Table Of Content

POLYNOMIALS REPRESENTING EYNARD-ORANTIN INVARIANTS 0 PAULNORBURYANDNICKSCOTT 1 0 2 Abstract. The Eynard-Orantin invariants of a plane curve are multilinear n differentialsonthecurve. Foraparticularclassofgenuszeroplanecurvesthese a invariantscanbeequivalentlyexpressedintermsofsimplerexpressionsgiven J bypolynomialsobtainedfromanexpansionoftheEynard-Orantininvariants around a point on the curve. This class of curves contains many interesting 4 examples. ] G A . 1. Introduction h at As a tool for studying enumerative problems in geometry Eynard and Orantin m [3] associate multilinear differentials to any compact Riemann surface C equipped with two meromorphic functions x and y with the property that the branch points [ of x are simple and the map 1 v C C2 → 9 p (x(p),y(p)) 4 7→ 4 is an immersion. For every (g,n) Z2 with g 0 and n > 0 they define a 0 multilineardifferential,i.e. atensorp∈roductofmero≥morphic1-formsontheproduct 1. Cn, notated by ωng(p1,...,pn) for pi ∈ C. When 2g −2+n > 0, ωng(p1,...,pn) 0 is defined recursively in terms of local information around branch points of x of 0 ωg′(p ,...,p ) for 2g′ 2+n′ < 2g 2+n. A dilaton relation between ωg and 1 ωng′ca1n be anpplied in t−he n=0 case−to define Fg (= “ωg”) for g 2, knownn+a1s the v: synmplectic invariants of the curve. The invariant Fg re0cursively≥uses all ωg′ with i n X g′+n g+1. ≤ r In principle Fg and ωg can be calculated explicitly from the recursion relations n a defining them, and implemented on a computer. In practice, the expressions ob- tained this way are unwieldy and computable only for small g and n. The main aim of this paper is to express ωg in a simpler form—essentially its inverse discrete n Laplace transform—and to develop methods to calculate general formulae for Fg in some examples. We consider the Eynard-Orantin invariants of the class of genus zero curves for which x is two-to-one and has two branch points. The case of one branch point x=z2 is dealt with in [3]. One can parametrise the domain so that the curve can be written: x=a+b(z+1/z) (1) C = (y =y(z) 1991 Mathematics Subject Classification. MSC(2010)32G15;30F30;05A15. 1 2 PAULNORBURYANDNICKSCOTT for constants a, b and any rational function y(z) with y′( 1) = 0. The definition ± 6 ofωg anditspropertiesrequirestheRiemannsurfacetobecompact. Nevertheless, n by taking sequences of compact Riemann surfaces one can extend the definition to allowy(z)tobeanyanalyticfunctiondefinedonadomaininCcontainingz = 1, ± e.g. y(z)=ln(z) defined on the complement of the negative imaginary axis. Theorem 1. Fortheplanecurve C defined in (1) and2g 2+n>0, ωg(z ,...,z ) − n 1 n has an expansion around z =0 given by i { } d d (2) ωg(z ,..,z )= ... Ng(b ,...,b )zb1...zbndz ...dz n 1 n dz dz n 1 n 1 n 1 n 1 n bXi>0 whereNg isaquasi-polynomialintheb2 ofhomogeneousdegree3g 3+n,dependent n i − on the parity of the b and is symmetric in all variables of the same parity. i The parity dependence means that there exists polynomials Ng (b ,...,b ) for n,k 1 n k =1,...,n such that Ng(b ,...,b ) decomposes n 1 n Ng(b ,...,b )=Ng (b ,...,b ), k =number of odd b . n 1 n n,k 1 n i ThepolynomialsrepresentingNg aresimplerexpressionsthanωg(z ,...,z )and n n 1 n canhavemeaningthemselves. Whenx=z+1/z andy =z,Ng arisesasasolution n of a Hurwitz problem [7, 8], with typical expression N0 = 1(b2+b2+b2+b2) 1 4,0 4 1 2 3 4 − (while the much larger expression ω0 can be expressed as the sum of 32 rational 4 functions.) The case x=z+1/z and y =lnz arises when studying partitions and the Plancherel measure [2], with N0 = 1 b2+b2+b2+b2 . 4,0 4 1 2 3 4 Theorem 2. The coefficients of the top h(cid:0)omogeneous degre(cid:1)e terms in the polynomial Ng (b ,...,b ), defined above, can be expressed in terms of intersection num- n,k 1 n bers of tautological line bundles over the moduli space L . For β = i → Mg,n i i 3g 3+n, the coefficient v of b2βi is − β i P y′(1)2−2g−n+( 1)Qky′( 1)2−2g−n v = − − c (L )β1...c (L )βn β x′′(1)2g−2+n23g−3+nβ! 1 1 1 n ZMg,n The invariants ωg satisfy string and dilaton equations—see Section 4—which n enable one to express Ng recursively in terms of Ng. They are most easily n+1 n expressed when y(z) is a monomial. The following two theorems apply to the polynomials Ng although we abuse notation and simply write Ng. n,k n Theorem 3. The polynomials associated to the plane curves x=z+1/z, y =zm/m, m=1,2,... satisfy the following recursion relations. n bj (3) Ng (m,b ,...b )= kNg(b ,...,b ) n+1 1 n n 1 n |bj=k j=1k=1 XX (4) (m+1)Ng (m+1,b ,...,b )+(m 1)Ng (m 1,b ,...,b ) n+1 1 n − n+1 − 1 n n bj n =2m kNg(b ,...,b ) m b Ng(b ,...,b ) n 1 n |bj=k− j n 1 n j=1k=1 j=1 XX X POLYNOMIALS REPRESENTING EYNARD-ORANTIN INVARIANTS 3 (5) (m 2)Ng (m 2,b ,...,b ) 2mNg (m,b ,...,b ) − n+1 − 1 n − n+1 1 n n +(m+2)Ng (m+2,b ,...,b )=m kNg(b ,...,b ) n+1 1 n n 1 n |bj=k Xj=1k=X1±bj (6) Ng (m+1,b ,...,b ) Ng (m 1,b ,...,b )=m(2g 2+n)Ng(b ,...,b ) n+1 1 n − n+1 − 1 n − n 1 n Corollary 1. Thesymplecticinvariantsofthecurvex=z+1/z,y =zm/msatisfy 1 Fg = (Ng(m+1) Ng(m 1)) m(2g 2) 1 − 1 − − for g 2. ≥ Thepolynomialscorrespondingtoy =lnz satisfyrecursionsobtainedbysetting m=0 into (3) and ((4) (6))/m. − Theorem3isaspecialcaseofthefollowingmoregeneraltheoremthatappliesto any analytic function y(z) defined on a domain in C containing z = 1 expanded ± as y(z) (a +zb )(1 z2)k. For example y(z)=lnz (1−z2)k. ∼ k k − ∼ −2k First we need the following notation. Define the operator on functions by P PD f(n)=f(n+1). D Further, define f(n) =f(a+1). As usual, for a polynomial p(z)= p zi, n=a i D{ } define p( ) f(n) = p f(a+i). Note that I is a discrete derivative and n=a i D { } D− P y( ) (a +zb )(1 2)k isasumoverpowersofthediscretederivative 2 I. k k D ∼ −DP D − Put b =(b ,...,b ) and b =b +...+b . S 1 n S 1 n P | | Theorem 4. For the plane curve x=z+1/z, y =y(z) n bj (7) y( ) mNg (m,b ) = kNg(b ) D D n+1 S m=−1 n S |bj=k (cid:8) (cid:9) Xj=1k6≡kX=bj1(2) n bj (8) (1+ 2)y( ) mNg (m,b ) =2 kNg(b ) b Ng(b ) D D n+1 S m=−1 n S |bj=k−| S| n S (cid:8) (cid:9) Xj=1k≡kX=bj1(2) n (9) (1 2)2y( ) mNg (m,b ) = kNg(b ) −D D n+1 S m=−2 n S |bj=k (cid:8) (cid:9) Xj=1k=X1±bj (10) (1 2)y( ) Ng (m,b ) =(2 2g n)Ng(b ) −D D n+1 S m=−1 − − n S Although y may n(cid:8)ot be a polyn(cid:9)omial the left hand sides of (7) - (10) are fi- nite sums, since large enough powers of a discrete derivative vanish on the quasi- polynomial Ng . See Section 2. n+1 Corollary 2. The symplectic invariants of the curve x=z+1/z, y =y(z) satisfy 1 Fg = (1 2)y( ) Ng(m) 2 2g −D D { 1 }m=−1 − for g 2. ≥ ThedefinitionoftheEynard-OrantininvariantsisgiveninSection2. Theproofs of Theorems 1 and 2 are in Section 3 and the proof of Theorems 3 and 4 is in Section 3. Section 5 contains examples. 4 PAULNORBURYANDNICKSCOTT 2. Eynard-Orantin invariants. For every (g,n) Z2 with g 0 and n > 0 the Eynard-Orantin invariant of ∈ ≥ a plane curve C is a multilinear differential ωg(p ,...,p ), i.e. a tensor product of n 1 n meromorphic 1-forms on the product Cn, where p C. When 2g 2+n > 0, i ∈ − ωg(p ,...,p ) is defined recursively in terms of local information around the poles n 1 n of ωg′(p ,...,p ) for 2g′+2 n′ <2g 2+n. Equivalently, the ωg′(p ,...,p ) are n′ 1 n − − n′ 1 n usedaskernelsontheRiemannsurfacetointegrateagainst. Thisisafamiliaridea, the main example being the Cauchy kernel which gives the derivative of a function in terms of the bilinear differential dwdz/(w z)2 as follows − f(w)dwdz f(w)dwdz f′(z)dz = Res = Res (w z)2 − (w z)2 w=z w=α − α − X where the sum is over all poles α of f(w). The Cauchy kernel generalises to a bilinear differential B(w,z) on any Riemann surface C given by the meromorphic differential η (z)dz unique up to scale which w has a double pole at w C and all A-periods vanishing. The scale factor can be ∈ chosen so that η (z)dz varies holomorphically in w, and transforms as a 1-form in w w and hence it is naturally expressed as the unique bilinear differential on C dwdz B(w,z)=η (z)dwdz, B =0, B(w,z) near w =z. w ∼ (w z)2 IAi − It is symmetric in w and z. We will call B(w,z) the Bergmann Kernel, following [3]. It is called the fundamental normalised differential of the second kind on C in [4]. Recall that a differential is normalised if its A-periods vanish and it is of the second kind if its residues vanish. It is used to express a normalised differential of the second kind in terms of local information around its poles. For2g 2+n>0,thepolesofωg(p ,...,p )occuratthebranchpointsofx,and − n 1 n theyareoforder6g 4+2n. Sinceeachbranchpointαofxissimple,foranypoint − p C close to α there is a unique point pˆ= p close to α such that x(pˆ) = x(p). ∈ 6 The recursive definition of ωg(p ,...,p ) uses only local information around branch n 1 n points of x and makes use of the well-defined map p pˆthere. The invariants are 7→ defined as follows. ω0 =ydx 1 (11) ω0 =B(w,z) 2 For 2g 2+n>0, − (12) ωg (z ,z )= ResK(z ,z) ωg−1(z,zˆ,z )+ ωg1 (z,z )ωg2 (zˆ,z ) n+1 0 S 0 n+2 S |I|+1 I |J|+1 J α z=α (cid:20) (cid:21) X g1+Xg2=g I⊔J=S where the sum is over branch points α of x, S = 1,...,n , I and J are non-empty { } and zB(z ,z′) K(z ,z)= − zˆ 0 0 2(y(z) y(zˆ))dx(z) R− POLYNOMIALS REPRESENTING EYNARD-ORANTIN INVARIANTS 5 is well-defined in the vicinity of each branch point of x. Note that the quotient of a differential by the differential dx(z) is a meromorphic function. The recursion (12) depends only on the meromorphic differential ydx and the map p pâround 7→ branch points of x. 2.1. Rationalcurveswithtwobranchpoints. Thesimplestexampleofaplane curvewithnon-trivialEynard-Orantininvariantsistherationalcurvey2 =xwhere the meromorphic function x defines a two-to-one branched cover with a single branch point. It is known as the Airy curve since the Eynard-Orantin invariants reproduceKontsevichsgeneratingfunction[6]forintersectionnumbersonthemod- uli space. In this paper we study rational curves such that x defines a two-to-one branched cover with two branch points. Lemma 2.1. If x is a two-to-one rational map on P1 with two branch points then we can parametrise the domain so that x=a+b(z+1/z). Proof. Using a conformal map we can arrange that the two branch points of x are z = 1. The conformal map z (z+λ)/(λz+1) which fixes z = 1 can be used ± 7→ ± to further arrange that x( ) = . Since x(z) x(1) has a double root at z = 1 ∞ ∞ − we have (z 1)2 x(z) x(1)= − , x( )= q =0, x′( 1)=0 q =0 − q z2+q z+q ∞ ∞⇒ 2 − ⇒ 0 2 1 0 so put a=x(1) 2/q and b=1/q , and the result follows. (cid:3) 1 1 − Thusx(z)=a+b(z+1/z)andy(z)isanyanalyticfunctiondefinedonadomain in C containing z = 1 and satisfying y′( 1) = 0. When y is not polynomial, for ± ± 6 example y is rational or transcendental, we expand it as a series of polynomials in the following non-standard way. Given such y(z), define the partial sums N y(N)(z)= (a +zb )(1 z2)k k k − k=0 X to agree with y(z) at z = 1 up to the Nth derivatives. One can achieve this ± by expressing y(z) = y (z)+y (z) where y (z) = 1/2(y(z) y( z)) and define + − ± ± − y(N)(z)= N a (1 z2)k where a are determined by the property + k=0 k − k P dky dky(N) +( 1)= + ( 1), k =0,...,N. dzk ± dzk ± Similarly define b from y (z)/z. k − The partial sums y(N)(z) do not necessarily converge to y(z). For example, { } (1 z2)k y(z)=lnz − ∼ 2k − is a divergent asymptotic expansion for lnX(z) at z =0 in the region Re(z2)>0. The partial sums y(N)(z) are used in the recursions defining ωg in place of y(z) n since they contain the same local information around z = 1 up to order N. More ± precisely, todefineωg forthecurve(x(z),y(z))itissufficienttouse(x(z),y(N)(z)) n for any N 6g 6+2n. ≥ − Therearevariouswaystoexpressatranscendentalfunctionasalimitofrational functions. The main benefit of the approach used here is the fact that the expres- sionswhichappearinthestringanddilatonequationsx(z)my(z)ωg havepolesonly n 6 PAULNORBURYANDNICKSCOTT at z = 1 and 0 and , allowing one to translate properties of ωg near z = 1 to ± ∞ n ± properties of ωg near z = , which is encoded by Ng. n ∞ n Theorem 4 involves the expression y( ) where is defined in the introduction, D D and in particular I is a discrete derivative. It is an easy fact (proved by − D induction) that for any degree d polynomial p(n), high enough discrete derivatives vanish: (1 )kp(n) 0 for k > d. Similarly, 1 2 is a discrete derivative, −D ≡ −D and for a parity dependent quasi-polynomial p(n) (so p(n)=p (n) for n even and + p(n)=p (n) for n odd, where p (n) are polynomials) − ± (1 2)kp(n) 0, k sufficiently large −D ≡ infactk >maximumdegreeofp (n). Tomakesenseof(7)-(10),onemustreplace ± y( ) with y(N)(D) for large enough N so that the left hand sides of (7) - (10) have D only finitely many terms. This procedure is well-defined, since the vanishing of discrete derivatives ensures that the left hand sides of (7) - (10) are independent of the choice of N when it is large enough. 3. Proofs Proof of Theorem 1. Theorem 1 reflects three main properties of the multilinear differential ωg(z ,...,z ) proven in [3]—it is meromorphic, with poles at z = 1 of n 1 n i ± order6g 4+2n=:2d+2andresidue0,andpossessessymmetryunderz 1/z . i i − 7→ Since all residues of ωg vanish, the integral n z1 zn g(z ,...,z )= ... ωg(z′,...,z′) Fn 1 n n 1 n Z0 Z0 isawell-definedmeromorphicfunctionthatvanisheswhenanyz =0andhaspoles i of order 2d+1 at z = 1. Write this rational function as i ± g(z ,...,z )= 0<ki<4d+2pkz1k1...znkn Fn 1 n n (1 z2)2d+1 P i=1 − i where the pk = pk1,...,kn ∈ C and the deQgree of the numerator is small enough to avoid a pole at infinity. The Taylor expansion 1 1 d2d ∞ ∞ m+2d = z2m = z2m (1 z2)2d+1 (2d)!d(z2)2d 2d − m=0 m=0(cid:18) (cid:19) X X has quasi-polynomial coefficients, meaning that the coefficients of zb are described by two polynomials in b—when b is odd the coefficient of zb is the zero polynomial and when b is even the coefficient of zb is a degree 2d polynomial in b. More generally, the Taylor expansion of g(z ,...,z ) about z =0 has quasi-polynomial Fn 1 n i coefficients, depending on parity. When n=1, p zk m+2d k = p z2m+k = Ng(b)zb. (1 z2)2d+1 k 2d 1 P− k,m (cid:18) (cid:19) b>0 X X The coefficient of zb consists of all terms where 2m+k = b, hence the odd part of p(z) = p zk gives rise to a degree 2d polynomial representing Ng(b) when b k 1 P POLYNOMIALS REPRESENTING EYNARD-ORANTIN INVARIANTS 7 is odd, and the even part of p(z) gives rise to a degree 2d polynomial representing Ng(b) when b is even. Similarly, 1 4d+2 n ∞ m +2d Fng(z1,...,zn)= pkz1k1...znkn i2d zi2mi k1,.X..,kn=0 iY=1mXi=0(cid:18) (cid:19) 4d+2 n m +2d = pk i z2mi+ki 2d i k1,.X..,kn=0mXi≥0 iY=1(cid:18) (cid:19) ∞ =: Ng(b ,...,b )zb1...zbn n 1 n 1 n bXi>0 expresses Ng(b ,...,b ) as the sum over the terms with 2m +k = b which is a n 1 n i i i quasi-polynomial depending on the parity of the b . By symmetry of the z , Ng i i n does not depend on which b are odd but only how many. Hence we write i Ng(b ,...,b )=Ng (b ,...,b ), k =number of odd b . n 1 n n,k 1 n i Each binomial coefficient and hence each polynomial Ng (b ,...,b ) is a polyno- n,k 1 n mial of degree 2d in each b . The stronger fact that they have homogeneous degree i 2d in the b is a consequence of Theorem 2. i ItremainstoshowthatNg isaquasi-polynomialintheb2. Equivalently,wewill n i show that b ...b Ng(b ,...,b ) is odd in each b using symmetries of 1 n n 1 n i ∞ ωg = b ...b Ng(b ,...,b )zb1−1...zbn−1dz ...dz . n 1 n n 1 n 1 n 1 n bXi>0 Lemma 3.1. A meromorphic 1-form on P1 with poles at z = 1 has the following ± related expansions around z =0 ∞ ∞ 1 (13) ω(z)= p(n)zn−1dz ω = p( n)zn−1dz ⇔ z − n=1 (cid:18) (cid:19) n=1 X X where p(n) is a quasi-polynomial depending on the parity of n. Proof. We can express ω as a rational function with numerator a polynomial of degreesmallenoughsothattherearenopolesatinfinity. Inparticular,bylinearity it is enough to prove the lemma when the numerator is a monomial so zkdz 1 z2m−kdz ω(z)= and ω =( 1)m . (1 z2)m+1 z − (1 z2)m+1 − (cid:18) (cid:19) − From the expansion 1 n+m = z2n (1 z2)m+1 m − (cid:18) (cid:19) X one gets zkdz n+m = zk+2ndz = p (b)zb−1dz (1 z2)m+1 m k − (cid:18) (cid:19) X X where 0, b k(mod 2) p (b)= ≡ . k (b−k−1)/2+m , b k(mod 2) (cid:26) m 6≡ (cid:0) (cid:1) 8 PAULNORBURYANDNICKSCOTT Also, (b 2m+k 1)/2+m (b+k 1)/2 p (b)= − − = − =( 1)mp ( b) 2m−k k m m − − (cid:18) (cid:19) (cid:18) (cid:19) (for b k(mod 2) and p (b)=0 for b k(mod 2) so the above equation holds 2m−k 6≡ ≡ for all b.) Hence (13) holds when ω has a monomial numerator and hence for all rational ω and the lemma is proven. (cid:3) An immediate corollary of the lemma is that if ω(z) = ω(1/z) then the quasi- polynomial p(n) is even in n and if ω(z) = ω(1/z) then p(n) is odd in n. A − consequence of a more general result in [3] is the symmetry ωg(z ,...,z )= ωg(1/z ,...,z ) n 1 n − n 1 n and similarly for each variable z . Hence b ...b Ng(b ,...,b ) is an odd quasi- i 1 n n 1 n polynomial in each b as required. (cid:3) i Remark. The expansion (2) of ωg(z ,...,z ) around (z ,...,z ) = (0,...,0) defines n 1 n 1 n Ng(b ,...,b ) only for (b ,...,b ) Zn. One can make sense of b = 0 using the ponlyn1omialnrepresentatio1n Ng n(b∈,...,+b ) of Ng(b ,...,b ) for k =inumber of odd n,k 1 n n 1 n b . In terms of ωg one has the following i n 0 Ng(0)= ωg(z) 1 1 Z∞ andmoregenerally, b ...b Ng(b ,...,b ,0,...,0)isthecoefficientofzb1−1...,zbk−1 in 1 k n 1 k 1 k the expansion around (z ,...,z )=(0,...,0) of 0 ... 0 ωg(z ,...,z ). 1 k zk+1=∞ zn=∞ n 1 n Proof of Theorem 2. The proof uses the behavRiour of ωgRnear the branch points n z = 1. Express ωg as a rational function i ± n ωg = k4d1+,..2.,kn=0ckz1k1...znkndz ...dz n n (1 z2)2d+2 1 n P i=1 − i forck C,andd=3g 3+n. QConsiderthechangeofvariableszi =ǫi+sxi where ǫ ∈1 , s R is sm−all and x is a local coordinate on the spectral curve. The i i ∈ {± } ∈ asymptoticbehaviourofωg nearz = 1correspondstos 0forallcombinations n i ± → of the ǫ . This change gives: i ωg = 4kd1+,..2.,kn=0ck(ǫ1+sx1)k1...(ǫn+sxn)knsnΠn dx , n s(2d+2)n n x2d+2(2ǫ +sx )2d+2 i=1 i P i=1 i i i and we must find a minimal qQ= q(ǫi) 0,1,...4d+2 so that the coefficient of ∈ { } sq in the numerator is the first non vanishing. For example, if q >0 this tells us 4d+2 n ck ǫki =0, i k1,.X..,kn=0 iY=1 and if q >1, then the coefficient of s1 =0. That is: 4d+2 n k x k x ck ǫki 1 1 + + n n =0 i 1 ǫ ··· 1 ǫ k1,.X..,kn=0 iY=1 (cid:18)(cid:18) (cid:19) 1 (cid:18) (cid:19) n(cid:19) POLYNOMIALS REPRESENTING EYNARD-ORANTIN INVARIANTS 9 and by equating coefficients of x , for 1 j n: j ≤ ≤ 4d+2 n k ck ǫki j =0. i ǫ j k1,.X..,kn=0 iY=1 For a general q and α=(α ,...,α ) 1 n 4d+2 k k ck α1 ... αn ǫ1k1−α1...ǫknn−αnxα11...xαnn =0 if |α|<q. k1,.X..,kn=0 (cid:18) 1(cid:19) (cid:18) n(cid:19) Thus inductively one gets 4d+2 (14) ckǫ1k1−α1...ǫknn−αnk1α1...knαn =0 if |α|<q. k1,.X..,kn=0 This means that the dominant asymptotic term as s 0 will look like: → 4d+2 ωg 1 ck Πn ǫki−αikαixαidx . n ∼ s(2d+1)n−q n x2d+2 2(2d+2)nα! i=1 i i i i i=1 i k1,.X..,kn=0|αX|=q In [2] it is shown thaQt as z ,...,z tends to the branch points ǫ ,...,ǫ , ǫ = 1 1 n 1 n j ± s6−6g−3n[1x′′(ǫ )y′(ǫ )]2−2g−nωg[Airy], for all ǫ the same ωg 2 i i n i n ∼( lower order asymptotics, for mixed ǫi where the Airy curve is given by y2 =x. Thus q =2d(n 1) if all ǫ are the same i − and q >2d(n 1) for all other combinations. − From [3] there is a relationship between ωg[Airy] and intersection numbers of n tautological line bundles over the moduli space. x′′(0)2−2g−n n (2β +1)! dz (15) ωg[Airy](z )= i i τ ...τ , n S 23g−3+n |βX|=diY=1 βi! zi2βi+2h β1 βni where we have used Witten’s [11] notation hτβ1...τβni = Mg,nc1(L1)β1...c1(Ln)βn and x′′(0)=2. From this we discover that if all ǫ =1: i R 4d+2 n 1 ck kαixαidx Qn x2d+2 2(2d+2)nα! i i i i=1 i k1,.X..,kn=0|α|=X2d(n−1) iY=1 4d+2 ck n kαixαi−2d−2 = i i dx 2(2d+2)n α ! i i k1,.X..,kn=0|α|=X2d(n−1) iY=1 x′′(1)y′(1) 2−2g−n n (2β +1)! dx i i = { } τ ...τ . 23g−3+n |βX|=diY=1 βi! x2iβi+2h β1 βni and so each α must be even and 0 α 2d. By equating powers of x (that is, i i i ≤ ≤ extracting the partition where α =2d 2β ) one gets the relation i i − (16) 4d+2 ck n ki2d−2βi = {x′′(1)y′(1)}2−2g−n n (2βi+1)! τ ...τ . 2(2d+2)n (2d 2β )! 23g−3+n β ! h β1 βni i i k1,.X..,kn=0 iY=1 − iY=1 10 PAULNORBURYANDNICKSCOTT Similarly, when all ǫ = 1 one gets i − 4d+2 ck n ( 1)kik2d−2βi (17) − i 2(2d+2)n (2d 2β )! i k1,.X..,kn=0 iY=1 − x′′( 1)y′( 1) 2−2g−n n (2β +1)! i = { − − } τ ...τ . 23g−3+n β ! h β1 βni i i=1 Y and when there is a mix of ǫ ’s, q will be greater, introducing more vanishing: i 4d+2 ck n ǫkik2d−2βi (18) i i =0 for β =d. 2(2d+2)n (2d 2β )! | | i k1,.X..,kn=0 iY=1 − Equations (16), (17) and (18) translate to an analogous equation (21) for coefficients of the polynomials Ng . To show this we now study the polynomials in n,k terms of the coefficients ck. As in the proof of Theorem 1, the Taylor expansion of ωg about z =0 can be written: n i 4d+2 ∞ n (b +1 k )/2+2d ωng = ck i 2−d+i1 zibi−1dzi k1,.X..,kn=0b1,.X..,bn=0 iY=1(cid:18) (cid:19) 2−(2d+1)n 4d+2 ∞ n 2d+1 = (2d+1)!n ck  σj(ki)bi2d+1−jzibi−1dzi k1,.X..,kn=0b1,.X..,bn=0 iY=1 Xj=0 ∞   = b ...b Ng(b ,...,b )zb1−1...zbn−1dz ...dz 1 n n 1 n 1 n 1 n bXi>0 whereforb even, respectivelyodd, weonlysumovertheodd, respectivelyeven, k i i and σ (k ) (= coefficient of b2d+1−j in (b +4d+1 k )...(b +3 k )(b +1 k )) j i i i − i i − i i − i is a degree j polynomial in k . i Thus the homogeneous degree 2q terms of the quasi-polynomial Ng are n n 2−(2d+1)n 4d+2 n (19) ( bi)Nng[degree 2q]= (2d+1)!n ck σ2d−2βi(ki)bi2βi+1 iY=1 k1,.X..,kn=0|βX|=q iY=1 where we are still summing over parity dependent k. The equations (14), (16), (17) and (18) give identities for sums over all k of i ck times monomials in ki and we wish to apply these to (19) which consists of coefficients that sum over only some of the k , depending on parity. To remedy i this, we add together the different polynomials representing Ng for every possible n parity. This removes the restriction on the k summand. For {i1,...,ik}⊂{1,...,n}, define vβ{i1,...,ik} to be the coefficient of b21β1...b2nβn in Ng with b ,...,b odd. For ǫ = (ǫ ,...,ǫ ) 1 n define ǫI = ǫ . n,k i1 ik 1 n ∈ {± } k∈I k Since σ (k ) is a polynomial of degree j in k , if β =q >d, then the homogeneous j i i | | Q degreeink ,k ,...,k oftheproductofσ (k )issmallenoughthat(14)implies 1 2 n 2d−2βi i the vanishing of each of the coefficients. In other words we have shown that (20) ǫIvI =0, β >d. β | | I⊂{X1,...,n} If β =q =d,thentheonlynonzerosumsresultingfromtheσ (k )arethe top p|ow|ers of the ki, that is ni=1(−ki)2d−2βi, since any compone2ndt−w2βiithia smaller Q