ELLIPTIC FUNCTIONS, GREEN FUNCTIONS AND THE MEAN FIELD …math.cts.nthu.edu.tw/Mathematics/preprints/prep2006-6... · 2006-09-08 · ELLIPTIC FUNCTIONS, GREEN FUNCTIONS AND THE MEAN

ELLIPTIC FUNCTIONS, GREEN FUNCTIONS AND THE MEAN

FIELD EQUATIONS ON TORI

CHANG-SHOU LIN AND CHIN-LUNG WANG

ABSTRACT. We show that the Green functions on flat tori can have either3 or 5 critical points only. There does not seem to be any direct method toattack this problem. Instead, we have to employ sophisticated non-linearpartial differential equations to study it.

We also study the distribution of number of critical points over themoduli space of flat tori through deformations. The functional equationsof special theta values provide important inequalities which lead to asolution for all rhombus tori. The general picture is also emerged, thoughsome of the necessary technicality is still to de developed.

1. INTRODUCTION AND STATEMENT OF RESULTS

The study of geometric or analytic problems on two dimensional tori isthe same as the study of problems on R2 with doubly periodic data. Suchsituations 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 ratherdeep 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 differential equations. Indeed, researches on doubly periodic problemsin mathematical physics or differential equations often restrict the study torectangular tori for simplicity. This leaves the impression that the theoryfor general tori may resemble much the same way as for the rectangularcase. However this turns out to be false. We will show that the solvabilityof the mean field equation depends on the shape of the Green function, whichin 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

2 CHANG-SHOU LIN AND CHIN-LUNG WANG

and∫

T G(z, w) dA = 0, where δw is the Dirac measure with singularity atz = 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) = − 1

2πlog

∣

∣

∣

∣

ϑ1(z)

ϑ′1(0)

∣

∣

∣

∣

+1

2by2

where z = x + iy and τ := ω2/ω1 = a + ib. The structure of G, especiallyits critical points and critical values, will be the fundamental objects thatinterest 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) := −∫ z

℘ and usingthe 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 criticalpoint 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 12 ω1, 1

2 ω2 and12 ω3 = (ω1 + ω2)/2 are the three obvious critical points and other criticalpoints 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 delicatequestion 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 geometryof the critical point set over the moduli space of flat tori M1 = 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] forthe 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 hassolutions, 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 criteriawhose proof is given in §3 by a detailed manipulation on elliptic functions:

MEAN FIELD EQUATIONS ON TORI 3

Theorem 1.1 (Existence). For ρ = 8π, the mean field equation on a flat torushas solutions if and only if the Green function has critical points other than thethree half period points. Moreover, each extra pair of critical points corresponds toan one parameter scaling family of solutions.

It is known that for rectangular tori G(z) has precisely the three obviouscritical 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/3 there are atleast 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 torushas at most one solution up to scaling.

In view of the correspondence in Theorem 1.1, an equivalent statementof 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 fromthe critical point equation (1.1). Instead, we will prove the uniqueness the-orem first, and then Theorem 1.3 is an immediate corollary. Our proof ofTheorem 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 adouble cover T → S2 via the Weierstrass ℘ function and to transform equa-tion (1.2) into a similar one on S2 but with three more delta singularitieswith negative coefficients. The condition ρ ≥ 4π is to guarantee that theoriginal 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 anduniqueness of solution (Theorem 3.2). For ρ ∈ [4π, 8π], the symmetriza-tion reduces the problem on the non-degeneracy of the linearized equationto the isoperimetric inequality on domains in R2 with respect to certainsingular measure:

Theorem 1.4 (Symmetrization Lemma). Let Ω ⊂ R2 be a simply connecteddomain and let v be a solution of

v + ev = ∑N

j=12παjδpj

in Ω. Suppose that the first eigenvalue of + ev is zero on Ω with ϕ the firsteigenfunction. 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∫

Ωev dx ≥ 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 veryfew instances that one needs to study a simple analytic equation, here thecritical 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 thenon-linear functional

(1.4) J8π(v) =1

2

∫

T|∇v|2 dA − 8π log

∫

Tev−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 rectangulartori was known in [6]. Here we sharpen the result to the non-existence ofsolutions. Also for ρ ∈ (4π, 8π) we sharpen the result on non-trivial degreeof equation (1.2) in [4] to the uniqueness of solutions within even functions.We expect the uniqueness holds true without the even assumption, but ourmethod only achieves this at ρ = 4π. Obviously, uniqueness without evenassumption fails at ρ = 8π due to the existence of scaling.

Naturally, the next question after Theorem 1.2 is to determine those toriwhose Green functions have five critical points. Certainly, it is the case ifthose three half-periods are all saddle points, and it is desirable to knowwhether the converse holds. The following theorem in [9] answers thisquestion. We present it separately since its proof requires computations ina different flavor:

Theorem A. Suppose that the Green function on the given torus has five criticalpoints. 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 threehalf 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 factwe will show in [9] that any solution of equation (1.2) must be a minimizerof the nonlinear functional J8π in (1.4). Thus we completely solve the exis-tence problem on minimizers, a question raised by Nolasco and Tarantelloin [11].

By Theorem A, we have reduced the question on detecting a given torusto 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

12(1 + i)

12 + b1i

FIGURE 1. Ω5 contains a neighborhood of the ∗ shape.

Theorem 1.5. Let z0 and z1 be 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, thenΩ3 ∪ ∞ is a closed subset containing i, Ω5 is an open subset contain-

ing eπi/3, both of them are simply connected and their common boundaryC := ∂Ω3 = ∂Ω5 is a curve homeomorphic to S1 containing ∞. Moreover,the extra critical points are split out from some half period point when thetori move from Ω3 to Ω5 across C.

Concerning with the experimental observation, we propose to prove itby the method of deformations in M1. The degeneracy analysis of criticalpoints, especially the half period points, is a crucial step. In this directionwe 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 equivalentto 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 1

2 ω1 is a degenerate critical

point of G(z; τ) if and only if b = b0 or b = b1. Moreover, 12 ω1 is a local

minimum point of G(z; τ) if b0 < b < b1 and it is a saddle point if b < b0

or b > b1.(2) Both 1

2 ω2 and 12 ω3 are non-degenerate saddle points of G.

(3) G(z; τ) has two more critical points ±z0(τ) when b < b0 or b > b1. Theyare 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-sition M1 = Ω3 ∪ Ω5. Part (3) implies that minimizers of J8π exist for tori

with τ = 12 + b with b < b0 or b > b1.


0.2 0.4 0.6 0.8 1 1.2 1.4

-20

-10

10

FIGURE 2. Graphs of η1 (the left one) and e1 in b where e1 is 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 12 e1 − η1.

Both functions are increasing in b and 12 e1 − η1 ր 0.

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 ellipticfunctions and we would like to single out the statements (recall that ei =℘( 1

2 ωi) and ηi = 2ζ( 12 ωi)):

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

db2log |ϑ3(0)| > 0. The same

holds for ϑ4(0) = ϑ3(0). In particular, e1 increases 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)):

4π2 det D2G(ω1

2

)

= (e1 + η1)(

e1 + η1 −2π

b

)

,

4π2 det D2G(ω2

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 anappropriate reference where these inequalities were studied. Part of (2),


namely 12 e1 − 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 provedin Theorem 8.1 and Theorem 9.1. The proofs make use of the modular-ity of special values of theta functions (Jacobi’s imaginary transformationformula) 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 ϑ2 to ϑ3 is still mysterious to us.

2. GREEN FUNCTIONS AND PERIODS INTEGRALS

We start with some basic properties of the Green functions that will beused in the proof of Theorem 1.1. Detailed behavior of the Green functionsand 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|and

∫

TG(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

2πlog |z − w|+ O(|z − w|2) + C

for a constant C which is independent of z and w. Moreover, due to thetranslation invariance of T we have that G(z, w) = G(z − w, 0). Hence itis also customary to call G(z) := G(z, 0) the Green function. It is an evenfunction 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 thetafunctions ϑj. Both are developed in this paper since they have differentadvantages. 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|∫

Tlog |℘(ξ) − ℘(z)| dA + C(τ),

It is straightforward to verify that the function of z, defined in the righthand 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 iselementary, also an equivalent form in theta functions will be proved inLemma 7.1, we skip the details here.

In view of Lemma 2.1, in order to analyze critical points of G(z), it isnatural to consider the following periods integral

(2.4) F(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 ≤ 1

2 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 ifthe integration lines cross z. Let T0\(L ∪ L∗) = T1 ∪ T2 ∪ T3, where T1 isthe region above L ∪ L∗, T2 is the region bounded by L and L∗ and T3 is theregion 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) =∫

L

d

dz

(

℘′(z)

℘(ξ) − ℘(z)

)

dξ.

Clearly, z and −z are the only (double) poles ofd

dz

(

℘′(z)

℘(ξ) − ℘(z)

)

as a

meromorphic function of z andd

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 seeF′(z) is a meromorphic function with the only singularity at 0.

By fixing L such that 0 6∈ L ∪ L∗, a straightforward computation showsthat

F(z) =2ω1

z− 2η1z + O(z2)

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|∫

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 pointsappear in pair, without loss of generality we may assume that z = tω1 +sω2 with s ≥ 0. We first integrate (2.6) along the ω1 direction and obtain

f (s0) : =∫

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 =∫ 1

2

− 12

∫

L1(s0)

−℘′(z)

℘(ζ) − ℘(z)dt0 ds0 =

∫ 12

− 12

f (s0) ds0

= ω−11 ((−2ω1ζ(z) + 2η1z)(1 − 2s) + (−2ω1ζ(z) + 2η1z − 2πi)2s)

= ω−11 (−2ω1ζ(z) + 2η1z − 4πsi)

= ω−11 (−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 12 ωk, k ∈ 1, 2, 3 are

critical points of G(z). Furthermore, if z is a critical point of G then both periodsintegrals

F1(z) := 2(ω1ζ(z) − η1z) and

F2(z) := 2(ω2ζ(z) − η2z)(2.9)

are purely imaginary numbers.

Proof. The half-periods 12 ω1, 1

2 ω2 and 12 ω3 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 = 12 ωi for some

i ∈ 1, 2, 3, then p = −p in T and so ∇G(p) = −∇G(p) = 0.If z = tω1 + sω2 is a critical point, then by Lemma 2.3,

ω1ζ(z) − η1z = ω1(tη1 + sη2) − η1(tω1 + sω2) = s(ω1η2 − ω2η1) = −2sπi.

The proof for F2 is similar.

Example 2.5. When τ = ω2/ω1 ∈ i R, by symmetry considerations it isknown (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 z 7→ ω2z is simply the counterclockwiserotation by angle π/3, which preserves the lattice Zω1 + Zω2, hence ℘satisfies

(2.10) ℘(ω2z) = ℘(z).


Similarly

g2 = 60 ∑′ 1

ω4= 60 ∑

′ 1

(ω2ω)4= ω2

2g2,

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 zerosat z0 and −z0 only. From here, it is easy to check that either z0 is one of

the half periods or z0 = ± 13 ω3. But z0 can not be a half period because

℘′′(z0) 6= 0 at any half period. Therefore, we conclude that z0 = ± 13 ω3 and

℘′′(± 13 ω3) = ℘(± 1

3 ω3) = 0.

We claim that 13 ω3 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) ζ(2ω3

3

)

= 2ζ(ω3

3

)

.

On the other hand,

ζ(2ω3

3

)

= ζ(−ω3

3

)

+ (η1 + η2) = −ζ(ω3

3

)

+ η1 + η2.

Together with (2.13) we get

ζ(ω3

3

)

=1

3(η1 + η2).

That is, 13 ω3 satisfies the critical point equation.

Thus G(z) has at least five critical points at 12 ωk, k = 1, 2, 3 and ± 1

3 ω3

when τ = ω2/ω1 = 12 (1 +

√3i).

By way of Theorem 1.2, these are precisely the five critical points, thoughwe 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

(2.15) v(z) + 8π(e−8πG(z)ev(z) − 1) = 0


in T. By (2.2), it is obvious that v(z) is a smooth solution of (2.15). Animportant 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) = maxT

vj → +∞ as j → ∞.

Then the limit p = limj→∞ pj is the only blow-up point of vj and p is infact 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. THE CRITERION FOR EXISTENCE VIA MONODROMIES

Consider the mean field equation

(3.1) u + ρeu = ρδ0, ρ ∈ R+

in a flat torus T, where δ0 is the Dirac measure with singularity at 0 and thevolume of T is normalized to be 1. A well known theorem due to Liouvillesays that any solution u of u + ρeu = 0 in a simply connected domainΩ ⊂ C must 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 mapof u. Given a torus T = C/Zω1 + Zω2, by gluing the f ’s among simplyconnected domains it was shown in [6] that for ρ = 4πl, l ∈ N, (3.2) holdson the whole C with f a meromorphic function. (The statement there is forrectangular 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

2u2

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 −qq p

)

∈ PSU(1)

(i.e. p, q ∈ C and |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 S1 and S2 in PSU(1) with

f (z + ω1) = S1 f ,

f (z + ω2) = S2 f .(3.5)

These relations also force that S1S2 = S2S1.


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 orderk = 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 z0 6≡ 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) thenk = 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 adeveloping map of u. By the above discussion, we may assume, after con-

jugating a matrix in PSU(1), that S1 =

(

eiθ 00 e−iθ

)

for some θ ∈ R. Let

S2 =

(

p −qq p

)

and then (3.5) becomes

f (z + ω1) = e2iθ f (z),

f (z + ω2) = S2 f (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 multiplicity2. On the other hand, f ′(z) has 4-th order poles at 0 and ω1. This yields acontradiction to the fact that an elliptic function has equal numbers of polesand zeros.

If f is regular and f (0) = 0, we can still apply the above argument to1/ 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. Sincef ′(z) 6= 0 for z 6≡ 0 (mod ω1, ω2), f (z)/ f ′(z) only has poles at 0 and ω2,


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 evenfunction, we have f (z)/ f ′(z) = f (−z)/ f ′(−z). Thus ( f (z) f (−z))′ = 0and 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 haspoles of multiplicity 2, we see that f ′ has precisely two double poles. So fhas only two simple poles on T′′, or equivalently f has four simple poleson T′. Therefore z | f (z) = f (0) has solutions with total multiplicity 4 onT′. Since f (z) − f (0) has zero of multiplicity 3 at 0, there exists an uniquesolution z0 6= 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. Butf ′(ωk) = 0 for k = 1, 2, 3 because ωk are half-periods in the torus T′. So themultiplicity of f (z) = f (0) at z = ωk is also 3, which yields a contradiction.Thus Case (1) is impossible.

Case (2). In this case we have

f (z + ω1) = e2iθ1 f (z),

f (z + ω2) = e2iθ2 f (z).(3.10)

Notice that if a meromorphic function f satisfies (3.10), then eλ f alsosatisfies (3.10) for any λ ∈ R. Thus

(3.11) uλ(z) = c1 + loge2λ| f ′(z)|2

(1 + e2λ| f (z)|2)2

is a scaling family of solutions of (3.1) and

(3.12) supT

|uλ(x)| → +∞ as λ → ∞.

This observation leads to a contradiction when the fundamental cell ofT is a rectangle. Indeed by Theorem 1.2 in [6], for T a rectangular torus, apriori 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 argumentabove to 1/ f , we again get a contradiction.

It remains to investigate the most delicate situation when f is regular at0 and f (0) 6= 0. By (3.10), f (z)/ f ′(z) is an elliptic function on T which has0 as its only pole. Thus there are constants A1 and A2 with

(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, z0 and −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 lossof generality we assume that f has a zero at z0. In particular z0 6= −z0 in Tand we conclude that z0 6= ωk/2 for any k ∈ 1, 2, 3.

As in the observation above, eλ f can be used as a developing map forsome solution uλ(x) defined by (3.11).

Clearly uλ(z) → −∞ as λ → +∞ for any z such that f (z) 6= 0 anduλ(z0) → +∞ as λ → +∞. Hence z0 is 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 S1 is the identity. So by another conju-gation in PSU(1) we may assume that S2 is in diagonal form. But this caseis 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= 12 ωk for any k ∈ 1, 2, 3.

For any closed curve C such that z0 and −z0 6∈ C, the residue theoremimplies that

(3.15)∫

C

℘′(z0)

℘(ξ) − ℘(z0)dξ = 2πmi

for some m ∈ 1, 0,−1. Hence

(3.16) f (z) := exp

(

∫ z

0

℘′(z0)

℘(ξ) − ℘(z0)dξ

)

is well-defined as a meromorphic function.Let L1 and L2 be 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

(

∫

Lj

℘′(z0)

℘(ξ) − ℘(z0)dξ

)

.


By Lemma 2.2,∫

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 + loge2λ| f ′(z)|2

(1 + e2λ| f (z)|2)2.

Then uλ(x) satisfies (3.1) for any λ ∈ R and uλ is doubly periodic by (3.18).Therefore, solutions have been constructed and the proof of Theorem 1.2 iscompleted.

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 tobe discussed. For Case (2), the subcases that f or 1/ f is singular at z = 0leads to contradiction as before. For the subcase that f is regular at z = 0and f (0) 6= 0 we see that f (z)/ f ′(z) is an elliptic function with 0 as its onlysimple 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 poleof f . By Lemma 3.1, if f or 1/ f is singular at z = 0 then g has no zero andwe get a contradiction. So f is regular at z = 0, f (0) 6= 0 and g has twosimple zeros at 0 and ω2.

Let σ(z) = exp∫ z

ζ(w) dw = z + · · · be the Weierstrass sigma functionon T′′. σ is odd with a simple zero at each lattice point. Then

(3.19) g(z) = Aσ(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 thelattice ω1, 2ω2, there is an unique solution of (a, b):

(3.20) a = −ω1

2; b =

ω1

2+ ω2.


Notice that the residue of g at a and b are both equal to 2πir where

r =σ( 1

2 ω1)σ( 12 ω1 + ω2)

σ(ω1 + ω2).

We claim that A = 1/r. Since

f (z) = exp∫ z

g(w) dw

is well defined, by the residue theorem, we must have A = m/r for somem ∈ 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 awell defined meromorphic function which gives rise to a solution of (3.1)for ρ = 4π. Hence the proof is completed.

4. AN UNIQUENESS THEOREM FOR ρ ∈ [4π, 8π] VIA SYMMETRIZATIONS

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 fand centered at a critical point other than the half periods. By choosingλ = − 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 inthis family of solutions. In order to prove the uniqueness up to scaling, it isequivalent to prove the uniqueness within the class of even functions.

The idea is to consider the following equation

(4.1)

u + ρeu = ρδ andu(−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 (Theorem3.2), we conclude the proof of Theorem 1.2 by the inverse function theorem.

We first prove Theorem 1.4, the Symmetrization Lemma. The proof willconsist of several Lemmas. The first step is an extension of the classicalisoperimetric inequality of Bol for domains in R2 with metric ew|dx|2 to thecase when the metric becomes singular.


Let Ω ⊂ R2 be a domain and w ∈ C2(Ω) which satisfies

(4.2)

w + ew ≥ 0 in Ω and∫

Ωew dx ≤ 8π.

This is equivalent to saying that the Gaussian curvature of ew|dx|2 is

K ≡ −1

2e−ww ≤ 1

2.

For any domain ω ⋐ Ω, we set

(4.3) m(ω) =∫

ωew dx and ℓ(∂ω) =

∫

∂ωew/2 ds.

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 = ∑Nj=1 2παjδpj

in Ω and∫

Ωew dx ≤ 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 wǫ by

w(x) = ∑j

αj log |x − pj|+ v(x),

wε(x) = ∑j

αj

2log(|x − pj|2 + ε2) + v(x).

By straightforward computations, we have

wε(x) + ewε(x)

= ∑j

(

αj

2

4ε2

(r2 + ε2)2+ (|x − pj|2 + ε2)αj/2ev − |x − pj|αj ev

)

≥ 0.

Let ℓε and mε be defined as in (4.3) with respect to the metric ewε(x)|dx|2.Then we have

2ℓ2ε (∂ω) ≥ mε(ω)(8π − mε(ω)).

By letting ε → 0 we obtain (4.6).


Next we consider the case that some of the αj’s are negative. For ourpurpose, it suffices to consider the case with only one singularity p1 withnegative α1 (and we only need the case that α1 = −1). In view of (the proofof) Lemma 4.2, the problem is reduced to the case with only one singularityp1. In other words, let w satisfy

(4.7) w + ew = −2πδp1in Ω.

Lemma 4.3. Let w satisfy (4.7) with Ω simply-connected. Suppose that

(4.8)∫

Ωewdx ≤ 4π,

then

(4.9) 2ℓ2(∂ω) ≥ m(ω)(4π − m(ω)).

Proof. We may assume that p1 = 0. If 0 6∈ 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|2 on 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) =∫

∂ωev/2 ds and m(ω) =

∫

ωev dx.

By Bol’s inequality, we have

4ℓ2(∂ω) ≥ ℓ2(∂ω) ≥ 1

2m(ω)(8π − m(ω)) = m(ω)(8π − 2m(ω)).

Thus 2ℓ2(∂ω) ≥ m(ω)(4π − m(ω)).


Lemma 4.4 (Symmetrization I). Let Ω ⊂ R2 be a simply-connected domainwith 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)∫

Ωev dx ≥ 2π.

Proof. Let ψ be the first eigenfunction of + ev:

(4.11)

ψ + evψ = 0 in Ω andψ = 0 on ∂Ω.

In R2, let U and ϕ be the radially symmetric functions

(4.12)

U(x) = log2

(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 andϕ + eU ϕ = 0 in R2.

For any t > 0, set Ωt = x ∈ Ω | ψ(x) > t and r(t) > 0 such that

(4.14)∫

Br(t)

eU(x) dx =∫

ψ>tev(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) = supt | r < r(t).

Obviously ψ∗(r) is decreasing in r and for t ∈ (0, max ψ), ψ∗(r) = t if andonly if r(t) = r. Thus by (4.14), we have a decreasing function

(4.15) f (t) :=∫

ψ∗>teU(x) dx =

∫

ψ>tev(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,

− d

dt

∫

Ωt

|∇ψ|2 dx =∫

∂Ωt

|∇ψ| ds and

− f ′(t) = − d

dt

∫

Ωt

ev dx =∫

∂Ωt

ev

|∇ψ| ds


hold almost everywhere in t. Thus

− d

dt

∫

Ωt

|∇ψ|2 dx =∫

ψ=t|∇ψ| ds

≥(

∫

ψ=tev/2 dx

)2(∫

ψ=t

ev

|∇ψ| dx

)−1

= −ℓ2(ψ = t) f ′(t)−1

≥ −1

2f (t)(4π − f (t)) f ′(t)−1.

(4.17)

The same procedure for ψ∗ leads to

− d

dt

∫

ψ∗>t|∇ψ∗|2 dx =

∫

ψ∗=t|∇ψ∗| ds

=

(

∫

ψ∗=teU/2 dx

)2(∫

ψ∗=t

eU

|∇ψ∗| ds

)−1

= −ℓ2(ψ∗ = t) f ′(t)−1

= −1

2f (t)(4π − f (t)) f ′(t)−1,

(4.18)

with all inequalities being equalities. Hence

− d

dt

∫

ψ>t|∇ψ|2 dx ≥ − d

dt

∫

ψ∗>t|∇ψ∗|2ds

holds almost everywhere in t. By integrating along t, we get∫

Ω|∇ψ|2 dx ≥

∫

BR0

|∇ψ∗|2 dx.

Since ψ and ψ∗ have the same distribution (or by looking at −∫

f ′(t)t2 dtdirectly), we have

∫

BR0

eU(x)ψ∗2 dx =∫

Ωev(x)ψ2 dx.

Therefore

0 =∫

Ω|∇ψ|2 dx −

∫

Ωevψ2 dx ≥

∫

BR0

|∇ψ∗|2 dx −∫

BR0

eUψ∗2 dx.

This implies that the linearized equation + eU(x) has non-positive firsteigenvalue. By (4.12), this happens if and only if R0 ≥ 1. Thus

∫

Ωev(x) dx =

∫

BR0

eU(x) dx ≥ 2π.


Remark 4.5. (See [2].) By applying symmetrization to v + ev = 0 in Ω withλ1( + ev) = 0, the corresponding radially symmetric functions are

U(x) = −2 log

(

1 +1

8|x|2

)

and ϕ(x) =8 − |x|28 + |x|2 .

The same computations leads to∫

Ωev dx ≥ 4π.

A closer look at the proof of Lemma 4.4 shows that it works for moregeneral situations as long as the isoperimetric inequality holds:

Lemma 4.6 (Symmetrization II). Let Ω ⊂ R2 be a simply-connected domainand let v be a solution of

v + ev = ∑N

j=12παjδpj

in Ω. Suppose that the first eigenvalue of + ev is zero on Ω with ϕ the firsteigenfunction. 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∫

Ωev dx ≥ 2π.

Notice that we do not need any further constraint on the sign of αj. Forthe last statement of Theorem 1.4, the limiting procedure of Lemma 4.2implies that the isoperimetric inequality (Lemma 4.3) and symmetrizationI (Lemma 4.4) both hold regardless on the presence of singularities withnon-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 thatwe must have

∫

Teu = 1. Suppose that ϕ(x) is a non-trivial solution of the

linearized equation at u:

(4.19)

ϕ + ρeu ϕ = 0 andϕ(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 anisometry:

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 = ∑3

j=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 → S2 via y = 1/℘(z) ∼ z2. The isometry conditionreads 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 p0 is a singularity with non-negative α0:

w + ρew = α0δ0 + ∑3

j=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 handside by 1 for simplicity. The total measure on T and R2 are then given by

∫

Teu dz = ρ ≤ 8π and

∫

R2ev dx =

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 simplyconnected component Ω which contains no pj’s, then the symmetrization(Remark 4.5) leads to

∫

Ωev dx ≥ 4π,

which is a contradiction because R2\Ω 6= ∅. If every simply connectedcomponent Ωi, i = 1, . . . , m, contains only one pj, then Lemma 4.4 impliesthat

∫

Ωi

ev dx ≥ 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 followingtwo 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, p2 and 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 p1 and p2, then Lemma 4.3 implies that

2ℓ2(∂ω) ≥ (4π − m(ω))m(ω).

If ω contains both p1 and p2, then R2\ω is simply connected which con-tains p3 only. Thus by Lemma 4.3

2ℓ2(∂ω) ≥ (4π − m(R2\ω))m(R

2\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 and∫

Ω+evdx ≥ 2π, we get

ρ

2=∫

R2ev ≥

∫

Ω+

ev +∫

ωev ≥ 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)mand 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

∫

Ω−ev dx = 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 onlyone component and it contains p1 and p2 for all t ∈ (min ψ, 0). But then ψattains its minimum along a connected set ψ−1(min ψ) containing p1 andp2, which is impossible.


In case (2), ρ < 8π again leads to a contradiction via the same argument.For ρ = 8π, we have

∫

Ω+

ev dx =∫

Ω−ev dx = 2π

and all inequalities in (4.17) are equalities. So under the notations there(

∫

∂Ωt

ev/2 dx

)2

=∫

∂Ωt

|∇ψ| ds∫

∂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 ont. 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 clearlyimpossible. 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 mosta solution at ρ = 8π, and it implies that the mean field equation (1.2) hasat most one solution up to scaling. Thus, Theorem 1.2 is proved and thenTheorem 1.3 follows immediately.

5. COMPARING CRITICAL VALUES OF GREEN FUNCTIONS

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 detectingminimum points of G(z). The following theorem is useful in this regard.

Theorem 5.1. Let z0 and z1 be two half-periods. Then G(z0) ≥ G(z1) if and onlyif |℘(z0)| ≥ |℘(z1)|.Proof. By integrating (2.7), the Green function G(z) can be represented by

(5.1) G(z) = − 1

2πRe∫

(ζ(z) − η1z) dz +1

2by2.

Thus

(5.2) G(ω2

2

)

− G(ω3

2

)

=1

2πRe∫

ω32

ω22

(ζ(z) − η1z) dz.


Set F(z) = ζ(z) − η1z. We have F(z + ω1) = F(z) and

ζ(

t +ω2

2

)

− η1

(

t +ω2

2

)

= −ζ(−ω2

2− t)

− η1

(

t +ω2

2

)

= −ζ(ω2

2− t)

+ η2 − η1

(

t +ω2

2

)

= −[

ζ(ω2

2− t)

− η1

(ω2

2− t)]

+ η2 − η1ω2

= −[

ζ(ω2

2− t)

− η1

(ω2

2− t)]

− 2πi,

hence Re F( 12 ω2 + t) is antisymmetric in t ∈ C.

To calculate the integral in (5.2), we use the addition theorem to get

℘′(z)

℘(z) − e1= ζ

(

z − ω1

2

)

+ ζ(

z +ω1

2

)

− 2ζ(z)

= F(

z − ω1

2

)

+ F(

z +ω2

2

)

− 2F(z).

Integrating it along the segment from 12 ω2 to 1

2 ω3, we get

2 loge3 − e1

e2 − e1=∫

L1

F(z) dz +∫

L2

F(z) dz − 2∫

L3

F(z) dz,

where L1 is the line from 12(ω2 − ω1) to 1

2 ω2, L2 is the line from 12 ω3 to

12 ω2 + ω1 and L3 is the line from 1

2 ω2 to 12 ω3. Since F(z) = F(z + ω1) and

Re F(z) is antisymmetric with respect to 12 ω2, we have

2 loge3 − e1

e2 − e1= 2

∫

LF(z) dz − 4

∫

L1

F(z) dz = 2πi − 4∫

L3

F(z) dz,

where L is the line from 12 (ω2 − ω1) to 1

2 ω3. Thus we have

log

∣

∣

∣

∣

e3 − e1

e2 − e1

∣

∣

∣

∣

= −2 Re∫

ω32

ω22

F(z) dz = −4π(

G(ω2

2

)

− G(ω3

2

))

.

That is,

(5.3) G(ω2

2

)

− G(ω3

2

)

=1

4πlog

∣

∣

∣

∣

e2 − e1

e3 − e1

∣

∣

∣

∣

.

Similarly, by integrating (2.7) in the ω2 direction, we get

(5.4) G(ω1

2

)

− G(ω3

2

)

=1

2πRe∫

ω32

ω12

(ζ(z) − η2z) dz.

The same proof then gives rise to

(5.5) G(ω1

2

)

− G(ω3

2

)

=1

4πlog

∣

∣

∣

∣

e1 − e2

e3 − e2

∣

∣

∣

∣

.

By combining the above two formulae we get also that

(5.6) G(ω1

2

)

− G(ω2

2

)

=1

4πlog

∣

∣

∣

∣

e1 − e3

e2 − e3

∣

∣

∣

∣

.


In order to compare, say, G( 12 ω1) and G( 1

2 ω3), 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 theoremfollows.

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) G(ω3

2

)

− G(ω2

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 = 12 ω2 then z = 1

2 ω2 = 12(1 − ω2) = 1

2 ω3 (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 = 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( 12 ω3) < G( 1

2 ω2) for τ = ib, we conclude that for τ ∈ Ω′ ∪ Ω,

|λ(τ) − 1| < 1 if and only if |Re τ| < 12 . Therefore for τ ∈ Ω′ ∪ Ω,

|Re τ| <1

2if and only if G

(ω3

2

)

< G(ω2

2

)

.

For |τ| = 1, using suitable Mobius 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 + 1z)

and so (compare (2.1))

(5.13) G(z) = G(τ + 1

τ + 1z)

.

Clearly, for z = 12 τ, (5.13) implies that G( 1

2 ω2) = G( 12 ω1). So

(5.14) |τ| = 1 if and only if

∣

∣

∣

∣

e2 − e3

e1 − e3

∣

∣

∣

∣

= 1,

(5.15) |τ| < 1 if and only if G(ω1

2

)

< G(ω2

2

)

.

Similarly,

(5.16) |τ − 1| < 1 if and only if G(ω1

2

)

< G(ω3

2

)

.

6. DEGENERACY ANALYSIS OF CRITICAL POINTS

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, η1 is 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)


We first consider the case z0 = 12 ω1. The degeneracy condition of G at

12 ω1 reads

e1 + η1 = 0 or e1 + η1 −2π

b= 0.

We will use the following two inequalities (Theorem 1.7) whose proofswill 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 1

2 ω1 is a degenerate

critical point of G(z; τ) if and only if b = b0 or b = b1. Moreover, 12 ω1 is a

local minimum point of G(z; τ) if b ∈ (b0, b1) and is a saddle point of G(z; τ) ifb ∈ (0, b0) or b ∈ (b1, ∞).

Proof. Let b0 and b1 be the zero of e1 + η1 = 0 and e1 + η1 − 2π/b = 0respectively. Then Lemma 6.1 follows from the explicit expression of theHessian of G by (6.4).

Numerically we know that b1 ≈ 0.7 <√

3/2. Now we analyze the

behavior of G near 12 ω1 for b > b1.

Lemma 6.2. Suppose that b > 12 , then 1

2 ω1 is the only critical point of G alongthe x-axis.

Proof. ℘(t; τ) is real if t ∈ R. Since ℘′(t; τ) 6= 0 for t 6= 12 ω1, ℘′(t; τ) < 0 for

0 < t < 12 ω1. Since b > 1

2 > b0, by (6.3) and Lemma 6.1,

Gxx(t) = ℘(t) + η1 > e1 + η1 > 0,

which implies that Gx(t) > Gx(12 ω1) = 0 if 0 < t < 1

2 ω1. Hence G has no

critical points on (0, 12 ω1). Since G(z) = G(−z), G can not have any critical

point on (− 12 ω1, 0).

By Lemma 6.1 and the conservation of local Morse indices, we know

that G(z; τ) has two more critical points near 12 ω1 when b is close to b1 and

b > b1. We denote these two extra points by z0(τ) and −z0(τ). In this

case, 12 ω1 becomes 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

G(ω1

2

)

< G(ω2

2

)

= G(ω3

2

)

if b1 < b <√

3/2.

Thus in this region ±z0(τ) must exist and they turn out to be the minimumpoint 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 > b1 and 12 ω2, 1

2 ω3 are all saddlepoints.

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 12 ω1 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 thesecond 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− ib

2

)

= G(

z − 1

2− ib

2

)

= G(1

2+

ib

2− z)

,

where the second and the third identity comes from G(z + 1) = G(z) andG(−z) = G(z) respectively. Thus (6.6) and then the lemma are proved.

Let G 12

be the restriction of G on the line Re z = 12 . Lemma 6.3 says that

any critical point of G 12

is automatically a critical point of G.

Lemma 6.4. G 12(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 −2π

b

)

.

Since ∂℘/∂y = −i℘′(z; τ) 6= 0 for z 6= 12 ω1 and Re z = 1

2 , Gyy(z) can haveone zero only. Let z0(τ) denote the critical point above when τ is close

to 12 + ib1. Thus Gy(z0(τ)) = Gy(

12 ω1) = 0, and then Gyy(z0) = 0 for

some z0 ∈ ( 12 ω1, z0(τ)). Since Gyy(

12 ω1) = −(e1 + η1 − 2π/b) < 0, we

have Gyy(z0(τ)) > 0. Hence z0(τ) is a non-degenerate minimum point ofG 1

2.

Remark 6.5. For b > b1, G 12(y) is increasing in y for y > 0 and

limy→b G 12(y) = +∞.

Theorem 6.6. For b > b1, ±z0(τ) are non-degenerate (local) minimum points ofG. 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),

2πGxx(z0(τ); τ) = ℘(z0(τ); τ) + η1.

We claim that if ℘′′(z0(τ); τ) = 0 then τ = (1 +√

3i)/2.To prove the claim, we use (2.13). Since ℘′′(z0(τ); τ) = 0, we have

ζ(2z0(τ)) = 2ζ(z0(τ)) = η1(2t0) + η2(2s0).

Thus 2z0(τ) is also a critical point. Note that Re 2z0(τ) = 1. Since G(z −ω2) = G(z), 2z0 − 1 + ω2 is also a critical point. Because Re (2z0 − 1 +ω2) = 1

2 , we have either 2z0 − 1 + ω2 = −z0 or 2z0 − 1 + ω2 = z0. Thelater leads to z0 = 1 − ω2 = ω3, which is not true. Thus we have 2z0 −1 + ω2 = −z0 and then 3z0 = ω2 − 1 = ω2, i.e., 2z0 = −z0. Therefore℘(2z0) = ℘(−z0) = ℘(z0). By the addition theorem for ℘, we have

℘(z0) = ℘(2z0) = −2℘(z0) +1

4

(

℘′′(z0)

℘′(z0)

)2

= −2℘(z0).

Therefore ℘(z0) = 0 and it leads to e1e2 + e1e3 + e2e3 = 0, which implies

that τ = (1 +√

3i)/2. Hence the claim is proved.Note that e1e2 + e2e3 + e1e3 = |e2|2 − e2

1 and

(6.8) ℘′′(z) = 2(3℘2(z) + |e2|2 − |e1|2).

In terms of λ(τ) and e1, e2 can be expressed by

(6.9) e2 =λ − 1

λ − 2e1.

Thus

d

dλ

(

e2

e1

)

=−1

(λ − 2)26= 0 for b >

1

2.

From here, we haved

db

∣

∣

∣

∣

e2

e1

∣

∣

∣

∣

2

6= 0 for b > 12 , which implies that

d

db

∣

∣

∣

∣

e2

e1

∣

∣

∣

∣

< 0

for b > 12 . Therefore

(6.10)d

db℘′′(z0(τ); τ) = |e1|2

d

db

|e2|2|e1|2

< 0 at τ =1

2+

√3i

2.

Since ℘′′(z0(τ); τ) = 0 at τ = (1 +√

3i)/2, we have ℘′′(z0(τ), τ) < 0 for

b >√

3/2 and sufficiently close to√

3/2. By the claim above, ℘′′(z0(τ); τ) <

0 for b >√

3/2. Thus

(6.11) |℘(z0(τ); τ)| ≤√

1

3(|e1|2 − |e2|2),


and

η1 + ℘(z0(τ); τ) ≥ η1 −√

1

3(|e1|2 − |e2|2)

> η1 −√

1

4|e1|2

= η1 −1

2e1

(6.12)

where |e2|2 = |e1|2/4 + |Im e2|2 > |e1|2/4 is used.Later we will show that η1 > 1

2 e1 always holds (this is part of Theorem1.7 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 >√

3/2 follows.

For b <√

3/2, we let z0(τ) = t0(τ) + (1 − 2t0(τ))τ. We claim that

t0(τ) > 1/3 if b <√

3/2. For if z0(τ) = 13 ω3 then 2z0(τ) = −z0(τ) is

also a critical point. By the addition formula of ζ and ℘, we conclude that

℘′′( 13 ω3) = 0 and ℘( 1

3 ω3) = 0, a contradiction. Hence

(6.13) ℘(z0(τ); τ) > ℘(ω3

3; τ)

.

Note that by the addition theorem for ℘, it is easy to see ℘( 13 ω3; τ) < 0 for

all τ = 12 + ib.

Let f (τ) := ℘( 13 ω3; τ) + 1

2 e1(τ). We have f ( 12 ) = ℘( 1

3 ω3; 12) < 0 and

f (√

3/2) = 12 e1(

√3/2) > 0. Therefore there exists a τ0 such that f (τ0) = 0.

By the addition theorem for ℘, we have

(6.14) 12℘(ω3

3; τ0

)

=

(

℘′′( 13 ω3; τ0)

℘′( 12 ω3; τ0)

)2

.

Since ℘′2 = 4(℘ − e1)(℘ − e2)(℘ − e3), we have

(6.15) ℘′′ = 2 ∑1≤i<j≤3(℘ − ei)(℘ − ej)

and

(6.16)℘′′2

℘′2 =℘′′

2

(

1

℘ − e1+

1

℘ − e2+

1

℘ − e3

)

,

where℘′′

2

1

℘ − ejcan be calculated by (6.15) as

1

2℘′′(℘ − ej)

−1 = (2℘ + ej) + (℘ − ej)−1(℘ − ej+1)(℘ − ej+2),

where the notations e4 = e1 and e5 = e2 are used. Therefore (6.16) is re-duced to

12℘(ω3

3

)

= 6℘ +(℘e2)(℘ − e3)

℘ − e1+

( −e3

℘ − e2+

℘ − e2

℘ − e3

)

(℘ − e1).


Plug in ℘( 13 ω3) = − 1

2 e1 and recall that e3 = e2 = − 12 e1 + 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

℘( 13 ω3) > − 1

2 e1 for b > b1. Hence together with (6.13)

℘(z0(τ), τ) + η1 > η1 −1

2e1 > 0

for b1 < b <√

3/2. With this, the proof of the non-degeneracy of z0(τ) iscompleted.

Next we discuss the non-degeneracy of G at 12 ω2 and 1

2 ω3. The localminimum property of z0(τ) is in fact global by

Theorem 6.7. For τ = 12 + ib, both 1

2 ω2 and 12 ω3 are non-degenerate saddle

points of G.

Proof. By (6.3), we have

2πGxx

(ω2

2

)

= Re e2 + η1 = η1 −1

2e1,

2πGxy

(ω2

2

)

= −Im e2,

2πGyy

(ω2

2

)

=2π

b+

1

2e1 − η1.

(6.18)

Hence the non-degeneracy of 12 ω2 for all b is equivalent to

(6.19) |Im e2|2 >(

η1 −1

2e1

)(2π

b+

1

2e1 − η1

)

for all b ∈ R. To prove (6.19), we need the following:

Lemma 6.8. ℘ maps [ 12 ω2, 1

2 ω3]∪ [ 12 (1−ω2), 1

2 ω2] 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 [ 12 ω2, 1

2 ω3] means segment 12 ω2 + t | 0 ≤ t ≤ 1

2, and [ 12 (1 −

ω2), 12 ω2] is 1

4 + it | |t| ≤ b2. Thus, the image of [ 1

2 ω2, 12 ω3] is the arc con-

necting e2 and e3 through ℘( 14 (ω2 + ω3)) and the image of [ 1

2 (1 − ω2), 12 ω2]

is the arc connecting e3 and e2 through ℘( 14). See Figure 4. Note our figure

is for the case e1 > 0, i.e., b > 12 . In this case, the angle ∠e3e1e2 is 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 −1

2e1 > 0


e1

℘( 14 (ω2 + ω3)) ℘( 1

4)

12 e1

e3

e2

FIGURE 4. The image of the mapping z 7→ ℘(z).

and

(6.21)2π

b+

1

2e1 − η1 > 0.

To prove (6.20), we have

(6.22) − η1 = 2∫ 1

2

0Re ℘

(ω2

2+ t)

dt,

where ℘( 12 ω2 + z) = ℘( 1

2 ω2 − t) is used. By Lemma 6.8,

Re ℘(ω2

2+ t)

≤ −1

2e1.

Hence

−η1 ≤ −1

2e1.

To prove (6.21), we have∫

ω22

1−ω22

℘(z) dz = ζ(1 − ω2

2

)

− ζ(ω2

2

)

=1

2(η1 − 2η2) = i(2π − bη1).

Therefore,

(2π − bη1) =∫ b

2

− b2

Re ℘(1

4+ it

)

dt ≥ −1

2e1b

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(

η1 −1

2e1

)(2π

b+

1

2e1 − η1

)

<(

|e2 − e1| −3

2e1

)(

|e2 − e1| +3

2e1

)

= |e2 − e1|2 −9

4e2

1

= |Im e2|2,

which is exactly (6.19). Therefore, the non-degeneracy of G at 12 ω2 is proved.

By (6.19), 12 ω2 is always a saddle point.

It remains to prove Lemma 6.8. First we have

(6.25) 4π(

G(z) − G(

z;ω1

2

))

= log∣

∣

∣℘(z) − ℘

(ω1

2

)∣

∣

∣+ C.

Since G( 12 ω2; 1

2 ω1) = G( 12 ω2 − 1

2 ω1) = G( 12ω1),

(6.26) 4π(

G(z) − G(

z;ω1

2

))

= log

∣

∣

∣

∣

℘(z) − e1

e2 − e1

∣

∣

∣

∣

.

Let z = 12 ω2 + t, t ∈ R. We have

G(

z;ω1

2

)

= G(ω2

2+ t − ω1

2

)

= G(ω1

2− ω2

2− t)

= G(ω2

2+ t)

= G(z).

By (6.26),

(6.27)

∣

∣

∣

∣

℘(z) − e1

e2 − e1

∣

∣

∣

∣

= 1 for z =ω2

2+ t.

Since ℘(z) is decreasing in y for z = 12 + iy,

℘(ω2 + ω3

4

)

= ℘(1

2+

ib

2

)

< ℘(1/2) = e1.

Thus ℘( 14 (ω2 + ω3)) = e1 − |e2 − e1| and the image of [ 1

2 ω2, 12 ω3] is exactly

the arc on the circle w | |w − e1| = |e2 − e1| connecting e2 and e3 through

℘( 14(ω2 + ω3)). It is one to one since ℘′(z) 6= 0 for z = 1

2 ω2 + t, t 6= 0.

Next let z = 14 + it. Then we have

G(

z;ω1

2

)

= G(

z − ω1

2

)

= G(

− 1

4+ it

)

= G(

− 1

4− it

)

= G(1

4+ it

)

= G(z).

Thus by (6.26) again,

|℘(z) − e1| = |e2 − e1| for z =1

4+ it.


Since ℘(t) is decreasing in t for t ∈ (0, 12), ℘( 1

4) > ℘( 12 ). So ℘( 1

4 ) =

e1 + |e2 − e1| and the image of [ 12(1 − ω2), 1

2 ω2] is the arc of w | |w − e1| =|e2 − e1| connecting e3 and e2 through e1 + |e2 − e1|. Therefore Lemma 6.8is proved and the proof of Theorem 6.7 is completed.

7. GREEN FUNCTIONS VIA THETA FUNCTIONS

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 setupfor the proof turns out has to be the theta functions. Formally this is easyto explain since the moduli variable τ is very explicit in theta functionsand differentiations in τ is much easier to be done than in the Weierstrasstheory. In order to avoid complicate cross references, we choose to writeeverything here independent of the previous sections and in particular sev-eral statements about Green functions and critical points are re-derived forcompleteness.

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 onC with 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 meromorphicfunctions (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+ 12 )2

e(2n+1)πiz

= 2∞

∑n=0

(−1)nq(n+ 12 )2

sin(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 followingheat 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 isgiven by

(7.4) G(z, w) = − 1

2πlog

∣

∣

∣

∣

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 havezR(z, w) = 1/b which integrates over T gives 1. Near z = w, R(z, w) hasthe correct behavior. So it remains to show that R(z, w) is indeed a functionon T. From the quasi-periodicity, R(z + 1, w) = R(z, w) is obvious. Also

R(z + τ, w) − R(z, w) = − 1

2πlog 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) = − 1

2πlog

∣

∣

∣

∣

ϑ1(z)

ϑ′1(0)

∣

∣

∣

∣

+1

2by2.

If we represent the torus T as centered at 0, then the symmetry z 7→ −zshows that G(z) = G(−z). By differentiation, we get ∇G(z) = −∇G(−z).If −z0 = z0 in T, that is 2z0 = 0 (mod Λ), then we get ∇G(z0) = 0. Hence

we obtain the half periods 12 , 1

2 τ and 12(1+ τ) as three obvious critical points

of G(z) for any T. By computing ∂G/∂z = 12(Gx − iGy) we find

Corollary 7.2. The equation of critical points z = x + iy of G(z) is given by

(7.5)∂G

∂z≡ −1

4π

(

(log ϑ1)z + 2πiy

b

)

= 0.

Remark 7.3. With ζ(z)− η1z = (log ϑ1(z))z understood (cf. (9.10)), this leadsto alternative simple proofs to Lemma 2.3 and Corollary 2.4.

We compute easily

Gx = − 1

2πRe (log ϑ1)z,

Gy = − 1

2πRe (log ϑ1)zi +

y

b=

1

2πIm (log ϑ1)z +

y

b,

Gxx = − 1

2πRe (log ϑ1)zz,

Gxy = +1

2πIm (log ϑ1)zz,

Gyy = − 1

2πRe (log ϑ1)zzi2 +

1

b=

1

2πRe (log ϑ1)zz +

1

b,

(7.6)

and hence that the Hessian

H =

∣

∣

∣

∣

Gxx Gxy

Gyx Gyy

∣

∣

∣

∣

=−1

4π2

[

(Re (log ϑ1)zz)2 +

2π

b(Re (log ϑ1)zz) + (Im (log ϑ1)zz)

2

]

=−1

4π2

[

∣

∣

∣(log ϑ1)zz +

π

b

∣

∣

∣

2−(π

b

)2]

.

(7.7)


To analyze the critical point of G(z) in general, we may try to use themethods 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 pointsgiven by the half periods (cf. [6], Lemma 2.1).

The idea is, new critical points should be born only at certain half periodpoints when it degenerates (as a critical point) under the deformation inτ. The heat equation provides a bridge between the degeneracy conditionand 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 THE LINE Re τ = 12

When τ = 12 + ib ∈ L, we have

ϑ1(z) = 2∞

∑n=0

(−1)neπi/8eπi n(n+1)2 e−πb(n+ 1

2 )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 theholomorphicity 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(n+1)

2 e−14 πb(2n+1)2 ∈ R

+.

To see this, notice that the right hand side is non-zero, real and positive forlarge 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)b is decreasing from positive infinity to −π/4. Hence that Gxx = 0 andGyy = 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 = −2π

4

∞

∑n=0

(−1)n(n+1)

2 (2n + 1)2h(2n+1)2

= −2hπ

4

∞

∑n=0

(−1)n(n+1)

2 (2n + 1)2rn(n+1)

2

|ϑ1|bb = 2π2

42

∞

∑n=0

(−1)n(n+1)

2 (2n + 1)4h(2n+1)2

= 2hπ2

42

∞

∑n=0

(−1)n(n+1)

2 (2n + 1)4rn(n+1)

2 .

(8.4)

Denote the arithmetic sum n(n + 1)/2 by An, then

(log |ϑ1|)bb =|ϑ1|bb|ϑ1| − |ϑ1|2b

|ϑ1|2

= h2 π2

4|ϑ1|−2

∞

∑n,m=0

(−1)An+Am((2n + 1)4 − (2n + 1)2(2m + 1)2)rAn+Am

= h2 π2

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|)bb which implies that the

case with 0 < b ≤ 12 is equivalent to the case b ≥ 1

2 .

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. Thenumber of terms with degree 2k is certainly no more than 2k, so a trivialupper bound for the positive part is given by

(8.6) A =∞

∑n=2

(2k)3r2k = 8r4 8 − 5r2 + 4r4 − r6

(1 − r2)4,

where the last equality is an easy exercise in power series calculations inCalculus. 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 − r(

1 − 64

125− 30

625

)

= −5r3 − 11

25r < 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 τ = (τ − 1)/(2τ − 1) = a + ib, it holds that

(8.8) (log ϑ1)b(1/2; τ) = −i(1 − 2τ) + (1 − 2τ)2(log ϑ1)b(1/2; τ).

When τ = 12 + ib, we have τ = 1

2 + i4b . As before we may then replace ϑ1

by |ϑ1|. Under τ → τ, [ 12 , ∞) is mapped onto (0, 1

2 ] with directions reversed.

Let f (b) = (log |ϑ1|)b(12 , 1

2 + ib). Then we get

(8.9) f (1/4b) = −2b − 4b2 f (b).

Plug in b = 12 we get that f ( 1

2 ) = − 12 . So − 1

2 = f ( 12) > f (b) for b > 1

2 .Then

f( 1

4b

)

= −2b + 2b2 + 4b2[

− 1

2− f (b)

]

= 2[

b − 1

2

]2− 1

2+ 4b2

[

− 1

2− f (b)

]

(8.10)

is strictly increasing in b > 12 . That is, f (b) is strictly decreasing in b ∈ (0, 1

2 ].The remaining statements are all clear.

Now we prove the functional equation. For this we need to use Jacobi’simaginary 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′)12 eπiτ′z2

ϑ1(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 τ = ST−2ST−1τ = (τ − 1)/(2τ − 1). Then

(8.12) (log ϑ1)τ(1/2; τ) = −(1 − 2τ) + (1 − 2τ)2(log ϑ1)τ(1/2; τ).

Proof. Let τ = 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

d

dτlog ϑ1(1/2; τ)

= τ21

d

dτ1

[

log(−iτ1)12 + πiτ1(1/2)2 + log ϑ1(τ1/2; τ1)

]

=τ1

2+

πiτ21

4+ τ2

1

d

dτ1log ϑ1(τ1/2; τ1)

=τ2 − 2

2+

πi(τ2 − 2)2

4+ τ2

2

d

dτ2log ϑ1(τ2/2; τ2)

=1

2

[−1

τ3− 2]

+πi

4

[−1

τ3− 2]2

+[−1

τ3− 2]2×

(

τ3

2+ πiτ2

3

d

dτ3

[

τ3(τ2/2)2]

+ τ23

d

dτ3log ϑ1(τ2τ3/2; τ3)

)

.

(8.13)

We plug in τ2τ3 = −1 and τ3 = τ − 1. It is clear that the second and thefourth terms are cancelled out, the first and the third terms are combinedinto

(8.14)1

2

1 − 2τ

τ − 1

[

1 +1 − 2τ

τ − 1(τ − 1)

]

= −(1 − 2τ).

This proves the result.

By the previous explicit computations,

Gxx

(1

2;

1

2+ ib

)

= −2(log ϑ1(1/2))b,

Gxy

(1

2;

1

2+ ib

)

= 0,

Gyy

(1

2;

1

2+ ib

)

= 2(log ϑ1(1/2))b +1

b.

(8.15)

A numerical computation shows that Gxx(b0) = 0 for b0 = 0.35 · · · andGyy(b1) = 0 for b1 = 0.71 · · · . Hence the sign of (Gxx, (b), Gyy(b)) is (−, +),(+, +) and (+,−) for b < b0, b0 < b < b1 and b1 < b respectively. That

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 near12 which come from the degeneracy of 1

2 at b = b1 and the conservation oflocal Morse index.

9. SECOND INEQUALITY ALONG THE LINE Re 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 laterin this section. Here we shall prove the inequality first. We recall the threeother theta functions.


ϑ2(z; τ) := ϑ1

(

z +1

2; τ)

ϑ4(z; τ) :=∞

∑n=−∞

(−1)nqn2e2πinz = 1 + 2

∞

∑n=1

(−1)nqn2cos 2nπz

ϑ3(z; τ) := ϑ4

(

z +1

2; τ)

= 1 + 2∞

∑n=1

qn2cos 2nπz =

∞

∑n=−∞

qn2e2πinz

(9.1)

It is readily seen that ϑ1(z) = −ieπiz+πiτ/4ϑ4(z + 12 τ). So the four theta

functions are translates of others by half periods.

We had seen previously that (log |ϑ1(12 )|)bb < 0 over the line Re τ = 1

2 .This is equivalent to that (log |ϑ2(0)|)bb < 0. We now discuss the case forϑ3(0) and ϑ4(0) where the situation is reversed(!) and it turns out the proofis easier and purely algebraic.

Theorem 9.1. Over the line Re τ = 12 , we have ϑ4(0) = ϑ3(0). Moreover,

(log |ϑ3(0)|)b < 0 and (log |ϑ3(0)|)bb > 0.

Proof. Since q = eπiτ = eπi/2e−πb = ie−πb, qn2= in2

e−πbn2, we see that

ϑ3(0) = ∑n∈2Z

rn2+ i ∑

m∈2Z+1

rm2,

ϑ4(0) = ∑n∈2Z

rn2 − i ∑m∈2Z+1

rm2,

(9.2)

where r = e−πb < 1. Then we compute directly that

|ϑ3(0)|2 =(

∑n∈2Zrn2)2

+(

∑m∈2Z+1rm2)2

= ∑n,n′∈2Z

rn2+n′2+ ∑

m,m′∈2Z+1

rm2+m′2= ∑

k∈2Z≥0

p2(k)rk,(9.3)

where p2(k) is the number of ways to represent k as the (ordered) sum oftwo squares of integers. Then

(9.4) (|ϑ3(0)|2)b = −π ∑k∈2Z≥0

p2(k)krk

and in particular (log |ϑ3(0)|2)b < 0.We also have

(9.5) (|ϑ3(0)|2)bb = π2 ∑k∈2Z≥0

p2(k)k2rk.


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 theWeierstrass ℘ function with periods ω1 = 1, ω2 = τ. Let ω3 = ω1 + ω2.

Then ℘′(z) is odd with three zeros at the half periods z = 12 ωi, i = 1, 2, 3.

Let ei = ℘( 12 ωi), then

(9.7) ℘′(z)2 = 4(℘(z) − e1)(℘(z) − e2)(℘(z) − e3).

Let ζ(z) = −∫ z

℘(w) dw which is odd with quasi-periods ηi = ζ(z +

ωi) − ζ(z), i = 1, 2, hence that ηi = 2ζ( 12 ωi). The invariants ei and ηi are all

holomorphic functions of τ.

Let σ(z) = exp∫ z

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+ 12 ωi)σ(z).

This is similar to the theta function transformation law. Indeed it is easilyseen that

(9.9) σ(z) = eη1z2/2 ϑ1(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 = − 2

3!

ϑ′′′1 (0)

ϑ′1(0)

= −4πi

3(log ϑ′

1(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]) assertsthat

ϑ′1(0) = πϑ2(0)ϑ3(0)ϑ4(0).

So

1

2e1 − η1 = 2πi

[

logϑ′

1(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 , 1

2 e1 − η1 < 0,

( 12 e1 − η1)b > 0 and 1

2 e1 − η1 increases to zero in b. Together with the firstestimate (9.12), we find also that e1 increases in b.

REFERENCES

[1] L. Alhfors; Complex Analysis, 2nd edition, McGraw-Hill Book Co., New York, 1966.[2] S.-Y. A. Chang, C.C. Chen and C.S. Lin; Extremal functions for a mean field equation in

two dimensions, in Lectures on Partial Differential Equations: Proceedings in honor ofLouis Nirenberg’s 75th Birthday, International Press 2003, 61-94.

[3] C.C. Chen and C.S. Lin; Sharp estimates for solutions of multi-bubbles in compact Riemannsurfaces, Comm. Pure Appl. Math., Vol. LV (2002), 0728-0771.

[4] ——; Topological degree for a mean field equation on a Riemann surfaces, Comm. Pure Appl.Math., Vol. LVI (2003), 1667-1727.

[5] ——; Topological degree for a mean field equation with singular data, in preparation.[6] C.C. Chen, C.S. Lin and G. Wang; Concentration phenomenon of two-vortex solutions in a

Chern-Simons model, Ann. Scuola Norm. Sup. Pisa CI. Sci. (5) Vol. III (2004), 367-379.[7] J. Jost, C.-S. Lin and G. Wang; Analytic aspects of the Toda system, Comm. Pure Appl.

Math. 59 (2006), no. 4, 526-558.[8] Y.Y. Li; Harnack type inequalities: the method of moving planes, Comm. Math. Phy. 200

(1999), 421-444.[9] C.-S. Lin and C.-L. Wang; On the minimality of extra critical points on flat tori, in prepara-

tion.[10] M. Nolasco and G. Tarantello; On a sharp Sobolev-type inequality on two dimensional com-

pact manifolds, Arch. Ration. Mech. Anal. 154 (1998), no.2, 161-195.[11] ——; Double vortex condensates in the Chern-Simons-Higgs theory, Calc. Var. and PDE 9

(1999), no. 1, 31-94.[12] ——; Vortex condensates for the SU(3) Chern-Simons theory, Comm. Math. Phys. 213

(2000), no.3, 599-639.[13] E.T. Whittaker and G.N. Watson; A Course of Modern Analysis, 4th edition, Cambridge

University Press, 1927.

CHANG-SHOU LIN: DEPARTMENT OF MATHEMATICS, NATIONAL CENTRAL UNIVER-SITY, CHUNG-LI, TAIWAN.

E-mail address: [email protected]

CHIN-LUNG WANG: DEPARTMENT OF MATHEMATICS, NATIONAL CENTRAL UNIVER-SITY, CHUNG-LI, TAIWAN., NATIONAL CENTER FOR THEORETIC SCIENCES, HSINCHU,TAIWAN.

E-mail address: [email protected] address: [email protected]

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