FIELD EQUATIONS ON TORI CHANG-SHOU LIN AND CHIN-LUNG WANG
ABSTRACT. We show that the Green functions on flat tori can have either 3 or 5 critical points only. There does not seem to be any direct method to attack this problem. Instead, we have to employ sophisticated non-linear partial differential equations to study it.
We also study the distribution of number of critical points over the moduli space of flat tori through deformations. The functional equations of special theta values provide important inequalities which lead to a solution for all rhombus tori. The general picture is also emerged, though some of the necessary technicality is still to de developed.
1. INTRODUCTION ANDSTATEMENT OFRESULTS
The study of geometric or analytic problems on two dimensional tori is the same as the study of problems on R2 with doubly periodic data. Such situations occur naturally in sciences and mathematics since early days. The mathematical foundation of elliptic functions was subsequently devel-oped in the 19th century. It turns out that these special functions are rather deep objects by themselves. Tori of different shape may result in very dif-ferent behavior of the elliptic functions and their associated objects. Arith-metic on elliptic curves is perhaps the eldest and the most vivid example.
In this paper, we show that this is also the case for certain non-linear par-tial differenpar-tial equations. Indeed, researches on doubly periodic problems in mathematical physics or differential equations often restrict the study to rectangular tori for simplicity. This leaves the impression that the theory for general tori may resemble much the same way as for the rectangular case. However this turns out to be false. We will show that the solvability of the mean field equation depends on the shape of the Green function, which in turn depends on the geometry of the tori in an essential way.
Recall that the Green function G(z, w)on a flat torus T=C/Zω1+Zω2
is the unique function on T×T which satisfies
−△zG(z, w) =δw(z) − 1
|T|
Date: May 30, 2004. Revised September 19, 2005, June 1st, 2006. 1
andR
TG(z, w)dA = 0, where δw is the Dirac measure with singularity at z = w. Because of the translation invariance of △z, we have G(z, w) = G(z−w, 0)and it is enough to consider the Green function G(z):=G(z, 0).
Not surprisingly, G can be explicitly solved in terms of elliptic functions. For example, using theta functions we have (cf. Lemma 2.1, Lemma 7.1)
G(z) = −2π1 log ϑ1(z) ϑ′1(0) + 1 2by 2
where z = x+iy and τ := ω2/ω1 = a+ib. The structure of G, especially its critical points and critical values, will be the fundamental objects that interest us. The critical point equation∇G(z) =0 is given by
∂G ∂z ≡ −1 4π (log ϑ1)z+2πiy b =0. In terms of Weierstrass’ elliptic functions℘(z), ζ(z) := −Rz
℘ and using the relation (log ϑ1)z = ζ(z) −η1z with ηi = ζ(z+ωi) −ζ(z)the quasi-periods, the equation takes the simpler form: z = tω1+sω2 is a critical point of G if and only if the following linear relation (Lemma 2.3) holds: (1.1) ζ(tω1+sω2) =tη1+sη2.
Since G is even, it is elementary to see that half periods 12ω1, 12ω2 and 1
2ω3 = (ω1+ω2)/2 are the three obvious critical points and other critical points must appear in pair. The question is: Are there other critical points? or How many critical points might G have?. It turns out that this is a delicate question and can not be attacked easily from the simple looking equation (1.1). One of our chief purposes in this paper is to understand the geometry of the critical point set over the moduli space of flat toriM1 = H/SL(2, Z) and to study its interaction with the non-linear mean field equation.
The mean field equation on a flat torus T takes the form (ρ∈R+)
(1.2) △u+ρeu=ρδ0.
This equation has its origin in the prescribed curvature problem in geom-etry like the Nirenberg problem, cone metrics etc.. It also comes from sta-tistical physics as the mean field limits of the Euler flow, hence the name. Recently it was shown to be related to the self dual condensation of Chern-Simons-Higgs model. We refer to [3], [4], [5], [6], [7], [10], [11] and [12] for the recent development of this subject.
When ρ 6= 8mπ for any m ∈ Z, it has been recently proved in [4], [5],
[6] that the Leray-Schauder degree is non-zero, so the equation always has solutions, regardless on the actual shape of T.
The first interesting case remained is when ρ=8π where the degree the-ory fails completely. Instead of the topological degree, the precise knowl-edge on the Green function plays a fundamental role in the investigation of (1.2). The first main result of this paper is the following existence criteria whose proof is given in§3 by a detailed manipulation on elliptic functions:
Theorem 1.1(Existence). For ρ = 8π, the mean field equation on a flat torus has solutions if and only if the Green function has critical points other than the three half period points. Moreover, each extra pair of critical points corresponds to an one parameter scaling family of solutions.
It is known that for rectangular tori G(z)has precisely the three obvious critical points, hence for ρ = 8π equation (1.2) has no solutions. However we will show in§2 that for the case ω1=1 and τ =ω2= eπi/3there are at least five critical points and the solutions of (1.2) exist.
Our second main result is the uniqueness theorem.
Theorem 1.2(Uniqueness). For ρ=8π, the mean field equation on a flat torus has at most one solution up to scaling.
In view of the correspondence in Theorem 1.1, an equivalent statement of Theorem 1.2 is the following result:
Theorem 1.3. The Green function has at most five critical points.
Unfortunately we were unable to find a direct proof of Theorem 1.3 from the critical point equation (1.1). Instead, we will prove the uniqueness the-orem first, and then Thethe-orem 1.3 is an immediate corollary. Our proof of Theorem 1.2 is based on the method of symmetrization applied to the lin-earized equation at a particularly chosen even solution in the scaling family. In fact we study in§4 the one parameter family
△u+ρeu=ρδ0, ρ∈ [4π, 8π]
on T within even solutions. This extra assumption allows us to construct a double cover T→S2via the Weierstrass℘function and to transform equa-tion (1.2) into a similar one on S2 but with three more delta singularities with negative coefficients. The condition ρ ≥ 4π is to guarantee that the original singularity at 0 still has non-negative coefficient of delta singular-ity.
The uniqueness is proved for this family via the method of continuity. For the starting point ρ=4π, by a construction similar to the proof of The-orem 1.1 we sharpen the result on nontrivial degree to the existence and uniqueness of solution (Theorem 3.2). For ρ ∈ [4π, 8π], the symmetriza-tion reduces the problem on the non-degeneracy of the linearized equasymmetriza-tion to the isoperimetric inequality on domains in R2 with respect to certain singular measure:
Theorem 1.4 (Symmetrization Lemma). Let Ω ⊂ R2 be a simply connected
domain and let v be a solution of
△v+ev=
∑
Nj=12παjδpjin Ω. Suppose that the first eigenvalue of△ +ev is zero on Ω with ϕ the first eigenfunction. If the isoperimetric inequality with respect to ds2 =ev|dx|2: (1.3) 2ℓ2(∂ω) ≥m(ω)(4π−m(ω))
holds for all level domains ω = {ϕ>t}with t>0, then
Z
Ωe
vdx ≥2π.
Moreover, (1.3) holds if there is only one negative αj and αj = −1.
The proof on the number of critical points appears to be one of the very few instances that one needs to study a simple analytic equation, here the critical point equation (1.1), by way of sophisticated non-linear analysis.
To get a deeper understanding of the underlying structure of solutions, we first notice that for ρ = 8π, (1.2) is the Euler-Lagrange equation of the non-linear functional (1.4) J8π(v) = 1 2 Z T|∇v| 2dA −8π log Z Te v−8πG(z)dA
on H12(T), the Sobolev space of functions with L2-integrable first deriva-tive. From this viewpoint, the non-existence of minimizers for rectangular tori was known in [6]. Here we sharpen the result to the non-existence of solutions. Also for ρ∈ (4π, 8π)we sharpen the result on non-trivial degree of equation (1.2) in [4] to the uniqueness of solutions within even functions. We expect the uniqueness holds true without the even assumption, but our method only achieves this at ρ=4π. Obviously, uniqueness without even assumption fails at ρ=8π due to the existence of scaling.
Naturally, the next question after Theorem 1.2 is to determine those tori whose Green functions have five critical points. Certainly, it is the case if those three half-periods are all saddle points, and it is desirable to know whether the converse holds. The following theorem in [9] answers this question. We present it separately since its proof requires computations in a different flavor:
Theorem A.Suppose that the Green function on the given torus has five critical points. Then the extra pair of critical points are minimum points. Furthermore, the Green function has more than three critical points if and only if all the three half period points are saddle points.
Together with Theorem 1.1, Theorem A implies that a minimizer of J8π exists if and only if the Green function has more than three critical points. In fact we will show in [9] that any solution of equation (1.2) must be a minimizer of the nonlinear functional J8π in (1.4). Thus we completely solve the exis-tence problem on minimizers, a question raised by Nolasco and Tarantello in [11].
By Theorem A, we have reduced the question on detecting a given torus to have five critical points to the technically much simpler criterion on (non)-local minimality of the three half period points. In this paper, how-ever, no reference to Theorem A is needed. Instead, it motivates the follow-ing comparison result, which also simplifies the criterion further:
M1 0 12 1 i 1 2(1+i) 1 2+b1i
FIGURE 1. Ω5contains a neighborhood of the∗shape.
Theorem 1.5. Let z0and z1be two half period points. Then G(z0) ≥G(z1) if and only if |℘(z0)| ≥ |℘(z1)|.
For general flat tori, a computer simulation suggests the following pic-ture: Let Ω3 (resp. Ω5) be the subset of the moduli space M1∪ {∞} ∼= S2 which corresponds to tori with three (resp. five) critical points, then Ω3∪ {∞} is a closed subset containing i, Ω5 is an open subset contain-ing eπi/3, both of them are simply connected and their common boundary C := ∂Ω3 = ∂Ω5is a curve homeomorphic to S1 containing ∞. Moreover, the extra critical points are split out from some half period point when the tori move from Ω3to Ω5across C.
Concerning with the experimental observation, we propose to prove it by the method of deformations inM1. The degeneracy analysis of critical points, especially the half period points, is a crucial step. In this direction we have the following partial result on tori corresponding to the line Re τ=
1/2. These are equivalent to the rhombus tori and τ = 1
2(1+i)is equivalent to the square torus where there are only three critical points.
Theorem 1.6(Moduli Dependence). Let ω1 = 1 and ω2 = τ = 12+ib with b>0. Then
(1) There exists b0 < 12 < b1 <
√
3/2 such that 12ω1is a degenerate critical point of G(z; τ)if and only if b= b0 or b= b1. Moreover, 12ω1is a local minimum point of G(z; τ)if b0<b<b1and it is a saddle point if b <b0 or b>b1.
(2) Both 12ω2and 12ω3are non-degenerate saddle points of G.
(3) G(z; τ)has two more critical points±z0(τ)when b<b0or b>b1. They are non-degenerate global minimum points of G and in the former case
Re z0(τ) = 1
2; 0<Im z0(τ) < b 2.
Part (1) gives a strong support of the conjectural shape of the decompo-sitionM1 = Ω3∪Ω5. Part (3) implies that minimizers of J8π exist for tori with τ = 1
0.2 0.4 0.6 0.8 1 1.2 1.4
-20 -10 10
FIGURE2. Graphs of η1(the left one) and e1in b where e1is increasing.
0.2 0.4 0.6 0.8 1 1.2 1.4 -40 -30 -20 -10 10
FIGURE 3. Graphs of e1+η1 (the upper one) and 12e1−η1. Both functions are increasing in b and 12e1−η1ր0.
The proofs are given in§6. Notably Lemma 6.1, 6.2 and Theorem 6.6, 6.7. They rely on two fundamental inequalities on special values of elliptic functions and we would like to single out the statements (recall that ei =
℘(12ωi)and ηi =2ζ(12ωi)):
Theorem 1.7(Fundamental Inequalities). Let ω1 = 1 and ω2 = τ = 12+ib with b>0. Then (1) d db(e1+η1) = −4π d2 db2log|ϑ2(0)| >0. (2) 1 2e1−η1 =4π d dblog|ϑ3(0)| <0 and d2
db2 log|ϑ3(0)| >0. The same holds for ϑ4(0) =ϑ3(0). In particular, e1increases in b.
These modular functions come into play due to the explicit computation of Hessian at half periods along Re τ= 12 (cf. (6.3) and (6.19)):
4π2det D2Gω1 2 = (e1+η1) e1+η1− 2π b , 4π2det D2Gω2 2 =η1−1 2e1 2π b + 1 2e1−η1 − |Im e2|2. Although ei’s and ηi’s are classical objects, we were unable to find an appropriate reference where these inequalities were studied. Part of (2),
namely 12e1−η1 < 0, can be proved within the Weierstrass theory (cf. (6.20)). The whole theorem, however, requires theta functions in an es-sential way. Theta functions are recalled in§7 and the theorem is proved in Theorem 8.1 and Theorem 9.1. The proofs make use of the modular-ity of special values of theta functions (Jacobi’s imaginary transformation formula) as well as the Jacobi triple product formula. Notice that the geo-metric meaning of these two inequalities has not yet been fully explored. For example, the variation on signs from ϑ2to ϑ3is still mysterious to us.
2. GREENFUNCTIONS ANDPERIODS INTEGRALS
We start with some basic properties of the Green functions that will be used in the proof of Theorem 1.1. Detailed behavior of the Green functions and their critical points will be studied in later sections.
Let T = C/Zω1+Zω2 be a flat torus. As usual we let ω3 = ω1+ω2.
The Green function G(z, w)is the unique function on T which satisfies (2.1) −△zG(z, w) =δw(z) − 1
|T|
andRTG(z, w)dA = 0. It has the property that G(z, w) =G(w, z)and it is smooth in(z, w)except along the diagonal z= w, where
(2.2) G(z, w) = − 1
2π log|z−w| +O(|z−w| 2) +C
for a constant C which is independent of z and w. Moreover, due to the translation invariance of T we have that G(z, w) = G(z−w, 0). Hence it is also customary to call G(z) := G(z, 0)the Green function. It is an even function with the only singularity at 0.
There are explicit formulae for G(z, w)in terms of elliptic functions, ei-ther in terms of the Weierstrass ℘ function or the Jacobi-Riemann theta functions ϑj. Both are developed in this paper since they have different advantages. We adopt the first approach in this section.
Lemma 2.1. There exists a constant C(τ), τ= ω2/ω1, such that (2.3) 8πG(z) = 2
|T| Z
Tlog|℘(ξ) − ℘(z)|dA+C(τ),
It is straightforward to verify that the function of z, defined in the right hand side of (2.3), satisfies the equation for the Green function. By com-paring with the behavior near 0 we obtain Lemma 2.1. Since the proof is elementary, also an equivalent form in theta functions will be proved in Lemma 7.1, we skip the details here.
In view of Lemma 2.1, in order to analyze critical points of G(z), it is natural to consider the following periods integral
(2.4) F(z) =
Z
L
℘′(z) ℘(ξ) − ℘(z)dξ,
where L is a line segment in T which is parallel to the ω1-axis.
Fix a fundamental domain T0 = {sω1+tω2| − 12 ≤ s, t ≤ 12}and set L∗ = −L. Then F(z)is an analytic function, except at 0, in each region of T0 divided by L∪L∗. Clearly, when ξ 6= z, ℘′(z)/(℘(ξ) − ℘(z))has residue
±1 at z= ±z. Thus for any fixed z, F(z)may change its value by±2πi if the integration lines cross z. Let T0\(L∪L∗) = T1∪T2∪T3, where T1 is the region above L∪L∗, T2is the region bounded by L and L∗and T3is the region below L∪L∗. Recall that ζ′(z) = −℘(z)and ηi = ζ(z+ωi) −ζ(z) for i ∈ {1, 2}. Then we have
Lemma 2.2. Let C1=2πi, C2=0 and C3 = −2πi. Then for z∈Tk, (2.5) F(z) =2ω1ζ(z) −2η1z+Ck.
Proof. For z∈T1∪T2∪T3, we have F′(z) = Z L d dz ℘′(z) ℘(ξ) − ℘(z) dξ. Clearly, z and −z are the only (double) poles of d
dz ℘′(z) ℘(ξ) − ℘(z) as a meromorphic function of z and d
dz
℘′(z) ℘(ξ) − ℘(z)
has zero residues at ξ=
z and−z. Thus the value of F′(z)is independent of L and it is easy to see F′(z)is a meromorphic function with the only singularity at 0.
By fixing L such that 0 6∈ L∪L∗, a straightforward computation shows that F(z) = 2ω1 z −2η1z+O(z 2) in a neighborhood of 0. Therefore F′(z) = −2ω1℘(z) −2η1. By integrating F′, we get F(z) =2ω1ζ(z) −2η1z+Ck.
Since F(ω2/2) = 0, F(ω1/2) = 0 and F(−ω2/2) = 0, Ck is as claimed. Here we have used the Legendre relation η1ω2−η2ω1 =2πi.
From (2.3), we have (2.6) 8πGz = 1 |T| Z T −℘′(z) ℘(ξ) − ℘(z)dA.
Lemma 2.3. Let G be the Green function. Then for z= tω1+sω2,
(2.7) Gz = − 1
4π(ζ(z) −η1t−η2s). In particular, z is a critical point of G if and only if (2.8) ζ(tω1+sω2) =tη1+sη2.
Proof. We shall prove (2.7) by applying Lemma 2.2. Since critical points appear in pair, without loss of generality we may assume that z = tω1+ sω2with s≥0. We first integrate (2.6) along the ω1direction and obtain
f(s0):= Z L1(s0) −℘′(z) ℘(ξ) − ℘(z)dξ = −2ζ(z) +2η1z if s0 >s, −2ζ(z) +2η1z−2πi if −s<s0< s, −2ζ(z) +2η1z if s0 <−s, where L1(s0) = {t0ω1+s0ω2| |t0| ≤ 12}. Thus, 8πGz = Z 1 2 −1 2 Z L1(s0) −℘′(z) ℘(ζ) − ℘(z)dt0ds0= Z 1 2 −1 2 f(s0)ds0 = ω1−1((−2ω1ζ(z) +2η1z)(1−2s) + (−2ω1ζ(z) +2η1z−2πi)2s) = ω1−1(−2ω1ζ(z) +2η1z−4πsi) = ω1−1(−2ω1ζ(z) +2η1tω1+2s(η1ω2−2πi)) = −2ζ(z) +2η1t+2sη2,
where the Legendre relation is used again.
Corollary 2.4. Let G(z) be the Green function. Then 12ωk, k ∈ {1, 2, 3} are critical points of G(z). Furthermore, if z is a critical point of G then both periods integrals
F1(z):=2(ω1ζ(z) −η1z) and F2(z):=2(ω2ζ(z) −η2z) (2.9)
are purely imaginary numbers.
Proof. The half-periods 12ω1, 12ω2 and 12ω3 are obvious solutions of (2.8). Alternatively, the half periods are critical points of any even functions. In-deed for G(z) =G(−z), we get∇G(z) = −∇G(−z). Let p = 1
2ωifor some i∈ {1, 2, 3}, then p = −p in T and so∇G(p) = −∇G(p) =0.
If z= tω1+sω2is a critical point, then by Lemma 2.3,
ω1ζ(z) −η1z=ω1(tη1+sη2) −η1(tω1+sω2) =s(ω1η2−ω2η1) = −2sπi.
The proof for F2is similar.
Example 2.5. When τ = ω2/ω1 ∈ i R, by symmetry considerations it is known (cf. [6], Lemma 2.1) that the half periods are all the critical points. Example 2.6. There are tori such that equation (2.8) has more than three solutions. One such example is the torus with ω1=1 and ω2 = 12(1+√3i). In this case, the multiplication map z7→ω2z is simply the counterclockwise rotation by angle π/3, which preserves the lattice Zω1+Zω2, hence ℘ satisfies
Similarly g2 =60
∑
′ 1 ω4 =60∑
′ 1 (ω2ω)4 = ω22g2,which implies that g2 =0. Hence ej’s satisfy (2.11) e1e2+e1e3+e2e3= g2 =0 and℘satisfies
℘′′ =6℘2.
Let z0 be a zero of℘(z). Then ℘′′(z0) = 0 too. By (2.10), ℘(ω2z0) = 0, hence either ω2z0 = z0 or ω2z0 = −z0 on T since ℘(z) = 0 has zeros at z0 and −z0 only. From here, it is easy to check that either z0 is one of the half periods or z0 = ±13ω3. But z0 can not be a half period because
℘′′(z0) 6=0 at any half period. Therefore, we conclude that z0 = ±13ω3and
℘′′(±13ω3) = ℘(±13ω3) =0.
We claim that 13ω3 is a critical point. To prove it, we use the addition formulas for ζ (2.12) ζ(2z) =2ζ(z) +1 2 ℘′′(z) ℘′(z) to obtain (2.13) ζ2ω3 3 =2ζω3 3 . On the other hand,
ζ2ω3 3 = ζ−ω3 3 + (η1+η2) = −ζ ω3 3 +η1+η2.
Together with (2.13) we get ζω3
3
= 1
3(η1+η2). That is, 13ω3satisfies the critical point equation.
Thus G(z) has at least five critical points at 12ωk, k = 1, 2, 3 and ±13ω3 when τ =ω2/ω1= 12(1+
√
3i).
By way of Theorem 1.2, these are precisely the five critical points, though we do not know how to prove this directly.
To conclude this section, let u be a solution of (1.2) with ρ=8π and set (2.14) v(z) =u(z) +8πG(z).
Then v(z)satisfies
in T. By (2.2), it is obvious that v(z) is a smooth solution of (2.15). An important fact which we need is the following: Assume that there is a blow-up sequence of solutions vj(z)of (2.15). That is,
vj(pj) =max
T vj → +∞ as j→∞.
Then the limit p = limj→∞pj is the only blow-up point of{vj}and p is in fact a critical point of G(z):
(2.16) ∇G(p) =0.
We refer the reader to [3] (p. 739, Estimate B) for a proof of (2.16). 3. THECRITERION FOR EXISTENCE VIA MONODROMIES
Consider the mean field equation
(3.1) △u+ρeu =ρδ0, ρ∈R+
in a flat torus T, where δ0is the Dirac measure with singularity at 0 and the volume of T is normalized to be 1. A well known theorem due to Liouville says that any solution u of △u+ρeu = 0 in a simply connected domain Ω⊂Cmust be of the form
(3.2) u=c1+log |f
′|2
(1+ |f|2)2,
where f is holomorphic in Ω. Conventionally f is called a developing map of u. Given a torus T = C/Zω1+Zω2, by gluing the f ’s among simply
connected domains it was shown in [6] that for ρ= 4πl, l∈ N, (3.2) holds
on the whole C with f a meromorphic function. (The statement there is for rectangular tori with l =2, but the proof works for the general case.)
It is straightforward to show that u and f satisfy
(3.3) uzz− 1 2u 2 z = f′′′ f′ − 3 2 f′′ f′ 2 .
The right hand side of (3.3) is the Schwartz derivative of f . Thus for any two developing maps f and ˜f of u, there exists S = p −¯q
q ¯p
∈ PSU(1)
(i.e. p, q∈Cand|p|2+ |q|2 =1) such that
(3.4) ˜f=S f := p f − ¯q
q f+ ¯p.
Now we look for the constraints. The first type of constraints are im-posed by the double periodicity of the equation. By applying (3.4) to f(z+
ω1)and f(z+ω2), we find S1and S2in PSU(1)with f(z+ω1) =S1f ,
f(z+ω2) =S2f . (3.5)
The second type of constraints are imposed by the Dirac singularity of (3.1) at 0. A straightforward local computation with (3.2) shows that
Lemma 3.1. (1) If f(z)has a pole at z0 ≡ 0 (mod ω1, ω2), then the order k=l+1.
(2) If f(z) = a0+zk+ · · · is regular at z0 ≡ 0 (mod ω1, ω2)then k = l+1.
(3) If f(z)has a pole at z06≡0 (mod ω1, ω2), then the order is 1.
(4) If f(z) =a0+ (z−z0)k+ · · · is regular at z0 6≡0 (mod ω1, ω2)then k=1.
Now we are in a position to prove Theorem 1.1, namely the case l=2. Proof. We first prove the “only if” part. Let u be a solution and f be a developing map of u. By the above discussion, we may assume, after con-jugating a matrix in PSU(1), that S1 = e
iθ 0
0 e−iθ
for some θ ∈ R. Let
S2 = pq −¯p¯q
and then (3.5) becomes
f(z+ω1) =e2iθf(z), f(z+ω2) =S2f(z). (3.6)
Since S1S2 = S2S1, a direct computation shows that there are three pos-sibilities:
(1) p=0 and eiθ = ±i; (2) q=0; and
(3) eiθ = ±1.
Case (1).By assumption we have
f(z+ω1) = −f(z), f(z+ω2) = −
(¯q)2 f(z).
(3.7)
If f(z) has singularity at z = 0 then the order of pole at 0 is 3. Now (3.7) implies that f(z)is an elliptic function on the torus T′ = C/Z2ω1+ Z2ω2 and f′(z)can have zeros only at ω2 and ω3, both with multiplicity
2. On the other hand, f′(z)has 4-th order poles at 0 and ω1. This yields a contradiction to the fact that an elliptic function has equal numbers of poles and zeros.
If f is regular and f(0) = 0, we can still apply the above argument to 1/ f to get a contradiction.
It remains to consider the case that f is regular and f(0) 6= 0. By (3.7), f(z)/ f′(z)is an elliptic function on the torus T′′ =C/Zω1+Z2ω2. Since
both of multiplicity 2. Thus there exist constants Aj, j∈ {1, 2, 3}such that (3.8) f(z) f′(z) = A1℘(z) + A2 ℘(z) − ℘(ω2) +A3,
where ℘ is the Weierstrass elliptic function on T′′. Since ℘(z) is an even function, we have f(z)/ f′(z) = f(−z)/ f′(−z). Thus (f(z)f(−z))′ = 0 and f(z)f(−z)is a constant function. So
(3.9) f(−z) = f(0)
2 f(z) .
Since f′has zeros only at 0 and ω2, both of multiplicity 2, and f′only has poles of multiplicity 2, we see that f′ has precisely two double poles. So f has only two simple poles on T′′, or equivalently f has four simple poles on T′. Therefore{z| f(z) = f(0)}has solutions with total multiplicity 4 on T′. Since f(z) − f(0)has zero of multiplicity 3 at 0, there exists an unique solution z06=0 of f(z0) = f(0)in T′. But by (3.9) we have
f(−z0) = f 2(0) f(z0)
= f(0).
Thus −z0 = z0 (mod 2ω1, 2ω2), i.e. z0 = ωk for some k ∈ {1, 2, 3}. But f′(ωk) =0 for k=1, 2, 3 because ωkare half-periods in the torus T′. So the multiplicity of f(z) = f(0)at z= ωkis also 3, which yields a contradiction. Thus Case (1) is impossible.
Case (2).In this case we have
f(z+ω1) =e2iθ1f(z), f(z+ω2) =e2iθ2f(z). (3.10)
Notice that if a meromorphic function f satisfies (3.10), then eλf also satisfies (3.10) for any λ∈ R. Thus
(3.11) uλ(z) =c1+log e
2λ|f′(z)|2
(1+e2λ|f(z)|2)2 is a scaling family of solutions of (3.1) and
(3.12) sup
T |
uλ(x)| → +∞ as λ→∞.
This observation leads to a contradiction when the fundamental cell of T is a rectangle. Indeed by Theorem 1.2 in [6], for T a rectangular torus, a priori bounds for solutions of (3.1) was obtained.
To prove the general case, as in case (1), we first assume that f has singu-larity at 0. Then f′(z)has no zeros in T and|f′(z)| ≥c>0 for z∈T. Since
|f′(z)|is well-defined on T by (3.10), we have |f′(z)| ≥ c > 0 for z ∈ C.
This implies that 1/ f′is a bounded holomorphic function and hence a con-stant, which is absurd.
Secondly, let f be regular at 0 with f(0) =0. By applying the argument above to 1/ f , we again get a contradiction.
It remains to investigate the most delicate situation when f is regular at 0 and f(0) 6=0. By (3.10), f(z)/ f′(z)is an elliptic function on T which has 0 as its only pole. Thus there are constants A1and A2with
(3.13) f(z)
f′(z) = A1℘(z) +A2. Since℘(z)is even, this yields as before that
(3.14) f(−z) = f(0)
2 f(z) .
By (3.13), f(z)/ f′(z)has two zeros, say, z0and−z0. By (3.14), one of z0 or−z0 is a zero of f and the other one is a simple pole of f . Without loss of generality we assume that f has a zero at z0. In particular z0 6= −z0 in T and we conclude that z06=ωk/2 for any k∈ {1, 2, 3}.
As in the observation above, eλf can be used as a developing map for some solution uλ(x)defined by (3.11).
Clearly uλ(z) → −∞ as λ → +∞ for any z such that f(z) 6= 0 and uλ(z0) → +∞ as λ→ +∞. Hence z0is the blow-up point and we have by (2.16) that
∇G(z0) =0.
Namely, it is a critical point other than the half periods.
Case (3). In this case we get that S1is the identity. So by another conju-gation in PSU(1)we may assume that S2is in diagonal form. But this case is then reduced to Case (2). The proof of the “only if” part is completed.
Now we prove the “if” part. Suppose that there is a critical point z0 of G(z)with z0 6= 12ωkfor any k∈ {1, 2, 3}.
For any closed curve C such that z0and −z0 6∈ C, the residue theorem implies that (3.15) Z C ℘′(z0) ℘(ξ) − ℘(z0)dξ =2πmi for some m∈ {1, 0,−1}. Hence
(3.16) f(z):=exp Z z 0 ℘′(z0) ℘(ξ) − ℘(z0) dξ
is well-defined as a meromorphic function.
Let L1and L2be lines in T which are parallel to the ω1-axis and ω2-axis respectively. Then for j =1, 2,
(3.17) f(z+ωj) = f(z)exp Z Lj ℘′(z0) ℘(ξ) − ℘(z0)dξ .
By Lemma 2.2, Z Lj ℘′(z0) ℘(ξ) − ℘(z0)dξ =Fj(z0) +Ck
for some Ck ∈ {2πi, 0,−2πi}. Also by Corollary 2.4, Fj(z0) =2(ωjζ(z0) −ηjz0) =2iθj ∈i R. Hence for j=1, 2, (3.18) f(z+ωj) = f(z)e2iθj. holds. Set uλ(x) =c1+log e 2λ|f′(z)|2 (1+e2λ|f(z)|2)2.
Then uλ(x)satisfies (3.1) for any λ ∈Rand uλis doubly periodic by (3.18). Therefore, solutions have been constructed and the proof of Theorem 1.2 is
completed.
A similar argument leads to
Theorem 3.2. For ρ =4π, there exists an unique solution of (3.1).
Proof. By the same procedure of the previous proof, there are three cases to be discussed. For Case (2), the subcases that f or 1/ f is singular at z = 0 leads to contradiction as before. For the subcase that f is regular at z = 0 and f(0) 6=0 we see that f(z)/ f′(z)is an elliptic function with 0 as its only simple pole (since now k−1 = l = 1). Hence Case (2) does not occur. Similarly Case (3) is not possible.
Now we consider Case (1). By (3.7), the function g= (log f)′ = f′
f
is elliptic on T′′ =C/Zω1+Z2ω2. g has a simple pole at each zero or pole
of f . By Lemma 3.1, if f or 1/ f is singular at z =0 then g has no zero and we get a contradiction. So f is regular at z = 0, f(0) 6= 0 and g has two simple zeros at 0 and ω2.
Let σ(z) = expRz
ζ(w)dw = z+ · · · be the Weierstrass sigma function on T′′. σ is odd with a simple zero at each lattice point. Then
(3.19) g(z) = A σ(z)σ(z−ω2) σ(z−a)σ(z−b)
for some a, b with a+b= ω2.
From (3.7), we have g(z+ω2) = −g(z). So a+ω2 = b (mod ω1, 2ω2). Since the representation of g in terms of sigma functions is unique up to the lattice{ω1, 2ω2}, there is an unique solution of(a, b):
(3.20) a= −ω21; b= ω1
Notice that the residue of g at a and b are both equal to 2πir where r = σ(
1
2ω1)σ(12ω1+ω2) σ(ω1+ω2) . We claim that A=1/r. Since
f(z) =exp
Z z
g(w)dw
is well defined, by the residue theorem, we must have A = m/r for some m ∈ Z. Moreover by Lemma 3.1 f has simple poles other than z = 0
(indeed at z =a and z=b), so we conclude that m=1.
Conversely, by picking up a, b and A as above, it is clear that f(z)is a well defined meromorphic function which gives rise to a solution of (3.1)
for ρ=4π. Hence the proof is completed.
4. ANUNIQUENESS THEOREM FORρ∈ [4π, 8π]VIASYMMETRIZATIONS
From the previous section, for ρ =8π, solutions to the mean field equa-tion exist in a one parameter scaling family in λ with developing map f and centered at a critical point other than the half periods. By choosing λ= −log|f(0)|we may assume that f(0) =1. Then we have
f(−z) = 1
f(z).
Consider the particular solution
u(z) =c1+log |f
′(z)|2
(1+ |f(z)|2)2.
It is easy to verify that u(−z) = u(z)and u is the unique even function in this family of solutions. In order to prove the uniqueness up to scaling, it is equivalent to prove the uniqueness within the class of even functions.
The idea is to consider the following equation (4.1)
△u+ρeu=ρδ and u(−z) =u(z) on T
where ρ∈ [4π, 8π]. We will use the method of symmetrization to prove
Theorem 4.1. For ρ∈ [4π, 8π], the linearized equation of (4.1) is non-degenerate. That is, the linearized equation has only trivial solutions.
Together with the uniqueness of solution in the case ρ = 4π (Theorem 3.2), we conclude the proof of Theorem 1.2 by the inverse function theorem. We first prove Theorem 1.4, the Symmetrization Lemma. The proof will consist of several Lemmas. The first step is an extension of the classical isoperimetric inequality of Bol for domains in R2with metric ew|dx|2to the case when the metric becomes singular.
Let Ω⊂R2be a domain and w∈C2(Ω)which satisfies (4.2) ( △w+ew≥0 in Ω and Z Ωe wdx≤8π.
This is equivalent to saying that the Gaussian curvature of ew|dx|2is K≡ −12e−w△w≤ 12.
For any domain ω⋐Ω, we set
(4.3) m(ω) = Z ωe wdx and ℓ(∂ω) =Z ∂ωe w/2ds.
Bol’s isoperimetric inequality says that if Ω is simply-connected then (4.4) 2ℓ2(∂ω) ≥m(ω)(8π−m(ω)).
We first extend it to the case when w acquires singularities: △w+ew =∑jN=12παjδpj in Ω and Z Ωe wdx ≤8π, (4.5) with αj >0, j=1, 2, . . . , N.
Lemma 4.2. Let Ω be a simply-connected domain and ω be a solution of (4.5). Then for any domain ω⋐Ω, we have
(4.6) 2ℓ2(∂ω) ≥m(ω)(8π−m(ω)). Proof. Define v and wǫby
w(x) =
∑
j αjlog|x−pj| +v(x), wε(x) =∑
j αj 2 log(|x−pj| 2+ε2) +v(x).By straightforward computations, we have
△wε(x) +ewε(x) =
∑
j α j 2 4ε2 (r2+ε2)2 + (|x−pj| 2+ε2)αj/2ev− |x−p j|αjev ≥0. Let ℓε and mε be defined as in (4.3) with respect to the metric ewε(x)|dx|2. Then we have2ℓ2ε(∂ω) ≥mε(ω)(8π−mε(ω)).
Next we consider the case that some of the αj’s are negative. For our purpose, it suffices to consider the case with only one singularity p1 with negative α1(and we only need the case that α1 = −1). In view of (the proof of) Lemma 4.2, the problem is reduced to the case with only one singularity p1. In other words, let w satisfy
(4.7) △w+ew= −2πδp1 in Ω.
Lemma 4.3. Let w satisfy (4.7) with Ω simply-connected. Suppose that (4.8) Z Ωe wdx≤4π, then (4.9) 2ℓ2(∂ω) ≥m(ω)(4π−m(ω)). Proof. We may assume that p1 =0. If 06∈ω then
2ℓ2(∂ω) ≥m(ω)(8π−m(ω)) >m(ω)(4π−m(ω))
by Bol’s inequality trivially. If 0∈ ω, we consider the double cover ˜Ω of Ω branched at 0. Namely we set ˜Ω= f−1(Ω)where
x= f(z) =z2 for z ∈C.
The induced metric ev|dz|2on ˜Ω satisfies
ev(z)|dz|2 =ew(x)|dx|2= ew(z2)4|z|2|dz|2. That is, v is the regular part
v(z):=w(x) +log|x| +log 4=w(z2) +2 log|z| +log 4. By construction, v satisfies
△v+ev=0 in ˜Ω\{0}.
Since v is bounded in a neighborhood, by the regularity of elliptic equa-tions, v(z)is smooth at 0. Hence v satisfies
△v+ev =0 in ˜Ω. Let ˜ω = f−1(ω). Clearly ∂ ˜ω ⊆ f−1(∂ω). Also
l(∂ ˜ω) ≤2l(∂ω) and m(w˜) =2m(ω), where l(∂ ˜ω) = Z ∂ ˜ωe v/2ds and m(ω˜) =Z ˜ ωe vdx. By Bol’s inequality, we have
4ℓ2(∂ω) ≥ ℓ2(∂ ˜ω) ≥ 1
2m(ω˜)(8π−m(ω˜)) =m(ω)(8π−2m(ω)).
Lemma 4.4 (Symmetrization I). Let Ω ⊂ R2 be a simply-connected domain
with 0∈Ω and let v be a solution of
△v+ev= −2πδ0 in Ω. If the first eigenvalue of△ +ev is zero on Ω then (4.10)
Z
Ωe
vdx≥2π. Proof. Let ψ be the first eigenfunction of△ +ev: (4.11)
△ψ+evψ=0 in Ω and ψ=0 on ∂Ω.
In R2, let U and ϕ be the radially symmetric functions
(4.12) U(x) =log 2 (1+ |x|)2|x| and ϕ(x) = 1− |x| 1+ |x|. From△ = ∂ 2 ∂r2 + 1 r ∂ ∂r + 1 r2 ∂2
∂θ2, it is easy to verify that U and ϕ satisfy (4.13)
△U+eU = −2πδ0 and
△ϕ+eUϕ=0 in R2.
For any t>0, set Ωt= {x∈Ω|ψ(x) >t}and r(t) >0 such that (4.14) Z Br(t) eU(x)dx= Z {ψ>t}e v(x)dx,
where Br(t)is the ball with center 0 and radius r(t). Clearly r(t)is strictly decreasing in t for t ∈ (0, max ψ). In fact, r(t) is Lipschitz in t. Denote by ψ∗(r)the symmetrization of ψ with respect to the measure eU(x)dx and ev(x)dx. That is,
ψ∗(r) =sup{t |r <r(t)}.
Obviously ψ∗(r)is decreasing in r and for t ∈ (0, max ψ), ψ∗(r) =t if and only if r(t) =r. Thus by (4.14), we have a decreasing function
(4.15) f(t):= Z {ψ∗>t} e U(x)dx= Z {ψ>t}e v(x)dx. By Lemma 4.3, for any t>0,
(4.16) 2ℓ2({ψ= t}) ≥ f(t)(4π− f(t)).
We will use inequality (4.16) in the following computation: For any t > 0, by the Co-Area formula,
−dtd Z Ωt |∇ψ|2dx=Z ∂Ωt |∇ψ|ds and −f′(t) = −dtd Z Ωt evdx= Z ∂Ωt ev |∇ψ|ds
hold almost everywhere in t. Thus − dtd Z Ωt |∇ψ|2dx=Z {ψ=t}|∇ψ|ds ≥ Z {ψ=t}e v/2dx !2 Z {ψ=t} ev |∇ψ|dx !−1 = −ℓ2({ψ= t})f′(t)−1 ≥ −12f(t)(4π− f(t))f′(t)−1. (4.17)
The same procedure for ψ∗leads to
− dtd Z {ψ∗>t}|∇ψ ∗|2dx=Z {ψ∗=t}|∇ψ ∗|ds = Z {ψ∗=t}e U/2dx !2 Z {ψ∗=t} eU |∇ψ∗|ds !−1 = −ℓ2({ψ∗ =t})f′(t)−1 = −12f(t)(4π− f(t))f′(t)−1, (4.18)
with all inequalities being equalities. Hence
−dtd Z {ψ>t}|∇ψ| 2dx ≥ −dtd Z {ψ∗>t}|∇ψ ∗|2ds holds almost everywhere in t. By integrating along t, we get
Z
Ω|∇ψ|
2dx≥Z
BR0|∇ψ ∗|2dx.
Since ψ and ψ∗ have the same distribution (or by looking at −R
f′(t)t2dt directly), we have Z BR0e U(x)ψ∗2dx= Z Ωe v(x)ψ2dx. Therefore 0= Z Ω|∇ψ| 2dx − Z Ωe vψ2dx ≥ Z BR0|∇ψ ∗|2dx − Z BR0e Uψ∗2 dx. This implies that the linearized equation △ +eU(x) has non-positive first eigenvalue. By (4.12), this happens if and only if R0≥1. Thus
Z Ωe v(x)dx=Z BR0e U(x)dx ≥2π.
Remark 4.5. (See [2].) By applying symmetrization to△v+ev =0 in Ω with λ1(△ +ev) =0, the corresponding radially symmetric functions are
U(x) = −2 log 1+ 1 8|x| 2 and ϕ(x) = 8− |x| 2 8+ |x|2. The same computations leads toRΩevdx ≥4π.
A closer look at the proof of Lemma 4.4 shows that it works for more general situations as long as the isoperimetric inequality holds:
Lemma 4.6 (Symmetrization II). Let Ω ⊂ R2 be a simply-connected domain
and let v be a solution of
△v+ev=
∑
Nj=12παjδpjin Ω. Suppose that the first eigenvalue of△ +ev is zero on Ω with ϕ the first eigenfunction. If the isoperimetric inequality with respect to ds2 =ev|dx|2:
2ℓ2(∂ω) ≥m(ω)(4π−m(ω))
holds for all level domains ω = {ϕ≥t}with t≥0, then
Z
Ωe
vdx≥2π.
Notice that we do not need any further constraint on the sign of αj. For the last statement of Theorem 1.4, the limiting procedure of Lemma 4.2 implies that the isoperimetric inequality (Lemma 4.3) and symmetrization I (Lemma 4.4) both hold regardless on the presence of singularities with non-negative α. Thus the proof of Theorem 1.4 is completed.
Proof. (of Theorem 4.1.) Let u be a solution of equation (4.1). It is clear that we must haveR
Teu =1. Suppose that ϕ(x)is a non-trivial solution of the linearized equation at u:
(4.19)
△ϕ+ρeuϕ=0 and ϕ(z) = ϕ(−z) in T. We will derive from this a contradiction.
Since both u and ϕ are even functions, by using x = ℘(z) as two-fold covering map of T onto S2 = C∪ {∞}, we may require that℘ being an isometry:
eu(z)|dz|2 =ev(x)|dx|2 =ev(x)|℘′(z)|2|dz|2. Namely we set
(4.20) v(x):=u(z) −log|℘′(z)|2 and ψ(x):= ϕ(z).
There are four branch points on C∪ {∞}, namely p0= ℘(0) =∞ and pj = ej := ℘(ωj/2)for j=1, 2, 3. Since℘′(z)2 =4 ∏3j=1(x−ej), by construction
v(x)and ψ(x)then satisfy (4.21)
(
△v+ρev =
∑
3j=1(−2π)δpj and△ψ+ρevψ=0 in R2.
To take care of the point at infinity, we use coordinate y=1/x or equiv-alently we consider T → S2via y = 1/℘(z) ∼ z2. The isometry condition reads as eu(z)|dz|2 =ew(y)|dy|2 =ew(y)|℘′(z)| 2 |℘(z)|4|dz| 2. Near y=0 we get w(y) =u(z) −log|℘′(z)| 2 |℘(z)|4 ∼ ρ 4π −1 log|y|. Thus ρ≥4π implies that p0is a singularity with non-negative α0:
△w+ρew= α0δ0+
∑
3j=1(−2π)δ1/pj.In dealing with equation (4.1) and the above resulting equations, by re-placing u by u+log ρ etc., we may (and will) replace the ρ in the left hand side by 1 for simplicity. The total measure on T and R2are then given by
Z Te udz=ρ ≤8π and Z R2e vdx= ρ 2 ≤4π.
The nodal line of ψ decomposes S2 into at least two connected compo-nents and at least two of them are simply connected. If there is a simply connected component Ω which contains no pj’s, then the symmetrization (Remark 4.5) leads to
Z
Ωe vdx
≥4π,
which is a contradiction because R2\Ω 6= ∅. If every simply connected component Ωi, i =1, . . . , m, contains only one pj, then Lemma 4.4 implies that
Z
Ωie
vdx≥2π for i=1, . . . , m.
The sum is at least 2mπ, which is again impossible unless m =2 and R2= Ω1∪Ω2. So without lost of generality we are left with one of the following two situations:
R2∪ {∞}\{ψ=0} =Ω+∪Ω− where
Ω+ ⊂ {x|ψ(x) >0} and Ω− ⊃ {x|ψ(x) <0}. Both Ω+and Ω−are simply-connected.
(1) Either Ω−contains p1, p2and p3∈Ω+or (2) p1 ∈Ω−, p3 ∈Ω+and p2∈ C= ∂Ω+= ∂Ω−.
Assume that we are in case (1). By Lemma 4.3, we have on Ω+ (4.22) 2ℓ2({ψ=t}) ≥m({ψ≥ t})(4π−m({ψ≥t}))
for t ≥0.
We will show that the similar inequality
(4.23) 2ℓ2({ψ=t}) ≥m({ψ≤ t})(4π−m({ψ≤t}))
holds on Ω−for all t≤0.
Let t≤0 and ω be a component of{ϕ≤t}.
If ω contains at most one point of p1and p2, then Lemma 4.3 implies that 2ℓ2(∂ω) ≥ (4π−m(ω))m(ω).
If ω contains both p1and p2, then R2\ω is simply connected which con-tains p3only. Thus by Lemma 4.3
2ℓ2(∂ω) ≥ (4π−m(R2\ω))m(R2\ω) = (4π−ρ/2+m(ω))(ρ/2−m(ω)) = (4π−m(ω))m(ω) + (4π−ρ/2)(ρ/2−2m(ω)). (4.24) Since ρ≤8π andR Ω+e vdx≥2π, we get ρ 2 = Z R2e v≥ Z Ω+ ev+ Z ωe v ≥2π+m(ω) ≥2m(ω). Then again 2ℓ2(∂ω) ≥ (4π−m(ω))m(ω)
with equality holds only when ρ=8π and m(ω) =2π.
Now it is a simple observation that domains satisfy the isoperimetric inequality (4.3) have the addition property. Indeed, if 2a2 ≥ (4π−m)m and 2b2≥ (4π−n)n, then
2(a+b)2=2a2+4ab+2b2 >(4π−m)m+ (4π−n)n
=4π(m+n) − (m+n)2+2mn>(4π− (m+n))(m+n). Hence (4.23) holds for all t≤0.
Now we can apply Lemma 4.6 to Ω−to conclude that
Z
Ω−e
vdx=2π
(which already leads to a contradiction if ρ<8π) and the equality 2ℓ2({ψ=t}) = (4π−m({ψ≤t}))m({ψ≤t})
in (4.23) holds for all t ∈ (min ψ, 0). This implies that{ψ ≤ t}has only one component and it contains p1 and p2for all t ∈ (min ψ, 0). But then ψ attains its minimum along a connected set ψ−1(min ψ) containing p1 and p2, which is impossible.
In case (2), ρ<8π again leads to a contradiction via the same argument. For ρ=8π, we have Z Ω+ evdx= Z Ω−e vdx=2π
and all inequalities in (4.17) are equalities. So under the notations there
Z ∂Ωt ev/2dx !2 = 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(p
2)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 2by 2. Thus (5.2) Gω2 2 −Gω3 2 = 1 2πRe Z ω3 2 ω2 2 (ζ(z) −η1z)dz.
Set F(z) =ζ(z) −η1z. We have F(z+ω1) =F(z)and ζt+ ω2 2 −η1 t+ ω2 2 = −ζ−ω2 2 −t −η1 t+ ω2 2 = −ζω2 2 −t +η2−η1 t+ ω2 2 = −hζω2 2 −t −η1 ω 2 2 −t i +η2−η1ω2 = −hζω2 2 −t −η1 ω 2 2 −t i −2πi, hence Re F(1 2ω2+t)is antisymmetric in t∈C.
To calculate the integral in (5.2), we use the addition theorem to get
℘′(z) ℘(z) −e1 =ζz− ω21+ζz+ ω1 2 −2ζ(z) = Fz−ω1 2 +Fz+ω2 2 −2F(z). Integrating it along the segment from 12ω2to 12ω3, we get
2 loge3−e1 e2−e1 = Z L1 F(z)dz+ Z L2 F(z)dz−2 Z L3 F(z)dz,
where L1 is the line from 12(ω2−ω1) to 12ω2, L2 is the line from 12ω3 to 1
2ω2+ω1and L3is the line from 12ω2 to 21ω3. Since F(z) = F(z+ω1)and Re F(z)is antisymmetric with respect to 12ω2, we have
2 loge3−e1 e2−e1 =2 Z LF(z)dz−4 Z L1 F(z)dz=2πi−4 Z L3 F(z)dz, where L is the line from 12(ω2−ω1)to 12ω3. Thus we have
log e3−e1 e2−e1 = −2 Re Z ω3 2 ω2 2 F(z)dz= −4πGω2 2 −Gω3 2 . That is, (5.3) Gω2 2 −Gω3 2 = 1 4π log e2−e1 e3−e1 . Similarly, by integrating (2.7) in the ω2direction, we get
(5.4) Gω1 2 −Gω3 2 = 1 2πRe Z ω3 2 ω1 2 (ζ(z) −η2z)dz. The same proof then gives rise to
(5.5) Gω1 2 −Gω3 2 = 1 4π log e1−e2 e3−e2 . By combining the above two formulae we get also that
(5.6) Gω1 2 −Gω2 2 = 1 4π log e1−e3 e2−e3 .
In order to compare, say, G(1
2ω1)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 −Gω2 2 = 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 = 1
2ω2 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 = −1 2e1 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= 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(1
2ω3) < G(12ω2) for τ = ib, we conclude that for τ ∈ Ω′∪Ω,
|λ(τ) −1| <1 if and only if|Re τ| < 1 2. Therefore for τ∈Ω′∪Ω, |Re τ| < 12 if and only if Gω3 2 <Gω2 2 .
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 ¯τ+1 2 ℘τ+1 ¯τ+1¯z and so (compare (2.1)) (5.13) G(z) =Gτ+1 ¯τ+1¯z . Clearly, for z= 1 2τ, (5.13) implies that G(12ω2) =G(12ω1). So (5.14) |τ| =1 if and only if e2−e3 e1−e3 =1, (5.15) |τ| <1 if and only if Gω1 2 < Gω2 2 . Similarly, (5.16) |τ−1| <1 if and only if Gω1 2 < Gω3 2 . 6. DEGENERACY ANALYSIS OFCRITICALPOINTS
By (2.7), the derivatives of G can be computed by 2πGx=Re(η1t+η2s−ζ(z)),
−2πGy=Im(η1t+η2s−ζ(z)). (6.1)
When τ= 12+ib, since℘(z)is real for z ∈R, η1is real and (6.1) becomes
2πGx=η1t+ 1
2η1s−Re ζ(z), 2πGy=Im ζ(z) + (2π−η1b)s. (6.2)
Thus the Hessian of G is given by
2πGxx=Re℘(z) +η1, 2πGxy = −Im℘(z), 2πGyy= − Re℘(z) +η1−2π b . (6.3)
