ELLIPTIC FUNCTIONS, GREEN FUNCTIONS AND THE MEAN FIELD EQUATIONS ON TORI

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1 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 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.. INTRODUCTION AND STATEMENT OF RESULTS The study of geometric or analytic problems on two dimensional tori is the same as the study of problems on R with doubly periodic data. Such situations occur naturally in sciences and mathematics since early days. The mathematical foundation of elliptic functions was subsequently developed in the 9th century. It turns out that these special functions are rather deep objects by themselves. Tori of different shape may result in very different behavior of the elliptic functions and their associated objects. Arithmetic 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 partial differential 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 Gz, w on a flat torus T = C/Zω + Zω is the unique function on T T which satisfies z Gz, w = δ w z T Date: May 30, 004. Revised September 9, 005, June st, 006.

2 CHANG-SHOU LIN AND CHIN-LUNG WANG and T Gz, w da = 0, where δ w is the Dirac measure with singularity at z = w. Because of the translation invariance of z, we have Gz, w = Gz w, 0 and it is enough to consider the Green function Gz := Gz, 0. Not surprisingly, G can be explicitly solved in terms of elliptic functions. For example, using theta functions we have cf. Lemma., Lemma 7. Gz = π log ϑ z ϑ 0 + b y where z = x + iy and τ := ω /ω = 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 Gz = 0 is given by G z 4π log ϑ z + πi y b = 0. In terms of Weierstrass elliptic functions z, ζz := z and using the relation log ϑ z = ζz η z with η i = ζz + ω i ζz the quasiperiods, the equation takes the simpler form: z = tω + sω is a critical point of G if and only if the following linear relation Lemma.3 holds:. ζtω + sω = tη + sη. Since G is even, it is elementary to see that half periods ω, ω and ω 3 = ω + ω / 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.. One of our chief purposes in this paper is to understand the geometry of the critical point set over the moduli space of flat tori M = H/SL, 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 +. u + ρe u = ρδ 0. This equation has its origin in the prescribed curvature problem in geometry like the Nirenberg problem, cone metrics etc.. It also comes from statistical 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], [0], [] and [] for the recent development of this subject. When ρ = 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 theory fails completely. Instead of the topological degree, the precise knowledge on the Green function plays a fundamental role in the investigation of.. 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:

3 MEAN FIELD EQUATIONS ON TORI 3 Theorem. 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 Gz has precisely the three obvious critical points, hence for ρ = 8π equation. has no solutions. However we will show in that for the case ω = and τ = ω = e πi/3 there are at least five critical points and the solutions of. exist. Our second main result is the uniqueness theorem. Theorem. 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., an equivalent statement of Theorem. is the following result: Theorem.3. The Green function has at most five critical points. Unfortunately we were unable to find a direct proof of Theorem.3 from the critical point equation.. Instead, we will prove the uniqueness theorem first, and then Theorem.3 is an immediate corollary. Our proof of Theorem. is based on the method of symmetrization applied to the linearized equation at a particularly chosen even solution in the scaling family. In fact we study in 4 the one parameter family u + ρe u = ρδ 0, ρ [4π, 8π] on T within even solutions. This extra assumption allows us to construct a double cover T S via the Weierstrass function and to transform equation. into a similar one on S 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 singularity. 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 Theorem. we sharpen the result on nontrivial degree to the existence and uniqueness of solution Theorem 3.. For ρ [4π, 8π], the symmetrization reduces the problem on the non-degeneracy of the linearized equation to the isoperimetric inequality on domains in R with respect to certain singular measure: Theorem.4 Symmetrization Lemma. Let Ω R be a simply connected domain and let v be a solution of v + e v = N j= πα jδ pj in Ω. Suppose that the first eigenvalue of + e v is zero on Ω with ϕ the first eigenfunction. If the isoperimetric inequality with respect to ds = e v dx :.3 l ω mω4π mω

4 4 CHANG-SHOU LIN AND CHIN-LUNG WANG holds for all level domains ω = {ϕ > t} with t > 0, then e v dx π. Moreover,.3 holds if there is only one negative α j and α j =. Ω 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., 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,. is the Euler-Lagrange equation of the non-linear functional.4 J 8π v = v da 8π log e v 8πGz da T T on H T, the Sobolev space of functions with L -integrable first derivative. 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. 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. 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., Theorem A implies that a minimizer of J 8π 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. must be a minimizer of the nonlinear functional J 8π in.4. Thus we completely solve the existence problem on minimizers, a question raised by Nolasco and Tarantello in []. 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, however, no reference to Theorem A is needed. Instead, it motivates the following comparison result, which also simplifies the criterion further:

5 MEAN FIELD EQUATIONS ON TORI 5 M i + i + b i 0 FIGURE. Ω 5 contains a neighborhood of the shape. Theorem.5. Let z 0 and z be two half period points. Then Gz 0 Gz if and only if z 0 z. For general flat tori, a computer simulation suggests the following picture: Let Ω 3 resp. Ω 5 be the subset of the moduli space M { } = S which corresponds to tori with three resp. five critical points, then Ω 3 { } is a closed subset containing i, Ω 5 is an open subset containing e πi/3, both of them are simply connected and their common boundary C := Ω 3 = Ω 5 is a curve homeomorphic to S containing. Moreover, the extra critical points are split out from some half period point when the tori move from Ω 3 to Ω 5 across C. Concerning with the experimental observation, we propose to prove it by the method of deformations in M. 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 = /. These are equivalent to the rhombus tori and τ = +i is equivalent to the square torus where there are only three critical points. Theorem.6 Moduli Dependence. Let ω = and ω = τ = + ib with b > 0. Then There exists b 0 < < b < 3/ such that ω is a degenerate critical point of Gz; τ if and only if b = b 0 or b = b. Moreover, ω is a local minimum point of Gz; τ if b 0 < b < b and it is a saddle point if b < b 0 or b > b. Both ω and ω 3 are non-degenerate saddle points of G. 3 Gz; τ has two more critical points ±z 0 τ when b < b 0 or b > b. They are non-degenerate global minimum points of G and in the former case Re z 0 τ = ; 0 < Im z 0τ < b. Part gives a strong support of the conjectural shape of the decomposition M = Ω 3 Ω 5. Part 3 implies that minimizers of J 8π exist for tori with τ = + b with b < b 0 or b > b.

6 6 CHANG-SHOU LIN AND CHIN-LUNG WANG FIGURE. Graphs of η the left one and e in b where e is increasing FIGURE 3. Graphs of e + η the upper one and e η. Both functions are increasing in b and e η ր 0. The proofs are given in 6. Notably Lemma 6., 6. 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 e i = ω i and η i = ζ ω i: Theorem.7 Fundamental Inequalities. Let ω = and ω = τ = + ib with b > 0. Then d db e + η = 4π d db log ϑ 0 > 0. e η = 4π d db log ϑ 30 < 0 and d db log ϑ 30 > 0. The same holds for ϑ 4 0 = ϑ 3 0. In particular, e increases in b. These modular functions come into play due to the explicit computation of Hessian at half periods along Re τ = cf. 6.3 and 6.9: 4π det D ω G = e + η e + η π, b 4π det D ω G = η π e b + e η Im e. Although e i s and η i s are classical objects, we were unable to find an appropriate reference where these inequalities were studied. Part of,

7 MEAN FIELD EQUATIONS ON TORI 7 namely e η < 0, can be proved within the Weierstrass theory cf The whole theorem, however, requires theta functions in an essential way. Theta functions are recalled in 7 and the theorem is proved in Theorem 8. and Theorem 9.. The proofs make use of the modularity of special values of theta functions Jacobi s imaginary transformation formula as well as the Jacobi triple product formula. Notice that the geometric meaning of these two inequalities has not yet been fully explored. For example, the variation on signs from ϑ to ϑ 3 is still mysterious to us.. GREEN FUNCTIONS AND PERIODS INTEGRALS We start with some basic properties of the Green functions that will be used in the proof of Theorem.. Detailed behavior of the Green functions and their critical points will be studied in later sections. Let T = C/Zω + Zω be a flat torus. As usual we let ω 3 = ω + ω. The Green function Gz, w is the unique function on T which satisfies. z Gz, w = δ w z T and Gz, w da = 0. It has the property that Gz, w = Gw, z and it is T smooth in z, w except along the diagonal z = w, where. Gz, w = π log z w + O z w + C for a constant C which is independent of z and w. Moreover, due to the translation invariance of T we have that Gz, w = Gz w, 0. Hence it is also customary to call Gz := Gz, 0 the Green function. It is an even function with the only singularity at 0. There are explicit formulae for Gz, w in terms of elliptic functions, either in terms of the Weierstrass function or the Jacobi-Riemann theta functions ϑ j. Both are developed in this paper since they have different advantages. We adopt the first approach in this section. Lemma.. There exists a constant Cτ, τ = ω /ω, such that.3 8πGz = log ξ z da + Cτ, T T It is straightforward to verify that the function of z, defined in the right hand side of.3, satisfies the equation for the Green function. By comparing with the behavior near 0 we obtain Lemma.. Since the proof is elementary, also an equivalent form in theta functions will be proved in Lemma 7., we skip the details here. In view of Lemma., in order to analyze critical points of Gz, it is natural to consider the following periods integral.4 Fz = L z ξ z dξ,

8 8 CHANG-SHOU LIN AND CHIN-LUNG WANG where L is a line segment in T which is parallel to the ω -axis. Fix a fundamental domain T 0 = { sω + tω s, t } and set L = L. Then Fz is an analytic function, except at 0, in each region of T 0 divided by L L. Clearly, when ξ = z, z/ ξ z has residue ± at z = ±z. Thus for any fixed z, Fz may change its value by ±πi if the integration lines cross z. Let T 0 \L L = T T T 3, where T is the region above L L, T is the region bounded by L and L and T 3 is the region below L L. Recall that ζ z = z and η i = ζz + ω i ζz for i {, }. Then we have Lemma.. Let C = πi, C = 0 and C 3 = πi. Then for z T k,.5 Fz = ω ζz η z + C k. Proof. For z T T T 3, we have F d z z = dξ. L dz ξ z d z Clearly, z and z are the only double poles of as a dz ξ z meromorphic function of z and d z has zero residues at ξ = dz ξ z 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 L L, a straightforward computation shows that Fz = ω η z + Oz z in a neighborhood of 0. Therefore By integrating F, we get F z = ω z η. Fz = ω ζz η z + C k. Since Fω / = 0, Fω / = 0 and F ω / = 0, C k is as claimed. Here we have used the Legendre relation η ω η ω = πi. From.3, we have.6 8πG z = T T z ξ z da. Lemma.3. Let G be the Green function. Then for z = tω + sω,.7 G z = 4π ζz η t η s. In particular, z is a critical point of G if and only if.8 ζtω + sω = tη + sη.

9 MEAN FIELD EQUATIONS ON TORI 9 Proof. We shall prove.7 by applying Lemma.. Since critical points appear in pair, without loss of generality we may assume that z = tω + sω with s 0. We first integrate.6 along the ω direction and obtain fs 0 : = = L s 0 z ξ z dξ ζz + η z if s 0 > s, ζz + η z πi if s < s 0 < s, ζz + η z if s 0 < s, where L s 0 = { t 0 ω + s 0 ω t 0 }. Thus, z 8πG z = ζ z dt 0 ds 0 = fs 0 ds 0 L s 0 = ω ω ζz + η z s + ω ζz + η z πis = ω ω ζz + η z 4πsi = ω ω ζz + η tω + sη ω πi = ζz + η t + sη, where the Legendre relation is used again. Corollary.4. Let Gz be the Green function. Then ω k, k {,, 3} are critical points of Gz. Furthermore, if z is a critical point of G then both periods integrals.9 are purely imaginary numbers. F z := ω ζz η z and F z := ω ζz η z Proof. The half-periods ω, ω and ω 3 are obvious solutions of.8. Alternatively, the half periods are critical points of any even functions. Indeed for Gz = G z, we get Gz = G z. Let p = ω i for some i {,, 3}, then p = p in T and so Gp = Gp = 0. If z = tω + sω is a critical point, then by Lemma.3, ω ζz η z = ω tη + sη η tω + sω = sω η ω η = sπi. The proof for F is similar. Example.5. When τ = ω /ω i R, by symmetry considerations it is known cf. [6], Lemma. that the half periods are all the critical points. Example.6. There are tori such that equation.8 has more than three solutions. One such example is the torus with ω = and ω = + 3i. In this case, the multiplication map z ω z is simply the counterclockwise rotation by angle π/3, which preserves the lattice Zω + Zω, hence satisfies.0 ω z = z.

10 0 CHANG-SHOU LIN AND CHIN-LUNG WANG Similarly g = 60 ω 4 = 60 ω ω 4 = ω g, which implies that g = 0. Hence e j s satisfy. e e + e e 3 + e e 3 = g = 0 and satisfies = 6. Let z 0 be a zero of z. Then z 0 = 0 too. By.0, ω z 0 = 0, hence either ω z 0 = z 0 or ω z 0 = z 0 on T since z = 0 has zeros at z 0 and z 0 only. From here, it is easy to check that either z 0 is one of the half periods or z 0 = ± 3 ω 3. But z 0 can not be a half period because z 0 = 0 at any half period. Therefore, we conclude that z 0 = ± 3 ω 3 and ± 3 ω 3 = ± 3 ω 3 = 0. We claim that 3 ω 3 is a critical point. To prove it, we use the addition formulas for ζ. ζz = ζz + z z to obtain.3 ζ On the other hand, ω3 ζ = ζ 3 ω3 Together with.3 we get 3 ζ ω3 ω3 = ζ. 3 3 ω3 + η + η = ζ + η + η. 3 ω3 = 3 3 η + η. That is, 3 ω 3 satisfies the critical point equation. Thus Gz has at least five critical points at ω k, k =,, 3 and ± 3 ω 3 when τ = ω /ω = + 3i. By way of Theorem., 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. with ρ = 8π and set.4 vz = uz + 8πGz. Then vz satisfies.5 vz + 8πe 8πGz e vz = 0

11 MEAN FIELD EQUATIONS ON TORI in T. By., it is obvious that vz is a smooth solution of.5. An important fact which we need is the following: Assume that there is a blowup sequence of solutions v j z of.5. That is, v j p j = max v j + as j. T Then the limit p = lim j p j is the only blow-up point of {v j } and p is in fact a critical point of Gz:.6 Gp = 0. We refer the reader to [3] p. 739, Estimate B for a proof of THE CRITERION FOR EXISTENCE VIA MONODROMIES Consider the mean field equation 3. u + ρe u = ρδ 0, ρ R + in a flat torus T, where δ 0 is the Dirac measure with singularity at 0 and the volume of T is normalized to be. A well known theorem due to Liouville says that any solution u of u + ρe u = 0 in a simply connected domain Ω C must be of the form f 3. u = c + log + f, where f is holomorphic in Ω. Conventionally f is called a developing map of u. Given a torus T = C/Zω + Zω, by gluing the f s among simply connected domains it was shown in [6] that for ρ = 4πl, l N, 3. holds on the whole C with f a meromorphic function. The statement there is for rectangular tori with l =, but the proof works for the general case. It is straightforward to show that u and f satisfy 3.3 u zz u z = f f 3 f f. The right hand side of 3.3 is the Schwartz derivative of f. Thus for any two developing maps f and f p q of u, there exists S = PSU q p i.e. p, q C and p + q = 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 imposed by the double periodicity of the equation. By applying 3.4 to fz + ω and fz + ω, we find S and S in PSU with 3.5 fz + ω = S f, fz + ω = S f. These relations also force that S S = S S.

12 CHANG-SHOU LIN AND CHIN-LUNG WANG The second type of constraints are imposed by the Dirac singularity of 3. at 0. A straightforward local computation with 3. shows that Lemma 3.. If fz has a pole at z 0 0 mod ω, ω, then the order k = l +. If fz = a 0 + z k + is regular at z 0 0 mod ω, ω then k = l +. 3 If fz has a pole at z 0 0 mod ω, ω, then the order is. 4 If fz = a 0 + z z 0 k + is regular at z 0 0 mod ω, ω then k =. Now we are in a position to prove Theorem., namely the case l =. 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 conjugating a matrix in PSU, that S = e iθ 0 0 e iθ for some θ R. Let p q S = and then 3.5 becomes q p 3.6 fz + ω = e iθ fz, fz + ω = S fz. Since S S = S S, a direct computation shows that there are three possibilities: 3.7 p = 0 and e iθ = ±i; q = 0; and 3 e iθ = ±. Case. By assumption we have fz + ω = fz, fz + ω = q fz. If fz has singularity at z = 0 then the order of pole at 0 is 3. Now 3.7 implies that fz is an elliptic function on the torus T = C/Zω + Zω and f z can have zeros only at ω and ω 3, both with multiplicity. On the other hand, f z has 4-th order poles at 0 and ω. This yields a contradiction to the fact that an elliptic function has equal numbers of poles and zeros. If f is regular and f0 = 0, we can still apply the above argument to / f to get a contradiction. It remains to consider the case that f is regular and f0 = 0. By 3.7, fz/ f z is an elliptic function on the torus T = C/Zω + Zω. Since f z = 0 for z 0 mod ω, ω, fz/ f z only has poles at 0 and ω,

13 MEAN FIELD EQUATIONS ON TORI 3 both of multiplicity. Thus there exist constants A j, j {,, 3} such that 3.8 fz f z = A A z + z ω + A 3, where is the Weierstrass elliptic function on T. Since z is an even function, we have fz/ f z = f z/ f z. Thus fz f z = 0 and fz f z is a constant function. So 3.9 f z = f0 fz. Since f has zeros only at 0 and ω, both of multiplicity, and f only has poles of multiplicity, 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 fz = f0} has solutions with total multiplicity 4 on T. Since fz f0 has zero of multiplicity 3 at 0, there exists an unique solution z 0 = 0 of fz 0 = f0 in T. But by 3.9 we have f z 0 = f 0 fz 0 = f0. Thus z 0 = z 0 mod ω, ω, i.e. z 0 = ω k for some k {,, 3}. But f ω k = 0 for k =,, 3 because ω k are half-periods in the torus T. So the multiplicity of fz = f0 at z = ω k is also 3, which yields a contradiction. Thus Case is impossible. Case. In this case we have 3.0 fz + ω = e iθ fz, fz + ω = e iθ fz. Notice that if a meromorphic function f satisfies 3.0, then e λ f also satisfies 3.0 for any λ R. Thus 3. u λ z = c + log is a scaling family of solutions of 3. and e λ f z + e λ fz 3. sup u λ x + as λ. T This observation leads to a contradiction when the fundamental cell of T is a rectangle. Indeed by Theorem. in [6], for T a rectangular torus, a priori bounds for solutions of 3. was obtained. To prove the general case, as in case, we first assume that f has singularity 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 3.0, we have f z c > 0 for z C. This implies that / f is a bounded holomorphic function and hence a constant, which is absurd.

14 4 CHANG-SHOU LIN AND CHIN-LUNG WANG Secondly, let f be regular at 0 with f0 = 0. By applying the argument above to / f, we again get a contradiction. It remains to investigate the most delicate situation when f is regular at 0 and f0 = 0. By 3.0, fz/ f z is an elliptic function on T which has 0 as its only pole. Thus there are constants A and A with 3.3 fz f z = A z + A. Since z is even, this yields as before that 3.4 f z = f0 fz. By 3.3, fz/ f z has two zeros, say, z 0 and z 0. By 3.4, one of z 0 or z 0 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 z 0. In particular z 0 = z 0 in T and we conclude that z 0 = ω k / for any k {,, 3}. As in the observation above, e λ f can be used as a developing map for some solution u λ x defined by 3.. Clearly u λ z as λ + for any z such that fz = 0 and u λ z 0 + as λ +. Hence z 0 is the blow-up point and we have by.6 that Gz 0 = 0. Namely, it is a critical point other than the half periods. Case 3. In this case we get that S is the identity. So by another conjugation in PSU we may assume that S is in diagonal form. But this case is then reduced to Case. The proof of the only if part is completed. Now we prove the if part. Suppose that there is a critical point z 0 of Gz with z 0 = ω k for any k {,, 3}. For any closed curve C such that z 0 and z 0 C, the residue theorem implies that 3.5 C z 0 dξ = πmi ξ z 0 for some m {, 0, }. Hence z 3.6 fz := z 0 exp ξ z 0 dξ 0 is well-defined as a meromorphic function. Let L and L be lines in T which are parallel to the ω -axis and ω -axis respectively. Then for j =,, z fz + ω j = fz exp L j ξ z 0 dξ.

15 By Lemma., MEAN FIELD EQUATIONS ON TORI 5 z 0 L j ξ z 0 dξ = F jz 0 + C k for some C k {πi, 0, πi}. Also by Corollary.4, Hence for j =,, F j z 0 = ω j ζz 0 η j z 0 = iθ j i R. 3.8 fz + ω j = fze iθ j. holds. Set u λ x = c + log e λ f z + e λ fz. Then u λ x satisfies 3. for any λ R and u λ is doubly periodic by 3.8. Therefore, solutions have been constructed and the proof of Theorem. is completed. A similar argument leads to Theorem 3.. For ρ = 4π, there exists an unique solution of 3.. Proof. By the same procedure of the previous proof, there are three cases to be discussed. For Case, the subcases that f or / f is singular at z = 0 leads to contradiction as before. For the subcase that f is regular at z = 0 and f0 = 0 we see that fz/ f z is an elliptic function with 0 as its only simple pole since now k = l =. Hence Case does not occur. Similarly Case 3 is not possible. Now we consider Case. By 3.7, the function g = log f = f is elliptic on T = C/Zω + Zω. g has a simple pole at each zero or pole of f. By Lemma 3., if f or / f is singular at z = 0 then g has no zero and we get a contradiction. So f is regular at z = 0, f0 = 0 and g has two simple zeros at 0 and ω. Let σz = exp z ζw dw = z + be the Weierstrass sigma function on T. σ is odd with a simple zero at each lattice point. Then 3.9 gz = A σzσz ω σz aσz b for some a, b with a + b = ω. From 3.7, we have gz + ω = gz. So a + ω = b mod ω, ω. Since the representation of g in terms of sigma functions is unique up to the lattice {ω, ω }, there is an unique solution of a, b: 3.0 a = ω ; b = ω + ω. f

16 6 CHANG-SHOU LIN AND CHIN-LUNG WANG Notice that the residue of g at a and b are both equal to πir where We claim that A = /r. Since r = σ ω σ ω + ω. σω + ω fz = exp z gw dw is well defined, by the residue theorem, we must have A = m/r for some m Z. Moreover by Lemma 3. f has simple poles other than z = 0 indeed at z = a and z = b, so we conclude that m =. Conversely, by picking up a, b and A as above, it is clear that fz is a well defined meromorphic function which gives rise to a solution of 3. 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 equation 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 f0 we may assume that f0 =. Then we have Consider the particular solution uz = c + log f z = fz. f z + fz. It is easy to verify that u z = uz 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 { u + ρe u = ρδ and 4. u z = uz on T where ρ [4π, 8π]. We will use the method of symmetrization to prove Theorem 4.. For ρ [4π, 8π], the linearized equation of 4. 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., we conclude the proof of Theorem. by the inverse function theorem. We first prove Theorem.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 R with metric e w dx to the case when the metric becomes singular.

17 MEAN FIELD EQUATIONS ON TORI 7 Let Ω R be a domain and w C Ω which satisfies { w + e w 0 in Ω and 4. e w dx 8π. Ω This is equivalent to saying that the Gaussian curvature of e w dx is K e w w. For any domain ω Ω, we set 4.3 mω = e w dx and l ω = ω ω e w/ ds. Bol s isoperimetric inequality says that if Ω is simply-connected then 4.4 l ω mω8π mω. We first extend it to the case when w acquires singularities: w + e w = N j= πα jδ pj in Ω and 4.5 e w dx 8π, Ω with α j > 0, j =,,..., N. Lemma 4.. 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 l ω mω8π mω. Proof. Define v and w ǫ by wx = α j log x p j + vx, j w ε x = j By straightforward computations, we have w ε x + e w εx = j αj α j log x p j + ε + vx. 4ε r + ε + x p j + ε αj/ e v x p j α j e v 0. Let l ε and m ε be defined as in 4.3 with respect to the metric e w εx dx. Then we have l ε ω m ε ω8π m ε ω. By letting ε 0 we obtain 4.6.

18 8 CHANG-SHOU LIN AND CHIN-LUNG WANG Next we consider the case that some of the α j s are negative. For our purpose, it suffices to consider the case with only one singularity p with negative α and we only need the case that α =. In view of the proof of Lemma 4., the problem is reduced to the case with only one singularity p. In other words, let w satisfy 4.7 w + e w = πδ p in Ω. Lemma 4.3. Let w satisfy 4.7 with Ω simply-connected. Suppose that 4.8 e w dx 4π, then 4.9 l ω mω4π mω. Proof. We may assume that p = 0. If 0 ω then Ω l ω mω8π mω > mω4π mω by Bol s inequality trivially. If 0 ω, we consider the double cover Ω of Ω branched at 0. Namely we set Ω = f Ω where x = fz = z for z C. The induced metric e v dz on Ω satisfies That is, v is the regular part e vz dz = e wx dx = e wz 4 z dz. vz := wx + log x + log 4 = wz + log z + log 4. By construction, v satisfies v + e v = 0 in Ω\{0}. Since v is bounded in a neighborhood, by the regularity of elliptic equations, vz is smooth at 0. Hence v satisfies v + e v = 0 in Ω. Let ω = f ω. Clearly ω f ω. Also where l ω l ω and l ω = ω By Bol s inequality, we have m w = mω, e v/ ds and m ω = ω e v dx. 4l ω l ω m ω8π m ω = mω8π mω. Thus l ω mω4π mω.

19 MEAN FIELD EQUATIONS ON TORI 9 Lemma 4.4 Symmetrization I. Let Ω R be a simply-connected domain with 0 Ω and let v be a solution of v + e v = πδ 0 in Ω. If the first eigenvalue of + e v is zero on Ω then 4.0 e v dx π. Proof. Let ψ be the first eigenfunction of + e v : { ψ + e 4. v ψ = 0 in Ω and ψ = 0 on Ω. In R, let U and ϕ be the radially symmetric functions Ux = log and 4. Ω ϕx = x + x. + x x From = r + r r + r, it is easy to verify that U and ϕ satisfy θ { U + e 4.3 U = πδ 0 and ϕ + e U ϕ = 0 in R. For any t > 0, set Ω t = {x Ω ψx > t} and rt > 0 such that 4.4 e Ux dx = e vx dx, B rt {ψ>t} where B rt is the ball with center 0 and radius rt. Clearly rt is strictly decreasing in t for t 0, max ψ. In fact, rt is Lipschitz in t. Denote by ψ r the symmetrization of ψ with respect to the measure e Ux dx and e vx dx. That is, ψ r = sup{t r < rt}. Obviously ψ r is decreasing in r and for t 0, max ψ, ψ r = t if and only if rt = r. Thus by 4.4, we have a decreasing function 4.5 ft := e Ux dx = e vx dx. {ψ >t} By Lemma 4.3, for any t > 0, {ψ>t} 4.6 l {ψ = t} ft4π ft. We will use inequality 4.6 in the following computation: For any t > 0, by the Co-Area formula, d ψ dx = ψ ds and dt Ω t Ω t f t = d e v e dx = dt Ω t Ωt v ψ ds

20 0 CHANG-SHOU LIN AND CHIN-LUNG WANG hold almost everywhere in t. Thus d ψ dx = dt Ω t e v/ dx 4.7 {ψ=t} = l {ψ = t} f t {ψ=t} {ψ=t} ft4π ft f t. The same procedure for ψ leads to d ψ dx = dt {ψ >t} = e U/ dx 4.8 {ψ =t} = l {ψ = t} f t ψ ds {ψ =t} {ψ =t} = ft4π ft f t, with all inequalities being equalities. Hence d ψ dx d dt dt {ψ>t} e v ψ dx ψ ds e U ψ ds {ψ >t} ψ ds holds almost everywhere in t. By integrating along t, we get ψ dx ψ dx. B R0 Ω Since ψ and ψ have the same distribution or by looking at f tt dt directly, we have e Ux ψ dx = e vx ψ dx. B R0 Ω Therefore 0 = Ω ψ dx e v ψ dx ψ dx e U ψ dx. Ω B R0 B R0 This implies that the linearized equation + e Ux has non-positive first eigenvalue. By 4., this happens if and only if R 0. Thus e vx dx = e Ux dx π. B R0 Ω

21 MEAN FIELD EQUATIONS ON TORI Remark 4.5. See []. By applying symmetrization to v+e v = 0 in Ω with λ + e v = 0, the corresponding radially symmetric functions are Ux = log + 8 x and ϕx = 8 x 8 + x. The same computations leads to Ω ev dx 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 R be a simply-connected domain and let v be a solution of v + e v = N j= πα jδ pj in Ω. Suppose that the first eigenvalue of + e v is zero on Ω with ϕ the first eigenfunction. If the isoperimetric inequality with respect to ds = e v dx : l ω mω4π mω holds for all level domains ω = {ϕ t} with t 0, then e v dx π. Ω Notice that we do not need any further constraint on the sign of α j. For the last statement of Theorem.4, the limiting procedure of Lemma 4. 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.4 is completed. Proof. of Theorem 4.. Let u be a solution of equation 4.. It is clear that we must have T eu =. Suppose that ϕx is a non-trivial solution of the linearized equation at u: { ϕ + ρe 4.9 u ϕ = 0 and ϕz = 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 S = C { }, we may require that being an isometry: Namely we set e uz dz = e vx dx = e vx z dz. 4.0 vx := uz log z and ψx := ϕz. There are four branch points on C { }, namely p 0 = 0 = and p j = e j := ω j / for j =,, 3. Since z = 4 3 j= x e j, by construction

22 CHANG-SHOU LIN AND CHIN-LUNG WANG vx and ψx then satisfy { v + ρe v = 4. 3 j= πδ p j ψ + ρe v ψ = 0 in R. and To take care of the point at infinity, we use coordinate y = /x or equivalently we consider T S via y = / z z. The isometry condition reads as e uz dz = e wy dy = e wy z z 4 dz. Near y = 0 we get wy = uz log z z 4 ρ 4π log y. Thus ρ 4π implies that p 0 is a singularity with non-negative α 0 : w + ρe w = α 0 δ j= πδ /p j. In dealing with equation 4. and the above resulting equations, by replacing u by u + log ρ etc., we may and will replace the ρ in the left hand side by for simplicity. The total measure on T and R are then given by e u dz = ρ 8π and e v dx = ρ R 4π. T The nodal line of ψ decomposes S into at least two connected components and at least two of them are simply connected. If there is a simply connected component Ω which contains no p j s, then the symmetrization Remark 4.5 leads to e v dx 4π, Ω which is a contradiction because R \Ω =. If every simply connected component Ω i, i =,..., m, contains only one p j, then Lemma 4.4 implies that Ω i e v dx π for i =,..., m. The sum is at least mπ, which is again impossible unless m = and R = Ω Ω. So without lost of generality we are left with one of the following two situations: R { }\{ψ = 0} = Ω + Ω where Ω + {x ψx > 0} and Ω {x ψx < 0}. Both Ω + and Ω are simply-connected. Either Ω contains p, p and p 3 Ω + or p Ω, p 3 Ω + and p C = Ω + = Ω.

23 MEAN FIELD EQUATIONS ON TORI 3 Assume that we are in case. By Lemma 4.3, we have on Ω + 4. l {ψ = t} m{ψ t}4π m{ψ t} for t 0. We will show that the similar inequality 4.3 l {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 p and p, then Lemma 4.3 implies that l ω 4π mωmω. If ω contains both p and p, then R \ω is simply connected which contains p 3 only. Thus by Lemma l ω 4π mr \ωmr \ω = 4π ρ/ + mωρ/ mω = 4π mωmω + 4π ρ/ρ/ mω. Since ρ 8π and Ω + e v dx π, we get ρ = e v e v + e v π + mω mω. R Ω + ω Then again l ω 4π mωmω with equality holds only when ρ = 8π and mω = π. Now it is a simple observation that domains satisfy the isoperimetric inequality 4.3 have the addition property. Indeed, if a 4π mm and b 4π nn, then a + b = a + 4ab + b > 4π mm + 4π nn = 4πm + n m + n + mn > 4π m + nm + n. Hence 4.3 holds for all t 0. Now we can apply Lemma 4.6 to Ω to conclude that Ω e v dx = π which already leads to a contradiction if ρ < 8π and the equality l {ψ = t} = 4π m{ψ t}m{ψ t} in 4.3 holds for all t min ψ, 0. This implies that {ψ t} has only one component and it contains p and p for all t min ψ, 0. But then ψ attains its minimum along a connected set ψ min ψ containing p and p, which is impossible.

24 4 CHANG-SHOU LIN AND CHIN-LUNG WANG In case, ρ < 8π again leads to a contradiction via the same argument. For ρ = 8π, we have e v dx = e v dx = π Ω + Ω and all inequalities in 4.7 are equalities. So under the notations there e v/ e dx = ψ ds Ω t Ω t Ωt v ψ ds for all t min Ω ψ, max Ω+ ψ. This implies that ψ x = Cte vx almost everywhere in ψ t for some constant C which depends only on t. By continuity we have 4.5 ψ x = Cψxe vx for all x except when ψx = max Ω+ ψ or ψx = min Ω ψ. By letting x = p ψ 0, we find C0 = ψ p e vp = 0. By 4.5 this implies that ψx = 0 for all x ψ 0, which is clearly impossible. Hence the proof of Theorem 4. is completed. Since equation 4. has an unique solution at ρ = 4π, by the continuation from ρ = 4π to 8π and Theorem 4., we conclude that 4. has a most a solution at ρ = 8π, and it implies that the mean field equation. has at most one solution up to scaling. Thus, Theorem. is proved and then Theorem.3 follows immediately. 5. COMPARING CRITICAL VALUES OF GREEN FUNCTIONS For simplicity, from now on we normalize all tori to have ω =, ω = τ. In section 3 we have shown that the existence of solutions of equation. is equivalent to the existence of non-half-period critical points of Gz. The main goal of this and next sections is to provide criteria for detecting minimum points of Gz. The following theorem is useful in this regard. Theorem 5.. Let z 0 and z be two half-periods. Then Gz 0 Gz if and only if z 0 z. Proof. By integrating.7, the Green function Gz can be represented by 5. Gz = π Re ζz η z dz + b y. Thus ω ω3 5. G G = ω3 π Re ζz η ω z dz.

25 MEAN FIELD EQUATIONS ON TORI 5 Set Fz = ζz η z. We have Fz + ω = Fz and ζ t + ω η t + ω ω = ζ t η t + ω ω = ζ t + η η t + ω [ ω = ζ t ω ] η t + η η ω [ ω = ζ t ω ] η t πi, hence Re F ω + t is antisymmetric in t C. To calculate the integral in 5., we use the addition theorem to get z = ζ z ω + ζ z + ω ζz z e = F z ω + F z + ω Fz. Integrating it along the segment from ω to ω 3, we get log e 3 e = Fz dz + Fz dz Fz dz, e e L L L 3 where L is the line from ω ω to ω, L is the line from ω 3 to ω + ω and L 3 is the line from ω to ω 3. Since Fz = Fz + ω and Re Fz is antisymmetric with respect to ω, we have log e 3 e e e = L Fz dz 4 Fz dz = πi 4 Fz dz, L L 3 where L is the line from ω ω to ω 3. Thus we have log e 3 e ω3 e e = Re ω Fz dz = 4π G G ω That is, ω ω3 5.3 G G = 4π log e e e 3 e. Similarly, by integrating.7 in the ω direction, we get ω ω3 5.4 G G = ω3 π Re ζz η ω z dz. The same proof then gives rise to ω ω3 5.5 G G = 4π log e e e 3 e. By combining the above two formulae we get also that ω ω 5.6 G G = 4π log e e 3 e e 3. ω3.

26 6 CHANG-SHOU LIN AND CHIN-LUNG WANG In order to compare, say, G ω and G ω 3, we may use 5.5. Let By using e + e + e 3 = 0, we get 5.7 λ = e 3 e e e. e 3 = λ e λ. It is easy to see that λ λ if and only if λ. Hence e 3 if and only if λ. e 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 function λτ = e 3 e. e e By 5.3, we have ω3 ω 5.8 G G = log λτ. 4π Therefore, it is important to know when λτ =. Lemma 5.. λτ = if and only if Re τ =. Proof. Let z; τ be the Weierstrass function with periods and τ, then z; τ = z; τ. For τ = + ib, τ = τ and then z; τ = z; τ. Thus 5.9 z; τ = z; τ for τ = + ib. Note that if z = ω then z = ω = ω = ω 3 mod ω, ω. Therefore ē = e 3 and ē = e. Since e + e + e 3 = 0, we have 5.0 Re e = e and Im e = Im e 3. Thus 5. λτ = e 3 e e e =. A classic result says that λ τ = 0 for all τ. By this and 5., it follows that λ maps { τ Re τ = } bijectively onto { λτ λτ = }.

27 MEAN FIELD EQUATIONS ON TORI 7 Let Ω be the fundamental domain for λτ, i.e., Ω = { τ C τ / /, 0 Re τ, Im τ > 0 }, and let Ω be the reflection of Ω with respect to the imaginary axis. Since G ω 3 < G ω for τ = ib, we conclude that for τ Ω Ω, λτ < if and only if Re τ <. Therefore for τ Ω Ω, Re τ < ω3 ω if and only if G < G. For τ =, using suitable Möbius transformations we may obtain similar results. For example, from the definition of, 5.9 implies that τ + τ + 5. z = τ + τ + z and so compare. τ Gz = G τ + z. Clearly, for z = τ, 5.3 implies that G ω = G ω. So 5.4 τ = if and only if e e 3 e e 3 =, 5.5 τ < if and only if ω ω G < G. Similarly, 5.6 τ < if and only if ω ω3 G < G DEGENERACY ANALYSIS OF CRITICAL POINTS By.7, the derivatives of G can be computed by πg x = Reη t + η s ζz, πg y = Imη t + η s ζz. When τ = + ib, since z is real for z R, η is real and 6. becomes Thus the Hessian of G is given by 6.3 πg x = η t + η s Re ζz, πg y = Im ζz + π η bs. πg xx = Re z + η, πg xy = Im z, πg yy = Re z + η π b.

28 8 CHANG-SHOU LIN AND CHIN-LUNG WANG We first consider the case z 0 = ω. The degeneracy condition of G at ω reads e + η = 0 or e + η π b = 0. We will use the following two inequalities Theorem.7 whose proofs will be given in 8 and 9 through theta functions: 6.4 e b + η b is increasing in b and 6.5 e b is increasing in b. Lemma 6.. There exists b 0 < < b < 3/ such that ω is a degenerate critical point of Gz; τ if and only if b = b 0 or b = b. Moreover, ω is a local minimum point of Gz; τ if b b 0, b and is a saddle point of Gz; τ if b 0, b 0 or b b,. Proof. Let b 0 and b be the zero of e + η = 0 and e + η π/b = 0 respectively. Then Lemma 6. follows from the explicit expression of the Hessian of G by 6.4. Numerically we know that b 0.7 < 3/. Now we analyze the behavior of G near ω for b > b. Lemma 6.. Suppose that b >, then ω is the only critical point of G along the x-axis. Proof. t; τ is real if t R. Since t; τ = 0 for t = ω, t; τ < 0 for 0 < t < ω. Since b > > b 0, by 6.3 and Lemma 6., G xx t = t + η > e + η > 0, which implies that G x t > G x ω = 0 if 0 < t < ω. Hence G has no critical points on 0, ω. Since Gz = G z, G can not have any critical point on ω, 0. By Lemma 6. and the conservation of local Morse indices, we know that Gz; τ has two more critical points near ω when b is close to b and b > b. We denote these two extra points by z 0 τ and z 0 τ. In this case, ω becomes a saddle point and z 0 τ and z 0 z are local minimum points. From Lemma 5., 5.5 and 5.6 we know that ω ω ω3 G < G = G if b < b < 3/. Thus in this region ±z 0 τ must exist and they turn out to be the minimum point of G since there are at most five critical points. In fact we will see below that this is true for all b < b 0 or b > b and ω, ω 3 are all saddle points. Lemma 6.3. The critical point z 0 τ is on the line Re z =. Moreover, the Green function Gz; τ is symmetric with respect to the line Re z =.

29 MEAN FIELD EQUATIONS ON TORI 9 Proof. Recall that G z = Gz. Since G has only two more critical points near ω and z R, z 0 τ = z 0 τ. Let z 0 τ = t 0 ω + s 0 ω. z 0 τ = t 0 + s 0 ω s 0 ω. Thus t 0 + s 0 =. Then Re z 0 z = t 0 + s 0 / =. Consider the rectangle: 0 Re z and Im z b/. Clearly the second statement follows from the symmetry of boundary value of G with respect to the line Re z =. Since Gz + = Gz, the symmetry of G on Re z = 0 and Re z = is obvious. Hence we have to show that 6.6 G z + + ib = G z + + ib for z R. By using G z = Gz, we have G z + + ib = G z + ib = G z ib = G + ib z, where the second and the third identity comes from Gz + = Gz and G z = Gz respectively. Thus 6.6 and then the lemma are proved. Let G be the restriction of G on the line Re z =. Lemma 6.3 says that any critical point of G is automatically a critical point of G. Lemma 6.4. G y has exactly one critical point in 0, b for b > b. Proof. Let z = tω + sω. Then Re z = is equivalent to t + s =, which implies z = t + sω sω = z. By 5.9 z, τ = z; z = z; τ = z; τ. Hence z; τ is real for Re z =. To prove Lemma 6.4, we apply 6.3, πg yy = + η π. b Since / y = i z; τ = 0 for z = ω and Re z =, G yyz can have one zero only. Let z 0 τ denote the critical point above when τ is close to + ib. Thus G y z 0 τ = G y ω = 0, and then G yy z 0 = 0 for some z 0 ω, z 0 τ. Since G yy ω = e + η π/b < 0, we have G yy z 0 τ > 0. Hence z 0 τ is a non-degenerate minimum point of G. Remark 6.5. For b > b, G y is increasing in y for y > 0 and lim y b G y = +. Theorem 6.6. For b > b, ±z 0 τ are non-degenerate local minimum points of G. Furthermore, < Im z 0 τ < b.

30 30 CHANG-SHOU LIN AND CHIN-LUNG WANG Proof. We want to prove G xx z 0 τ; τ > 0 and Im z 0 τ < b/. By 6.3, πg xx z 0 τ; τ = z 0 τ; τ + η. We claim that if z 0 τ; τ = 0 then τ = + 3i/. To prove the claim, we use.3. Since z 0 τ; τ = 0, we have ζz 0 τ = ζz 0 τ = η t 0 + η s 0. Thus z 0 τ is also a critical point. Note that Re z 0 τ =. Since Gz ω = Gz, z 0 + ω is also a critical point. Because Rez 0 + ω =, we have either z 0 + ω = z 0 or z 0 + ω = z 0. The later leads to z 0 = ω = ω 3, which is not true. Thus we have z 0 + ω = z 0 and then 3z 0 = ω = ω, i.e., z 0 = z 0. Therefore z 0 = z 0 = z 0. By the addition theorem for, we have z 0 = z 0 = z z 0 = z 0. z 0 Therefore z 0 = 0 and it leads to e e + e e 3 + e e 3 = 0, which implies that τ = + 3i/. Hence the claim is proved. Note that e e + e e 3 + e e 3 = e e and 6.8 z = 3 z + e e. In terms of λτ and e, e can be expressed by 6.9 e = λ λ e. Thus d e = dλ e λ = 0 for b >. From here, we have d e db e = 0 for b >, which implies that d db for b >. Therefore e e < d db z 0 τ; τ = e d e db e < 0 at τ = + 3i. Since z 0 τ; τ = 0 at τ = + 3i/, we have z 0 τ, τ < 0 for b > 3/ and sufficiently close to 3/. By the claim above, z 0 τ; τ < 0 for b > 3/. Thus 6. z 0 τ; τ 3 e e,

31 MEAN FIELD EQUATIONS ON TORI 3 and 6. η + z 0 τ; τ η 3 e e > η 4 e = η e where e = e /4 + Im e > e /4 is used. Later we will show that η > e always holds this is part of Theorem.7 to be proved in 9, but we will give another direct proof of it in 6.0. Thus the non-degeneracy of z 0 τ for b > 3/ follows. For b < 3/, we let z 0 τ = t 0 τ + t 0 ττ. We claim that t 0 τ > /3 if b < 3/. For if z 0 τ = 3 ω 3 then z 0 τ = z 0 τ is also a critical point. By the addition formula of ζ and, we conclude that 3 ω 3 = 0 and 3 ω 3 = 0, a contradiction. Hence ω3 6.3 z 0 τ; τ > 3 ; τ. Note that by the addition theorem for, it is easy to see 3 ω 3; τ < 0 for all τ = + ib. Let fτ := 3 ω 3; τ + e τ. We have f = 3 ω 3; < 0 and f 3/ = e 3/ > 0. Therefore there exists a τ 0 such that fτ 0 = 0. By the addition theorem for, we have ω ; τ 3 0 = ω 3; τ 0 ω. 3; τ 0 Since = 4 e e e 3, we have 6.5 = i<j 3 e i e j and 6.6 where = + +, e e e 3 e j can be calculated by 6.5 as e j = + e j + e j e j+ e j+, where the notations e 4 = e and e 5 = e are used. Therefore 6.6 is reduced to ω3 = 6 + e e 3 e3 + + e e 3 e e. e 3

32 3 CHANG-SHOU LIN AND CHIN-LUNG WANG Plug in 3 ω 3 = e and recall that e 3 = ē = e + iim e 3, we get 6.7 e e = Numerically we know that e / e 3.6 < 37 4 at b = b. Thus 3 ω 3 > e for b > b. Hence together with 6.3 z 0 τ, τ + η > η e > 0 for b < b < 3/. With this, the proof of the non-degeneracy of z 0 τ is completed. Next we discuss the non-degeneracy of G at ω and ω 3. The local minimum property of z 0 τ is in fact global by Theorem 6.7. For τ = + ib, both ω and ω 3 are non-degenerate saddle points of G. Proof. By 6.3, we have ω πg xx = Re e + η = η e, ω 6.8 πg xy = Im e, ω πg yy = π b + e η. Hence the non-degeneracy of ω for all b is equivalent to 6.9 Im e > η π e b + e η for all b R. To prove 6.9, we need the following: Lemma 6.8. maps [ ω, ω 3] [ ω, ω ] one to one and onto the circle {w w e = e e }, where e = ē 3, 4 ω + ω 3 = e e e < 0 and 4 = e + e e > 0. Here [ ω, ω 3] means segment { ω + t 0 t }, and [ ω, ω ] is { 4 + it t b }. Thus, the image of [ ω, ω 3] is the arc connecting e and e 3 through 4 ω + ω 3 and the image of [ ω, ω ] is the arc connecting e 3 and e through 4. See Figure 4. Note our figure is for the case e > 0, i.e., b >. In this case, the angle e 3e e is less than π. For the case e < 0, the angle is greater than π. We will derive 6.9 from Lemma 6.8. First we prove 6.0 η e > 0

33 MEAN FIELD EQUATIONS ON TORI 33 e 3 e 4 ω + ω 3 4 e e and 6. FIGURE 4. The image of the mapping z z. To prove 6.0, we have 6. η = π b + e η > 0. 0 ω Re + t dt, where ω + z = ω t is used. By Lemma 6.8, ω Re + t e. Hence To prove 6., we have ω Therefore, ω η e. ω z dz = ζ π bη = ζ ω = η η = iπ bη. b b Re 4 + it dt e b and the inequality 6. follows. To prove 6.9, we need two more inequalities. By 6., ω + ω η > = e e e, 4 and by Lemma 6.8, 6.4 π bη < /4b = e + e e b.