Counting lattice points in the moduli space of curves

0.2 MB·English

by  Paul Norbury

COUNTING LATTICE POINTS IN THE MODULI SPACE OF
CURVES.
8
0 PAULNORBURY
0
2
Abstract. Weshowhowtodefineandcountlatticepointsinthemodulispace
n Mg,n of genus g curves with n labeled points. This produces a polynomial
a withcoefficientsthatincludetheEulercharacteristicofthemodulispace,and
J
tautological intersectionnumbersonthecompactifiedmodulispace.

0
3

]
G 1. Introduction
A
Let M be the moduli space of genus g curves with n labeled points. The
g,n
. decorated moduli space M ×Rn equips the labeled points with positive num-
h g,n +
t bers (b1,...,bn). It has a cell decomposition due to Penner, Harer, Mumford and
a
Thurston
m
[ (1) Mg,n×Rn+ ∼= PΓ

Γ∈F[atg,n
1
where the indexing set Fat is the space of labeled fatgraphs of genus g and n
v g,n
0 boundary components. See Section 2 for definitions of a fatgraph Γ, its automor-
9 phism group AutΓ and the cell decomposition (1) realised as the space of labeled
5 fatgraphs with metrics. Restricting this homeomorphism to a fixed n-tuple of pos-
4
itive numbers (b ,...,b ) yields a space homeomorphic to M decomposed into
. 1 n g,n
1 compact convex polytopes P (b ,...,b ). When the b are positive integers the
Γ 1 n i
0
polytope P (b ,...,b ) is an integral polytope and we define N (b ,...,b ) to be
8 Γ 1 n Γ 1 n
its number of positive integer points. The weighted sum of N over all labeled
0 Γ
: fatgraphs of genus g and n boundary components is the lattice count polynomial:
v
i 1
X Definition 1. Ng,n(b1,...,bn)= NΓ(b1,...,bn)
|AutΓ|
r Γ∈XFatg,n
a
EachintegralpointinthepolytopeP (b ,...,b )correspondstoaDessind’enfants
Γ 1 n
definedbyGrothendieck[3]whichrepresentsacurveinM definedoverQ¯. Thus
g,n
the lattice count polynomial N (b ,...,b ) counts curves defined over Q¯. This is
g,n 1 n
described in Section 2 where the integral points in P (b ,...,b ) represent metrics
Γ 1 n
onlabeledfatgraphswithintegeredgelengths,orequivalentlycurvesequippedwith
a canonical meromorphic quadratic (Strebel) differential with integral residues.
Quite generallythe number of integer points in a convexpolytope is a piecewise
definedpolynomial. Neverthelessthefollowingtheoremshowsthataweightedsum
of the piecewise defined polynomials N (b ,...,b ) is a polynomial.
Γ 1 n
Theorem 1. The number of lattice points N (b ,...,b ) is a degree 3g−3+n
g,n 1 n
polynomial in the integers (b2,...,b2) depending on the parity of the b .
1 n i
1991 Mathematics Subject Classification. MSC(2000) 32G15; 11P21;57R20.
1
2 PAULNORBURY
The dependence on the parity means that N (b ,...,b ) is represented by 2n
g,n 1 n
polynomials (by symmetry at most [n] +2 are different.) The polynomials are
2
symmetric under permutations of b of the same parity. If the number of odd b
i i
is odd then N (b ,...,b ) = 0. Otherwise, the top degree homogeneous part of
g,n 1 n
N (b ,...,b )isindependentoftheparity. Table1showsthesimplestpolynomials.
g,n 1 n
ThefactorisationsareexpectedfromthevanishingresultofLemma2inSection2.3.
Harer and Zagier [5] calculated the orbifold Euler characteristic χ(M ) and
g,1
Penner [10] calculated χ(M ) for general n. This information is encoded in the
g,n
lattice count polynomial for all even b .
i
Theorem 2. N (0,...,0)=χ(M ).
g,n g,n
Kontsevich [6] defined the volume polynomial
1
V (b ,...,b )= Vol (b ,...,b )
g,n 1 n Γ 1 n
|AutΓ|
Γ∈XFatg,n
where Vol (b ,...,b ) is the volume of the convex polytope P (b ,...,b ). (The
Γ 1 n Γ 1 n
LaplacetransformofV appearsasI in[6].) Heshowedthatthecoefficientsgive
g,n g
intersection numbers of Chern classes of the tautological line bundles L over the
i
compactified moduli space M . By considering finer and finer meshes it follows
g,n
that the homogeneous top degree part of the lattice point count polynomial is the
volume polynomial.
Theorem 3. N (b ,...,b )=V (b ,...,b )+ lower order terms.
g,n 1 n g,n 1 n
Corollary 1. For |d| = idi = 3g−3+n and d! = di! the coefficient cd of
b2d = b2di in N (b ,...,b ) is the intersection number
i g,n 1 Pn Q
Q 1
cd = 26g−6+2n−gd! c1(L1)d1...c1(Ln)dn.
ZMg,n
Kontsevich proved that these tautological intersection numbers satisfy a recur-
sion relation conjectured by Witten [12] that determine the intersection numbers.
The lattice count polynomials satisfy a recursion relation that uniquely determine
the polynomialsandwhenrestrictedtothe topdegreeterms imply Witten’s recur-
sion.
Table 1. Lattice count polynomials for even b
i
g n N (b ,...,b )
g,n 1 n
0 3 1
1 1 1 b2−4
48 1
0 4 14 b21+b22(cid:0)+b23+(cid:1)b24−4
1 2 3814 (cid:0)b21+b22−4 b21+b22−(cid:1)8
2 1 2161335 b21−(cid:0)4 b21−16(cid:1)(cid:0)b21−36 5(cid:1)b21−32
(cid:0) (cid:1)(cid:0) (cid:1)(cid:0) (cid:1)(cid:0) (cid:1)
COUNTING LATTICE POINTS IN THE MODULI SPACE OF CURVES. 3
Theorem 4. The lattice count polynomials satisfy the following recursion relation
which determines the polynomials uniquely from N and N .
0,3 1,1
n
b N (b ,...,b ) = pqN (p,b ,..,ˆb ,..,ˆb ,..,b )
i g,n 1 n g,n−1 1 i j n
!
Xi=1 Xi6=jp+qX=bi+bj
(2) + pqr N (p,q,b ,..,ˆb ,..,b )
g−1,n+1 1 i n
Xi p+qX+r=bi (cid:20)
+ N (p,b )N (q,b )
g1,|I| I g2,|J| J
(cid:21)
g1+Xg2=g
I⊔J={1,..,ˆi,..,n}
TheproofofTheorem4iselementary. Therecursionrelation(2)isusedtoprove
Theorem 1. It resembles Mirzakhani’s recursion relation [7] between polynomials
givingtheWeil-Peterssonvolumeofthemodulispace. Infactthetophomogeneous
degree part of N (b ,...,b ) coincides with the top homogeneous degree part of
g,n 1 n
Mirzakhani’sWeil-Peterssonvolumepolynomial(after multiplying by anappropri-
ate power of 2) since both of these coincide with Kontsevich’s volume. Mirzakhani
[8]alreadyshowedthe coefficientsofthe Weil-Peterssonvolumepolynomialarethe
intersectionnumbersgiveninCorollary1. DoandSafnuk[2]usefatgraphstogivea
simpler proof of Mirzakhani’s recursionrelation restricted to the top homogeneous
degree part and show that it is a rescaled version of Mirzakhani’s proof.
Although Table 1 shows only evenb , the recursionrelationneeds the odd cases
i
too. We willfill in the cases of oddb here. When b is odd, N (b ,...,b )≡0.
i i g,n 1 n
ThepolynomialN (b ,...,b )isthesameasinthetablewhenb ,...,b areallodd,
0,4 1 4 1 4
and when exactly two of the b are odd N (b ,...P,b )= 1 b2+b2+b2+b2−2 .
i 0,4 1 4 4 1 2 3 4
For genus 1 when b and b are odd N (b ,b )= 1 b2+b2−2 b2+b2−10 .
1 2 1,2 1 2 384 1 (cid:0) 2 1 2 (cid:1)
Section 2 contains preliminaries on fatgraphs and lattice point counting. The-
(cid:0) (cid:1)(cid:0) (cid:1)
orems 1 and 4 are proven in Section 2.2. Section 2.3 contains a simple vanishing
result for N (b ,...,b ) which has powerful consequences. In Section 3 we prove
g,n 1 2
Theorem 2 and treat the special case of n=1 labeled points.
Acknowledgements. The authorwouldliketo thankNormanDoformanyuseful
conversations.
2. Fatgraphs
A fatgraph is a graph Γ with vertices of valency > 2 equipped with a cyclic
ordering of edges at each vertex. In Figure 1 we use the projection to define the
cyclic ordering to be anticlockwise at each vertex. The two pictured fatgraphs are
Figure 1. Fatgraphs
4 PAULNORBURY
different, although the underlying graphs are the same. A fatgraph structure on
a graph is equivalent to an embedding of a graph into a surface Γ → Σ such that
Σ−Γisaunionofdisks. Thisgivesagenusg andnumberofboundarycomponents
n to Γ. The examples in Figure 1 have genus 0 and 1 shown in Figure 2.
Figure 2. Graphs embedded in genus 0 and 1 surfaces
A labeled fatgraphis a fatgraphwith boundarycomponents labeled1,...,n. The
set of all labeled fatgraphs of genus g and n boundary components is notated by
Fat .
g,n
Itisusefultodescribeafatgraphinthefollowingequivalentway[6]whichmakes
the automorphisms transparent. Given a graph Γ with vertices of valency >2, let
X be the set of oriented edges, so each edge of Γ appears in X twice. Define the
map τ : X → X that flips the orientation of each edge. A fatgraph, or ribbon,
1
structureonΓisamapτ :X →X thatpermutescyclicallytheorientededgeswith
0
a common source vertex. Let X , X and X be the vertices, edges and boundary
0 1 2
components of the fatgraph Γ. Then X = X/τ , X = X/τ and X = X/τ for
0 0 1 1 2 2
τ =τ τ . Anautomorphism ofthelabeledfatgraphΓisapermutationφ:X →X
2 0 1
that commutes with τ and τ and acts trivially on X . The examples in Figure 1
0 1 2
given any labeling have automorphism groups {1} and Z .
6
A metric on a labeled fatgraph Γ assigns positive numbers—lengths—to each
edge of the fatgraph. If Γ∈Fat then the valency >2 conditions on the vertices
g,n
ensures that the number of edges e(Γ) of Γ is bounded e(Γ)≤6g−6+3n. Let P
Γ
be the 6g−6+3n cell consisting of all metrics on Γ. Construct the cell-complex
Mcombinatorial = P
g,n Γ
Γ∈F[atg,n
where we identify isometric metrics on fatgraphs, and when the length of an edge
l →0weidentifythiswiththemetriconthefatgraphwiththeedgeE contracted.
E
By the existence and uniqueness of meromorphic quadratic differentials with foli-
ations having compact leaves, known as Strebel differentials, the cell complex is
homeomorphic to the decorated moduli space Mcgo,nmbinatorial ∼=Mg,n×Rn+ [4].
Denote by P (b ,...,b ) ⊂ P the metrics on Γ with fixed boundary lengths
Γ 1 n Γ
b = (b ,...,b ) ∈ Rn or equivalently with specified residues of the (square root of
1 n +
the) associated Strebel differential. Then
(3) Mcgo,nmbinatorial(b1,...,bn)= PΓ(b1,...,bn)∼=Mg,n.
Γ∈F[atg,n
2.1. Countinglatticepointsinconvexpolytopes. AconvexpolytopeP ⊂Rn
can be defined as the convex hull of a finite set of vertices in Rn. We will consider
integral polytopes P where the vertices lie in Zn. Define the number of integral
points in P by N = #{P ∩Zn} and N (k) = #{kP ∩Zn} where kP rescales
P P
COUNTING LATTICE POINTS IN THE MODULI SPACE OF CURVES. 5
λ 7→ kλ . Also, define N0(k) to be the number of integral points in the interior
j j P
of kP.
Theorem 2.1 (Ehrhart). If P ⊂Rn is an n-dimensional convex polytope then
N (k)=Vol(P)kn+...
P
is a degree n polynomial in k with top coefficient the volume of P. Furthermore,
N0(k)=(−1)nN (−k).
P P
We can define a convex polytope with positive codimension as follows. Given a
linear map A:RN →Rn and b∈Rn define
P (b)={x∈RN|Ax=b}.
A +
If A and b have integer entries (with respect to the standard bases) then P (b) is
A
integral and we define N (b)=#{P ∩ZN}. In this case N (b) is a piecewise
PA A PA
defined polynomial in b - for example, N (b) may be zero for some values of b.
PA
The set P (b) in (3) is a convex polytope defined by solutions x∈Re(Γ) of
Γ +
A x=b
Γ
whereA isthe incidence matrixthat mapsthe vectorspace generatedbyedges of
Γ
Γ to the vector space generated by boundary components of Γ—an edge maps to
the sum ofits two incident boundary components. In the examples in Figure 1 the
incidence matrices are
1 1 0
AΓ = 1 0 1 , AΓ′ = 2 2 2 .
 
0 1 1
(cid:0) (cid:1)
 
We define
N (b)=#{P ∩ZN}.
Γ A +
Itisnaturaltoallownon-negativesolutionsalthoughwe allowonlypositive integer
solutions. This is justified by the fact that if some of the x vanish then this will
i
be countedusing afatgraphobtainedby collapsingedgesofΓ. (If the collapsingof
edges of Γ does not yield a fatgraph,for example collapsinga loop, then we do not
want to count such solutions.)
Since each edge is incident to exactly two (not necessarly distinct) boundary
componentsthecolumnsofA addto2,orequivalently(1,1,...,1)·A =(2,2,...,2).
Γ Γ
Thus,
b =(1,1,...,1)·b=(1,1,...,1)·A x=(2,2,...,2)·x∈2Z
i Γ
so N (b)X= 0 if b is odd. Hence the lattice count polynomial N (b ,...,b )
Γ i g,n 1 n
given in Definition 1 also vanishes when b is odd.
i
P
If we relax the condition on fatgraphs that the valency of each vertex must be
P
> 2 then Grothendieck [3] showed that fatgraphs with all edge lengths 1 possess
branched covers of P1 branched over 0, 1 and ∞. By a theorem of Belyi these
correspond to curves defined over Q¯. When the length of each edge is a positive
integer this is the same as a string of length 1 edges joined by valency 2 vertices.
Thus, N (b ,...,b ) counts curves defined over Q¯ branched over of 0,1,∞ ∈ P1
g,n 1 n
with all points over 1 of ramification 2, and all points over 0 of ramification >2.
6 PAULNORBURY
For a convex polytope P ⊂ RN and a polynomial φ on RN define the following
generalisationof counting lattice points.
N (φ,k)= φ(x)
P
x∈kXP∩ZN
and N0(φ,k) the sum over interior integer points of kP. Later when applying the
P
recursion relation we will need to calculate sums with a parity restriction as in
Lemma 1 because the polynomials N vanish if the sum of the arguments is odd.
g,n
Lemma 1.
(4) Sm(k)= p2m+1q, Rm,m′(k)= p2m+1q2m′+1r
p+Xq=k p+Xq+r=k
q even r even
are odd polynomials in k of degree 2m+3, respectively 2m+2m′+5, depending
on the parity of k.
Proof. ThedependenceontheparitymeansthattherearetwopolynomialsSeven(k)
m
andSmodd(k)dependingonwhetherkisevenorodd. ThesameistrueforRm.m′(k).
Notice that
S (k)=2N (φ ,k)
m P 1
for P ={(x,y)∈R2|x+2y =1} and φ =x2m+1y (substitute q =2Q.) Similarly,
+ 1
Rm,m′(k)=2NP′(φ2,k)
for P′ ={(x,y,z)∈R3|x+y+2z =1} and φ =x2m+1y2m′+1z.
+ 2
The polytopes P and P′ are rational, not integral. They can be expressed in
terms of the integral convex polytopes of higher dimension
P ={x≥0,y ≥0,x+2y ≤2}, P ={x≥0,y ≥0,z ≥0,x+y+2z ≤2}.
1 2
For k even
k k k k
Smeven(k)=NP1(φ1, 2)−NP01(φ1, 2), Rmev,emn′(k)=NP2(φ2, 2)−NP02(φ2, 2).
A generalisation [1] of Ehrhart’s theorem states that for a dimension n integral
convex polytope P ⊂Rn, N (φ,k) is a degree degφ+n polynomial in k and
P
N0(φ,k)=(−1)degφ+nN (φ,−k).
P P
For the cases at hand, degφ+n is even so the right hand side is N (φ,−k) and
P
Seven(k) and Reven (k) are odd polynomials in k of degree 2m +3, respectively
m m,m′
2m+2m′+5. For k odd,
k+1 k−1
Sodd(k)=N0 (φ , )−N (φ , )
m P1 1 2 P1 1 2
andRodd (k)isthesameexpressionwithP inplaceofP . OnceagainSodd(k)and
m,m′ 2 1 m
Rodd (k)areoddpolynomialsink ofdegree2m+3,respectively2m+2m′+5. (cid:3)
m,m′
COUNTING LATTICE POINTS IN THE MODULI SPACE OF CURVES. 7
2.2. Recursion.
Proof of Theorem 4. The lattice count polynomial N (b ,...,b ) counts labeled
g,n 1 n
fatgraphs with positive integer edge lengths which we call integer fatgraphs in
P (b ,...,b ). We can produce an integer fatgraph in P (b ,...,b ) from simpler
Γ 1 n Γ 1 n
integerfatgraphsinthe three waysshowninFigures3, 4and5. Choosea graphin
PΓ′(p,b3,...,bn)andaddanedgeoflengthq/2insidetheboundaryoflengthpasin
Figure3sothatp+q =b +b . Similarly,attachanedgeandaloopoftotallength
1 2
Γ
q
p 2
b1 b2
p+q=b +b
1 2
Figure 3. Γ is obtained from a simpler fatgraph by adding the
broken line.
q/2 inside the boundary of length p as in Figure 4 so that p+q =b +b . In both
1 2
Γ
p b
2
q
b
1 2
p+q=b +b
1 2
Figure 4. Γ is obtained by adding a line and loop of total length q/2.
cases for each Γ′ there are p possible ways to attach the edge so this construction
contributes pN (p,b ,...,b ) to N (b ,...,b ). However we have overcounted,
g,n−1 3 n g,n 1 n
particularly when we repeat this construction for any pair b and b , since each
i j
integer fatgraph in P (b ,...,b ) can be produced in many ways like this. To deal
Γ 1 n
with this, we overcount even further by taking pqN (p,b ,...,b ), i.e. taking
g,n−1 3 n
each constructed fatgraph q times. But now we see that for each edge that we
attachoflengthq/2wehaveovercountedq times. Ifwewereto useallofthe edges
of Γ in this way then we would have overcountedby
n
l(E)= b .
i
E∈Γ i=1
X X
Indeed all of the edges of Γ are used, exactly once, when we include one further
construction of the integer fatgraph Γ.
8 PAULNORBURY
Choose an integer fatgraph in PΓ′(p,q,b2,...,bn) for Γ′ ∈ Fatg−1,n+1 or choose
twointegerfatgraphsinP (p,b ,...,b )andP (q,b ,...,b )forΓ ∈Fat and
Γ1 2 j Γ2 j+1 n 1 g1,j
Γ ∈ Fat where g +g = g and attach an edge of length r/2 connecting
2 g2,n+1−j 1 2
these two boundary components as in Figure 5 so that p+q+r =b .
1
Γ
p b1 q
r
2
p+q+r=b
1
Figure 5. Γ is obtained from a single fatgraph or two disjoint
fatgraphs by adding the broken line.
In the diagram, the two boundary components of lengths p and q are part
of a fatgraph that may or may not be connected. There are pq possible ways
to attach the edge so this construction contributes pqN (p,q,b ,...,b ) and
g−1,n+1 2 n
pqN (p,b ,...,b )N (q,b ,...,b ) to N (b ,...,b ) and again we have
g1,j 2 j g2,n+1−j j+1 n g,n 1 n
overcounted. WeovercountfurtherbyafactorofrtogetpqrN (p,q,b ,...,b )
g−1,n+1 2 n
andpqrN (p,b ,...,b )N (q,b ,...,b ). Werepeatthisforeachg +g =
g1,j 2 j g2,n+1−j j+1 n 1 2
g and I ⊔J ={2,...n} and then for each b in place of b .
j 1
As previewed above, each edge of Γ has been attached to construct Γ and
N (b ,...,b ) has been overcounted n b times yielding (2). (cid:3)
g,n 1 n i=1 i
Remark. The idea in the proof aboPveto overcountby the length of each edge of
the graph Γ comes from the similar idea introduced by Mirzakhani [7] where she
unfolds a function on Teichmu¨ller space that sums to the analogue of b .
1
To apply the recursion we need to first calculate N (b ,b ,b ) and N (b ).
0,3 1 2 3 1,1 1
TherearesevenlabeledfatgraphsinFat comingfromthreeunlabeledfatgraphs.
0,3
ItiseasytoseethatN (b ,b ,b )=1ifb +b +b iseven(and0otherwise.) This
0,3 1 2 3 1 2 3
is because for each (b ,b ,b ) there is exactly one of the seven labeled fatgraphs
1 2 3
Γ with a unique solution of A x = b while the other six labeled fatgraphs yield
Γ
no solutions. For example, if b > b +b then only the fatgraph Γ with A =
1 2 3 Γ
2 1 1
0 1 0 has a solution and that solution is unique.
 
0 0 1
 To calculate N1,1(b1), note that AΓ = [2 2 2] or [2 2] for the 2-vertex and
1-vertex fatgraphs. Hence
b1 −1 b1 −1
N (b )=a 2 +a 2
1,1 1 1 2
2 1
(cid:18) (cid:19) (cid:18) (cid:19)
where a is the number of trivalent fatgraphs (weighted by automorphisms) and
1
a is the number of 1-vertex fatgraphs. The genus 1 graph Γ from Figure 1 has
2
|AutΓ| = 6 so a = 1/6, and a uses the genus 1 figure 8 fatgraph which has
1 2
COUNTING LATTICE POINTS IN THE MODULI SPACE OF CURVES. 9
automorphism group Z hence a =1/4. Thus
4 2
1 b1 −1 1 b1 −1 1
N (b )= 2 + 2 = b2−4 .
1,1 1 6 2 4 1 48 1
(cid:18) (cid:19) (cid:18) (cid:19)
(cid:0) (cid:1)
We can also calculate N (b ) using a version of the recursion
1,1 1
1
b N (b )= pq.
1 1,1 1
2
p+qX+p=b
b even
We will calculate N [b ,b ,b ,b ] to demonstrate the recursion relation and the
0,4 1 2 3 4
parity issue.
4
b N (b ,b ,b ,b )= pq.
i 0,4 1 2 3 4
!
i=1 i6=j
X X p+qX=bi+bj
q even
If all b are even, or all b are odd, then b +b is always even so the sum is over p
i i i j
and q even. We have
k +1
Seven(k)= pq =4 2
0 3
i6=j (cid:18) (cid:19)
X p+Xq=k
q even
so
4 bi+bj +1 4 1
b N (b)= 4 2 = b b2+b2+b2+b2−4
i 0,4 3 i 4 1 2 3 4
i=1 ! i6=j (cid:18) (cid:19) i=1 !
X X X (cid:0) (cid:1)
agreeing with Table 1. If b and b are odd and b and b are even then we need
1 2 3 4
1 k+1
Sodd(k)= pq =
0 2 3
i6=j (cid:18) (cid:19)
X p+Xq=k
q even
so
4 bi+bj +1 1 b +b +1
b N (b) = 4 2 + i j
i 0,4
3 2 3
Xi=1 ! (i,j)=(1X,2) or (3,4)(cid:18) (cid:19) (i,j)6=(1X,2) or (3,4) (cid:18) (cid:19)
4
1
= b b2+b2+b2+b2−2
i 4 1 2 3 4
!
i=1
X (cid:0) (cid:1)
so we see that the polynomial representatives of N (b) agree up to a constant
0,4
term.
Proof of Theorem 1. We can use the recursion (2) to prove that N (b ,...,b ) is
g,n 1 n
a polynomial of the right degree but to prove that it is a polynomial in b2 we need
i
a different recursion formula (5). For simplicity we use (5) to prove each part of
10 PAULNORBURY
Theorem 1.
1
(5) b N (b ,...,b ) = pqN (p,b ,..,ˆb ,..,b )
1 g,n 1 n g,n−1 2 j n
2
Xj>1 p+qX=b1+bj

+ pqN (p,b ,..,ˆb ,..,b )
g,n−1 2 j n

p+qX=b1−bj

+ pqr N (p,q,b ,...,b )
g−1,n+1 2 n
p+qX+r=b1 (cid:20)
+ N (p,b )N (q,b )
g1,|I| I g2,|J| J
(cid:21)
g1+Xg2=g
I⊔J={2,...,n}
This differs from the recursion formula (2) by breaking the symmetry around b .
1
The sum over the term p + q = b − b needs to be interpreted as follows. If
1 j
b −b >0itisreadaswritten,whereasifb −b <0thenreplaceb −b byb −b
1 j 1 j 1 j j 1
and negate the sum. (This is not the same as sending (p,q) to (−p,−q).)
We willprovethe recursion(5) below. Beforethat wewill provethatgivenN
0,3
and N then (5) determines polynomials N′ (b ,...,b ) of degree 3g−3+n in
1,1 g,n 1 n
b2. By induction, the simpler polynomials are polynomials in b2 so monomials on
i i
the right hand side of the recursion are of the form
Sm(k)= p2m+1q, Rm,m′(k)= p2m+1q2m′+1r
p+Xq=k p+Xq+r=k
q even r even
as in (4). In Lemma 1 it is proven that Sm(k) and Rm,m′(k) are odd polynomials
ink. Inparticular,S (b −b )=−S (b −b )explainingthe interpretationofthe
m 1 j m j 1
sum over b −b <0
1 j
The sums over p+q+r = b1 yield terms which are odd in b1 from Rm,m′(b1)
and even in b for i > 1 hence 1/b times these terms is even in all b2. The sums
i 1 i
overp+q =b +b andp+q =b −b havethe samesummands soeachmonomial
1 j 1 j
occurs with the same coefficient. Hence the terms involving b are of the form
1
S (b +b )+S (b −b )andsinceS isodd,this sumisoddinb andeveninb ,
m 1 j m 1 j m 1 j
and even in the all other b . Again 1/b times these terms is even in all b2. Thus
i 1 i
byinductionthepolynomialsgeneratedbythe recursionrelation(5)fromN and
0,3
N are polynomials in b2.
1,1 i
Wewillnowcalculatethe degreeinb2. ByinductiondegN =3g−3+n−1
i g,n−1
andbyLemma2S (k)takesatermp2m+1q andproducesadegree2m+3polyno-
m
mial,i.e.itincreasesthedegreeby1. Inthiscase3g−3+n−1+1=3g−3+nasre-
quired. Similarly,byinductiondegN =3g−3+n−2anddegN N =
g−1,n+1 g1,|I| g2,|J|
3g−3+n−2. By Lemma 2 Rm,m′(k) increases the degree of its summand by 2.
Since3g−3+n−2+2=3g−3+nthe resultis provenbyinductionstartingfrom
the degrees of N and N .
0,3 1,1
As above, write N′ for the polynomials produced from the recursion (5). To
g,n
prove the recursion (5) we use the fact that both (2) and (5) uniquely determine