Polynomials representing Eynard-Orantin invariants

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by  Paul Norbury

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 polyno-
mial 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 pˆaround
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 coef-
ficients 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