Z
∂Ωt
|∇ψ|ds
Z
∂Ωt
ev
|∇ψ|ds for all t∈ (minΩ−ψ, maxΩ+ψ). This implies that
|∇ψ|2(x) =C(t)ev(x)
almost everywhere in ψ−1(t)for some constant C which depends only on t. By continuity we have
(4.25) |∇ψ|2(x) =C(ψ(x))ev(x)
for all x except when ψ(x) =maxΩ+ψ or ψ(x) =minΩ−ψ.
By letting x= p2 ∈ψ−1(0), we find
C(0) = |∇ψ|2(p2)e−v(p2)=0.
By (4.25) this implies that|∇ψ(x)| =0 for all x ∈ ψ−1(0), which is clearly impossible. Hence the proof of Theorem 4.1 is completed. Since equation (4.1) has an unique solution at ρ = 4π, by the continua-tion from ρ=4π to 8π and Theorem 4.1, we conclude that (4.1) has a most a solution at ρ = 8π, and it implies that the mean field equation (1.2) has at most one solution up to scaling. Thus, Theorem 1.2 is proved and then Theorem 1.3 follows immediately.
5. COMPARING CRITICALVALUES OFGREENFUNCTIONS
For simplicity, from now on we normalize all tori to have ω1 =1, ω2= τ.
In section 3 we have shown that the existence of solutions of equation (1.2) is equivalent to the existence of non-half-period critical points of G(z). The main goal of this and next sections is to provide criteria for detecting minimum points of G(z). The following theorem is useful in this regard.
Theorem 5.1. Let z0and z1be two half-periods. Then G(z0) ≥G(z1)if and only if|℘(z0)| ≥ |℘(z1)|.
Proof. By integrating (2.7), the Green function G(z)can be represented by (5.1) G(z) = −2π1 Re
Z
(ζ(z) −η1z)dz+ 1 2by2. Thus
(5.2) Gω2
2
−Gω23= 1 2πRe
Z ω3
2 ω22
(ζ(z) −η1z)dz.
Set F(z) =ζ(z) −η1z. We have F(z+ω1) =F(z)and
To calculate the integral in (5.2), we use the addition theorem to get
℘′(z) Similarly, by integrating (2.7) in the ω2direction, we get (5.4) Gω1
The same proof then gives rise to
(5.5) Gω1 By combining the above two formulae we get also that
(5.6) Gω1
In order to compare, say, G(12ω1)and G(12ω3), we may use (5.5). Let λ= e3−e2
e1−e2. By using e1+e2+e3 =0, we get
(5.7) e3
e1
= 2λ−1 2−λ .
It is easy to see that|2λ−1| ≥ |2−λ|if and only if|λ| ≥1. Hence
e3
e1
≥1 if and only if |λ| ≥1.
The same argument applies to the other two cases too and the theorem
follows.
It remains to make the criterion effective in τ. Recall the modular func-tion
λ(τ) = e3−e2 e1−e2. By (5.3), we have
(5.8) Gω3
2
−Gω22= 1
4πlog|λ(τ) −1|. Therefore, it is important to know when|λ(τ) −1| =1.
Lemma 5.2. |λ(τ) −1| =1 if and only if Re τ= 12.
Proof. Let℘(z; τ)be the Weierstrass℘function with periods 1 and τ, then
℘(z; τ) = ℘(¯z; ¯τ).
For τ = 12+ib, ¯τ =1−τ and then℘(z; ¯τ) = ℘(z; τ). Thus (5.9) ℘(z; τ) = ℘(¯z; τ) for τ= 1
2+ib.
Note that if z = 12ω2 then ¯z = 12ω¯2 = 12(1−ω2) = 12ω3 (mod ω1, ω2). Therefore
¯e2=e3 and ¯e1=e1. Since e1+e2+e3=0, we have
(5.10) Re e2 = −12e1 and Im e2 = −Im e3. Thus
(5.11) |λ(τ) −1| =
e3−e1
e2−e1
=1.
A classic result says that λ′(τ) 6=0 for all τ. By this and (5.11), it follows that λ maps{τ|Re τ= 12}bijectively onto{λ(τ) | |λ(τ) −1| =1}.
Let Ω be the fundamental domain for λ(τ), i.e.,
Ω= {τ∈ C| |τ−1/2| ≥1/2, 0≤ Re τ≤1, Im τ>0}, and let Ω′be the reflection of Ω with respect to the imaginary axis.
Since G(12ω3) < G(12ω2) for τ = ib, we conclude that for τ ∈ Ω′∪Ω,
For|τ| =1, using suitable M ¨obius transformations we may obtain simi-lar results. For example, from the definition of℘, (5.9) implies that
(5.12) ℘(¯ z) =τ+1 6. DEGENERACY ANALYSIS OFCRITICALPOINTS
By (2.7), the derivatives of G can be computed by 2πGx=Re(η1t+η2s−ζ(z)),
Thus the Hessian of G is given by
2πGxx=Re℘(z) +η1,
We first consider the case z0 = 12ω1. The degeneracy condition of G at
1
2ω1reads
e1+η1 =0 or e1+η1− 2πb =0.
We will use the following two inequalities (Theorem 1.7) whose proofs will be given in§8 and §9 through theta functions:
(6.4) e1(b) +η1(b) is increasing in b and (6.5) e1(b) is increasing in b.
Lemma 6.1. There exists b0 < 12 < b1 < √
3/2 such that 12ω1 is a degenerate critical point of G(z; τ) if and only if b = b0 or b = b1. Moreover, 12ω1 is a local minimum point of G(z; τ)if b ∈ (b0, b1)and is a saddle point of G(z; τ)if b∈ (0, b0)or b∈ (b1, ∞).
Proof. Let b0 and b1 be the zero of e1+η1 = 0 and e1+η1−2π/b = 0 respectively. Then Lemma 6.1 follows from the explicit expression of the
Hessian of G by (6.4).
Numerically we know that b1 ≈ 0.7 < √
3/2. Now we analyze the behavior of G near 12ω1for b>b1.
Lemma 6.2. Suppose that b > 12, then 12ω1 is the only critical point of G along the x-axis.
Proof. ℘(t; τ)is real if t∈R. Since℘′(t; τ) 6=0 for t6= 12ω1,℘′(t; τ) <0 for 0<t< 12ω1. Since b> 12 >b0, by (6.3) and Lemma 6.1,
Gxx(t) = ℘(t) +η1 >e1+η1 >0,
which implies that Gx(t) > Gx(12ω1) =0 if 0 < t < 12ω1. Hence G has no critical points on(0,12ω1). Since G(z) =G(−z), G can not have any critical
point on(−12ω1, 0).
By Lemma 6.1 and the conservation of local Morse indices, we know that G(z; τ)has two more critical points near 12ω1when b is close to b1and b > b1. We denote these two extra points by z0(τ) and −z0(τ). In this case, 12ω1becomes a saddle point and z0(τ)and−z0(z)are local minimum points. From Lemma 5.2, (5.15) and (5.16) we know that
Gω1
2
<Gω2
2
= Gω3
2
if b1< b<√ 3/2.
Thus in this region±z0(τ)must exist and they turn out to be the minimum point of G since there are at most five critical points. In fact we will see below that this is true for all b < b0 or b > b1and 12ω2, 12ω3are all saddle points.
Lemma 6.3. The critical point z0(τ)is on the line Re z= 12. Moreover, the Green function G(z; τ)is symmetric with respect to the line Re z= 12.
Proof. Recall that G(¯z) = G(z). Since G has only two more critical points near 12ω1 and z 6∈ R, z0(τ) = −z0(τ). Let z0(τ) = t0ω1+s0ω2. z0(τ) = (t0+s0)ω1−s0ω2. Thus 2t0+s0 =1. Then Re z0(z) =t0+s0/2= 12.
Consider the rectangle: 0 ≤ Re z ≤ 1 and |Im z| ≤ b/2. Clearly the second statement follows from the symmetry of boundary value of G with respect to the line Re z= 12. Since G(z+1) = G(z), the symmetry of G on Re z=0 and Re z=1 is obvious. Hence we have to show that
(6.6) G
z+1 2+ ib
2
= G
−z+ 1 2+ib
2
for z∈ R. By using G(¯z) =G(z), we have
G z+ 1
2+ ib 2
= G z+ 1
2−ib2
=G z− 1
2− ib 2
= G1 2+ ib
2 −z,
where the second and the third identity comes from G(z+1) = G(z)and G(−z) =G(z)respectively. Thus (6.6) and then the lemma are proved.
Let G1
2 be the restriction of G on the line Re z = 12. Lemma 6.3 says that any critical point of G1
2 is automatically a critical point of G.
Lemma 6.4. G1
2(y)has exactly one critical point in(0, b)for b>b1.
Proof. Let z= tω1+sω2. Then Re z= 12 is equivalent to 2t+s = 1, which implies ¯z= (t+s)ω1−sω2 = −z. By (5.9)
℘(z, τ) = ℘(¯z; z) = ℘(−z; τ) = ℘(z; τ).
Hence℘(z; τ)is real for Re z= 12. To prove Lemma 6.4, we apply (6.3), 2πGyy = −
℘+η1−2πb
.
Since ∂℘/∂y= −i℘′(z; τ) 6= 0 for z 6= 12ω1and Re z = 12, Gyy(z)can have one zero only. Let z0(τ) denote the critical point above when τ is close to 12 +ib1. Thus Gy(z0(τ)) = Gy(12ω1) = 0, and then Gyy(˜z0) = 0 for some ˜z0 ∈ (12ω1, z0(τ)). Since Gyy(12ω1) = −(e1+η1−2π/b) < 0, we have Gyy(z0(τ)) > 0. Hence z0(τ)is a non-degenerate minimum point of G1
2.
Remark 6.5. For b> b1, G1
2(y)is increasing in y for y>0 and limy→bG1
2(y) = +∞.
Theorem 6.6. For b> b1,±z0(τ)are non-degenerate (local) minimum points of G. Furthermore,
(6.7) 0<Im z0(τ) < b 2.
Proof. We want to prove Gxx(z0(τ); τ) >0 and Im z0(τ) <b/2. By (6.3),
3i)/2. Hence the claim is proved.
Note that e1e2+e2e3+e1e3 = |e2|2−e21and
3/2 and sufficiently close to√
3/2. By the claim above,℘′′(z0(τ); τ) <
and 1.7 to be proved in§9, but we will give another direct proof of it in (6.20)).
Thus the non-degeneracy of z0(τ)for b>√ also a critical point. By the addition formula of ζ and℘, we conclude that
℘′′(13ω3) =0 and℘(13ω3) =0, a contradiction. Hence
By the addition theorem for℘, we have
(6.14) 12℘ω3
Plug in℘(13ω3) = −12e1and recall that e3= ¯e2= −12e1+i(Im e3), we get
(6.17) |e2|2
|e1|2 = 37 4 .
Numerically we know that |e2|2/|e1|2 ≈ 3.126 < 374 at b = b1. Thus
℘(13ω3) > −12e1for b>b1. Hence together with (6.13)
℘(z0(τ), τ) +η1 >η1−12e1>0 for b1 < b < √
3/2. With this, the proof of the non-degeneracy of z0(τ)is
completed.
Next we discuss the non-degeneracy of G at 12ω2 and 12ω3. The local minimum property of z0(τ)is in fact global by
Theorem 6.7. For τ = 12 +ib, both 12ω2 and 12ω3 are non-degenerate saddle points of G.
Proof. By (6.3), we have 2πGxx
ω2
2
=Re e2+η1=η1−12e1, 2πGxy
ω2
2
= −Im e2, 2πGyy
ω2
2
= 2π b +1
2e1−η1. (6.18)
Hence the non-degeneracy of 12ω2for all b is equivalent to (6.19) |Im e2|2>η1− 12e1
2π b + 1
2e1−η1
for all b∈R. To prove (6.19), we need the following:
Lemma 6.8. ℘maps[12ω2,12ω3] ∪ [12(1−ω2),12ω2]one to one and onto the circle {w| |w−e1| = |e2−e1|}, where e2= ¯e3,℘(14(ω2+ω3)) =e1− |e2−e1| <0 and℘(14) =e1+ |e2−e1| >0.
Here [12ω2,12ω3] means segment {12ω2+t | 0 ≤ t ≤ 12}, and [12(1− ω2),12ω2]is{14+it | |t| ≤ 2b}. Thus, the image of[12ω2,12ω3]is the arc con-necting e2and e3through℘(14(ω2+ω3))and the image of[12(1−ω2),12ω2] is the arc connecting e3and e2through℘(14). See Figure 4. Note our figure is for the case e1 >0, i.e., b > 12. In this case, the angle∠e3e1e2is less than π. For the case e1 <0, the angle is greater than π.
We will derive (6.19) from Lemma 6.8. First we prove
(6.20) η1− 12e1 >0
e1
℘(14(ω2+ω3)) ℘(14)
1 2e1
e3
e2
FIGURE 4. The image of the mapping z7→ ℘(z). and
(6.21) 2π
b + 1
2e1−η1>0.
To prove (6.20), we have
(6.22) −η1 =2
Z 1
2
0 Re℘ω2 2 +t
dt, where℘(12ω2+z) = ℘(12ω2−t)is used. By Lemma 6.8,
Re℘ω2 2 +t
≤ −12e1. Hence
−η1≤ −12e1. To prove (6.21), we have
Z ω22
1−ω2 2
℘(z)dz=ζ1−ω2
2
−ζω22
= 1
2(η1−2η2) =i(2π−bη1). Therefore,
(2π−bη1) =
Z b
2
−2bRe℘1 4+it
dt≥ −12e1b and the inequality (6.21) follows.
To prove (6.19), we need two more inequalities. By (6.22), (6.23) −η1 > ℘ω2+ω3
4
=e1− |e2−e1|, and by Lemma 6.8,
(6.24) (2π−bη1) < ℘(1/4)b= (e1+ |e2−e1|)b.
Thus
which is exactly (6.19). Therefore, the non-degeneracy of G at12ω2is proved.
By (6.19), 12ω2is always a saddle point.
It remains to prove Lemma 6.8. First we have (6.25) 4π
Since ℘(t) is decreasing in t for t ∈ (0,12), ℘(14) > ℘(12). So℘(14) = e1+ |e2−e1|and the image of[12(1−ω2),12ω2]is the arc of{w| |w−e1| =
|e2−e1| }connecting e3and e2through e1+ |e2−e1|. Therefore Lemma 6.8 is proved and the proof of Theorem 6.7 is completed.
7. GREENFUNCTIONSVIATHETAFUNCTIONS
The purpose of §7 to §9 is to prove the two fundamental inequalities (Theorem 1.7) that have been used in previous sections. The natural setup for the proof turns out has to be the theta functions. Formally this is easy to explain since the moduli variable τ is very explicit in theta functions and differentiations in τ is much easier to be done than in the Weierstrass theory. In order to avoid complicate cross references, we choose to write everything here independent of the previous sections and in particular sev-eral statements about Green functions and critical points are re-derived for completeness.
In the following sections we consider a torus T = C/Λ with Λ= (Z+ Zτ), a lattice with τ = a+bi, b > 0. A function f on T is a function on Cwith f(z+1) = f(z)and f(z+τ) = f(z). No such f exists holomor-phically by Liouville’s theorem. Hence one considers either meromorphic functions (e.g. Weierstrass℘(z)) or quasi-periodic holomorphic functions (e.g. Riemann’s theta functions ϑi(z)) instead.
We firstly recall the definition and basic properties of the theta functions.
We take [13] as our general reference. Let q = eπiτ with |q| = e−πb < 1.
Then we have the exponentially convergent series ϑ1(z; τ) = −i
∞
∑
n=−∞(−1)nq(n+21)2e(2n+1)πiz
=2
∞
∑
n=0
(−1)nq(n+21)2sin(2n+1)πz.
(7.1)
For simplicity we also write it as ϑ1(z). It is entire with ϑ1(z+1) = −ϑ1(z),
ϑ1(z+τ) = −q−1e−2πizϑ1(z), (7.2)
thus it has a simple zero at the lattice point (and no others). The following heat equation is also clear from the definition
(7.3) ∂2ϑ1
∂z2 =4πi∂ϑ1
∂τ .
As usual we use z= x+iy. Here comes the starting point:
Lemma 7.1. The Green’s function G(z, w) for the Laplace operator △ on T is given by
(7.4) G(z, w) = −2π1 log
ϑ1(z−w) ϑ′1(0)
+ 1
2b(Im(z−w))2.
Proof. Let R(z, w) be the right hand side. Clearly for z 6= w we have
△zR(z, w) = 1/b which integrates over T gives 1. Near z=w, R(z, w)has the correct behavior. So it remains to show that R(z, w)is indeed a function on T. From the quasi-periodicity, R(z+1, w) =R(z, w)is obvious. Also
R(z+τ, w) −R(z, w) = −2π1 log eπb−2πy+ 1
2b((y+b)2−y2) =0.
These properties uniquely characterize the Green’s function. By the translation invariance of G, it is enough to consider w=0. Let
G(z) =G(z, 0) = −2π1 log
If we represent the torus T as centered at 0, then the symmetry z7→ −z shows that G(z) = G(−z). By differentiation, we get∇G(z) = −∇G(−z). to alternative simple proofs to Lemma 2.3 and Corollary 2.4.
We compute easily
To analyze the critical point of G(z) in general, we may try to use the methods of continuity to connect τ to a standard model like the square, that is τ =i, which under the modular groups SL(2, Z)is equivalent to the point τ = 12(1+i) by τ 7→ 1/(1−τ). On this special torus, elementary symmetry consideration shows that there are precisely three critical points given by the half periods (cf. [6], Lemma 2.1).
The idea is, new critical points should be born only at certain half period points when it degenerates (as a critical point) under the deformation in τ. The heat equation provides a bridge between the degeneracy condition and deformations in τ. In the following, we shall focus on the critical point z = 12 and in particular analyze its degeneracy behavior along the half line L given by 12+ib, b∈R.
8. FIRST INEQUALITY ALONG THELINERe τ = 12 When τ= 12+ib∈ L, we have
ϑ1(z) =2
∞
∑
n=0
(−1)neπi/8eπin(n2+1)e−πb(n+21)2sin(2n+1)πz, hence the important observation that
eπi/8ϑ1(z) ∈R when z ∈R.
Similarly this holds for any derivatives of ϑ1(z)in z. In particular, (log ϑ1)zz= ϑ1zzϑ1− (ϑ1z)2
ϑ21
=4πiϑ1τ
ϑ1 − (log ϑ1)2z =4π(log ϑ1)b− (log ϑ1)2z (8.1)
is real-valued for all z ∈ R and τ ∈ L. Here the heat equation and the holomorphicity of(log ϑ1)have been used.
Now we focus on the critical point z = 12. The critical point equation implies that(log ϑ1)z(12) = −2πiy/b=0 since now y =0. Thus
(8.2) (log ϑ1)zz =4π(log ϑ1)b
as real functions in b. In this case, the point 12 is a degenerate critical point (H(b) =0) if and only if that
(8.3) 4π(log ϑ1)b=0 or 4π(log ϑ1)b+2π b =0.
Notice that as functions in b>0,
|ϑ1| =e−πi/8ϑ1
1 2;1
2+ib
=2
∞
∑
n=0
(−1)n(n2+1)e−14πb(2n+1)2 ∈R+. To see this, notice that the right hand side is non-zero, real and positive for large b, hence positive for all b. Clearly(log|ϑ1|)b= (log ϑ1)b.
Theorem 8.1. Over the line L, (log ϑ1)bb = (log|ϑ1|)bb < 0. Namely that (log ϑ1)bis decreasing from positive infinity to−π/4. Hence that Gxx = 0 and Gyy =0 occur exactly once on L respectively.
Proof. Denote e−πb/4 by h and r = h8 = e−2πb. Since (2n+1)2 −1 = 4n(n+1), we get
|ϑ1|b= −2π4
∞
∑
n=0
(−1)n(n2+1)(2n+1)2h(2n+1)2
= −2hπ4
∞
∑
n=0
(−1)n(n2+1)(2n+1)2rn(n2+1)
|ϑ1|bb =2π2 42
∞
∑
n=0
(−1)n(n2+1)(2n+1)4h(2n+1)2
=2hπ2 42
∞
∑
n=0
(−1)n(n2+1)(2n+1)4rn(n2+1). (8.4)
Denote the arithmetic sum n(n+1)/2 by An, then
(log|ϑ1|)bb = |ϑ1|bb|ϑ1| − |ϑ1|2b
|ϑ1|2
=h2π2 4 |ϑ1|−2
∞
∑
n,m=0
(−1)An+Am((2n+1)4− (2n+1)2(2m+1)2)rAn+Am
=h2π2
4 |ϑ1|−2
∑
n>m(−1)An+Am((2n+1)2− (2m+1)2)2rAn+Am
=16h2π2|ϑ1|−2
∑
n>m(−1)An+Am(An−Am)2rAn+Am
=16h2π2|ϑ1|−2(−r−9r3+4r4+36r6−25r7−9r9+100r10+ · · · ). (8.5)
We will prove (log|ϑ1|)bb < 0 in two steps. First we show by direct estimate that this is true for b≥ 12(indeed the argument holds for b>0.26).
Then we derive a functional equation for(log|ϑ1|)bbwhich implies that the case with 0< b≤ 12 is equivalent to the case b≥ 12.
Step 1: (Direct Estimate). The point is to show that in the above expres-sion the sum of positive (even degree) terms is small. So let 2k ∈ 2N. The number of terms with degree 2k is certainly no more than 2k, so a trivial upper bound for the positive part is given by
(8.6) A=
∞
∑
n=2
(2k)3r2k =8r48−5r2+4r4−r6 (1−r2)4 ,
where the last equality is an easy exercise in power series calculations in Calculus. For r ≤1/5 we compute
(−r−9r3+A)(1−r2)4
= −r−5r3+64r4+30r5−40r6−50r7+32r8+35r9−8r10−9r11
<−r−5r3+64r4+30r5
<−5r3−r1− 12564 − 62530 = −5r3−1125r <0.
(8.7)
So(log|ϑ1|)bb <0 for b= −(log r)/2π >(log 5)/2π ∼0.25615.
Step 2: (Functional Equation). By the Lemma to be proved below, we have for ˆτ = (τ−1)/(2τ−1) = ˆa+iˆb, it holds that
(8.8) (log ϑ1)ˆb(1/2; ˆτ) = −i(1−2τ) + (1−2τ)2(log ϑ1)b(1/2; τ). When τ= 12+ib, we have ˆτ= 12+4bi . As before we may then replace ϑ1
by|ϑ1|. Under τ→ ˆτ,[12, ∞)is mapped onto(0,12]with directions reversed.
Let f(b) = (log|ϑ1|)b(12,12+ib). Then we get (8.9) f(1/4b) = −2b−4b2f(b).
Plug in b = 12 we get that f(12) = −12. So−12 = f(12) > f(b)for b > 12. Then
f 1 4b
= −2b+2b2+4b2h
−12 −f(b)i
=2h
b− 12i2−12 +4b2h
−12− f(b)i (8.10)
is strictly increasing in b > 12. That is, f(b)is strictly decreasing in b∈ (0,12].
The remaining statements are all clear.
Now we prove the functional equation. For this we need to use Jacobi’s imaginary transformation formula, which explains the modularity for cer-tain special theta values (cf. p.475 in [13]). It reads that for ττ′ = −1, (8.11) ϑ1(z; τ) = −i(iτ′)12eπiτ′z2ϑ1(zτ′; τ′).
Recall the two generators of SL(2, Z) are Sτ = −1/τ and Tτ = τ+1.
Since ϑ1(z; τ+1) =eπi/4ϑ1(z; τ), T plays no role in(log ϑ1(z; τ))τ. Lemma 8.2. Let ˆτ=ST−2ST−1τ= (τ−1)/(2τ−1). Then
(8.12) (log ϑ1)ˆτ(1/2; ˆτ) = −(1−2τ) + (1−2τ)2(log ϑ1)τ(1/2; τ). Proof. Let ˆτ = Sτ1 = −1/τ1, τ1 = T−2τ2 = τ2−2, τ2 = Sτ3 = −1/τ3
and finally τ3 = T−1τ =τ−1. Notice that for ττ′ = −1 we have d/dτ =
τ′2d/dτ′. Then fourth terms are cancelled out, the first and the third terms are combined into is, 12 is a saddle point, local minimum point and saddle point respectively.
This implies that for b = b1+ǫ > b1, there are more critical points near
1
2 which come from the degeneracy of 12 at b = b1and the conservation of local Morse index.
9. SECOND INEQUALITY ALONG THELINERe τ= 12
The analysis of extra critical points split from 12 relies on other though similar inequalities. We shall need Weierstrass’ elliptic functions to moti-vate their origin, as we have done in§5, and we will go back to that later in this section. Here we shall prove the inequality first. We recall the three other theta functions.
ϑ2(z; τ):=ϑ1 functions are translates of others by half periods.
We had seen previously that(log|ϑ1(12)|)bb < 0 over the line Re τ = 12.
Hence that
(|ϑ3(0)|2)bb|ϑ3(0)|2− (|ϑ3(0)|2)2b
=π2
∑
k,l∈2Z≥0
p2(k)p2(l)(k2−kl)rk+l
=π2
∑
k<l
p2(k)p2(l)(k−l)2rk+l >0.
(9.6)
This implies that(log|ϑ3(0)|)bb >0. The proof is complete. Now we relate these to Weierstrass’ elliptic functions. Let℘(z) be the Weierstrass℘function with periods ω1 = 1, ω2 = τ. Let ω3 = ω1+ω2. Then℘′(z)is odd with three zeros at the half periods z = 12ωi, i = 1, 2, 3.
Let ei = ℘(12ωi), then
(9.7) ℘′(z)2 =4(℘(z) −e1)(℘(z) −e2)(℘(z) −e3). Let ζ(z) = −Rz
℘(w)dw which is odd with quasi-periods ηi = ζ(z+ ωi) −ζ(z), i=1, 2, hence that ηi =2ζ(12ωi). The invariants eiand ηiare all holomorphic functions of τ.
Let σ(z) =expRz
ζ(w)dw, that is(log σ(z))′ =ζ(z), then σ(z)is entire, odd with a simple zero on lattice points. σ(z)satisfies
(9.8) σ(z+ωi) = −eηi(z+12ωi)σ(z).
This is similar to the theta function transformation law. Indeed it is easily seen that
(9.9) σ(z) =eη1z2/2ϑ1(z) ϑ′1(0). Hence
(9.10) ζ(z) −η1z=
logϑ1(z) ϑ′1(0)
z
= (log ϑ1(z))z and
(9.11) ℘(z) +η1 = −(log ϑ1(z))zz = −4πi(log ϑ1(z))τ+ [(log ϑ1(z))z]2. For z = 12, this simplifies to e1+η1 = −4πi(log ϑ1(12))τ. Thus our first estimate simply says that on the line Re τ= 12,
(9.12) (e1+η1)b= −4π
log ϑ1
1 2
bb = −4π(log ϑ2(0))bb >0.
From the Taylor expansion of σ(z)and ϑ1(z), it is known that (9.13) η1 = −3!2 ϑ′′′1 (0)
ϑ1′(0) = −4πi3 (log ϑ1′(0))τ, hence
(9.14) e1= −4πi(log ϑ2(0))τ+ 4πi
3 (log ϑ′1(0))τ.
The celebrated Jacobi Triple Product Formula (cf. p.490 in [13]) asserts that
ϑ′1(0) =πϑ2(0)ϑ3(0)ϑ4(0). So
1
2e1−η1 =2πi
logϑ1′(0) ϑ2(0)
τ
=2πi(log ϑ3(0)ϑ4(0))τ =4π(log|ϑ3(0)|)b. (9.15)
Our second estimate then says that on the line Re τ = 12, 12e1−η1 < 0, (12e1−η1)b > 0 and 12e1−η1increases to zero in b. Together with the first estimate (9.12), we find also that e1increases in b.
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CHANG-SHOULIN: DEPARTMENT OFMATHEMATICS, NATIONALCENTRALUNIVER
-SITY, CHUNG-LI, TAIWAN.
E-mail address:cslin@math.ncu.edu.tw
CHIN-LUNGWANG: DEPARTMENT OFMATHEMATICS, NATIONALCENTRALUNIVER
-SITY, CHUNG-LI, TAIWAN., NATIONALCENTER FOR THEORETICSCIENCES, HSINCHU, TAIWAN.
E-mail address:dragon@math.ncu.edu.tw E-mail address:dragon@math.cts.nthu.edu.tw