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A Unified Quantum SO(3) Invariant for Rational Homology 3-Spheres

Given a rational homology 3-sphere M with the first integral homology of rank b and a link L inside M, colored by odd numbers, we construct a unified invariant I_{M,L} belonging to a modification of the Habiro ring where b is inverted. Our unified invariant dominates the whole set of the SO(3) Witten-Reshetikhin-Turaev invariants of the pair (M,L). If b=1 and L is empty, I_M coincides with Habiro's invariant of integral homology 3-spheres. For b>1, the unified invariant defined by the third author is determined by I_M. One of the applications are the new Ohtsuki series (perturbative expansions of I_M at roots of unity) dominating all quantum SO(3) invariants.


Introduction
Background. The SU (2) Witten-Reshetikhin-Turaev (WRT) invariant is defined for any closed oriented 3-manifold M and any root of unity ξ [30]. Kirby and Melvin [13] introduced the SO(3) version of the invariant τ M (ξ) ∈ Q(ξ) for roots of unity ξ of odd order. If the order of ξ is prime, then by the results of Murakami [23] (also Masbaum-Roberts [21]), τ M (ξ) is an algebraic integer. This integrality result was the starting point for the construction of finite type 3-manifold invariants, Ohtsuki series [25], integral TQFTs, representations of the mapping class group over Z[ξ] [6], and categorification of quantum 3-manifold invariants [11]. The proofs in [23] and [21] depend heavily on the arithmetic of Z[ξ] for a root of unity ξ of prime order and do not extend to other roots of unity.
Is it true that τ M (ξ) is always an algebraic integer (belongs to Z[ξ]), even when the order of ξ is not a prime? The positive answer to this question was given first for integral homology spheres by Habiro [7], and then for arbitrary 3-manifolds by the first and third author [3], in connection with the study of "strong integrality". What Habiro proved for integral homology 3-spheres is actually much stronger than integrality. For any integral homology 3-sphere M , Habiro [7] constructed a unified invariant J M whose evaluation at any root of unity coincides with the value of the Witten-Reshetikhin-Turaev invariant at that root. Habiro's unified invariant J M is an element of the following ring (Habiro's ring) ((q; q) k ) , where (q; q) k = k j=1 (1 − q j ).
Every element f (q) ∈ Z[q] can be written as an infinite sum with f k (q) ∈ Z[q]. When q = ξ, a root of unity, only a finite number of terms on the right hand side are not zero, hence the right hand side gives a well-defined value, called the evaluation ev ξ (f (q)). Since f k (q) ∈ Z[q], ev ξ (f (q)) ∈ Z[ξ] is an algebraic integer. The fact that the unified invariant belongs to Z[q] is stronger than just integrality of τ M (ξ). We will refer to it as "strong" integrality. The Habiro ring has beautiful arithmetic properties. Every element f (q) ∈ Z[q] can be considered as a function whose domain is the set of roots of unity. Moreover, there is a natural Taylor series for f at every root of unity. Two elements f, g ∈ Z[q] are the same if and only if their Taylor series at a root of unity coincide. In addition, each function f (q) ∈ Z[q] is totally determined by its values at, say, infinitely many roots of order 3 n , n ∈ N. Due to these properties the Habiro ring is also called a ring of "analytic functions at roots of unity" [7]. Thus belonging to Z[q] means that the collection of the SO(3) WRT invariants is far from a random collection of algebraic integers; together they form a nice function.
Perturbative expansion at 1 of WRT invariants for rational homology 3-spheres was first constructed by Ohtsuki in the case when the order of the quantum parameter ξ is prime [26]. General properties of the Habiro ring imply that for any integral homology 3-sphere M , the Taylor expansion of the unified invariant J M at q = 1 coincides with the Ohtsuki series and dominates WRT invariants of M at all roots of unity (not only of prime order).
To generalize Habiro's results to rational homology 3-spheres, new ideas and techniques are required. Strong integrality of quantum invariants for rational homology 3-spheres was studied in [15] and [3]. Among other things, in [15] a unified invariant was constructed for the case when the order r of the quantum parameter ξ is coprime with b. In [3], it was proved that for any 3-manifold M (not necessary a rational homology 3-sphere), the SO(3) WRT invariant τ M (ξ) is always an algebraic integer, i.e. τ M (ξ) ∈ Z[ξ] with no restriction on the order of ξ at all. There we used a (2nd order) Laplace transform method [2] and a difficult identity of Andrews [1] in q-calculus, generalizing those of Rogers-Ramanujan.
Thus, although we have had integrality of all SO(3) WRT invariants, we still lacked a "strong integrality" for the case when (r, b) = 1. This is the main object of this paper.
In this paper we will generalize Habiro's construction of the unified invariant to all rational homology 3-spheres. Our new unified invariant I M dominates SO(3) WRT invariants also in the case when the order r of the quantum parameter is not coprime with b = |H 1 (M, Z)|. Although this includes the case (r, b) = 1 of [15], the ring our invariant belongs to is simpler than the one obtained in [15] and [3]. In particular, we don't need any fractional power of q. We show that the Taylor expansion of our unified invariant at a root of unity of order c (new Ohtsuki series) dominates all WRT invariants with r = cl and (l, b) = 1.
Results. The WRT or quantum SO(3) invariant τ M,L (ξ) is defined for a pair of a closed 3manifold M and a link L in it, with link components colored by integers. Here ξ is a root of unity of odd order. We will recall the definitions in Section 1.
Suppose M is a rational homology 3-sphere, i.e. where L(b, a) denotes the (b, a) lens space. We will see that τ L(b,1) (ξ) is always nonzero. For any positive integer b, we define the cyclotomic completion ring R b to be .
For any f (q) ∈ R b and a root of unity ξ of odd order, the evaluation ev ξ (f (q)) := f (ξ) is welldefined. Similarly, we put ((q; q) k ) .
Here the evaluation at any root of unity is well-defined. For odd b, there is a natural embedding S b → R b , see Section 3. Let us denote by M b the set of rational homology 3-spheres such that |H 1 (M, Z)| divides b n for some n. The main result of this paper is the following. If b = 1 and L is the empty link, I M coincides with Habiro's unified invariant J M . The proof of Theorem 1 uses the Laplace transform method and Andrew's identity. We also construct a Frobenius type isomorphism to get rid of the formal fractional power of q that appeared in [15], [3]. As a byproduct, we generalize the deep integrality result of Habiro (Theorem 8.2 in [7]), underlying the construction of the unified invariants, to a union of an algebraically split link with any odd colored one.
The rings R b and S b have properties similar to those of the Habiro ring. An element f (q) ∈ R b is totally determined by the values at many infinite sets of roots of unity (see Section 3), one special case is the following.
Proposition 2. Let p be an odd prime not dividing b and T the set of all integers of the form p k b ′ with k ∈ N and b ′ any odd divisor of b n for some n. Any element f (q) ∈ R b , and hence also {τ M (ξ)}, is totally determined by the values at roots of unity with orders in T .
The Ohtsuki series [26,16], originally defined through some arithmetic congruence property of the SO(3) invariant, can be identified with the Taylor expansion of I M at q = 1 [7,15]. We will also investigate the Taylor expansions of I M at roots of unity and show that these Taylor expansions satisfy congruence relations similar to the original definition of the Ohtsuki series, see Section 4.
Plan of the paper. In Section 1 we recall known results and definitions. In the next section we explain the strategy of our proof of Theorem 1. In Sections 3 and 5, we develop properties of cyclotomic completions of polynomial rings. New Ohtsuki series are discussed in Section 4. The unified invariant of lens spaces, needed for the diagonalization, is defined in Section 6. The main technical result of the paper based on Andrew's identity is proved in Section 7. The Appendix is devoted to the proof of the generalization of Habiro's integrality theorem.
Throughout this paper, ξ will be a primitive root of unity of odd order r and e n := exp(2πI/n). All 3-manifolds in this paper are supposed to be closed and oriented. Every link in a 3-manifold is framed, oriented, and has components ordered.
In this paper, L ⊔ L ′ denotes a framed link in S 3 with disjoint sublinks L and L ′ , with m and l components, respectively. Surgery along the framed link L transforms (S 3 , L ′ ) into (M, L ′ ). We use the same notation L ′ to denote the link in S 3 and the corresponding one in M .
1.2. The colored Jones polynomial. Suppose L is a framed, oriented link in S 3 with m ordered components. For positive integers n 1 , . . . , n m , called the colors of L, one can define the quantum invariant J L (n 1 , . . . , n m ) ∈ Z[q ±1/4 ], known as the colored Jones polynomial of L (see e.g. [30,22]). Let us recall here a few well-known formulas. For the unknot U with 0 framing one has If L 1 is obtained from L by increasing the framing of the ith component by 1, then (4) J L1 (n 1 , . . . , n m ) = q (n 2 i −1)/4 J L (n 1 , . . . , n m ).
If all the colors n i are odd, then J L (n 1 , . . . , n m ) ∈ Z[q ±1 ].
1.3. Evaluation and Gauss sums. For each root of unity ξ of odd order r, we define the evaluation map ev ξ by replacing q with ξ. Suppose f (q; n 1 , . . . , n m ) is a function of variables q ±1 and integers n 1 , . . . , n m . In quantum topology, the following sum plays an important role ni odd ev ξ f (q; n 1 , . . . , n m ) where in the sum all the n i run over the set of odd numbers between 0 and 2r. In particular, the following variation of the Gauss sum is well-defined, since for odd n, 4 | n 2 − 1. It is known that, for odd r, |γ b (ξ)| is never 0.
1.4. Definition of the WRT invariant. Suppose the components of L ′ are colored by fixed integers j 1 , . . . , j l . Let An important special case is when L = U b , the unknot with framing b = 0, and L ′ = ∅. In that case F U b (ξ) can be calculated using the Gauss sum and is nonzero, see Section 6 below.
Let σ + (respectively σ − ) be the number of positive (negative) eigenvalues of the linking matrix of L. Then the quantum SO(3) invariant of the pair (M, L ′ ) is defined by (see e.g. [13,30]) The invariant τ M,L ′ (ξ) is multiplicative with respect to the connected sum.
For example, the SO(3) invariant of the lens space L(b, 1), obtained by surgery along U b , is were sn(b) is the sign of the integer b. Let us focus on the special case when the linking matrix of L is diagonal, with b 1 , b 2 , . . . , b m on the diagonal. Assume each b i is a power of a prime up to sign. Then Thus from the definitions (5), (6) and (1) we have

1.5.
Habiro's cyclotomic expansion of the colored Jones polynomial. Recall that L and L ′ have m and l components, respectively. Let us color L ′ by fixed j = (j 1 , . . . , j l ) and vary the colors n = (n 1 , . . . , n m ) of L. For non-negative integers n, k we define For k = (k 1 , . . . , k m ) let A(n j , k j ).
Note that A(n, k) = 0 if k j ≥ n j for some index j. Also The colored Jones polynomial J L⊔L ′ (n, j), when j is fixed, can be repackaged into the invariant C L⊔L ′ (k, j) as stated in the following theorem.
Theorem 3. Suppose L ⊔ L ′ is a link in S 3 , with L having zero linking matrix. Assume the components of L ′ have fixed odd colors j = (j 1 , · · · j l ). Then there are invariants such that for every n = (n 1 , . . . , n m ) When L ′ = ∅, this is Theorem 8.2 in [7]. This generalization, essentially also due to Habiro, can be proved similarly as in [7]. For completeness we give a proof in the Appendix. Note that the existence of C L⊔L ′ (k, j) as rational functions in q satisfying (9) is easy to establish. The difficulty here is to show the integrality of (8).
Since A(n, k) = 0 unless k < n, in the sum on the right hand side of (9) one can assume that k runs over the set of all m-tuples k with non-negative integer components. We will use this fact later.

Strategy of the proof of the main theorem
Here we give the proof of Theorem 1 using technical results that will be proved later. As before, L⊔L ′ is a framed link in S 3 with disjoint sublinks L and L ′ , with m and l components, respectively. Assume that L ′ is colored by fixed j = (j 1 , . . . , j l ), with j i 's odd. Surgery along the framed link L transforms (S 3 , L ′ ) into (M, L ′ ). We will define I M,L ′ ∈ R b , such that (10) τ ′ M,L ′ (ξ) = ev ξ (I M,L ′ ) for any root of unity ξ of odd order. This unified invariant is multiplicative with respect to the connected sum.
The following observation is important. By Proposition 2, there is at most one element f (q) ∈ R b such that for every root ξ of odd order one has τ ′ M,L (ξ) = ev ξ (f (q)) . That is, if we can find such an element, it is unique, and we put I M,L ′ := f (q).
2.1. Laplace transform. The following is the main technical result of the paper. A proof will be given in Section 7.
Theorem 4. Suppose b = ±1 or b = ±p l where p is a prime and l is positive. For any nonnegative integer k, there exists an element Q b,k ∈ R b such that for every root ξ of odd order r one has

2.2.
Definition of the unified invariant: diagonal case. Suppose that the linking number between any two components of L is 0, and the framing on components of L are b i = ±p ki i for i = 1, . . . , m, where each p i is prime or 1. Let us denote the link L with all framings switched to zero by L 0 .
Using (9), taking into account the framings b i 's, we have By the definition of F L⊔L ′ , we have From (7) and Theorem 4, we get where the unified invariant of the lens space I L(bi,1) ∈ R b , with ev ξ (I L(bi,1) ) = τ ′ L(bi,1) (ξ), exists by Lemma 6 below. Thus if we define 2.3. Diagonalization using lens spaces. The general case reduces to the diagonal case by the well-known trick of diagonalization using lens spaces. We say that M is diagonal if it can be obtained from S 3 by surgery along a framed link L with diagonal linking matrix, where the diagonal entries are of the form ±p k with p = 0, 1 or a prime. The following lemma was proved in [15,  To define the unified invariant for a general rational homology sphere M , one first adds to M lens spaces to get a diagonal M ′ , for which the unified invariant I M ′ had been defined in Subsection 2.2. Then I M is the quotient of I M ′ by the unified invariants of the lens spaces. But unlike the simpler case of [15], the unified invariant of lens spaces are not invertible in general. To overcome this difficulty we insert knots in lens spaces and split the unified invariant into different components. This will be explained in the remaining part of this section.

2.4.
Splitting of the invariant. Suppose p is a prime divisor of b, then it's clear that R p ⊂ R b .
In Section 3 we will see that there is a decomposition then ev ξ (f ) can be defined when the order of ξ is coprime with p; and in this case ev ξ (g) = ev ξ (π p 0 (g)) for every On the other hand, if f ∈ R p,0 b then ev ξ (f ) can be defined when the order of ξ is divisible by p, and one has ev ξ (g) = ev ξ (π p 0 (g)) for every g ∈ R b . It also follows from the definition that R p,ε p ⊂ R p,ε b for ε = 0 or0. For S b , there exists a completely analogous decomposition. For any odd divisor p of b, an element x ∈ R b (or S b ) determines and is totally determined by the pair (π p 0 (x), π p 0 (x)). If p = 2 divides b, then for any x ∈ R b , x = π p 0 (x). Hence, to define I M it is enough to fix I 0 M = π p 0 (I M ) and I0 M = π p 0 (I M ). The first part I 0 M = π p 0 (I M ), when b = p, was defined in [15] (up to normalization), where the third author considered the case when the order of roots of unity is coprime with b. We will give a self-contained definition of I 0 M , and show that it is coincident (up to normalization) with the one introduced in [15].
where ε = 0 if the order of ξ is not divisible by p, and ε =0 otherwise. Moreover, if p is odd, then I ε M ε (b,a) belongs to and is invertible in S p,ε p . A proof of Lemma 6 will be given in Section 6.

Cyclotomic completions of polynomial rings
In this section we adapt the results of Habiro on cyclotomic completions of polynomial rings [9] to our rings.

On cyclotomic polynomial.
Recall that e n := exp(2πI/n) and denote by Φ n (q) the cyclo- The degree of Φ n (q) ∈ Z[q] is given by the Euler function ϕ(n). Suppose p is a prime and n an integer. Then (see e.g. [24]) It follows that Φ n (q p ) is always divisible by Φ np (q).
The ideal of Z[q] generated by Φ n (q) and Φ m (q) is well-known, see e.g. [15,Lemma 5.4]: (a) If m n = p e for any prime p and any integer e = 0, then . Note that in a commutative ring R, (x) + (y) = (1) if and only if x is invertible in R/(y). Also (x) + (y) = (1) implies (x k ) + (y l ) = (1) for any integers k, l ≥ 1.

3.2.
Habiro's results. Let us summarize some of Habiro's results on cyclotomic completions of polynomial rings [9]. Let R be a commutative integral domain of characteristic zero and R[q] the polynomial ring over R. For any S ⊂ N, Habiro defined the S-cyclotomic completion ring R[q] S as follows: where Φ * S denotes the multiplicative set in Z[q] generated by Φ S = {Φ n (q) | n ∈ S} and directed with respect to the divisibility relation.
For example, since the sequence (q; q) n , n ∈ N, is cofinal to Φ * N , we have [9]. Two positive integers n, n ′ are called adjacent if n ′ /n = p e with a nonzero e ∈ Z and a prime p, such that the ring R is p-adically separated, i.e. ∞ n=1 (p n ) = 0 in R. A set of positive integers is R-connected if for any two distinct elements n, n ′ there is a sequence n = n 1 , n 2 , . . . , n k−1 , n k = n ′ in the set, such that any two consecutive numbers of this sequence are adjacent. Theorem 4.1 of [9] says that if S is R-connected, then for any subset S ′ ⊂ S the natural map ρ R S,S ′ : R[q] S ֒→ R[q] S ′ is an embedding. If ζ is a root of unity of order in S, then for every f (q) ∈ R[q] S the evaluation ev ζ (f (q)) ∈ R[ζ] can be defined by sending q → ζ. For a set Ξ of roots of unity whose orders form a subset T ⊂ S, one defines the evaluation ev Ξ : Theorem 6.1 of [9] shows that if R ⊂ Q, S is R-connected, and there exists n ∈ S that is adjacent to infinitely many elements in T , then ev Ξ is injective.
3.3. Taylor expansion. Fix a natural number n, then we have .
is injective, by Proposition 13 below. Taking the inverse limit, we see that there is a natural injective algebra homomorphism Suppose n ∈ S. Combining h and ρ S,{n} : R[q] S → R[q] {n} , we get an algebra map If f ∈ R[q] S , then t n (f ) is called the Taylor expansion of f at e n .

3.4.
Splitting of S p and evaluation. For every integer a, we put N a := {n ∈ N | (a, n) = 1}.
Suppose p is a prime. Analogously to (13), we have Note that for every f ∈ S p , the evaluation ev ξ (f ) can be defined for every root ξ of unity. For f ∈ S p,j , the evaluation ev ξ (f ) can be defined when ξ is a root of unity of order in p j N p .
Proposition 8. For every prime p one has Proof. Suppose n i ∈ p ji N p for i = 1, . . . , m, with distinct j i 's. Then n i /n s , with i = s, is either not a power of a prime or a non-zero power of p, hence by Lemma 7 (and the remark right after Lemma 7), for any positive integers k 1 , . . . , k m , we have By the Chinese remainder theorem, we have Taking the inverse limit, we get (14).
Let π j : S p → S p,j denote the projection onto the jth component in the above decomposition.
Lemma 9. Suppose ξ is a root of unity of order r = p j r ′ , with (r ′ , p) = 1. Then for any x ∈ S p , one has ev ξ (x) = ev ξ (π j (x)).
We have similarly If ξ is a root of unity of order coprime with p, then ev ξ (f ) = ev ξ (π p 0 (f )). Similarly, if the order of ξ is divisible by p, then ev ξ (f ) = ev ξ (π p 0 (f )).
3.6. Properties of the ring R b . For any b ∈ N, we have Here N 2 is the set of all odd numbers. Let {p i | i = 1, . . . , m} be the set of all distinct odd prime divisors of b. For n = (n 1 , . . . , n m ), a tuple of numbers n i ∈ N, let p n = i p ni i . Let S n := p n N 2b . Then An argument similar to that for Equation (14) gives Let T be an infinite set of powers of an odd prime not dividing b and let P be an infinite set of odd primes not dividing b.
Proposition 10. With the above notations, one has the following.
(a) For any l ∈ S n , the Taylor map t l : Since both sets contain infinitely many numbers adjacent to p n , the claim follows from Theorem 6.1 in [9].

On the Ohtsuki series at roots of unity
The Ohtsuki series was defined for SO(3) invariants by Ohtsuki [26] and extended to all other Lie algebras by the third author [17,16].
In the works [26,17,16], it was proved that the sequence of quantum invariants at e p , where p runs through the set of primes, obeys some congruence properties that allow to define uniquely the coefficients of the Ohtsuki series. The proof of the existence of such congruence relations is difficult. In [7], Habiro proved that Ohtsuki series coincide with the Taylor expansion of the unified invariant at q = 1 in the case of integral homology spheres; this result was generalized to rational homology spheres by the third author [15].
Here, we prove that the sequence of SO(3) invariants at the prth roots e r e p , where r is a fixed odd number and p runs through the set of primes, obeys some congruence properties that allow to define uniquely the coefficients of the "Ohtsuki series" at e r , which is coincident with the Taylor expansion at e r .
is an isomorphism. We put x = q − e r and get ) .
Note that x n e p−n−1 r is a monic polynomial in x of degree p − 1, and the coefficient of By Theorem 1, for any root of unity ξ of order pr .
Hence we can write The following proposition shows that the coefficients a p,n stabilize as p → ∞.
Proposition 11. Suppose M is a rational homology 3-sphere with |H 1 (M, Z)| = b, and r is an odd positive integer. For every non-negative integer n, there exists a unique invariant a n = a n (M ) ∈ Z[1/b, e r ] such that for every prime p > max(b, r), we have Moreover, the formal series n a n (q −e r ) n is equal to the Taylor expansion of the unified invariant I M at e r .
Proof. The uniqueness of a n follows from the easy fact that if a ∈ Z[1/b, e r ] is divisible by infinitely many rational primes p, then a = 0. Assume Theorem 1 holds. We define a n to be the coefficient of (q − e r ) n in the Taylor series of I M at e r , and will show that Equation (18) holds true. Recall is commutative. Here the middle and the right vertical maps are the quotient maps by the corresponding ideals. Note that I M belongs to the upper left corner ring, its Taylor series is the image in the upper right corner ring, while the evaluation (17) is in the lower middle ring. Using the commutativity at the lower right corner ring, we see that Since the coefficients of f p (x + e r ) up to degree p − 2 are divisible by p, we get the congruence (18).
Remark 12. Proposition 11, when r = 1, was the main result of Ohtsuki [26], which leads to the development of the theory of finite type invariant and the LMO invariant.
When (r, b) = 1, then Taylor series at e r determines and is determined by the Ohtsuki series. But when, say, r is a divisor of b, a priori the two Taylor series, one at e r and the other at 1, are independent. We suspect that the Taylor series at e r , with r | b, corresponds to a new type of LMO invariant.

Frobenius maps
The proof of Theorem 4, and hence of the main theorem, uses the Laplace transform method. The aim of this section is to show that the image of the Laplace transform, defined in Section 7, belongs to R b , i.e. that certain roots of q exist in R b .
). Since cyclotomic completions are built from modules like Z[q]/(Φ k n (q)), we first consider these modules. Fix n, k ≥ 1. Let , The following is probably well-known.
(a) Both E and G are free Z-modules of the same rank kϕ(n).
descends to a well-defined algebra homomorphism, also denoted by h, from E to G. Moreover, the algebra homomorphism h : E → G is injective.
Proof. (a) Since Φ k n (q) is a monic polynomial in q of degree kϕ(n), it is clear that E = Z[q]/(Φ k n (q)) is a free Z-module of rank kϕ(n). Since G = Z[e n ] ⊗ Z Z[x]/(x k ), we see that G is free over Z of rank kϕ(n).
(b) To prove that h descends to a map E → G, one needs to verify that h(Φ k n (q)) = 0. Note that h(Φ k n (q)) = Φ k n (x + e n ) = (j,n)=1 (x + e n − e j n ) k .

A Frobenius homomorphism.
We use E and G of the previous subsection. Suppose b is a positive integer coprime with n. If ξ is a primitive nth root of 1, i.e. Φ n (ξ) = 0, then ξ b is also a primitive nth root of 1, i.e. Φ n (ξ b ) = 0. It follows that Φ n (q b ) is divisible by Φ n (q). Therefore the algebra map F b : Z[q] → Z[q], defined by F b (q) = q b , descends to a well-defined algebra map, also denoted by F b , from E to E. We want to understand the image F b (E).
Proof. Using Proposition 13 we identify E with its image h(E) in G.
It is easy to see thatF b is a well-defined algebra homomorphism, and thatF b restricted to E is exactly F b . Since E is a lattice of maximal rank in G ⊗ Q, it follows that the index of F b is exactly the determinant ofF b , acting on G ⊗ Q.
A basis of G is e j n x l , with (j, n) = 1, 0 < j < n and j = 0, and 0 ≤ l < k. Note that F b (e j n x l ) = b l e jb n e (b−1)l n x l + O(x l+1 ).
Since (b, n) = 1, the set e jb n , with (j, n) = 1 is the same as the set e j n , with (j, n) = 1. Let f 1 : G → G be the Z-linear map defined by f 1 (e jb n x l ) = e j n x l . Since f 1 permutes the basis elements, its determinant is ±1. Let f 2 : G → G be the Z-linear map defined by f 2 (e j n x l ) = e j n (e 1−b n x) l . The determinant of f 2 is again ±1. This is because, for any fixed l, f 2 restricts to the automorphism of Z[e n ] sending a to e s n a, each of these maps has a well-defined inverse: a → e −s n a. Now f 1 f 2Fb (e j n x l ) = b l e j n x l + O(x l+1 ) can be described by an upper triangular matrix with b l 's on the diagonal; its determinant is equal to b k(k−1)ϕ(n)/2 .
From the proposition we see that if b is invertible, then the index is equal to 1, hence we have Proposition 15. For any n coprime with b and k ∈ N, the Frobenius homomorphism F b : For finitely many n i ∈ N b and k i ∈ N, the Frobenius endomorphism sending q to q b , is again well-defined. Taking the inverse limit, we get an algebra endomorphism Proof. For finitely many n i ∈ N b and k i ∈ N, consider the natural algebra homomorphism .

This map is injective, because in the unique factorization domain
Since the Frobenius homomorphism commutes with J and is an isomorphism on the target of J by Proposition 15, it is an isomorphism on the domain of J. Taking the inverse limit, we get the claim. Proof. It suffices to show that for any n 1 , n 2 . . . n m ∈ T , the ring Q[q]/(Φ k1 n1 . . . Φ km nm ) does not contains a bth root of 1 except possibly for ±1. Using the Chinese remainder theorem, it suffices to consider the case where m = 1.
In contrast with Lemma 18, we have Proof. Let us first consider the case T = N b . Since F b is an isomorphism by Theorem 16, we can define a bth root of q by If y 1 and y 2 are two bth root of the same element, then their ratio y 1 /y 2 is a bth root of 1. From Lemma 18 it follows that if b is odd, there is only one bth root of q in Z[1/b][q] N b , and if b is even, there are 2 such roots, one is the minus of the other. We will denote them ±q 1/b .
Further it is known that q is invertible in Z[q] N (see [9]). Actually, there is an explicit expression In a commutative ring, if x | y and y is invertible, then so is x. Hence any root of q is invertible.
In the general case of T ⊂ N b , we use the natural map Relation with [15]. By Proposition 19, S b,0 is isomorphic to the ring Λ [15]. Furthermore, our invariant π 0 I M and the one defined in [15] belong to S b,0 . This follows from Theorem 1 for b odd, and from Proposition 10(c) for b even. Finally, the invariant defined in [15]

5.5.
Another Frobenius homomorphism. We define another Frobenius type algebra homomorphism. The difference of the two types of Frobenius homomorphisms is in the target spaces of these homomorphisms. Suppose m is a positive integer. Define the algebra homomorphism Since Φ mr (q) always divides Φ r (q m ), G m is well-defined.

5.6.
Realization of q a 2 /b in S p . Throughout this subsection, let p be a prime or 1. Suppose b = ±p l for an l ∈ N and let a be an integer. Let B p,j = G p j (S p,0 ). Note that B p,j ⊂ S p,j . If b is odd, by Proposition 19 there is a unique bth root of q in S p,0 ; we denote it by x b;0 . If b is even, by Proposition 19 there are exactly two bth root of q, namely ±q 1/b . We put x b;0 = q 1/b . We define an element z b,a ∈ S p as follows. If b | a, let z b,a := q a 2 /b ∈ S p . If b = ±p l ∤ a, then z b,a ∈ S p is defined by specifying its projections π j (z b,a ) := z b,a;j ∈ S p,j as follows. Suppose a = p s e, with (e, p) = 1. Then s < l. For j > s let z b,a;j := 0. For 0 ≤ j ≤ s let Similarly, for b = ±p l we define an element x b ∈ S p as follows. We put π 0 (x b ) := x b;0 . For j < l, π j (x b ) := [G p j (x b;0 )] p j . If j ≥ l, π j (x b ) := q b . Notice that for c = (b, p j ) we have Proposition 20. Suppose ξ is a root of unity of order r = cr ′ , where c = (r, b). Then Proof. Let us compute ev ξ (z b,a ). The case of ev ξ (x b ) is completely analogous. If b | a, then c | a, and the proof is obvious. Suppose b ∤ a. Let a = p s e and c = p i . Then s < l. Recall that z b,a = ∞ j=0 z b,a;j . By Lemma 9, ev ξ (z b,a ) = ev ξ (z b,a;i ).
If c ∤ a, then i > s. By definition, z b,a;i = 0, hence the statement holds true. It remains the case c | a, or i ≤ s. Note that ζ = ξ c is a primitive root of order r ′ and (p, r ′ ) = 1.
From the definition of z b,a;i it follows that (z b,a;i ) b/c = (q c ) a 2 /c 2 , hence after evaluation we have Using Lemma 17 we conclude ev ξ (z b,a;i ) = (ξ c ) a 2 Since ev 1 (q 1/b ) = 1 and therefore ev ξ (q 1/b ) = ξ b * (and not −ξ b * ) we get the claim.

Invariant of lens spaces
The purpose of this section is to prove Lemma 6. Throughout this section we will use the following notations.
Let r be a fixed odd integer (the order of ξ). For l ∈ Z coprime to r, let l * denote the inverse of l modulo r. If (b, r) = c, let b ′ * denote the inverse of b ′ := b c modulo r ′ := r c . Notice that for c = 1, we have b * = b ′ * . Further, we denote by x y the Jacobi symbol and by s(a, b) the Dedekind sum (see e.g. [12]).

Invariants of lens spaces. Let us compute the SO(3) invariant of the lens space M (b, a; d).
Recall that M (b, a; d) is the lens space L(b, a) together with a knot K inside colored by d (see Figure 1).
Proof. We consider first the case where b, a > 0. Since two lens spaces L(b, a 1 ) and L(b, a 2 ) are homeomorphic if a 1 ≡ a 2 (mod b), we can assume a < b. Let b/a be given by a continued fraction b a = m n − 1 Using the Lagrange identity The τ M(b,a;d) (ξ) can be computed in the same way as the invariant ξ r (L(b, a), A) in [20], after replacing A 2 (respectively A) by ξ 2 * (respectively ξ 4 * ). Representing the b/a-framed unknot in Figure 1 by a Hopf chain (as e.g. in Lemma 3.1 of [3]), we have Using Lemmas 4.11, 4.12 and 4.20 of [20] 1 (and replacing e r by ξ 4 * , c n by c, N n,1 = p by b, N n−1,1 = q by a, N n,2 = q * byâ and −N n−1,2 = p * byb), we get Inserting the last formula into the Definition (5) we get where we used that σ + = n and σ − = 0 (compare [12, p. 243]). From r j=1 ξ 4 * j 2 = ǫ(r) √ r, we Applying the following formulas for the Dedekind sum (compare [12, Theorem 1.12]) and dividing the formula for τ M(b,a;d) (ξ) by the formula for τ L(b,1) (ξ) we get There are misprints in Lemma 4.21: q * ± n should be replaced by q * ∓ n for n = 1, 2. where Notice, that u ∈ Z. Further observe, that by using aâ + bb = 1, we get This implies the result for 0 < a < b.
For ε = 0, this follows directly from Propositions 20 and 21 with c = d = 1. For ε =0, we have c = (p j , b) > 1 and we get the claim by using Proposition 21 and where for the second equality we use c ≡ bb ′ * (mod r). Notice that due to part (2) of Lemma 22 below, b and c divideâ − sn(a)d(0) and therefore all powers of ξ in (20) are integers.
The following Lemma is used in the proof of Lemma 6.
Lemma 22. We have Proof. The first claim follows from the formulas (19) for the Dedekind sum. The second claim follows from the fact that (a, b) = 1 and since d is chosen such that sn(a)ad ≡ 1 (mod b). For the third claim, notice that for odd d we have 4 | (1 − d 2 ) and 4 | 2(sn(a)d − sn(b)).

Laplace transform
This section is devoted to the proof of Theorem 4 by using Andrew's identity. Throughout this section, let p be a prime or p = 1, and b = ±p l for an l ∈ N.

Definition. The Laplace transform is a Z[q ±1 ]-linear map defined by
In particular, we put L b;j := π j • L b and have L b;j (z a ) = z b,a;j ∈ S p,j .
Further, for any f ∈ Z[z ±1 , q ±1 ] and n ∈ Z, we definê Then for a root of unity ξ of odd order r, Proof. It is sufficient to consider the case f = z a . Then, by the same arguments as in the proof of [3, Lemma 1.3], we have The result follows now from Proposition 20.

Proof of Theorem 4. Recall that
We have to show that there exists an element Q b,k ∈ R b such that for every root of unity ξ of odd order r one has [n] 2 , we get for c = (b, r) where as usual, χ(c) = 1 if c = 1 and is zero otherwise. We will prove that for an odd prime p and any number j ≥ 0 there exists an element Q k (q, x b , j) ∈ S p,j such that If p = 2 we will prove the claim for j = 0 only, since S 2,0 ≃ R 2 . The case p = ±1 was done e.g. in [2]. Theorem 4 follows then from Lemma 23 and (22) where Q b,k is defined by its projections We split the proof of (23) into two parts. In the first part we will show that there exists an element Q k (q, x b , j) such that Equality (23) holds. In the second part we show that Q k (q, x b , j) lies in S p,j .
Part 1, b odd case. Assume b = ±p l with p = 2. We split the proof into several lemmas.
Observe that for n > k+1 c the term (q −k−1 ; q) cn is zero and therefore the sum in (24) is finite.
and the q-binomial theorem (e.g. see [5], II.3) gives Notice that L b;j (z a ) = 0 if and only if c | a. Applying L b;j to the RHS of (25), only the terms with c | i survive and therefore Separating the case n = 0 and combining positive and negative n this is equal to where we use the convention that x −1 is put to be zero for positive x. Further, 2k + 2 k + cn + 1 (q cn/2 + q −cn/2 ) and Using z b,cn;j = (z b,c;j ) n 2 = x n 2 b;j we get the result.
Lemma 25. S b;j (k, q) is equal to the LHS of Andrew's identity with the parameters fixed below.
Proof. Since it is enough to look at the case when b > 0. Define b ′ := b c and let ω be a b ′ th primitive root of unity. For simplicity, put N := k + 1 and t := x b;j . Using the following identities where the later is true due to t b ′ = x b ′ b;j = q c for all j, and choosing a cth root of t denoted by t 1 c we can see that Then where g = a + cd and s = (c + 1) b ′ 2 + 1. We now calculate the LHS of Andrew's identity. Using the notation (ω ±1 t x ) n = (ωt x ) n (ω −1 t x ) n and the identities lim c→∞ (c) n c n = (−1) n t n(n−1) 2 and lim c→∞ t c n = 1 we get Taking all the results together, we see that the LHS is equal to S b;j (k, q).
Let us now calculate the RHS of Andrew's identity with parameters chosen as above. For simplicity, we put δ j := n j+1 − n j . Then the RHS is given by For c = 1 or d = 0, we use the convention that empty products are set to be 1 and empty sums are set to be zero. Let us now have a closer look at the RHS. Notice, that lim bs,cs→∞ The term (t −1 ) δg+i is zero unless δ g+i ∈ {0, 1}. Therefore, we get Due to the term (t −m ) ns , we have n s ≤ m and therefore n i ≤ m for all i. Multiplying the numerator and denominator of each term of the RHS by gives in the denominator The term (t −N +m ) δs−1 is zero unless δ s−1 ≤ N − m and therefore Taking the above calculations into account, we get and x ′ := x + ns(ns−1) 2 + N n s . We now define the element Q k (q, x b , j) by By Lemmas 24 and 25, Equation (26) and the following Lemma 26, we see that this element satisfies Equation (23).
Lemma 26. The following formula holds.
Part 1, b even case. Let b = ±2 l . We have to prove Equality (23) only for j = 0, i.e. we have to show The calculation works similar to the odd case. Note that we have c = 1 here. This case was already done in [3] and [15]. Since their approaches are slightly different and for the sake of completeness, we will give the parameters for Andrew's identity and the formula for Q k (q, x b , 0) nevertheless. We put t := x b;0 , d := b 2 − 1, ω a bth root of unity and choose a primitive square root ν of ω. Define the parameters of Andrew's identity by where s = b + 2. Now we can define the element n d+i + ns−1(ns−1−1) 2 + (N + 1)(n b + n s−1 ). We use the notation (a; b) −1 = 1 1−ab −1 . Part 2. We have to show that Q k (q, x b , j) ∈ S p,j , where j ∈ N ∪ {0} if p is odd, and j = 0 for p = 2. The following two lemmas do the proof.
Proof. Let us first look at the case b odd and positive. Since for a = 0, (t a ) n is always divisible by (t) n , it is easy to see that the denominator of each term of T k (q, t) divides its numerator. Therefore we proved that T k (q, t) ∈ Z[t ±1/c , ω]. Since , g0 . This implies that T k (q, t) ∈ Z[q ±1 , t ±1 ] since f 0 and g 0 do not depend on ω and the cth root of t.
The proofs for the even and the negative case work similar.
Theorem A.1. Suppose L ⊔ L ′ is a colored framed link in S 3 such that L has zero linking matrix and L ′ has odd colors. Then for k = max{k 1 , . . . , k m } we have In the case L ′ = ∅, this statement was proved in [7,Theorem 8.2]. Since our proof is a modification of the original one, we first sketch Habiro's original proof for the reader's convenience.
A.1. Sketch of the proof of Habiro's integrality theorem.
Geometric part. Let us first recall the notion of a bottom tangle, introduced by Habiro in [8].
An n-component bottom tangle T = T 1 ⊔ · · · ⊔ T n is a framed tangle consisting of n arcs Algebraic part.
There is Z/2Z-grading, U q = U 0 q ⊕ U 1 q , where U 0 q (resp. U 1 q ) is spanned byF (i) K 2j e k (resp. F (i) K 2j+1 e k ). We call this the ε-grading, and U 0 q (resp. U 1 q ) the even (resp. odd) part. The two-sided ideal F p in U q generated by e p induces a filtration on (U q ) ⊗n , n ≥ 1, by Let (Ũ q )⊗ n be the image of the homomorphism where⊗ is the h-adically completed tensor product. By using F p (U ε q ) := F p (U q ) ∩ U ε q one defines (Ũ ε q )⊗ n for ε ∈ {0, 1} in a similar fashion. By definition (Section 4.2 of [7]), the universal sl 2 invariant J T of an n-component bottom tangle T is an element of U⊗ n h . Theorem 4.1 in [7] states that, in fact, for any bottom tangle T with zero linking matrix, J T is even, i.e.
(A.1) J T ∈ (Ũ 0 q )⊗ n . Further, using the fact that J K of a 0-framed bottom knot K (i.e. a 1-component bottom tangle) belongs to the center ofŨ 0 q , Habiro showed that the quantum Casimir operator. The σ n provide a basis for the even part of the center. From this, Habiro deduced that J K (P ′ n ) ∈ (q n+1 ;q)n+1 . The case of n-component bottom tangles reduces to the 1-component case by partial trace, using certain integrality of traces of even element (Lemma 8.5 of [7]) and the fact that J T is invariant under the adjoint action.
Algebro-geometric part. The proof of (A.1) uses Proposition A.2, which allows to build any bottom tangle T with zero linking matrix from simple parts, i.e. T = W (B ⊗k ).
On the other hand, the construction of the universal invariant J T extends to the braided functor J : B → Mod U h from B to the category of U h -modules. This means that J W (B ⊗k ) = J W (J B ⊗k ). Therefore, in order to show (A.1), we need to check that J B ∈ (Ũ 0 q )⊗ 3 , and then verify that J W maps the even part to itself. The first check can be done by a direct computation [7,Section 4.3]. The last verification is the content of Corollary 3.2 in [7].
It is easy to see that the definitions do not depend on the orientation of L.
The meaning of gr q (L) is the following: The colored Jones polynomial of L, a priori a Laurent polynomial of q 1/4 , is actually a Laurent polynomial of q after dividing by q gr q (L)/4 ; see [18] for this result and its generalization to other Lie algebras.
Recall that the universal invariant J X can also be defined when X is the union of a bottom tangle and a colored link (see [8,Section 7.3]). In [8], it is proved that J X is adjoint invariant. The generalization of Theorem 4.1 of [7] is the following.
Theorem A.3. Suppose X = T ⊔ L, where T is a n-component bottom tangle with zero linking matrix and L is a framed unoriented colored link with gr ε (X) = (ε 1 , . . . , ε n ). Then Corollary A.4. Suppose L is colored by a tuple of odd numbers, then Since J X is invariant under the adjoint action, Theorem A.1 follows from Corollary A.4 by repeating Habiro's arguments.
Hence it remains to prove Theorem A.3. In the proof we will need a notion of a good morphism.
Good morphisms. Let I m := 1 b ⊗m ∈ B(m, m) be the identity morphism of b ⊗m in the cube C. A framed link L in the complement C \ I m is good if L is geometrically disjoint from all the up arrows of b ⊗m , i.e. there is a plane dividing the cube into two halves, such that all the up arrows are in one half, and all the down arrows and L are in the other. Equivalently, there is a diagram in which all the up arrows are above all components of L. The union W of I m and a colored framed good link L is called a good morphism. If Y is any bottom tangle so that we can compose X = W Y , then it is easy to see that gr ε (X) does not depend on Y , and we define gr ε (W ) := gr ε (X). Also define gr q (W ) := gr q (L). As in the case with L = ∅, the universal invariant extends to a map J W : U ⊗m h → U ⊗m h . Proof of Theorem A.3. The strategy here is again analogous to the Habiro case: In Proposition A.5 we will decompose X into simple parts: the top is a bottom tangle with zero linking matrix, the next is a good morphism, and the bottom is a morphism obtained by pasting copies of µ b . Since, any bottom tangle with zero linking matrix satisfies Theorem A.3 and µ b is the product in U q , which preserves the gradings, it remains to show that any good morphism preserves the gradings. This is done in Proposition A.6 below.
Proposition A.5. Assume X = T ⊔ L where T is a n-component bottom tangle with zero linking matrix and L is a link. Then there is a presentation X = W 2 W 1 W 0 , where W 0 is a bottom tangle with zero linking matrix, W 1 is a good morphism, and W 2 is obtained by pasting copies of µ b .
Proof. Let us first defineγ ± ∈ B(i, i + 1) for any i ∈ N as follows.
If a copy of µ b is directly above ψ ±1 b,b or γ ± , one can move µ b down by isotopy and represent the result by pasting copies of ψ ±1 b,b andγ ± . It is easy to see that after the isotopy γ ± gets replaced bỹ γ ± and ψ ±1 b,b by two copies of ψ ±1 b,b . Using Proposition A.2 and reordering the basic morphisms so that the µ's are at the bottom, one can see that T admits the following presentation: where B is the Borromean tangle, W 2 is obtained by pasting copies of µ b andW 1 is obtained by pasting copies of ψ ±1 b,b ,γ ± and η b . Let P be the horizontal plane separatingW 1 from W 2 . Let P + (P − ) be the upper (lower, respectively) half-space. Note that W 0 =W 1 (B ⊗k ) is a bottom tangle with zero linking matrix lying in P + and does not have any minimum points. Hence the pair (P + , W 0 ) is homeomorphic to the pair (P + , l trivial arcs). Similarly, W 2 does not have any maximum points; hence L can be isotoped off P − into P + . Since the pair (P + , W 0 ) is homeomorphic to the pair (P + , l trivial arcs) one can isotope L in P + to the bottom end points of down arrows. We then obtain the desired presentation.
The rest of the appendix is devoted to the proof of Proposition A.6.
Proof of Proposition A.6. We proceed as follows. Since J X is invariant under cabling and skein relations, and by Lemma A.8 below, both relations preserve gr ε and gr q , we consider the quotient ofS(D 3 \ T ) by these relations known as a skein module of D 3 \ T . For T = I n , this module has a natural algebra structure, with good morphisms forming a subalgebra. By Lemma A.7 (see also Figure A.3), the basis elements W γ of this subalgebra are labeled by n-tuples γ = (γ 1 , . . . , γ n ) ∈ (Z/2Z) n . It's clear that if the proposition holds for W γ1 and W γ2 , then it holds for W γ1 W γ2 . Hence it remains to check the claim for W γ 's. This is done in Corollary A.10 for basic good morphisms corresponding to γ whose non-zero γ j 's are consecutive. Finally, any W γ can be obtained by pasting a basic good morphism with few copies of ψ ± b,b . Since J ψ ± preserves gradings (compare (3.15), (3.16) in [7]), the claim follows from Lemmas A.7, A.8 and Proposition A.9 below.
Cabling and skein relations. Let us introduce the following relations inS(D 3 \ T ).
Cabling relations: (a) Suppose n i = 1 for some i. The first cabling relation is L =L, whereL is obtained from L by removing the ith component.  We denote by S(D 3 \ T ) the quotient ofS(D 3 \ T ) by these relations. It is known as the skein module of D 3 \ T (compare [27], [28] and [4]). Recall that the ground ring is Z[q ±1/4 ].
Using the cabling relations, we can reduce all colors of L in S(D 3 \ T ) to be 2. Note that the skein module S(C \ I n ) has a natural algebra structure, given by putting one cube on the top of the other. Let us denote by A n the subalgebra of this skein algebra generated by good morphisms.
For a set γ = (γ 1 , . . . , γ n ) ∈ (Z/2Z) n let W γ be a simple closed curve encircling the end points of those downward arrows with γ i = 1. See Figure A.3 for an example.
Similarly to the case of Kauffman bracket skein module [4], one can easily prove the following. Lemma A.7. The algebra A n is generated by 2 n curves W γ .
Using linearity, we can extend the definition of J X to X = T ⊔ L, where L is any element of S(D 3 \ T ). It is known that J X is invariant under the cablings and skein relations (Theorem 4.3 of [13]), hence J X is defined for L ∈ S(D 3 \ T ). Moreover, we have Lemma A.8. Both gradings gr ε and gr q are preserved under the cabling and skein relations.
Proof. The statement is obvious for the ε-grading. For the q-grading, notice that gr q (L) = 2 1≤i<j≤m p ij n ′ i n ′ j + 1≤j≤m p jj n ′2 j + 2 1≤j≤m (p jj + 1)n ′ j , and therefore gr q (L ′′ ) ≡ gr q (L ′ ) ≡ gr q (L) (mod 4). This takes care of the cabling relations.
Let us now assume that all colors of L are equal to 2 and therefore gr q (L) = 2 The statement is obvious for the first skein relation. For the second skein relation, choose an arbitrary orientation on L. Let us first assume that the two strands in the crossing depicted in Figure  It is obvious, that this does not depend on the orientation of L R . If the crossing of L R is negative or the two strands do not belong to the same component of L R , the proof works similar.
A.3. Proof of Proposition A.9. The statement holds true for J Z1 = C ⊗ id ↑ . Now Lemma 7.4 in [8] states that applying ∆ to the ith component of the universal quantum invariant of a tangle is the same as duplicating the ith component. Using this fact we represent where Φ is defined as follows. For x ∈ U q with ∆(x) = x (1) ⊗ x (2) , we put Φ(x) := (x),m,n x (1) ⊗ β m S(β n ) ⊗ α n x (2) α m where the R-matrix is given by R = l α l ⊗ β l . See Figure below for a picture. In U q , in addition to the ε-grading, there is also the K-grading, defined by |K| = |K −1 | = 0, |e| = 1, |F | = −1. In general, the co-product ∆ does not preserve the ε-grading. However, we have the following.
Lemma A.11. Suppose x ∈ U q is homogeneous in both ε-grading and K-grading. Then we have a presentation where each x (1) , x (2) are homogeneous with respect to the ε-grading and K-grading. In addition, for x ∈ U ε q , we have x (2) ∈ U ε q and x (1) K −|x (2) | ∈ U ε q . Proof. If the statements hold true for x, y ∈ U q , then they hold true for xy. Therefore, it is enough to check the statements for the generators e,F (1) , and K, for which they follow from explicit formulas of the co-product.
Lemma A.12. Suppose x ∈ U q is homogeneous in both ε-grading and K-grading. There is a presentation such that each x ij is homogeneous in both ε-grading and K-grading, and for x ∈ U ε q , x i2 and x i0 x i1 belong to U ε q .
Proof. We put D = D ′ ⊗ D ′′ := v where we used (id ⊗S)D = D −1 and D ±1 (1 ⊗ x) = (K ±|x| ⊗ x)D ±1 for homogeneous x ∈ U q with respect to the K-grading. Now, the claim follows from Lemma A.11.
By induction on n in (A.2), given that C ∈ v U 1 q , Lemma A.12 implies Proposition A.9.