Let ρ=4πl, l ∈N. Let u be a type I solution and f be a developing map of u. In this section we will prove Theorem 0.4 stated in the introduction.
Proposition 1.5.1 proves that if l is odd then the solution is of type I. We will start by proving the converse in Theorem 2.2, i.e., if the solution is of type I then l must be odd. At the same time the evenness of u is deduced.
2.1. The evenness of solutions. Recall the logarithmic derivative g= (log f)′ = f′
f
which is elliptic on E′ = C/Λ′ with Λ′ = Zω1+Z2ω2. For the ease of notations we will use ω1′ = ω1 and ω′2 = 2ω2. In the following all the elliptic functions are with respect to the torus E′.
Since g has zero at z = 0 of order l, it also has zero of order l at z = ω2. There are no other zeros hence it has simple poles at p1, . . . , pland q1, . . . , ql
where pi’s are simple zeros of f and qi’s are simple poles of f modulo Λ′. Thus we may assume that
qi = pi+ω2, i=1, . . . , l.
From
f(z) = f(0)exp Z z
0 g(w)dw,
the residues of g are 1 at pi’s and−1 at qi’s. Thus we may write g as (2.1.1) g(z) =
l
∑
i=1
(ζ(z−pi) −ζ(z−pi−ω2)) +c By (1.4.1), it is easily seen that c=lη2/2.
There are also other useful equivalent forms of g:
g(z) = 1 2
l
∑
i=1
(2ζ(z−pi) −ζ(z−pi−ω2) −ζ(z−pi+ω2))
= −1 2
l
∑
i=1
℘′(z−pi)
℘(z−pi) −e2
= −12
l
∑
i=1
d
dzlog(℘(z−pi) −e2) by the addition formula.
Remark 2.1.1. The middle formula says that up to a constant g(z) is the sum of slopes of the l lines from the point (℘(ω2),℘′(ω2)) = (e2, 0) to the points(℘(z−pi),℘′(z−pi))of the torus E′ under the standard cubic embedding into C2∪ {∞}, for i=1, . . . , l.
The only constraint remained is the zero order of g at z=0. Namely 0= g(0) =g′(0) = · · · =g(l−1)(0).
The proof starts by noticing that 2g(0) =
∑
℘(℘p′(pi)i) −e2 =:
∑
s(pi)is the first (degree one) symmetric polynomial of the slops s(pi). It is rea-sonable to expect that some of the higher derivatives g(m)(0)are also higher
degree symmetric polynomials of slops. The expectation turns out to be true only for m even and for odd degree polynomials:
Proposition 2.1.2. The even order differentiation g(2j)(0), j= 0, . . . ,[l−21]from a basis of the odd degree symmetric polynomials in si’s up to degree l for l being odd and up to degree l−1 for l being even.
Proof. Consider the slop function s(z) = d
dzlog(℘(z) −e2) = ℘′(z)
℘(z) −e2
= −2ζ(z) +ζ(z+ω2) +ζ(z−ω2)
= −2(ζ(z) −ζ(z−ω2) −η2/2). (2.1.2)
By differentiating the last equation, we get 1
2s′(z) = ℘(z) − ℘(z−ω2)
= ℘(z) −e2− ℘(z) −µ e2
(2.1.3)
where we have used the half period formula with
µ= (e1−e2)(e3−e2) =e1e3− (e1+e3)e2+e22 =2e22+e1e3. Also
1
2s′′ = ℘′+ µ℘′ (℘ −e2)2
=s
℘−e2+ µ
℘−e2
.
(Notice the variations on signs with (2.1.3).) Then we have Lemma 2.1.3. The slope satisfies the ODE:
(2.1.4) s′′ = 1
2s3−6e2s.
Proof. We will compute s′′in a different way, namely (2.1.5) s′ = ℘′′
℘−e2 − ℘′℘′
(℘ −e2)2 = 6℘2− 12g2
℘−e2 −s2. It is elementary to see
6℘2− g22 =6(℘ −e2)2+12e2(℘ −e2) +6e22− g22 and
6e22− g22 =6e22+2(e1e2+e3e2+e1e3) =2(2e22+e1e3) =2µ.
Thus (2.1.5) becomes
s′ =12e2−s2+6(℘ −e2) + 2µ
℘−e2.
Then
s′′ = −2ss′+6℘′− ℘2µs−e2
= −24e2s+2s3−12s(℘ −e2) −℘4µs−e2
+6s(℘ −e2) − ℘2µs−e2
= −24e2s+2s3−6s℘−e2+ µ
℘−e2
= −24e2s+2s3−3s′′,
where the last equality is by (2.1.4). The lemma follows. To proceed to higher even derivatives, we notice that
(2.1.6) (sk)′′ = (ksk−1s′)′ = k(k−1)sk−2(s′)2+ksk−1s′′. By (2.1.3) and (2.1.4),
(s′)2=4
℘−e− ℘µ−e
2
=4
℘−e+ µ
℘−e
2
−16µ= s′′
s
2
−16µ
which is an even degree polynomial in s of degree 4 by Lemma 2.1.3. Thus (sk)′′ is odd in s of degree k+2 if k is odd. By induction we then have that s(2j)is a degree 2j+1 odd polynomial in s.
The proposition now follows easily from (2.1.7) 2g(2j)(0) =
l
∑
i=1
s(2j)(pi)
and general facts on symmetric polynomials.
Now we are ready to prove
Theorem 2.2. Let ρ = 4πl. If the developing map f satisfies the type I relation (1.3.2), then l is odd. Furthermore g(−z) = −g(z)and u(−z) =u(z).
Proof. Consider the polynomial S(x) =
l
∏
i=1
(x−s(pi)). By Proposition 2.1.2, the relations
0= g(0) =g′′(0) = · · · =g(2[l−21])(0)
lead to the vanishing of all odd symmetric polynomials of s(pi)’s in the expansion of S(x).
If l =2n, then S(x)consists of only even degrees and its roots s(pi)must appears in pairs. Without loss of generality we may assume that
(2.2.1) s(p1) = −s(pn+1), s(p2) = −s(pn+2), . . . , s(pn) = −s(p2n).
Notice that the slope equation
℘′(a)
℘(a) −e2
=s(a) = −s(b) = − ℘′(b)
℘(b) −e2
leads to b = −a or b = a+ω2. To see this, notice that under the cubic embedding z 7→ (℘(z),℘′(z)), s(a)is slope of the line ℓa connecting the images of z= ω2and z =a, with the unique third intersection point being z = −a−ω2 and s(−a−ω2) = s(a). Thus the slope function defines a branched double cover
s : E′ →P1(C).
(From (2.1.3), it has 4 branch points given by℘(z) =e2±√µ.)
In particular the line with slope −s(a) = s(−a) and passing through (e2, 0)must beℓ−a ≡ ℓa+ω2. That is, b= −a or b= a+ω2as claimed.
In our case (2.2.1), we must conclude pn+1 = −p1since p1+ω2= q1can not appear in pi’s. In the same way we conclude that
(2.2.2) pi = −pi+n, i=1, . . . , n.
In particular ∑ pi = 0. But this violates ∑ pi ≡ 12ω1 modulo Λ′ (which follows from g(z+ω2) = −g(z)in Lemma 1.4.5), hence l is odd.
For l = 2n+1, S(x)is a polynomial in odd degrees only. In particular there is a root x= 0 of S(x)and we may assume that s(p2n+1) =0 (namely p2n+1 = 12ω1or 12(ω1′ +ω′2) = 12ω1+ω2).
Consider the polynomial S(x)/x in pure even degrees, then in exactly the same manner as above we conclude that (2.2.2) still holds and
S(x) =x
n
∏
i=1
(x−s(pi))(x+s(pi)).
It is clear that now g(−z) = −g(z). Then f(−z) = f(z), which implies
that u is an even function.
2.3. The polynomial system. The remaining statements in Theorem 0.4 which have not been proved yet are that these p1, . . . , pn are determined by polynomial equations in℘(pi)’s.
Philosophically this follows easily from (2.1.1) and (2.1.3). Indeed it is clear that the odd order derivatives of g at z= 0 will involve only rational expressions with denominator being powers of℘(pi) −e2and with at most even derivatives℘(z)(2j)(pi) in the numerator (all expressions in −pi are transformed into expressions in pi). The latter can be written into polyno-mials in℘(pi)and thus the polynomial system is obtained.
Proof of Theorem 0.4. To write down the complete set of polynomial
The lemma follows from the cubic relations.
Now we set xi = ℘(pi), ˜xi = ℘(˜ pi) = ℘(pi+ω2)for i = 1, . . . , n. It is clear that(xi−e2)(˜xi−e2) =µ for all i=1, . . . , n.
During the following computations, we assume that p2n+1 = 12ω1 and pn+i = −pi for i= 1, . . . , n. For the other case p2n+1 = 12ω1+ω2, we could
the equation g′′′(0) =0 becomes
n
∑
i=1
x2i −
n
∑
i=1
˜x2i = −12(e21−e23).
This is the degree two equation (m=2) with c2 = −12(e21−e23).
The general case follows from Lemma 2.3.1. Suppose that g(2j+1)(0) =0 gives rise to a new polynomial relation ∑ni=1xji−∑ni=1 ˜xji = cj. A further double differentiation increases the degree of the polynomial in℘by one, hence it gives rise to a new relation ∑ni=1xij+1−∑ni=1 ˜xij+1 = cj+1, with the universal constant cj+1being determined by c1, c2, g2, g3recursively.
Therefore, we conclude that xi = ℘(pi), ˜xi = ℘(pi+ω2), i = 1, . . . , n, satisfy the polynomial system:
n
∑
i=1
xji−
n
∑
i=1
˜xij =cj, j=1, . . . , n, (xi−e2)(˜xi−e2) =µ, i=1, . . . , n, which is easily seen to be equivalent to the system (0.4.1).
Conversely, any solution of the polynomial system gives rise to a func-tion g which satisfies
g(j)(0) =0, j=0, 1, . . . , 2n.
From g, the developing map f is then constructed by Proposition 1.4.8. Remark 2.3.2. In the next section we will prove that except for a finite set of tori, the mean field equation (0.1.3) has exactly n+1 solutions for ρ = 4πl with l = 2n+1. This implies that, except for those tori, the above polynomial system has exactly n+1 solutions up to permutation symmetry by Sn. Equivalently it has(n+1)! solutions.
Since cj(τ)’s are all holomorphic in τ, solutions(xi(τ), ˜xi(τ))of the poly-nomial system, hence the developing map f(z; τ), should then depend on τ holomorphically. It is not so obvious how to prove the holomorphic de-pendence of f(z; τ)in the moduli space of tori by other methods.
Example 2.4. For ρ=4π, l =1 and n =0. Then p1 = 12ω1. The polynomial system is empty and the solution u is unique. This was first proved in [42].
Example 2.5. Consider the case ρ = 12π, i.e. l = 3 and n = 1. Let p1 = a.
p2 = −a and p3= 12ω1. Then the equation g′(0) =0 becomes 2
(℘(a) −e2) −℘(a) −µ e2
+ (e1−e3) =0.
That is, we get a degree 2 polynomial in℘(a):
(℘(a) −e2)2+12(e1−e3)(℘(a) −e2) −µ=0
and then
℘(a) =e2+14(e3−e1) ± 14 q
(e3−e1)2+16(e1−e2)(e3−e2).
These are exactly the solutions obtained in [43] via a different method.
In particular there are precisely two solutions of the mean field equation on any torus E with non-zero discriminant(e3−e1)2+16(e1−e2)(e3−e2) 6=0 for the double cover E′, and with ρ=12π. The case with zero discriminant will be discussed in Example 3.6.
Example 2.6. Consider the case ρ = 20π, i.e. l = 5 and n = 2. The full set of polynomial equations in xi’s and ˜xi’s is given by
x1+x2− ˜x1− ˜x2=c1= −12(e1−e3), x21+x22− ˜x21− ˜x22=c2= −12(e21−e23), (x1−e2)(˜x1−e2) =µ,
(x2−e2)(˜x2−e2) =µ.
Now the number of solutions Nn′ (here n = 2) for x1.x2, ˜x1, ˜x2 can be calculated by the Bezout theorem to be N2′ = 1×2×2×2−r∞2 = 8−r2∞
where r∞2 is the number of solutions at ∞, counted with multiplicity, of the projectivized system of polynomial equations. The projective system is
X1+X2−X˜1−X˜2=c1X0, X12+X22−X˜21−X˜22=c2X02, (X1−e2X0)(X˜1−e2X0) =µX02, (X2−e2X0)(X˜2−e2X0) =µX02. And the infinity solutions are given by setting X0 =0:
X1+X2= X˜1+X˜2, X21+X22= X˜12+X˜22, X1X˜1=0, X2X˜2=0.
This shows that{X1, X2} = {X˜1, ˜X2}. Since these four variables are not all zero, it is easy to see that there are precisely two solutions given by
P1: X1=0=X˜2, X2 =X˜1 6=0, P2: X2=0=X˜1, X1 =X˜2 6=0.
It remains to compute the multiplicity of P1 and P2. Consider P1 first.
Since it is in the chart ˜U1 := {X˜1 6= 0}, in terms of yi = Xi/ ˜X1, i = 1, 2,
˜y2=X˜2/ ˜X1and y0 =X0/ ˜X1, P1has coordinates(y0, y1, y2, ˜y2) = (0, 0, 1, 0)
and the system at point P1reads as fi =0, i=1, . . . , 4, where f1=y1+y2−1− ˜y2−c1y0,
f2=y21+ (y2−1)2+2(y2−1) −˜y22−c2y20, f3= (y1−e2y0)(1−e2y0) −µy20=y1+ · · · ,
f4= (y2−e2y0)(˜y2−e2y0) −µy20= (y2−1)˜y2+ ˜y2+ · · · .
From these expressions, the appearance of degree one monomial in each fi shows that the local analytic coordinates (y0, y1, y2−1, ˜y2)at the point P1 can be replaced by f1, f3, f2, f4 accordingly, and thus the multiplicity is one. Indeed P1 = (0, 0, 1, 0)is a simple point of{fi =0}by computing the Jacobian
det ∂(f1, f2, f3, f4)
∂(y0, y1, y2, ˜y2)(0, 0, 1, 0) =e1−e36=0.
Similarly the multiplicity at P2is one. Thus r∞2 =2 and N2′ =8−2=6.
Since any reordering of pi’s leads to the same solution, also it is easy to see that for generic tori we do not have any solution with x1 = x2, so finally
N2= N2′/2!=3=2+1.
Remark 2.7. The above method can be extended to the case n=3, ρ =28π to show that N3 = 4 since in this case the infinity solutions are still zero dimensional. It fails for n ≥ 4 since positive dimensional intersections at infinity do occur and excess intersection theory is needed. The cases n =4 and n =5 were recently settled in [41] where the infinity solutions are one dimensional.
3. Lam´e for type I: Finite monodromies In this section we prove Theorem 0.4.1 (c.f. Theorem 3.5).
3.1. From mean field equations to Lam´e. The second order equation (3.1.1) Lη,Bw := w′′(z) − (η(η+1)℘(z) +B)w(z) =0
is known as the Lam´e equation with two parameters η and B; the parameter η is called the index and B is the called the “accessary parameter”.
3.1.1. Recall that for any two linearly independent solutions w1and w2of a general second order ODE w′′ = Iw, the Schwarzian derivative
S(h) = h′′′
h′ −32
h′′
h′
2
of h = w1/w2 satisfies S(h) = −2I, hence for any two linear independent local solutions w1, w2of the Lam´e equation (3.1.1) we have
S(ww1
2) = −2(η(η+1)℘(z) +B).
Conversely if h1is meromorphic function with S(h1) = −2(η(η+1)℘(z) + B, then S(h1) = S(ww1
2)for a chosen pair of linearly independent solutions
w1, w2of (3.1.1), therefore h1 is equal to a linear fractional transformation of ww12, or equivalently there exists a pair of linearly independent solutions w3, w4of (3.1.1) such that h1 =w3/w4.
3.1.2. Suppose that u is a solution o f the mean field equation
(3.1.2) △u+eu =ρδ0
on a flat torus E= C/Λ, Λ =Zω1+Zω2, and f is a developing map of u on a covering space of the punctured torus Er{0}. Locally u is expressed in f via
u(z) =log 8|f′(z)|2 (1+ |f(z)|2)2. Let η :=ρ/8π. By (1.1.4), we have
S(f):= f′′′
f′ − 3 2
f′′
f′
2
=uzz−12u2z = −2η(η+1)1
z2 +O(1), (3.1.3)
where the last equality follows from the asymptotic expansion u∼4η log|z|
at z = 0 and that u is smooth outside z = 0 in E. The expression of S(f) in u shows that it is a meromorphic function on E, which is holomorphic outside{0} and its polar part at z = 0 is given by the last expression in (3.1.3). Therefore there exists a constant B=B(E, η, u)such that
(3.1.4) S(f) =uzz− 12u2z = −2(η(η+1)℘(z) +B).
It follows that there exists two linearly independents solutions w1and w2
of the Lam´e equation (3.1.1) with accessary parameter B(E, η, u)such that f =w1/w2.
3.1.3. The Lam´e equation (3.1.1) had been studied in the classical litera-ture in two special cases, very extensively in case when the index η is a positive integer, and somewhat less so in the case when the index η is a half-integer, i.e. 2η = 2n+1 is an odd positive integer. We have seen in the previous sections that the former case corresponds to type II solutions while the latter case is for type I solutions. The main objective of this sec-tion is to prove that for any odd positive integer 2n+1, on all but a finite number of isomorphism classes of elliptic curves, there are precisely n+1 solutions to the mean field equation △u+eu =4π(2n+1)δ0.
The following theorem is due to Brioschi [7], Halphen [26, pp. 471–473]
and Crawford [18] in the late nineteenth century; see for [18] for a complete proof. See also [53, pp. 162–164] for a succinct presentation of Halphen’s transformation and Crawford’s procedure for analyzing Brioschi’s solu-tion; c.f. [67, p. 570].
Theorem 3.2. Let n be a non-negative integer.
(a) There exists a monic polynomial pn(B; Λ) = pn(B, g2(Λ), g3(Λ))of de-gree n+1 in B with coefficients in Z[g2(4Λ); g3(4Λ)] such that the Lam´e equation Ln+1/2,Bw=0 on C/Λ has all solutions free from logarithm at z = 0 if and only if pn(B) = 0. This polynomial pn(B, g2, g3) ∈ Z[12][B, g2, g3] is homogeneous of weight n if B, g2, g3 are given weights 1, 2, 3 respectively.
(b) For any lattice Λ outside a finite subset Snof homothety classes of lattices in C, the polynomial pn(B; Λ)has n+1 distinct roots.
Proof. The logarithm-free solutions of the Lam´e equation Ln+1/2,Bw=0 were first discovered by Brioschi [7, p. 314], but the underlying structure are more transparently exhibited using Halphen’s transformation [26, p. 471]
as carried out in detail by Crawford [18]. The statement (a) is proved in [18]; see also [53, p. 164] for a presentation of Crawford’s proof. A slightly different proof of (a) following the same train of ideas can be found in [3, p. 26–28].
Crawford’s proof provides a recursive formula for pn(B; Λ). When Λ is of the form Z+√
−1aZ with a ∈ R>0, this recursive formula also pro-duces a Sturm sequence starting with pn(B), therefore pn(B; Λ) has n+1 distinct real roots; see [18, p. 94].16This implies that the discriminant of the polynomial pn(B; Λ), which is a modular form for SL2(Z), is not identi-cally 0. The statement (b) follows. See§3.3 for remarks on Sturm’s theorem
used in Crawford’s proof.
Remark 3.2.1. We will give an alternative proof of part (a) of Theorem 3.2 in§3.4, which is essentially local near z = 0. Our proof not only provides a new construction of the polynomial pn(B), it also generalizes to the case with multiple singular sources. This generalization will be presented in a later work; c.f. [10].
3.3. Remark on Sturm’s theorem. Crawford’s proof in [18, p. 94] that the polynomial pn(B; Λ) has n+1 distinct real roots for rectangular tori uses a fact closely related to Sturm’s theorem on real roots of polynomials over R, not found in standard treatment of this topic, such as [64, 11.3] and [32, 5.2].17We have been able to find only one reference of this fact, as a “starred
16The statement that pn(B) has n+1 distinct real roots was proved in [18, p. 94] un-der the condition that the x-coordinates of the three non-trivial two-torsion points, ei =
℘(ωi/2; Λ)for i=1, 2, 3, are real numbers. This is the case when the lattice Λ is of the form Λτwith τ∈√
−1R>0.
17This fact must be familiar to all educated scientists in the late nineteenth and early twentieth century, often used freely without comments in mathematical writings at the time. This is the case for the proofs in [67, p. 557] and [53, p. 163] for the existence of 2m+1 distinct real roots of the polynomial lm(B)corresponding to 2m+1 Lam´e functions for the equation
d2w
dz2 − (m(m+1)℘(z; Λ) +B)w=0
exercise” in [63, p. 149 ex. 30]. In Proposition 3.3.3 below we provide a mild generalization of the usual form of Sturm’s theorem for the convenience of the readers. Its corollary 3.3.4 is equivalent to [63, p. 149 ex. 30].
Definition 3.3.1. A sequence of non-zero polynomials f0(x), f1(x), . . . , fm(x) ∈R[x]
is a Sturm sequence on(a, b]if the following two properties hold.
(i) fm(x)is either positive definite or negative definite on(a, b].
(ii) Suppose that ξ ∈ (a, b] and fi(ξ) = 0 for some i with 1 ≤ i ≤ m−1.
Then fi−1(ξ)and fi+1(ξ)have opposite signs (in the sense that either they are both non-zero with opposite signs, or are both zero.18)
Remark. There is an extra condition in the conventional definition of a Sturm sequence: f1(ξ) and f0(ξ) have the same sign for every root ξ of f0(x) in(a, b]. This condition has been dropped in Definition 3.3.1 above.
3.3.2. Definition. Let f0(x), f1(x), . . . , fm(x)be a Sturm sequence.
(1) For every real number ξ, define σ(ξ) to be the total number of changes of signs in the sequence(f0(ξ+), f1(ξ+), . . . , fm−1(ξ+), fm(ξ)).19
(2) Define a{−1, 0, 1}-valued “local index” function ǫf0(x)on R attached to a real polynomial f0(x) ∈R[x]as follows.
• Suppose that f0(ξ) =020and multx=ξ f0(x)is odd.21Define ǫf0(x)(ξ):=
1 if f0(ξ+)and f1(ξ)have the same sign
−1 if f0(ξ+)and f1(ξ)have opposite signs
• ǫf0(x)(ξ) = 0 if multx=ξ f0(x) is even. In particular ǫf0(x)(ξ) = 0 if f0(ξ) 6= 0.
(3) Define Zf0(x)((a, b]) ∈Zby
Zf0(x)((a, b]):=
∑
ξ∈(a,b]
ǫf0(x)(ξ).
This number Zf0(x)((a, b]) counts the number of zeros of f0(x) with odd multiplicity with a signed weight given by ǫf0(x). It can be thought of as some sort of “total Lefschetz number” for f0(x)|(a,b].
when Λ=Z+√
−1aZ for some a∈R>0and m∈N>0. However this then-well-known fact is no longer part of the general education for mathematicians today.
18The latter possibility is ruled out by condition (i).
19Here we used fi(ξ+)to make sure that each term has a well-defined sign. In view of condition (ii), we could have used the sequence(f0(ξ+), f1(ξ), . . . , fm−1(ξ), fm(ξ))in the definition, suppress zeros when counting the number of variations of signs in it.
20f1(ξ) 6=0 if f0(ξ) =0, by (i) and (ii).
21For a zero ξ of f0(x), the sign of f0(x)changes when x moves across ξ if and only if multx=ξf0(x)is odd.
Proposition 3.3.3. Let f0(x), f1(x), . . . , fm(x) be a Sturm sequence on (a, b]. Then
Zf0(x)((a, b]) =σ(a) −σ(b),
i.e. σ(a) −σ(b)is the number of zeros of f0(x)in the half-open interval(a, b]with odd multiplicity, counted with the sign ǫf0(x).
Proof. Condition (ii) ensures that crossing a zero in[a, b)of any of the inter-nal members f1(x), . . . , fm−1(x)of the Sturm chain makes no contribution to changes of σ(ξ). Each time a zero ξ0 of f0(x)with odd multiplicity is crossed, σ(ξ) decreases by ǫf0(x)(ξ) as ξ moves from the left of ξ0 to its right. On the other hand, moving across a zero of f0(x)with even multi-plicity does not change the value of σ. So the σ(b) −σ(a)is equal to the total number of zeros of f0(x)in(a, b]with odd multiplicity, counted with
the sign ǫf0(x).
Corollary 3.3.4. Let f0(x), . . . , fm(x) be a Sturm sequence on(a, b]. Let n ∈ N be a non-negative integer. If σ(a) −σ(b) = ±n and f0(x) has at most n distinct real roots in(a, b], then f0(x)has exactly n distinct real roots in the half-open interval (a, b]. In particular if a = −∞, b = ∞, deg(f0(x)) = n and σ(−∞) −σ(∞) = ±n, then f0(x)has n distinct real roots.
3.4. A proof of Theorem 3.2 (a). Let’s start with any f as the quotient of two independent solutions of Lam´e equation Ln+1/2,Bw = 0 at z = 0 and consider v(z) =log f′(z). It is readily seen that
v′′−12(v′)2 =f′′
f′
′
− 12ff′′′2= S(f).
We remark that the function v satisfies the similar equation as u in (3.1.3), but v is analytic in nature while u is only a real function.
The indicial equation at z = 0 is given by λ2−λ−η(η+1) = (λ− (η+1))(λ+η) = 0. If there are logarithmic solutions, the fundamental solutions are given as
(3.4.1) w1(z) =zη+1h1(z), w2(z) =ξw1(z)log z+z−ηh2(z), where ξ 6=0 and h1, h2are holomorphic and non-zero at z=0. But then
f = aw1+bw2
cw1+dw2
is easily seen to be logarithmic as well if ad−bc 6=0, thus the Lam´e equa-tion has no logarithmic soluequa-tions at z = 0 if and only if we have one non-trivial solution quotient f to be logarithmic free at z=0.
Now suppose that the Lam´e equation has no solutions with logarithmic term. Let f be a ratio of two independent solutions. Without lose of gener-ality, we may assume that f is regular at 0. Since η= n+ 12 and
(3.4.2) S(f) = −2((n+12)(n+32)℘(z) +B),
to require that f is logarithmic free at z = 0 is equivalent to that f(z) = c0+c2n+2z2n+2+ · · · near z=0 with c06=0.
Recall that
℘(z) = 1 z2 +
∑
k≥1
(2k+1)Gk+1z2k
where Gk = ∑ω∈Λ∗1/ω2k is the standard Eisenstein series of weight 2k for SL2(Z). It is customary to write g2 = 60G2 and g3 = 140G3. It is also well known that all Gk’s are expressible as polynomials in g2, g3.
We will show that the solvability of the Schwarzian equation (3.4.2) for f being of the proposed form is equivalent to that B satisfies pn(B) =0 for some universal polynomial pn(B, g2, g3)of degree n+1. Indeed,
v=log f′ =log c2n+2(2n+2) + (2n+1)log z+
∑
j≥1
djzj.
For convenience we set ej = (j+1)dj+1 for j≥0 and then v′ = 2n+1
z +
∑
j≥0
ejzj
The degree z−1terms in
v′′−12(v′)2 = −2((n+ 12)(n+ 32)℘(z) +B)
match by our choice. There is no z−1 term in the RHS shows that e0 = 0.
Then the constant terms give e1− 122(2n+1)e1 = −2B, i.e. ne1 = B. For n=0, we must conclude B=0. Thus we set p0(B) =B.
Similarly, for j≥1, the degree j terms in the LHS give (j+1)ej+1−122(2n+1)ej+1−12
j−1
∑
i=1
eiej−i.
Since there is no odd degree terms in the RHS, by considering j=1, 3, 5, . . . we first conclude inductively that ei =0 for i even.
Next we consider degree j = 2, 4, 6, . . . terms inductively. Write Ek = e2k−1for k ≥1. Then j=2k leads to
(3.4.3) 2(k−n)Ek+1−12
k
∑
i=1
EiEk+1−i = −2(n+ 12)(n+32)(2k+1)Gk+1. We have just seen that nE1 = B. If we assign degree k to Gk, then (3.4.3) shows inductively that Ek = Ek(B, g2, g3) is a degree k polynomial in B which is homogeneous in B, g2, g3of degree k up to k≤ n.
Now put k=n in (3.4.3), the first term vanishes and we must have
˜pn(B, g2, g3):=
n
∑
i=1
EiEn+1−i−8(n+12)2(n+32)Gn+1
vanishes too. Up to a multiplicative constant, this ˜pn(B) is the degree n+1 polynomial in B we search for. Indeed, by our inductive construc-tion through (3.4.3), the leading coefficients cnof ˜pn(B)depends only on n.
Hence pn(B, g2, g3) := c−n1˜pn(B, g2, g3)is monic in B and homogeneous of degree n+1 in B, g2, g3.
Conversely, if ˜pn(B) = 0, then E1, . . . , En can be solved by (3.4.3) up to k = n−1. For k = n−1, ˜pn(B) = 0 is equivalent to (3.4.3) at k = n. By assigning any value to En+1, we can use (3.4.3) for k ≥ n+1 to find Ej, j≥ n+2. Thus this f is a solution to the Schwarzian equation (3.4.2) and is free from logarithmic terms. The proof is complete. Remark 3.4.1. Notice that En+1 =e2n+1 = (2n+2)d2n+2is a free parameter.
All Ek’s are determined by B and En+1. For any B with pn(B) =0, the three constants c0, c2n+2and En+1provide the three dimensional freedom for f due to the freedom of SL2(C)action on f .
Remark 3.4.2. We have seen that the type I developing map f(z)is even.
This also follows form our proof of Theorem 3.2 since we do not assume the a priori evenness during the proof.
To apply Theorem 3.2 to study mean field equations for ρ=4π(2n+1), the essential point is the following theorem.
Theorem 3.5(=Theorem 0.4.1). Let n be a non-negative integer. The projective monodromy group of the Lam´e equation Ln+(1/2),Bw=0 is isomorphic to Klein’s four-group(Z/2Z)2if and only if there exists two meromorphic solutions w1, w2
on C of the above Lam´e equations such that ww12 is a type I developing map of a so-lution of the mean field equation △u+eu=4π(2n+1)δ0. Moreover, each such value of the accessary parameter B with the above property gives rise to exactly one type I solution.
Proof. Let u be a type I solution of the mean field equation △u+eu = ρ δ0
on C/Λ and let f be a normalized developing map of u satisfying the type I transformation rules (1.3.2). We know from Theorem 2.2 that there exists a non-negative integer n such that ρ =4π(2n+1), and we have seen that there exists a complex number B such that the Schwarzian derivative S(f) of f is equal to −2 n+ 12)(n+32)℘(z; Λ) +B. Then local solutions of the Lam´e equation Ln+1/2,Bw = 0 are free of logarithmic solutions, and there exists two solutions w1, w2over C such that f = ww1
2. The projective mon-odromy group of the equation Ln+1/2,Bw= 0 is canonically isomorphic to the monodromy group of the meromorphic function ww12, which is a Klein four group. We have proved the “only if” part of Theorem 0.4.1.
Conversely, suppose that the projective monodromy group of a Lam´e equation Ln+1/2,Bw = 0 is a Klein-four group. Then all local solutions of this Lam´e equation are free of logarithmic singularities, and there are for two linearly independent solutions w1, w2 of this equation which are
meromorphic functions over C. It is easy to check from basic theory of linear ODE’s with regular singularities that the holomorphic map ww12 : C → P1(C) has no critical point outside Λ, and has multiplicity 2n+2 at points of Λ.
Let ρ : Λ → GL2(C) be the monodromy representation of the differ-ential equation Ln+1/2,Bw = 0 attached to the basis w1, w2 of solutions of Ln+1/2,Bw = 0. Let ¯ρ : Λ → PSL2(C)be the composition of ρ with the
Let ρ : Λ → GL2(C) be the monodromy representation of the differ-ential equation Ln+1/2,Bw = 0 attached to the basis w1, w2 of solutions of Ln+1/2,Bw = 0. Let ¯ρ : Λ → PSL2(C)be the composition of ρ with the