q^{3} + \dots, E_{6} = 1 - 504 q - 16632 q^{2} - 122976 q^{3} + \dots,$ $Δ = q - 24 q^{2} + 252 q^{3} - 1472 q^{4} + \dots, j = q^{- 1} + 744 + 196884 q + 21493760 q^{2} + \dots .$ The coefficients are read from $E_{k} = 1 - \frac{2 k}{B_{k}} \sum σ_{k - 1} (n) q^{n}$ : $240 = - 8 / B_{4}$ , $- 504 = - 12 / B_{6}$ , and the higher coefficients from $σ_{3} (n), σ_{5} (n)$ and the binomial expansions.

Facts & Assumptions

Given: The expansion $E_{k} = 1 - \frac{2 k}{B_{k}} \sum_{n \geq 1} σ_{k - 1} (n) q^{n}$ of Eisenstein series are modular forms; their Fourier coefficients with $B_{4} = - 1 / 30$ , $B_{6} = 1 / 42$ (The Bernoulli numbers are defined by the generating series $t / (e^{t} - 1)$ ), the divisor sums of The divisor power sums $σ_{k}$ , and $Δ = q \prod (1 - q^{n})^{24} = (E_{4}^{3} - E_{6}^{2}) / 1728$ , $j = E_{4}^{3} / Δ$ (The Jacobi product formula for the discriminant, The discriminant is a nonvanishing cusp form of weight 12, The modular discriminant and the j-invariant).

[F1]

$σ_{1} (1) = 1$ and for prime powers $σ_{k} (p^{e}) = 1 + p^{k} + \dots + p^{k e}$ ; in particular $σ_{3} (1) = 1$ , $σ_{3} (2) = 9$ , $σ_{3} (3) = 28$ , $σ_{3} (4) = 73$ and $σ_{5} (1) = 1$ , $σ_{5} (2) = 33$ , $σ_{5} (3) = 244$ (The divisor power sums $σ_{k}$ ).

[F2]

Multiplication of modular forms adds weights and multiplies $q$ -expansions as absolutely convergent Cauchy products; the product formula for $Δ$ is available (Eisenstein series are modular forms; their Fourier coefficients, The Jacobi product formula for the discriminant).

Verification

1.1F1givenalgebra

By [F1] and the Bernoulli values, $240 = - \frac{8}{- 1 / 30}$ and $- 504 = - \frac{12}{1 / 42}$ , so $E_{4} = 1 + 240 q + 240 \cdot 9 q^{2} + 240 \cdot 28 q^{3} + \dots = 1 + 240 q + 2160 q^{2} + 6720 q^{3} + \dots$ and $E_{6} = 1 - 504 q - 504 \cdot 33 q^{2} - 504 \cdot 244 q^{3} + \dots = 1 - 504 q - 16632 q^{2} - 122976 q^{3} + \dots$ .

2.1F2step 1.1givenalgebra

Squaring and cubing these expansions by the binomial theorem [F2], $E_{4}^{3} = 1 + 720 q + 179280 q^{2} + 16954560 q^{3} + \dots$ and $E_{6}^{2} = 1 - 1008 q + 220752 q^{2} + 16519104 q^{3} + \dots$ ; hence $Δ = (E_{4}^{3} - E_{6}^{2}) / 1728 = q - 24 q^{2} + 252 q^{3} + O (q^{4})$ . The product formula gives $Δ = q \prod (1 - q^{n})^{24} = q (1 - 24 q + 252 q^{2} - 1472 q^{3} + \dots) = q - 24 q^{2} + 252 q^{3} - 1472 q^{4} + \dots$ , in agreement through $q^{3}$ and supplying the fourth coefficient.

3.1F2step 2.1givenalgebra∎

Dividing $E_{4}^{3}$ by $Δ = q P$ with $P = 1 - 24 q + 252 q^{2} - 1472 q^{3} + O (q^{4})$ , whose inverse begins $P^{- 1} = 1 + 24 q + 324 q^{2} + 3200 q^{3} + O (q^{4})$ , gives $q j = E_{4}^{3} P^{- 1} = 1 + 744 q + 196884 q^{2} + 21493760 q^{3} + O (q^{4})$ , that is $j = q^{- 1} + 744 + 196884 q + 21493760 q^{2} + \dots$ .

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There are no nonzero odd-weight level-one modular forms

Example

If $k$ is odd then $M_{k} = {0}$ and $S_{k} = {0}$ .

Facts & Assumptions

Given: An odd integer $k$ and an $f \in M_{k}$ (Level-one modular forms and cusp forms).

[F1]

$- I \in S L_{2} (Z)$ acts on $H$ as the identity and has $c = 0$ , $d = - 1$ , hence factor $(c τ + d)^{k} = (- 1)^{k}$ in the weight- $k$ transformation law (The modular group and its action on the upper half-plane, Level-one modular forms and cusp forms).

Verification

1.1F1givenalgebra

Applying the modular transformation law to $γ = - I$ gives $f (τ) = (- 1)^{k} f (τ) = - f (τ)$ for every $τ \in H$ , since $k$ is odd.

2.1step 1.1givenalgebra∎

Hence $2 f (τ) = 0$ for every $τ$ , so $f = 0$ ; thus $M_{k} = {0}$ for odd $k$ . Since a cusp form of weight $k$ is in particular a modular form of weight $k$ , also $S_{k} = {0}$ .

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The square and hexagonal tori have j-invariants 1728 and 0

Example

For the square lattice $Λ_{i} = Z + Z i$ and the hexagonal lattice $Λ_{ω} = Z + Z ω$ , $ω = e^{2 π i / 3}$ : $j (Λ_{i}) = 1728, j (Λ_{ω}) = 0.$ Moreover $j (Λ) = 1728$ if and only if $Λ$ is homothetic to $Z [i]$ , and $j (Λ) = 0$ if and only if $Λ$ is homothetic to $Z [ω]$ . Multiplication by $i$ (respectively $ω$ ) induces an automorphism of the corresponding torus of order $4$ (respectively $3$ ).

Facts & Assumptions

Given: The lattices $Λ_{i} = Z + Z i$ and $Λ_{ω} = Z + Z ω$ with $ω = e^{2 π i / 3}$ , their tori $T_{Λ} = C / Λ$ and class maps $π_{Λ}$ (Complex lattice and quotient torus, The quotient $C / Λ$ is a compact Riemann surface); the modular function $j$ of The modular discriminant and the j-invariant; and the zeros of $E_{4}, E_{6}$ (The zeros of E4 and E6 at the elliptic points).

[F1]

For a full lattice $Λ$ with oriented basis $(ω_{1}, ω_{2})$ the value $j (Λ) : = j (ω_{2} / ω_{1})$ is independent of the choice of oriented basis; moreover $Λ^{'} = c Λ$ for some $c \in C^{\times}$ , biholomorphy of $T_{Λ}$ and $T_{Λ^{'}}$ , and $j (Λ^{'}) = j (Λ)$ are equivalent (The j-invariant classifies complex tori).

[F2]

$j (i) = 1728$ and $j (ω) = 0$ ; on $H$ the function $j$ is holomorphic and $P S L_{2} (Z)$ -invariant, and $j (τ) = j (τ^{'})$ only when $τ, τ^{'}$ lie in one $P S L_{2} (Z)$ -class (The modular discriminant and the j-invariant, The j-invariant uniformizes X(1), The zeros of E4 and E6 at the elliptic points).

[F3]

The class map $π_{Λ} : C \to T_{Λ}$ is a holomorphic covering; for $a \in C^{\times}$ with $a Λ \subseteq Λ$ the formula $m_{a} ([z]) : = [a z]$ is well defined on $T_{Λ}$ because $z - w \in Λ$ implies $a (z - w) \in Λ$ , and $m_{a}$ is holomorphic since $π_{Λ} \circ (a \cdot) = m_{a} \circ π_{Λ}$ ; when $a Λ = Λ$ it is a biholomorphism with inverse $m_{a^{- 1}}$ (The quotient $C / Λ$ is a compact Riemann surface, Complex lattice and quotient torus).

Verification

1.1F1F2givenalgebra

Oriented bases. The pair $(1, i)$ is an oriented basis of $Λ_{i} = Z + Z i$ : $i \notin R$ and $Im (i / 1) = 1 > 0$ . Hence $Λ_{i}$ has parameter $τ = i$ and, by [F1] and the first value in [F2], $j (Λ_{i}) = j (i) = 1728$ . Likewise $(1, ω)$ is an oriented basis of $Λ_{ω} = Z + Z ω$ , since $Im ω = \sin (2 π / 3) = \sqrt{3} / 2 > 0$ , so $Λ_{ω}$ has parameter $ω$ and $j (Λ_{ω}) = j (ω) = 0$ by [F2].

2.1F1F2step 1.1givenalgebra

The level sets of $1728$ and $0$ . Let $Λ$ be a full lattice. By [F1], $j (Λ) = j (Λ_{i})$ holds if and only if $Λ$ is homothetic to $Λ_{i}$ , and by 1.1 the value on the right is $j (Λ_{i}) = 1728$ ; hence $j (Λ) = 1728$ if and only if $Λ$ is homothetic to $Z [i] = Λ_{i}$ . The same argument with $Λ_{ω}$ gives: by [F1] and 1.1, $j (Λ) = j (Λ_{ω}) = 0$ if and only if $Λ$ is homothetic to $Λ_{ω} = Z [ω]$ . Equivalently, both statements say that the level set of each of the two special values is a single homothety class, as also follows from the injectivity of $j$ on $P S L_{2} (Z)$ -classes recorded in [F2].

3.1F3givenalgebra∎

Automorphisms of order $4$ and $3$ . Multiplication by $i$ preserves $Λ_{i}$ : $i (m + n i) = - n + m i$ with $- n, m \in Z$ ; hence $i Λ_{i} = Λ_{i}$ (equality, since $i$ is invertible and $i Λ_{i} \subseteq Λ_{i}$ with the same argument applied to $- i$ ), and [F3] provides the biholomorphic automorphism $m_{i} ([z]) = [i z]$ of $T_{Λ_{i}}$ . Its fourth power is the identity, $m_{i}^{4} = m_{i^{4}} = m_{1} = id$ , while $m_{i}^{2} = m_{- 1}$ is not the identity because $[1 / 3] \neq [- 1 / 3]$ : their difference would require $2 / 3 \in Λ_{i}$ , and no element $m + n i$ of $Λ_{i}$ with $n = 0$ equals $2 / 3 \notin Z$ ; so the order of $m_{i}$ divides $4$ but not $2$ , hence is exactly $4$ . Similarly $ω Λ_{ω} \subseteq Λ_{ω}$ because $ω \cdot 1 = ω$ and $ω \cdot ω = ω^{2} = - 1 - ω$ lie in $Λ_{ω}$ , and $ω Λ_{ω} = Λ_{ω}$ by the same inverse argument with $ω^{- 1} = ω^{2}$ ; so [F3] gives the automorphism $m_{ω} ([z]) = [ω z]$ . Its cube is $m_{ω}^{3} = m_{ω^{3}} = m_{1} = id$ since $ω^{3} = 1$ , and $m_{ω}$ is not the identity because $[ω / 2] \neq [1 / 2]$ : $(ω - 1) / 2 = - \frac{1}{2} + \frac{1}{2} ω$ is not of the form $m + n ω$ with $m, n \in Z$ ; so the order of $m_{ω}$ divides $3$ but is not $1$ , hence is exactly $3$ .

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The principal congruence subgroup Gamma(2)

Definition

The principal congruence subgroup of level $2$ is

$Γ (2) : = {γ \in S L_{2} (Z) : γ \equiv I (m o d 2)},$

the congruence $γ \equiv I (m o d 2)$ being entrywise (Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ , The congruence class $[a]_{n}$ and the quotient set $Z / n$ , The modular group and its action on the upper half-plane).

It is the kernel of the entrywise reduction homomorphism $ρ : S L_{2} (Z) \to S L_{2} (F_{2})$ : reduction of entries is a group homomorphism because addition and multiplication of residues are compatible with the operations on $Z$ , and it lands in $S L_{2} (F_{2})$ because $\det γ = 1$ reduces to $1$ (Monoid homomorphism and group homomorphism, The kernel and image of a group homomorphism). Hence $Γ (2)$ is a normal subgroup of $S L_{2} (Z)$ (First isomorphism theorem for groups: $G / \ker f ≅ im f$ ) and it contains $\pm I$ , since $- 1 \equiv 1 (m o d 2)$ .

The reduction is surjective. Indeed $S L_{2} (F_{2}) = G L_{2} (F_{2})$ (the only nonzero scalar in $F_{2}$ is $1$ ), and it has order $6$ : its first column is any of the three nonzero vectors of $F_{2}^{2}$ , and then the second column is any of the two vectors outside the span of the first, the resulting matrix being automatically invertible. The images $\overset{ˉ}{S}, \overset{ˉ}{T}$ of $S, T \in S L_{2} (Z)$ lie in the image of $ρ$ and generate $S L_{2} (F_{2})$ : ${\overset{ˉ}{T}}^{2} = 1$ , $(\overset{ˉ}{S} \overset{ˉ}{T})^{3} = 1$ by reduction of $S^{2} = (S T)^{3} = - I \equiv I$ , and $\overset{ˉ}{S} \notin ⟨ \overset{ˉ}{T} ⟩$ , so $⟨ \overset{ˉ}{S}, \overset{ˉ}{T} ⟩$ has order divisible by $2$ and $3$ and at most $6$ , hence equals $S L_{2} (F_{2})$ (The modular group and its action on the upper half-plane). So $ρ$ is onto, and the first isomorphism theorem identifies $S L_{2} (Z) / Γ (2)$ with $S L_{2} (F_{2})$ ; therefore

$[S L_{2} (Z) : Γ (2)] = ∣ S L_{2} (F_{2}) ∣ = 6$

(First isomorphism theorem for groups: $G / \ker f ≅ im f$ , If $[G : N]$ is finite then $∣ G / N ∣ = [G : N]$ ; for finite $G$ this equals $∣ G ∣ / ∣ N ∣$ , For $N ⊴ G$ , the cosets form a group with identity $N$ and inverse $(g N)^{- 1} = g^{- 1} N$ ).

Finally let $\overset{ˉ}{Γ} (2) \leq P S L_{2} (Z)$ be the image of $Γ (2)$ under the quotient map $S L_{2} (Z) \to P S L_{2} (Z)$ . Since $ρ$ kills $- I$ , it factors through that quotient and defines a surjective homomorphism $P S L_{2} (Z) \to S L_{2} (F_{2})$ whose kernel is exactly $\overset{ˉ}{Γ} (2)$ . As $Γ (2)$ contains $\ker (S L_{2} (Z) \to P S L_{2} (Z)) = {\pm I}$ , the correspondence of subgroups in the quotient gives $[P S L_{2} (Z) : \overset{ˉ}{Γ} (2)] = [S L_{2} (Z) : Γ (2)] = 6$ , and $\overset{ˉ}{Γ} (2)$ is the kernel of that reduction (First isomorphism theorem for groups: $G / \ker f ≅ im f$ ).

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The projective group $\overset{ˉ}{Γ} (2)$ is torsion-free and acts freely

Statement

The only torsion elements of $Γ (2)$ are $\pm I$ ; equivalently $\overset{ˉ}{Γ} (2) = Γ (2) / {\pm I}$ is torsion-free and has no elliptic fixed points on $H$ . In particular $Γ (2)$ has no elliptic points and every point of $H$ has trivial stabiliser in $\overset{ˉ}{Γ} (2)$ .

Facts & Assumptions

Given: $Γ (2) = {γ \in S L_{2} (Z) : γ \equiv I (m o d 2)}$ with image $\overset{ˉ}{Γ} (2) \leq P S L_{2} (Z)$ (The principal congruence subgroup Gamma(2), Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ); the action of $P S L_{2} (Z)$ on $H$ (The modular group and its action on the upper half-plane).

[F1]

A matrix $γ = (\begin{matrix} a & b \\ c & d \end{matrix}) \in Γ (2)$ satisfies $a \equiv d \equiv 1 (m o d 2)$ and $b \equiv c \equiv 0 (m o d 2)$ , and $a d - b c = 1$ (The principal congruence subgroup Gamma(2), Congruence modulo an integer: $a \equiv b (m o d n)$ when $n ∣ (a - b)$ , including the moduli $0$ and $1$ ).

[F2]

A nonidentity element of $P S L_{2} (R)$ represented by $A \in S L_{2} (R)$ is either parabolic (conjugate to a nonzero translation) or, with two fixed points on $\hat{C}$ , is conjugate to $z \mapsto λ z$ , $λ \neq 0, 1$ , and $tr (A)^{2} = λ + 2 + λ^{- 1}$ ; it is elliptic exactly when $∣ λ ∣ = 1$ (Nonidentity Möbius transformations are parabolic or conjugate to a dilation, with the projective trace invariant).

[F3]

Order and torsion in a group; a nonidentity element of $P S L_{2} (R)$ with a fixed point in $H$ is conjugate to a rotation $z \mapsto e^{i θ} z$ , and every point of $H$ with nontrivial stabiliser in $P S L_{2} (Z)$ is $P S L_{2} (Z)$ -equivalent to $i$ , $ω$ or $ω + 1$ (The order $∣ G ∣$ of a finite group and the order $ord (g)$ of an element, with $ord (g) = \infty$ when no positive power of $g$ is the identity, The standard fundamental domain, boundary identifications and elliptic stabilisers, Nonidentity Möbius transformations are parabolic or conjugate to a dilation, with the projective trace invariant).

Proof

1.1F1F2givenalgebra

Suppose $γ \in Γ (2)$ has a nontrivial finite-order class in $P S L_{2} (Z)$ . By [F2] it cannot be parabolic, since a nonzero translation has infinite order. Thus it is conjugate to $z \mapsto λ z$ with $λ \neq 1$ a root of unity, and $(tr γ)^{2} = λ + 2 + λ^{- 1} = 2 + 2 \cos θ < 4$ . The trace $a + d$ is an even integer by [F1], so $∣ a + d ∣ < 2$ forces $a + d = 0$ . Hence $d = - a$ , where $a$ is odd, and the determinant equation gives $b c = - a^{2} - 1 \equiv 2 (m o d 4)$ . But $b, c$ are even, so $b c \equiv 0 (m o d 4)$ , a contradiction. Therefore a finite-order projective class is trivial and its representative is $\pm I$ . In particular the only torsion matrices in $Γ (2)$ are $\pm I$ .

2.1F3step 1.1givenalgebra∎

For any $τ \in H$ , its $P S L_{2} (Z)$ -stabiliser is conjugate to a subgroup of the finite stabilisers in the standard domain [F3]. Thus every matrix in $Γ (2)$ fixing $τ$ has a finite-order projective class. By 1.1 it is $\pm I$ , so its class in $\overset{ˉ}{Γ} (2)$ is the identity. Hence $\overset{ˉ}{Γ} (2)$ is torsion-free and acts freely on $H$ , and there are no elliptic points modulo scalars.

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The modular lambda function

Definition

For $τ \in H$ let $Λ_{τ} : = Z + Z τ$ (Complex lattice and quotient torus) and let $℘_{Λ_{τ}}$ be its Weierstrass function (Weierstrass p function). Put

$e_{1} : = ℘_{Λ_{τ}} (\frac{1}{2}), e_{2} : = ℘_{Λ_{τ}} (\frac{τ}{2}), e_{3} : = ℘_{Λ_{τ}} (\frac{1 + τ}{2}),$

the three finite branch values of $℘_{Λ_{τ}}$ (Degree two of ℘ and its four branch points, clause 3). Those values are pairwise distinct: the discriminant $Δ (Λ_{τ}) = g_{2}^{3} - 27 g_{3}^{2}$ is nonzero and $4 x^{3} - g_{2} x - g_{3}$ has the three distinct roots $e_{1}, e_{2}, e_{3}$ (Nonvanishing of the lattice discriminant). The modular lambda function is

$λ (τ) : = \frac{e_{3} - e_{2}}{e_{1} - e_{2}} \in C ∖ {0, 1} .$

The value lies in $C ∖ {0, 1}$ because $e_{1}, e_{2}, e_{3}$ are pairwise distinct, so numerator $e_{3} - e_{2}$ and difference $e_{1} - e_{2}$ are nonzero and $e_{3} - e_{2} \neq e_{1} - e_{2}$ . Equivalently, in the cross-ratio convention of The cross-ratio of an ordered quadruple of sphere points,

$λ (τ) = [\infty, e_{2}; e_{1}, e_{3}] = \frac{e_{2} - e_{3}}{e_{2} - e_{1}},$

matching the displayed formula: the ordered quadruple $(\infty, e_{1}, e_{2}, e_{3})$ of branch points of the associated Weierstrass cubic determines $λ (τ)$ up to the Möbius transformations fixing $\infty$ . The half-plane conventions are those of The unit disc, the upper half-plane, and Blaschke factors.

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Transformation laws and S_3-action of the modular lambda function

Statement

$λ$ is invariant under $Γ (2)$ , and under the generators of $P S L_{2} (Z)$ it satisfies $λ (τ + 1) = \frac{λ (τ)}{λ (τ) - 1}, λ (- 1 / τ) = 1 - λ (τ) .$ Consequently $λ (γ τ)$ takes, as $γ$ ranges over $P S L_{2} (Z)$ , the values of the six expressions $λ, \frac{1}{λ}, 1 - λ, \frac{1}{1 - λ}, \frac{λ}{λ - 1}, \frac{λ - 1}{λ},$ these expressions may coincide at special parameters, and the substitution action on rational functions defines an isomorphism $P S L_{2} (Z) / \overset{ˉ}{Γ} (2) ≅ S_{3}$ . Moreover $λ (i) = 1 / 2$ , and for $τ = i y$ with $y > 0$ one has $λ (τ) \in (0, 1)$ .

Facts & Assumptions

Given: $λ (τ) = \frac{e_{3} - e_{2}}{e_{1} - e_{2}}$ with $e_{1} = ℘_{Λ_{τ}} (1 / 2)$ , $e_{2} = ℘_{Λ_{τ}} (τ / 2)$ , $e_{3} = ℘_{Λ_{τ}} ((1 + τ) / 2)$ , $Λ_{τ} = Z + Z τ$ , and the $e_{j}$ pairwise distinct (The modular lambda function, Degree two of ℘ and its four branch points, Nonvanishing of the lattice discriminant, Complex lattice and quotient torus).

[F1]

$℘_{Λ}$ is even and $Λ$ -periodic, and its convergence is normal in the point for a fixed lattice; its parameter continuity used below is established by a local compact bound (Weierstrass p function, Normal convergence, parity and periodicity of the Weierstrass p function, Degree two of ℘ and its four branch points).

[F2]

For $c \in C^{\times}$ , $℘_{c Λ} (c z) = c^{- 2} ℘_{Λ} (z)$ : substituting $ω = c ω^{'}$ in the defining series scales every corrected summand by $c^{- 2}$ , and the family is absolutely summable (Weierstrass p function).

[F3]

$℘^{'}$ vanishes exactly at the nonzero half-periods, and $4 x^{3} - g_{2} x - g_{3} = 4 (x - e_{1}) (x - e_{2}) (x - e_{3})$ has the three distinct roots $e_{1}, e_{2}, e_{3}$ , so $e_{1} + e_{2} + e_{3} = 0$ and $g_{3} = 4 e_{1} e_{2} e_{3}$ (Weierstrass cubic differential equation, Nonvanishing of the lattice discriminant).

[F4]

$Γ (2) = {γ \equiv I (m o d 2)}$ has index $6$ in $S L_{2} (Z)$ and image $\overset{ˉ}{Γ} (2)$ of index $6$ in $P S L_{2} (Z) = ⟨ S, T ⟩$ ; $S^{2} = (S T)^{3} = 1$ (The principal congruence subgroup Gamma(2), The standard fundamental domain, boundary identifications and elliptic stabilisers, The modular group and its action on the upper half-plane).

[F5]

An action of a group by permutations defines a homomorphism with kernel the intersection of all point stabilisers; isomorphic groups satisfy the usual group-isomorphism conditions (Actions of $G$ on $X$ correspond exactly to homomorphisms $G \to Sym (X)$ , Group isomorphisms, automorphisms and the set $Aut (G)$ ).

[F6]

$∣ ℘ (z) ∣$ and $℘ (z)$ itself are continuous in the pair (lattice, $z$ ) on compacta away from the lattice, by the following compact estimate: for $τ$ in a compact subset of $H$ , the corrected lattice summands at each half-period are bounded by $C \max (∣ m ∣, ∣ n ∣)^{- 3}$ outside finitely many pairs. This follows from $∣ m τ + n ∣ \geq c \max (∣ m ∣, ∣ n ∣)$ and expanding $(z - ω)^{- 2} - ω^{- 2}$ for bounded $z$ . Summing over shells gives a uniform $\sum_{j} O (j^{- 2})$ bound; each finite term is holomorphic in $τ$ and no half-period meets the lattice, proving parameter holomorphy and hence continuity by the Weierstrass convergence theorem (Locally uniform limits of holomorphic functions are holomorphic and their derivatives converge locally uniformly); conjugation of the lattice $Λ_{i y}$ to itself gives $℘ (\overset{ˉ}{z}) = \overline{℘ (z)}$ for that lattice (Normal convergence, parity and periodicity of the Weierstrass p function, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

Proof

1.1F1F2givenalgebra

Let $γ = (\begin{matrix} a & b \\ c & d \end{matrix}) \in Γ (2)$ , so $a \equiv d \equiv 1$ and $b \equiv c \equiv 0 (m o d 2)$ . Put $ω_{1} : = a τ + b$ , $ω_{2} : = c τ + d$ , so $γ τ = ω_{1} / ω_{2}$ and $Λ_{γ τ} = (1 / ω_{2}) Λ_{τ}$ because $Z + Z (ω_{1} / ω_{2}) = (1 / ω_{2}) (Z ω_{1} + Z ω_{2}) = (1 / ω_{2}) Λ_{τ}$ . By [F2], $℘_{Λ_{γ τ}} (z) = ω_{2}^{2} ℘_{Λ_{τ}} (ω_{2} z)$ . Now $ω_{1} - τ = (a - 1) τ + b \in 2 Λ_{τ}$ and $ω_{2} - 1 = c τ + (d - 1) \in 2 Λ_{τ}$ ; hence the half-periods $1 / 2$ , $γ τ / 2 = ω_{1} / (2 ω_{2})$ and $(1 + γ τ) / 2 = (ω_{1} + ω_{2}) / (2 ω_{2})$ of $Λ_{γ τ}$ correspond under the scaling to $ω_{2} / 2$ , $ω_{1} / 2$ and $(ω_{1} + ω_{2}) / 2$ , which are congruent modulo $Λ_{τ}$ to $\frac{1}{2}$ , $\frac{τ}{2}$ and $\frac{1 + τ}{2}$ by [F1] and the evenness of $℘$ . Therefore $λ (γ τ) = \frac{ω_{2}^{2} (e_{3} - e_{2})}{ω_{2}^{2} (e_{1} - e_{2})} = λ (τ)$ .

1.2F1F2givenalgebra

For $T$ one has $Λ_{τ + 1} = Λ_{τ}$ and the half-period values of the basis $(1, τ + 1)$ are $e_{1}$ , $℘ ((τ + 1) / 2) = e_{3}$ , $℘ ((τ + 2) / 2) = e_{2}$ , so $λ (τ + 1) = \frac{e_{2} - e_{3}}{e_{1} - e_{3}} = \frac{e_{3} - e_{2}}{(e_{3} - e_{2}) - (e_{1} - e_{2})} = \frac{λ (τ)}{λ (τ) - 1}$ . For $S$ one has $S τ = - 1 / τ$ and $Λ_{S τ} = (1 / τ) Λ_{τ}$ , so by [F2] with $c = 1 / τ$ the scaled $℘$ is $℘_{Λ_{S τ}} (z) = τ^{2} ℘_{Λ_{τ}} (τ z)$ ; the half-periods $1 / 2 = c \cdot (τ / 2)$ , $S τ / 2 = c \cdot (- 1 / 2)$ , $(1 + S τ) / 2 = c \cdot ((τ - 1) / 2)$ of $Λ_{S τ}$ therefore give the values $e_{1}^{'} = τ^{2} ℘ (τ / 2) = τ^{2} e_{2}$ , $e_{2}^{'} = τ^{2} ℘ (- 1 / 2) = τ^{2} e_{1}$ and $e_{3}^{'} = τ^{2} ℘ ((τ - 1) / 2) = τ^{2} ℘ ((1 + τ) / 2) = τ^{2} e_{3}$ ; hence $λ (- 1 / τ) = \frac{e_{3} - e_{1}}{e_{2} - e_{1}} = 1 - λ (τ)$ .

1.3F1F2F3F6givenalgebra

The square lattice $Λ_{i} = Z + Z i$ is invariant under multiplication by $i$ , and the substitution $(m, n) \mapsto (- n, m)$ is a bijection of $Z^{2}$ sending $m i + n$ to $i (m i + n)$ , so the absolutely summable family $((m i + n)^{- 6})$ [F3] equals its negative and $g_{3} = 140 G_{6} = 0$ for $τ = i$ ; also $℘_{Λ_{i}} (i z) = - ℘_{Λ_{i}} (z)$ by [F2] with $c = i$ , so $e_{2} = ℘ (i / 2) = - ℘ (1 / 2) = - e_{1}$ , and then [F3] gives $e_{3} = 0$ . Hence $λ (i) = \frac{e_{3} - e_{2}}{e_{1} - e_{2}} = \frac{e_{1}}{2 e_{1}} = \frac{1}{2}$ . For $τ = i y$ , $y > 0$ , the lattice $Λ_{i y}$ is invariant under conjugation: the conjugate of $m i y + n$ is $- m i y + n \in Λ_{i y}$ , so $℘ (\overset{ˉ}{z}) = \overline{℘ (z)}$ [F6]; the half-periods $1 / 2$ , $i y / 2$ , $(1 + i y) / 2$ are each congruent to their conjugates modulo $Λ_{i y}$ , so $e_{1}, e_{2}, e_{3}$ are real and $λ (i y)$ is real, while $λ (i y) \neq 0, 1$ because the $e_{j}$ stay distinct [F3]. By [F6] each $e_{j} (i y)$ is continuous in $y$ ; $λ (i y)$ is therefore a continuous real function of $y \in (0, \infty)$ avoiding $0$ and $1$ , so it lies in a single connected component of $R ∖ {0, 1}$ ; since $λ (i) = 1 / 2 \in (0, 1)$ , it follows that $λ (i y) \in (0, 1)$ for every $y > 0$ .

2.1F4F5F6step 1.1step 1.2step 1.3givenalgebra∎

Let $X$ be the set of the six rational functions $x$ , $1 / x$ , $1 - x$ , $1 / (1 - x)$ , $x / (x - 1)$ , $(x - 1) / x$ of an indeterminate $x$ ; these are pairwise distinct functions on $C ∖ {0, 1}$ , and $σ (x) : = x / (x - 1)$ and $τ (x) : = 1 - x$ satisfy $σ^{2} = τ^{2} = 1$ and generate a group of order $6$ acting transitively on $X$ (the orbit of $x$ is exactly $X$ ), hence isomorphic to $S_{3}$ . By 1.2, $λ (S τ) = 1 - λ (τ) = τ (λ (τ))$ and $λ (T τ) = σ (λ (τ))$ , and for a word $γ$ in $S, T$ induction gives $λ (γ τ) = f_{γ} (λ (τ))$ with $f_{γ}$ the corresponding composition in this group; since $λ (i) = 1 / 2$ and $λ (i + 1) = - 1$ by 1.2 and 1.3, it is not constant; its real restriction $λ (i y)$ cannot be constant by the identity theorem, so its continuous image is an interval with more than one point by the intermediate value theorem (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ). Thus distinct $f_{γ}$ give distinct functions $f_{γ} \circ λ$ , so the assignment $γ \mapsto f_{γ}$ is a homomorphism $P S L_{2} (Z) \to S_{3}$ [F5]. Its kernel is ${γ : λ (γ τ) = λ (τ)}$ , which contains $\overset{ˉ}{Γ} (2)$ by 1.1 and therefore has index at most $6$ ; the image is generated by $σ, τ$ and has order $6$ , so the index is exactly $6$ and the kernel is $\overset{ˉ}{Γ} (2)$ , giving $P S L_{2} (Z) / \overset{ˉ}{Γ} (2) ≅ S_{3}$ [F4]. Hence $λ (γ τ)$ runs over the displayed expressions, with coincidences allowed (for example $λ (i) = 1 / 2$ gives the three values $1 / 2, 2, - 1$ ) as $γ$ runs over $P S L_{2} (Z)$ .

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The fibres of lambda are exactly the Gamma(2)-orbits

Statement

If $τ, τ^{'} \in H$ and $λ (τ) = λ (τ^{'})$ , then $τ^{'} = γ \cdot τ$ for some $γ \in Γ (2)$ . In particular $λ$ separates the $Γ (2)$ -orbits on $H$ .

Facts & Assumptions

Given: $λ (τ) = \frac{e_{3} - e_{2}}{e_{1} - e_{2}}$ with $e_{1} = ℘_{Λ_{τ}} (1 / 2)$ , $e_{2} = ℘_{Λ_{τ}} (τ / 2)$ , $e_{3} = ℘_{Λ_{τ}} ((1 + τ) / 2)$ , and $e_{1}, e_{2}, e_{3}$ pairwise distinct with $e_{1} + e_{2} + e_{3} = 0$ (The modular lambda function, Degree two of ℘ and its four branch points, Nonvanishing of the lattice discriminant, Complex lattice and quotient torus).

[F1]

$℘ (z) = ℘ (w)$ if and only if $w \equiv z$ or $w \equiv - z$ modulo the lattice (Degree two of ℘ and its four branch points); in particular on the $2$ -torsion classes $\pm h$ coincide.

[F2]

The invariants $g_{2} = 60 G_{4}$ , $g_{3} = 140 G_{6}$ satisfy $℘^{' 2} = 4 ℘^{3} - g_{2} ℘ - g_{3}$ , and $4 x^{3} - g_{2} x - g_{3} = 4 (x - e_{1}) (x - e_{2}) (x - e_{3})$ with $e_{1} + e_{2} + e_{3} = 0$ , so $g_{2} = - 4 (e_{1} e_{2} + e_{1} e_{3} + e_{2} e_{3})$ and $g_{3} = 4 e_{1} e_{2} e_{3}$ ; moreover $G_{4} (c Λ) = c^{- 4} G_{4} (Λ)$ and $G_{6} (c Λ) = c^{- 6} G_{6} (Λ)$ for $c \in C^{\times}$ , directly from the defining absolutely summable series (Weierstrass cubic differential equation, Weierstrass p function).

[F3]

The map $[z] \mapsto [℘_{Λ} (z) : ℘_{Λ}^{'} (z) : 1]$ , extended at $[0]$ to $O = [0 : 1 : 0]$ , is a biholomorphism from the torus to the smooth cubic $Y^{2} Z = 4 X^{3} - g_{2} X Z^{2} - g_{3} Z^{3}$ (The torus is biholomorphic to its Weierstrass cubic). A nonzero complex number has a square root (Every complex number has a square root, by an explicit Cartesian formula).

[F5]

$γ \in Γ (2)$ exactly when $γ \equiv I (m o d 2)$ , i.e. its two columns are congruent to $(1, 0)$ and $(0, 1)$ modulo $2$ ; and $λ (γ τ) = λ (τ)$ (The principal congruence subgroup Gamma(2), Transformation laws and S_3-action of the modular lambda function).

[F6]

The torus class maps are holomorphic coverings; maps from the simply connected plane lift uniquely after a basepoint is fixed, and every entire biholomorphism is affine (The quotient $C / Λ$ is a compact Riemann surface, Every nonempty convex subset of $R^{n}$ is simply connected, Lifting criterion for maps from path-connected locally path-connected spaces, Every biholomorphic self-map of the complex plane is affine).

Proof

1.1F2givenalgebra

Put $L_{j} = e_{j} (τ)$ and $L_{j}^{'} = e_{j} (τ^{'})$ . If $λ (τ) = λ (τ^{'})$ then with $u : = L_{3} - L_{2}$ , $v : = L_{1} - L_{2}$ , $u^{'} : = L_{3}^{'} - L_{2}^{'}$ , $v^{'} : = L_{1}^{'} - L_{2}^{'}$ we have $u / v = u^{'} / v^{'}$ , so $u^{'} v = u v^{'}$ . Hence $(L_{3}^{'} - L_{2}^{'}) (L_{1} - L_{2}) = (L_{3} - L_{2}) (L_{1}^{'} - L_{2}^{'})$ , i.e. $(L_{3}^{'} - L_{2}^{'}) (L_{1} - L_{2}) - (L_{3} - L_{2}) (L_{1}^{'} - L_{2}^{'}) = 0$ , a determinant condition; the affine map $Φ (x) = α x + β$ with $α : = \frac{L_{2}^{'} - L_{1}^{'}}{L_{2} - L_{1}}$ and $β : = L_{1}^{'} - α L_{1}$ satisfies $Φ (L_{1}) = L_{1}^{'}$ and $Φ (L_{2}) = L_{2}^{'}$ , and the displayed identity says exactly $Φ (L_{3}) = L_{3}^{'}$ . Since $\sum L_{j} = \sum L_{j}^{'} = 0$ we get $β = \frac{1}{3} (\sum L_{j}^{'}) - α \frac{1}{3} (\sum L_{j}) = 0$ , so $L_{j}^{'} = α L_{j}$ for $j = 1, 2, 3$ with $α \neq 0$ .

2.1F2step 1.1givenalgebra

Choose a square root $a$ of $α^{- 1}$ and put $Λ^{*} : = a Λ_{τ}$ . By [F2], $G_{4} (Λ^{*}) = a^{- 4} G_{4} (Λ_{τ}) = α^{2} G_{4} (Λ_{τ})$ and $G_{6} (Λ^{*}) = a^{- 6} G_{6} (Λ_{τ}) = α^{3} G_{6} (Λ_{τ})$ , hence $g_{2} (Λ^{*}) = α^{2} g_{2} (τ)$ and $g_{3} (Λ^{*}) = α^{3} g_{3} (τ)$ . On the other hand $e_{j} (τ^{'}) = α e_{j} (τ)$ by 1.1 and the invariants are the elementary symmetric functions of the three branch values [F2], so $g_{2} (τ^{'}) = α^{2} g_{2} (τ)$ and $g_{3} (τ^{'}) = α^{3} g_{3} (τ)$ as well; therefore $Λ^{*}$ and $Λ_{τ^{'}}$ have the same invariants $g_{2}, g_{3}$ .

3.1F1F2F3F6step 2.1givenconstruct

By 2.1 the lattices $Λ^{*}$ and $Λ_{τ^{'}}$ have the same invariants, so their Weierstrass cubics are identical. Their biholomorphisms [F3] to this cubic induce a biholomorphism of tori fixing the origin and matching the three labelled half-periods: the branch values for $Λ^{*} = a Λ_{τ}$ are $a^{- 2} L_{j} = α L_{j} = L_{j}^{'}$ . Lift this map and its inverse to based maps of $C$ using the holomorphic lattice coverings, the lifting criterion and simple connectedness of $C$ ; uniqueness of based lifts makes the lifts inverse biholomorphisms. The entire-biholomorphism theorem gives a lift $z \mapsto b z$ , $b \neq 0$ . Therefore multiplication by $A : = b a$ carries $Λ_{τ}$ onto $Λ_{τ^{'}}$ and matches the labelled half-periods modulo these lattices. This labelled homothety, rather than the false Laurent recursion previously recorded, is sufficient for the final congruence calculation.

4.1F1F2F5step 3.1givenalgebra∎

Since $Λ_{τ^{'}} = A Λ_{τ}$ , the numbers $A$ and $A τ$ form a positively oriented basis of $Λ_{τ^{'}}$ (multiplication by $A$ preserves orientation), so $A = p + q τ^{'}$ and $A τ = r + s τ^{'}$ for integers $p, q, r, s$ forming a matrix $γ_{0} = (\begin{matrix} p & r \\ q & s \end{matrix}) \in S L_{2} (Z)$ . The labelled homothety from 3.1 gives $A h_{j} (τ) \equiv h_{j} (τ') (m o d Λ_{τ'})$ for $j = 1, 2, 3$ . Taking $j = 1$ gives $p + q τ^{'} = A \equiv 1 (m o d 2 Λ_{τ^{'}})$ , so $p$ is odd and $q$ even; taking $j = 2$ gives $A τ = r + s τ^{'} \equiv τ^{'} (m o d 2 Λ_{τ^{'}})$ , so $r$ is even and $s$ odd; hence $γ_{0} \equiv I (m o d 2)$ [F5]. Finally $A τ = r + s τ^{'}$ and $A = p + q τ^{'}$ give $τ^{'} = \frac{p τ - r}{s - q τ}$ , that is $τ^{'} = γ \cdot τ$ for $γ = (\begin{matrix} p & - r \\ - q & s \end{matrix})$ , which has determinant $p s - q r = 1$ and entries congruent to $I$ modulo $2$ , so $γ \in Γ (2)$ [F5]. Conversely $λ$ is $Γ (2)$ -invariant [F5], so the fibres of $λ$ are exactly the $Γ (2)$ -orbits.

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The j-invariant of the Legendre normal form

Statement

For a full lattice $Λ_{τ}$ with invariants $g_{2}, g_{3}$ one has $g_{2} = \frac{4 π^{4}}{3} E_{4} (τ)$ and $g_{3} = \frac{8 π^{6}}{27} E_{6} (τ)$ , hence $\frac{1728 g_{2}^{3}}{g_{2}^{3} - 27 g_{3}^{2}} = j (τ) .$ Writing $λ = λ (τ)$ and putting $J_{L e g} (x) : = \frac{256 (x^{2} - x + 1)^{3}}{x^{2} (x - 1)^{2}}$ for $x \in C ∖ {0, 1}$ , the affine normalisation of the associated Legendre cubic gives $\frac{1728 g_{2}^{3}}{g_{2}^{3} - 27 g_{3}^{2}} = J_{L e g} (λ (τ)) = j (τ);$ in particular $J_{L e g} (1 - λ) = J_{L e g} (1 / λ) = J_{L e g} (λ)$ .

Facts & Assumptions

Given: $Λ_{τ} = Z + Z τ$ , its invariants $g_{2} = 60 G_{4}$ , $g_{3} = 140 G_{6}$ , the cubic relation $℘^{' 2} = 4 ℘^{3} - g_{2} ℘ - g_{3} = 4 (℘ - e_{1}) (℘ - e_{2}) (℘ - e_{3})$ with distinct $e_{j}$ , and $λ (τ) = \frac{e_{3} - e_{2}}{e_{1} - e_{2}}$ (Weierstrass cubic differential equation, Degree two of ℘ and its four branch points, The modular lambda function).

[F1]

$G_{k} = 2 ζ (k) E_{k}$ . The unnormalised Fourier coefficient computed in Eisenstein series are modular forms; their Fourier coefficients, Proof 1.2, is $(- 2 π i)^{k} / ((k - 1)! ζ (k))$ ; comparing its values $240$ for $k = 4$ and $- 504$ for $k = 6$ gives $ζ (4) = π^{4} / 90$ and $ζ (6) = π^{6} / 945$ (The level-one Eisenstein series E_k and the weight-two series E_2, The Lipschitz formula for the reciprocal-power sums).

[F2]

The lattice discriminant $D_{Λ} : = g_{2}^{3} - 27 g_{3}^{2}$ is nonzero because the roots $e_{j}$ are distinct. The normalised modular discriminant is $Δ = (E_{4}^{3} - E_{6}^{2}) / 1728$ , with $j = E_{4}^{3} / Δ$ ; these are different normalisations, related in step 1.1 (Degree two of ℘ and its four branch points, The discriminant is a nonvanishing cusp form of weight 12, The modular discriminant and the j-invariant).

[F3]

Writing the Weierstrass cubic $Y^{2} = 4 x^{3} - g_{2} x - g_{3}$ as $y^{2} = x^{3} + A x + B$ uses $Y = 2 y$ , $A = - g_{2} / 4$ , $B = - g_{3} / 4$ . Hence $1728 \cdot 4 A^{3} / (4 A^{3} + 27 B^{2}) = 1728 g_{2}^{3} / (g_{2}^{3} - 27 g_{3}^{2})$ . Choose $h$ with $h^{2} = d \neq 0$ (Every complex number has a square root, by an explicit Cartesian formula). Under $x = d X$ , $y = h^{3} \tilde{y}$ , the coefficients become $A / d^{2}, B / d^{3}$ , so both numerator and denominator acquire $d^{- 6}$ and this ratio is unchanged.

[F4]

The affine map $x \mapsto (x - e_{2}) / (e_{1} - e_{2})$ carries the branch triple $(e_{1}, e_{2}, e_{3})$ to $(1, 0, λ)$ ; cross-ratios and the labelling of branch points are preserved by affine maps (The modular lambda function, The cross-ratio is invariant under Möbius transformations, Degree two of ℘ and its four branch points).

Proof

1.1F1F2givenalgebra

By [F1], $g_{2} = 60 G_{4} = 120 ζ (4) E_{4} = 120 \cdot \frac{π^{4}}{90} E_{4} = \frac{4 π^{4}}{3} E_{4}$ and $g_{3} = 140 G_{6} = 280 ζ (6) E_{6} = 280 \cdot \frac{π^{6}}{945} E_{6} = \frac{8 π^{6}}{27} E_{6}$ ; hence $g_{2}^{3} = \frac{64 π^{12}}{27} E_{4}^{3}$ and $27 g_{3}^{2} = 27 \cdot \frac{64 π^{12}}{729} E_{6}^{2} = \frac{64 π^{12}}{27} E_{6}^{2}$ , so $g_{2}^{3} - 27 g_{3}^{2} = \frac{64 π^{12}}{27} (E_{4}^{3} - E_{6}^{2}) = \frac{64 π^{12}}{27} \cdot 1728 Δ$ by [F2]. Dividing, $\frac{1728 g_{2}^{3}}{g_{2}^{3} - 27 g_{3}^{2}} = \frac{E_{4}^{3}}{Δ} = j (τ)$ .

2.1F3F4step 1.1givenalgebra

Put $d = e_{1} - e_{2} \neq 0$ and choose $h$ with $h^{2} = d$ . In $Y^{2} = 4 (x - e_{1}) (x - e_{2}) (x - e_{3})$ , the substitutions $x = e_{2} + d u$ , $Y = 2 h^{3} v$ give $v^{2} = u (u - 1) (u - λ)$ by [F4]. After the original Weierstrass cubic is written with leading coefficient one, the translation by $e_{2}$ and subsequent centring cancel each other, while the dilation divides the centred coefficients $- g_{2} / 4, - g_{3} / 4$ by $d^{2}, d^{3}$ . Thus its invariant ratio is unchanged by [F3]. Completing the cube by $x = X + \frac{1 + λ}{3}$ gives $y^{2} = X^{3} + A X + B$ with $A = λ - \frac{(1 + λ)^{2}}{3} = - \frac{λ^{2} - λ + 1}{3}$ and $B = \frac{(1 + λ) (9 λ - 2 (1 + λ)^{2})}{27} = - \frac{(λ^{2} - λ - 2) (2 λ - 1)}{27}$ . Therefore $4 A^{3} = - \frac{4 (λ^{2} - λ + 1)^{3}}{27}$ and $27 B^{2} = \frac{(λ^{2} - λ - 2)^{2} (2 λ - 1)^{2}}{27}$ , and the algebraic identity $4 (λ^{2} - λ + 1)^{3} - (λ^{2} - λ - 2)^{2} (2 λ - 1)^{2} = 27 λ^{2} (λ - 1)^{2}$ , which holds for all $λ$ by expanding both sides, gives $4 A^{3} + 27 B^{2} = - λ^{2} (λ - 1)^{2}$ . Hence $J : = \frac{1728 \cdot 4 A^{3}}{4 A^{3} + 27 B^{2}} = \frac{1728 \cdot (- 4 (λ^{2} - λ + 1)^{3} / 27)}{- λ^{2} (λ - 1)^{2}} = \frac{256 (λ^{2} - λ + 1)^{3}}{λ^{2} (λ - 1)^{2}} = J_{L e g} (λ)$ .

3.1F3step 2.1givenalgebra∎

Invariance under the cross-ratio substitutions: $J_{L e g} (1 - λ) = J_{L e g} (λ)$ because $(1 - λ)^{2} - (1 - λ) + 1 = λ^{2} - λ + 1$ and $(1 - λ)^{2} ((1 - λ) - 1)^{2} = λ^{2} (λ - 1)^{2}$ ; and $J_{L e g} (1 / λ) = J_{L e g} (λ)$ because $λ^{- 2} - λ^{- 1} + 1 = (λ^{2} - λ + 1) λ^{- 2}$ and $(λ^{- 2}) (λ^{- 1} - 1)^{2} = λ^{2} (λ - 1)^{2} λ^{- 6}$ . Combining 1.1 and 2.1, $\frac{1728 g_{2}^{3}}{g_{2}^{3} - 27 g_{3}^{2}} = j (τ) = J_{L e g} (λ (τ))$ , which is the assertion.

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The modular lambda function: Y(2) biholomorphic to the twice-punctured plane, and the slit-plane quadrilateral

Example

The modular lambda function induces a biholomorphism $\overset{ˉ}{λ} : Y (2) = H / Γ (2) ⟶ C ∖ {0, 1} .$ Let $Q^{\circ} = {τ \in H : ∣ ℜ τ ∣ < 1, ∣ τ - 1 / 2 ∣ > 1 / 2, ∣ τ + 1 / 2 ∣ > 1 / 2}$ be the interior of the standard ideal quadrilateral with vertices $\infty, - 1, 0, 1$ . Its restriction is a biholomorphism $λ : Q^{\circ} ⟶ C ∖ ((- \infty, 0] \cup [1, \infty)) .$ Moreover $λ : H \to C ∖ {0, 1}$ is a regular covering whose deck group is $\overset{ˉ}{Γ} (2) = Γ (2) / {\pm I}$ acting freely and simply transitively on every fibre.

Facts & Assumptions

Given: $λ (τ) = \frac{e_{3} - e_{2}}{e_{1} - e_{2}}$ with the $e_{j}$ the half-period values of $℘_{Λ_{τ}}$ (The modular lambda function); the quadrilateral $Q^{\circ}$ and the slit plane $C ∖ ((- \infty, 0] \cup [1, \infty))$ . The matrices $T^{2} = (\begin{matrix} 1 & 2 \\ 0 & 1 \end{matrix})$ , $γ_{+} = (\begin{matrix} 1 & 0 \\ 2 & 1 \end{matrix})$ , $γ_{-} = (\begin{matrix} - 1 & 0 \\ 2 & - 1 \end{matrix})$ lie in $Γ (2)$ , and $L_{\pm} = (\begin{matrix} 1 & 0 \\ \pm 1 & 1 \end{matrix})$ satisfy $L_{+} (i y) = \frac{i y}{1 + i y}$ , $L_{-} (i y) = \frac{i y}{1 - i y}$ on the imaginary axis (The principal congruence subgroup Gamma(2), The modular group and its action on the upper half-plane).

[F1]

On compact subsets of $H$ the normally convergent $℘$ -series is uniformly controlled by the lattice estimate $∣ m τ + n ∣ \geq C \max (∣ m ∣, ∣ n ∣)$ , so the half-period values $e_{j} (τ)$ and hence $λ (τ)$ are holomorphic functions of $τ$ ; conjugation of the same series gives $λ (- \overset{ˉ}{τ}) = \overline{λ (τ)}$ (Normal convergence, parity and periodicity of the Weierstrass p function, Locally uniform limits of holomorphic functions are holomorphic and their derivatives converge locally uniformly, Reduction of orbits to the standard domain, The modular lambda function).

[F2]

$λ$ is $Γ (2)$ -invariant, satisfies $λ (τ + 1) = \frac{λ}{λ - 1}$ , $λ (- 1 / τ) = 1 - λ$ , takes the values of the six expressions (which may coincide) $λ, \frac{1}{λ}, 1 - λ, \frac{1}{1 - λ}, \frac{λ}{λ - 1}, \frac{λ - 1}{λ}$ under $P S L_{2} (Z)$ , and $0 < λ (i y) < 1$ for $y > 0$ (Transformation laws and S_3-action of the modular lambda function).

[F3]

$λ (τ) = λ (τ^{'})$ implies $τ^{'} \in Γ (2) τ$ (The fibres of lambda are exactly the Gamma(2)-orbits); $\overset{ˉ}{Γ} (2)$ is torsion-free and acts freely with local quotient charts (The projective group $\overset{ˉ}{Γ} (2)$ is torsion-free and acts freely, Local charts and the Riemann surface structure of a modular quotient, 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, Covering maps, evenly covered neighbourhoods, fibres, sheets, and trivial coverings).

[F4]

$j : X (1) \to \hat{C}$ is a biholomorphism and $j (τ) = J_{L e g} (λ (τ)) = R (λ (τ))$ with $R (x) = J_{L e g} (x) = \frac{256 (x^{2} - x + 1)^{3}}{x^{2} (x - 1)^{2}}$ (The j-invariant uniformizes X(1), The j-invariant of the Legendre normal form).

[F5]

An injective holomorphic map of Riemann surfaces is biholomorphic onto its open image (An injective holomorphic map has no critical point and is biholomorphic onto its image, Local power-map normal form on Riemann surfaces, Biholomorphic maps between complex domains).

[F6]

For a covering with connected total space, deck transformations are determined by their value at one point and act freely (On a connected covering space, a deck transformation is determined by one point and the deck action is free, Deck transformations and the deck-transformation group of a covering).

Verification

1.1F1F2F3givenalgebra

$λ$ is holomorphic on $H$ and satisfies $λ (- \overset{ˉ}{τ}) = \overline{λ (τ)}$ by [F1]; by the $Γ (2)$ -invariance of [F2] and the local quotient charts of [F3], it descends to a holomorphic function $\overset{ˉ}{λ}$ on $Y (2) = H / Γ (2)$ , which takes values in $C ∖ {0, 1}$ because the $e_{j}$ are always distinct.

1.2F2F3givenalgebra

Reduction to the quadrilateral. Fix $τ \in H$ . The set $S = {∣ c τ + d ∣ : γ = (\begin{matrix} a & b \\ c & d \end{matrix}) \in Γ (2)}$ contains $1$ (the identity), and the pairs with $∣ c τ + d ∣ \leq 1$ are finite by Reduction of orbits to the standard domain; hence $S$ has a least positive element $m$ , realized by some $γ_{0} \in Γ (2)$ , and $τ_{0} = γ_{0} τ$ has maximal imaginary part in the $Γ (2)$ -orbit. Applying a power of $T^{2} \in Γ (2)$ , which adds an even integer and does not change the height, we may assume $∣ ℜ τ_{0} ∣ \leq 1$ . Maximality forces $∣ 2 τ_{0} + 1 ∣ \geq 1$ and $∣ 2 τ_{0} - 1 ∣ \geq 1$ , since otherwise $γ_{+} τ_{0}$ or $γ_{-} τ_{0}$ would have strictly larger imaginary part; thus $τ_{0}$ lies in the closure of $Q^{\circ}$ , and every $Γ (2)$ -orbit meets that closure.

2.1F3F4step 1.1givenalgebra

$\overset{ˉ}{λ}$ is injective by [F3]: if $\overset{ˉ}{λ}$ agrees at two classes, the underlying $λ$ -values agree and the points lie in one $Γ (2)$ -orbit. It is surjective: let $z \in C ∖ {0, 1}$ and put $R (x) = \frac{256 (x^{2} - x + 1)^{3}}{x^{2} (x - 1)^{2}}$ . Since $j : X (1) \to \hat{C}$ is onto [F4], there is $τ \in H$ with $j (τ) = R (z)$ (the value is finite, so it is attained off the cusp); then $R (λ (τ)) = j (τ) = R (z)$ by [F4]. Put $H (x) = (x^{2} - x + 1)^{3}$ and $D (x) = x^{2} (x - 1)^{2}$ . For $u \neq 0, 1$ , clearing denominators gives $D (u) H (v) - H (u) D (v) = 0$ , a degree-six polynomial in $v$ with nonzero leading coefficient $D (u)$ . Let $f_{1} (u), \dots, f_{6} (u)$ be the six substitutions of [F2]. The identity $D (u) H (v) - H (u) D (v) = D (u) \prod_{r = 1}^{6} (v - f_{r} (u))$ holds first for generic $u$ , where the six roots are distinct by direct substitution and the leading coefficients agree. It then holds for every $u \neq 0, 1$ , since each coefficient is a rational function of $u$ and an identity outside finitely many exceptional values is a rational-function identity. Thus the same factorisation handles the repeated roots at special parameters, and its root set is exactly the displayed substitutions; so $z$ is one of $λ (τ), \frac{1}{λ (τ)}, 1 - λ (τ), \frac{1}{1 - λ (τ)}, \frac{λ (τ)}{λ (τ) - 1}, \frac{λ (τ) - 1}{λ (τ)}$ . By [F2] each of these is $λ (γ τ)$ for some $γ \in P S L_{2} (Z)$ , so $z$ lies in the image of $\overset{ˉ}{λ}$ .

2.2F2F3step 1.2givenalgebra

Uniqueness and the interior. Let $γ = (\begin{matrix} a & b \\ c & d \end{matrix}) \in Γ (2)$ with $c \neq 0$ , so $c$ is even and $d$ is odd, and consider the open disc ${τ : ∣ c τ + d ∣ < 1} = {τ : ∣ τ + d / c ∣ < 1 / ∣ c ∣}$ . Its centre $- d / c$ is not $- 1, 0$ or $1$ (it is not an integer), and if it lies in $(- 1, 0)$ or in $(0, 1)$ then its distances to the endpoints of the corresponding interval are at least $1 / ∣ c ∣$ , since each is a nonzero integer divided by $∣ c ∣$ , so the disc lies inside the boundary disc with diameter $[- 1, 0]$ or $[0, 1]$ ; if the centre lies outside $[- 1, 1]$ , its centre is at distance at least $1 / ∣ c ∣$ from the strip $∣ ℜ τ ∣ \leq 1$ . Hence $∣ c τ + d ∣ \geq 1$ on the closure of $Q^{\circ}$ and $∣ c τ + d ∣ > 1$ on $Q^{\circ}$ . If $c = 0$ then $γ = \pm T^{2 k}$ is translation by an even integer, and two points of the closure with $∣ ℜ ∣ \leq 1$ differ by $2 k$ with $∣ k ∣ \leq 1$ ; for $k = \pm 1$ both have real part $\mp 1$ , a boundary value. Therefore no two distinct points of $Q^{\circ}$ are $Γ (2)$ -equivalent, and no point of $Q^{\circ}$ is equivalent to a boundary point: two interior points related by $γ$ with $c \neq 0$ would give $Im (γ τ) < Im τ$ and, applying the same to $γ^{- 1}$ and $γ τ$ in the closure, the reverse weak inequality. This also excludes an interior-to-boundary identification.

3.1F2F3F5F6step 1.1step 2.1givenalgebra

By 1.1 and 2.1 the holomorphic map $\overset{ˉ}{λ} : Y (2) \to C ∖ {0, 1}$ is bijective, hence biholomorphic by [F5] (injectivity forces local degree one everywhere). The quotient map $π_{2} : H \to Y (2)$ is a covering by [F3], so $λ = \overset{ˉ}{λ} \circ π_{2}$ is a covering. The total space $H$ is connected: the straight segment between any two points stays in $H$ . Every $\overset{ˉ}{Γ} (2)$ element is a deck transformation by [F2]. Conversely, for a deck transformation $h$ and a fixed $τ_{0} \in H$ , [F3] gives $γ \in \overset{ˉ}{Γ} (2)$ with $h (τ_{0}) = γ τ_{0}$ ; connected-cover uniqueness [F6] then gives $h = γ$ . Thus the deck group is exactly $\overset{ˉ}{Γ} (2)$ , acting transitively on each fibre by [F3] and freely by [F6], hence simply transitively; the covering is regular.

3.2F2F5step 1.1step 1.2step 2.2givenalgebra

Boundary values and the slit plane. The vertical boundary edges are $T^{\pm 1} (i y) = i y \pm 1$ . The rational substitution $s (x) = x / (x - 1)$ is its own inverse, so both edges have value $s (λ (i y)) < 0$ . The semicircular edges are $L_{\pm} (i y)$ , where $L_{+} = T S T$ in $P S L_{2} (Z)$ and $L_{-} = L_{+}^{- 1}$ . The substitution for $L_{+}$ is $s \circ (1 - x) \circ s = 1 / x$ , also its own inverse; hence both semicircular edges have value $1 / λ (i y) > 1$ . Hence boundary values avoid the slit plane. Now let $w$ lie in the slit plane. Since $\overset{ˉ}{λ}$ is onto, $w = λ (τ^{'})$ for some $τ^{'}$ , whose orbit meets the closure of $Q^{\circ}$ by 1.2; choose $τ$ in that closure with $λ (τ) = w$ by $Γ (2)$ -invariance. By the boundary computation, $τ$ is not on the vertical or semicircular boundary (those values are negative or greater than $1$ , while $w \notin (- \infty, 0] \cup [1, \infty)$ ), so $τ \in Q^{\circ}$ . Thus $λ (Q^{\circ})$ contains the slit plane. Conversely, if $τ \in Q^{\circ}$ and $λ (τ)$ is real, then $- \overset{ˉ}{τ} \in Q^{\circ}$ and $λ (- \overset{ˉ}{τ}) = \overline{λ (τ)} = λ (τ)$ by 1.1, so $τ = - \overset{ˉ}{τ}$ by 2.2, that is $τ = i y$ , and then $λ (τ) \in (0, 1)$ . Hence $λ (Q^{\circ})$ is contained in the slit plane, the restriction is bijective onto it, and being injective holomorphic it is biholomorphic onto the slit plane by [F5].

4.1F2step 3.2givenalgebra∎

Consistency check: $λ (1 + i) = λ (T (i)) = \frac{λ (i)}{λ (i) - 1} = \frac{1 / 2}{- 1 / 2} = - 1$ , a value on the vertical boundary, so $1 + i$ is not an interior point of $Q^{\circ}$ and its image is not in the slit plane; this is exactly the boundary behaviour that distinguishes the image of the quadrilateral from the twice-punctured plane.

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FALSE: the weight-two Eisenstein series E_2 is a modular form

Statement

The function $E_{2} (τ) = 1 - 24 \sum_{n \geq 1} σ_{1} (n) q^{n}$ satisfies the weight-two transformation law $f (τ + 1) = f (τ)$ , and its completion $E_{2}^{*} (τ) = E_{2} (τ) - \frac{3}{π ℑ τ}$ is real-analytic and transforms like a weight-two form, so one might expect $E_{2}$ itself to be a modular form of weight $2$ . This is false.

Facts & Assumptions

Given: $E_{2} (τ) = 1 - 24 \sum_{n \geq 1} σ_{1} (n) q^{n}$ on $H$ (The level-one Eisenstein series E_k and the weight-two series E_2), and the weight-two transformation law of Level-one modular forms and cusp forms.

[F1]

$E_{2} (- 1 / τ) = τ^{2} E_{2} (τ) - \frac{6 i}{π} τ$ for every $τ \in H$ (The transformation law of the weight-two Eisenstein series E_2, The modular group and its action on the upper half-plane).

[F2]

A modular form $f$ of weight $2$ satisfies $f (- 1 / τ) = τ^{2} f (τ)$ (Level-one modular forms and cusp forms).

Refutation

1.1F1givenalgebra

Evaluating [F1] at $τ = i$ , where $- 1 / i = i$ and $i^{2} = - 1$ , gives $E_{2} (i) = i^{2} E_{2} (i) - \frac{6 i}{π} i = - E_{2} (i) + \frac{6}{π}$ , because $1 / i = - i$ and $i^{2} = - 1$ ; hence $2 E_{2} (i) = \frac{6}{π}$ and therefore $E_{2} (i) = \frac{3}{π} \neq 0$ .

2.1F1F2step 1.1givenalgebra∎

A weight-two modular form would satisfy, by [F2] at $τ = i$ , the equation $f (i) = i^{2} f (i) = - f (i)$ , hence $f (i) = 0$ ; but $E_{2} (i) = 3 / π \neq 0$ by 1.1. Moreover [F1] shows directly that $E_{2} (- 1 / i) = E_{2} (i) = 3 / π$ while $i^{2} E_{2} (i) = - 3 / π$ , so $E_{2} (- 1 / τ) \neq τ^{2} E_{2} (τ)$ at $τ = i$ . Hence $E_{2}$ is not a modular form of weight $2$ ; the correctly transforming object is the non-holomorphic completion $E_{2}^{*}$ .

Level-One Modular Forms and the j-Invariant — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

The standard fundamental domain tessellates the upper half-plane

Example

Facts & Assumptions

Verification

The elliptic points of the modular group and their images under j

Example

Facts & Assumptions

Verification

The first Fourier coefficients of E4, E6, Delta and j

Example

Facts & Assumptions

Verification

There are no nonzero odd-weight level-one modular forms

Example

Facts & Assumptions

Verification

The square and hexagonal tori have j-invariants 1728 and 0

Example

Facts & Assumptions

Verification

The principal congruence subgroup Gamma(2)

Definition

The projective group Γˉ(2)Γˉ(2) is torsion-free and acts freely

Statement

Facts & Assumptions

Proof

The modular lambda function

Definition

Transformation laws and S_3-action of the modular lambda function

Statement

Facts & Assumptions

Proof

The fibres of lambda are exactly the Gamma(2)-orbits

Statement

Facts & Assumptions

Proof

The j-invariant of the Legendre normal form

Statement

Facts & Assumptions

Proof

The modular lambda function: Y(2) biholomorphic to the twice-punctured plane, and the slit-plane quadrilateral

Example

Facts & Assumptions

Verification

FALSE: the weight-two Eisenstein series E_2 is a modular form

Statement

Facts & Assumptions

Refutation

Sources

The projective group $\overset{ˉ}{Γ} (2)$ is torsion-free and acts freely