The Weierstrass Function
Fix β, γ ∈ (0, ∞). Then
1
℘(z) = 2 +
z
X
ω∈γZZ⊕iβZZ
ω6=0
1
1
−
:C→C
2
2
(z − ω)
ω
is the Weierstrass function with primitive periods γ, iβ.
It obeys
a) ℘(z) is analytic on C \ (γZZ ⊕ iβZZ).
b) ℘(z + ζ) = ℘(z) for all ζ ∈ γZZ ⊕ iβZZ.
c) ℘(−z) = ℘(z) and ℘(z) = ℘(z̄)
⇒ ℘(x + in β2 ), ℘(iy + n γ2 ) are real ∀x, y ∈ IR, n ∈ ZZ
d) Let c ∈ C. Then ℘(z) = c for exactly two z’s in
each fundamental domain.
⇒ ℘(z) = ℘(z 0 ) if and only if z − z 0 ∈ γZZ ⊕ iβZZ or
z + z 0 ∈ γZZ ⊕ iβZZ.
If z ∈
/ γZZ ⊕ iβZZ but 2z ∈ γZZ ⊕ iβZZ, ℘0 (z) = 0.
WF 1
Theorem W.2 Let f (z) be a nonconstant meromorphic
function that is periodic with respect to Ω = ω1 ZZ + ω2 ZZ.
Suppose that f (z) has poles of order n1 , · · · , nk at p1 +
Ω, · · · , pk + Ω and is analytic elsewhere. Let c be any
complex number. Suppose that f (z) − c has zeroes of
order m1 , · · · , mh at z1 + Ω, · · · , zh + Ω and is nonzero
elsewhere. Then
h
X
i=1
Idea of Proof.
mi =
k
X
nk
i=1
For any nonconstant meromorphic
function f (z) and any domain D
Z
f 0 (z)
f (z) dz = 2πi # zeroes in D − # poles in D
∂D
Choose D of the form
z0 + ω 2
z0
z0 + ω 1
with no zeroes or poles on D.
WF 2
Weierstrass Function Relatives
Define
ζ(z) =
1
z
+
X
1
z−ω
+
1
ω
+
z
ω2
ω∈γZZ⊕iβZZ
ω6=0
σ(z) = z
Y
1−
z
ω
ω∈γZZ⊕iβZZ
ω6=0
e
1 z2
z
+
ω
2 ω2
They obey
a) ζ(z) is analytic on C \ (γZZ ⊕ iβZZ). σ(z) is entire
and vanishes if and only if z ∈ γZZ ⊕ iβZZ.
b) ζ(z) =
σ 0 (z)
σ(z)
and ℘(z) = −ζ 0 (z).
c) There are constants η1 ∈ IR , η2 ∈ iIR satisfying
η1 iβ − η2 γ = 2πi such that
ζ(z + γ) = ζ(z) + η1
ζ(z + iβ) = ζ(z) + η2
σ(z + iβ) = −σ(z) e
η2 (z+i β
2)
σ(z + γ) = −σ(z) e
η1 (z+ γ2 )
d) ζ(−z) = −ζ(z) and ζ(z) = ζ(z̄).
⇒ ζ(x) ∈ IR ∀x ∈ IR and ζ(iy) ∈ iIR ∀y ∈ IR
σ(−z) = −σ(z) and σ(z) = σ(z̄).
WF 3
e) ℘(u + v) + ℘(u) + ℘(v) = ζ(u + v) − ζ(u) − ζ(v)
2
Idea of Proof. For each fixed v ∈ C \ (γZZ ⊕ iβZZ),
both the left and right hand sides are periodic and
have double poles, with the same singular part, at
each u ∈ γZZ ⊕ iβZZ and each u ∈ −v + γZZ ⊕ iβZZ.
Define
k(z) = −i ζ(z) −
z ηγ1
It obeys
a) k(z) is analytic on C \ (γZZ ⊕ iβZZ).
b) k(z + γ) = k(z) and k(z + iβ) = k(z) −
2π
γ .
c) k(−z) = −k(z) and k(z) = −k(z̄).
γ
⇒ k(iy), k iy + 2 ∈ IR for all y ∈ IR.
k(x) ∈ iIR, k(x + i β2 ) ∈
π
γ
+ iIR for all x ∈ IR.
WF 4
Set, for z ∈ C \ γZZ ⊕ iβZZ ,
ϕ(z, x) = e
ζ(z)x
σ z
σ
λ(z) = −℘(z)
k(z) = −i ζ(z) −
− x − i β2
β
x + i2
z ηγ1
ξ(z) = eγik(z) = eγζ(z)−zη1
Lemma S.11
a)
d2
− dx2 ϕ(z, x)
+ 2℘ x +
i β2
ϕ(z, x) = λ(z)ϕ(z, x)
b) ϕ(z, x + γ) = ξ(z) ϕ(z, x)
c) ξ(z + γ) = ξ(z)
ξ(z + iβ) = ξ(z)
LM 1
a) First observe that
β
d σ z −x−i2
dx σ x + i β2
0
β β
β
0
σ x + i2 σ z − x− i2
σ z −x−i2
=−
+
β
β
β
σ z −x−i2
σ x + i2
σ x + i2
β
h
i
σ z −x−i2
β
β
= − ζ z − x − i2 + ζ x + i2
β
σ x+ i2
β
β
d
⇒ dx ϕ(z, x) = ζ(z)−ζ z − x − i 2 −ζ x + i 2 ϕ(z, x)
Proof:
As ζ(u + v) − ζ(u) − ζ(v)
d2
dx2 ϕ(z, x)
0
i β2
2
0
= ℘(u + v) + ℘(u) + ℘(v)
i β2
= ζ z−x−
−ζ x+
ϕ(z, x)
β
β 2
+ ζ(z) − ζ z − x − i 2 − ζ x + i 2 ϕ(z, x)
β
β
= − ℘ z − x − i 2 − ℘ x + i 2 ϕ(z, x)
β
β 2
+ ζ(z) − ζ z − x − i 2 − ζ x + i 2 ϕ(z, x)
β
β
= − ℘ z − x − i 2 − ℘ x + i 2 ϕ(z, x)
β
β
+ ℘(z) + ℘ z − x − i 2 + ℘ x + i 2 ϕ(z, x)
β
= ℘(z) + 2℘ x + i 2 ϕ(z, x)
LM 2
b)
β
ζ(z)(x+γ) σ z − x − γ − i 2
ϕ(z, x + γ) = e
β
σ x+γ +i2
β
ζ(z)(x+γ) σ − z + x + γ + i 2
= −e
β
σ x+γ +i2
η (−z+x+i β + γ )
β
1
2
2
ζ(z)(x+γ) σ − z + x + i 2 e
= −e
η (x+i β + γ )
β
2
2
σ x + i2 e 1
β
ζ(z)(x+γ) −η1 z σ z − x − i 2
=e
e
β
σ x + i2
ζ(z)γ−η1 z
=e
ϕ(z, x)
c)
ξ(z + γ) = eγζ(z+γ)−(z+γ)η1 = eγζ(z)−zη1 = ξ(z)
ξ(z + iβ) = eγζ(z+iβ)−(z+iβ)η1 = eγη2 −iβη1 eγζ(z)−zη1
= ξ(z)
LM 3
Set Γ = γZZ and
∞
(IR/Γ)
V (x) = 2℘(x + i β2 ) ∈ CIR
d 2
H = i dx + V (x)
The Lamé equation is
d2
− dx2 φ
+ 2℘(x + i β2 )φ = λφ
(S.8)
A solution φ(k, x) of (S.8) that satisfies
φ(k, x + γ) = eiγk φ(k, x)
(S.9)
is called a Bloch solution with energy λ and quasimomentum k.
Lemma S.11 says that, for each z ∈ C \ γZZ ⊕ iβZZ ,
ϕ(z, x) is a Bloch solution of the Lamé equation with
energy λ = λ(z) and quasimomentum k = k(z).
LM 4
The energy λ and multiplier ξ = eγik are fully parameterized by
λ(z) = −℘(z)
ξ(z) = eγζ(z)−zη1
That is, the boundary value problem (S.8), (S.9) has a
nontrivial solution if and only if (λ, eiγk ) = (λ(z), ξ(z)),
for some z ∈ C \ γZZ ⊕ iβZZ .
Idea of Proof. Unless 2z ∈ γZZ ⊕ iβZZ, the functions
ϕ(z, x) and ϕ(−z, x) are linearly independent (Lemma
S.12) solutions of (S.8) for λ(z) = λ(−z). As a second
order ordinary differential equation, (S.8) only has two
linearly independent solutions for each fixed value of λ.
For z ∈ γZZ ⊕ iβZZ, λ(z) is not finite.
For 2z ∈ γZZ ⊕ iβZZ with z ∈
/ γZZ ⊕ iβZZ, λ0 (z) = 0 and
the second linearly independent solution is
∂
∂z ϕ(z, x).
LM 5
Theorem S.13 Set
Λ1 = −℘( γ2 )
Λ2 = −℘( γ2 + i β2 )
Λ3 = −℘(i β2 )
Then Λ1 , Λ2 , Λ3 are real, Λ1 < Λ2 < Λ3 and the spec
d 2
trum of H = i dx + 2℘(x + i β2 ) is [Λ1 , Λ2 ] ∪ [Λ3 , ∞).
Proof:
If, for given values of λ and k, the boundary
value problem (S.8), (S.9) has a nontrivial solution and
if k is real then λ is in the spectrum of H. We know
that all such λ’s are also real.
iβ
γ + iβ
γ
0
Imagine walking along the path in the z–plane that follows the four line segments from 0 to
γ
2
to
γ
2
+ i β2 to
LM 6
i β2 and back to 0. As ℘(z) = ℘(z̄), ℘(−z) = ℘(z) and
℘(z − γ) = ℘(z − iβ) = ℘(z), λ(z) = −℘(z) remains real
throughout the entire excursion. Near z = 0,
λ(z) = −℘(z) ≈ − z12
so λ starts out near −∞ at the beginning of the walk
and moves continuously to +∞ at the end of the walk.
Furthermore, as
℘(z) = ℘(z 0 ) if and only if z − z 0 ∈ γZZ ⊕ iβZZ
or z + z 0 ∈ γZZ ⊕ iβZZ.
λ never takes the same value twice on the walk, because
no two distinct points z, z 0 on the walk obey z ± z 0 ∈
γZZ ⊕ iβZZ.
• On the first quarter of the walk, from z = 0 to
z =
γ
2,
λ(z) increases from −∞ to Λ1 = −℘( γ2 ).
But we cannot put these λ’s into the spectrum of H
because k(z) is pure imaginary on this part of the
walk.
LM 7
• On the second quarter of the walk, from z =
γ
2 +
−℘( γ2 + i β2 ).
to z =
γ
2
i β2 , λ(z) increases from Λ1 to Λ2 =
As k(z) is pure real on this part of the
walk, so these λ’s are in the spectrum of H.
• On the third quarter of the walk, from z =
γ
2
+ i β2
to z = i β2 , λ(z) increases from Λ2 to Λ3 = −℘(i β2 ).
These λ’s do not go into the spectrum of H, because
k(z) has nonzero imaginary part.
• On the last quarter of the walk, from z = i β2 back
to zero, λ(z) increases from Λ3 to +∞. These λ’s
are in the spectrum of H, because k(z) is pure real.
LM 8