aa r X i v : . [ m a t h . C O ] D ec Enumerative and asymptotic analysis of a moduli space
Margaret A. READDY
To Doron Zeilberger on the occasion of his 60th birthday
Abstract
We study exact and asymptotic enumerative aspects of the Hilbert series of the cohomologyring of the moduli space of stable pointed curves of genus zero. This manifold is related to theWDVV (Witten-Dijkgraaf-Verlinde-Verlinde) equations of string theory.
Keywords: asymptotic enumeration, cohomology of moduli space, graded Hilbert series, integral oper-ator, Lambert W function. MSC: primary 05E40, secondary 05A16.
The study of moduli spaces has been a source of much research activity in physics, topology andalgebraic topology. See for example [10, 12]. The classical example is the moduli space of smooth n -pointed stable curves of genus g , denoted M g,n . This moduli space was introduced by Deligne,Mumford and Knudsen [4, 9, 16] and gives a natural compactification of Mumford’s [15] modulispace of nonsingular curves of genus g . To show how rich this family is, the case M ,n correspondsto elliptic curves when n = 1 and to abelian varieties for n > We begin by considering exact enumerative data of the Hilbert series of the cohomology ring ofthe moduli space M ,n . For completeness, we give the physicist’s definition of this cohomologyring based on Keel’s work. See [11, Section 0.10] and [8]. For n ≥ P n the set of stable -partitions of { , . . . , n } , that is, the set of all unorderedpartitions of the elements { , . . . , n } into two blocks σ = S | S with | S i | ≥
2. For each σ ∈ P n , theelement x σ corresponds to a cohomology class of H ∗ ( M ,n ). For σ = S | S and τ = T | T from theindex set P n , let a ( σ, τ ) be the number of nonempty pairwise distinct sets among S i ∩ T j , where1 ≤ i, j ≤
2. Define the ideal I n in the polynomial ring k[ x σ : σ ∈ P n ] to be generated by therelations: 1 11 1 21 5 1 71 16 16 1 341 42 127 42 1 2131 99 715 715 99 1 16301 219 3292 7723 3292 219 1 14747(a) (b)Table 1: The coefficients of the Hilbert series of the cohomology ring of the moduli space M ,n for n = 0 , . . . ,
6: (a) The triangular array of the graded dimensions α i,j for 0 ≤ i + j ≤ σ n for 0 ≤ n ≤ i, j, k, l distinct: R ijkl : X ijσkl x σ − X kjτil x τ , (2.1)where the summand ijσkl means to sum over all stable 2-compositions σ = S | S with theelements i, j ∈ S and the elements k, l ∈ S .2. (quadratic relations) For each pair σ and τ with a ( σ, τ ) = 4, x σ · x τ . (2.2)The cohomology ring of the moduli space of n -pointed stable curves of genus 0 is given by thequotient ring H ∗ ( M ,n ) = k[ x σ : σ ∈ P n ] /I n .See Table 1 for some values. One immediately sees the coefficients satisfy Poincar´e duality. Proposition 2.1
The coefficients of the Hilbert series of the moduli space M ,n are symmetric. This follows from the fact that the genus g moduli space M g,n is a smooth, complete and compactvariety [4, 9, 16]. For a proof in the genus 0 case, see [8]. By Proposition 2.1 we thus can write theHilbert series as H ( H ∗ ( M ,n )) = X ( i,j ) α i,j t i , where the sum is over all pairs ( i, j ) of nonnegative integers with i + j = n .We next present a recursion for the coefficients of the Hilbert series of the cohomology ring ofthe moduli space M ,n . The recursion we give here is a symmetrized version of Keel’s [8, page 550]. Theorem 2.2
The coefficients of the Hilbert series series of cohomology ring of the moduli space M ,n satisfy the recursion: α i +1 ,j +1 = α i +1 ,j + α i,j +1 + 12 i X p =0 j X q =0 (cid:18) i + j + 4 p + q + 2 (cid:19) α p,q · α i − p,j − q , (2.3)2 ( x ) = e x − x − f ( x ) = 2 e x − (cid:18) x + 2 x + 2 (cid:19) e x f ( x ) = 272 e x − (cid:0) x + 20 x + 22 (cid:1) e x + (cid:18) x + 116 x + 172 x + 15 x + 172 (cid:19) e x f ( x ) = 5123 e x − (cid:18) x + 324 x + 378 (cid:19) e x + (cid:18) x + 1603 x + 240 x + 432 x + 262 (cid:19) e x − (cid:18) x + 23 x + 18524 x + 41 x + 4234 x + 126 x + 1643 (cid:19) e x Table 2: Expressions for f j ( x ), j = 0 , . . . , where i, j ≥ and with the initial conditions α i, = α ,i = 1 for i ≥ . Define the exponential generating function f j ( x ) = X i ≥ α i,j x i +2 ( i + 2)! . These are series formed by taking diagonal entries in the α i,j triangle with the power of t slightlyshifted. See Table 2 for the first few values of f j ( x ). Let I be the integral operator I ( f ( x )) = Z x f ( t ) dt and I j denote the j th iterated integral operator. The following integral operator identity holdsamong these series. Lemma 2.3
The generating function f j ( x ) satisfies the integral operator identity I j +1 ( f j +1 ( x )) = I j +2 ( f j +1 ( x )) + I j +1 ( f j ( x )) + 12 j X q =0 I q ( f q ( x )) · I j − q ( f j − q ( x )) , (2.4) where f is given by f = e x − x − . Proof:
Multiply the identity (2.3) with x i + j +4 / ( i + j + 4)! and sum over i ≥ X i ≥ α i +1 ,j +1 ( i + j + 4)! · x i + j +4 = X i ≥ α i,j +1 ( i + j + 4)! · x i + j +4 + X i ≥ α i +1 ,j ( i + j + 4)! · x i + j +4 (2.5)+ 12 X i ≥ i X p =0 j X q =0 α p,q ( p + q + 2)! · α i − p,j − q ( i − p + j − q + 2)! · x i + j +4 (2.6)The left-hand side of (2.5) can be written as X i ≥ α i +1 ,j +1 ( i + j + 4)! · x i + j +4 = X i ≥ α i,j +1 · x i + j +3 ( i + j + 3)! − α ,j +1 · x j +3 ( j + 3)!= I j +1 ( f j +1 ( x )) − x j +3 ( j + 3)! .
3n a similar manner, the right-hand side of (2.5) becomes X i ≥ ( α i,j +1 + α i +1 ,j ) · x i + j +4 ( i + j + 4)! = I j +2 ( f j +1 ( x )) + I j +1 ( f j ( x )) − x j +3 ( j + 3)! . Finally, the triple sum in (2.6) can be written as j X q =0 X p ≥ α p,q · x p + q +2 ( p + q + 2)! · X p ≥ α p,j − q · x p + j − q +2 ( p + j − q + 2)! = j X q =0 I q ( f q ( x )) · I j − q ( f j − q ( x )) . The result now follows by combining these identities. ✷ Notice the degrees of the polynomials preceding the e kx terms in f j ( x ) depend only on j and k . This motivates one to define the degree sequence for functions having the expansion f ( x ) = P kj =0 p j ( x ) · e jx where the p j ( x ) are polynomials in x by deg( f ) = (deg( p j ( x ))) kj =0 . Herewe define the degree of the zero polynomial to be deg(0) = −∞ .It is straightforward to verify the following proposition. Proposition 2.4
Let f and g have series expansions f = P kj =0 p j ( x ) · e jx and g = P mj =0 q j ( x ) · e jx with deg( f ) = ( a , a , . . . , a k ) and deg( g ) = ( b , b , . . . , b m ) . Then(a) deg( f + g ) ≤ ( c , . . . , c n ) , where c i = max( a i , b i ) and n = max( k, m ) .(b) deg( f · g ) ≤ ( d , . . . , d k + m ) , where d i = max p + q = i ( a p + b q ) .(c) deg( dfdx ) = ( a − , a , . . . , a k ) with the convention − −∞ .(d) If f is non-zero then deg( I ( f )) = ( b , a , . . . , a k ) with b = max( a + 1 , .(e) When f ≡ , that is, deg( f ) = ( −∞ , −∞ , . . . , −∞ ) then deg( I ( f )) = (0 , −∞ , . . . , −∞ ) . (f ) The solution y to the differential equation y ′ = y + f ( x ) has degree sequence deg( y ) = ( a , max( a + 1 , , a , . . . , a k ) . The degree sequence properties are now used to derive an upper bound for the degree sequencesof the series f j . 4 roposition 2.5 For j ≥ , the series f j ( x ) has the expansion f j ( x ) = j +1 X k =1 p j,k ( x ) e kx , where p j,k ( x ) is a polynomial of degree at most j − k + 1) , that is, deg( f j ( x )) ≤ ( −∞ , j, j − , . . . , , . Proof:
It is straightforward to verify that f = e x − − x and f = 2 e x − ( x + 2 x + 2) e x . Hencedeg( f ) = (1 ,
0) and deg( f ) = ( −∞ , , j , and we have just verified the induction basis j = 1. Apply thedifferential operator d j +2 dx j +2 to equation (2.4) to obtain ddx ( f j +1 ( x )) = f j +1 ( x ) + d j +2 dx j +2 I j +1 ( f j ( x )) + 12 j X q =0 I q ( f q ( x )) · I j − q ( f j − q ( x )) . (2.7)Assume that deg( f k ) ≤ ( −∞ , k, k − , . . . , ,
0) for all 1 ≤ k ≤ j . By Proposition 2.4 we have(i) deg( I j +1 ( f j )) ≤ ( j, j, j − , . . . , , I q ( f q ) · I j − q ( f j − q )) ≤ ( j − , j + max( q, j − q ) − , j, j − , . . . , ,
0) for 1 ≤ q ≤ j − I j ( f j ) · f ) ≤ ( j, j + 1 , j, j − , . . . , , j, j + 1 , j, j − , . . . , , d j +2 /dx j +2 , we have the upper bound( −∞ , j + 1 , j, j − , . . . , ,
0) for the rightmost term in the differential equation (2.7). Hence byProposition 2.4 (f), the degree sequence of f j +1 has the desired form. ✷ In order to prove the next two theorems, we will need a lemma.
Lemma 2.6
The following two identities hold for t ≥ : X c + d = tc,d ≥ (cid:18) tc, d (cid:19) c c − · d d − = 2 · ( t − · t t − (2.8) X c + d = tc,d ≥ (cid:18) tc, d (cid:19) c c · d d − = ( t − · t t − (2.9) Proof:
Observe the left-hand side of equation (2.8) enumerates the number of pairs of rootedlabeled trees on a t -element set, that is, the coefficient of x t /t ! in the generating function f ( x ) ,where f ( x ) = P t ≥ t t − · x t /t !. Noticing that f ( x ) satisfies the functional equation f ( x ) = x · e f ( x ) ,the coefficient can also be determined by Lagrange inversion formula [20, Theorem 5.4.2] as follows: (cid:20) x t t ! (cid:21) f ( x ) = t ! · t (cid:2) x t − (cid:3) ( e x ) t = 2 · ( t − · t t − ( t − · ( t − · t t − . p n ( x ) = x · ( x − na ) n − form a sequence of binomial type , that is, they satisfy the relation p n ( x + y ) = P nk =0 (cid:0) nk (cid:1) p k ( x ) p n − k ( y ) . See [19]. The desired identity then follows from the substitution x = y = 1, a = − n = t − c + d = t and use that the two resulting sums are equal. ✷ We now show the bounds given in Proposition 2.5 are sharp.
Theorem 2.7
For s ≥ , t ≥ , the coefficient of x s e tx in the series f s + t − is given by [ x s e tx ] f s + t − = ( − s s · s ! · t s + t − t ! . (2.10) Furthermore, the degree sequence of the series f j for j ≥ is equal to deg( f j ( x )) = ( −∞ , j, j − , . . . , , . Proof:
Let γ s,t = [ x s e tx ] f s + t − . Note γ s, = 0 for s ≥
0. Observe that for t ≥ x s e tx ]( I k ( f s + t − )) = γ s,t t k . Extract the coefficient of x s e tx from Lemma 2.3 in the case j + 1 = s + t − γ s,t t s + t − = γ s,t t s + t + 0 + 12 · · [ x s e tx ] f · I s + t − ( f s + t − )+ 12 s + t − X q =1 [ x s e tx ] I q ( f q ( x )) · I s + t − − q ( f s + t − − q ( x )) . We obtain γ s,t t s + t − · (cid:18) − t (cid:19) = 12 X a + b = s X c + d = tc,d ≥ γ a,c c a + c − · γ b,d d b + d − (2.11)because the sum of the inequalities a + c ≤ q +1 and b + d ≤ s + t − − q +1 implies a + b + c + d ≤ s + t .Furthermore, since a + b + c + d = s + t , these two inequalities are actually equalities, giving theindex of summation as above.Equation (2.11) can be viewed as a recursion for the coefficients γ s,t . Hence we can prove thetheorem by induction on the quantity s + t . The induction basis is s = 0 and t = 1, which isstraightforward to verify. The induction step is to verify( − s s · s ! t s + t − t ! (cid:18) − t (cid:19) = 12 X a + b = s X c + d = tc,d ≥ ( − a a a ! · c a + c − c ! · ( − b b · b ! · d b + d − d ! . Multiply by 2 · ( − s · s ! · t ! to obtain2( t − · t s + t − = X a + b = s X c + d = tc,d ≥ (cid:18) sa, b (cid:19)(cid:18) tc, d (cid:19) c a + c − · d b + d − = X c + d = tc,d ≥ (cid:18) tc, d (cid:19) c c − · d d − X a + b = s (cid:18) sa, b (cid:19) c a · d b = t s · X c + d = tc,d ≥ (cid:18) tc, d (cid:19) c c − · d d − , γ s,t are non-zero. ✷ Corollary 2.8
The asymptotic behavior of α i,j as i → ∞ is given by α i,j ∼ ( j + 1) j +1 j ! · ( j + 1) i . By a similar argument, we can find the next coefficient in the f j ( x ). Theorem 2.9
For s, t ≥ , the coefficient of x s − e tx in the series f s + t − is given by [ x s − e tx ] f s + t − ( x ) = ( − s · (5 s + 9 t − · t s +2 t − · s − ( s − t − . (2.12) Proof:
Using integration by parts we have that for t = 0 I (cid:0) ( γ · x s + δ · x s − ) · e tx ) (cid:1) = (cid:18) γt · x s + (cid:18) δt − s · γt (cid:19) · x s − (cid:19) · e tx + lower order terms.By iterating operator I k times, we obtain I k (cid:0) ( γ · x s + δ · x s − ) · e tx ) (cid:1) = (cid:18) γt k · x s + (cid:18) δt k − · k · s · γt k +1 (cid:19) · x s − (cid:19) · e tx +lower order terms.Let δ s,t denote the coefficient δ s,t = [ x s − e tx ] f s + t − ( x ). Note that δ , = − δ s, = 0 for s ≥ δ ,t = 0 for t ≥ x s − e tx in Lemma 2.3 where j = s + t − δ s,t t s + t − − · ( s + t − · s · γ s,t t s + t = δ s,t t s + t − · ( s + t ) · s · γ s,t t s + t +1 + 0+[ x s − e tx ] 12 · s + t − X q =0 I q ( f q ( x )) · I s + t − − q ( f s + t − − q ( x ))We now focus our attention on the last term in the right-hand side of the above identity. Forshorthand, we call this term X . X = 12 · s + t − X q =0 X a + b = s X c + d = t [ x a − e cx ] I q ( f q ( x )) · [ x b e dx ] I s + t − − q ( f s + t − − q ( x ))+ 12 · s + t − X q =0 X a + b = s X c + d = t [ x a e cx ] I q ( f q ( x )) · [ x b − e dx ] I s + t − − q ( f s + t − − q ( x ))In order for [ x a e cx ] I q ( f q ( x )) to be non-zero, we need a + c ≤ q +1. Similarly, for [ x a − e cx ] I q ( f q ( x ))to be non-zero, we need a + c ≤ q + 1. Thus the variables of summation must satisfy a + c ≤ q + 1and b + d ≤ s + t − q −
1. Adding these two inequalities gives an equality. Hence q = a + c − q . Furthermore, the two double summation expressionsare equal, so we obtain X = X a + b = s X c + d = t [ x a − e cx ] I a + c − ( f a + c − ( x )) · [ x b e dx ] I b + d − ( f b + d − ( x ))= − γ s − ,t t s + t − + X a + b = s X c + d = tc,d ≥ (cid:18) δ a,c c a + c − − a + c − · a · γ a,c c a + c (cid:19) · γ b,d d b + d − Observe the terms corresponding to d = 0 all vanish. The only surviving term corresponding to c = 0 is when a = 1, yielding the term − γ s − ,t /t s + t − above.To summarize, we have δ s,t t s + t − − · ( s + t − · s · γ s,t t s + t = δ s,t t s + t − · ( s + t ) · s · γ s,t t s + t +1 − γ s − ,t t s + t − + X a + b = s X c + d = tc,d ≥ (cid:18) δ a,c c a + c − − a + c − · a · γ a,c c a + c (cid:19) · γ b,d d b + d − Using Theorem 2.7, we note − γ s − ,t t s + t − = 2 s · γ s,t t s + t , thus simplifying the identity to δ s,t t s + t − · (cid:18) − t (cid:19) − · ( s + t ) · s · γ s,t t s + t · (cid:18) − t (cid:19) = X a + b = s X c + d = tc,d ≥ (cid:18) δ a,c c a + c − − a + c − · a · γ a,c c a + c (cid:19) · γ b,d d b + d − . Observe that this identity is a recursion for δ s,t . Hence we prove the theorem by induction on s + t .The induction basis s = t = 1 is straightforward. The induction step is to verify the followingidentity: (cid:18) ( − s · (5 s + 9 t − · t s + t − · s − · ( s − t − − · ( s + t ) · s · ( − s · t s + t − s · s ! t ! (cid:19) · (cid:18) − t (cid:19) = X a + b = s X c + d = tc,d ≥ (cid:18) ( − a · a · (5 a + 9 c − · c a + c − · a · a ! c ! − ( − a · a · ( a + c − · c a + c − a · a ! c ! (cid:19) · ( − b d b + d − b · b ! d != X a + b = s X c + d = tc,d ≥ ( − a · a · c a + c − a · a ! c ! (cid:18) (5 a + 9 c − − ( a + c − (cid:19) · ( − b d b + d − b · b ! d !Multiplying by 3 · ( − s · s ! · t ! and simplifying gives(2 s + 6 t − · s · t s + t − · ( t −
1) = X a + b = s X c + d = tc,d ≥ (cid:18) sa, b (cid:19)(cid:18) tc, d (cid:19) a · c a + c − · d b + d − · (2 a + 6 c − X c + d = tc,d ≥ (cid:18) tc, d (cid:19) · c c − · d d − X a + b = s a · (2 a + 6 c − · (cid:18) sa, b (cid:19) · c a · d b .
8y applying the operator c · ∂∂c once, respectively twice, to the binomial theorem, we obtain thetwo identities X a + b = s a · (cid:18) sa, b (cid:19) · c a · d b = s · c · ( c + d ) s − , (2.13) X a + b = s a · (cid:18) sa, b (cid:19) · c a · d b = s · c · ( c + d ) s − + s · ( s − · c · ( c + d ) s − . (2.14)We now have(2 s + 6 t − · s · t s + t − · ( t −
1) = X c + d = tc,d ≥ (cid:18) tc, d (cid:19) · c c − · d d − (cid:0) (6 c − · s · c · ( c + d ) s − +2 · s · c · ( c + d ) s − + 2 · s · ( s − · c · ( c + d ) s − (cid:1) = s · t s − · X c + d = tc,d ≥ (cid:18) tc, d (cid:19) · c c − · d d − ((6 c − · t + 2 t + 2( s − · c )= s · t s − · (cid:0) − t · t − t t − + (6 t + 2( s − · ( t − t t − (cid:1) = s · t s − · ( t − · t t − · ( − t + 2 s − , where we have applied identities (2.13) and (2.13) in the first step and Lemma 2.6 in the third step.This completes the induction. ✷ Mathematical physicists often refer to the sum of the coefficients of a finite Hilbert series as the “di-mension” of a space. For example, using techniques from combinatorial commutative algebra [18],it was proved that the dimension of the Losev–Manin ring [13], that is, the cohomology ring arisingfrom the Commutativity Equations of physics, is given by n !.Let σ n = X i + j = n α i,j denote the total graded dimension of the cohomology of the moduli space M ,n . Proposition 3.1
The total graded dimension σ n satisfies the recursion σ n +2 = 2 σ n +1 + 12 X i + j = n (cid:18) n + 4 i + 2 , j + 2 (cid:19) σ i · σ j , (3.1) with the initial condition σ = 1 . Proof:
Sum the recursion (2.3) for α p,q in the case p + q = n + 2. ✷ The Lambert W function is the multivalued complex function which satisfies W( z ) · e W( z ) = z .The principal branch W ( z ) and the − − ( z ) are the only two branches which take9n real values. The Lambert W function has applications to classical tree enumeration, signalprocessing and fluid mechanics. For more details, see [1, 3].For our purposes we will need to consider two branches of the Lambert W function whichtogether form an analytic function near − /e . Lemma 3.2
Let Ω be the function defined by Ω( z ) = (cid:26) W − ( z ) Im( z ) ≥ ,W ( z ) Im( z ) < . Then Ω is analytic in C − ( −∞ , − /e ] − [0 , + ∞ ) . The Puiseux series of Ω is given by Ω( z ) = X k ≥ µ k p k = − p − p + 1172 p − · · · where p = − p ez + 1) and the coefficients µ k are computed by the recurrences µ k = k − k + 1 (cid:16) µ k − α k − (cid:17) − α k − µ k − k + 1 ,α k = k − X j =2 µ j · µ k +1 − j , with µ = − , µ = 1 , α = 2 and α = − . Proof:
Analyticity follows from piecing together the analytic parts of the Lambert W function.The Puiseux series expansion is due to D. Coppersmith. See [3, Section 4]. ✷ Form the exponential generating function for the total graded dimension: g ( x ) = X n ≥ σ n x n +2 ( n + 2)! . Proposition 3.3
The function g ( x ) satisfies the differential equation g ′ ( x ) = x + 2 g ( x )1 − g ( x ) (3.2) with boundary condition g (0) = 0 . Proof:
Multiplying (3.1) by x n / ( n + 4)! and summing over n ≥ X n ≥ σ n +2 · x n ( n + 4)! = 2 X n ≥ σ n +1 · x n ( n + 4)! + 12 (cid:18) x g ( x ) (cid:19) . (3.3)Observe g ′ ( x ) = P n ≥ σ n · x n +1 / ( n + 1)! and I ( g ( x )) = x X m ≥ σ m +1 · x m ( m + 4)! + x g ( x ) − x − x
3! = 2 Z x g ( t ) dt − x
3! + 12 g ( x ) . (3.4)Differentiating (3.4) gives the desired differential equation. ✷ This generating function is related to the Lambert W function.
Theorem 3.4
The function g ( z ) is given by g ( z ) = 1 − ( z + 2) (cid:18) − e − ( z + 2)) (cid:19) . (3.5) It is analytic in C − ( −∞ , − − [ e − , + ∞ ) . Proof:
Substitute g ( x ) = 1 − ( x + 2)( u + 1) in the differential equation (3.2). We obtain theequation dxx + 2 = − (cid:18) u + 1 u (cid:19) du. Integrating gives log | x + 2 | + C = − log | u | + 1 u . The boundary condition g (0) = 0 implies u (0) = − / C = −
2. Hence − e − ( x + 2) = 1 u · e u . As the Lambert W function satisfies the functional equation W( z ) · e W( z ) = z , we have that W ( − ( x + 2) /e ) = 1 /u . Recall that the value x = 0 implies u = − /
2, so W ( − e − ) = − − is the correct real branch of the Lambert W function to select. See [3]. Hencewe obtain the generating function g ( x ) = 1 − ( x + 2)(1 + 1 / W − ( − e − ( x + 2))). The analyticcontinuation to C − ( −∞ , − − [ e − , + ∞ ) follows from Lemma 3.2. ✷ In order to obtain asymptotic behavior of the total graded dimension, one needs to apply resultsof Flajolet and Odlyzko. See [6, Section 4.2] and the overview article [17]. We follow Flajolet andOdlyzko’s notation and terminology. Write f ( z ) ∼ g ( z ) as z → w to mean that f ( z ) g ( z ) → z → w .For r, η > < ϕ < π/ r, ϕ, η ) = { z : | z | ≤ r + η, | arg( z − r ) | ≥ ϕ } . A function L ( u ) is of slow variation at ∞ if it satisfies ( i ) there exists a positive real number u andan angle ϕ with 0 < ϕ < π/ L ( u ) = 0 and is analytic for { u : − ( π − ϕ ) ≤ arg( u − u ) ≤ π − ϕ } and ( ii ) there exists a function ǫ ( x ) defined for x ≥ x → + ∞ ǫ ( x ) = 0 and forall θ ∈ [ − ( π − ϕ ) , π − ϕ ] and u ≥ u we have (cid:12)(cid:12)(cid:12)(cid:12) L ( ue iθ ) L ( u ) − (cid:12)(cid:12)(cid:12)(cid:12) < ǫ ( u ) and (cid:12)(cid:12)(cid:12)(cid:12) L ( u log ( u )) L ( u ) − (cid:12)(cid:12)(cid:12)(cid:12) < ǫ ( u ) . The following asymptotic theorem appears in [6, Theorem 5].11 heorem 3.5 (Flajolet and Odlyzko)
Assume f ( z ) is analytic on ∆( r, ϕ, η ) − { r } and L ( u ) isa function of slow variation at ∞ . If α ∈ R and α
6∈ { , , , . . . } and f ( z ) ∼ ( r − z ) α · L (cid:18) r − z (cid:19) uniformly as z → r for z ∈ ∆( r, ϕ, η ) − { r } , then [ z n ] f ( z ) ∼ r − n · n − α − Γ( − α ) · L ( n ) . Applying Theorem 3.5 to the Puiseux series of the analytic function Ω gives the following result.
Theorem 3.6
The total dimension of the cohomology ring of the moduli space M ,n has asymptoticbehavior σ n ∼ r e π · ( e − − n − · n − / · ( n + 2)! as n → ∞ . Proof:
We will apply Theorem 3.5 to the function f ( z ) = g ( z ) − g ( e −
2) = g ( z ) −
1. By Lemma 3.2Ω( ζ ) = − p + O (cid:0) p (cid:1) , where p = − p eζ + 1). By inverting this relation we have1Ω( ζ ) = − − p + O (cid:0) p (cid:1) = − √ e · r ζ + 1 e + O (cid:18) ζ + 1 e (cid:19) . In the above, substitute ζ = − e − ( z + 2), add 1 and then multiply with − ( z + 2) = − e + ( e − − z ).We obtain f ( z ) = − e · √ e · r ( e − − z ) e + O ( e − − z ) . In other words, f ( z ) ∼ −√ e · √ e − − z uniformly as z → e − z in a deleted pie-shaped neighborhood ∆( r, ϕ, η ) of r = e −
2. By letting r = e − α = 1 / L be the constant function −√ e , Theorem 3.5 applies. We conclude that[ x n ] g ( x ) = [ x n ] f ( x ) ∼ − ( e − − n · n − / Γ( − / · √ e = r e π · ( e − − n · n − / as n −→ ∞ . Since [ x n ] g ( x ) = σ n − /n !, we obtain σ n ∼ r e π · ( e − − n − · ( n + 2) − / · ( n + 2)! . Since ( n + 2) − / ∼ n − / as n → ∞ , the result follows. ✷ An equivalent asymptotic expression appears without proof in [14, Chapter 4, page 194]. See [7]for a recursion for the total dimension of a generalization related to configuration spaces.12
Concluding remarks
It remains to find the complete sequence of polynomials to describe the coefficients in the series f j .We make the following conjecture. Conjecture 4.1
For k a non-negative integer and s, t ≥ k , the coefficient of x s − k e tx in f s + t − ( x ) is given by [ x s − k e tx ] f s + t − ( x ) = ( − s t s +2 t − k − s − k · ( s − k )!( t − k )! · Q k ( s, t ) , (4.1) where Q k ( s, t ) is a polynomial in the variables s and t of degree k . As a special case we have already proved Q ( s, t ) = 1 and Q ( s, t ) = (5 s + 9 t − / M ,n is related to n -pointed rooted trees of one-dimensionalprojective spaces. It is interesting to note that trees also emerge in our asymptotic study of theHilbert series of the cohomology ring. See recent work of Chen, Gibney and Krashen [2] for furtherwork on trees of higher-dimensional projective spaces. The author would like to thank the Institute for Advanced Study where this paper was com-pleted while the author was a Member of the School of Mathematics during 2010–2011, as well asRobert Corless and David Jeffrey for information about the Lambert W function.
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