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Elliptic Functions and Complex Tori: Examples and Counterexamples

1 · Prerequisites

2 · Summary

These computations make the constructions of the companion page explicit on lattices that can be manipulated by hand. The examples begin with lattice bookkeeping: the reduced ratio $Im τ > 0$ , $- \frac{1}{2} < Re τ \leq \frac{1}{2}$ , $∣ τ ∣ \geq 1$ , with $Re τ \geq 0$ when $∣ τ ∣ = 1$ , is shown to exist and be unique for every lattice, with stabiliser counts two, four and six at the square and hexagonal ratios, and the oriented bases of $Z + i Z$ are enumerated with their ${S L}_{2} (Z)$ change-of-basis data. A shifted fundamental parallelogram is exhibited whose boundary misses every pole and every zero of a given nonconstant elliptic function, so the divisor laws of the companion page can always be applied without a boundary degeneracy.

The invariant computations are symmetry arguments rather than numerical evaluations: $i Λ_{s q} = Λ_{s q}$ forces $g_{3} (Z + i Z) = 0$ and $ρ Λ_{h e x} = Λ_{h e x}$ forces $g_{2} (Z + Z ρ) = 0$ , while the nonvanishing of the complementary invariant follows from $Δ \neq 0$ . On the square lattice the scaling identity $℘ (i z) = - ℘ (z)$ then gives $℘ (\frac{1 + i}{2}) = 0$ , and the remaining two half-period values are the two nonzero roots of $4 x^{3} - g_{2} x$ ; the four branch values of the associated degree-two map are $0, \pm \sqrt{g_{2}} / 2, \infty$ . The same identities specialise to the duplication formula for $℘ (2 z)$ , the simple zero of $σ$ at each lattice point, and the degenerate singular cubic with $g_{2} = 3$ , $g_{3} = 1$ , whose node at $(- 1 / 2, 0)$ shows that $Δ \neq 0$ cannot be dropped.

The rectangular case turns the real locus of $℘$ into an explicit conformal map: the boundary of a half-period rectangle is carried onto the extended real line, the interior onto a half-plane, and the inverse is an elliptic integral $\int d ζ / \sqrt{4 ζ^{3} - g_{2} ζ - g_{3}}$ with the classical relation to the Jacobi function $s n$ . Finally the rank-one analogue $π \cot (π z)$ is uniformised by the conic $Y Z = X^{2} + π^{2} Z^{2}$ through the bijection $C / Z \to C^{*}$ , $z \mapsto e^{2 π i z}$ — the rank-one counterpart of the cubic uniformisation of the companion page.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Oriented bases and ${S L}_{2} (Z)$

Example

Let $Λ = Z + i Z = {m + n i : m, n \in Z}$ .

The pairs $(1, i)$ and $(1 + i, i)$ are bases of $Λ$ of positive complex orientation and differ by an integer change-of-basis matrix of determinant $1$ ;
the pair $(i, 1)$ is a basis of $Λ$ of negative orientation, and it differs from $(1, i)$ by an integer matrix of determinant $- 1$ ;
all three pairs are bases of the same lattice, so they all define the same complex torus $C / Λ$ and the same compact Riemann surface.

Facts & Assumptions

Given: The lattice $Λ = Z + i Z$ , whose elements are the numbers $m + n i$ with $m, n \in Z$ and whose real-linear independence datum is $i / 1 = i \notin R$ .

[F1]

A basis of a lattice $Λ$ is a pair $(ω_{1}, ω_{2})$ with $Λ = Z ω_{1} + Z ω_{2}$ and $ω_{1}, ω_{2}$ real-linearly independent; it is oriented when $Im (ω_{2} / ω_{1}) > 0$ ; two bases of the same lattice differ by a matrix in ${G L}_{2} (Z)$ , and two oriented bases by a matrix in ${S L}_{2} (Z)$ ; the quotient torus and all its structure depend on the set $Λ$ alone (Complex lattice and quotient torus).

[F2]

For a full lattice $Λ$ the quotient $T_{Λ} = C / Λ$ is a compact Riemann surface with the quotient topology of the class map $π$ , which is a holomorphic covering map (The quotient $C / Λ$ is a compact Riemann surface).

Verification

1.1F1

The pair $(1, i)$ is a basis of $Λ$ : by definition $Λ = Z \cdot 1 + Z \cdot i$ , and $1, i$ are real-linearly independent since $i \notin R$ ; its orientation is positive because $Im (i / 1) = 1 > 0$ .

1.2F1algebra

The pair $(1 + i, i)$ is a basis of the same lattice: $1 + i, i \in Λ$ , so $Z (1 + i) + Z i \subseteq Λ$ , while $1 = (1 + i) - i$ and $i = i$ show the reverse inclusion; the change-of-basis matrix, whose columns are the new vectors in the old basis, expressing $(1 + i, i) = (ω_{1} + ω_{2}, ω_{2})$ in the basis $(1, i)$ is $(\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix})$ , of determinant $1$ , so the orientation is positive as well, and independently $Im (i / (1 + i)) = Im ((1 + i) / 2) = \frac{1}{2} > 0$ .

1.3F1algebra

The pair $(i, 1)$ is a basis of $Λ$ with change-of-basis matrix $(\begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix})$ relative to $(1, i)$ , of determinant $- 1$ ; its orientation is negative because $Im (1 / i) = Im (- i) = - 1 < 0$ .

2.1F1F2step 1.1step 1.2step 1.3∎

By steps 1.1, 1.2 and 1.3 the three pairs are bases of the same lattice $Λ$ , so they give the same quotient $C / Λ$ and the same lattice sums, and by [F2] this quotient is a compact Riemann surface with holomorphic covering map $π$ , independently of which basis is used to describe $Λ$ .

The example illustrates that orientation is a property of an ordered basis, not of the lattice: $(1, i)$ and $(1 + i, i)$ are related by the unipotent matrix $(\begin{matrix} 1 & 0 \\ 1 & 1 \end{matrix}) \in {S L}_{2} (Z)$ , whereas swapping the two vectors multiplies the orientation sign by $- 1$ .

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Moving the boundary of a fundamental parallelogram

Example

Let $Λ = Z + i Z$ and let $℘ = ℘_{Λ}$ be its Weierstrass function. For $b \in C$ put

$P_{b} : = {b + s + t i : 0 \leq s, t \leq 1}, \partial P_{b} = {b + s + t i : 0 \leq s, t \leq 1, s \in {0, 1} or t \in {0, 1}} .$

Then:

the translate with $a = (1 + i) / 4$ has no pole of $℘$ on $\partial P_{a}$ : the points of $\partial P_{a}$ have real part or imaginary part equal to $1 / 4$ or $5 / 4$ , so none of them is a lattice point;
the unshifted parallelogram $P_{0}$ has poles of $℘$ on its boundary: its four vertices $0, 1, i, 1 + i$ are lattice points and belong to $\partial P_{0}$ ;
for every nonzero $Λ$ -elliptic meromorphic function $f$ the translates whose boundary avoids the zeros and poles of $f$ are generic: in every nonempty open ball of basepoints $b$ there is a $b$ with $\partial P_{b} \cap (Zer (f) \cup Pol (f)) = \emptyset$ . In particular the divisor of $f$ on the torus $T_{Λ}$ has only finitely many classes modulo $Λ$ , and the parallelogram hypotheses of the divisor law can always be met.

Facts & Assumptions

Given: The lattice $Λ = Z + i Z$ , the parallelograms $P_{b}$ and their boundaries $\partial P_{b}$ as displayed, the Weierstrass function $℘ = ℘_{Λ}$ , and a nonzero $Λ$ -elliptic meromorphic function $f$ with zero set $Zer (f)$ and pole set $Pol (f)$ ; also $Z : = Zer (f) \cup Pol (f)$ and $P_{0} = P_{0}$ .

[F1]

$Λ = Z + i Z = {m + n i : m, n \in Z}$ is a full complex lattice, since $i \notin R$ gives real-linear independence; for all $z, w \in C$ one has $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ , $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ , and $∣ z ∣ = 0$ exactly for $z = 0$ (Complex lattice and quotient torus, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive); $C$ is the real coordinate plane, so its closed bounded subsets are compact and its compact subsets are bounded ( $C$ is the real coordinate plane, with coordinate arithmetic, 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 openness and continuity are the metric notions for $d (z, w) = ∣ z - w ∣$ (The Euclidean metric, convergence, Cauchy sequences, and continuity on the complex plane).

[F2]

$℘$ is holomorphic on $C ∖ Λ$ ; at each lattice point it has a double pole with principal part $(z - λ)^{- 2}$ and it has no other poles (Normal convergence, parity and periodicity of the Weierstrass p function).

[F3]

Let $g$ be a nonconstant $Λ$ -elliptic meromorphic function and let $P = {c + s + t i : 0 \leq s, t \leq 1}$ be a translate whose boundary contains no zero and no pole of $g$ . Then the numbers of zeros and of poles of $g$ in the interior of $P$ , counted with multiplicity, are finite and equal, the sum of the residues of $g$ at its poles in the interior vanishes, and both numbers are independent of the translate among such $P$ ; a $Λ$ -elliptic function with no poles is constant (Divisor and residue laws for elliptic functions).

[F4]

A $Λ$ -elliptic function is a meromorphic $f : C \to \hat{C}$ with $f (z + λ) = f (z)$ for all $z, λ$ , and a meromorphic function on $C$ is one holomorphic on the complement of its pole set, every point of which is a pole (Elliptic function for a lattice).

[F5]

Let $g$ be meromorphic on a plane domain with pole set $P$ . Then every point of the domain has a neighbourhood meeting $P$ in at most one point, and $P$ is closed in the domain (Poles of a meromorphic function form a closed discrete set and are at most countable). If $g$ is not identically zero then $1 / g$ is meromorphic, with pole set the zero set of $g$ (Meromorphic functions on a connected plane domain form a field).

[F6]

For every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ , its integer part (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

Verification

1.1F1F2algebra

(The shifted parallelogram is boundary-free for $℘$ .) Every point of $\partial P_{a}$ with $a = (1 + i) / 4$ has the form $z = a + s + t i$ with $(s, t) \in {0, 1} \times [0, 1] \cup [0, 1] \times {0, 1}$ , so $Re z \in {1 / 4, 5 / 4}$ or $Im z \in {1 / 4, 5 / 4}$ . A lattice point of $Λ$ has integer real and imaginary parts, hence cannot have real part or imaginary part equal to $1 / 4$ or $5 / 4$ ; so $\partial P_{a} \cap Λ = \emptyset$ . By [F2] the poles of $℘$ are exactly the lattice points, so $℘$ has no pole on $\partial P_{a}$ .

1.2F2given

(The unshifted parallelogram meets the poles.) The four vertices of $P_{0}$ are $0, 1, i, 1 + i$ , all of which lie in $Λ$ and in $\partial P_{0}$ ; by [F2] each is a double pole of $℘$ .

1.3F1F4F5F6given

(Finitely many divisor classes.) Suppose first that $f$ is nonconstant. By [F5] applied to $f$ , the pole set $Pol (f)$ is closed and every point of $C$ has a neighbourhood meeting it in at most one point; by [F5] applied to $1 / f$ , whose pole set is $Zer (f)$ , the same holds for the zero set. Hence every point of $C$ has a neighbourhood meeting $Z$ in at most two points, and since $P_{0}$ is compact by [F1] finitely many of these neighbourhoods cover $P_{0}$ ; therefore $Z \cap P_{0} = {z_{1}, \dots, z_{N}}$ is finite. By [F4] the function $f$ is $Λ$ -periodic, so $Z$ is $Λ$ -invariant; conversely every $z \in Z$ can be written $z = s + t i$ with real $s, t$ , and subtracting the integer parts $m \leq s < m + 1$ , $n \leq t < n + 1$ given by [F6] puts $z - (m + n i) \in Z \cap P_{0}$ . Hence $Z = ⋃_{j = 1}^{N} (z_{j} + Λ)$ : the zeros and poles of $f$ fall into the finitely many classes of $z_{1}, \dots, z_{N}$ modulo $Λ$ .

2.1F1F4givenstep 1.3choose

(Generic translates avoid the divisor.) Let $U$ be a nonempty open ball of basepoints; we show that some $b \in U$ has $\partial P_{b} \cap Z = \emptyset$ , treating the constant case at the end. Fix the finite list of step 1.3 and put $B_{j} : = z_{j} - \partial P_{0}$ for $j = 1, \dots, N$ ; each $B_{j}$ is closed, being a translate of $\partial P_{0}$ , and contains no ball: if a ball $B (x, r)$ were contained in a segment with direction vector $q \neq 0$ , then with $u : = i q \neq 0$ perpendicular to $q$ the two points $x$ and $x + \frac{r}{2 ∣ u ∣} u$ of the ball would both lie on the line of the segment, whose direction is $q$ , forcing the nonzero vector $\frac{r}{2 ∣ u ∣} u$ to be a real multiple of $q$ and contradicting $u ⊥ q$ . Consequently no ball is contained in $\partial P_{0}$ or in any of its translates, which are finite unions of segments: if the closed sets $F_{1}, \dots, F_{m}$ each contain no ball and a ball $B \subseteq F_{1} \cup \dots \cup F_{m}$ , then either $B \subseteq F_{1}$ or there is $y \in B ∖ F_{1}$ , and since $F_{1}$ is closed some ball $B (y, ρ) \subseteq B ∖ F_{1} \subseteq F_{2} \cup \dots \cup F_{m}$ ; iterating gives a ball inside some $F_{i}$ , a contradiction. Hence each $B_{j}$ is closed with no ball inside it. For a basepoint $b \in U$ one has $\partial P_{b} \cap Z \neq \emptyset$ iff $b \in B_{j} + λ$ for some $j$ and some $λ \in Λ$ : indeed $z \in \partial P_{b} \cap Z$ means $z = z_{j} + λ = b + w$ with $w \in \partial P_{0}$ , that is $b = z_{j} + λ - w$ . For $b \in U$ such a $λ$ must satisfy $∣ λ ∣ \leq ∣ b - z_{j} ∣ + ∣ w ∣ \leq C$ for a constant $C$ depending only on $U$ and the finitely many $z_{j}$ by [F1], so only finitely many $λ \in Λ$ occur, and the set of bad basepoints in $U$ is the finite union $(⋃_{j = 1}^{N} ⋃_{λ \in Λ, ∣ λ ∣ \leq C} (B_{j} + λ)) \cap U$ of closed sets containing no ball. A finite union of sets each containing no ball does not contain $U$ : successively avoiding each closed member leaves a nonempty open subball, as in the induction just described. Hence there is $b \in U$ with $\partial P_{b} \cap Z = \emptyset$ , which is the claim. If $f$ is a nonzero constant then $Z = \emptyset$ and every $b \in U$ works.

3.1F2F3givenstep 1.1step 1.2step 2.1∎

(The divisor law applies, and the explicit cases.) For a nonconstant $f$ , step 2.1 provides a translate $P_{b}$ whose boundary avoids $Zer (f) \cup Pol (f)$ , so the hypotheses of [F3] are met and the zero count, the pole count and the vanishing residue sum for $f$ hold for that parallelogram; for nonzero constant $f$ those counts are $0$ and the conclusions are trivial. Steps 1.1 and 1.2 are the explicit statements for $℘$ on $Λ = Z + i Z$ : the boundary of $P_{(1 + i) / 4}$ carries no pole, while the boundary of the unshifted $P_{0}$ contains the four lattice vertices. This proves the three clauses of the Example.

Remarks

The point of the example is bookkeeping rather than computation: the divisor law is stated only for translates whose boundary avoids the zeros and poles, and the verification shows that such translates are never in short supply, because the bad basepoints in any bounded ball form a finite union of translated parallelogram boundaries, none of which contains a ball. The explicit translate $a = (1 + i) / 4$ moves the four vertices of $P_{0}$ off its own boundary: $1 + i$ becomes the interior point $1 + i \in P_{a}^{\circ}$ , while $0$ , $1$ and $i$ lie outside $P_{a}$ , and the interior of $P_{a}$ contains exactly one lattice point, namely $1 + i$ . The same argument applies to any full lattice once one knows that a bounded set meets the lattice in finitely many points, which is the uniform-gap estimate used in the convergence proof for the $℘$ -series.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-10-02Open item page →

Square and hexagonal lattice invariants

Example

For the square lattice $Λ_{s q} = Z + i Z$ one has $g_{3} (Λ_{s q}) = 0$ and $g_{2} (Λ_{s q}) \neq 0$ . For the hexagonal lattice $Λ_{h e x} = Z + Z ρ$ with $ρ = e^{2 π i / 3}$ one has $g_{2} (Λ_{h e x}) = 0$ and $g_{3} (Λ_{h e x}) \neq 0$ . In both cases $Δ \neq 0$ . The argument is a symmetry argument under multiplication by $i$ respectively by $ρ$ : no numerical value of any Eisenstein sum is evaluated.

Facts & Assumptions

Given: The lattices $Λ_{s q} : = Z + i Z$ and $Λ_{h e x} : = Z + Z ρ$ with $ρ : = e^{2 π i / 3}$ , their invariants $g_{2} = 60 G_{4}$ , $g_{3} = 140 G_{6}$ and discriminants $Δ = g_{2}^{3} - 27 g_{3}^{2}$ (Weierstrass p function, Weierstrass cubic differential equation, Nonvanishing of the lattice discriminant).

[F1]

A full complex lattice is a subgroup $Λ = Z ω_{1} + Z ω_{2} \subseteq C$ with $ω_{1}, ω_{2}$ real-linearly independent, and $(ω_{1}, ω_{2})$ is oriented when $Im (ω_{2} / ω_{1}) > 0$ ; for either ordering real-linear independence is equivalent to $Im (ω_{2} / ω_{1}) \neq 0$ , so one of the two orderings is oriented (Complex lattice and quotient torus). The sums defining the Weierstrass data depend only on the lattice $Λ$ , not on the oriented basis chosen to describe it (Weierstrass p function).

[F2]

$G_{4} (Λ) = \sum_{ω \in Λ ∖ {0}} ω^{- 4}$ and $G_{6} (Λ) = \sum_{ω \in Λ ∖ {0}} ω^{- 6}$ are the unordered finite-subset sums over the lattice; both families are absolutely summable, so $G_{4} (Λ)$ and $G_{6} (Λ)$ are well-defined complex numbers, and the invariants are $g_{2} = 60 G_{4}$ , $g_{3} = 140 G_{6}$ (Weierstrass cubic differential equation).

[F3]

For every full complex lattice the discriminant $Δ = g_{2}^{3} - 27 g_{3}^{2}$ satisfies $Δ \neq 0$ (Nonvanishing of the lattice discriminant).

[F4]

The complex exponential satisfies $\exp (z + w) = \exp (z) \exp (w)$ for all $z, w$ , is nowhere zero, has kernel $\ker (\exp) = 2 π i Z$ , and $∣ \exp (i y) ∣ = 1$ for every real $y$ (The complex exponential by its power series, $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential, $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ , $\exp (x + i y) = e^{x} (\cos y + i \sin y)$ , $∣ \exp (x + i y) ∣ = e^{x}$ , and $e^{i π} + 1 = 0$ ).

[F5]

$C$ is a field with $i^{2} = - 1$ , every complex number has a unique form $a + b i$ with $a, b \in R$ , and a product of two nonzero complex numbers is nonzero ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ , The complex numbers as $R [x] / (x^{2} + 1)$ , with the real embedding and imaginary unit $i$ ).

[F6]

For $z \in C$ one has $z = Re z + i Im z$ with $∣ z ∣^{2} = (Re z)^{2} + (Im z)^{2}$ , and $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ for all $z, w$ (Real and imaginary parts, complex conjugation, and modulus, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F7]

For nonzero $z, w \in C$ and integers $m, n$ one has $z^{m + n} = z^{m} z^{n}$ , $(z w)^{n} = z^{n} w^{n}$ and $z^{- n} = (z^{n})^{- 1}$ (Integer powers in the complex field).

Verification

1.1F4F5F6F7algebra

(The element $ρ$ , its inverse power and non-reality.) By [F4] and [F5], $ρ = e^{2 π i / 3}$ satisfies $ρ^{3} = e^{2 π i} = 1$ , and $ρ \neq 1$ because $2 π / 3 \notin 2 π Z$ ; also $ρ \neq - 1$ , since $(- 1)^{3} = - 1 \neq 1$ . Expanding $t^{3} - 1 = (t - 1) (t^{2} + t + 1)$ and using that $C$ is a field gives $0 = ρ^{3} - 1 = (ρ - 1) (ρ^{2} + ρ + 1)$ with $ρ - 1 \neq 0$ , hence $ρ^{2} + ρ + 1 = 0$ and $ρ^{2} = - 1 - ρ$ . Moreover $1 - ρ^{2} \neq 0$ : if $ρ^{2} = 1$ then $(ρ - 1) (ρ + 1) = 0$ , and a field has no zero divisors, so $ρ = 1$ or $ρ = - 1$ , both excluded. Finally $ρ \notin R$ : if $ρ$ were real, then $∣ ρ ∣^{2} = ρ^{2}$ by [F6], while $∣ ρ ∣ = ∣ \exp (2 π i / 3) ∣ = 1$ by [F4], so $ρ^{2} = 1$ and again $ρ = \pm 1$ , a contradiction. For the power law, $ρ^{- 4} = ρ^{- 4} (ρ^{3})^{2} = ρ^{- 4} ρ^{6} = ρ^{2}$ by [F7] and $ρ^{3} = 1$ .

1.2F2F7algebra

(Scaling identity for the lattice sums.) Let $Λ = Z ω_{1} + Z ω_{2}$ be a full lattice and $c \in C^{\times}$ . Then $c Λ = Z (c ω_{1}) + Z (c ω_{2})$ is again a full lattice: real-linear independence is preserved because $a (c ω_{1}) + b (c ω_{2}) = c (a ω_{1} + b ω_{2})$ vanishes only if $a ω_{1} + b ω_{2} = 0$ . For $k \in {4, 6}$ and every finite $E \subseteq Λ ∖ {0}$ one has $\sum_{ω \in E} (c ω)^{- k} = c^{- k} \sum_{ω \in E} ω^{- k}$ by [F7]; the map $E \mapsto c E$ is an order-isomorphism from the directed set of finite subsets of $Λ ∖ {0}$ onto that of $c Λ ∖ {0}$ , and the sum over $c Λ$ is the net of these finite sums by [F2]. Hence the net for $c Λ$ is the constant multiple $c^{- k}$ of the convergent net for $Λ$ , so it converges and $G_{k} (c Λ) = c^{- k} G_{k} (Λ)$ .

2.1F1F5step 1.1algebra

(The two lattices.) Since every complex number is uniquely $a + b i$ , the pair ${1, i}$ is a real basis of $C$ , so $Λ_{s q}$ is a full complex lattice with oriented basis $(1, i)$ ( $Im (i / 1) = 1 > 0$ ); and $i Λ_{s q} = Λ_{s q}$ : indeed $i \cdot 1 = i \in Λ_{s q}$ and $i \cdot i = - 1 \in Λ_{s q}$ , so $i Λ_{s q} \subseteq Λ_{s q}$ , while multiplication by $i$ is a bijection of $C$ with inverse multiplication by $i^{- 1} = - i$ , which likewise preserves $Λ_{s q}$ . Also $i^{- 6} = (i^{2})^{- 3} = (- 1)^{- 3} = - 1$ . For the hexagonal lattice: ${1, ρ}$ is real-linearly independent because $ρ \notin R$ by step 1.1, so $Λ_{h e x}$ is a full complex lattice and one of the orderings of $(1, ρ)$ is oriented by [F1]; and $ρ Λ_{h e x} = Λ_{h e x}$ because $ρ \cdot 1 = ρ \in Λ_{h e x}$ and $ρ \cdot ρ = ρ^{2} = - 1 - ρ \in Λ_{h e x}$ show $ρ Λ_{h e x} \subseteq Λ_{h e x}$ , while multiplication by $ρ$ is invertible on $C$ with inverse multiplication by $ρ^{- 1} = ρ^{2}$ , and $ρ^{2} Λ_{h e x} \subseteq Λ_{h e x}$ since $ρ^{2} \cdot 1 = ρ^{2} = - 1 - ρ \in Λ_{h e x}$ and $ρ^{2} \cdot ρ = ρ^{3} = 1 \in Λ_{h e x}$ .

3.1F2F5step 1.2step 2.1algebra

( $g_{3} (Λ_{s q}) = 0$ .) By step 2.1, $i Λ_{s q} = Λ_{s q}$ , hence $G_{6} (Λ_{s q}) = G_{6} (i Λ_{s q}) = i^{- 6} G_{6} (Λ_{s q}) = - G_{6} (Λ_{s q})$ by step 1.2 with $c = i$ ; therefore $2 G_{6} (Λ_{s q}) = 0$ and, since $C$ is a field, $G_{6} (Λ_{s q}) = 0$ , so $g_{3} (Λ_{s q}) = 140 \cdot 0 = 0$ .

3.2F2F5step 1.1step 1.2step 2.1algebra

( $g_{2} (Λ_{h e x}) = 0$ .) By step 2.1, $ρ Λ_{h e x} = Λ_{h e x}$ , hence $G_{4} (Λ_{h e x}) = G_{4} (ρ Λ_{h e x}) = ρ^{- 4} G_{4} (Λ_{h e x}) = ρ^{2} G_{4} (Λ_{h e x})$ by step 1.2 with $c = ρ$ and step 1.1; therefore $(1 - ρ^{2}) G_{4} (Λ_{h e x}) = 0$ , and since $1 - ρ^{2} \neq 0$ by step 1.1 and $C$ is a field, $G_{4} (Λ_{h e x}) = 0$ , so $g_{2} (Λ_{h e x}) = 60 \cdot 0 = 0$ .

4.1F3F5step 3.1step 3.2algebra

(The complementary invariants and the discriminant.) By [F3] the discriminant is nonzero for both lattices. For $Λ_{s q}$ step 3.1 gives $Δ (Λ_{s q}) = g_{2}^{3} - 27 \cdot 0^{2} = g_{2}^{3}$ , which is nonzero, so $g_{2} (Λ_{s q}) \neq 0$ in the field $C$ . For $Λ_{h e x}$ step 3.2 gives $Δ (Λ_{h e x}) = 0^{3} - 27 g_{3}^{2} = - 27 g_{3}^{2}$ with $- 27 \neq 0$ , and a nonzero discriminant forces $g_{3}^{2} \neq 0$ , hence $g_{3} (Λ_{h e x}) \neq 0$ .

5.1

(Assembly.) Step 3.1 gives $g_{3} (Λ_{s q}) = 0$ and step 4.1 gives $g_{2} (Λ_{s q}) \neq 0$ ; step 3.2 gives $g_{2} (Λ_{h e x}) = 0$ and step 4.1 gives $g_{3} (Λ_{h e x}) \neq 0$ ; step 4.1 also gives $Δ \neq 0$ in both cases. These are exactly the assertions of the example. ∎

Remarks

The two symmetries are the only inputs: $i Λ_{s q} = Λ_{s q}$ reindexes the $G_{6}$ -sum into its negative, and $ρ Λ_{h e x} = Λ_{h e x}$ reindexes the $G_{4}$ -sum into $ρ^{2}$ times itself. The complementary invariant is then forced to be nonzero by $Δ \neq 0$ , so no Eisenstein sum is evaluated numerically and no transcendental input about the elliptic integral is used. The hexagonal case is the equianharmonic one: its cubic $4 x^{3} - g_{3}$ of Nonvanishing of the lattice discriminant has no $x$ -term, while the square lattice is the one whose cubic has no constant term.

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Half-period values of the square lattice

Example

Let $Λ_{s q} = Z + i Z$ be the square lattice with its oriented basis $(1, i)$ , let $℘ = ℘_{Λ_{s q}}$ with invariants $g_{2}, g_{3}$ , put $h : = (1 + i) / 2$ , and let $e_{1} : = ℘ (1 / 2)$ , $e_{2} : = ℘ (i / 2)$ , $e_{3} : = ℘ (h)$ be the three half-period values. Then:

$℘ (h) = 0$ , so $e_{3} = 0$ ;
the other two half-period values $e_{1}$ and $e_{2}$ are the two elements of the pair $\pm \sqrt{g_{2}} / 2$ of roots of $4 x^{3} - g_{2} x$ ; with the normalisation $\sqrt{g_{2}} : = 2 e_{1}$ one has $e_{1} = \sqrt{g_{2}} / 2$ and $e_{2} = - \sqrt{g_{2}} / 2$ literally, and $g_{2} \neq 0$ ;
the four branch values of the torus form $\overset{ˉ}{℘} : T_{Λ_{s q}} \to \hat{C}$ of $℘$ — the images of its critical points (Ramification index, ramification order and branch value) — are exactly $0$ , $\pm \sqrt{g_{2}} / 2$ and $\infty$ .

The verification below evaluates no Eisenstein sum: it uses the scaling identity $℘_{c Λ} (c z) = c^{- 2} ℘_{Λ} (z)$ at $c = i$ , the symmetry $i Λ_{s q} = Λ_{s q}$ , and the cubic differential equation.

Facts & Assumptions

Given: The square lattice $Λ_{s q} = Z + i Z$ with its basis $(1, i)$ , the Weierstrass function $℘ = ℘_{Λ_{s q}}$ and its derivative $℘^{'}$ , the invariants $g_{2} = 60 G_{4}$ , $g_{3} = 140 G_{6}$ , the torus $T_{Λ_{s q}} = C / Λ_{s q}$ with class map $π$ , the torus form $\overset{ˉ}{℘}$ characterised by $\overset{ˉ}{℘} \circ π = ℘$ , the half-periods $h_{1} = 1 / 2$ , $h_{2} = i / 2$ , $h_{3} = (1 + i) / 2 = h$ and the values $e_{j} = ℘ (h_{j})$ .

[F1]

$C$ is a field; every complex number has a unique form $a + b i$ with $a, b \in R$ ; and $(a + b i) (u + v i) = (a u - b v) + (a v + b u) i$ for real $a, b, u, v$ ( $C = R [x] / (x^{2} + 1)$ is a field, every element is uniquely $a + b i$ , and every nonzero element has inverse $(a - b i) / (a^{2} + b^{2})$ ).

[F2]

[F3]

$℘_{Λ} (z) = z^{- 2} + \sum_{ω \in Λ ∖ {0}} ((z - ω)^{- 2} - ω^{- 2})$ on $C ∖ Λ$ , the sum being the unordered finite-subset sum over the directed set of finite subsets of $Λ ∖ {0}$ ; the value depends only on the lattice (Weierstrass p function).

[F4]

For every full lattice the sum defining $℘_{Λ}$ converges absolutely at every $z \in C ∖ Λ$ and uniformly on compact subsets; $℘_{Λ}$ is holomorphic on $C ∖ Λ$ and $Λ$ -periodic, $℘_{Λ} (z + λ) = ℘_{Λ} (z)$ for every $λ \in Λ$ and every $z \in C$ with poles matched (Normal convergence, parity and periodicity of the Weierstrass p function).

[F5]

With the invariants $g_{2} = 60 G_{4}$ and $g_{3} = 140 G_{6}$ one has $(℘^{'})^{2} = 4 ℘^{3} - g_{2} ℘ - g_{3}$ on $C ∖ Λ$ (Weierstrass cubic differential equation).

[F6]

For the square lattice $g_{3} (Λ_{s q}) = 0$ and $g_{2} (Λ_{s q}) \neq 0$ (Square and hexagonal lattice invariants).

[F7]

For a full lattice $Λ = Z ω_{1} + Z ω_{2}$ with $h_{1} = ω_{1} / 2$ , $h_{2} = ω_{2} / 2$ , $h_{3} = (ω_{1} + ω_{2}) / 2$ one has $℘^{'} (h) = 0$ for every $h \in C ∖ Λ$ with $2 h \in Λ$ , and the zeros of $℘^{'}$ are precisely the $Λ$ -translates of $h_{1}, h_{2}, h_{3}$ , each of order one (Degree two of ℘ and its four branch points).

[F8]

For the same data the classes $[h_{1}], [h_{2}], [h_{3}]$ are three distinct nonzero half-period classes, the values $e_{1}, e_{2}, e_{3}$ are three distinct complex numbers, and the torus form $\overset{ˉ}{℘} : T_{Λ} \to \hat{C}$ of $℘$ has critical points exactly $[0], [h_{1}], [h_{2}], [h_{3}]$ , with branch values $\overset{ˉ}{℘} ([0]) = \infty$ and $e_{1}, e_{2}, e_{3}$ (Degree two of ℘ and its four branch points).

[F9]

For a nonconstant holomorphic map $f : X \to Y$ of Riemann surfaces, a branch value of $f$ is a point $y \in Y$ for which there is a critical point $x \in X$ with $f (x) = y$ , and the set of all branch values is the branch locus of $f$ (Ramification index, ramification order and branch value).

Verification

1.1F1F2F8givenalgebra

(The square lattice, its symmetry and its half-periods.) Since every complex number is uniquely $a + b i$ with $a, b \in R$ , the pair ${1, i}$ spans $C$ over $R$ and $a + b i = 0$ forces $a = b = 0$ , so $1, i$ are real-linearly independent and $Λ_{s q} = Z + i Z$ is a full complex lattice with oriented basis $(1, i)$ , as $Im (i / 1) = 1 > 0$ ; multiplication by $i$ maps $Λ_{s q}$ into itself because $i \cdot 1 = i$ and $i \cdot i = - 1$ both lie in it, and multiplication by $i$ is a bijection of $C$ with inverse multiplication by $i^{- 1} = - i$ (as $i \cdot (- i) = 1$ ), which also preserves $Λ_{s q}$ , so $i Λ_{s q} = Λ_{s q}$ ; with the oriented basis $(1, i)$ the half-periods of the degree-two lemma are $h_{1} = 1 / 2$ , $h_{2} = i / 2$ , $h_{3} = (1 + i) / 2 = h$ , and these are nonzero classes, so $h \notin Λ_{s q}$ ; moreover $2 h_{1} = 1$ , $2 h_{2} = i$ and $2 h_{3} = 1 + i$ lie in $Λ_{s q}$ , and $h - i h = \frac{1}{2} (1 + i) (1 - i) = \frac{1}{2} (1 - i^{2}) = 1$ .

1.2F1F2F3F4algebra

(The scaling identity for the $℘$ -series.) Let $Λ = Z ω_{1} + Z ω_{2}$ be a full lattice, $c \in C^{\times}$ and $z \in C ∖ Λ$ ; then $c Λ = Z (c ω_{1}) + Z (c ω_{2})$ is again a full lattice, because $a (c ω_{1}) + b (c ω_{2}) = c (a ω_{1} + b ω_{2})$ vanishes for real $a, b$ only if $a ω_{1} + b ω_{2} = 0$ , and $ω \mapsto c ω$ is a bijection $Λ \to c Λ$ ; for every finite $F \subseteq Λ ∖ {0}$ one has $\sum_{ω \in F} ((c z - c ω)^{- 2} - (c ω)^{- 2}) = c^{- 2} \sum_{ω \in F} ((z - ω)^{- 2} - ω^{- 2})$ by the multiplication formula and $(c ζ)^{- 2} = c^{- 2} ζ^{- 2}$ ; the finite subsets of $c Λ ∖ {0}$ are exactly the image sets $c F$ , so the finite-subset net defining $℘_{c Λ} (c z)$ is the constant $(c z)^{- 2} = c^{- 2} z^{- 2}$ plus $c^{- 2}$ times the finite-subset net defining $℘_{Λ} (z)$ , which converges at $z \notin Λ$ by the absolute-convergence clause; hence $℘_{c Λ} (c z) = c^{- 2} ℘_{Λ} (z)$ .

2.1step 1.1step 1.2algebra

(The scaling identity at $c = i$ .) Taking $c = i$ and $Λ = Λ_{s q}$ in step 1.2, and using $i Λ_{s q} = Λ_{s q}$ from step 1.1 as well as $i^{- 2} = (i^{2})^{- 1} = (- 1)^{- 1} = - 1$ , gives $℘ (i z) = - ℘ (z)$ for every $z \in C ∖ Λ_{s q}$ .

2.2F5F6F7givenalgebra

(The cubic relation at each half-period.) Fix $j \in {1, 2, 3}$ : by step 1.1, $h_{j} \notin Λ_{s q}$ and $2 h_{j} \in Λ_{s q}$ , so the degree-two lemma gives $℘^{'} (h_{j}) = 0$ ; since $h_{j} \in C ∖ Λ_{s q}$ , the differential equation may be evaluated there, giving $0 = (℘^{'} (h_{j}))^{2} = 4 ℘ (h_{j})^{3} - g_{2} ℘ (h_{j}) - g_{3} = 4 e_{j}^{3} - g_{2} e_{j} - g_{3}$ , and with $g_{3} = 0$ for the square lattice this reads $e_{j} (4 e_{j}^{2} - g_{2}) = 0$ .

3.1F1F4step 1.1step 2.1algebra

(The vanishing $℘ (h) = 0$ .) By step 1.1, $h \notin Λ_{s q}$ and $h = i h + 1$ with $1 \in Λ_{s q}$ ; hence $h$ lies in the domain of $℘$ and the periodicity and step 2.1 give $℘ (h) = ℘ (i h + 1) = ℘ (i h) = - ℘ (h)$ , so $2 ℘ (h) = 0$ , and since $C$ is a field in which $2 \neq 0$ this forces $℘ (h) = 0$ ; thus $e_{3} = ℘ (h_{3}) = ℘ (h) = 0$ .

4.1F1F6F8step 3.1step 2.2algebra

(The two nonzero half-period values and the factorisation of the cubic.) By step 3.1, $e_{3} = 0$ , and by the degree-two lemma $e_{1}, e_{2}, e_{3}$ are pairwise distinct, so $e_{1}, e_{2} \neq 0$ ; step 2.2 for $j = 1, 2$ then gives $e_{j} (4 e_{j}^{2} - g_{2}) = 0$ with $e_{j} \neq 0$ , hence $4 e_{1}^{2} = g_{2} = 4 e_{2}^{2}$ and $e_{1}^{2} = e_{2}^{2}$ , that is $(e_{1} - e_{2}) (e_{1} + e_{2}) = 0$ in the field $C$ , so $e_{1} - e_{2} \neq 0$ forces $e_{1} + e_{2} = 0$ and $e_{2} = - e_{1}$ ; consequently $g_{2} = 4 e_{1}^{2} \neq 0$ , and substituting $g_{2} = 4 e_{1}^{2}$ and $g_{3} = 0$ gives the polynomial identity $4 x^{3} - g_{2} x - g_{3} = 4 x^{3} - 4 e_{1}^{2} x = 4 x (x - e_{1}) (x + e_{1})$ , so the cubic $4 x^{3} - g_{2} x - g_{3}$ has exactly the roots $0, e_{1}, - e_{1}$ , which are pairwise distinct; with the normalisation $\sqrt{g_{2}} : = 2 e_{1}$ one has $(\sqrt{g_{2}})^{2} = 4 e_{1}^{2} = g_{2}$ and $\pm \sqrt{g_{2}} / 2 = \pm e_{1}$ , so the other two half-period values $e_{1} = ℘ (1 / 2)$ and $e_{2} = ℘ (i / 2) = - e_{1}$ are exactly the two distinct roots $\pm \sqrt{g_{2}} / 2$ of $4 x^{3} - g_{2} x$ .

5.1F8F9step 4.1algebra

(The four branch values.) By the degree-two lemma the torus form $\overset{ˉ}{℘}$ of $℘$ has critical points exactly $[0], [h_{1}], [h_{2}], [h_{3}]$ , with branch values $\overset{ˉ}{℘} ([0]) = \infty$ and $e_{1}, e_{2}, e_{3}$ , so by the definition of a branch value its branch locus is the four-element set ${\infty, e_{1}, e_{2}, e_{3}}$ ; by step 4.1 this set is ${\infty, 0, e_{1}, - e_{1}} = {\infty, 0, \pm \sqrt{g_{2}} / 2}$ with $e_{1} \neq 0$ , so the branch locus of $\overset{ˉ}{℘}$ consists exactly of the four distinct values $\infty, 0, \sqrt{g_{2}} / 2, - \sqrt{g_{2}} / 2$ .

6.1step 3.1step 4.1step 5.1

(Assembly.) Step 3.1 proves $℘ (h) = 0$ for $h = (1 + i) / 2$ , so the half-period value $e_{3}$ is $0$ ; step 4.1 proves that $e_{1}$ and $e_{2}$ are the two distinct roots $\pm \sqrt{g_{2}} / 2$ of $4 x^{3} - g_{2} x$ for the normalisation $\sqrt{g_{2}} = 2 e_{1}$ , and that $g_{2} \neq 0$ ; step 5.1 proves that the branch values of the torus form of $℘$ are exactly $0, \pm \sqrt{g_{2}} / 2, \infty$ . These are the three assertions of the example. ∎

Remarks

The sign in $℘ (h) = 0$ comes from the scaling identity alone: $i$ is a similarity of the square lattice, and under it $℘$ is multiplied by $i^{- 2} = - 1$ , while $h$ and $i h$ differ by the period $1$ . The remaining half-period values are then forced by the cubic: everything is a root of $4 x^{3} - g_{2} x$ because $g_{3} = 0$ , and the two nonzero roots sum to zero, matching $e_{2} = - e_{1}$ . The four branch values of the degree-two map $\overset{ˉ}{℘} : T_{Λ_{s q}} \to \hat{C}$ are consequently $\infty$ and the three finite values $0, \pm \sqrt{g_{2}} / 2$ — the two-element pair beyond $0$ being exactly the pair of nonzero roots, without any need to evaluate $g_{2}$ numerically.

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Addition and duplication for $℘$

Example

Let $Λ$ be a full complex lattice with Weierstrass function $℘$ , and let $z \in C ∖ Λ$ satisfy $℘^{'} (z) \neq 0$ . Then $℘ (2 z) = - 2 ℘ (z) + \frac{1}{4} {(\frac{℘^{''} (z)}{℘^{'} (z)})}^{2} .$ Moreover $℘^{''} = 6 ℘^{2} - \frac{1}{2} g_{2}$ on $C ∖ Λ$ , so on the same locus the duplication value is the rational expression $- 2 ℘ (z) + \frac{(6 ℘ (z)^{2} - \frac{1}{2} g_{2})^{2}}{4 ℘^{'} (z)^{2}}$ in $℘ (z)$ and $℘^{'} (z)$ . The duplication formulas agree as meromorphic functions on $C$ . At a nonzero half-period both sides have a genuine double pole, since $2 z \in Λ$ ; no finite value is asserted there.

Facts & Assumptions

Given: A full complex lattice $Λ = Z ω_{1} + Z ω_{2}$ with oriented basis, its Weierstrass function $℘$ and derivative $℘^{'}$ , the invariants $g_{2} = 60 G_{4}$ , $g_{3} = 140 G_{6}$ , and a point $z \in C ∖ Λ$ .

[F1]

$℘$ is holomorphic on $C ∖ Λ$ , is even and $Λ$ -periodic, and at each lattice point has a double pole with principal part $(z - λ)^{- 2}$ and no other poles; $℘^{'} (z) = - 2 \sum_{ω \in Λ} (z - ω)^{- 3}$ on $C ∖ Λ$ , this series being normally convergent there, and $℘^{'}$ is odd and $Λ$ -periodic with a pole of order $3$ at each lattice point (Normal convergence, parity and periodicity of the Weierstrass p function, Weierstrass p function).

[F2]

$℘ (z) = ℘ (w)$ if and only if $w \equiv z$ or $w \equiv - z$ modulo $Λ$ ; the zeros of $℘^{'}$ are exactly the $Λ$ -translates of the three nonzero half-periods, and each of them is of order one; consequently, for $w \in C ∖ Λ$ , $℘^{'} (w) = 0$ if and only if $2 w \in Λ$ (Degree two of ℘ and its four branch points).

[F3]

$(℘^{'})^{2} = 4 ℘^{3} - g_{2} ℘ - g_{3}$ on $C ∖ Λ$ (Weierstrass cubic differential equation).

[F4]

The addition formula holds meromorphically in $(z, w)$ : it holds as an equality of values wherever the displayed quotient is defined, and all apparent exceptional cases are interpreted by meromorphic continuation, without assigning a finite value at a genuine pole (Addition formula for $℘$ ).

[F5]

A holomorphic function with a zero of order $m$ at $a$ factors near $a$ as $(z - a)^{m} g (z)$ with $g (a) \neq 0$ ; a holomorphic function on a domain that is not identically zero has isolated zeros; and two holomorphic functions on a domain agreeing on a set with an accumulation point in the domain agree everywhere (The order of a zero is the exponent in its local holomorphic factorization, Zeros of a nonzero holomorphic function are isolated, Identity theorem for holomorphic functions).

[F6]

Complex derivatives are linear, satisfy the product rule and the chain rule, and a complex-differentiable function is continuous (Linearity, product, reciprocal, and quotient rules for complex derivatives, The chain rule for complex derivatives, Complex differentiability at a point implies continuity there).

[F7]

Meromorphic functions on a connected plane domain form a field: sums, products and quotients with nonzero denominator are meromorphic, and the pole set of a meromorphic function is discrete and closed; a meromorphic function on a domain that vanishes on a nonempty open subset is identically zero (Meromorphic functions on a plane domain, Meromorphic functions on a connected plane domain form a field, Poles of a meromorphic function form a closed discrete set and are at most countable).

[F8]

The complex plane is a connected plane domain (Complex lattice and quotient torus).

Verification

1.1F1F2F3F5F6F7algebra

(The second-derivative identity on $C ∖ Λ$ .) Differentiating [F3] with the rules of [F6] gives $2 ℘^{'} ℘^{''} = (12 ℘^{2} - g_{2}) ℘^{'}$ on $C ∖ Λ$ ; at every point with $℘^{'} \neq 0$ this gives $℘^{''} = 6 ℘^{2} - \frac{1}{2} g_{2}$ . Let $h \notin Λ$ with $℘^{'} (h) = 0$ : by [F2] the zero of $℘^{'}$ at $h$ is of order one, so [F5] gives $℘^{'} (w) = (w - h) g (w)$ with $g (h) \neq 0$ , and hence $℘^{'} (w) \neq 0$ for $0 < ∣ w - h ∣ < ρ$ with some $ρ > 0$ ; shrinking $ρ$ if necessary, the pole set of the meromorphic function $℘$ is discrete by [F7], so the disc $D (h, ρ)$ contains no lattice point. Both $℘^{''}$ and $6 ℘^{2} - \frac{1}{2} g_{2}$ are holomorphic on $D (h, ρ)$ and agree on the punctured disc $D (h, ρ) ∖ {h}$ , which is a nonempty connected open set, so [F5] makes them agree on all of $D (h, ρ)$ , in particular at $h$ . Every point of $C ∖ Λ$ either has $℘^{'} \neq 0$ or is such a zero $h$ by [F2], so $℘^{''} = 6 ℘^{2} - \frac{1}{2} g_{2}$ on all of $C ∖ Λ$ .

1.2F1F2F4F5F6F7algebra

(The duplication identity where $℘^{'} (z) \neq 0$ .) Fix $z \in C ∖ Λ$ with $℘^{'} (z) \neq 0$ ; then $2 z \notin Λ$ by [F2], and $℘$ is holomorphic near $2 z$ . Choose $ρ > 0$ so small that $D (z, ρ)$ avoids $Λ$ , $(z + Λ) ∖ {z}$ and $- z + Λ$ , and that $z + w \notin Λ$ for $w \in D (z, ρ)$ . For $0 < ∣ w - z ∣ < ρ$ the addition formula [F4] applies and gives $℘ (z + w) = - ℘ (z) - ℘ (w) + \frac{1}{4} Q (w)^{2}$ with $Q (w) = (℘^{'} (z) - ℘^{'} (w)) / (℘ (z) - ℘ (w))$ . Both numerator and denominator vanish at $w = z$ , and the denominator has a simple zero there because its derivative is $- ℘^{'} (z) \neq 0$ ; the numerator has a zero of at least order one. Thus $Q$ extends holomorphically with $Q (z) = ℘^{''} (z) / ℘^{'} (z)$ , whether or not $℘^{''} (z)$ vanishes. Letting $w \to z$ gives $℘ (2 z) = - 2 ℘ (z) + \frac{1}{4} (℘^{''} (z) / ℘^{'} (z))^{2}$ .

2.1step 1.1step 1.2algebra

(Rational expression in $℘$ and $℘^{'}$ .) Substituting the identity of step 1.1 into the duplication identity of step 1.2 gives, for every $z \in C ∖ Λ$ with $℘^{'} (z) \neq 0$ , $℘ (2 z) = - 2 ℘ (z) + \frac{1}{4} {(\frac{6 ℘ (z)^{2} - \frac{1}{2} g_{2}}{℘^{'} (z)})}^{2} = - 2 ℘ (z) + \frac{(6 ℘ (z)^{2} - \frac{1}{2} g_{2})^{2}}{4 ℘^{'} (z)^{2}},$ a value of the rational function $R (x, y) = - 2 x + \frac{(6 x^{2} - \frac{1}{2} g_{2})^{2}}{4 y^{2}}$ of the two variables $x, y$ with coefficients in the field generated by $g_{2}$ over $C$ (here the denominator $4 y^{2}$ does not vanish because $℘^{'} (z) \neq 0$ ).

3.1F1F2F7F8step 1.1step 1.2step 2.1algebra

(Meromorphic extension and pole set.) The function $F (z) : = ℘ (2 z)$ is meromorphic on $C$ : it is holomorphic off $\frac{1}{2} Λ = {z : 2 z \in Λ}$ , and if $z_{0} \in \frac{1}{2} Λ$ with $λ : = 2 z_{0} \in Λ$ , then [F1] gives that $℘ (w) - (w - λ)^{- 2}$ is holomorphic near $λ$ , so substituting $w = 2 z$ shows $F (z) - \frac{1}{4} (z - z_{0})^{- 2}$ holomorphic near $z_{0}$ : each $z_{0} \in \frac{1}{2} Λ$ is a double pole of $F$ , and there are no others. The right-hand side $R (℘ (z), ℘^{'} (z))$ is meromorphic on $C$ by the field property [F7], because $℘$ and $℘^{'}$ are meromorphic and $℘^{'}$ is not the zero function by [F2]. By steps 1.2 and 2.1 the two meromorphic functions agree on ${z \in C ∖ Λ : ℘^{'} (z) \neq 0} = C ∖ \frac{1}{2} Λ$ , a nonempty open subset of the connected domain $C$ [F8]; hence their difference vanishes on a nonempty open set and is identically zero by [F7]. Therefore the duplication identity is an identity of meromorphic functions on $C$ : it holds wherever both sides are finite, and at the points of $\frac{1}{2} Λ$ , where $F$ has a double pole and the right-hand side likewise has a pole (at half-periods because $℘^{'}$ has a zero of order one and $℘^{''}$ is nonzero there, and at lattice points by the equality of the two meromorphic functions), no finite value is asserted.

4.1

(Assembly.) Step 1.1 gives the identity $℘^{''} = 6 ℘^{2} - \frac{1}{2} g_{2}$ on $C ∖ Λ$ , extended holomorphically across the half-periods where the division by $℘^{'}$ was only apparently problematic; step 1.2 gives the duplication identity for $℘^{'} (z) \neq 0$ ; step 2.1 exhibits it as the rational expression in $℘ (z)$ and $℘^{'} (z)$ ; and step 3.1 upgrades the duplication identity to an identity of meromorphic functions on $C$ , with the genuine poles retained. These are exactly the assertions of the example. ∎

Remarks

The only point of substance is that the addition formula becomes $0 / 0$ when $w = z$ : the secant through two coincident points has to be replaced by the tangent, and in the formula that means replacing the difference quotient by the derivative quotient $℘^{''} (z) / ℘^{'} (z)$ . The second identity is what makes the result algebraic: $(℘^{'})^{2} = 4 ℘^{3} - g_{2} ℘ - g_{3}$ can be differentiated and solved for $℘^{''}$ wherever $℘^{'} \neq 0$ , and the apparent failure of that solution at the half-periods is repaired by the identity theorem, since $℘^{''}$ and $6 ℘^{2} - \frac{1}{2} g_{2}$ are holomorphic across them. Both formulas are used in The chord-tangent group law and elliptic uniformization, where the tangent case of the chord-tangent law is exactly the limiting case $w \to z$ used here; note that the theorem derives its own copy of the differentiated differential equation locally, so this example carries no load for it.

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A simple zero of the Weierstrass sigma function on the square lattice

Example

Let $Λ = Z + i Z$ , with oriented basis $ω_{1} = 1$ , $ω_{2} = i$ , and let $ζ = ζ_{Λ}$ , $σ = σ_{Λ}$ be the Weierstrass zeta and sigma functions. Then $σ (0) = σ (1) = 0$ , $σ^{'} (0) = 1$ and $σ^{'} (1) = - \exp (η_{1} / 2) \neq 0$ where $η_{1} = 2 ζ (1 / 2)$ : both zeros of $σ$ at $0$ and at $1$ are simple. Moreover $ζ$ has residue $1$ at both points.

Facts & Assumptions

Given: The square lattice $Λ = Z + i Z = Z ω_{1} + Z ω_{2}$ with $ω_{1} = 1$ , $ω_{2} = i$ , its Weierstrass functions $ζ = ζ_{Λ}$ and $σ = σ_{Λ}$ (Weierstrass $ζ$ and $σ$ functions), and $η_{1} = 2 ζ (ω_{1} / 2) = 2 ζ (1 / 2)$ .

[F1]

$Λ = Z ω_{1} + Z ω_{2}$ is a full complex lattice with oriented basis $ω_{1} = 1$ , $ω_{2} = i$ , and $ζ, σ$ are its Weierstrass zeta and sigma functions (Complex lattice and quotient torus, Weierstrass $ζ$ and $σ$ functions).

[F2]

$ζ$ is meromorphic on $C$ , holomorphic exactly on $C ∖ Λ$ , odd, and at every lattice point $λ \in Λ$ has a simple pole with principal part $(z - λ)^{- 1}$ and residue $1$ , with no other poles. $σ$ is entire and odd, its zero set is exactly $Λ$ and every zero is simple, $σ^{'} (0) = 1$ , and $σ^{'} (z) / σ (z) = ζ (z)$ for $z \in C ∖ Λ$ . Moreover the quasi-period laws hold for all $z \in C$ with poles matched: $ζ (z + ω_{j}) = ζ (z) + η_{j}$ and $σ (z + ω_{j}) = - \exp (η_{j} (z + ω_{j} / 2)) σ (z)$ for $j = 1, 2$ , where $η_{j} = 2 ζ (ω_{j} / 2)$ (Convergence, zeros and quasi-periods of the Weierstrass zeta and sigma functions).

[F4]

The complex exponential satisfies $\exp (0) = 1$ and $\exp (z + w) = \exp (z) \exp (w)$ ; consequently $\exp (z) \exp (- z) = 1$ and $\exp (z) \neq 0$ for every $z \in C$ (The complex exponential by its power series, $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential). Its defining series also proves continuity: for $∣ u ∣ \leq 1$ , $n! \geq 2^{n - 1}$ for $n \geq 1$ gives $∣ \exp (u) - 1 ∣ \leq ∣ u ∣ \sum_{n \geq 1} 1 / n! \leq 2 ∣ u ∣$ . The addition law then gives $∣ \exp (v + u) - \exp (v) ∣ \leq 2 ∣ \exp (v) ∣ ∣ u ∣$ , so $\exp$ is continuous at every $v$ .

Verification

1.1F1F2

(Values at $0$ .) Since $0 \in Λ$ , [F2] gives $σ (0) = 0$ and $σ^{'} (0) = 1$ , and the zero of $σ$ at $0$ is simple; also $ζ$ has at $0$ a simple pole with residue $1$ .

1.2F1F2

(Residues of $ζ$ .) Both $0$ and $1$ lie in $Λ = Z ω_{1} + Z ω_{2}$ ; by [F2] the only poles of $ζ$ are the lattice points and each is simple with residue $1$ , so $ζ$ has residue $1$ at $0$ and at $1$ .

2.1F1F2step 1.1

( $σ (1) = 0$ .) Put $z = 0$ in the sigma quasi-period law for $j = 1$ : $ω_{1} = 1$ and $η_{1} = 2 ζ (1 / 2)$ , so $σ (1) = - \exp (η_{1} \cdot \frac{1}{2}) σ (0) = - \exp (η_{1} / 2) \cdot 0 = 0$ .

3.1F2F4step 1.1step 2.1

( $σ^{'} (1) = - \exp (η_{1} / 2) \neq 0$ .) For $u \neq 0$ the quasi-period law and step 2.1 give [F2, step 2.1] $\frac{σ (1 + u) - σ (1)}{u} = - \exp ⁣ (η_{1} (u + 1 / 2)) \frac{σ (u) - σ (0)}{u} .$ As $u \to 0$ , the second factor tends to $σ^{'} (0) = 1$ by step 1.1, and the exponential tends to $\exp (η_{1} / 2)$ by the continuity derived in [F4]. Thus $σ^{'} (1) = - \exp (η_{1} / 2) \neq 0$ by [F4]; simplicity and the residue $1$ of $ζ$ at $1$ also follow directly from [F2].

4.1

(Assembly.) Steps 1.1, 2.1 and 3.1 give $σ (0) = σ (1) = 0$ with simple zeros, $σ^{'} (0) = 1$ and $σ^{'} (1) = - \exp (η_{1} / 2) \neq 0$ ; step 1.2 gives residue $1$ of $ζ$ at both points. This is the asserted statement. ∎

Remarks

The whole example is a computation with the transformation law alone: the zero of $σ$ at the lattice point $1$ is inherited from the zero at $0$ through $σ (z + 1) = - \exp (η_{1} (z + 1 / 2)) σ (z)$ , and taking its difference quotient at $z = 0$ gives the derivative at $1$ , using continuity of the exponential from its defining series. The residue statement is the local form of $ζ = σ^{'} / σ$ at a simple zero of $σ$ , which is how the normalization $σ^{'} (0) = 1$ enters. This is the concrete display of the lattice-zero convention used in Convergence, zeros and quasi-periods of the Weierstrass zeta and sigma functions.

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A singular cubic outside the lattice family

Example

For the coefficient pair $(g_{2}, g_{3}) = (3, 1)$ the number $Δ : = g_{2}^{3} - 27 g_{3}^{2} = 3^{3} - 27 \cdot 1^{2} = 0$ , and the projective cubic $Y^{2} Z = 4 X^{3} - 3 X Z^{2} - Z^{3}$ has the affine singular point $(x, y) = (- 1 / 2, 0)$ : the affine equation $y^{2} = 4 x^{3} - 3 x - 1$ factors as $y^{2} = (x - 1) (2 x + 1)^{2}$ , and near that point the curve is the union of the two smooth branches $y = \pm (x + \frac{1}{2}) u (x + \frac{1}{2})$ meeting transversally. Consequently no full complex lattice has these invariants, and this cubic is a degeneration outside the lattice family; it cannot be used as a supplier for any lattice statement.

Facts & Assumptions

Given: The coefficient pair $(g_{2}, g_{3}) = (3, 1)$ , its associated projective cubic $C : = {[X : Y : Z] \in {C P}^{2} : Y^{2} Z = 4 X^{3} - 3 X Z^{2} - Z^{3}}$ and the affine chart ${Z \neq 0}$ of ${C P}^{2}$ with coordinates $x = X / Z$ , $y = Y / Z$ , containing the affine curve $F (x, y) : = y^{2} - (4 x^{3} - 3 x - 1) = 0$ and the point $p : = (- 1 / 2, 0)$ .

[F1]

For a full complex lattice $Λ$ with Weierstrass invariants $g_{2} = 60 G_{4}$ , $g_{3} = 140 G_{6}$ and discriminant $Δ (Λ) : = g_{2}^{3} - 27 g_{3}^{2}$ one has $Δ (Λ) \neq 0$ , and the projective cubic $C_{Λ} = {[X : Y : Z] \in {C P}^{2} : Y^{2} Z = 4 X^{3} - g_{2} X Z^{2} - g_{3} Z^{3}}$ is nonsingular in the Jacobian-rank sense at every point, including its unique point at infinity $O = [0 : 1 : 0]$ (Nonvanishing of the lattice discriminant).

[F2]

(Jacobian-rank nonsingularity.) If a complex algebraic curve near $q$ in $C^{N}$ is the common zero set of exactly $N - 1$ holomorphic functions whose complex Jacobian matrix at $q$ has rank $N - 1$ , then after permuting the ambient coordinates so that the $j$ -th comes first the curve agrees near $q$ with the graph ${(z, φ (z)) : z \in A}$ of a holomorphic $φ$ on a plane domain $A$ , the projection to the first coordinate being a homeomorphism onto $A$ (Local holomorphic charts on nonsingular complex algebraic curves). In particular, the graph representation holds in one of the two coordinate directions when $N = 2$ .

[F3]

(Implicit function theorem.) If $G$ is holomorphic near $(a, b) \in C^{2}$ , $G (a, b) = 0$ and $\partial G / \partial w (a, b) \neq 0$ , then on a product of discs around $(a, b)$ the zero set of $G$ is the graph $w = ψ (z)$ of a unique holomorphic function $ψ$ with $ψ (a) = b$ (The holomorphic implicit function theorem).

[F4]

A function complex differentiable at a point is continuous there; and for all complex numbers $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ and $∣ z ∣ \geq 0$ with $∣ z ∣ = 0$ only for $z = 0$ (Complex differentiability at a point implies continuity there, Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive). Hence if $u$ is holomorphic near $0$ with $u (0) \neq 0$ , then $∣ u (s) - u (0) ∣ \leq ∣ u (0) ∣ / 2$ on a neighbourhood of $0$ , and there $∣ u (s) ∣ \geq ∣ u (0) ∣ / 2 > 0$ .

[F5]

${C P}^{2} = (C^{3} ∖ {0}) / \sim$ with $a \sim b$ exactly when $b = λ a$ for some $λ \in C^{\times}$ , classes written $[X : Y : Z]$ , and the sets where one homogeneous coordinate is nonzero are the standard affine charts with the remaining ratios as coordinates: on ${Z \neq 0}$ one uses $(x, y) = (X / Z, Y / Z)$ (projective space points).

Verification

1.1givenalgebra

(The discriminant vanishes.) For the coefficient pair $(g_{2}, g_{3}) = (3, 1)$ the number displayed in the statement is $Δ = g_{2}^{3} - 27 g_{3}^{2} = 27 - 27 = 0$ , since $3^{3} = 27$ and $27 \cdot 1^{2} = 27$ .

1.2givenF5algebra

(The point lies on the affine curve.) With the chart coordinates of [F5], the cubic of the statement has affine equation $y^{2} = 4 x^{3} - 3 x - 1$ at $Z = 1$ ; writing $F (x, y) : = y^{2} - (4 x^{3} - 3 x - 1)$ , at $p = (- 1 / 2, 0)$ one has $x^{2} = 1 / 4$ and $4 x^{3} - 3 x - 1 = 4 (- 1 / 8) - 3 (- 1 / 2) - 1 = - 1 / 2 + 3 / 2 - 1 = 0$ , so $F (p) = 0 - 0 = 0$ and $p$ lies on the affine curve.

1.3givenalgebra

(The differential vanishes at the point.) The partial derivatives of $F$ are $\partial F / \partial y = 2 y$ and $\partial F / \partial x = - (12 x^{2} - 3)$ ; at $p$ these are $2 \cdot 0 = 0$ and $- (12 \cdot \frac{1}{4} - 3) = - (3 - 3) = 0$ . Thus $d F (p) = 0$ , so the plane curve $F = 0$ has vanishing differential at $p$ ; this is the elementary singularity criterion of the affine chart.

1.4F3F4algebra

(Factorisation and the unit square root.) Expanding $(x - 1) (2 x + 1)^{2} = (x - 1) (4 x^{2} + 4 x + 1) = 4 x^{3} - 3 x - 1$ gives $4 x^{3} - 3 x - 1 = (x - 1) (2 x + 1)^{2}$ . Put $s : = x + \frac{1}{2}$ , so that $2 x + 1 = 2 s$ and $x - 1 = s - \frac{3}{2}$ , and hence $4 x^{3} - 3 x - 1 = (s - \frac{3}{2}) (2 s)^{2} = 4 s^{3} - 6 s^{2} = s^{2} h (s)$ with $h (s) : = 4 s - 6$ . Apply [F3] to $G (s, w) : = w^{2} - h (s) = w^{2} - 4 s + 6$ at $(0, w_{0})$ with $w_{0} : = i \sqrt{6}$ : $G (0, w_{0}) = - 6 + 6 = 0$ and $\partial G / \partial w (0, w_{0}) = 2 w_{0} = 2 i \sqrt{6} \neq 0$ ; hence there is a holomorphic $u$ on a disc around $0$ with $u (0) = w_{0}$ and $u (s)^{2} = h (s) = 4 s - 6$ . By [F4] there is $ε > 0$ with $u (s) \neq 0$ for $∣ s ∣ < ε$ .

2.1F3step 1.4algebra

(Two smooth branches crossing at $p$ .) With $u$ as in step 1.4, the identity $y^{2} - s^{2} h (s) = (y - s u (s)) (y + s u (s))$ exhibits the affine curve near $p$ (which is $s = 0$ , $y = 0$ ) as the union of the two graphs $Σ_{\pm} = {(s, y) : y = \pm s u (s)}$ over the $s$ -coordinate, $∣ s ∣ < ε$ . Each $Σ_{\pm}$ is smooth with parametrisation $s \mapsto (s, \pm s u (s))$ , and the two branches meet exactly at $s = 0$ : for $0 < ∣ s ∣ < ε$ one has $s u (s) \neq 0$ by step 1.4, so the two points $(s, s u (s))$ and $(s, - s u (s))$ are distinct. The tangent directions at the meeting point are $(1, w_{0})$ and $(1, - w_{0})$ with $w_{0} \neq 0$ , hence distinct, so the branches cross transversally. Moreover $φ (s) : = s u (s)$ satisfies $φ (0) = 0$ and $φ^{'} (0) = u (0) = w_{0} \neq 0$ ; applying [F3] to $(y, s) \mapsto φ (s) - y$ at $(0, 0)$ gives a holomorphic inverse branch $ψ$ with $φ (ψ (y)) = y$ for small $y$ . The inverse branches of the two curve graphs are $s = ψ (y)$ and $s = ψ (- y)$ .

3.1F2step 2.1algebra

(The point is not a holomorphic graph in either direction.) Let $P = D_{s} \times D_{y}$ be any small polydisc around $(0, 0)$ contained in the domain of $u$ and $ψ$ , with $u$ nowhere zero on $D_{s}$ and $φ (D_{s}) \subset D_{y}$ . (i) For small $0 \neq s \in D_{s}$ , both $(s, s u (s))$ and $(s, - s u (s))$ are points of the curve in $P$ with the same $s$ -coordinate and distinct $y$ -coordinates; a graph over the $s$ -coordinate would contain exactly one point over $s$ , so the curve is not a holomorphic graph over $s$ . (ii) For small $0 \neq y \in D_{y}$ with $ψ (\pm y) \in D_{s}$ , the points $(ψ (y), y) \in Σ_{+}$ and $(ψ (- y), y) \in Σ_{-}$ are distinct points of the curve in $P$ with the same $y$ -coordinate, because $ψ$ is injective on $D_{y}$ and $y \neq - y$ ; so the curve is not a holomorphic graph over $y$ either. By [F2] a Jacobian-rank nonsingular point of a plane curve germ is a holomorphic graph over one of the two coordinates, so $p$ is not nonsingular in the Jacobian-rank sense.

4.1F1step 1.1step 1.2step 3.1

(No lattice has these invariants.) Suppose a full complex lattice $Λ$ had invariants $g_{2} = 3$ , $g_{3} = 1$ . Then its associated cubic $C_{Λ}$ of [F1] is exactly the projective cubic of the statement, and [F1] asserts that $C_{Λ}$ is nonsingular in the Jacobian-rank sense at every point. But the affine point $p$ is a point of $C_{Λ}$ by step 1.2 and is not Jacobian-rank nonsingular by step 3.1, a contradiction. The same conclusion is visible in the numbers alone: [F1] gives $Δ (Λ) \neq 0$ , while step 1.1 computes $Δ = 0$ for the pair $(3, 1)$ . Hence no full complex lattice realizes the invariants $(3, 1)$ , so the cubic of the statement is a degeneration outside the lattice family.

5.1

(Assembly.) Steps 1.1, 1.2 and 2.1 show that the projective cubic $Y^{2} Z = 4 X^{3} - 3 X Z^{2} - Z^{3}$ has $Δ = 3^{3} - 27 \cdot 1^{2} = 0$ and has at $(x, y) = (- 1 / 2, 0)$ an affine singular point at which the two smooth branches $y = \pm (x + \frac{1}{2}) u (x + \frac{1}{2})$ cross transversally, with vanishing differential recorded in step 1.3; step 4.1 shows that this coefficient pair is excluded for every full lattice, by both the nonsingularity clause and the nonvanishing-discriminant clause of [F1]. This is the asserted degeneration. ∎

Remarks

The factorisation $4 x^{3} - 3 x - 1 = (x - 1) (2 x + 1)^{2}$ is what makes the cubic a nodal curve: the affine polynomial has a double root at $x = - 1 / 2$ , so the two branches $y = \pm (x + \frac{1}{2}) u (x + \frac{1}{2})$ cross rather than osculate, and the same vanishing differential that produces the node also annihilates the discriminant $Δ = g_{2}^{3} - 27 g_{3}^{2}$ with the coefficient pair $(g_{2}, g_{3}) = (3, 1)$ . The lattice theorem Nonvanishing of the lattice discriminant is the statement that $Δ \neq 0$ for every lattice, and it mentions this example only as a contrast: no proof step of any item in the pair cites this example, so it is terminal and contributes no dependency. The unique point at infinity $O = [0 : 1 : 0]$ is nonsingular even for this cubic: in the chart ${Y \neq 0}$ with coordinates $u = X / Y$ , $v = Z / Y$ the equation is $v - 4 u^{3} + 3 u v^{2} + v^{3} = 0$ , whose partial derivative in $v$ equals $1 + 6 u v + 3 v^{2} = 1$ at the origin.

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Rectangular lattices, real mapping, and inverse elliptic integrals

Example

Let $Λ = Z α + Z β$ with $α > 0$ real and $β = i b$ with $b > 0$ , put $c_{1} = α / 2$ , $c_{2} = β / 2$ , $c_{3} = (α + β) / 2$ and $e_{j} = ℘ (c_{j})$ , ordered $e_{2} < e_{3} < e_{1}$ . Then for $z \in C ∖ Λ$ , $℘ (z) \in R$ exactly when $Re z \in (α / 2) Z$ or $Im z \in (b / 2) Z$ . These are the horizontal and vertical lines through $\frac{1}{2} Λ$ ; their lattice points are poles of $℘$ , not finite real values. The restriction to a half-period rectangle maps conformally onto one half-plane. Every inverse branch satisfies $d z / d ζ = (4 ζ^{3} - g_{2} ζ - g_{3})^{- 1 / 2}$ , giving $\frac{α}{2} = \int_{e_{1}}^{\infty} \frac{d x}{\sqrt{4 x^{3} - g_{2} x - g_{3}}} = \int_{e_{2}}^{e_{3}} \frac{d x}{\sqrt{4 x^{3} - g_{2} x - g_{3}}}, \frac{β}{2 i} = \int_{- \infty}^{e_{2}} \frac{d x}{\sqrt{g_{3} + g_{2} x - 4 x^{3}}} = \int_{e_{3}}^{e_{1}} \frac{d x}{\sqrt{g_{3} + g_{2} x - 4 x^{3}}},$ with positive real square roots. In the corresponding Jacobi example, for $0 < k < 1$ put $K (a) = \int_{0}^{1} d s / \sqrt{(1 - s^{2}) (1 - a^{2} s^{2})}$ , $K = K (k)$ and $K^{'} = K (\sqrt{1 - k^{2}})$ . Then $I_{k} (z) = \int_{0}^{z} d ζ / \sqrt{(1 - ζ^{2}) (1 - k^{2} ζ^{2})}$ maps the upper half-plane conformally onto the rectangle with vertices $- K, K, K + i K^{'}, - K + i K^{'}$ , and its inverse sn extends by reflection to a doubly periodic meromorphic function with periods $4 K$ and $2 i K^{'}$ .

Facts & Assumptions

Given: A rectangular lattice $Λ = Z α + Z β$ with $α > 0$ real and $β = i b$ , $b > 0$ , the half-periods $c_{1} = α / 2$ , $c_{2} = β / 2$ , $c_{3} = (α + β) / 2$ , the values $e_{j} = ℘ (c_{j})$ , the rectangle $S = {s α / 2 + t β / 2 : 0 \leq s, t \leq 1}$ , and a parameter $0 < k < 1$ .

[F1]

$Λ = Z α + Z β$ is a full complex lattice with oriented basis, $T_{Λ} = C / Λ$ its torus, and $℘ = ℘_{Λ}$ is the Weierstrass function $z^{- 2} + \sum_{ω \neq 0} ((z - ω)^{- 2} - ω^{- 2})$ of the lattice, with derivative $℘^{'}$ ; $℘$ is holomorphic on $C ∖ Λ$ , even, $Λ$ -periodic, and at every $λ \in Λ$ has a double pole with principal part $(z - λ)^{- 2}$ and no other poles (Complex lattice and quotient torus, Weierstrass p function, Normal convergence, parity and periodicity of the Weierstrass p function).

[F2]

$℘^{'}$ is odd and $Λ$ -periodic with poles of order three exactly at the lattice points; $℘ (z) = ℘ (w)$ holds if and only if $w \equiv \pm z$ modulo $Λ$ ; and the zeros of $℘^{'}$ are exactly the $Λ$ -translates of $c_{1}, c_{2}, c_{3}$ , each of order one (Normal convergence, parity and periodicity of the Weierstrass p function, Degree two of ℘ and its four branch points).

[F3]

$(℘^{'})^{2} = 4 ℘^{3} - g_{2} ℘ - g_{3} = 4 (℘ - e_{1}) (℘ - e_{2}) (℘ - e_{3})$ with $g_{2} = 60 G_{4}$ , $g_{3} = 140 G_{6}$ , and $e_{1}, e_{2}, e_{3}$ are the three distinct roots of the cubic (Weierstrass cubic differential equation, Nonvanishing of the lattice discriminant); differentiating the identity gives $2 ℘^{'} ℘^{''} = (12 ℘^{2} - g_{2}) ℘^{'}$ , hence $℘^{''} = 6 ℘^{2} - g_{2} / 2$ first where $℘^{'} \neq 0$ and then at its isolated zeros by continuity.

[F4]

Complex conjugation is a continuous real-field automorphism, so it commutes with sums, products, quotients and limits of convergent nets of complex numbers (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive); $z$ is real exactly when $z = \overline{z}$ , and $Re z = (z + \overline{z}) / 2$ , $Im z = (z - \overline{z}) / (2 i)$ (Real and imaginary parts, complex conjugation, and modulus, $C$ is the real coordinate plane, with coordinate arithmetic).

[F5]

Every $z \in C$ has a unique representation $z = s α + t β$ with $s, t \in R$ ; subtracting integer parts (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ) gives representatives in the full-period rectangle ${s α + t β : 0 \leq s, t \leq 1}$ (Complex lattice and quotient torus, $C$ is the real coordinate plane, with coordinate arithmetic).

[F6]

For $f$ meromorphic on an open $Ω$ , an admissible cycle $Γ$ , and $f$ not identically zero on any component and nonzero on its trace, the argument principle gives $\frac{1}{2 π i} \int_{Γ} f^{'} / f = Z (f, Γ) - P (f, Γ)$ (The argument principle for an admissible null-homologous cycle). For $w$ avoided on the trace and $f - w$ not identically zero on any component, $\frac{1}{2 π i} \int_{Γ} f^{'} / (f - w) = N_{w} (f, Γ) - P (f, Γ)$ ; when $f$ is holomorphic, $P = 0$ (The argument principle counts preimages of a target value). The winding number is $n (γ, p) = \frac{1}{2 π i} \int_{γ} d ζ / (ζ - p)$ (The winding number of a closed contour about a point off its trace). Positively oriented boundaries of rectangles and of truncated half-discs have index $1$ inside and $0$ outside (Index of the boundary of a graph-bounded plane region). Endpoint-fixed homotopic rectifiable paths have equal integrals of a holomorphic function (Endpoint-fixed homotopic paths have equal holomorphic line integrals); applying this to $(ζ - p)^{- 1}$ proves winding invariance under homotopies avoiding $p$ . If a loop's basepoint moves, insert the basepoint path and its reversal to obtain a fixed-basepoint homotopy; their integrals cancel. Uniformly close closed contours avoiding $p$ are linearly homotopic while still avoiding $p$ , so have the same winding number.

[F7]

A function holomorphic on a half-disc, continuous on its closure and real on its diameter extends holomorphically by complex conjugation across that diameter (Harmonic and holomorphic Schwarz reflection across the real axis). Translating, rotating and rescaling the domain gives the same assertion at a straight side. At a boundary pole, apply this holomorphic result to a holomorphic reciprocal vanishing on the boundary, then invert the extension.

[F8]

The upper half-plane $H$ is a complex domain; its complement in $\hat{C}$ is ${Im z \leq 0} \cup {\infty}$ , the closure in the sphere of the convex set ${Im z \leq 0}$ . That convex set is contractible, hence path-connected, hence connected (Every nonempty convex subset of $R^{n}$ is contractible, Every nonempty contractible space is path-connected, Every path-connected space is connected, and every path component lies inside a component), and the closure of a connected set is connected, so $\hat{C} ∖ H$ is connected (If $A$ is connected and $A \subseteq B \subseteq \overline{A}$ then $B$ is connected; in particular the closure of a connected set is connected, A complex domain is a nonempty connected open subset of $C$ ). A complex domain whose complement in $\hat{C}$ is connected has every cycle null-homologous in it, i.e. is homologically simply connected (A connected spherical complement forces every cycle in the domain to be null-homologous, Homologically simply connected complex domains); on such a domain every holomorphic function has a primitive, and every nowhere-zero holomorphic function has a holomorphic square root (Equivalent characterisations of a homologically simply connected domain, A nonvanishing holomorphic function on such a domain has holomorphic roots of every positive order).

[F9]

If $g$ is continuous on an interval and nowhere zero, then $g$ has constant sign there (Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ ); a positive continuous integrand on an interval gives a strictly increasing integral, and improper integrals at a finite endpoint and at infinity converge or diverge as evaluated there (Improper integrals at a finite singular endpoint, Improper integrals over unbounded intervals); the change-of-variables formula holds for such integrals with the absolute derivative (In one dimension the compact-Jordan formula is substitution over the unoriented image interval with the absolute derivative).

[F10]

An injective holomorphic map on a complex domain is biholomorphic onto its open image, with nowhere-zero derivative (An injective holomorphic map has no critical point and is biholomorphic onto its image).

[F11]

Holomorphic functions have local Taylor expansions and obey the chain and product rules (A holomorphic function equals its Taylor series throughout the largest centred disc in its domain, The chain rule for complex derivatives, Linearity, product, reciprocal, and quotient rules for complex derivatives). A holomorphic function with nonzero derivative has a holomorphic local inverse (Holomorphic inverse function theorem and local-degree criterion). If $g (a) \neq 0$ is holomorphic near $a$ , the implicit function theorem applied to $v^{2} - g (z)$ supplies a nonvanishing holomorphic square root of $g$ near $a$ (The holomorphic implicit function theorem).

Verification

1.1F1F4F5

(Conjugation symmetry of $℘$ and $℘^{'}$ .) Since $\overline{α} = α \in Λ$ and $\overline{β} = - β \in Λ$ , the lattice is conjugation-invariant. Conjugating the defining net of $℘$ termwise for $z \notin Λ$ and using continuity of conjugation and the local uniform convergence of the net ([F1, F4]) gives $℘ (\overline{z}) = \overline{℘ (z)}$ ; differentiating this identity gives $℘^{'} (\overline{z}) = \overline{℘^{'} (z)}$ . In particular $℘$ is real on the real axis, and for $z = x + β / 2$ one has $\overline{z} = z - β \equiv z$ modulo $Λ$ , so $℘ (\overline{z}) = ℘ (z)$ and $℘ (z) = \overline{℘ (z)}$ : $℘$ is real on the two lines $Im z = 0$ and $Im z = b / 2$ .

1.2F1F2F5

(The half-period rectangle: no poles and no critical points inside.) Put $S = {s \frac{α}{2} + t \frac{β}{2} : 0 \leq s, t \leq 1}$ ; its interior $S^{\circ}$ is ${0 < s, t < 1}$ in these coordinates. An interior point $z = s α / 2 + t β / 2$ is not in $Λ$ : if $z = m α + n β$ with $m, n \in Z$ , then $s = 2 m$ and $t = 2 n$ by uniqueness of coordinates ([F5]), but $0 < s, t < 1$ excludes integers. It is not a zero of $℘^{'}$ : such a zero would satisfy $z \equiv c_{j}$ modulo $Λ$ ([F2]) for one of $c_{1} = α / 2$ , $c_{2} = β / 2$ , $c_{3} = (α + β) / 2$ , i.e. $(s, t)$ would differ from $(1, 0)$ , $(0, 1)$ , $(1, 1)$ by even integers, forcing $s \in 2 Z + 1$ or $t \in 2 Z + 1$ , again impossible for $0 < s, t < 1$ . Consequently $℘$ is holomorphic on $S^{\circ}$ , $℘^{'} \neq 0$ there, and for $z, w \in S^{\circ}$ the equality $℘ (z) = ℘ (w)$ forces $w \equiv \pm z$ by [F2]: a lattice shift $w = z + λ$ forces $s^{'} - s, t^{'} - t \in 2 Z$ with $∣ s^{'} - s ∣ < 1$ , hence $s = s^{'}$ , $t = t^{'}$ ; and a sign choice $w = - z + λ$ forces $s + s^{'} \in 2 Z$ with $0 < s + s^{'} < 2$ , hence $s + s^{'} = 2$ (impossible, as $s, s^{'} < 1$ ) or $s + s^{'} = 0$ (impossible). Thus $℘$ is injective on $S^{\circ}$ .

1.3F8F9F11

(The Jacobi integrand: a normalized square root on $H$ .) Let $H = {z : Im z > 0}$ and $P (ζ) = (1 - ζ^{2}) (1 - k^{2} ζ^{2})$ . Its zeros are $\pm 1, \pm 1 / k$ , all real, so $P$ is holomorphic and nowhere zero on the simply connected domain $H$ ; by [F8] there is a holomorphic square root $q$ of $P$ on $H$ . For each real $x$ with $P (x) \neq 0$ , [F11] supplies a local nonvanishing holomorphic square root $v$ of $P$ on a disc about $x$ . The quotient $q / v$ on the connected upper half-disc has square $1$ , so is a constant sign; thus $q$ extends holomorphically across $x$ . On $(- 1, 1)$ the extended function $q^{2} = P$ is positive, so $q$ is continuous and nowhere zero there with $q (x) = \pm \sqrt{P (x)}$ : the sign is constant by [F9], and after replacing $q$ by $- q$ if necessary we may and do assume $q (x) > 0$ for $- 1 < x < 1$ . Then $1 / q$ is holomorphic on $H$ and, by [F8], has a primitive $I_{k} (z) = \int_{0}^{z} d ζ / q (ζ)$ whose extension at $0$ is normalized by $I_{k} (0) = 0$ ; it satisfies $I_{k}^{'} (z) = 1 / q (z) \neq 0$ on $H$ .

2.1F1F2F4F5step 1.1

(The exact real locus.) For $z \in C ∖ Λ$ : $℘ (z) \in R$ iff $℘ (z) = \overline{℘ (z)} = ℘ (\overline{z})$ iff $\overline{z} \equiv \pm z$ modulo $Λ$ by the fibre criterion ([F2]); and $\overline{z} \equiv z$ means $2 i Im z \in Λ$ , i.e. $Im z \in \frac{b}{2} Z$ , while $\overline{z} \equiv - z$ means $2 Re z \in Λ$ , i.e. $Re z \in \frac{α}{2} Z$ ([F4, F5]). Thus on $C ∖ Λ$ the real-value locus is exactly the union of the horizontal lines $Im z \in \frac{b}{2} Z$ and the vertical lines $Re z \in \frac{α}{2} Z$ ; the excluded lattice points on those lines are poles by [F1].

2.2F1F2F3F9step 1.1

(Monotonicity along the four edges and the order $e_{2} < e_{3} < e_{1}$ .) The boundary $\partial S$ consists of the images of the four segments $0 \to c_{1} \to c_{3} \to c_{2} \to 0$ . On the open segment from $0$ to $c_{1}$ (a real interval), $℘$ is real (step 1.1), has no pole and no critical point, so its derivative is continuous and nowhere zero, hence of constant sign ([F9]); since $℘ (x) \to + \infty$ as $x \to 0^{+}$ by the principal part in [F1] and $℘ (c_{1}) = e_{1}$ , this sign is negative and $℘$ decreases strictly from $+ \infty$ to $e_{1}$ . Likewise along the open segment from $c_{1}$ to $c_{3}$ and from $c_{3}$ to $c_{2}$ and from $c_{2}$ to $0$ , all lying on the real locus lines, $℘$ takes real values with nowhere-zero derivative and is thus strictly monotone; the segment from $c_{2}$ to $0$ is the imaginary interval because $c_{2} = i b / 2$ , and $℘ (i y) \to - \infty$ as $y \to 0^{+}$ by [F1], so there $℘$ decreases to $- \infty$ . Because $℘^{'} (c_{j}) = 0$ ([F2]), Taylor expansion at $c_{j}$ using [F3] gives $℘ (c_{j} + w) = e_{j} + \frac{1}{2} ℘^{''} (c_{j}) w^{2} + O (∣ w ∣^{3})$ with $℘^{''} (c_{j}) = 6 e_{j}^{2} - g_{2} / 2$ and, comparing the expansions of both sides of $(℘^{'})^{2} = 4 (℘ - e_{1}) (℘ - e_{2}) (℘ - e_{3})$ , $℘^{''} (c_{j}) = 2 \prod_{i \neq j} (e_{j} - e_{i})$ . At $c_{1}$ the incident edge towards $0$ has values $> e_{1}$ and the incident edge towards $c_{3}$ has values $< e_{1}$ (both signs by the second-order expansion, whose quadratic coefficient for the first edge is $+ ℘^{''} (c_{1}) / 2$ and for the second $- ℘^{''} (c_{1}) / 2$ ), so $e_{3} < e_{1}$ . At $c_{2}$ the incident edge towards $0$ has values $< e_{2}$ and the incident edge towards $c_{3}$ has values $> e_{2}$ , so $e_{3} > e_{2}$ . Hence $e_{2} < e_{3} < e_{1}$ .

2.3F4F8F9F11step 1.3

(Boundary values of $q$ and $I_{k}$ .) Step 1.3 extends $q$ across every real point where $P \neq 0$ . At a simple root $a$ , write $P (z) = (z - a) g_{a} (z)$ with $g_{a} (a) = P^{'} (a) \neq 0$ and use [F11] to choose a holomorphic unit $v_{a}$ with $v_{a}^{2} = g_{a}$ . On the upper half-disc, $q (z) = \pm v_{a} (z) \sqrt{z - a}$ , since the quotient has square $1$ and is constant on that connected set. As one passes from the interval to the left of $a$ to the interval to its right through the upper half-plane, $\sqrt{z - a}$ changes from $i \sqrt{∣ x - a ∣}$ to $\sqrt{∣ x - a ∣}$ ; the unit $v_{a}$ retains its sign continuously. Starting with $q > 0$ on $(- 1, 1)$ , this gives $q = - i \sqrt{∣ P ∣}$ on $(1, 1 / k)$ , $q = + i \sqrt{∣ P ∣}$ on $(- 1 / k, - 1)$ , and $q = - \sqrt{P}$ on both tails. Here $P^{'} (1) = - 2 (1 - k^{2})$ , $P^{'} (- 1) = 2 (1 - k^{2})$ , $P^{'} (1 / k) = 2 (1 - k^{2}) / k$ , and $P^{'} (- 1 / k) = - 2 (1 - k^{2}) / k$ . Moreover $∣ 1 / q (z) ∣ \leq C_{a} ∣ z - a ∣^{- 1 / 2}$ near $a$ in $H$ . Integrating on radial segments and circular arcs gives a finite boundary limit of $I_{k}$ with $∣ I_{k} (z) - I_{k} (a) ∣ \leq C_{a}^{'} ∣ z - a ∣^{1 / 2}$ ; thus the primitive is continuous at each of the four branch points. Hence, using $I_{k}^{'} (x + i 0) = 1 / q (x + i 0)$ and $I_{k} (0) = 0$ : $I_{k}$ is strictly increasing on $(- 1, 1)$ with $I_{k} (\pm 1) = \pm K$ ; $I_{k} (x) = K + i \int_{1}^{x} d s / \sqrt{(s^{2} - 1) (1 - k^{2} s^{2})}$ for $1 \leq x \leq 1 / k$ , so $I_{k} (1 / k) = K + i K^{'} (k)$ where $K^{'} (k) = \int_{1}^{1 / k} d s / \sqrt{(s^{2} - 1) (1 - k^{2} s^{2})}$ ; $I_{k} (x) = - K + i \int_{x}^{- 1} d s / \sqrt{(s^{2} - 1) (1 - k^{2} s^{2})}$ for $- 1 / k \leq x \leq - 1$ , so $I_{k} (- 1 / k) = - K + i K^{'} (k)$ ; and on the tails $I_{k} (x) = I_{k} (1 / k) - \int_{1 / k}^{x} d s / \sqrt{(s^{2} - 1) (k^{2} s^{2} - 1)}$ and $I_{k} (x) = I_{k} (- 1 / k) + \int_{x}^{- 1 / k} d s / \sqrt{(s^{2} - 1) (k^{2} s^{2} - 1)}$ , with $\int_{1 / k}^{\infty} d s / \sqrt{(s^{2} - 1) (k^{2} s^{2} - 1)} = K$ and $\int_{- \infty}^{- 1 / k} d s / \sqrt{(s^{2} - 1) (k^{2} s^{2} - 1)} = K$ , so both tails tend to $i K^{'} (k)$ . Here $\int_{1}^{1 / k} d s / \sqrt{(s^{2} - 1) (1 - k^{2} s^{2})} = K (\sqrt{1 - k^{2}}) = K^{'}$ under the substitution $s = (1 - k^{' 2} u^{2})^{- 1 / 2}$ with $k^{'} = \sqrt{1 - k^{2}}$ , and $\int_{1 / k}^{\infty} d s / \sqrt{(s^{2} - 1) (k^{2} s^{2} - 1)} = K$ under $s = 1 / (k u)$ , both by the change-of-variables formula ([F9]). Finally, for $∣ z ∣ = R > 1 / k$ in $H$ one has $∣ P (z) ∣ \geq (R^{2} - 1) (k^{2} R^{2} - 1)$ , so the integral of $1 / q$ along the semicircle of radius $R$ is $O (1 / R)$ ; since $I_{k} (R) \to i K^{'} (k)$ along the real axis, $I_{k} (z) \to i K^{'} (k)$ uniformly as $∣ z ∣ \to \infty$ in $H$ .

3.1F4step 2.1step 1.2

(The image lies in one half-plane.) $S^{\circ}$ is convex, hence connected, and $℘ (S^{\circ})$ is connected; since $℘^{'} \neq 0$ on $S^{\circ}$ , $℘$ is an open map there and $℘ (S^{\circ})$ is open ([F10]); and $℘ (S^{\circ}) \cap R = \emptyset$ by step 2.1, because no interior point lies on any of the lines $Re z \in \frac{α}{2} Z$ , $Im z \in \frac{b}{2} Z$ . An open connected subset of $C ∖ (R \cup {\infty})$ is contained in the upper or in the lower half-plane.

3.2F1step 2.2

(The boundary maps onto $R \cup {\infty}$ .) By step 2.2 the four open edges have images $(e_{1}, \infty)$ , the interval between $e_{1}$ and $e_{3}$ , the interval between $e_{3}$ and $e_{2}$ , and $(- \infty, e_{2})$ , and with $e_{2} < e_{3} < e_{1}$ these intervals are $(e_{1}, \infty)$ , $(e_{3}, e_{1})$ , $(e_{2}, e_{3})$ and $(- \infty, e_{2})$ , which together with the endpoint values and $℘ (0) = \infty$ cover $R \cup {\infty}$ exactly once on the boundary circle.

3.3

( $I_{k}$ maps $H$ biholomorphically onto the rectangle.) Let $R_{0} = (- K, K) \times (0, K^{'})$ and $R > 1 / k$ . The real-segment path $γ_{R} : = I_{k} ([- R, R])$ starts at $i K^{'} - T_{R}$ and ends at $i K^{'} + T_{R}$ , where $T_{R} = \int_{R}^{\infty} d s / \sqrt{(s^{2} - 1) (k^{2} s^{2} - 1)} > 0$ tends to zero by step 2.3. It follows the boundary of $R_{0}$ counterclockwise except for the short top segment joining those endpoints. The image $δ_{R}$ of the upper semicircle runs from $i K^{'} + T_{R}$ back to $i K^{'} - T_{R}$ and lies in a disc of radius $O (1 / R)$ about $i K^{'}$ by step 2.3. For a fixed $w$ away from $\partial R_{0}$ , choose $R$ large enough that both $δ_{R}$ and the missing top segment lie in a disc about $i K^{'}$ disjoint from $w$ . A straight-line homotopy in that disc deforms $δ_{R}$ to the missing segment, so the closed full image contour $γ_{R} * δ_{R} = I_{k} (\partial D_{R})$ has winding number $1$ about $w \in R_{0}$ and $0$ about $w \notin \overline{R_{0}}$ .

To apply the argument principle without crossing the four branch points on the real boundary, use $D_{R, ϵ} : = {z \in H : ∣ z ∣ < R, Im z > ϵ}$ . The function $I_{k}$ is holomorphic with $I_{k}^{'} = 1 / q \neq 0$ on a neighbourhood of its closure. As $ϵ ↓ 0$ , the image of its closed boundary tends uniformly to $γ_{R} * δ_{R}$ , since step 2.3 gives continuous boundary values, including the integrable square-root endpoints. For $w$ off $\partial R_{0}$ , winding number is stable for small $ϵ$ ; the argument principle [F6], with zero pole count because $I_{k}$ is holomorphic and with index $1$ on the truncated half-disc, therefore gives exactly one preimage in $D_{R, ϵ}$ when $w \in R_{0}$ , and none when $w \notin \overline{R_{0}}$ . Letting $ϵ ↓ 0$ and then $R \to \infty$ proves the same counts on $H$ : any preimage lies in some such truncated half-disc, and step 2.3 gives $I_{k} (z) \to i K^{'}$ at infinity. Thus every $w \in R_{0}$ has exactly one preimage, and no $w \notin \overline{R_{0}}$ has one. An image point on $\partial R_{0}$ would, by $I_{k}^{'} \neq 0$ , have an open image neighbourhood containing a point outside $\overline{R_{0}}$ , impossible. Hence $I_{k} (H) = R_{0}$ , and the injective holomorphic map $I_{k} : H \to R_{0}$ is biholomorphic by [F10]. [F6, F10, step 1.3, step 2.3, algebra]

4.1F10step 3.1step 3.2

(The image of $S$ is exactly one half-plane and $℘ ∣_{S}$ is conformal.) Every boundary point of $℘ (S^{\circ})$ is a limit $℘ (z_{n})$ with $z_{n} \in S^{\circ}$ ; by compactness of $\overline{S}$ pass to a subsequence with $z_{n} \to z_{0} \in \overline{S}$ and $℘ (z_{0}) = w$ with the value $℘ (0) = \infty$ allowed. If $z_{0} \in S^{\circ}$ then $w \in ℘ (S^{\circ})$ , which is impossible for a boundary point because $℘ (S^{\circ})$ is open; hence $z_{0} \in \partial S$ and, by step 3.2, $w \in ℘ (\partial S) \subseteq R \cup {\infty}$ . Thus $\partial ℘ (S^{\circ}) \subseteq R \cup {\infty}$ , while by step 3.1 the set $℘ (S^{\circ})$ is a nonempty open connected subset of one half-plane $U_{\pm}$ , and $U_{\pm} \cap (R \cup {\infty}) = \emptyset$ . If $w \in U_{\pm} ∖ ℘ (S^{\circ})$ , join $w$ to a point $q \in ℘ (S^{\circ})$ by the segment $γ$ inside the convex set $U_{\pm}$ and let $t_{0} = \inf {t : γ (t) \in ℘ (S^{\circ})}$ ; then $γ (t_{0}) \in \partial ℘ (S^{\circ}) \cap U_{\pm}$ , contradicting $\partial ℘ (S^{\circ}) \subseteq R \cup {\infty}$ . Hence $℘ (S^{\circ}) = U_{\pm}$ , and $℘ : S^{\circ} \to U_{\pm}$ is injective with nowhere-zero derivative, hence biholomorphic, and in particular conformal.

5.1F3step 4.1

(Derivative of the inverse branch.) Let $U = U_{\pm}$ and $f = ℘ ∣_{S^{\circ}}^{- 1} : U \to S^{\circ}$ . Then $f$ is holomorphic, $℘ (f (ζ)) = ζ$ for $ζ \in U$ , and the chain rule gives $℘^{'} (f (ζ)) f^{'} (ζ) = 1$ , so $f^{'} (ζ) = 1 / ℘^{'} (f (ζ))$ . Substituting $℘ (f (ζ)) = ζ$ in the differential equation [F3] gives $℘^{'} (f (ζ))^{2} = 4 ζ^{3} - g_{2} ζ - g_{3}$ , and since $f^{'}$ is nowhere zero its reciprocal $ζ \mapsto 1 / f^{'} (ζ) = ℘^{'} (f (ζ))$ is a holomorphic square root of $4 ζ^{3} - g_{2} ζ - g_{3}$ on $U$ ; writing $(4 ζ^{3} - g_{2} ζ - g_{3})^{- 1 / 2}$ for that reciprocal root, every inverse branch satisfies $f^{'} (ζ) = (4 ζ^{3} - g_{2} ζ - g_{3})^{- 1 / 2}$ .

6.1F1F2F3F9F11step 2.2step 3.2step 5.1

(The four period integrals.) On each of the four real intervals between consecutive roots the polynomial $4 x^{3} - g_{2} x - g_{3} = 4 (x - e_{1}) (x - e_{2}) (x - e_{3})$ has constant sign; $4 x^{3} - g_{2} x - g_{3} > 0$ on $(e_{1}, \infty)$ and on $(e_{2}, e_{3})$ , while $g_{3} + g_{2} x - 4 x^{3} > 0$ on $(- \infty, e_{2})$ and on $(e_{3}, e_{1})$ . Each of these intervals is the image under $℘$ of one open edge of $\partial S$ . Since $℘^{'}$ is nonzero on each open edge, the local inverse theorem [F11] extends the inverse branch across the corresponding real interval, mapping it bijectively onto that edge with $∣ f^{'} (x) ∣ = 1 / \sqrt{∣ 4 x^{3} - g_{2} x - g_{3} ∣}$ by step 5.1, so the integral of the positive square root over the interval equals the length of the corresponding displacement: writing the inverse branch as $g$ , $\int_{a}^{b} d x / \sqrt{∣ 4 x^{3} - g_{2} x - g_{3} ∣} = ∣ \lim_{x ↓ a} g (x) - \lim_{x ↑ b} g (x) ∣$ by the change-of-variables formula and strict monotonicity of $g$ ([F9]). The four displacements are: from $0$ to $c_{1}$ , length $α / 2$ ; from $c_{2}$ to $c_{3}$ , length $α / 2$ ; from $c_{2}$ to $0$ , length $∣ β ∣ / 2 = β / (2 i)$ ; from $c_{1}$ to $c_{3}$ , length $β / (2 i)$ . The improper endpoints converge: at a simple root the integrand is $O (∣ x - e_{j} ∣^{- 1 / 2})$ , and at infinity it is $O (∣ x ∣^{- 3 / 2})$ . Hence the four displayed identities hold with the positive real square roots.

7.1F7F10F11step 2.3step 3.3algebra∎

(The inverse sn and its double periodicity.) At a finite boundary branch point $a$ of $I_{k}$ , use $z = a + t^{2}$ . The expression $q (a + t^{2}) / t$ is a holomorphic unit by the factorization in step 2.3; hence $J (t) : = I_{k} (a + t^{2})$ extends holomorphically to $t = 0$ and $J^{'} (0) = 2 / (q (a + t^{2}) / t) ∣_{t = 0} \neq 0$ . By [F11] its local inverse makes $z = a + t (w)^{2}$ holomorphic at the associated rectangle vertex. On the tails, the branch sign in step 2.3 gives $q (z) = - k z^{2} (1 + O (z^{- 2}))$ : this follows by applying [F11] to the square root of $1 - (1 + k^{2}) / (k^{2} z^{2}) + 1 / (k^{2} z^{4})$ near $1 / z = 0$ . Thus $I_{k} (z) = i K^{'} + 1 / (k z) + O (z^{- 3})$ , and $t \mapsto I_{k} (1 / t)$ extends holomorphically at $0$ with derivative $1 / k \neq 0$ . Its local inverse shows that $1 / s n$ is holomorphic with a simple zero at $i K^{'}$ . Let $s n = I_{k}^{- 1} : R_{0} \to H$ ; it is holomorphic ([F10]) and continuous on $\overline{R_{0}} ∖ {i K^{'}}$ with real boundary values: the bottom edge $[- K, K]$ maps into $(- 1, 1)$ with $s n (0) = 0$ , $s n (\pm K) = \pm 1$ ; the left edge $- K + i (0, K^{'})$ maps into $(- 1 / k, - 1)$ with $s n (- K + i K^{'}) = - 1 / k$ (at the boundary branch value $- 1 / k$ ); the right edge $K + i (0, K^{'})$ maps into $(1, 1 / k)$ with $s n (K + i K^{'}) = 1 / k$ ; and the top edge $(- K, K) + i K^{'}$ maps into $R$ with $∣ s n ∣ > 1 / k$ , with a simple pole at $i K^{'}$ (from step 2.3, $I_{k} (z) = i K^{'} + 1 / (k z) + O (z^{- 2})$ for large $z$ , so near the omitted value $i K^{'}$ the inverse behaves like $1 / (k (w - i K^{'}))$ ). Since $s n$ is holomorphic on $R_{0}$ and real on each of the four open sides, the Schwarz reflection principle [F7] extends it across each side by reflection there: away from $i K^{'}$ this is the stated holomorphic reflection, and at $i K^{'}$ one applies the same principle to the reciprocal $u = 1 / s n$ , which is holomorphic near $i K^{'}$ with $u (i K^{'}) = 0$ and real boundary values, so $u$ reflects and $s n = 1 / u$ reflects meromorphically; the reflected copies tile the plane, because the four reflections $σ_{b} (z) = \overline{z}$ , $σ_{t} (z) = \overline{z} + 2 i K^{'}$ , $σ_{l} (z) = - 2 K - \overline{z}$ , $σ_{r} (z) = 2 K - \overline{z}$ have compositions $σ_{b} σ_{t} : z \mapsto z - 2 i K^{'}$ and $σ_{r} σ_{l} : z \mapsto z + 4 K$ , and their reflected rectangle copies tile the plane. Across an open edge the extensions agree by reflection; at a vertex they agree with the holomorphic squared inverse just constructed, and at a reflected copy of $i K^{'}$ they agree with its meromorphic pole extension. Thus the pieces glue on every edge and vertex, producing a single meromorphic function $s n$ on $C$ satisfies $s n (σ_{j} z) = \overline{s n (z)}$ for each reflection $σ_{j}$ , so $s n (z - 2 i K^{'}) = \overline{s n (σ_{t} z)} = s n (z)$ and $s n (z + 4 K) = \overline{s n (σ_{l} z)} = s n (z)$ : $s n$ is doubly periodic with periods $4 K$ and $2 i K^{'}$ (and period $2 i K^{'}$ implies period $- 2 i K^{'}$ ).

Remarks

The sign analysis is the only delicate point. The normalized square root $q$ of $(1 - ζ^{2}) (1 - k^{2} ζ^{2})$ is positive on $(- 1, 1)$ , and the local factorizations at $\pm 1$ and $\pm 1 / k$ force the boundary values $- i \sqrt{∣ P ∣}$ on $(1, 1 / k)$ , $+ i \sqrt{∣ P ∣}$ on $(- 1 / k, - 1)$ , and $- \sqrt{P}$ on both tails; the two tails then approach the same point $i K^{'}$ , which is what closes the image of the real axis into the boundary of the rectangle. The same computation for $℘$ gives $e_{2} < e_{3} < e_{1}$ directly from the second-order expansions at $c_{1}$ and $c_{2}$ ; no numerical evaluation of any elliptic integral is used, only the two standard substitutions reducing $K^{'} (k)$ to $K (\sqrt{1 - k^{2}})$ and the tail integral to $K$ itself.

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The rank-one cotangent and its conic

Example

Put

$P (z) : = π \cot (π z) (z \in C ∖ Z), q : = e^{2 π i z} .$

Then:

the map $z \mapsto q$ induces a bijection $C / Z \to C^{*} = C ∖ {0}$ , and $P (z) = π i (q + 1) / (q - 1)$ for every $z \in C ∖ Z$ ;
$P$ is holomorphic on $C ∖ Z$ , its poles are exactly the integers and they are simple of residue $1$ , its period group is exactly $Z$ , its principal part at $0$ is $1 / z$ , and $- P^{'} (z) = P (z)^{2} + π^{2} (z \notin Z);$
writing $x = P (z)$ and $y = - P^{'} (z)$ , the map $z \mapsto [x : y : 1]$ extends at the class of $0$ to $[0 : 1 : 0]$ and identifies $C / Z$ bijectively with the conic $Y Z = X^{2} + π^{2} Z^{2}$ minus its two points $[- i π : 0 : 1]$ and $[i π : 0 : 1]$ ; off the class of $0$ the identification is holomorphic with nowhere-vanishing derivative.

This is the rank-one analogue of the double-periodic uniformization of the companion page: there the period lattice $Λ$ has rank two, here the period group is $Z$ , and the cubic is replaced by a conic.

Facts & Assumptions

Given: the functions $\sin, \cos$ of Complex sine, cosine, hyperbolic sine, and hyperbolic cosine from the complex exponential, the cotangent $\cot x = \cos x / \sin x$ of Tangent, cotangent, secant, and cosecant on their exact natural domains, the complex exponential $\exp$ of The complex exponential by its power series, and the function $P (z) = π \cot (π z)$ on $C ∖ Z$ .

[F1]

For $z \in C$ , $\sin z = \frac{\exp (i z) - \exp (- i z)}{2 i}$ and $\cos z = \frac{\exp (i z) + \exp (- i z)}{2}$ (Complex sine, cosine, hyperbolic sine, and hyperbolic cosine from the complex exponential).

[F2]

$\cot x = \frac{\cos x}{\sin x}$ for every $x \neq m π$ with $m \in Z$ (Tangent, cotangent, secant, and cosecant on their exact natural domains).

[F3]

The functions $\sin, \cos$ are entire and satisfy $\sin^{'} = \cos$ and $\cos^{'} = - \sin$ (Complex sine, cosine, hyperbolic sine, and hyperbolic cosine are entire with their standard derivatives).

[F4]

$\sin z = 0$ exactly for $z = k π$ with $k \in Z$ , and $\cos z = 0$ exactly for $z = (k + \frac{1}{2}) π$ with $k \in Z$ (The zeros of complex sine are the integer multiples of pi, and the zeros of complex cosine are the odd half-integer multiples of pi).

[F5]

For every $z \in C ∖ Z$ , $π \cot (π z) = \frac{1}{z} + \sum_{n \geq 1} \frac{2 z}{z^{2} - n^{2}}$ , the series converging locally uniformly on $C ∖ Z$ (The Mittag-Leffler expansion of pi cotangent).

[F6]

If $f$ is complex differentiable at $a$ and $g$ at $f (a)$ , then $(g \circ f)^{'} (a) = g^{'} (f (a)) f^{'} (a)$ (The chain rule for complex derivatives).

[F7]

The sum, product, reciprocal and quotient rules displayed in Linearity, product, reciprocal, and quotient rules for complex derivatives hold at every point where the functions are complex differentiable and the denominators do not vanish; in particular $(f / g)^{'} = (f^{'} g - f g^{'}) / g^{2}$ .

[F8]

$\exp z = \sum_{n \geq 0} z^{n} / n!$ for every $z \in C$ , and $\exp (z + w) = \exp z \exp w$ for all $z, w \in C$ (The complex exponential by its power series, $\exp (z + w) = \exp z \exp w$ , and the complex exponential extends the real exponential).

[F9]

$\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ( $\ker (\exp) = 2 π i Z$ , and $\exp z = \exp w$ exactly when $z - w \in 2 π i Z$ ).

[F10]

The exponential maps $C$ onto $C ∖ {0}$ (The complex exponential maps $C$ onto $C ∖ {0}$ ).

Every object below is given by an explicit formula and no choice principle is used.

Verification

technique · direct

Steps 1.1-1.5 compute the quotient description, the Möbius form of $P$ , its principal part, the derivative identity and the conic; steps 2.1-2.4 prove holomorphy and zero-freeness, injectivity modulo the period, the exact period group and the extension across the class of $0$ ; steps 3.1-4.1 record the pole bookkeeping and identify the image.

1.1F8F9F10algebra

The map $Q ([z]) : = e^{2 π i z}$ on $C / Z$ is well defined because $e^{2 π i (z + m)} = e^{2 π i z} e^{2 π i m} = e^{2 π i z}$ for $m \in Z$ , as $2 π i m \in \ker (\exp)$ ; it is injective because $e^{2 π i z} = e^{2 π i w}$ forces $2 π i (z - w) \in 2 π i Z$ , that is $z - w \in Z$ ; it is surjective because every $v \neq 0$ is $e^{w}$ for some $w$ and then $v = Q ([w / (2 π i)])$ ; and it is holomorphic with derivative $2 π i e^{2 π i z} \neq 0$ at every point.

1.2F1F2F4algebra

For $z \in C ∖ Z$ put $s = e^{i π z}$ , so that $s^{2} = q$ and $s^{- 1} = e^{- i π z}$ ; then $\sin (π z) = \frac{s - s^{- 1}}{2 i}$ and $\cos (π z) = \frac{s + s^{- 1}}{2}$ by [F1], and $\sin (π z) \neq 0$ by [F4], so by [F2] and multiplication of numerator and denominator by $s$ , $P (z) = π \frac{\cos (π z)}{\sin (π z)} = π i \frac{s + s^{- 1}}{s - s^{- 1}} = π i \frac{s^{2} + 1}{s^{2} - 1} = π i \frac{q + 1}{q - 1}$ .

1.3F5algebra

The expansion of [F5] shows that near $z = 0$ the sum $\sum_{n \geq 1} \frac{2 z}{z^{2} - n^{2}}$ tends to $\sum_{n \geq 1} (\frac{1}{0 - n} + \frac{1}{0 + n}) = 0$ , so $P (z) - \frac{1}{z}$ extends holomorphically to $0$ with value $0$ ; equivalently $P$ has principal part $1 / z$ at $0$ .

1.4F3F6F7algebra

Differentiating $P (z) = π \cos (π z) / \sin (π z)$ by the chain rule [F3, F6] and the quotient rule [F7], and using $\sin (π z) \neq 0$ , gives $P^{'} (z) = - π^{2} / \sin^{2} (π z)$ , so $- P^{'} (z) = π^{2} / \sin^{2} (π z)$ ; since $\sin^{2} + \cos^{2} = 1$ , dividing by $\sin^{2}$ gives $\csc^{2} = 1 + \cot^{2}$ and hence $- P^{'} (z) = π^{2} (1 + \cot^{2} (π z)) = π^{2} + P (z)^{2}$ .

1.5algebragiven

The projective curve $Y Z = X^{2} + π^{2} Z^{2}$ has a unique point with $Z = 0$ , namely $[0 : 1 : 0]$ , because $Z = 0$ forces $X^{2} = 0$ ; in the chart $Y = 1$ its equation is $v = u^{2} + π^{2} v^{2}$ with $u = X / Y$ , $v = Z / Y$ , and $\partial_{v} (v - u^{2} - π^{2} v^{2}) = 1 - 2 π^{2} v$ equals $1$ at the origin, so $u$ is a local coordinate there. In the chart $Z = 1$ the equation is the parabola $y = x^{2} + π^{2}$ , whose projection to the $x$ -line is a bijection onto $C$ ; hence every conic point is $[x : x^{2} + π^{2} : 1]$ for a unique $x \in C$ , or $[0 : 1 : 0]$ , and the two points with vanishing $y$ are $[\pm i π : 0 : 1]$ .

2.1F4F7F8step 1.2algebra

The function $P$ is holomorphic on $C ∖ Z$ : there it is the composite of $z \mapsto e^{2 π i z}$ with the rational function $q \mapsto π i (q + 1) / (q - 1)$ , holomorphic on $C^{*} ∖ {1}$ . By [F4], its zeros are exactly $\frac{1}{2} + Z$ .

2.2F9step 1.2step 1.4algebra

If $P (z) = P (w)$ for $z, w \in C ∖ Z$ , then step 1.2 gives $π i (q + 1) / (q - 1) = π i (q^{'} + 1) / (q^{'} - 1)$ with $q = e^{2 π i z} \neq 1$ and $q^{'} = e^{2 π i w} \neq 1$ ; cross-multiplying, $q q^{'} - q + q^{'} - 1 = q^{'} q - q^{'} + q - 1$ , hence $2 q^{'} = 2 q$ , $q = q^{'}$ , and $z - w \in Z$ by [F9]. Thus $P$ is injective on the coset space $(C ∖ Z) / Z$ ; since $P^{'} \neq 0$ there by step 1.4, $P$ is locally biholomorphic on $C ∖ Z$ .

2.3F8F9step 1.2step 1.4

For $z \notin Z$ and $T \in C$ with $z + T \notin Z$ , step 1.2 gives $P (z + T) = P (z)$ if and only if $e^{2 π i T} = 1$ , which by [F9] holds exactly when $T \in Z$ ; since $P$ is nonconstant by step 1.4, the period group of $P$ is exactly $Z$ .

2.4F1F8step 1.2step 1.4algebra

On $C ∖ Z$ the point $[P (z) : - P^{'} (z) : 1]$ equals $[u (z) : 1 : v (z)]$ with $u = P / (- P^{'}) = \sin (2 π z) / (2 π)$ and $v = 1 / (- P^{'}) = \sin^{2} (π z) / π^{2}$ , using step 1.4 and $\sin (2 π z) = 2 \sin (π z) \cos (π z)$ ; writing $q = 1 + 2 π i z h (z)$ with $h (z) = \sum_{k \geq 1} (2 π i z)^{k - 1} / k!$ entire and $h (0) = 1$ by [F8], step 1.2 gives $P (z) = \frac{1}{z} \cdot \frac{1 + π i z h (z)}{h (z)}$ , so $u$ and $v$ are holomorphic near $0$ with $u (0) = v (0) = 0$ and $v = u^{2} + π^{2} v^{2}$ by step 1.4. Thus $z \mapsto [u (z) : 1 : v (z)]$ extends holomorphically to $z = 0$ with value $[0 : 1 : 0]$ .

3.1F5step 2.1algebra

By [F5] the function $P (z) - \frac{1}{z} = \sum_{n \geq 1} \frac{2 z}{z^{2} - n^{2}}$ is holomorphic on $C ∖ Z$ ; near $z = m \in Z ∖ {0}$ the summand with $n = ∣ m ∣$ contributes the only singularity, a simple pole of residue $1$ , so the poles of $P$ are exactly the integers, all simple of residue $1$ . Consequently the only class of $C / Z$ at which $P$ is not defined is the class of $0$ .

4.1step 1.1step 1.2step 1.5step 2.2step 2.3step 2.4step 3.1∎

By steps 2.2 and 2.3 the map is invariant under $Z$ and injective on $(C ∖ Z) / Z$ , so adding the class of $0$ gives a bijection onto its image; by step 3.1 every other class is in the domain of $P$ ; for $z \notin Z$ the finite image $[x : y : 1]$ has $y = - P^{'} (z) = π^{2} / \sin^{2} (π z) \neq 0$ , so it avoids $[\pm i π : 0 : 1]$ , and $x = P (z)$ runs exactly once over $C ∖ {\pm i π}$ because $q$ runs exactly once over $C^{*} ∖ {1}$ by step 1.1 and $q \mapsto π i (q + 1) / (q - 1)$ is a bijection $C^{*} ∖ {1} \to C ∖ {\pm i π}$ with inverse $x \mapsto (x + π i) / (x - π i)$ by step 1.2. Hence the extended map is a bijection of $C / Z$ onto the conic minus $[\pm i π : 0 : 1]$ , holomorphic with nowhere-vanishing derivative off the class of $0$ .

The excluded points $[- i π : 0 : 1]$ and $[i π : 0 : 1]$ correspond to $q = 0$ and $q = \infty$ , respectively, approached as $Im z \to + \infty$ and $Im z \to - \infty$ . At the real half-periods $z = \pm \frac{1}{2}$ , one has $P (z) = 0$ .

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-10-02Open item page →

A canonical reduced basis for a complex lattice

Example

Call a ratio $τ$ reduced when

$Im τ > 0, - \frac{1}{2} < Re τ \leq \frac{1}{2}, ∣ τ ∣ \geq 1, and Re τ \geq 0 whenever ∣ τ ∣ = 1,$

and let $R$ be the set of reduced ratios. Every full complex lattice $Λ$ admits an oriented basis $(ω_{1}, ω_{2})$ with $τ = ω_{2} / ω_{1} \in R$ , this reduced ratio is uniquely determined by $Λ$ , and the number of oriented bases of $Λ$ realizing it is two in general, four when $τ = i$ , and six when $τ = e^{i π / 3} = \frac{1}{2} + \frac{\sqrt{3}}{2} i$ . The lattices $Z + i Z$ and $Z + Z e^{i π / 3}$ realize the exceptional ratios $i$ and $e^{i π / 3}$ .

Facts & Assumptions

Given: A full complex lattice $Λ = Z ω_{1} + Z ω_{2}$ with $ω_{1}, ω_{2}$ real-linearly independent, and $τ : = ω_{2} / ω_{1}$ .

[F1]

$ω_{1}, ω_{2}$ are real-linearly independent, the pair $(ω_{1}, ω_{2})$ is an oriented basis when $Im (ω_{2} / ω_{1}) > 0$ , two oriented bases of one lattice differ by a matrix in ${S L}_{2} (Z)$ , and all lattice-theoretic structure depends on the set $Λ$ alone (Complex lattice and quotient torus).

[F2]

Every $z \in C$ has unique real coordinates $z = a + b i$ ; $Re z = a$ , $Im z = b$ , $\overline{z} = a - b i$ and $∣ z ∣ = \sqrt{a^{2} + b^{2}}$ (Real and imaginary parts, complex conjugation, and modulus).

[F3]

For all $z, w \in C$ : $z \overline{z} = ∣ z ∣^{2}$ , $∣ z ∣ \geq 0$ , $∣ z ∣ = 0$ exactly when $z = 0$ , $∣ z w ∣ = ∣ z ∣ ∣ w ∣$ and $∣ z + w ∣ \leq ∣ z ∣ + ∣ w ∣$ ; conjugation is an involutive real-field automorphism (Conjugation is an involutive real-field automorphism, $z \overline{z} = ∣ z ∣^{2}$ , and modulus is definite, multiplicative, and subadditive).

[F4]

For every real $x$ there is exactly one integer $m = ⌊ x ⌋$ with $m \leq x < m + 1$ , hence an integer $m$ with $∣ x - m ∣ \leq \frac{1}{2}$ , namely $m = ⌊ x + \frac{1}{2} ⌋$ (Integer part: for every real $x$ there is exactly one integer $m$ with $m \leq x < m + 1$ ).

Verification

technique · direct

1.1F1algebra

Real-linear independence of $ω_{1}, ω_{2}$ gives $τ \notin R$ , so after exchanging the two vectors if necessary one has $Im τ > 0$ and $(ω_{1}, ω_{2})$ is an oriented basis of $Λ$ ; every oriented basis of $Λ$ is $(a ω_{1} + b ω_{2}, c ω_{1} + d ω_{2})$ with integers $a, b, c, d$ satisfying $a d - b c = 1$ , and conversely every such tuple yields an oriented basis, the coefficients being unique because $ω_{1}, ω_{2}$ are real-linearly independent.

1.2F2F3algebra

Writing $τ = x + i y$ with $x, y \in R$ and $y > 0$ , the form $q (a, b) : = ∣ a + b τ ∣^{2} = a^{2} + 2 a b x + b^{2} (x^{2} + y^{2})$ equals $(a + x b)^{2} + y^{2} b^{2}$ and also $x^{2} + y^{2}$ times a square plus $y^{2} a^{2} / (x^{2} + y^{2})$ , so $q (a, b) \geq y^{2} b^{2}$ and $q (a, b) \geq y^{2} a^{2} / (x^{2} + y^{2})$ ; hence $q (a, b) \geq ε^{2} \max (a^{2}, b^{2})$ with $ε^{2} : = y^{2} / \max (1, x^{2} + y^{2}) > 0$ for all real $a, b$ .

1.3F2F3algebra

A matrix $A \in {S L}_{2} (Z)$ fixes $τ_{0}$ exactly when $b τ_{0}^{2} + (a - d) τ_{0} - c = 0$ ; if $b = 0$ this equation and $a d = 1$ give $a = d = \pm 1$ , $c = 0$ , so $A = \pm I$ . If $b \neq 0$ , the discriminant $(a - d)^{2} + 4 b c = (a + d)^{2} - 4$ is negative, so the trace $t = a + d$ lies in ${- 1, 0, 1}$ ; writing $τ_{0} = u + i v$ with $v > 0$ , the fixed point equation reads $b (u^{2} - v^{2}) + (a - d) u - c + (2 b u + a - d) v i = 0$ , so $u = (d - a) / (2 b)$ and $v^{2} = - ((a - d)^{2} + 4 b c) / (4 b^{2}) = (4 - t^{2}) / (4 b^{2})$ .

2.1F2F3step 1.1algebra

For the basis of step 1.1 the ratio is $τ_{A} = (c + d τ) / (a + b τ)$ with $A = (\begin{matrix} a & b \\ c & d \end{matrix}) \in {S L}_{2} (Z)$ , and $Im τ_{A} = Im τ / ∣ a + b τ ∣^{2}$ .

2.2step 1.3F2F3algebra

For trace $t = 0$ , $d = - a$ , so step 1.3 gives $u = - a / b$ and $v^{2} = 1 / b^{2}$ . At a reduced fixed point, $∣ u ∣ \leq \frac{1}{2}$ and $∣ τ_{0} ∣ \geq 1$ , whence $4 a^{2} \leq b^{2} \leq a^{2} + 1$ . Thus $3 a^{2} \leq 1$ , forcing the integer $a = 0$ , then $∣ b ∣ = 1$ and $τ_{0} = i$ . Now $d = 0$ and $- b c = 1$ , giving precisely $A = \pm (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix}) = \mp S$ , where $S = (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$ ; both fix $i$ directly.

2.3step 1.3F3algebra

For trace $t = \pm 1$ : replacing $A$ by $- A$ changes the sign of $t$ and leaves the fixed points unchanged, so take $t = 1$ ; then $d = 1 - a$ and $v^{2} = 3 / (4 b^{2})$ , so with $B = ∣ b ∣ \geq 1$ and $σ = b / B \in {- 1, 1}$ one has $τ_{0} = σ (1 - 2 a) / (2 B) + \frac{\sqrt{3}}{2 B} i$ , and the constraints $∣ Re τ_{0} ∣ \leq \frac{1}{2}$ and $∣ τ_{0} ∣ \geq 1$ give $∣ 1 - 2 a ∣ \leq B$ and $B^{2} \leq a^{2} - a + 1$ , hence $3 a^{2} - 3 a \leq 0$ and $a \in {0, 1}$ . For $a = 0$ the determinant condition gives $c = - σ$ and $τ_{0} = σ / 2 + \frac{\sqrt{3}}{2} i$ , which lies in $R$ only for $σ = 1$ , giving $τ_{0} = e^{i π / 3}$ and $A = (\begin{matrix} 0 & 1 \\ - 1 & 1 \end{matrix})$ ; for $a = 1$ it gives $c = - σ$ , $τ_{0} = - σ / 2 + \frac{\sqrt{3}}{2} i$ , which lies in $R$ only for $σ = - 1$ , giving again $τ_{0} = e^{i π / 3}$ and $A = (\begin{matrix} 1 & - 1 \\ 1 & 0 \end{matrix})$ . Together with their negatives and $\pm I$ these six matrices form the stabiliser of $e^{i π / 3}$ , and both displayed matrices are checked directly to fix $e^{i π / 3}$ .

3.1step 2.1step 1.2algebra

Only finitely many values $Im τ_{A}$ satisfy $Im τ_{A} \geq \frac{1}{2} Im τ$ : by steps 2.1 and 1.2 that condition implies $∣ a + b τ ∣^{2} \leq 2$ , hence $a^{2} + b^{2} \leq 4 / ε^{2}$ , which has only finitely many integer solutions $(a, b)$ . The value $Im τ_{A} = Im τ / ∣ a + b τ ∣^{2}$ depends only on $(a, b)$ ; the determinant equation may have infinitely many solutions $(c, d)$ .

3.2step 2.1F2F3algebra

For uniqueness let $τ, τ^{'} \in R$ and suppose $τ^{'} = (c + d τ) / (a + b τ)$ with $A = (\begin{matrix} a & b \\ c & d \end{matrix}) \in {S L}_{2} (Z)$ ; replacing $A$ by $A^{- 1} = (\begin{matrix} d & - b \\ - c & a \end{matrix})$ , which also lies in ${S L}_{2} (Z)$ and expresses $τ$ through $τ^{'}$ in the same form, we may assume $Im τ^{'} \geq Im τ$ , and then step 2.1 gives $∣ a + b τ ∣ \leq 1$ ; moreover $Im τ \geq \sqrt{3} / 2$ , because ${Im}^{2} τ = ∣ τ ∣^{2} - {Re}^{2} τ \geq 1 - \frac{1}{4}$ .

3.3step 1.1step 2.1algebra

Fix an oriented basis $(ω_{1}, ω_{2})$ of $Λ$ with reduced ratio $τ_{0} \in R$ . By step 1.1 the oriented bases of $Λ$ are exactly the $(a ω_{1} + b ω_{2}, c ω_{1} + d ω_{2})$ with $A \in {S L}_{2} (Z)$ , and by step 2.1 such a basis again has ratio $τ_{0}$ exactly when $A$ lies in the stabiliser $Stab (τ_{0}) = {A \in {S L}_{2} (Z) : (c + d τ_{0}) / (a + b τ_{0}) = τ_{0}}$ ; the assignment $A \mapsto (a ω_{1} + b ω_{2}, c ω_{1} + d ω_{2})$ is injective, so the oriented bases of $Λ$ with reduced ratio $τ_{0}$ are in bijection with $Stab (τ_{0})$ .

4.1step 2.1step 3.1algebra

Some oriented basis of $Λ$ has maximal imaginary part of its ratio: the set of values $Im τ_{A}$ over $A \in {S L}_{2} (Z)$ contains $Im τ$ (take $A = I$ ), so it meets $[\frac{1}{2} Im τ, \infty)$ , and by step 3.1 the values in that interval form a nonempty finite set; its maximum is attained at some matrix $A_{0}$ and dominates every value, because a value outside the interval is $< \frac{1}{2} Im τ \leq Im τ$ .

4.2step 3.2algebra

If $b = 0$ , then $a d = 1$ forces $a = d = \pm 1$ and $τ^{'} = (c + a τ) / a = τ + c / a$ ; both $Re τ$ and $Re τ^{'}$ lie in $(- \frac{1}{2}, \frac{1}{2}]$ , so $c / a = 0$ , hence $τ^{'} = τ$ .

4.3step 3.2algebra

If $b \neq 0$ , then replacing $A$ by $- A$ leaves $τ^{'}$ unchanged, so we may assume $b \geq 1$ ; by step 3.2, $b Im τ \leq ∣ a + b τ ∣ \leq 1$ , so $b \leq 1 / Im τ \leq 2 / \sqrt{3} < 2$ and therefore $b = 1$ .

5.1F4step 1.1step 4.1algebra

Let $(ω_{1}^{*}, ω_{2}^{*})$ realize the maximum of step 4.1, with ratio $τ^{*}$ . Replacing $ω_{2}^{*}$ by $k ω_{1}^{*} + ω_{2}^{*}$ changes the ratio to $τ^{*} + k$ without changing its imaginary part or orientation. Choose $k = - ⌊ Re τ^{*} + \frac{1}{2} ⌋$ ; if the resulting real part is $- \frac{1}{2}$ , add one more copy of $ω_{1}^{*}$ . Thus we may suppose $- \frac{1}{2} < Re τ^{*} \leq \frac{1}{2}$ , still with maximal imaginary part.

5.2step 4.2step 4.3F2F3algebra

With $b = 1$ the bound $∣ a + τ ∣ \leq 1$ of step 3.2 reads $a^{2} + 2 a Re τ + ∣ τ ∣^{2} \leq 1$ , hence $a (a + 2 Re τ) \leq 1 - ∣ τ ∣^{2} \leq 0$ , and we distinguish three cases. If $a \geq 1$ , then $a + 2 Re τ \leq 0$ gives $Re τ \leq - \frac{1}{2}$ , contradicting $τ \in R$ . If $a = 0$ , then $∣ τ ∣ = 1$ ; the determinant condition $a d - b c = 1$ gives $c = - 1$ and $τ^{'} = (c + d τ) / τ = d - \overline{τ}$ , and $τ^{'} \in R$ forces $d = 0$ , $τ = i$ or $d = 1$ , $τ = e^{i π / 3}$ , in both cases $τ^{'} = τ$ . If $a \leq - 1$ , then $a + 2 Re τ \geq 0$ gives $Re τ \geq - \frac{a}{2} \geq \frac{1}{2}$ , so $Re τ = \frac{1}{2}$ ; then $a^{2} + a + ∣ τ ∣^{2} \leq 1$ and $∣ τ ∣ \geq 1$ give $a \in {- 1, 0}$ and $∣ τ ∣ = 1$ , so $a = - 1$ , $τ = e^{i π / 3}$ , and with $d = - c - 1$ one computes $τ^{'} = - c + e^{2 π i / 3} \in R$ , which forces $c = - 1$ and $τ^{'} = τ \in R$ . Hence $τ^{'} = τ$ in every case of $b \neq 0$ .

6.1step 5.1F3algebra

If $∣ τ^{*} ∣ < 1$ , then $(- ω_{2}^{*}, ω_{1}^{*})$ is an oriented basis of $Λ$ whose ratio $- 1 / τ^{*}$ has $Im (- 1 / τ^{*}) = Im τ^{*} / ∣ τ^{*} ∣^{2} > Im τ^{*}$ , contradicting maximality; hence $∣ τ^{*} ∣ \geq 1$ .

7.1step 5.1step 6.1F3algebra

The ratio now has positive imaginary part, $- \frac{1}{2} < Re τ^{*} \leq \frac{1}{2}$ and $∣ τ^{*} ∣ \geq 1$ . It is reduced unless $∣ τ^{*} ∣ = 1$ and $Re τ^{*} < 0$ . In that case the oriented basis $(- ω_{2}^{*}, ω_{1}^{*})$ has ratio $- 1 / τ^{*} = - \overline{τ^{*}}$ , with real part in $(0, \frac{1}{2})$ , modulus $1$ and the same positive imaginary part, hence lies in $R$ .

8.1step 7.1step 4.2step 5.2

Steps 4.2 and 5.2 prove that two reduced ratios related by a basis change are equal; with step 7.1 this gives existence and uniqueness of the reduced ratio of $Λ$ , and shows it is realized by at least one oriented basis.

9.1step 3.3step 2.2step 2.3algebra∎

By steps 1.3, 2.2 and 2.3 the stabiliser of $τ_{0} \in R$ is ${\pm I, \pm S}$ , of order four, when $τ_{0} = i$ , the six-element set ${\pm I, \pm (\begin{matrix} 0 & 1 \\ - 1 & 1 \end{matrix}), \pm (\begin{matrix} 1 & - 1 \\ 1 & 0 \end{matrix})}$ when $τ_{0} = e^{i π / 3}$ , and ${\pm I}$ , of order two, for every other reduced $τ_{0}$ ; by step 3.3 these are exactly the numbers of oriented bases of $Λ$ with reduced ratio $τ_{0}$ . Finally $Z + i Z$ has the reduced oriented basis $(1, i)$ with ratio $i$ , and $Z + Z e^{i π / 3}$ has the reduced oriented basis $(1, e^{i π / 3})$ with ratio $e^{i π / 3}$ , so the exceptional cases occur.

The reduction uses the basis changes $τ \mapsto τ + k$ and $τ \mapsto - 1 / τ$ . Positive definiteness of $∣ a + b τ ∣^{2}$ makes the relevant denominator pairs finite; step 5.2 handles the boundary of the modular fundamental domain.

Elliptic Functions and Complex Tori: Examples and Counterexamples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, theorems and proofs

5 · Examples, counterexamples and false statements

Oriented bases and SL2(Z)SL2​(Z)

Example

Facts & Assumptions

Verification

Moving the boundary of a fundamental parallelogram

Example

Facts & Assumptions

Verification

Remarks

Square and hexagonal lattice invariants

Example

Facts & Assumptions

Verification

Remarks

Half-period values of the square lattice

Example

Facts & Assumptions

Verification

Remarks

Addition and duplication for ℘℘

Example

Facts & Assumptions

Verification

Remarks

A simple zero of the Weierstrass sigma function on the square lattice

Example

Facts & Assumptions

Verification

Remarks

A singular cubic outside the lattice family

Example

Facts & Assumptions

Verification

Remarks

Rectangular lattices, real mapping, and inverse elliptic integrals

Example

Facts & Assumptions

Verification

Remarks

The rank-one cotangent and its conic

Example

Facts & Assumptions

Verification

A canonical reduced basis for a complex lattice

Example

Facts & Assumptions

Verification

Sources

Oriented bases and ${S L}_{2} (Z)$

Addition and duplication for $℘$