ON DENOMINATORS OF THE KONTSEVICH INTEGRAL AND

THE UNIVERSAL PERTURBATIVE INVARIANT OF 3-MANIFOLDS

THANG T. Q. LE

Abstract. The integrality of the Kontsevich integral and perturbative invariants is discussed. We show that the denominator of the degree n part of the Kontsevich integral of any knot or link is a divisor of (2!3! … n!)^{4}(n + 1)!. We also show that the denominator of of the degree n part of the universal perturbative invariant of homology 3-spheres is not divisible by any prime greater than 2n + 1.

0. Introduction

The aim of this paper is to study the denominators of the Kontsevich integral and the universal perturbative invariants of homology 3-spheres. The Kontsevich integral is a very interesting knot invariant (see [Ko1]) which contains in itself all finite type invariants, as well as the Jones polynomial and its numerous generalizations (see, for example, [BL, BN1]).

The Kontsevich integral has values in a graded algebra of chord diagrams and is expressed by a formula involving iterated integrals. These iterated integrals, when calculated separately, are mostly transcendental numbers, and can be expressed as rational linear combinations of multiple zeta values. However, if we collect terms together using the relations between chord diagrams, then J. Murakami and the author showed that the coefficients are rational, see [LM2]. The proof follows Drinfeld’s work on quasi-Hopf algebras [Dr2].

In many problems concerning the Kontsevich integrals and quantum and perturbative invariants of 3-manifolds, one needs to know the prime factors of the denominators of the Kontsevich integral. Examples of such problems are the conjecture 7.3 in [LMO] about the relation between perturbative and quantum invariants of 3-manifolds and the conjecture of Lawrence in [Law] about the integrality and padic convergence of perturbative invariants (see also [LR]). In this paper we show that the denominator of the degree n part of the Kontsevich integral is a divisor of (2! … n!)^{4}(n+1)!, hence it can not have any prime factor greater than n+1. Example shows that the denominators of the degree 6 part have prime factor 7. This result plays important role in Ohtsuki’s proof (see [Oh2]) of the conjecture 7.3 of [LMO] for the sl_{2} case.

This work is partially supported by NSF grant DMS-9626404. This and related preprints can be obtained at http://www.math.buffalo.edu/∼letu/ .

In the joint work [LMO] with Murakami and Ohtsuki, using the Kontsevich integral and the Kirby calculus, we constructed an invariant Ω of 3-manifolds with values in a graded algebra of 3-valent graphs. We call it a universal perturbative invariant. Later the author in [Le] showed that Ω is a universal finite type invariant of integral homology 3-spheres, hence it plays the role of the Kontsevich integral for homology 3spheres. For the theory of finite type invariants of homology 3-spheres see [Oh1, GO]. Here we show that the denominator of the degree n part of Ω of rational homology 3-spheres does not have prime factor greater than 2n+1. This result is closely related to the integrality property of quantum invariants which says that quantum invariants (see [Tu]) of homology 3-spheres at prime root of unity are cyclotomic integer. This had been conjectured by Kontsevich, and was proved in various cases by Murakami, Takata-Yokota, Masbaum-Wenzl (see [Mur, TY, MW]).

The idea of the proof of the main result can be explained as follows. First we reduce the proof to establishing the existence of an associator, whose denominator has some specific properties. Associators are solution of a system of equations, important among them are the so-called pentagon and hexagon equations. The well-known Knizhnik-Zamolodchikov associator (found by Drinfeld) is not good, since its coefficients are not even rational. An explicit formula for this associator is given in [LM2]. So we search for another associator. Drinfeld used perturbative method to find associators. In this method, one first finds an associator up to degree n, and then tries to extend it to degree n + 1. Drinfeld observed that the obstruction to the extension is in the cohomology of a certain complex. He then showed that the cohomology is equal to 0, hence there is no obstruction at all. Bar-Natan in [BN2] carried this program over to the space of chord diagrams. In Drinfeld’s and Bar-Natan’s papers, the mentioned cohomology groups vanish, over the nationals. We follow Drinfeld’s method, trying to solve the hexagons and pentagons equations and keeping track of the denominators. This leads to the problem of calculating the cohomology groups over the integers. It turns out that this cohomology group is a torsion group, annihilated by (n + 1)! in degree n. Hence we can estimate the denominator in each step.

There is, however, another difficulty to overcome. In Drinfeld’s and Bar-Natan’s papers, in order to solve the hexagon and pentagon equation, one has to assume some freedom for the so-called R-matrix. But in order to get the Kontsevich integral, the R-matrix must be fixed and equal to the simplest one. The purely combinatorial method to solve the hexagon and pentagon equations does not work if the R-matrix is fixed. We show here that one can still use the perturbative method to solve these equations when the R-matrix is fixed and equal to the simplest one. This is done by using a result about the uniqueness of the associator up to gauge transformations (see [LM2]) and the existence of a special associator (proved in [Dr2]; the proof used analysis).

The paper is organized as follows. In §1 we recall basic definitions of Chinese character diagrams (chord diagrams). In §2 we discuss the cobar complex of Chinese character diagrams. Associators are discussed in §3. We proved the main result about the existence of an associator with special denominators in §4. The results about the denominators of the Kontsevich integral and Ω are proved in §5 and 6. Finally in §7 we prove some technical results.

Acknowledgments: Much of this work was carried out while the author was visiting the Mathematical Sciences Research Institute in Berkeley. Research at MSRI is supported in part by NSF grant DMS-9022140. The author thanks P. Melvin, J. Murakami, T. Ohtsuki, S. Schack and D. Thurston for helpful discussions.

1. Chinese character diagrams

1.1. Preliminaries. A uni-trivalent graph is a graph every vertex of which is either univalent or trivalent. In this paper we always assume that each connected component of a uni-trivalent graph contains at least one univalent vertex. A uni-trivalent graph is vertex-oriented if at each trivalent vertex a cyclic order of edges is fixed. Trivalent vertices are also called internal vertices, and univalent vertices – external. A vertexoriented uni-trivalent graph is also known a Chinese character.

Let X be a compact oriented 1-dimensional manifold whose components are ordered. A Chinese character diagram with support X is the manifold X together with a vertex-oriented uni-trivalent graph whose univalent vertices are on X. A Chinese character diagram is g-connected if the graph is connected.

In all figures the components of X will be depicted by solid lines, the graph by dashed lines, and the orientation at every internal vertex is given by the counterclockwise direction. For this reason the graph is also called the dashed graph.

Chinese characters and Chinese character diagrams are regarded up to homeomorphisms preserving components of the support and orientations at vertices.

Let Ã(X) be the vector space over ℚ spanned by Chinese character diagrams with support X. Let A(X) be the quotient space of Ã(X) by dividing out the STU relation shown in Figure 1.

[figure]  =  \overline{}−  \overline{}

Figure 1. The STU relation

The degree of a Chinese character diagram is half the number of vertices of the dashed graph. Since the relation STU respects the degree, there is a grading on A(X) induced by this degree. We also use A(X) to denote the completion of A(X) with respect to the degree.

We define a co-multiplication Δ̂ in A(X) as follows. A Chinese character subdiagram of a Chinese character diagram D with dashed graph G is any Chinese character diagram obtained from D by removing some (possibly empty) connected components of G. The complement Chinese character sub-diagram of a Chinese character sub-diagram D′ is the Chinese character sub-diagram obtained by removing components of G which are in D′. We define

Δ̂(D) = ∑ D′ ⊗ D′′.

Here the sum is over all Chinese character sub-diagrams D′ of D, and D′′ is the complement of D′. This co-multiplication is co-commutative.

Suppose that X, X′ have distinguished components ℓ, ℓ′, and that X consists of loop components only. Let D ∈ A(X) and D′ ∈ A(X′) be two Chinese character diagrams. From each of ℓ, ℓ′ we remove a small arc which does not contain any vertices. The remaining part of ℓ is an arc which we glue to ℓ′ in the place of the removed arc such that the orientations are compatible. The new Chinese character diagram is called the connected sum of D, D′ along the distinguished components. It does not depend on the locations of the removed arcs, which follows from the STU relation and the fact that all components of X are loops. The proof is the same as in case X = X′ = S^{1} as in [BN1].

In case when X = X′ = S^{1}, the connected sum defines a multiplication which turns A(S^{1}) into an algebra.

1.2. Algebra structure. In special cases we can equip A(X) with an algebra structure. Suppose that X is n ordered lines on the plane pointing downwards. The space A(X) will be denoted by P_{n}; and a connected component of X will be called a string. If D_{1} and D_{2} are two Chinese character diagrams in P_{n}, let D_{1} × D_{2} be the Chinese character diagram obtained by placing D_{1} on top of D_{2}. The unit of this algebra is is the Chinese character diagram without dashed graph. Let P_{0} = ℚ. It is known that the algebra P_{1} is commutative (see [BN1]).

The algebra and co-algebra structure are compatible, and P_{n} becomes a Hopf algebra. It is not hard to see that every primitive element, i.e. element x such that Δ̂(x) = 1 ⊗ x + x ⊗ 1, is a linear combination of g-connected Chinese character diagrams.

We now introduce a couple of operators acting on P_{n}.

Suppose D is a Chinese character diagram in P_{n} with the dashed graph G. Replace the i-th string by two strings, the left and the right, very close to the old one, and renumber all the strings from left to right. Attach the graph G to the new set of strings in the same way as in D; this would cause no problem if there is no univalent vertex on the i-th string of D. If there is a univalent vertex of G on the i-th string

[figure: ξ_{1}; ξ_{2}; ξ_{3}] Figure 2.

then it yields two possibilities, attaching to the left or to the right string. Summing up all 2^{m}, where m is the number of univalent vertices of D on the i-th string, possible Chinese character diagrams of this type, we get Δ_{i}(D) ∈ P_{n+1}. Using linearity we can define Δ_{i} : P_{n} → P_{n+1}, for i = 1, 2, …, n.

Define ε_{i} by ε_{i}(D) = 0 if the Chinese character diagram D ∈ P_{n} has a univalent vertex on the i-th string. Otherwise let ε_{i}(D) be the Chinese character diagram in P_{n−1} obtained by removing the i-th string and renumbering the remaining strings from left to right. We continue ε_{i} to a linear map from P_{n} to P_{n−1}.

1.3. Chinese characters. An n-marked Chinese character ξ is a Chinese character whose external vertices are colored by 1, 2, …, n. Two n-marked Chinese characters are considered the same if there is a homeomorphism between them which preserves the colors.

Let B_{n} be the vector space over ℚ spanned by all n-marked Chinese characters subject to the following identities:

(1) the antisymmetry identity: ξ_{1} +ξ_{2} = 0, for every two Chinese characters ξ_{1} and ξ_{2} identical identical everywhere except for the orientation at one internal vertex.

(2) the Jacobi identity: ξ_{1} = ξ_{2} + ξ_{3}, for every three Chinese characters identical outside a ball in which they differ as in Figure 2.

Now we define a linear mapping χ : B_{n} → P_{n} as follows. Suppose an n-marked Chinese character ξ has k_{i} external vertices of color i. There are k_{i}! ways to put vertices of color i on the i-th string and each of the k_{1}! … k_{n}! possibilities gives us a Chinese character diagram in P_{n}. Summing up all such elements and dividing by 1/(k_{1}!k_{2}! … k_{n}!), we get χ(ξ). It is well-known that χ is an isomorphism between the vector spaces B_{n} and P_{n}. A proof for the case n = 1 is presented in [BN1]; the statement itself is Kontsevich’s. The proof can be easily generalized to any n. For an explicit description of χ^{−1}, see §7.3.

2. The cobar complex of Chinese character diagrams

2.1. The general complex. Let us define

C^{n}(P) = P_{n},
s_{i}^{n} = ε_{i} : C^{n} → C^{n−1},

and

d_{i}^{n} : C^{n} → C^{n+1}

with 0 ≤ i ≤ n + 1 by d_{i}^{n} = Δ_{i} if 1 ≤ i ≤ n, and

d_{0}^{n}(x) = 1 ⊗ x, d_{n+1}^{n} = x ⊗ 1,

where 1 ⊗ x (respectively, x ⊗ 1) is obtained from x by adding a string to the left (respectively, right) of x and renumbering all the strings from left to right.

It was noticed in [BN2] that (C^{n}, d_{i}^{n}, s_{i}^{n}) form a co-simplicial set. It is natural to consider the following differential complex

0 → C^{0} →^{d} C^{1} →^{d} C^{2} →^{d} … C^{n} →^{d} C^{n+1} …

where

d(x) = ∑_{i=0}^{n+1}(−1)^{i}d_{i}^{n} = 1 ⊗ x − Δ_{1}(x) + Δ_{2}(x) + ⋯ + (−1)^{n}Δ_{n}(x) + (−1)^{n+1}(x ⊗ 1).

We call it the cobar complex of Chinese character diagrams. The cohomology of this complex and its subcomplexes will play important role.

2.2. Subcomplexes. The symmetric group 𝔖_{n} acts on C^{n} = P_{n} on the left by permuting the strings of the support. An element x ∈ P_{n} is said to be symmetric if

x + (−1)^{n(n+1)/2}σ(x) = 0,

where σ is the permutation sending 1 to n, 2 to n − 1, 3 to n − 2, etc.

Let C_{sym}^{n}(P) be the subspace of all symmetric elements of C^{n}(P). It is easy to see that (C_{sym}^{∗}, d) is a differential subcomplex of (C^{∗}, d).

Let us define the Harrison sub-complex of P_{n} (see [BN2]). Let p, q be positive integers. A permutation σ of {1, 2, …, p + q} is called a (p, q)-shuffle if σ preserves the order of {1, …, p} and the order of {p + 1, …, p + q}. Define: sh_{p,q} ∈ ℚ[𝔖_{p+q}] by

sh_{p,q} = ∑_{all (p, q)-shuffles σ}(−1)^{σ}σ.

Let C_{Harr}^{n}(P) be the subset of C^{n}(P) consisting of elements x ∈ C^{n} such that sh_{p,q}(x) = 0 whenever p + q = n. As in the usual theory of Harrison cohomology, one can show that (C_{Harr}^{∗}, d) is a subcomplex of (C^{n}, d).

Another way to define the Harrison complex is to use the Eulerian idempotents. It was proved in [GS] that there exist idempotents e_{n}^{(l)} ∈ ℚ[𝔖_{n}], for l = 1, 2, …, n such that (e_{n}^{(l)})^{2} = e_{n}^{(l)}, e_{n}^{(l)}e_{n}^{(k)} = 0 if k ≠ l, and e_{n}^{(1)} + … e_{n}^{(n)} = 1. These idempotents, as elements of ℚ[𝔖_{n}], acts on P_{n} as projections. The important point is that all these idempotents commutes with the differential operator d. Hence for each fixed l, we have a subcomplex (e_{n}^{(l)}C^{n}, d) of (C^{n}, d), if we put e_{n}^{(l)} = 0 for l > n.

The subcomplex (e_{n}^{(1)}C^{n}, d) is exactly the above defined Harrison subcomplex. We record here the formula of e_{n}^{(1)}:

e_{n}^{(1)} = ∑_{σ∈𝔖_{n}} \frac{(−1)^{r(σ)}}{n\binom{n−1}{r(σ)}} (−1)^{σ}σ,  (2.1)

where r(σ) is the number of k ∈ {1, 2, …, n − 1} such that σ(k) > σ(k + 1).

2.3. Non-degeneracy, integral lattices. An element x ∈ P_{n} is said to be nondegenerate if ε_{i}(x) = 0 for every i = 1, 2, …, n. We will be interested in g-connected, non-degenerate Chinese character diagrams.

Let Q_{n}^{ℤ} be the set of all elements in P_{n} which are linear combinations of g-connected non-degenerate Chinese character diagrams with integer coefficients. Then Q_{n}^{ℤ} is a free ℤ-module, and let Q_{n} be the vector spanned by Q_{n}^{ℤ}: Q_{n} = Q_{n}^{ℤ} ⊗ ℚ. From the STU relation one can easily prove the following.

Lemma 2.1. If x, y are in Q_{n}^{ℤ}, then so is the commutator xy − yx.

Definition . Suppose V is a vector space over ℚ which contains a fixed integral lattice V^{ℤ}, i.e. a ℤ-module such that V = V^{ℤ} ⊗_{ℤ} ℚ. We say that an element x ∈ V has denominator N if Nx is in V^{ℤ}.

We will consider Q_{n}^{ℤ} the integral lattice of Q_{n}.

Note that the operator d preserves Q_{n} and Q_{n}^{ℤ}, hence we can speak about the subcomplexes (C^{∗}(Q), d) and (C^{∗}(Q^{ℤ}), d), where C^{n}(Q) = Q_{n} and C^{n}(Q^{ℤ}) = Q_{n}^{ℤ}. The latter complex is a complex over ℤ. One can also consider the subcomplexes (C_{sym}^{∗}(Q), d), (C_{Harr}^{∗}(Q), d), (C_{sym}^{∗}(Q^{ℤ}), d) and (C_{Harr}^{∗}(Q^{ℤ}), d) by taking the intersections with C_{sym}^{∗}(P) and C_{Harr}^{∗}(P).

It is known that (see [BN2]) the even-dimensional cohomology groups of (C_{sym}^{∗}(P), d) vanish. This fact is fundamental in solving the pentagon equation (see below) in [BN2, Dr2]. The proof can be modified easily to show that the even-dimensional cohomology groups of (C_{sym}^{∗}(Q), d) vanish.

Actually, to solve the pentagon equation, one needs the result only for the fourdimensional cohomology group. For the purpose of this paper we need to calculate the four-dimensional cohomology group of the ℤ-complex (C_{sym}^{∗}(Q^{ℤ}), d), which must have rank 0 as a ℤ-module, but may have some non-trivial torsion part.

Note that d preserves the degree of Chinese character diagrams, hence the complex (C_{sym}^{n}(Q^{ℤ}), d) can be decomposed further by degree. The result, whose proof will be given in §7, is

Proposition 2.2. The degree m part of the four-dimensional cohomology group of the complex (C_{sym}^{n}(Q^{ℤ}), d) is annihilated by 2(m + 1)!(m!)^{2}[(m − 1)!]^{2}.

3. The Drinfeld associator

3.1. Associators. Drinfeld defined associators and R-matrix for quasi-triangular quasi-Hopf algebras, see [Dr1, Dr2]. We recall here the definition, adapted for the case of Chinese character algebras in [LM1, LM2] (see also [BN2]). Note that we don’t have any quasi-Hopf algebra here. The algebra P_{n} will play the role of the n-th power of a quasi-Hopf algebra.

Let r ∈ P_{2} be the Chinese character diagram whose dashed graph is a line connecting the two strings of the support. Let r^{ij} ∈ P_{3} be the Chinese character diagram whose dashed graph is a line connecting the i-th and j-th strings of the support. Define R = exp(r/2) ∈ P_{2} and R^{ij} = exp(r^{ij}/2) ∈ P_{3}.

Definition . An associator is an element Φ ∈ P_{3} satisfying the following equations:

Δ_{3}(Φ) × Δ_{1}(Φ) = (1 ⊗ Φ) × Δ_{2}(Φ) × (Φ ⊗ 1),  (A1)
Δ_{1}(R) = Φ^{312} × R^{13} × (Φ^{132})^{−1} × R^{23} × Φ,  (A2)
Φ^{−1} = Φ^{321},  (A3)
ε_{1}(Φ) = ε_{2}(Φ) = ε_{3}(Φ) = 1.  (A4)

Here Φ^{ijk} is the element of P_{3} obtained from Φ by permuting the strings: the first to the i-th, the second to the j-th, the third to the k-th. Equation (A1) holds in P_{4}, equations (A2,A3) in P_{3}, and equation (A4) in P_{2}. There are some redundancy in this system of equations. Equation (A1) is known as the pentagon equation, (A2) – the hexagon equation. Due to (A3), one has that (Φ^{132})^{−1} = Φ^{231}.

Remark . Our definition of associator is stricter than that of [BN2, Dr2] since the R-matrix is exp(r/2).

An element in A(X) is said to be even if it is a linear combination of Chinese character diagrams of even degree. It follows from Drinfeld [Dr2] that there is an even associator Φ_{e}. Evenness has a nice topological interpretation, see [LM3].

3.2. Symmetric twisting. Suppose that F ∈ P_{2} satisfies the following conditions: (T1) ε_{1}(F) = ε_{2}(F) = 1, (T2) F is symmetric, i.e. F is a linear combination of symmetric Chinese character diagrams.

Then there exists the inverse F^{−1} in P_{2}. If Φ is an element of P_{3}, then the element

Φ^{F} := [1 ⊗ F] Δ_{2}(F) Φ Δ_{1}(F^{−1}) [F^{−1} ⊗ 1]

is said to be obtained from Φ by twisting via F, or by a gauge transformation (see [Dr1]).

If Φ ∈ P_{3} is an associator, i.e. a solution of (A1-A4), then it is not difficult to check that Φ^{F} is also an associator. Condition (T2) guarantees that Φ^{F} satisfies (A3).

In [LM2], following Drinfeld’s treatment of the Lie algebra case, it was proved that any two associators Φ, Φ′ are related by a twist F. The proof also shows that if both Φ, Φ′ are even, then F can be chosen to be even. Moreover, if Φ, Φ′ are “even associators up to degree 2k”, i.e. they satisfy the equations (A1–A4) up to degree 2k, then they are related by a twist via an even F.

Lemma 3.1. Suppose that Φ ∈ P_{3} is an even element which satisfies the hexagon and pentagon equations (A1), (A2) up to (and including) degree 2k. Then Φ also satisfies the same equations up to degree 2k + 1.

Proof. The pentagon equation is obviously satisfied, since there is nothing of odd degree. For the hexagon equation, we need the fact that there exits an even associator Φ_{e}.

Since both Φ and Φ_{e} are associator up to degree 2k, there is an even element F ∈ P_{2}, satisfying (T1) and (T2), such that Φ_{e}^{F} = Φ up to degree 2k. Since both Φ_{e}^{F} and Φ don’t have terms of odd degree, they are the same up to degree 2k + 1. The element Φ_{e}^{F} is still an associator, hence it satisfies the hexagon equation. It follows that Φ satisfies the hexagon equation up to degree 2k + 1.

Remark . In the proof we used the existence of Φ_{e}, which was established by Drinfeld using analysis. It is still an open problem to prove this lemma using only algebra.

4. Solving the pentagon and hexagon equations

4.1. The existence of a special associator. To solve the pentagon and hexagon equations we will follow [BN2, Dr2]. Lemma 3.1 makes the procedure much simpler. Let

d_{n} = (2!3! … n!)^{4}(n + 1)!

Theorem 4.1. There exists an even associator Φ ∈ P_{3} of the form Φ = exp(ϕ), where ϕ is a linear combination of even, non-degenerate, g-connected, symmetric Chinese character diagrams. In other words, ϕ is

even and in C_{sym}^{3}(Q).  (4.1)

Moreover, the degree 2m part of ϕ has denominator d_{2m}.

This is the main result. Recall that ϕ ∈ C^{3}(Q) = Q_{3} is symmetric means that ϕ^{321} = −ϕ. The proof of this theorem will occupy the rest of this section.

If ϕ satisfies (4.1), then Φ = exp(ϕ) is even and satisfies (A3) and (A4). There are only the hexagon and pentagon equations to worry about.

For a graded algebra A let Grad_{m}A (resp. Grad_{≤m}) be the subspace spanned by elements of grading m (of grading ≤ m).

We will solve the pentagon and hexagon equations up to degree 2m, and then show that the solution can be extended so that it solves these equations up to degree 2m + 2.

Suppose that there exists Φ_{2m} = exp(ϕ_{2m}) satisfying (A1) and (A2) up to degree 2m, where ϕ_{2m} ∈ Grad_{≤2m}P_{3} satisfies (4.1) and the part of degree 2k of ϕ_{2m} has denominator d_{2k}. We know that for m = 2 such a Φ_{2m} exists, see [BN2].

We will find ϕ_{2m+2} = ϕ_{2m} + φ, where φ is of degree 2m + 2 and satisfying (4.1) such that Φ_{(2m+2)} := exp(ϕ_{2m+2}) satisfies the pentagon and hexagon equations up to degree 2m + 2. In addition, φ, and hence ϕ_{2m+2}, has denominator d_{2m+2}.

By Lemma 3.1, Φ_{2m} also satisfies the pentagon and hexagon equations up to degree 2m + 1. Now focus on degree 2m + 2.

4.2. The hexagon equation. Let ψ be the degree 2m + 2 mistake in the hexagon equation when using Φ_{2m} in place of Φ, i.e.

1 + ψ = Φ_{2m}^{312} × (R)^{13} × (Φ_{2m}^{−1})^{132} × (R)^{23} × Φ_{2m} × Δ_{1}(R^{−1})  (4.2)

(the equation is taken modulo degree ≥ 2m + 3)

It is conceivable that with the knowledge of the denominators of the terms of the right hand side, one can estimate the denominator of ψ. In fact, in §7 we we will show that

Lemma 4.2. The mistake ψ is an element in Q_{3} having denominators [(2m+2)!]^{2}d_{2m}.

Suppose Φ′ = exp(ϕ_{2m} + u), where u satisfies (4.1) and is of degree 2m + 2. If we replace Φ_{2m} by Φ′ in Equation (4.2), then it’s easy to see that the new mistake ψ′ is

ψ′ = ψ + u + u^{312} + u^{231}.  (4.3)

Hence if ψ + u + u^{312} + u^{231} = 0, then Φ′ solves the hexagon equation up to degree 2m + 2.

The following fact was proved in [BN2, Dr1].

Lemma 4.3. The mistake ψ is totally antisymmetric, i.e. for every permutation σ of 3 numbers {1, 2, 3} one has σ(ψ) = sign(σ)ψ.

Let u = −ψ/3, then ψ + u + u^{312} + u^{231} = 0. Hence Φ̄ := exp(ϕ_{≤2m} − ψ/3) solves the hexagon equation up to degree 2m + 2. Note that u satisfies (4.1), is of degree 2m + 2, and has denominator 3 × [(2m + 2)!]^{2}d_{2m}.

4.3. The pentagon equation. Let μ be the mistake of degree 2m + 2 in the pentagon equation if Φ is replaced by Φ̄, i.e.

1 + μ = Δ_{1}(Φ̄^{−1}) × Δ_{3}(Φ̄^{−1}) × (1 ⊗ Φ̄) × Δ_{2}(Φ̄) × (Φ̄ ⊗ 1)  (4.4)

(this equation is taken modulo degrees ≥ 2m + 3). Again, one can easily estimate the denominator of μ. In §7 we will show

Lemma 4.4. The mistake μ is in C^{4}(Q) = Q_{4} having denominator 3[(2m+2)!]^{2}d_{2m}.

If we replace Φ̄ by Φ̄′ = exp(ϕ_{≤2m} −ψ/3+v), where v satisfies (4.1) and is of degree 2m + 2, then the new mistake μ′ is

μ′ = μ − Δ_{1}(v) − Δ_{3}(v) + Δ_{2}(v) + v ⊗ 1 + 1 ⊗ v = μ + dv.

The new mistake of the hexagon equation, by (4.3), is v^{123} + v^{312} + v^{231}. Hence, we need to find v of degree 2m+2 satisfying (4.1) and the following equations

dv + μ = 0,  (4.5)
v^{123} + v^{312} + v^{231} = 0.  (4.6)

The condition (4.1) says that v ∈ C^{3}(Q) and v is symmetric. That v is symmetric and satisfying (4.6) is equivalent to v is symmetric and annihilated by (1,2)- and (2,1)-shuffles. So, we need to find v in the intersection of C_{sym}^{3}(Q) and C_{Harr}^{3}(Q) and satisfying dv = −μ.

The following has been proved in [BN2].

Lemma 4.5. The mistake μ is in the intersection of C_{sym}^{4}(Q) and C_{Harr}^{4}(Q), and dμ = 0.

So μ ∈ C_{sym}^{4}(Q) and dμ = 0. From Proposition 2.2 it follows that there exists v′ ∈ C_{sym}^{3}(Q) having denominator 2[(2m + 1)!(2m + 2)!]^{2}(2m + 3)! times that of μ, such that dv′ = −μ. This v′ may not be in C_{Harr}^{3}, i.e. may not satisfy (4.6). Put v = e_{3}^{(1)}(v′). Explicitly one has (see the equation (2.1))

v = \frac{2}{3}v′ − \frac{1}{3}(v′)^{312} − \frac{1}{3}(v′)^{231}.

Then v is in both C_{sym}^{3}(Q) and C_{Harr}^{3}(Q). The commutativity of e_{n}^{(1)} and d shows that dv = d(e_{3}^{(1)}v′) = e_{4}^{(1)}dv′ = e_{4}^{(1)}(−μ) = −μ. So v is an element satisfies all (4.1), (4.6) and (4.5). Note that v has denominator 6 × 3[(2m + 1)!]^{2}[(2m + 2)!]^{4}(2m + 3)!d_{2m}, which is a divisor of d_{2m+2} when m ≥ 2. This completes the induction step, and hence the proof of Theorem 4.1.

5. Denominators of the Kontsevich integral

5.1. The Kontsevich integral. We briefly recall here the (modification of the) Kontsevich integral for framed links and framed tangles (see [LM1, LM2]). First we recall the definition of framed tangles.

We fix an oriented 3-dimensional Euclidean space ℝ^{3} with coordinates (x, y, t). A tangle is a smooth one-dimensional compact oriented manifold L ⊂ ℝ^{3} lying between two horizontal planes {t = a}, {t = b}, a < b such that all the boundary points are lying on two lines {t = a, y = 0}, {t = b, y = 0}, and at every boundary point L is orthogonal to these two planes. These lines are called the top and the bottom lines of the tangle.

A normal vector field on a tangle L is a smooth vector field on L which is nowhere tangent to L (and, in particular, is nowhere zero) and which is given by the vector (0, −1, 0) at every boundary point. A framed tangle is a tangle enhanced with a normal vector field. Two framed tangles are isotopic if they can be deformed by a 1-parameter family of diffeomorphisms into one another within the class of framed tangles.

Framed oriented links are special framed tangles when there is no boundary point. The empty link, or empty tangle, by definition, is the empty set.

One assigns a symbol + or − to all the boundary points of a tangle according to whether the tangent vector at this point directs downwards or upwards. Then on the top boundary line of a tangle diagram we have a word of symbols consisting of + and −. Similarly on the bottom boundary line there is another word of symbols + and −.

A non-associative word on +, − is an element of the free magma generated by +, − (see the definition of a free magma in [Ser]). For a non-associative word one w one defines its length as the number of its letters.

A q-tangle (or non-associative tangle) is a tangle together with two non-associative words w_{t}, w_{b} such that if we ignore the non-associative structure, from w_{t}, w_{b} we get the words on the top and bottom lines.

If T_{1}, T_{2} are framed q-tangles such that w_{b}(T_{1}) = w_{t}(T_{2}) we can define the product T = T_{1} T_{2} by placing T_{1} on top of T_{2}. In this case, if ξ_{1} ∈ A(T_{1}), ξ_{2} ∈ A(T_{2}) are Chinese character diagrams, then the product ξ_{1}ξ_{2} is a chord diagram in A(T) obtained by placing ξ_{1} on top of ξ_{2}.

For any two framed q-tangles T_{1}, T_{2} with the same top and bottom lines, we can define their tensor product T_{1} ⊗T_{2} by putting T_{2} to the right of T_{1}. The non-associate structure of the boundaries are the natural composition of those of T_{1}, T_{2}. Similarly, if ξ_{1} ∈ P(T_{1}), ξ_{2} ∈ P(T_{2}) are chord diagrams, then one defines ξ_{1} ⊗ ξ_{2} ∈ P(T_{1} ⊗ T_{2}) by the same way.

It is easy to see that every framed q-tangle T can be obtained from elementary q-tangles, using the product and tensor product. Here an elementary q-tangles is one of the following:

a) a trivial framed q-tangle, i.e. a bunch of vertical lines pointing downwards, the framing everywhere is given by the vector (0, −1, 0).

b) one of the framed q-tangle depicted in Figures 3. Again the framing is given by the vector (0, −1, 0).

c) the framed q-tangle T_{w_{1},w_{2},w_{3}} which has trivial underlying framed tangle, but the non-associative word of the top is w_{1}(w_{2}w_{3}), the one of the bottom is (w_{1}w_{2})w_{3}. Here w_{1}, w_{2}, w_{3} are three arbitrary non-associative words.

d) any of the above with reversing orientation on some of the components.

[figure: X^{+}; X^{−}; Y^{+}; Y^{−}] Figure 3

We will define an invariant Ẑ(T) ∈ A(T) for every framed q-tangle such that

Ẑ(T_{1}T_{2}) = Ẑ(T_{1})Ẑ(T_{2}),
Ẑ(T_{1} ⊗ T_{2}) = Ẑ(T_{1}) ⊗ Ẑ(T_{2})

With these requirements, One needs only to define Ẑ of the elementary framed qtangles. For a trivial framed q-tangle T let Ẑ(T) = T, the Chinese character diagram in A(T) without dashed graph. Put

Ẑ(X^{±}) = 1 + \frac{±1}{2}(x)^{(1)} + ⋯ + \frac{(±1)^{n}}{n!2^{n}} x^{(n)} + …,

where x^{(n)} is the Chinese character diagrams in A(X^{±}) whose dashed graph consisting of n parallel horizontal dashed lines connecting the two solid strings of the support. Let

Ẑ(T_{w_{1}w_{2}w_{3}}) = Δ^{(|w_{1}|)} ⊗ Δ^{(|w_{2}|)} ⊗ Δ^{(|w_{3}|)}(Φ).

Here |w| is the length of the word w and Δ^{(1)} = id, Δ^{(2)} = Δ, Δ^{(n)} = Δ_{1}∘Δ_{1}∘⋯∘Δ_{1} (n − 1 times). The operation Δ^{(n)} replaces one string of the support by n strings. The right hand side of the above equation means that we apply Δ^{(|w_{1}|)} to the first string, Δ^{(|w_{2}|)} to the second, and Δ^{(|w_{3}|)} to the third string of the support of Φ.

If T′ is obtained from T by reversing the orientation of a component C, then we put Ẑ(T′) = S_{C}[Ẑ(T)], where S_{C} : A(T) → A(T′) is the linear map defined as follows. Suppose D ∈ A(T) is a Chinese character diagram with m univalent vertices on C. Reversing the orientation of C, then multiplying by (−1)^{m}, from D we get S_{C}(D).

Finally, Ẑ(Y^{±}) = \sqrt{ν} where ν ∈ P_{1} is obtained from S_{C_{2}}Φ by identifying the terminating point of the first string with the beginning point of the second string, and the terminating point of the second with the beginning point of the third string. Here C_{2} is the second string.

These requirements define Ẑ(T) uniquely. It is known (see [LM1, LM2, BN2]) that Ẑ is well-defined and is an isotopy invariant of framed q-tangles. In fact, Ẑ(T) is a universal finite type invariant of framed q-tangles. For more properties of Ẑ, see [LM2]. For a knot K, the natural projection from A(S^{1}) to A(S^{1})/ ≈ takes Ẑ(K) to the Kontsevich integral of K. Here ≈ is the equivalence relation generated by: any Chinese character diagram with an isolated dashed chord is equivalent to 0. The original Kontsevich integral is given by an explicit formula (see [Ko1]).

It was proved in [LM2] that Ẑ(T) does not depend on the associator Φ if T is a link.

5.2. Denominators of the Kontsevich integral. Let A^{ℤ}(X) be the set of elements in A(X) which are linear combinations of Chinese character diagrams with integer coefficients. We consider A^{ℤ}(X) as the integral lattice of A(X).

Lemma 5.1. Suppose a q-tangle T is decomposed as T = T_{1}T_{2} … T_{k}, and that x_{i} ∈ A(T_{i}), i = 1, .., k satisfy

the denominator of the degree m part has denominator d_{m}.  (5.1)

Then the element (x_{1} … x_{k})/k! ∈ A(T) also satisfies 5.1. In particular, the product x_{1} … x_{k} satisfies 5.1.

Proof. An element of degree n in (x_{1} … x_{k})/k! has denominator k!d_{n_{1}} … d_{n_{k}}, where n_{1} + ⋯ + n_{k} = n. By Corollary 7.8 (proved in §7) k!d_{n_{1}} … d_{n_{k}} is a divisor of d_{n}.

By Theorem 4.1, Φ = exp(ϕ) = ∑_{k} ϕ^{k}/k!. Since ϕ satisfies (5.1), the previous lemma shows that Φ also satisfies (5.1). The values of Ẑ of elementary q-tangles satisfy (5.1). Hence we have

Theorem 5.2. For every framed q-tangle T, the degree m part of Ẑ(T) has denominator d_{m} = [2! … m!]^{4}(m + 1)!. In particular, the degree m part of the Kontsevich integral of any knot has denominator d_{m}.

A framed string link is a framed tangle containing no loops such that the endpoint of any component on the top line projects vertically to the other endpoint of the same component. In what follows, every framed string link will be considered as a framed q-tangle, where the non-associative words of the top and the bottom are be the same.

Let L be framed string link with l components as follows. Closing the framed string link in the same way as one closes the braids, we obtain an l component link. We assume that the framing of the closing part is given by the vector (0, −1, 0) everywhere. We define the linking number of two components of L to be the linking number of their closures. Similarly, the self-liking number of a component is the selflinking number of its closure. The linking numbers and self-linking numbers form the linking matrix.

Definition . An element x ∈ A(X) is said to have i-filter k if it is a linear sum of Chinese character diagrams, each has at least k internal vertices.

If x has i-filter k, then certainly x has i-filter k − 1.

Proposition 5.3. Suppose L is a framed string link whose matrix of linking numbers is 0. Then Ẑ(L) can be represented as a linear sum ∑_{k} z_{m}, where z_{m} has i-filter m and has denominator not divisible by any prime greater than m + 2.

Proof. It is known that Ẑ(L) ∈ P_{l} is a group-like element (see [LM3]). Hence Ẑ(L) = exp ξ, where ξ is primitive. The element ξ is a linear sum of g-connected Chinese character diagrams. Let ξ = ξ_{1} + ξ_{2} + …, where ξ_{k} has degree k. Since the linking matrix is 0, we have ξ_{1} = 0. The g-connectedness implies that ξ_{k} has i-filter k − 1.

We have ξ = ln Ẑ(L). From the formula of expansion of ln(1 + x) and Lemma 7.7 (proved in §7), it follows that ξ_{k} has denominator d_{k}.

Return to the formula

Ẑ(T) = exp(ξ_{2} + ξ_{3} + …).

Expanding the right hand side, we get a sum of terms of the form x_{n_{1}} … x_{n_{k}}/k!. This term has i-filter (n_{1} − 1) + (n_{2} − 1) + … (n_{k} − 1), and has denominator not divisible by any prime greater than the maximum q of {k, n_{1} + 1, n_{2} + 1, …, n_{k} + 1}. Let m = q − 2. Then m ≤ (n_{1} − 1) + (n_{2} − 1) + … (n_{k} − 1), since each n_{i} is ≥ 2. Hence this term has i-filter m and denominator not divisible by any prime greater than m + 2.

6. The universal perturbative invariant of homology 3-spheres

6.1. The invariant. In [LMO] we used the Kontsevich integral and the Kirby calculus to construct an invariant Ω of 3-manifolds with values in the algebra of 3-valent graphs. We briefly recall here the definition of Ω.

Let D_{n} be the vector space (over ℚ) generated by vertex-oriented 3-valent graphs with 2n vertices, subject to the anti-symmetry and the Jacobi relations. We don’t allow loop components in these 3-valent graphs, so each component contains at least one, and hence two, vertices. If loops are allowed as components of vertex oriented 3-valent graphs, we denote the vector space by D^{∘}_{n}. For n = 0 let D_{n} = ℚ. Set D = ∏_{n=0}^{∞} D_{n}; the D_{n} is the subspace of degree n. Similarly let D^{∘} = ∏_{n=0}^{∞} D^{∘}_{n}. In D we define the product of two 3-valent graph as their disjoint union. The unit is the empty graph which is in D_{0}.

We define first a couple of linear operators. Suppose x ∈ B_{l} is an l-marked Chinese character. If the number of vertices of any color is different from 2n, or if the degree of x is greater than (l + 1)n, put j′_{n}(x) = 0. Otherwise, partitioning the 2n vertices of the same color into n pairs and identifying points of each pair, from x we get a 3-valent graph which may contains some loops and which depends on the partition. Summing over all possible partitions, we get j′_{n}(x) ∈ D^{∘}.

There is a map

pr : P_{l} → A(⊔^{l}S^{l})

defined by identifying the two endpoints of each (solid) string. For x ∈ A(⊔^{l}S^{l}), let y ∈ P_{l} such that x = pr(y). Define j′_{n}(x) = j′_{n}(χ^{−1}(y)). It is easy to see that j′_{n} is well-defined. Note that j′_{n} lowers the degree by nl, and the values of j′_{n} is in

Grad_{≤n}D^{∘}.

Finally, we define ι_{n}(x) by replacing every loop in j′_{n}(x) by −2n; the result is in Grad_{≤n}D.

For a framed link L let Ž(L) be obtained by taking connected sum of Ẑ(L) with ν along every component of L. It was proved in [LMO] that ι_{n}(Ž(L), where L is a framed link, does not depend on the orientations of components of L, and does not change under the second Kirby move. Hence the element

Ω_{n}(L) = \frac{ι_{n}(L)}{ι_{n}(U_{+})^{σ_{+}}ι_{n}(U_{−})^{σ_{−}}} ∈ Grad_{≤n}D

is an invariant of the 3-manifold M obtained from S^{3} by surgery along L. Here U_{±} are the trivial knots with framing ±1, and σ_{±} are the numbers of positive and negative eigenvalues of the linking matrix of L.

This invariant takes value in Grad_{≤n}D, for a fixed n. Without loss of information, we can combine all the invariants Ω_{n} into one by putting

Ω(M) = 1 + ∑_{n=1}^{∞} Grad_{n}(Ω_{n}(M)).

The invariant Ω has the following important property: for integral homology 3spheres, it’s degree n part is a universal invariant of degree 3n (see [Le]). For the theory of finite type invariants of integral homology 3-spheres, see [Oh1]. The degrees of finite type invariants of homology 3-spheres are always multiples of 3 (see [GO]). From Ω(M), together with weight system coming from Lie algebras (see [BN1]) or from symplectic geometry (see [Kap, Ko2]) one can construct invariants of 3-manifolds with values in the space of formal power series in one variable.

6.2. Denominators of Ω. Note that if D is a Chinese character diagram with less than 2n external vertices on one solid component, then j′_{n}(D) = 0. Since j′_{n} annihilates any Chinese character diagram in A(⊔^{l}(S^{1}) of degree > (l + 1)n, it follows that if D has > 4n external vertices on one solid component, then j′_{n}(D) = 0.

An element x in D or in D^{∘} is said to have denominator N if Nx is a linear combination of 3-valent graphs with integer coefficients.

Proposition 6.1. Suppose that L is a framed link with diagonal linking matrix. Then j′_{n}(Ž(L)) has denominator not divisible by any prime greater than 2n + 1.

Proof. Recall that if T′ is obtained from T by increasing the framing of a component by 1, then Ẑ(T) is obtained from Ẑ(T) by taking connected sum with exp(θ/2) along that component, where θ is the Chinese character diagram in A(S^{1}) whose dashed graph is a dashed line (see [LM2]).

Suppose that L is obtained from a framed string link T by closing. In [LM2] we proved that

Ẑ(L) = pr[Ẑ(T) × Δ^{(l)}(ν)].

Changing the framing of each component to 0, from T we get T′. Then T′ has 0 linking matrix. One has

Ẑ(T) = Ẑ(T′) × (e^{k_{1}θ_{1}/2} ⊗ ⋯ ⊗ e^{k_{l}θ/2}),

where k_{1}, …, k_{l} are the framings of the components of T. Write Ẑ(T′) = ∑_{m} z_{m} as in Proposition 5.3. Similarly, we can write ν = ∑_{q} y_{q} where y_{q} has i-filter q and denominator not divisible by any prime greater than q + 2. We have

Ẑ(L) = pr[(∑_{m} z_{m}) × Δ^{(l)}(∑_{q} y_{q}) × (e_{1}^{k}θ_{1}/2 ⊗ ⋯ ⊗ e^{k_{l}θ/2}].

If we expand the right hand side, then we get a sum of terms of the form

pr[z_{m} × Δ^{(l)}(y_{q}) × (\frac{θ^{n_{1}}}{n_{1}!2^{n_{1}}} ⊗ ⋯ ⊗ \frac{θ^{n_{l}}}{n_{l}!2^{n_{l}}})].

Since j′_{n} annihilates any Chinese character with i-filter 2n + 1, we can restrict to the case with m, q ≤ 2n. And since j′_{n} annihilates any Chinese character diagram with with more than 4n external vertices on one components, we may assume that all the n_{k} are less than or equal to 2n. This means the above term has denominator not divisible by any prime greater than 2n + 1. It remains to use Lemma 6.4 (proved below) to conclude that j′_{n} of this term has denominator not divisible by any prime greater than 2n + 1.

Theorem 6.2. Suppose that M is a rational homology 3-spheres, i.e. H_{1}(M, ℚ) = 0. Then the degree n part of Ω(M) has denominator not divisible by any prime greater than 2n + 1.

Proof. It’s sufficient to show that Ω_{n}(M) has denominator not divisible by any prime greater than 2n + 1. Suppose M is obtained from S^{3} by surgery along a framed link L with diagonal linking matrix. Then by the previous proposition, j′_{n}(Ž(L)) has denominator not divisible by any prime greater than 2n + 1 , hence so do ι_{n}(Ž(L)) and Ω_{n}(M).

For an arbitrary rational homology 3-sphere M, Ohtsuki [Oh2] showed there are lens spaces M_{i} of type (k_{i}, 1), i = 1, ..., k such that the connected of M and all the M_{i} is obtained from S^{3} by surgery along a framed link with diagonal linking matrix.

Since (see [LMO])

Ω_{n}(M#M′) = Ω_{n}(M)Ω_{n}(M′),

it’s sufficient to consider the case when M is obtained from S^{3} by surgery along a framed link L with diagonal matrix.

6.3. On the map j′_{n}. Let A(m), for positive number m, be the subspace of B_{m} spanned by m-marked Chinese characters which have exactly 1 vertex of each color {1, 2, …, m}. Let A(0) = ℚ. The symmetric group 𝔖_{m} acts on A(m) by permuting the colors. For an element τ in the symmetric group 𝔖_{m−2} acting on the set {2, 3, …, m − 1}, let T_{τ} ∈ A(m) be the graph shown in Figure 4.

1  m τ(2) τ(3)  τ(m − 1) ⋯

Figure 4. The definition of T_{τ}

We define T_{m} ∈ A(m) by

T_{m} = ∑_{τ∈𝔖_{m−2}} \frac{(−1)^{r(τ)}}{(m − 1)\binom{m−2}{r(τ)}}T_{τ},

where we denote by r(τ) the number of k which satisfies τ(k) > τ(k + 1). The coefficients here look very similar to those of the Eulerian idempotent e_{m−1}^{(1)}, see (2.1).

There is a shuffle product in the space ∏_{m=0}^{∞} A(m) defined as follows. Suppose D is a Chinese character in A(m), D′ – in A(m′). Change the colors of external vertices of D′: 1 to m + 1, 2 to m + 2, etc., m′ to m + m′. The union of D and D′ now is an element of A(m + m′). Define

D • D′ := ∑_{(m, m′)-shuffles σ ∈ 𝔖_{m+m′}} σ(D ∪ D′).

Let T = ∑_{m=0}^{∞} T_{m}. Let T^{•n} be the n-th power of T in the shuffle product. Denote by T_{m}^{n} the part of (T^{•n}/n!) with m external vertices. In other words,

T_{m}^{n} = \frac{1}{n!} ∑_{m_{1}+···+m_{n}=m} T_{m_{1}} • ⋯ • T_{m_{n}}.  (6.1)

Note that if m < 2n, then, by definition, T_{m}^{n} = 0. The first non-trivial element T_{2n}^{n} ∈ A(2n) is the following. Partition 2n points {0, 1, …, 2n − 1} into n pairs (there are (2n−1)!! ways to do this), and then connect the two points of each pair by a dashed line, we get an element of A(2n). Summing up all such possible elements, we get T_{2n}^{n}.

In [LMO] it was proved that the T_{m}^{n} satisfy the following important properties:

1) T_{m}^{n} is invariant under the cyclic permutation of the m external vertices.

2) For every n, m we have

T_{m}^{n} − (k k + 1)(T_{m}^{n}) = T_{m−1}^{n} ⋆_{k} Y,  (*)

where (k k + 1) is the permutation which interchanges k and k + 1 (1 ≤ k ≤ m − 1), and T_{m−1}^{n} ⋆_{k} Y denotes the element obtained from T_{m−1} by attaching a Y-shaped graph to the vertex k and then re-numbering the vertices so that the remaining two vertices of Y are k and k + 1. Equation (*) is also presented in Figure 5.

[figure: T_{m−1}^{n}; =] [figure: T_{m}^{n}; −] [figure: T_{m}^{n}]

Figure 5. The dual of the STU relation

For a fixed number n, we define a linear map j_{n} : A(⊔^{l}S^{1}) → D^{∘} as follows. Suppose that D ∈ A(⊔^{l}S^{1}) is a Chinese character diagram and that C is a solid circle of D with m external vertices on it. Number the vertices, beginning at any vertex and following the orientation of C, by 1,2,...,m. Now remove the solid circle C, and glue the external vertices to the corresponding vertices of T_{m}^{n}. Do this with all solid circles of the Chinese character diagram; and we get j_{n}(D).

The well-definedness (because of the STU relation) of this map follows from equation (*), and this equation can be regarded as the dual to the STU.

It follows from the definition that j_{n} lowers the degree of a chord diagram by ln, where l is the number of solid circles of D.

Lemma 6.3. Suppose x ∈ A(⊔^{l}S^{1}) is a Chinese character diagram of degree less than or equal to n(l + 1). Then j′_{n}(x) = j_{n}(x).

Proof. Since χ is an isomorphism, we can assume that x = pr(χ(y)), where y is an l-marked Chinese character. Suppose that y has exactly 2n vertices of each color. Then j_{n}(x) is obtained by gluing T_{2n}^{n} to each group of vertices of the same color of y, which is, by definition, j′_{n}(x). Hence in this case j_{n}(x) = j′_{n}(x).

Now suppose one of the color, say k, of y has m vertex, where m ≠ 2n. In this case j′_{n} = 0. If m < 2n, then j_{n}(x) = 0, since T_{m}^{n} = 0. Suppose m > 2n. Then T_{m}^{n} is a linear combination of Chinese character diagrams, each has at least one internal vertex, and each is a union of several Chinese character of the form T_{τ} (see Figure 4). Hence

∑_{σ∈𝔖_{m}} σ(T_{m}^{n}) = 0,

due to the anti-symmetry relation. On the other hand, the vertices of color k of the element χ(y) is invariant under actions of 𝔖_{m}. Hen when we glue T_{m}^{n} to the set of vertices of color k, we get 0.

Proposition 6.4. Suppose D is a Chinese character diagram in A(⊔^{l}S^{1}). Then j′_{n}(D) has denominator not divisible by any prime greater than 2n + 1.

Proof. We can assume that the degree of D is less than or equal to n(l+1), since otherwise j′_{n}(D) = 0. By the previous proposition, j′_{n}(D) = j_{n}(D) which is constructed using T_{m}^{n}, where m is between 2n and 4n. In (6.1), m_{1} + ⋯ + m_{n} = m, hence the maximum of {m_{1}, …, m_{n}} is less than or equal to 2n + 2. From the definition it follows that the denominator of these T_{m_{i}} is not divisible by any prime greater than 2n + 1.

7. Proof of Propositions 2.2, 4.2 and 4.4

7.1. The cobar complex of Chinese characters. The isomorphism χ between P_{n} and B_{n} carries the maps Δ_{i}, ε_{i} over to B_{n}. These maps can be described as follows.

Suppose x is an n-marked Chinese character, with m vertices of color i. Here i ≤ n is a fixed number. There are 2^{m} ways of partition the set of vertices of color i into an ordered pair of subsets, the first and the second subsets. For each such partition, form an (n + 1)-marked Chinese character by first changing the color k to k + 1 for every k > i, then coloring vertices in the first subset by i, in the second by i + 1, and leave alone the vertices of color < i. Summing up, over all possible partitions, such (n + 1)-marked Chinese characters, we get Δ_{i}(x).

If there is at least one vertex of color i, let ε_{i}(x) = 0. Otherwise ε_{i}(x) is the (n − 1)-marked Chinese character obtained from x by changing the color k to k − 1 for every k > i.

It is easy to check that Δ_{i}, ε_{i} commute with χ.

For an n-marked Chinese character, the number of external vertices is called the e-grading, while half the number of all vertices is called the degree. Note that all the mappings Δ_{i}, ε_{i} preserve both the e-grading and degree, and χ preserves the degree.

The operator d acting on P_{n} is carried by χ over B_{n} to d : B_{n} → B_{n+1}, where

d(ξ) = 1 ⊗ ξ − Δ_{1}(ξ) + Δ_{2}(ξ) − ⋯ + (−1)^{n}Δ_{n}(ξ) + (−1)^{n+1}ξ ⊗ 1.

Here 1 ⊗ ξ is the (n + 1)-marked Chinese character obtained from x by changing the color k to k + 1 (for every k), and ξ ⊗ 1 – by leaving alone the colors.

7.2. Subcomplexes. An element x ∈ B_{n} is non-degenerate if ε_{i}(x) = 0 for i = 1, 2, …, n. Let F_{n} be the subspace of B_{n} spanned by connected non-degenerate Chinese characters. Let F_{n}^{ℤ} be the subset of F_{n} consists of elements which are linear combinations of connected non-degenerated Chinese character with integer coefficients. Then

F_{n} = F_{n}^{ℤ} ⊗_{ℤ} ℚ,

and we regard F_{n} as the integral lattice of F_{n}. Note that χ maps F_{n} isomorphically on Q_{n}. Since χ preserves the degree, we can split this isomorphisms into smaller ones.

Let F_{n}^{ℤ}(m) (respectively, F_{n}^{ℤ}(m, k)) be the subset of F_{n}^{ℤ} consisting of elements of degree m (respectively, degree m and e-grading k). Every connected Chinese character of degree m has at most m + 1 external vertices. Hence

F_{n}^{ℤ}(m) = ⊕_{k=1}^{m+1}F_{n}^{ℤ}(m, k).

It’s important that the above summation is up to k = m + 1. One has

F_{n}(m, k) = F_{n}^{ℤ}(m, k) ⊗ ℚ.

Let

C^{n}(F^{ℤ}(m)) = F_{n}^{ℤ}(m),
C^{n}(F^{ℤ}(m, k)) = F_{n}^{ℤ}(m, k).

7.3. The inverse of χ. Let H(m) be the subspace of B_{m+1} spanned by (m + 1)- marked Chinese characters which have one vertex of each color 1,2, …, m. The other vertices must have color m + 1; and we recolor these ones so that they have color 0. We define a shuffle product H(m_{1}) ⊗ H(m_{2}) →^{•} H(m_{1} + m_{2}) as follows. Suppose D, D′ are Chinese characters in A(m_{1}), A(m_{2}), respectively. Change the colors of D′: 1 to m_{1} + 1, 2 to m_{2} + 1, etc, and m_{2} to m_{1} + m_{2}, leaving the 0 color alone. Now D ∪ D′ is an element of H(m_{1} + m_{2}). Let

D • D′ = ∑_{(m_{1}, m_{2})-shuffles σ ∈ 𝔖_{m_{1}+m_{2}}} σ(D ∪ D′).

Lemma 7.1. If x ∈ ∏_{m=0}^{∞} H(m) has denominator 1, then so does x^{•n}/n!.

Proof. In the shuffle product x^{•n} we have repeated terms obtained by permuting the factor x. This cancels the denominator n!.

Consider the element T_{m+1} ∈ A_{m+1} defined in §6.3. Changing the color m + 1 to 0, from T_{m+1} we get t_{m} ∈ H_{m}. Let

t = t_{1} + t_{2} + ⋯ + t_{m} + ⋯ ∈ ∏_{m=0}^{∞} H(m).

Then e^{t} ∈ H, where we use the shuffle product. Let t(m) be the part of e^{t} lying in H_{m} (m ≥ 1). In other words,

t(m) = ∑_{n=1}^{∞}[\frac{1}{n!}( ∑_{m_{1}+···+m_{n}=m} t_{m_{1}} • ⋯ • t_{m_{n}})].

The inverse of χ can be expressed by e^{t} as follows. Suppose D ∈ P_{l} is a Chinese character diagram. Suppose on the i-th string there are m external vertices. Remove the i-th string, then glue the vertices of colors 1,2,...,m of t(m) to the external vertices of the i-th string, and finally change the color of the other vertices of t_{m} from 0 to i. Do this with all the strings. The result is χ^{−1}(D). The well-definedness follows from (*). The fact that χ(χ^{−1}(D)) = D is easy to verify.

Proposition 7.2. a) If x is a Chinese character in B_{l} having m external vertices, then χ(x) has denominator m!.

If D is a Chinese character diagram in P_{n} with m external vertices then χ^{−1}(D) has denominator m!.

b) If x is a Chinese character in F_{l} having m external vertices, where l ≥ 2, then χ(x) has denominator [(m − 1)!]^{2}.

If D is a Chinese character diagram in Q_{l} having m external vertices, where l ≥ 2, then χ^{−1}(x) has denominator [(m − 1)!]^{2}.

Proof. a) The first statement follows directly from the definition of χ. Let us prove the second statement. As discussed above, χ^{−1} can be constricted explicitly using e^{t}. By Lemma 7.1 t^{n}/n! has the same denominator as t. From the formula of t_{m} = T_{m+1} in §6.3 it follows that has denominator m!.

b) follows from the proof of a) with the following observation: if both p, q are positive integers, then p!q! is a divisor of (p + q − 1)!(p + q − 2)!.

7.4. Some cohomology. Instead of the complex (C^{∗}(Q^{ℤ}(m)), d), we will study the complex C^{∗}(F^{ℤ}(m)), d) which is the direct sum of the complexes C^{∗}(F^{ℤ}(m, k)), d), with k = 1, 2, …, m + 1:

C^{∗}(F^{ℤ}(m)), d) = ∑_{k=1}^{m+1} C^{∗}(F^{ℤ}(m, k)), d)  (7.1)

For a fix number k consider the following ℤ-complex E(k). Let C^{n}(E(k)) be the ℤ-module spanned by partitions θ_{1}, …, θ_{n} of {1, 2, …, k}, such that each θ_{i} is a non-empty subset of {1, 2, …, k}. So if n > k then C^{n}(E(k)) = 0. Define d : C^{n}(E(k)) → C^{n+1}(E(k)) by

d(θ_{1}, …, θ_{n}) = (dθ_{1}, θ_{2}, …, θ_{n}) − (θ1, dθ_{2}, θ_{3}, …, θ_{n}) + ⋯ + (−1)^{k−1}(θ_{1}, …, dθ_{n}),

where for a non-empty set θ we set dθ = ∑(θ′, θ′′), the sum is over all possible partition of θ into an order pair θ′, θ′′ of non-empty subsets. Actually, C^{∗}(E(k)), d) is the cochain complex of a triangulation of the k-dimensional punctured sphere (see, for example, [LM2], §9.2), and we have

Proposition 7.3. One has that H^{k}(C^{∗}(E(k))) = ℤ, and H^{l}(C^{∗}(E(k))) = 0 for l ≠ k.

The symmetric group 𝔖_{k} acts on the left on the complex C^{∗}(E(k)) by permuting the numbers in the partitions. The action is compatible with the operator d. We will see that C^{∗}(F^{ℤ}(m, k)), d) is isomorphic to the tensor product the complex C^{∗}(E(k)), d) and a 𝔖_{k}-module which we are going to describe.

Let Γ^{ℤ}(m, k) be the subset of F_{k} of ℤ-linear combination of non-degenerate kmarked Chinese character of degree m with exactly one vertex of each color {1, 2, …, k}. And let Γ(m, k) = Γ^{ℤ}(m, k) ⊗_{ℤ} ℚ. The symmetric group 𝔖_{k} acts on the right on Γ^{ℤ}(m, k) by permuting the colors of external vertices.

Lemma 7.4. For fixed m, k, there is an isomorphism between complexes

(C^{∗}(F^{ℤ}(m, k)), d) and Γ(m, k) ⊗_{𝔖_{k}} (C^{∗}(E(k)), d).

Proof. Consider a k-marked Chinese character ξ in Γ^{ℤ}(m, k). The k external vertices are colored by {1, 2, …, k}. We map the element ξ ⊗ (θ_{1}, …, θ_{n}) to the element η of C^{n}(F^{ℤ}(m, k)) obtained from ξ by changing the colors in θ_{i} to i. It can be verified at once that this is an isomorphism between the two complexes.

Lemma 7.5. The cohomology groups of the complex Γ^{ℤ}(m, k) ⊗_{𝔖_{k}} (C^{∗}(E(k)), d) is annihilated by k!, except for the k-th cohomology group.

Proof. First let us consider the tensor product over ℤ: Γ^{ℤ}(m, k)⊗_{ℤ}C^{∗}(E(k)), d). This complex has 0 cohomology, except for the cohomology of dimension k, by Proposition 7.3 and the universal coefficient formula for homology.

There is a natural projection

p_{n} : Γ^{ℤ}(m, k) ⊗_{ℤ} C^{n}(E(k)) → Γ^{ℤ}(m, k) ⊗_{𝔖_{k}} C^{n}(E(k)).

The kernel ker(p_{n}) is a finite ℤ-module whose cardinality is a divisor of k!. Consider the short exact sequence of cochain complexes:

0 → ker(p_{n}) → Γ^{ℤ}(m, k) ⊗_{ℤ} C^{n}(E(k)) → Γ^{ℤ}(m, k) ⊗_{𝔖_{k}} C^{n}(E(k)) → 0.

By the above observation, the cohomology of the middle complex vanishes unless in dimension k, while the cohomology of the left one is annihilated by k!. From the long exact sequence derived from this short exact sequence, we get the lemma.

Lemma 7.6. If x ∈ C^{4}(F^{ℤ}(m, k)) is symmetric and dx = 0, then there is a symmetric y ∈ C^{3}(F(m, k)), such that dy = x. Moreover, y has denominator 2 k!.

Proof. By Lemma 7.4 we may suppose that x ∈ Γ^{ℤ}(m, k) ⊗_{𝔖_{k}} C^{4}(E(k)). Consider two cases: k = 4 and k ≠ 4.

Suppose k = 4. Then x is equal to γ ⊗ e, where γ ∈ Γ^{ℤ}(m, 4) and

e = ({1}, {2}, {3}, {4}),

a partition of {1, 2, 3, 4}.

Let

y′ = (12, 3, 4) − (2, 13, 4) + (2, 3, 14) − (2, 34, 1) + (24, 3, 1) − (4, 23, 1),

where, for example, (2, 13, 4) means the element γ ⊗({2}, {1, 3}, {4}) in Γ^{ℤ}(m, 4)⊗_{𝔖_{4}} C^{3}(E(4)). One can readily check that dy′ = x − x^{4321} = 2x.

Let y = (1/4)[y′ − (y′)^{321}]; we see that y is symmetric and dy = x.

Now consider the case k ≠ 4. By Lemma 7.5 there is y′ ∈ Γ^{ℤ}(m, k) ⊗_{𝔖_{k}} C^{3}(E(k)) such that dy′ = (k!)x. Now y = (1/2 k!)[y′ − (y′)^{321}] is the element to find.

7.5. Proof of Proposition 2.2. Suppose ξ ∈ C_{sym}^{4}(Q^{ℤ}) is of degree m and dξ = 0. Then ξ′ = χ^{−1}(ξ) is symmetric and in C^{4}(F^{ℤ}(m)). In addition, χ(ξ) has denominator (m!)^{2}, since ξ, as an element of degree m, can have at most m + 1 external vertices. (Here we use the fact that ξ is a linear combination of g-connected elements, and Proposition 7.2, part b)).

Using the decomposition (7.1), we can assume that ξ′ is in C^{4}(F(m, k)) with some k ≤ m+1. Besides ξ′ is symmetric, has denominator (m!)^{2} and dξ′ = 0. From Lemma 7.6 it follows that there is y ∈ C^{3}(F(m, k)), symmetric, with denominator 2(m!)^{2}k! such that dy = ξ′. Since k ≤ m + 1, the number 2m!k! is a divisor of 2m!(m + 1)!. Now let η = χ^{−1}y. Then dη = χ^{−1}(ξ′) = ξ, and η is symmetric. By Proposition 7.2, η has denominator 2(m!)^{2}(m + 1)!. This completes the proof of Proposition 2.2.

7.6. On the numbers d_{n}. We will write p ⊲ q if p is a divisor of q.

Lemma 7.7. a) For any positive integers p, q one has (p + q)! d_{p}d_{q} ⊲ d_{p+q}. b) For any positive integers p, q one has d_{p}d_{q} ⊲ 2d_{p+q−1}.

c) For any integers p, q ≥ 2 one has d_{p}d_{q} ⊲ 96d_{p+q−2}.

Proof. We will prove a). The others can be proved in a similar way.

Note that d_{p+1} = d_{p} × [(p + 1)!]^{3}(p + 2)!. We use induction on q. The statement is true for q = 1. Suppose it has been true for q. Then,

(p + q + 1)! d_{p}d_{q+1} = (p + q + 1)!d_{p}d_{q}[(q + 1)!]^{3}(q + 2)!

By the induction hypothesis, the latter is a divisor of (p + q + 1)d_{p+q}[(q + 1)!]^{3}(q + 2)!

Since (p + q + 1)(q + 1)! ⊲ (p + q + 1)!, the number (p + q + 1)d_{p+q}[(q + 1)!]^{3}(q + 2)! is a divisor of d_{p+q}[(p + q + 1)!]^{3}(p + q + 2)! which is d_{p+q+1}. This completes the proof of a).

Corollary 7.8. For every positive integers n_{1}, n_{2}, … n_{k}, the number k!d_{n_{1}}d_{n_{2}} … d_{n_{k}} is a divisor of d_{n_{1}+n_{2}+...n_{k}}.

Proof. From part a) of the previous lemma, one has that

(n_{1} + n_{2} + ⋯ + n_{k})!d_{n_{1}}d_{n_{2}} … d_{n_{k}} ⊲ d_{n_{1}+n_{2}+...n_{k}}.

It remains to notice that k ≤ (n_{1}+n_{2}+⋯+n_{k}), since each n_{i} is a positive integer.

We recall here the Campbell-Hausdorff formula. Let B = ℚ⟨⟨x^{(1)}, x^{(2)}, …, x^{(l)}⟩⟩ be the algebra of formal power series in l non-commuting variables. This algebra is graded by the degree of monomials in x^{(j)}, j = 1, 2..., l. The free Lie algebra L over ℤ (the set of integers) generated by x^{(j)}, j = 1, 2..., l is a subset of B, and an element in B is called a Lie polynomial if it is a ℚ-linear combination of elements in L.

The Campbell-Hausdorff formula says that

exp(x^{(1)}) … exp(x^{(l)}) = exp[∑_{k=1}^{∞} f_{k}(x^{(1)}, …, x^{(l)})],

where f_{k} is a homogeneous Lie polynomial of total degree k. Moreover from the Dynkin’s form of the Campbell-Hausdorff formula (see, for example [Ser]), one can easily see that each f_{k} is a ℚ-linear combination of elements in L whose coefficients have denominators (k!)^{2}.

7.7. Proof of Lemmas 4.2 and 4.4. We have that (modulo part of degree ≥ 2m + 3)

1 + ψ = exp(ϕ_{2m}^{312}) × exp(\frac{r^{13}}{2} ) × exp(−ϕ_{2m}^{132}) × exp(\frac{r^{23}}{2} ) × exp(ϕ_{2m}) × exp(\frac{−r^{13} − r^{23}}{2} ).

Hence

ψ = ∑_{k} f_{k}(ϕ_{2m}^{312}, \frac{r^{13}}{2} , −ϕ_{2m}^{132}, \frac{r^{23}}{2} , ϕ_{2m}, \frac{−r^{13} − r^{23}}{2} ).  (7.2)

Recall that ϕ_{2m} is a sum of even degree parts, the part of degree 2k has denominator d_{2k}. The element r^{ij}/2 has degree 1, and has denominator d_{1} = 2. Replacing ϕ_{2m} by the sum of its degree 2k parts, k = 1, …, m, then expanding the right hand side, we see that ψ is a sum of elements having denominator (k!)^{2}d_{n_{1}} … d_{n_{k}}, where n_{1} + ⋯ + n_{k} = 2m + 2. So we need to show that (k!)^{2}d_{n_{1}} … d_{n_{k}} is a divisor of [(2m + 2)!]^{2}d_{2m} if n_{1} + ⋯ + n_{k} = 2m + 2. Since all n_{i} ≤ 2m, the number k is greater than 1. We assume that n_{1} ≤ n_{2} ≤ ⋯ ≤ n_{k}. Consider several cases.

The case k = 2. Then each of n_{1}, n_{2} must be greater than 1. By Lemma 7.7, part c), (k!)^{2}d_{n_{1}}d_{n_{2}} = 4d_{n_{1}}d_{n_{2}} is a divisor of 4 × 96d_{2m}, which, in turn, is a divisor of [(2m + 2)!]^{2}d_{2m} if m ≥ 2.

The case 2m + 1 ≥ k ≥ 3. Applying Lemma 7.7, part b), repeatedly, we have that

d_{n_{1}}d_{n_{2}} … d_{n_{k}} ⊲ 2^{k−1}d_{n_{1}+n_{2}+···+n_{k}−k+1} = 2^{k−1}d_{2m+3−k}.

If k = 3, then (k!)^{3}d_{n_{1}}d_{n_{2}}d_{n_{3}} ⊲ (3!)^{2} × 2^{2} × d_{2m}, which is a divisor of [(2m + 2)!]^{2}d_{2m}. If k ≥ 4, then 2m + 3 − k ≤ 2m − 1. Noting that k ≤ 2m + 1, one has

(k!)^{2}d_{n_{1}} … d_{n_{k}} ⊲ 2^{k−1}(k!)^{2}d_{2m−1} ⊲ 2^{2m}[(2m + 1)!]^{2}d_{2m−1}.

The latter is a divisor of [(2m + 2)!]^{2}d_{2m}.

The case k = 2m + 2. Then each n_{i}, i = 1, 2, ..., 2m + 2 is 1. One has

(k!)^{2}d_{n_{1}} … d_{n_{2m+2}} = 2^{2m+2}[(2m + 2)!]^{2}

is a divisor of [(2m + 2)!]^{2}d_{2m}. This completes the proof of Lemma 4.2. The proof of Lemma 4.4 is similar, even easier.

26 THANG LE

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