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

Example

Let Λ=Z+iZ, with oriented basis ω1=1, ω2=i, and let ζ=ζΛ, σ=σΛ be the Weierstrass zeta and sigma functions. Then σ(0)=σ(1)=0, σ′(0)=1 and σ′(1)=−exp⁡(η1/2)≠0 where η1=2ζ(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+iZ=Zω1+Zω2 with ω1=1, ω2=i, its Weierstrass functions ζ=ζΛ and σ=σΛ (Weierstrass ζ and σ functions), and η1=2ζ(ω1/2)=2ζ(1/2).

[F1]

Λ=Zω1+Zω2 is a full complex lattice with oriented basis ω1=1, ω2=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∈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∈C∖Λ. Moreover the quasi-period laws hold for all z∈C with poles matched: ζ(z+ωj)=ζ(z)+ηj and σ(z+ωj)=−exp⁡(ηj(z+ωj/2))σ(z) for j=1,2, where ηj=2ζ(ωj/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)≠0 for every z∈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∣≤1, n!≥2n−1 for n≥1 gives ∣exp⁡(u)−1∣≤∣u∣∑n≥11/n!≤2∣u∣. The addition law then gives ∣exp⁡(v+u)−exp⁡(v)∣≤2∣exp⁡(v)∣ ∣u∣, so exp⁡ is continuous at every v.

Verification

1.1F1F2

(Values at 0.) Since 0∈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=Zω1+Zω2; 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: ω1=1 and η1=2ζ(1/2), so σ(1)=−exp⁡(η1⋅12)σ(0)=−exp⁡(η1/2)⋅0=0.

3.1F2F4step 1.1step 2.1

(σ′(1)=−exp⁡(η1/2)≠0.) For u≠0 the quasi-period law and step 2.1 give [F2, step 2.1] σ(1+u)−σ(1)u=−exp⁡ ⁣(η1(u+1/2))σ(u)−σ(0)u. As u→0, the second factor tends to σ′(0)=1 by step 1.1, and the exponential tends to exp⁡(η1/2) by the continuity derived in [F4]. Thus σ′(1)=−exp⁡(η1/2)≠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⁡(η1/2)≠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⁡(η1(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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