---
abstract: 'Given a rational homology 3–sphere $M$ with $|H_1(M,\Z)|=b$ 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 $L=\emptyset$, $I_M$ 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.'
address:
- 'Institut für Mathematik, Universität Zurich, Winterthurerstrasse 190, 8057 Zürich, Switzerland'
- 'Department of Mathematics, Georgia Institute of Technology, Atlanta, GA 30332–0160, USA '
author:
- Anna Beliakova
- Irmgard Bühler
- Thang Le
title: |
    A unified quantum SO(3) invariant\
    for rational homology 3–spheres
---

Introduction {#introduction .unnumbered}
============

Background {#background .unnumbered}
----------

The $SU(2)$ Witten–Reshetikhin–Turaev (WRT) invariant is defined for any closed oriented 3–manifold $M$ and any root of unity $\xi$ [@Tu]. Kirby and Melvin [@KM] introduced the $SO(3)$ version of the invariant $\tau_M(\xi)\in \Q(\xi)$ for roots of unity $\xi$ of odd order. If the order of $\xi$ is prime, then by the results of Murakami [@Mu] (also Masbaum–Roberts [@MR]), $\tau_M(\xi)$ is an algebraic integer. This integrality result was the starting point for the construction of finite type 3–manifold invariants, Ohtsuki series [@Ohtsukibook], integral TQFTs, representations of the mapping class group over $\Z[\xi]$ [@GM], and categorification of quantum 3–manifold invariants [@Kho]. The proofs in [@Mu] and [@MR] depend heavily on the arithmetic of $\Z[\xi]$ for a root of unity $\xi$ of [*prime*]{} order and do not extend to other roots of unity.

Is it true that $\tau_M(\xi)$ is always an algebraic integer (belongs to $\Z[\xi]$), even when the order of $\xi$ is not a prime? The positive answer to this question was given first for [*integral homology spheres*]{} by Habiro [@Ha], and then for arbitrary 3–manifolds by the first and third author [@BL], 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 [@Ha] 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) $$\Habiro:=\lim_{\overleftarrow{\hspace{2mm}k\hspace{2mm}}}
\frac{\Z[q]}{
((q;q)_k)},  \qquad \text{ where} \quad (q;q)_k = \prod_{j=1}^k (1-q^j).$$ Every element $f(q)\in \Habiro$ can be written as an infinite sum $$f(q)= \sum_{k\ge 0} f_k(q)\, (1-q)(1-q^2)...(1-q^k),$$ with $f_k(q)\in \Z[q]$. When $q=\xi$, 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 $\ev_\xi(f(q))$. Since $f_k(q)\in \Z[q]$, $\ev_\xi(f(q))\in \Z[\xi]$ is an algebraic integer. The fact that the unified invariant belongs to $\Habiro$ is stronger than just integrality of $\tau_M(\xi)$. We will refer to it as “strong” integrality.

The Habiro ring has beautiful arithmetic properties. Every element $f(q) \in \Habiro$ 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 $f,g \in \Habiro$ are the same if and only if their Taylor series at a root of unity coincide. In addition, each function $f(q) \in \Habiro$ is totally determined by its values at, say, infinitely many roots of order $3^n,\, n\in \N$. Due to these properties the Habiro ring is also called a ring of “analytic functions at roots of unity” [@Ha]. Thus belonging to $\Habiro$ 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 $\xi$ is prime [@Oh]. 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 [@Le] and [@BL]. Among other things, in [@Le] a unified invariant was constructed for the case when the order $r$ of the quantum parameter $\xi$ is coprime with $b$. In [@BL], it was proved that for any 3–manifold $M$ (not necessary a rational homology 3–sphere), the $SO(3)$ WRT invariant $\tau_M(\xi)$ is always an algebraic integer, i.e. $\tau_M(\xi)\in \Z[\xi]$ with no restriction on the order of $\xi$ at all. There we used a (2nd order) Laplace transform method [@BBL] and a difficult identity of Andrews [@And] 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 $(r,b) \neq 1$. 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 [@Le], the ring our invariant belongs to is simpler than the one obtained in [@Le] and [@BL]. 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 $(l,b)=1$.

For rational homology 3–spheres the universal finite type invariant was constructed by Le, Murakami and Ohtsuki [@LMO]. It determines Ohtsuki series and, hence, $\{\tau_M(\xi)\,|\,({\operatorname{ord}}(\xi),b)=1\}$ [@Le]. An interesting open question is whether the Le–Murakami–Ohtsuki invariant determines $I_M$.

Results {#results .unnumbered}
-------

The WRT or quantum $SO(3)$ invariant $\tau_{M,L}(\xi)$ is defined for a pair of a closed 3–manifold $M$ and a link $L$ in it, with link components colored by integers. Here $\xi$ is a root of unity of odd order. We will recall the definitions in Section \[defs\].

Suppose $M$ is a rational homology 3–sphere, i.e. $|H_1(M,\Z)|:={\rm card}\,H_1 (M,\Z) < \infty$. There is a unique decomposition $ H_1(M,\Z)=\bigoplus_{i}
\Z/{b_{i}\Z}$, where each $b_i$ is a prime power. We renormalize the $SO(3)$ WRT invariant of the pair $(M, L)$ as follows: $$\tau'_{M, L}(\xi)=\frac{\tau_{M,L}(\xi)}
{\prod\limits_{i}\;\,
\tau_{L(b_{i},1)}(\xi)}\; ,
\label{0910}$$ where $L(b,a)$ denotes the $(b,a)$ lens space. We will see that $\tau_{L(b,1)}(\xi)$ is always nonzero.

For any positive integer $b$, we define the cyclotomic completion ring $\R_b$ to be \[ab\] \_b:=\_ , (q;q\^2)\_k = (1-q)(1-q\^3) …(1-q\^[2k-1]{}). For any $f(q)\in \R_b$ and a root of unity $\xi$ of [*odd*]{} order, the evaluation $\ev_\xi (f(q)):= f(\xi)$ is well–defined. Similarly, we put $$\calS_b :=\lim_{\overleftarrow{\hspace{2mm}k\hspace{2mm}}}
\frac{\Z[1/b][q]}
{((q;q)_k)}\; .$$ Here the evaluation at any root of unity is well–defined. For odd $b$, there is a natural embedding $\calS_b\to \R_b$, see Section \[cyc\].

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.

\[main\] Suppose the components of a framed oriented link $L \subset M$ have odd colors, and $M\in \cM_b$. Then there exists an invariant $I_{M,L} \in \R_b$, such that for any root of unity $\xi$ of odd order $$\ev_\xi(I_{M,L})=\tau'_{M,L}(\xi)\, .$$ In addition, if $b$ is odd, then $I_{M,L}\in \calS_b$.

If $b=1$ and $L$ is the empty link, $I_{M}$ coincides with Habiro’s unified invariant $J_M$.

The proof of Theorem \[main\] 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 [@Le], [@BL]. As a byproduct, we generalize the deep integrality result of Habiro (Theorem 8.2 in [@Ha]), 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 $\calS_b$ have properties similar to those of the Habiro ring. An element $f(q) \in \R_b$ is totally determined by the values at many infinite sets of roots of unity (see Section \[cyc\]), one special case is the following.

\[main-cor\] Let $p$ be an odd prime not dividing $b$ and $T$ the set of all integers of the form $p^k b'$ with $k\in \N$ and $b'$ any odd divisor of $b^n$ for some $n$. Any element $f(q) \in \R_b$, and hence also $\{\tau_M(\xi)\}$, is totally determined by the values at roots of unity with orders in $T$.

The Ohtsuki series [@Oh; @Le3], originally defined through some arithmetic congruence property of the $SO(3)$ invariant, can be identified with the Taylor expansion of $I_M$ at $q=1$ [@Ha; @Le]. 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 \[Oh-strategy\].

Plan of the paper {#plan-of-the-paper .unnumbered}
-----------------

In Section \[defs\] we recall known results and definitions. In the next section we explain the strategy of our proof of Theorem \[main\]. In Sections \[cyc\] and \[map-frob\], we develop properties of cyclotomic completions of polynomial rings. New Ohtsuki series are discussed in Section \[Oh-strategy\]. The unified invariant of lens spaces, needed for the diagonalization, is defined in Section \[diag\]. The main technical result of the paper based on Andrew’s identity is proved in Section \[laplace\]. The Appendix is devoted to the proof of the generalization of Habiro’s integrality theorem.

Acknowledgments {#acknowledgments .unnumbered}
---------------

The authors would like to thank Kazuo Habiro and Christian Krattenthaler for helpful remarks and stimulating conversations.

Quantum (WRT) invariants {#defs}
========================

Notations and conventions
-------------------------

We will consider $q^{1/4}$ as a free parameter. Let $$\{n\} = q^{n/2}-q^{-n/2},
 \quad  \{n\}!=
\prod_{i=1}^n \{i\} ,\quad  [n] =\frac{\{n\}}{\{1\}}, \quad
{\text{$\left[\begin{array}{c}n\\ k\end{array}
\right]$}} = \frac{\{n\}!}{\{k\}!\{n-k\}!}.$$ We denote the set $\{1,2,3,\ldots\}$ by $\N$. We also use the following notation from $q$–calculus: $$(x;q)_n := \prod_{j=1}^n (1-x q^{j-1}).$$ Throughout this paper, $\xi$ will be a primitive root of unity of [*odd*]{} order $r$ and $e_n:=\exp(2\pi I/n)$.

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, $L\sqcup L'$ 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 $M$.

The colored Jones polynomial
----------------------------

Suppose $L$ is a framed, oriented link in $S^3$ with $m$ ordered components. For positive integers $n_1,\dots,n_m$, called the colors of $L$, one can define the quantum invariant $J_L(n_1,\dots,n_m)\in \Z[q^{\pm 1/4}]$, known as the colored Jones polynomial of $L$ (see e.g. [@Tu; @MM]). Let us recall here a few well–known formulas. For the unknot $U$ with 0 framing one has $$J_U(n) = [n]. \label{unknot}$$ If $L_1$ is obtained from $L$ by increasing the framing of the $i$th component by 1, then $$\label{framing}
J_{L_1}(n_1,\dots,n_m) = q^{(n_i^2-1)/4} J_{L}(n_1,\dots,n_m).$$ If all the colors $n_i$ are odd, then $J_{L}(n_1,\dots,n_m) \in \Z[q^{\pm 1}]$.

Evaluation and Gauss sums
-------------------------

For each root of unity $\xi$ of odd order $r$, we define the evaluation map $\ev_\xi$ by replacing $q$ with $\xi$. Suppose $f(q;n_1,\dots,n_m)$ is a function of variables $q^{\pm 1}$ and integers $n_1,\dots,n_m$. In quantum topology, the following sum plays an important role $${\sum_{n_i}}^\xi f := \sum_{\substack{0< n_i< 2r\\ n_i \text{ odd}}}
\ev_\xi  f(q; n_1,\dots, n_m)$$ 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 $$\gamma_b(\xi):= {\sum_{n}}^\xi q^{b\frac{n^2-1}{4}}$$ is well–defined, since for odd $n$, $4\mid n^2-1$. It is known that, for odd $r$, $|\gamma_b(\xi)|$ is never 0.

Definition of the WRT invariant
-------------------------------

Suppose the components of $L'$ are colored by fixed integers $j_1,\dots,j_l$. Let $$ F_{L\sqcup L'}(\xi):= {\sum_{n_i}}^\xi\;
\left \{  J_{L\sqcup L'}(n_1,\dots,n_m, j_1,\dots,j_l)\prod_{i=1}^m [n_i]\right \}.$$

An important special case is when $L=U^b$, the unknot with framing $b \neq 0$, and $L'=\emptyset$. In that case $F_{U^{b}}(\xi)$ can be calculated using the Gauss sum and is nonzero, see Section \[diag\] below.

Let $\sigma_+ $ (respectively $\sigma_-$) 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. [@KM; @Tu]) $$\tau_{M,L'}(\xi) =
\frac{F_{L\sqcup L'}(\xi)}{(F_{U^{+1}}(\xi))^{\sigma_+}\,
(F_{U^{-1}}(\xi))^{\sigma_-} }\, .
\label{def_qi}$$ The invariant $\tau_{M,L'}(\xi)$ 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 $$\tau_{L(b,1)} (\xi)= \frac{ F_{U^b}(\xi)}{F_{U^{{\operatorname{sn}}(b)}}(\xi) },
\label{2005}$$ were ${\operatorname{sn}}(b)$ is the sign of the integer $b$.

Let us focus on the special case when the linking matrix of $L$ is diagonal, with $b_1, b_2, \dots, b_m$ on the diagonal. Assume each $b_i$ is a power of a prime up to sign. Then $H_1(M,\Z) = \oplus_{i=1}^m \Z/|b_i|$, and $$\sigma_+ = {\rm card}\, \{ i\mid b_i >0\}, \quad \sigma_- =
{\rm card}\, \{ i \mid b_i < 0\}.$$ Thus from the definitions , and we have $$\tau'_{M,L'}(\xi) = \left( \prod_{i=1}^m  \tau'_{L(b_i,1)}(\xi)
 \right)\,
\frac{F_{L\sqcup L'}(\xi)}
{\prod_{i=1}^m F_{U^{b_i}}(\xi) }
\,  ,
\label{0077}$$ with $$\tau'_{L(b_i,1)}(\xi)= \frac{\tau_{L(b_i,1)}(\xi)}{\tau_{L(|b_i|,1)}(\xi)}\, .$$

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 $\bj=(j_1,\dots,j_l)$ and vary the colors $\bn=(n_1,\dots,n_m)$ of $L$.

For non–negative integers $n,k$ we define $$A(n,k) := \frac{\prod^{k}_{i=0}
\left(q^{n}+q^{-n}-q^i -q^{-i}\right)}{(1-q) \, (q^{k+1};q)_{k+1}}.$$ For $\bk=(k_1,\dots,k_m)$ let $$A(\bn,\bk):= \prod_{j=1}^m \; A(n_j,k_j).$$ Note that $A(\bn,\bk)=0$ if $k_j \ge n_j$ for some index $j$. Also

$$A(n,0)= q^{-1} J_U(n)^2.$$

The colored Jones polynomial $J_{L\sqcup L'}
(\n, \bj)$, when $\bj$ is fixed, can be repackaged into the invariant $C_{L\sqcup L'}
(\bk, \bj)$ as stated in the following theorem.

\[GeneralizedHabiro\] Suppose $L\sqcup L'$ is a link in $S^3$, with $L$ having zero linking matrix. Assume the components of $L'$ have fixed [*odd*]{} colors $\bj = (j_1,\dotsm j_l)$. Then there are invariants $$\label{Jones2}
C_{L\sqcup L'}(\bk,\bj) \in \frac{(q^{k+1};q)_{k+1}}{(1-q)}
\,\,\BZ[q^{\pm 1}] ,\quad \text{where  $k=\max\{k_1,\dots, k_m\}$}$$ such that for every $\bn =(n_1,\dots, n_m)$ $$\label{Jones}
 J_{L\sqcup L'}
(\n, \bj)  \,  \prod^m_{i=1}\; [n_i] = \sum_{0\le k_i \le n_i-1}
C_{L\sqcup L'}(\bk,\bj)\;
  A(\bn, \bk).$$

When $L'=\emptyset$, this is Theorem 8.2 in [@Ha]. This generalization, essentially also due to Habiro, can be proved similarly as in [@Ha]. For completeness we give a proof in the Appendix. Note that the existence of $C_{L\sqcup L'}(\bk,\bj)$ as rational functions in $q$ satisfying is easy to establish. The difficulty here is to show the integrality of .

Since $A(\bn, \bk) =0$ unless $ \bk < \bn$, in the sum on the right hand side of one can assume that $\bk$ runs over the set of all $m$–tuples $\bk$ with non–negative integer components. We will use this fact later.

Strategy of the proof of the main theorem  {#strategy}
==========================================

Here we give the proof of Theorem \[main\] using technical results that will be proved later.

As before, $ L\sqcup 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 $\bj=(j_1,\dots, j_l)$, with $j_i$’s odd. Surgery along the framed link $L$ transforms $(S^3,L')$ into $(M,
L')$. We will define $I_{M,L'}\in \cR_{b}$, such that $$\tau'_{M,L'}(\xi)\;=\; \ev_{\xi}\left(I_{M,L'}\right)
\label{0080}$$ for any root of unity $\xi$ of odd order. This unified invariant is multiplicative with respect to the connected sum.

The following observation is important. By Proposition \[main-cor\], there is [*at most one*]{} element $f(q)\in \R_b$ such that for every root $\xi$ of odd order one has $$\tau'_{M,L} (\xi) = \ev_\xi\left( f(q)\right).$$ That is, if we can find such an element, it is unique, and we put $I_{M,L'} := f(q)$.

Laplace transform
-----------------

The following is the main technical result of the paper. A proof will be given in Section \[laplace\].

Suppose $b=\pm 1$ or $b= \pm p^l$ where $p$ is a prime and $l$ is positive. For any non–negative integer $k$, there exists an element $Q_{b,k} \in \R_b $ such that for every root $\xi$ of odd order $r$ one has $$\frac{{\sum\limits_n}^\xi \, q^{b\frac{n^2-1}{4}} A(n,k) }{F_{U^b}(\xi)}
= \ev_\xi (Q_{b,k}).$$ \[0078\] In addition, if $b$ is odd, $Q_{b,k} \in \calS_b $.

Definition of the unified invariant: diagonal case {#2501}
--------------------------------------------------

Suppose that the linking number between any two components of $L$ is 0, and the framing on components of $L$ are $b_i=\pm p_i^{k_i}$ for $i=1,\dots, m$, where each $p_i$ is prime or 1. Let us denote the link $L$ with all framings switched to zero by $L_0$.

Using , taking into account the framings $b_i$’s, we have $$J_{L\sqcup L'}(\bn,\bj)\prod_{i=1}^m [n_i] = \sum_{\bk\ge 0} C_{L_0 \sqcup L'}
(\bk,\bj) \, \prod_{i=1}^m q^{b_i \frac{n_i^2-1}{4}} A(n_i,k_i).$$ By the definition of $F_{L\sqcup L'}$, we have $$F_{L\sqcup L'}(\xi)= \sum_{\bk \ge 0}
\ev_\xi(C_{L_0 \sqcup L'}(\bk,\bj)) \, \prod_{i=1}^m
{\sum_{n_i}}^\xi \,  q^{b_i \frac{n_i^2-1}{4}} A(n_i,k_i).$$ From and Theorem \[0078\], we get $$\tau'_{M,L'}(\xi) = \ev_\xi \left \{  \prod_{i=1}^m  I_{L(b_i,1)} \,
 \sum_{\bk} C_{L_0 \sqcup L'}(\bk,\bj) \, \prod_{i=1}^m  Q_{b_i,k_i} \right \},$$ where the unified invariant of the lens space $I_{L(b_i,1)}\in \R_b$, with $\ev_\xi(I_{L(b_i,1)})=\tau'_{L(b_i,1)}(\xi)$, exists by Lemma \[0079\] below. Thus if we define $$I_{(M,L')}:= \prod_{i=1}^m  I_{L(b_i,1)} \, \sum_{\bk}
C_{L_0 \sqcup L'}(\bk,\bj) \, \prod_{i=1}^m  Q_{b_i,k_i}\, ,$$ then is satisfied. By Theorem \[GeneralizedHabiro\], $C_{L_0 \sqcup L'}(\bk,\bj)$ is divisible by $(q^{k+1};q)_{k+1}/(1-q)$, which is divisible by $(q;q)_k$, where $k = \max k_i$. It follows that $I_{(M,L')} \in \R_b$. In addition, if $b$ is odd, then $I_{(M,L')} \in \calS_b$.

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 $\pm p^k$ with $p=0,1$ or a prime. The following lemma was proved in [@Le Proposition 3.2 (a)].

For every rational homology sphere $M$, there are lens spaces $L(b_i,a_i)$ such that the connected sum of $M$ and these lens spaces is diagonal. Moreover, each $b_i$ is a prime power divisor of $|H_1(M,\Z)|$. \[diagonalization\]

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 \[2501\]. Then $I_M$ is the quotient of $I_{M'}$ by the unified invariants of the lens spaces. But unlike the simpler case of [@Le], 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.

Splitting of the invariant
--------------------------

Suppose $p$ is a prime divisor of $b$, then it’s clear that $\R_p \subset \R_b$.

In Section \[cyc\] we will see that there is a decomposition

$$\R_b = \R_{b}^{p,0} \times \R_{b}^{p,{{\bar 0}}},$$ with canonical projections $\pi^p_0: \R_b \to \R_{b}^{p,0}$ and $\pi^p_{{\bar 0}}:\R_b \to \R_{b}^{p,{{\bar 0}}}$. If $f\in \R_{b}^{p,0}$ then $\ev_\xi(f)$ can be defined when the order of $\xi$ is coprime with $p$; and in this case $ \ev_\xi(g) = \ev_\xi(\pi^p_0(g))$ for every $g\in \R_b$.

On the other hand, if $f\in \R_{b}^{p,{{\bar 0}}}$ then $\ev_\xi(f)$ can be defined when the order of $\xi$ is divisible by $p$, and one has $ \ev_\xi(g) = \ev_\xi(\pi^p_{{\bar 0}}(g))$ for every $g\in \R_b$.

It also follows from the definition that $\R_p^{p,\ve}
\subset \R_b^{p,\ve}$ for $\ve = 0$ or ${{\bar 0}}$.

For $\calS_b$, there exists a completely analogous decomposition. For any odd divisor $p$ of $b$, an element $ x \in \R_b$ (or $\calS_b$) determines and is totally determined by the pair $(\pi^p_0(x), \pi^p_{{\bar 0}}(x))$. If $p=2$ divides $b$, then for any $x\in \R_b$, $x=\pi^p_0(x)$.

Hence, to define $I_M$ it is enough to fix $I^0_M= \pi^p_0(I_M)$ and $I^{{\bar 0}}_M=\pi^p_{{\bar 0}}(I_M)$. The first part $I^0_M= \pi^p_0(I_M)$, when $b=p$, was defined in [@Le] (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 [@Le].

Lens spaces
-----------

Suppose $b,a,d$ are integers with $(b,a)=1$ and $b\neq 0$. Let $M(b,a;d)$ be the pair of a lens space $L(b,a)$ and a knot $K\subset L(b,a)$, colored by $d$, as described in Figure \[figure:LensSpaceKnot\].

Among these pairs we want to single out some whose quantum invariants are invertible. For $\ve\in \{0,{{\bar 0}}\}$, let $M^\ve(b,a) := M(b,a;d(\ve))$, where $d(0):=1$ and $d({{\bar 0}})$ is the smallest odd positive integer such that $|a|d({{\bar 0}}) \equiv 1 \pmod {b}$. Note that if $|a|=1$, $d(0)=d({{\bar 0}})=1$.

It is known that if the color of a link component is $1$, then the component can be removed from the link without affecting the value of quantum invariants. Hence $$\tau_{M(b,a;1)} = \tau_{L(b,a)}.$$

Suppose $b=\pm p^{l}$ is a prime power. For $\ve \in \{0,{{\bar 0}}\}$, there exists an invertible invariant $I^\ve_{M^\ve(b,a)} \in \R^{p,\ve}_p$ such that $$\tau'_{M^\ve(b,a)}(\xi)\;=\; \ev_{\xi}\left(I^\ve_{M^\ve(b,a)}\right)$$ where $\ve=0$ if the order of $\xi$ is not divisible by $p$, and $\ve={{\bar 0}}$ otherwise. Moreover, if $p$ is odd, then $I^\ve_{M^\ve(b,a)}$ belongs to and is invertible in $\calS^{p,\ve}_p$. \[0079\]

A proof of Lemma \[0079\] will be given in Section \[diag\].

Definition of the unified invariant: general case
-------------------------------------------------

Now suppose $(M,L')$ is an arbitrary pair of a rational homology 3–sphere with a link $L'$ in it colored by odd numbers $j_1,\dots, j_l$. Let $L(b_i,a_i)$ for $i=1,\dots, m$ be the lens spaces of Lemma \[diagonalization\]. We use induction on $m$. If $m=0$, then $M$ is diagonal and $I_{M,L'}$ has been defined in Subsection \[2501\].

Since $(M,L') \# M(b_1,a_1;d)$ becomes diagonal after adding $m-1$ lens spaces, the unified invariant of $(M,L') \# M(b_1,a_1;d)$ can be defined by induction, for any odd integer $d$. In particular, one can define $I_{M^\ve}$, where $M^\ve := (M,L') \# M^\ve(b_1,a_1)$. Here $\ve = 0$ or $\ve ={{\bar 0}}$ and $b_1$ is a power of a prime $p$ dividing $b$. It follows that the components $\pi^p_\ve(I_{M^\ve}) \in \R^{p,\ve}_b$ are defined.

By Lemma \[0079\], $I^\ve_{M^\ve(b_1,a_1)}$ is defined and invertible. Now we put $$I^\ve_{M,L'} := I^\ve_{M^\ve} \cdot  (I^\ve_{M^\ve(b_1,a_1)})^{-1}.$$ It is easy to see that $I_{M,L'}:= (I^0_{M,L'}, I^{{\bar 0}}_{M,L'})$ satisfies . This completes the construction of $I_{M,L'}$. It remains to prove Lemma \[0079\] and Theorem \[0078\].

Cyclotomic completions of polynomial rings {#cyc}
==========================================

In this section we adapt the results of Habiro on cyclotomic completions of polynomial rings [@Ha1] to our rings.

On cyclotomic polynomial
------------------------

Recall that $e_n := \exp(2\pi I/n)$ and denote by $\Phi_n(q)$ the cyclotomic polynomial $$\Phi_n(q) = \prod_{\substack{(j,n)=1\\0<j<n}} (q - e_n^j).$$ The degree of $\Phi_n(q)\in \Z[q]$ is given by the Euler function $\varphi(n)$. Suppose $p$ is a prime and $n$ an integer. Then (see e.g. [@Na]) $$\Phi_n(q^p)= \begin{cases}    \Phi_{np}(q)  & \text{ if } p \mid n \\
\Phi_{np}(q) \Phi_n(q)  & \text{ if } p \nmid n.
\end{cases}$$ It follows that $\Phi_n(q^p)$ is always divisible by $\Phi_{np}(q)$.

The ideal of $\Z[q]$ generated by $\Phi_n(q)$ and $\Phi_m(q)$ is well–known, see e.g. [@Le Lemma 5.4]:

$\text{ }$

-   If $\frac{m}{n} \neq p^e$ for any prime $p$ and any integer $e\neq 0$, then $(\Phi_n)+ (\Phi_m)=(1)$ in $\Z[q]$.

-   If $\frac{m}{n} = p^e$ for a prime $p$ and some integer $e \neq 0$, then $(\Phi_n)+ (\Phi_m)=(1)$ in $\Z[1/p][q]$. \[0911\]

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 $k,l \ge 1$.

Habiro’s results
----------------

Let us summarize some of Habiro’s results on cyclotomic completions of polynomial rings [@Ha1]. Let $R$ be a commutative integral domain of characteristic zero and $R[q]$ the polynomial ring over $R$. For any $S\subset \N$, Habiro defined the $S$–cyclotomic completion ring $R[q]^S$ as follows: \[rs\] R\[q\]\^S:=\_ where $\Phi^*_S$ denotes the multiplicative set in $\Z[q]$ generated by $\Phi_S=\{\Phi_n(q)\mid n\in S\}$ and directed with respect to the divisibility relation.

For example, since the sequence $(q;q)_n$, $n\in \N$, is cofinal to $\Phi^*_\N$, we have \[cofinal\] \^.

Note that if $S$ is finite, then $R[q]^S$ is identified with the $(\prod \Phi_S)$–adic completion of $R[q]$. In particular, $$R[q]^{\{1\}}\simeq R[[q-1]], \quad
R[q]^{\{2\}}\simeq R[[q+1]].$$ Suppose $S' \subset S$, then $\Phi^*_{S'}\subset \Phi^*_S$, hence there is a natural map $$\rho^R_{S, S'}: R[q]^S \to R[q]^{S'}.$$

Recall important results concerning $R[q]^S$ from [@Ha1]. Two positive integers $n, n'$ are called [*adjacent*]{} if $n'/n=p^e$ with a nonzero $e\in \Z$ and a prime $p$, such that the ring $R$ is $p$–adically separated, i.e. $\bigcap_{n=1}^\infty (p^n) =0$ in $R$. A set of positive integers is [*$R$–connected*]{} if for any two distinct elements $n,n'$ there is a sequence $n=n_1, \,n_2, \dots,\, n_{k-1},\, n_k= n'$ in the set, such that any two consecutive numbers of this sequence are adjacent. Theorem 4.1 of [@Ha1] says that if $S$ is $R$–connected, then for any subset $S'\subset S$ the natural map $ \rho^R_{S,S'}: R[q]^S \hookrightarrow R[q]^{S'}$ is an embedding.

If $\zeta$ is a root of unity of order in $S$, then for every $f(q)\in R[q]^S$ the evaluation $\ev_\zeta(f(q))\in R[\zeta]$ can be defined by sending $q\to\zeta$. For a set $\Xi$ of roots of unity whose orders form a subset $\cT\subset S$, one defines the evaluation $$\ev_\Xi: R[q]^S \to \prod_{\zeta \in \Xi} R[\zeta].$$ Theorem 6.1 of [@Ha1] shows that if $R\subset \Q$, $S$ is $R$–connected, and there exists $n\in S$ that is adjacent to infinitely many elements in $\cT$, then $\ev_\Xi$ is injective.

Taylor expansion
----------------

Fix a natural number $n$, then we have $$R[q]^{\{n\}} =
\lim_{\overleftarrow{\hspace{2mm}k\hspace{2mm}}}
\frac{R[q]}{(\Phi^k_n(q))}\; .$$ Suppose $ \Z \subset R \subset \Q$, then the natural algebra homomorphism

$$h: \frac{R[q]}{(\Phi^k_n(q))} \to \frac{R[e_n][q]}{((q-e_n)^k)}$$ is injective, by Proposition \[h\_kInjektive\] below. Taking the inverse limit, we see that there is a natural injective algebra homomorphism

$$h : R[q]^{\{n\}} \to R[e_n][[q-e_n]].$$

Suppose $n \in S$. Combining $h$ and $\rho_{S, \{n\}}: R[q]^S \to R[q]^{\{n\}}$, we get an algebra map

$${{\mathfrak t}}_n: R[q]^S \to R[e_n][[q-e_n]].$$ If $f\in R[q]^S$, then ${{\mathfrak t}}_n(f)$ is called the Taylor expansion of $f$ at $e_n$.

Splitting of $\calS_p$ and evaluation
-------------------------------------

For every integer $a$, we put $\N_a := \{ n \in \N \mid (a,n)=1\}$.

Suppose $p$ is a prime. Analogously to , we have $$\calS_p \simeq \Z[1/p][q]^\N\,.$$

Observe that $\N$ is not $\Z[1/p]$–connected. In fact one has $\N =\amalg_{j=0}^\infty \; p^j \N_p$, where each $p^j\N_p$ is $\Z[1/p]$–connected. Let us define $$\calS_{p,j}:= \Z[1/p][q]^{p^j \N_p}.$$ Note that for every $f \in \calS_p$, the evaluation $\ev_\xi(f)$ can be defined for every root $\xi$ of unity. For $f\in \calS_{p,j}$, the evaluation $\ev_\xi(f)$ can be defined when $\xi$ is a root of unity of order in $p^j\N_p$.

For every prime $p$ one has $$\calS_p\simeq \prod_{j=0}^\infty \calS_{p,j}.
\label{0912}$$

Suppose $n_i \in p^{j_i}\N_p$ for $ i=1,\dots, m$, with distinct $j_i$’s. Then $n_{i}/n_{s}$, with $i \neq s$, is either not a power of a prime or a non–zero power of $p$, hence by Lemma \[0911\] (and the remark right after Lemma \[0911\]), for any positive integers $k_1,\dots, k_m$, we have $$(\Phi_{n_i}^{k_i})+ (\Phi_{n_s}^{k_s})=(1) \quad \text{ in }\Z[1/p][q].$$ By the Chinese remainder theorem, we have $$\frac{ \Z[1/p][q]}{\left(\prod_{i=1}^m \Phi_{n_i}^{k_i}\right)}\;\simeq\;
 \prod_{i=1}^m
\frac{\Z[1/p][q]}{\left(\Phi_{n_i}^{k_i}\right)} .$$ Taking the inverse limit, we get .

Let $\pi_j: \calS_p \to \calS_{p,j}$ denote the projection onto the $j$th component in the above decomposition.

Suppose $\xi$ is a root of unity of order $r= p^j r'$, with $(r',p)=1$. Then for any $x\in \calS_p$, one has $$\ev_\xi(x) = \ev_\xi(\pi_j(x)).$$ If $i\neq j$ then $\ev_\xi(\pi_i(x))=0$.

\[1100\]

Note that $\ev_\xi(x)$ is the image of $x$ under the projection $\calS_p \to \calS_p/(\Phi_r(q))= \Z[1/p][\xi]$. It remains to notice that $\calS_{p,i}/(\Phi_r(q))=0$ if $i\neq j$.

Splitting of $\calS_b$
----------------------

Suppose $p$ is a prime divisor of $b$. Let $$\calS_{b}^{p,0} := \Z[1/b][q]^{\N_p} \qquad \text {and } \quad
\calS_{b}^{p,{{\bar 0}}} := \Z[1/b][q]^{p\N}\simeq \prod_{j>0}\Z[1/b][q]^{p^j\N_p}$$

We have similarly $$\calS_b = \calS_{b}^{p,0} \times \calS_{b}^{p,{{\bar 0}}}$$ with canonical projections $\pi^p_0 : \calS_b \to \calS_b^{p,0}$ and $\pi^p_{{\bar 0}}: \calS_b \to \calS_b^{p,{{\bar 0}}}$. Note that if $b=p$, then $\calS_{p}^{p,0}=\calS_{p,0}$ and $\calS_{p}^{p,{{\bar 0}}}=\prod_{j>0} \calS_{p,j}$. As before we set $\calS_{b,0}:=\Z[1/b][q]^{\N_b}$ and $\pi_0:\calS_b\to\calS_{b,0}$.

Suppose $f \in \calS_b$. If $\xi$ is a root of unity of order coprime with $p$, then $\ev_\xi(f) = \ev_\xi(\pi^p_0(f))$. Similarly, if the order of $\xi$ is divisible by $p$, then $\ev_\xi(f) = \ev_\xi(\pi^p_{{\bar 0}}(f))$.

Properties of the ring $\cR_b$
------------------------------

For any $b\in \N$, we have $$\R_b\simeq\Z[1/b][q]^{\N_2}$$ since the sequence $(q;q^2)_k,\, k\in \N$, is cofinal to $\Phi^*_{\N_2}$. Here $\N_2$ is the set of all odd numbers. Let $\{p_i\,|\, i=1,\dots,m\}$ be the set of all distinct [*odd*]{} prime divisors of $b$. For $\bn=(n_1,\dots,n_m)$, a tuple of numbers $n_i\in \N$, let $\bp^{\bn}=\prod_{i}p_i^{n_i}$. Let $S_{\bn}:=\bp^{\bn}\N_{2b}$. Then $\N_2=\amalg_{\bn}\, S_\bn$. Moreover, for $a\in S_{\bn},$ $a'\in S_{\bn'}$, we have $(\Phi_{a}(q),\Phi_{a'}(q))=(1)$ in $\Z[1/b]$ if $\bn\neq\bn'$. In addition, each $S_\bn$ is $\Z[1/b]$–connected. An argument similar to that for Equation gives $$\cR_b\simeq\prod_{\bn}\Z[1/b][q]^{S_\bn}.$$ In particular, $\R^{p_i,0}_b:=\Z[1/b][q]^{\N_{2p_i}}$ and $\R^{p_i,{{\bar 0}}}_b:=\Z[1/b][q]^{p_i\N_2}$ for any $1\leq i\leq m$. If $2\mid b$, then $\R^{2,0}_b$ coincides with $\R_b$.

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 $b$.

\[main-corProof\] With the above notations, one has the following.

-   For any $l\in S_\bn$, the Taylor map ${{\mathfrak t}}_l : \Z[1/b][q]^{S_\bn}\;\to  \;\Z[1/b][e_{l}][[q-e_{l}]]$ is injective.

-   Suppose $f, g\in \Z[1/b][q]^{S_{\bn}}$ such that $\ev_\xi(f)=\ev_\xi(g)$ for any root of unity $\xi$ with $\text{ord}(\xi)\in \bp^\n T$, then $f=g$. The same holds true if $\bp^\n T$ is replaced by $\bp^\n P$.

-   For odd $b$, the natural homomorphism $\rho_{\N,\N_2}:\calS_b\to \R_b$ is injective. If $2\mid b$, then the natural homomorphism $\calS_{b}^{2,0}\to \R_{b}$ is an isomorphism.

\(a) Since each $S_\bn$ is $\Z[1/b]$–connected in Habiro sense, by [@Ha1 Theorem 4.1], for any $l\in S_\bn$ \[inj\] \_[S, {l}]{}: \^[S\_]{}\^[{ l}]{} is injective. Hence ${{\mathfrak t}}_l = h \circ \rho_{S, \{l\}}$ is injective too.

\(b) Since both sets contain infinitely many numbers adjacent to $\bp^\n$, the claim follows from Theorem 6.1 in [@Ha1].

\(c) Note that for odd $b$ $$\calS_b\simeq \prod_{\bn}\Z[1/b][q]^{S'_\bn}$$ where $S'_{\bn}:=\bp^{\bn}\N_{b}$. Further observe that $S'_{\bn}$ is $\Z[1/b]$–connected if $b$ is odd. Then by [@Ha1 Theorem 4.1] the map $$\Z[1/b][q]^{S'_\bn}\hookrightarrow \Z[1/b][q]^{S_{\bn}}$$ is an embedding. If $2\mid b$, then $\calS^{2,0}_b:=\Z[1/b][q]^{\N_2}\simeq\R_b$.

Assuming Theorem \[main\], Proposition \[main-corProof\] (b) implies Proposition \[main-cor\].

On the Ohtsuki series at roots of unity {#Oh-strategy}
=======================================

The Ohtsuki series was defined for $SO(3)$ invariants by Ohtsuki [@Oh] and extended to all other Lie algebras by the third author [@Le2; @Le3].

In the works [@Oh; @Le2; @Le3], 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 [@Ha], 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 [@Le].

Here, we prove that the sequence of $SO(3)$ invariants at the $pr$th 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 $e_r$.

Extension of $\Z[1/b][e_r]$
---------------------------

Fix an odd positive integer $r$. Assume $p$ is a prime bigger than $b$ and $r$. The cyclotomic rings $\Z[1/b][e_{pr}]$ and $\Z[1/b][e_r]$ are extensions of $\Z[1/b]$ of degree $\varphi(rp)=\varphi(r) \varphi(p)$ and $\varphi(r)$, respectively. Hence $\Z[1/b][e_{pr}]$ is an extension of $\Z[1/b][e_r]$ of degree $\varphi(p)=p-1$. Actually, it is easy to see that for $$f_p(q):= \frac{q^p - e_r^p}{q-e_r},$$ the map $$\frac{\Z[1/b,e_r][q]}{(f_p(q))} \;\to\; \Z[1/b][e_{pr}], \hspace{5mm}
q\mapsto e_p e_r,$$ is an isomorphism. We put $x=q-e_r$ and get $$\Z[1/b][e_{pr}]\simeq \frac{\Z[1/b,e_r][x]}{(f_p(x+e_r))}\; .
\label{5500}$$ Note that $$f_p(x+ e_r) = \sum_{n=0}^{p-1} \binom{p}{n+1} x^n e_r^{p-n-1}$$ is a monic polynomial in $x$ of degree $p-1$, and the coefficient of $x^n$ in $f_p(x+e_r)$ is divisible by $p$ if $n \le p-2$.

Arithmetic expansion of $\tau'_M$
---------------------------------

Suppose $M$ is a rational homology 3–sphere with $|H_1(M,\Z)|=b$. By Theorem \[main\], for any root of unity $\xi$ of order $pr$ $$\tau'_M(\xi)\in \Z[1/b][e_{pr}] \simeq \frac{\Z[1/b,e_r][x]}{(f_p(x+e_r))}\; .$$ Hence we can write \[Ohtsuki\] ’\_M(e\_[r]{}e\_[p]{})= \_[n=0]{}\^[p-2]{} a\_[p,n]{} x\^n where $a_{p,n}\in \Z[1/b,e_r]$. The following proposition shows that the coefficients $a_{p,n}$ stabilize as $p\to \infty$.

\[main-cor1\] 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 $a_n= a_n(M) \in \Z[1/b,e_r]$ such that for every prime $p > \max (b,r)$, we have $$a_n\equiv a_{p,n} \pmod p\;\;\; \text{in $ \Z[1/b,e_r]$ for} \;\;\;
0\le n \le  p-2.
\label{5501}$$ Moreover, the formal series $\sum_{n} a_n (q-e_r)^n$ is equal to the Taylor expansion of the unified invariant $I_M$ at $e_r$.

The uniqueness of $a_n$ follows from the easy fact that if $a \in \Z[1/b,e_r]$ is divisible by infinitely many rational primes $p$, then $a=0$.

Assume Theorem \[main\] 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 holds true.

Recall that $x= q-e_r$. The diagram $$\begin{CD} \Z[\frac{1}{b}][q]^{\N_2} @>>> \Z[\frac{1}{b}, e_r][q]^{r\N_2} @>>> \Z[\frac{1}{b},e_r][[x]] \\
@VV q \to e_re_pV     @VV /(f_p(q))V   @VV /(f_p(x+e_r))V \\
\Z[\frac{1}{b}][e_{rp}] @> e_re_p \to q >> \frac{\Z[\frac{1}{b}, e_r][q]}{(f_p(q))} @>>>\frac{ \Z[\frac{1}{b},e_r][[x]]}{(f_p(x+e_r))}
\end{CD}$$ 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 is in the lower middle ring. Using the commutativity at the lower right corner ring, we see that

$$\sum_{n=0}^{p-2} a_{p,n}x^n  = \sum_{n=0}^{\infty} a_{n}x^n  \pmod{f_p(x+e_r)}
\quad \text{in} \quad \Z[1/b,e_r][[x]].$$ Since the coefficients of $f_p(x+e_r)$ up to degree $p-2$ are divisible by $p$, we get the congruence .

Proposition \[main-cor1\], when $r=1$, was the main result of Ohtsuki [@Oh], 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 $r\mid b$, corresponds to a new type of LMO invariant.

Frobenius maps {#map-frob}
==============

The proof of Theorem \[0078\], 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 \[laplace\], belongs to $\R_b$, i.e. that certain roots of $q$ exist in $\R_b$.

On the module $\Z[q]/(\Phi^k_{n}(q))$
-------------------------------------

Since cyclotomic completions are built from modules like $\Z[q]/(\Phi^k_{n}(q))$, we first consider these modules. Fix $n,k\ge 1$. Let $$E  := \frac{\Z[q]}{(\Phi^k_{n}(q))}, \quad \text { and } \quad G:=
\frac{\Z[e_{n}][x]}{(x^k)}\; .$$

The following is probably well–known.

\[h\_kInjektive\]$\text{ }$

-   Both $E$ and $G$ are free $\Z$–modules of the same rank $k \varphi(n)$.

-   The algebra map $h: \Z[q] \to \Z[e_n][x]$ defined by

    $$h (q) = e_n + x$$ descends to a well–defined algebra homomorphism, also denoted by $h$, from $E$ to $G$. Moreover, the algebra homomorphism $h: E \to G$ is injective.

\(a) Since $\Phi^k_n(q)$ is a monic polynomial in $q$ of degree $k \varphi(n)$, it is clear that $$E=  \Z[q]/(\Phi^k_n(q))$$ is a free $\Z$–module of rank $k\varphi(n)$. Since $G = \Z[e_n]\otimes_\Z \Z[x]/(x^k)$, we see that $G$ is free over $\Z$ of rank $k\varphi(n)$.

\(b) To prove that $h$ descends to a map $E \to G$, one needs to verify that $h(\Phi^k_n(q))=0$. Note that $$h(\Phi^k_n(q))= \Phi^k_{n}(x + e_n) =
\prod_{(j,n)=1} (x+e_n - e_n^j)^k.$$ When $j=1$, the factor is $x^k$, which is 0 in $\Z[e_n][x]/(x^k)$. Hence $h(\Phi^k_n(q))=0$.

Now we prove that $h$ is injective. Let $f(q)\in\Z[q]$. Suppose $h(f(q))=0$, or $f(x+e_n)=0$ in $\Z[e_n][x]/(x^k)$. It follows that $f(x+e_n)$ is divisible by $x^k$; or that $f(x)$ is divisible by $(x-e_n)^k$. Since $f$ is a polynomial with coefficients in $\Z$, it follows that $f(x)$ is divisible by all Galois conjugates $(x-e_n^j)^k$ with $(j,n)=1$. Then $f$ is divisible by $\Phi^k_n(q)$. In other words, $f=0$ in $E=  \Z[q]/(\Phi^k_n(q))$.

A Frobenius homomorphism
------------------------

We use $E$ and $G$ of the previous subsection. Suppose $b$ is a positive integer coprime with $n$. If $\xi$ is a primitive $n$th root of 1, i.e. $\Phi_n(\xi)=0$, then $\xi^b$ is also a primitive $n$th root of $1$, i.e. $\Phi_n(\xi^b)=0$. It follows that $\Phi_n(q^b)$ is divisible by $\Phi_n(q)$.

Therefore the algebra map $F_b: \Z[q] \to \Z[q]$, 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 $F_b(E)$.

The image $F_b(E)$ is a free $\Z$–submodule of $E$ of maximal rank, i.e. ${\operatorname{rk}}(F_b(E)) = {\operatorname{rk}}(E)$. Moreover, the index of $F_b(E)$ in $E$ is $b^{k(k-1)\varphi(n)/2}$.

Using Proposition \[h\_kInjektive\] we identify $E$ with its image $h(E)$ in $G$.

Let $\tilde F_b: G \to G$ be the $\Z$–algebra homomorphism defined by $\tilde F_b(e_n) =e_n^b, \tilde F_b(x)=
(x+e_n)^b - e_n^b$.

Note that $\tilde F_b(x) = b e_n^{b-1} x + O(x^2)$, hence $\tilde F_b(x^k)=0$. It is easy to see that $\tilde F_b$ is a well–defined algebra homomorphism, and that $\tilde F_b$ restricted to $E$ is exactly $F_b$. Since $E$ is a lattice of maximal rank in $G\otimes \Q$, it follows that the index of $F_b$ is exactly the determinant of $\tilde F_b$, acting on $G\otimes \Q$.

A basis of $G$ is $e_n^j x^l$, with $(j,n)=1, 0<j<n$ and $j=0$, and $0\le l < k$. Note that $$\tilde F_b(e_n^j x^l) = b^le_n^{jb}  e_n^{(b-1)l} x^l + O(x^{l+1}).$$ Since $(b,n)=1$, the set $e_n^{jb}$, with $(j,n)=1$ is the same as the set $e_n^{j}$, with $(j,n)=1$. Let $f_1: G\to G$ be the $\Z$–linear map defined by $f_1(e_n^{jb} x^l) =
e_n^{j} x^l$. Since $f_1$ permutes the basis elements, its determinant is $\pm 1$. Let $f_2: G\to G$ be the $\Z$–linear map defined by $f_2(e_n^{j} x^l) = e_n^{j} (e_n^{1-b}x)^l$. The determinant of $f_2$ is again $\pm 1$. 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: $a \mapsto e^{-s}_n a$. Now $$f_1 f_2 \tilde F_b(e_n^j x^l) = b^l e_n^j x^l + O(x^{l+1})$$ can be described by an upper triangular matrix with $b^l$’s on the diagonal; its determinant is equal to $b^{k(k-1)\varphi(n)/2}$.

From the proposition we see that if $b$ is invertible, then the index is equal to 1, hence we have

\[onen\] For any $n$ coprime with $b$ and $k\in \N$, the Frobenius homomorphism $F_b:  \Z[1/b][q]/\left(\Phi^k_n(q)\right) \to \Z[1/b][q]/\left(\Phi^k_n(q)\right)$, defined by $F_b(q)= q^b$, is an isomorphism.

Frobenius endomorphism of $\calS_{b,0}$
---------------------------------------

For finitely many $n_i\in\N_b$ and $k_i\in \N$, the Frobenius endomorphism

$$F_b :\frac{\Z[1/b][q]}{\left(\prod_i\Phi^{k_i}_{n_i}(q)\right)}\to
\frac{\Z[1/b][q]}{\left(\prod_i\Phi^{k_i}_{n_i}(q)\right)}$$ sending $q$ to $q^b$, is again well–defined. Taking the inverse limit, we get an algebra endomorphism $$F_b: \Z[1/b][q]^{\N_b} \to \Z[1/b][q]^{\N_b}.$$

\[frob\] The Frobenius endomorphism $F_b: \Z[1/b][q]^{\N_b} \to \Z[1/b][q]^{\N_b}$, sending $q$ to $q^b$, is an isomorphism.

For finitely many $n_i\in\N_b$ and $k_i\in \N$, consider the natural algebra homomorphism $$J:\frac{\Z[1/b][q]}{\left(\prod_i\Phi^{k_i}_{n_i}(q)\right)}
\to
\prod_i \frac{ \Z[1/b][q]}{\left(\Phi^{k_i}_{n_i}(q)\right)} .$$ This map is injective, because in the unique factorization domain $\Z[1/b][q]$, one has $$(\Phi_{n_1}(q)^{k_1} \dots \Phi_{n_s}(q)^{k_s})
= \bigcap_{j=1}^s \Phi_{n_j}(q)^{k_j} \, .$$ Since the Frobenius homomorphism commutes with $J$ and is an isomorphism on the target of $J$ by Proposition \[onen\], it is an isomorphism on the domain of $J$. Taking the inverse limit, we get the claim.

Existence of $b$th root of $q$ in $\calS_{b,0}$ {#qthroot}
-----------------------------------------------

Suppose $n$ and $b$ are coprime positive integers and $y \in \Q[e_n]$ such that $y^b=1$. Then $y =\pm 1$. If $b$ is odd then $y=1$. \[0923\]

Let $d\mid b$ be the order of $y$, i.e. $y$ is a primitive $d$th root of $1$. Then $\Q[e_n]$ contains $y$, and hence $e_d$. Since $(n,d)=1$, one has $\Q[e_n]\cap \Q[e_d]=\Q$ (see e.g. [@Lang Corollary of IV.3.2]). Hence if $e_d \in \Q[e_n]$, then $e_d\in\Q$, it follows that $d=1$ or $2$. Thus $y=1$ or $y=-1$. If $b$ is odd, then $y$ cannot be $-1$.

Let $b$ be a positive integer, $T\subset \N_b$, and $y \in \Q[q]^T$ satisfying $y^b=1$. Then $y =\pm 1$. If $b$ is odd then $y=1$. \[0913\]

It suffices to show that for any $n_1, n_2 \dots n_m \in T$, the ring $\Q[q]/(\Phi^{k_1}_{n_1}\dots\Phi^{k_m}_{n_m})$ does not contains a $b$th root of $1$ except possibly for $\pm 1$. Using the Chinese remainder theorem, it suffices to consider the case where $m=1$. The ring $\Q[q]/(\Phi_{n}^k(q))$ is isomorphic to $\Q[e_n][x]/(x^k)$, by Proposition \[h\_kInjektive\]. If

$$y= \sum_{j=0}^{k-1} a_j x^j, \quad  a_j \in \Q[e_n]$$ satisfies $y^b=1$, then it follows that $a_0^b=1$. By Lemma \[0923\] we have have $a_0=\pm1$. One can easily see that $a_1=\dots =a_{k-1}=0$. Thus $y=\pm 1$.

In contrast with Lemma \[0913\], we have

\[cor9\] For any odd positive $b$, and any subset $T\subset \N_b$, the ring $\Z[1/b][q]^{T}$ contains a unique $b$th root of $q$, which is invertible in $\Z[1/b][q]^T$.

For any even positive $b$, and any subset $T\subset \N_b$, the ring $\Z[1/b][q]^{T}$ contains two $b$th roots of $q$, which are invertible in $\Z[1/b][q]^T$; one is the negative of the other.

Let us first consider the case $T=\N_b$. Since $F_b$ is an isomorphism by Theorem \[frob\], we can define a $b$th root of $q$ by $$q^{1/b}:=F^{-1}_b (q)  \in \calS_{b,0}\,.$$ If $y_1$ and $y_2$ are two $b$th root of the same element, then their ratio $y_1/y_2$ is a $b$th root of 1. From Lemma \[0913\] it follows that if $b$ is odd, there is only one $b$th 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 $\pm q^{1/b}$.

Further it is known that $q$ is invertible in $\Z[q]^\N$ (see [@Ha1]). Actually, there is an explicit expression $q^{-1}=\sum_n q^n (q;q)_n $. Hence $q^{-1}\in \Z[1/b][q]^{\N_b}$, since the natural homomorphism from $\Z[q]^\N$ to $\Z[1/b][q]^{\N_b}$ maps $q$ to $q$. In a commutative ring, if $x\mid y$ and $y$ is invertible, then so is $x$. Hence any root of $q$ is invertible.

In the general case of $T \subset \N_b$, we use the natural map $\Z[1/b][q]^{\N_b}\hookrightarrow \Z[1/b][q]^T$.

 Relation with [@Le] {#relation-with .unnumbered}
--------------------

By Proposition \[cor9\], $\calS_{b,0}$ is isomorphic to the ring $\Lambda^{\N_b}_b:=\Z[1/b][q^{1/b}]^{\N_b}$ used in [@Le]. Furthermore, our invariant $\pi_0 I_M$ and the one defined in [@Le] belong to $\calS_{b,0}$. This follows from Theorem \[main\] for $b$ odd, and from Proposition \[main-corProof\](c) for $b$ even. Finally, the invariant defined in [@Le] for $M$ divided by the invariant of $\#_i L(b^{k_i}_i,1)$ (which is invertible in $\calS_{b,0}$ [@Le Subsection 4.1]) coincides with $\pi_0 I_M$ up to factor $q^{\frac{1-b}{4}}$ by Theorem \[main\], [@Le Theorem 3] and Proposition \[main-corProof\](b).

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 $$G_m : R[q]^T \to R[q]^{mT} \quad \text{ by } \qquad G_m(q) = q^m.$$ Since $\Phi_{mr}(q)$ always divides $\Phi_{r}(q^m)$, $G_m$ is well–defined.

Realization of $q^{a^2/b}$ in $\calS_p$ {#defxb}
---------------------------------------

Throughout this subsection, let $p$ be a prime or $1$. Suppose $b= \pm p^l$ for an $l\in \N$ and let $a$ be an integer. Let $B_{p,j} = G_{p^j}(\calS_{p,0})$. Note that $B_{p,j}\subset \calS_{p,j}$. If $b$ is odd, by Proposition \[cor9\] there is a unique $b$th root of $q$ in $\calS_{p,0}$; we denote it by $x_{b;0}$. If $b$ is even, by Proposition \[cor9\] there are exactly two $b$th root of $q$, namely $\pm q^{1/b}$. We put $x_{b;0}=q^{1/b}$. We define an element $z_{b,a} \in \calS_p$ as follows.

If $b\mid a$, let $z_{b,a} := q^{a^2/b} \in \calS_p$.

If $ b=\pm p^l \nmid a$, then $z_{b,a}\in \calS_p$ is defined by specifying its projections $\pi_j(z_{b,a}):=z_{b,a;j}\in \calS_{p,j}$ 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\le j \le s$ let $$z_{b,a;j} := [G_{p^j}(x_{b;0})]^{a^2/p^j} = [G_{p^j}(x_{b;0})]^{e^2 \, p^{2s-j}}
\in B_{p,j} \subset \calS_{p,j}.$$

Similarly, for $b=\pm p^l$ we define an element $x_b \in \calS_p$ as follows. We put $\pi_0(x_b): =x_{b;0}$. For $j<l$, $\pi_j(x_b):= [G_{p^j}(x_{b;0})]^{p^j}$. If $j\geq l$, $\pi_j(x_b):=q^{b}$. Notice that for $c=(b,p^j)$ we have $$\pi_j(x_b)=z_{b,c;j}.$$

\[eval\_z\] Suppose $\xi$ is a root of unity of order $r = c r'$, where $c= (r,b)$. Then $$\ev_{\xi}(z_{b,a}) = \begin{cases} 0 & \text{ if } c \nmid a \\
 (\xi^c)^{a_1^2 b'_*} &  \text{ if } a=ca_1,
\end{cases}$$ where $b'_*$ is the unique element in $\Z/r'\Z$ such that $  b'_\ast (b/c) \equiv 1 \pmod{r'}$. Moreover, $$\ev_\xi(x_b)=(\xi^c)^{b'_\ast}\; .$$

Let us compute $\ev_\xi(z_{b,a})$. The case of $\ev_\xi(x_b)$ is completely analogous.

If $b\mid a$, then $c\mid a$, and the proof is obvious.

Suppose $b\nmid a$. Let $a=p^s e$ and $c= p^i$. Then $s <l$. Recall that $z_{b,a} = \prod_{j=0}^\infty z_{b,a;j}$. By Lemma \[1100\], $$\ev_\xi(z_{b,a}) = \ev_\xi(z_{b,a;i}).$$

If $c\nmid a$, then $i > s$. By definition, $z_{b,a;i}=0$, hence the statement holds true.

It remains the case $c\mid a$, or $ i \le s$. Note that $\zeta = \xi^c$ is a primitive root of order $r'$ and $(p, r')=1$. Since $z_{b,a;i} \in B_{p,i}$, $$\ev_\xi(z_{b,a;i}) \in \Z[1/p][\zeta].$$ 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 $$[\ev_\xi(z_{b,a;i})]^{b/c} =  (\zeta)^{a^2_1}.$$ Note also that $$[(\xi^c)^{a^2_1 b'_*}]^{b/c} = (\zeta)^{a^2_1}.$$ Using Lemma \[0923\] we conclude $\ev_\xi(z_{b,a;i}) =
(\xi^c)^{a^2_1 b'_*}$ if $b$ is odd, and $\ev_\xi(z_{b,a;i}) =(\xi^c)^{a^2_1 b'_*}$ or $\ev_\xi(z_{b,a;i}) =-(\xi^c)^{a^2_1 b'_*}$ if $b$ is even. Since $\ev_{1}(q^{1/b})=1$ and therefore $\ev_{\xi}(q^{1/b})=\xi^{b_*}$ (and not $-\xi^{b*}$) we get the claim.

Invariant of lens spaces {#diag}
========================

The purpose of this section is to prove Lemma \[0079\]. Throughout this section we will use the following notations.

Let $a$ and $b$ be coprime integers. Choose $\hat{a}$ and $\hat{b}$ such that $b\hat{b}+a\hat{a}=1$ with $0<{\operatorname{sn}}(a)\hat{a}<|b|$. Notice that for $a=1$ we have $\hat{1}=1$ and $\hat{b}=0$. Let $r$ be a fixed odd integer (the order of $\xi$). For $l\in\Z$ coprime to $r$, let $l_*$ denote the inverse of $l$ modulo $r$. If $(b,r)=c$, let $b'_*$ denote the inverse of $b':=\frac{b}{c}$ modulo $r':=\frac{r}{c}$. Notice that for $c=1$, we have $b_*=b'_*$.

Further, we denote by $\left(\frac{x}{y}\right)$ the Jacobi symbol and by $s(a,b)$ the Dedekind sum (see e.g. [@KM-Dedekind]).

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).

\[1409\] Suppose $c=(b,r)$ divides $d-{\operatorname{sn}}(a)\hat{a}$. Then $$\begin{aligned}
\tau'_{M(b,a;d)}(\xi)&= (-1)^{\frac{c+1}{2}\,\frac{{\operatorname{sn}}(ab)-1}{2}}
\left(\frac{|a|}{c}\right)
\left(\frac{1-\xi^{-{\operatorname{sn}}(a)db'_*}}{1-\xi^{-{\operatorname{sn}}(b)b'_*}}\right)^{\chi(c)}
\xi^{
4_*u-4_*b'_* \frac{a(\hat{a}-{\operatorname{sn}}(a)d)^2}{c}
}\end{aligned}$$ where $$u=12s(1,b)-12 {\operatorname{sn}}(b)s(a,b)+
\frac{1}{b}\left(a(1-d^2)+2({\operatorname{sn}}(a)d-{\operatorname{sn}}(b))
+a(\hat{a}-{\operatorname{sn}}(a)d)^2\right) \in \Z$$ and $\chi(c)=1$ if $c=1$ and is zero otherwise. If $c\nmid (\hat{a}\pm d)$, $\tau_{M(b,a;d)}(\xi)=0\; .$ \[lens\]

In particular, it follows that $\tau_{L(b,a)}(\xi)=0$ if $c\nmid \hat{a}\pm 1$.

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 $a_1 \equiv a_2 \pmod{b}$, we can assume $a<b$. Let $b/a$ be given by a continued fraction $$\frac{b}{a}=m_n-\frac{1}{\displaystyle m_{n-1}-\frac{1}{\displaystyle
    m_{n-2}-\dots \frac{1}{\displaystyle m_2-\frac{1}{\displaystyle
        m_1}}}}.$$ Using the Lagrange identity $$a-\frac{1}{b}=(a-1)+\frac{1}{\displaystyle 1+\frac{1}{\displaystyle
    (b-1)}}$$ we can assume $m_i\geq 2$ for all $i$.

The $\tau_{M(b,a;d)}(\xi)$ can be computed in the same way as the invariant $\xi_r(L(b,a),A)$ in [@LiLi], after replacing $A^2$ (respectively $A$) by $\xi^{2_*}$ (respectively $\xi^{4_*}$). Representing the $b/a$–framed unknot in Figure \[figure:LensSpaceKnot\] by a Hopf chain (as e.g. in Lemma 3.1 of [@BL]), we have $$F_{L\sqcup K}(\xi,d)={\sum_{j_i}}^{\xi}\prod_{i=1}^{n}
q^{m_i  \frac{j_i^2-1}{4}}
\prod_{i=1}^{n-1} [j_i j_{i+1}]\cdot[j_n d][j_1]
=\frac{S_n(d)}{(\xi^{2_*}-\xi^{-2_*})^{n+1}}\cdot \xi^{-4_*\sum_{i=1}^{n}m_i}$$ where $$S_n(d)=
{\sum_{\substack{j_i=1 \\ \text{odd}}}^{2r}}
\xi^{\sum m_ij^2_i}
(\xi^{2_*j_1}-\xi^{-2_*j_1})
(\xi^{2_*j_1j_2}-\xi^{-2_*j_1j_2})\dots
(\xi^{2_*j_{n-1}j_n}-\xi^{-2_*j_{n-1}j_n})
(\xi^{2_*j_n d}-\xi^{-2_*j_n d})\, .$$

Using Lemmas 4.11, 4.12 and 4.20 of [@LiLi][^1] (and replacing $e_r$ by $\xi^{4_*}$, $c_n$ by $c$, $N_{n,1}=p$ by $b$, $N_{n-1,1}=q$ by $a$, $N_{n,2}=q^*$ by $\hat{a}$ and $-N_{n-1,2}=p^*$ by $\hat{b}$), we get $$S_n(d)=(-2)^n (\sqrt{r} \epsilon(r))^n \sqrt{c} \epsilon(c)
\left(\frac{\frac{b}{c}}{\frac{r}{c}}\right)\left(\frac{a}{c}\right)
(-1)^{\frac{r-1}{2}\frac{c-1}{2}}
\cdot\sum_{\pm}\chi^{\pm}(d)
\xi^{-ca4_*b'_*
\left(\frac{d\mp \hat{a}}{c}\right)^2
\pm 2_*\hat{b}(d\mp \hat{a}) +4_*\hat{a} \hat{b}}$$ where $\chi^{\pm}(d)=\pm 1$ if $c\mid d\mp \hat{a}$ and is zero otherwise. Further $\epsilon(x)=1$ if $x \equiv 1 \pmod 4$ and $ \epsilon(x)=I$ if $x \equiv 3 \pmod 4$. This implies the second claim of the lemma.

Note that when $c=1$, both $\chi^\pm(d)$ are nonzero. If $c>1$ and $c \mid (d-\hat{a})$, $\chi^+(d)=1$, but $\chi^- (d)=0$. Indeed, for $c$ dividing $d-\hat{a}$, $c \mid (d+\hat{a})$ if and only if $c\mid \hat{a}$ which is impossible, because $c\mid b$ but $(b,\hat{a})=1$.

Inserting the last formula into the Definition we get $$\tau_{M(b,a;d)}(\xi)=
\frac{S_n(d)}{\xi^{2_*}-\xi^{-2_*}}
\left(-2 \xi^{-3\cdot4_*} \sum_{j=1}^{r} \xi^{4_*j^2}\right)^{-n}
\xi^{-4_* \sum_{i=1}^{n}m_i}$$ where we used that $\sigma_+=n$ and $\sigma_-=0$ (compare [@KM-Dedekind p. 243]). From $\sum_{j=1}^{r}\xi^{4_*j^2}=\epsilon(r)\sqrt{r}$, we obtain $$\tau_{M(b,a;d)}(\xi)=(-1)^{\frac{(c-1)(r-1)}{4}}\epsilon(c)
\left(\frac{b'}{r'}\right)
\left(\frac{a}{c}\right)
\sqrt{c}
\; \frac{(1-\xi^{- db'_*})^{\chi(c)}}{\xi^{2_*}-\xi^{-2_*}}
\; \xi^{4_*(3n-\sum_i m_i)
-4_*\hat{b}(\hat{a}-2d)
-4_*b'_*\frac{a(d-\hat{a})^2}{c}}\,.$$ Applying the following formulas for the Dedekind sum (compare [@KM-Dedekind Theorem 1.12]) \[Dedekind\] 3n- \_i m\_i=-12 s(a,b)+, -3+b=12 s(1,b)- and dividing the formula for $\tau_{M(b,a;d)}(\xi)$ by the formula for $\tau_{L(b,1)}(\xi)$ we get $$\tau'_{M(b,a;d)}(\xi)=
\left(\frac{a}{c}\right)
\left(\frac{1-\xi^{-db'_*}}{1-\xi^{-b'_*}}\right)^{\chi(c)}
\xi^{4_*u-4_*b'_*\frac{a(d-\hat{a})^2}{c}}$$ where $$u=
-12s(a,b)+12s(1,b)+
\frac{1}{b}\left(a+\hat{a}
-2
-\hat{b}b(\hat{a}-2d)\right).$$ Notice, that $u\in\Z$. Further observe, that by using $a\hat{a}+b\hat{b}=1$, we get $$a+\hat{a}-2-\hat{b}b(\hat{a}-2d)=
2(d-1)+a(1-d^2)+a(\hat{a}-d)^2.$$ This implies the result for $0<a<b$.

To compute $\tau_{M(-b,a;d)}(\xi)$, observe that $\tau_{M(b,-a;d)}=\tau_{M(-b,a;d)}$ is equal to the complex conjugate of $\tau_{M(b,a;d)}$. The ratio $$\tau'_{M(-b,a;d)}(\xi)=\frac{\;\;\overline{\tau_{M(b,a;d)}(\xi)}\;\;}
{\tau_{L(b,1)}(\xi)}$$ can be computed analogously. Using $\overline{\epsilon(c)}=(-1)^{\frac{c-1}{2}}\epsilon(c)$, we have for $a,b>0$ $$\tau'_{M(-b,a,d)}(\xi) =(-1)^{\frac{c+1}{2}}
\left(\frac{a}{c}\right)
\left(\frac{1- \xi^{db'_*}}{1-\xi^{-b'_*}}\right)^{\chi(c)}
\xi^{4_*\tilde{u}+4_*b'_*\frac{a(d-\hat{a})^2}{c}}$$ where $$\tilde{u}=12s(a,b)+12s(1,b)+\frac{1}{b}
\left(
-a-\hat{a}-2+\hat{b}b(\hat{a}-2d)
\right)$$ Using $s(a,b)=s(a,-b)=-s(-a,b)$, we get the result.

For $b>0$, we have $$\tau'_{L(-b,1)}(\xi)
=
(-1)^{\frac{c+1}{2}-{\chi(c)}}\,
\xi^{2_*(b-3)+b_*\chi(c)}
\; .$$

Proof of Lemma \[0079\]
-----------------------

Assume $b=\pm p^l$ and $p$ is prime. We have to define the unified invariant of $M^\ve(b,a) := M(b,a;d(\ve))$, where $d(0)=1$ and $d({{\bar 0}})$ is the smallest odd positive integer such that ${\operatorname{sn}}(a)ad({{\bar 0}}) \equiv 1 \pmod {b}$. First observe that such $d({{\bar 0}})$ always exists. Indeed, if $p$ is odd, we can achieve this by adding $b$, otherwise the inverse of any odd number modulo $2^l$ is odd.

Recall that we denote the unique positive $b$th root of $q$ in $S_{p,0}$ by $q^{\frac{1}{b}}$. We define the unified invariant $I_{M^\ve(b,a)}\in \R_b$ as follows. If $p\neq 2$, then $I_{M^\ve(b,a)}\in \calS_p$ is defined by specifying its projections $$\pi _j I_{M^\ve(b,a)}
:=
\,
\begin{cases}
q^{3 s(1,b)-3{\operatorname{sn}}(b)\, s(a,b)}
&\text{if } j=0, \;\ve =0
\\&\\
(-1)^{\frac{p^j+1}{2}\,\frac{{\operatorname{sn}}(ab)-1}{2}}
\left(\frac{|a|}{p}\right)^j \,
q^{\frac{u'}{4}}
&\text{if } 0<j<l, \;\ve={{\bar 0}}\\&\\
(-1)^{\frac{p^l+1}{2}\,\frac{{\operatorname{sn}}(ab)-1}{2}}
\left(\frac{|a|}{p}\right)^l q^{\frac{u'}{4}}
&\text{if } j\geq l, \;\ve={{\bar 0}}\end{cases}$$ where $u':=u-\frac{a(\hat{a}-{\operatorname{sn}}(a)d({{\bar 0}}))^2}{b}$ and $u$ is defined in Proposition \[1409\]. If $p=2$, then only $\pi_0 I_{M(b,a)}\in \calS_{2,0}=\R_2$ is non–zero and it is defined to be $q^{3s(1,b)-3 {\operatorname{sn}}(b)\; s(a,b)}$.

The $I_{M^{\varepsilon}(b,a)}$ is well–defined due to Lemma \[0622\] below, i.e. all powers of $q$ in $I_{M^{\ve}(b,a)}$ are integers for $j>0$ or lie in $\frac{1}{b}\Z$ for $j=0$. Further, for b odd (respectively even) $I_{M^\ve(b,a)}$ is invertible in $\calS_{p}^{p,\varepsilon}$ (respectively $\R_p^{p,\varepsilon}$) since $q$ and $q^{\frac{1}{b}}$ are invertible in these rings.

In particular, for odd $b=p^l$, we have $I_{L(b,1)}=1$, and $$\pi _j I_{L(-b,1)}
=
\,
\begin{cases}
q^{\frac{b-3}{2}+\frac{1}{b}}
&\text{if } j=0
\\&\\
(-1)^{\frac{p^j+1}{2}}q^{\frac{b-3}{2}}
&\text{if } 0<j<l, \; p \text{ odd} \\&\\
(-1)^{\frac{p^l+1}{2}}q^{\frac{b-3}{2}}
&\text{if } j\geq l, \; p \text{ odd} \, .
\end{cases}$$

It is left to show, that for any $\xi$ of order $r$ coprime with $p$, we have $$\ev_\xi ( I_{M^0(b,a)})=\tau'_{M^0(b,a)}(\xi)\,$$ and if $r=p^j k$ with $j>0$, then $$\ev_\xi ( I_{M^{{{\bar 0}}}(b,a)})=\tau'_{M^{{\bar 0}}(b,a)}(\xi)\,.$$ For $\ve=0$, this follows directly from Propositions \[eval\_z\] and \[1409\] with $c=d=1$. For $\ve={{\bar 0}}$, we have $c=(p^j,b)>1$ and we get the claim by using Proposition \[1409\] and $$\label{0623}
\xi^{\frac{a(\hat{a}-{\operatorname{sn}}(a)d({{\bar 0}}))^2}{b}}=
\xi^{c\,\frac{a(\hat{a}-{\operatorname{sn}}(a)d({{\bar 0}}))^2}{bc}}=
\xi^{bb'_*\,\frac{a(\hat{a}-{\operatorname{sn}}(a)d({{\bar 0}}))^2}{bc}}=
\xi^{b'_*\,\frac{a(\hat{a}-{\operatorname{sn}}(a)d({{\bar 0}}))^2}{c}},$$ where for the second equality we use $c\equiv bb'_*\pmod{r}$. Notice that due to part (2) of Lemma \[0622\] below, $b$ and $c$ divide $\hat{a}-{\operatorname{sn}}(a)d({{\bar 0}})$ and therefore all powers of $\xi$ in (\[0623\]) are integers.

The following Lemma is used in the proof of Lemma \[0079\].

\[0622\] We have

-   $ 3 s(1,b) - 3{\operatorname{sn}}(b)\, s(a,b) \in \frac{1}{b}\Z$,

-   $b\mid \hat{a}-{\operatorname{sn}}(a)d({{\bar 0}})$ and therefore $u'\in\Z$, and

-   $4\mid u'$ for $d=d({{\bar 0}})$.

The first claim follows from the formulas (\[Dedekind\]) for the Dedekind sum. The second claim follows from the fact that $(a,b)=1$ and $$a(\hat{a}-{\operatorname{sn}}(a)d)=1-{\operatorname{sn}}(a)ad-b\hat{b}\equiv 0\pmod{b},$$ since $d$ is chosen such that ${\operatorname{sn}}(a)ad\equiv 1\pmod{b}$. For the third claim, notice that for odd $d$ we have $$4\mid (1-d^2) \;\;\text{ and }\;\;
4\mid 2({\operatorname{sn}}(a)d-{\operatorname{sn}}(b)).$$

Laplace transform {#laplace}
=================

This section is devoted to the proof of Theorem \[0078\] by using Andrew’s identity. Throughout this section, let $p$ be a prime or $p=1$, and $b= \pm p^l$ for an $l\in \N$.

Definition
----------

The Laplace transform is a $\Z[q^{\pm 1}]$–linear map defined by $$\begin{aligned}
\cL_{b}: \Z[z^{\pm 1}, q^{\pm 1}] &\to& \calS_p \\
z^a &\mapsto& z_{b,a}.\end{aligned}$$ In particular, we put $\cL_{b;j}:=\pi_j \circ \cL_{b}$ and have $\cL_{b;j}(z^a)=z_{b,a;j} \in \calS_{p,j}$.

Further, for any $f \in \BZ[z^{\pm 1},q^{\pm 1}]$ and $n\in \Z$, we define $$\hat f:=f|_{z=q^n}\in \Z[q^{\pm n}, q^{\pm 1}]\,.$$

Suppose $f \in \BZ[z^{\pm 1},q^{\pm 1}]$. Then for a root of unity $\xi$ of odd order $r$, $${\sum_{n}}^\xi q^{b\frac{n^2-1}{4}}  \hat f = \gamma_{b}(\xi) \,
\ev_\xi(\cL_{-b}(f)).$$ \[1001\]

It is sufficient to consider the case $f=z^a$. Then, by the same arguments as in the proof of [@BL Lemma 1.3], we have \[four\] [\_[n]{}]{}\^q\^[b]{} q\^[ na]{} =

0 &\
(\^c)\^[-[a\_1\^2 b’\_\*]{}]{} \_b() & .

The result follows now from Proposition \[eval\_z\].

Proof of Theorem \[0078\]
-------------------------

Recall that $$A(n,k) = \frac{\prod^{k}_{i=0}
\left(q^{n}+q^{-n}-q^i -q^{-i}\right)}{(1-q) \, (q^{k+1};q)_{k+1}}.$$ We have to show that there exists an element $Q_{b,k} \in \R_b $ such that for every root of unity $\xi$ of odd order $r$ one has $$\frac{{\sum_n}^{\hspace{-1.8mm}\xi } \; q^{b\frac{n^2-1}{4}} A(n,k) }{F_{U^b}(\xi)}
= \ev_\xi (Q_{b,k}).$$ Applying Lemma \[1001\] to $F_{U^b}(\xi)={\sum\limits_ n}^\xi q^{b\frac{n^2-1}{4}}
[n]^2$, we get for $c=(b,r)$ \[1122\] F\_[U\^b]{}()=2\_b() \_() ,where as usual, $\chi(c)=1$ if $c=1$ and is zero otherwise. We will prove that for an odd prime $p$ and any number $j\geq 0$ there exists an element $Q_k(q,x_b,j) \in \calS_{p,j}$ such that \[imp\] \_[b;j]{}( \_[i=0]{}\^k (z+z\^[-1]{} - q\^i -q\^[-i]{}) ) = 2 Q\_k(q\^[(b)]{}, x\_[b]{},j). If $p=2$ we will prove the claim for $j=0$ only, since $\calS_{2,0}\simeq\cR_{2}$. The case $p=\pm 1$ was done e.g. in [@BBL]. Theorem \[0078\] follows then from Lemma \[1001\] and where $Q_{b,k}$ is defined by its projections $$\pi_j Q_{b,k}:=\;
\frac{1-q^{-1}}{(1-x_{-b})^{\chi(p^j)}}\; Q_k(q^{-{\operatorname{sn}}(b)},x_{-b},j).$$ We split the proof of (\[imp\]) into two parts. In the first part we will show that there exists an element $Q_{k}(q,x_b,j)$ such that Equality holds. In the second part we show that $Q_k(q,x_b,j)$ lies in $\calS_{p,j}$.

Part 1, $b$ odd case {#part-1-b-odd-case .unnumbered}
--------------------

Assume $b=\pm p^l$ with $p\not=2$. We split the proof into several lemmas.

\[S\_[b;j]{}(k,q)\] For $x_{b;j}:=\pi_j(x_b)$ and $c=(b,p^j)$, $$\cL_{b;j}\left( \prod_{i=0}^k (z+z^{-1} - q^i -q^{-i})   \right)=
2\, (-1)^{k+1} \, {\text{$\left[\begin{array}{c}2k+1\\ k\end{array}
\right]$}} \, S_{b;j}(k,q),$$ where $$\label{unif-s}
S_{b;j}(k,q):=1+\sum_{n=1}^{\infty}\frac{q^{(k+1)cn}(q^{-k-1};q)_{cn}}
{(q^{k+2};q)_{cn}}
(1+q^{cn}) x_{b;j}^{n^2}.$$

Observe that for $n> \frac{k+1}{c}$ the term $(q^{-k-1};q)_{cn}$ is zero and therefore the sum in is finite.

Since $\cL_{b}$ is invariant under $z \to z^{-1}$ one has $$\cL_b\left(\prod_{i=0}^k (z+z^{-1} - q^i -q^{-i})\right) =
-2\cL_{b} (z^{-k}(zq^{-k};q)_{2k+1}),$$ and the $q$–binomial theorem (e.g. see [@GR], II.3) gives $$\label{qbinomial}
z^{-k}(zq^{-k};q)_{2k+1}=
(-1)^k \sum_{i=-k}^{k+1}(-1)^i{\text{$\left[\begin{array}{c}2k+1\\ k+i\end{array}
\right]$}}z^i.$$ Notice that $\cL_{b;j}(z^a)\not=0$ if and only if $c \mid a$. Applying $\cL_{b;j}$ to the RHS of (\[qbinomial\]), only the terms with $c \mid i$ survive and therefore $$\cL_{b;j}\left(z^{-k}(zq^{-k};q)_{2k+1}\right)=
(-1)^{k} \sum_{n=-\lfloor k/c \rfloor}^{\lfloor (k+1)/c \rfloor} (-1)^{cn}
{\text{$\left[\begin{array}{c}2k+1\\ k+cn\end{array}
\right]$}} z_{b,cn;j}.$$ Separating the case $n=0$ and combining positive and negative $n$ this is equal to $$(-1)^{k}{\text{$\left[\begin{array}{c}2k+1\\ k\end{array}
\right]$}}
+(-1)^k \sum_{n=1}^{\lfloor (k+1)/c \rfloor} (-1)^{cn}\left({\text{$\left[\begin{array}{c}2k+1\\ k+cn\end{array}
\right]$}}+{\text{$\left[\begin{array}{c}2k+1\\ k-cn\end{array}
\right]$}}\right) z_{b,cn;j},$$ where we use the convention that ${\text{$\left[\begin{array}{c}x\\ -1\end{array}
\right]$}}$ is put to be zero for positive $x$. Further, $${\text{$\left[\begin{array}{c}2k+1\\ k+cn\end{array}
\right]$}} +{\text{$\left[\begin{array}{c}2k+1\\ k-cn\end{array}
\right]$}} =\frac{\{k+1\}}{\{2k+2\}}{\text{$\left[\begin{array}{c}2k+2\\ k+cn+1\end{array}
\right]$}}(q^{cn/2}+q^{-cn/2})$$ and $$\frac{\{k+1\}}{\{2k+2\}}{\text{$\left[\begin{array}{c}2k+2\\ k+cn+1\end{array}
\right]$}}{\text{$\left[\begin{array}{c}2k+1\\ k\end{array}
\right]$}}^{-1}=
(-1)^{cn}q^{(k+1)cn+\frac{cn}{2}}\frac{(q^{-k-1};q)_{cn}}{(q^{k+2};q)_{cn}}
.$$ Using $z_{b,cn;j}=(z_{b,c;j})^{n^2}=x_{b;j}^{n^2}$ we get the result.

To define $Q_k(q,x_b,j)$ we will need Andrew’s identity (3.43) of [@And]: $$\begin{aligned}
&&\hspace{-6mm}\sum_{n\geq 0} (-1)^n\alpha_n t^{-\frac{n(n-1)}{2}+sn+Nn} \frac{(t^{-N})_n}{(t^{N+1})_n}
\prod_{i=1}^{s}\frac{(b_i)_n (c_i)_n}{b_i^nc_i^n(\frac{t}{b_i})_n(\frac{t}{c_i})_n}
=\\
&&
\hspace{-4mm}
\frac{(t)_N(\frac{q}{b_sc_s})_N}{(\frac{t}{b_s})_N(\frac{t}{c_s})_N}
\sum_{n_s\geq \cdots\geq n_2\geq n_1 \geq 0}
\beta_{n_1}
\frac{t^{n_s}(t^{-N})_{n_s}(b_s)_{n_s}(c_s)_{n_s}}{(t^{-N} b_s c_s)_{n_s}}
\prod_{i=1}^{s-1}
\frac{t^{n_i}}{b_i^{n_{i}} c_i^{n_{i}}}
\frac{(b_i)_{n_{i}}(c_i)_{n_{i}}}{(\frac{t}{b_i})_{n_{i+1}}(\frac{t}{c_i})_{n_{i+1}}}
\frac{(\frac{t}{b_i c_i})_{n_{i+1}-n_i}}{(t)_{n_{i+1}-n_i}}\;.\end{aligned}$$ Here and in what follows we use the notation $(a)_n:=(a;t)_n$ . The special Bailey pair $(\alpha_n,\beta_n)$ is chosen as follows $$\begin{array}{rclcrcl}
\alpha_0&=&1, &&\alpha_n&=&(-1)^nt^{\frac{n(n-1)}{2}}(1+t^n)\\
\beta_0&=&1, &&\beta_n&=&0 \;\;\;\text{ for } n\geq 1.
\end{array}$$

\[LHS\] $S_{b;j}(k,q)$ is equal to the LHS of Andrew’s identity with the parameters fixed below.

Since $$S_{b;j}(k,q)=S_{-b;j}(k,q^{-1}),$$ it is enough to look at the case when $b>0$. Define $b':=\frac{b}{c}$ and let $\omega$ be a $b'$th primitive root of unity. For simplicity, put $N:=k+1$ and $t:=x_{b;j}$. Using the following identities $$\begin{aligned}
(q^{y};q)_{cn}&=&\prod_{l=0}^{c-1}(q^{y+l};q^c)_n\\
(q^{yc};q^c)_n&=&\prod_{i=0}^{b'-1}(\omega^i t^y;t)_n,\end{aligned}$$ where the later is true due to $t^{b'}=x_{b;j}^{b'}=q^c$ for all $j$, and choosing a $c$th root of $t$ denoted by $t^{\frac{1}{c}}$ we can see that $$S_{b;j}(k,q)=1+\sum_{n=1}^{\infty}
\prod_{i=0}^{b'-1}\prod_{l=0}^{c-1}
\frac{
(\omega^i t^{\frac{-N+l}{c}})_{n}
}{
(\omega^i t^{\frac{N+1+l}{c}})_{n}
}
(1+t^{b'n})t^{n^2+b'Nn}.$$

Now we choose the parameters for Andrew’s identity as follows. We put $a:=\frac{c-1}{2}$, $d:=\frac{b'-1}{2}$ and $m:=\lfloor \frac{N}{c}\rfloor$. For $l\in\{1,\ldots,c-1\}$ there exist unique $u_l,v_l\in\{0,\ldots,c-1\}$ such that $u_l\equiv N+l\pmod{c}$ and $v_l\equiv N-l\pmod{c}$. Note that $v_l=u_{c-l}$. We define $U_l:=\frac{-N+u_l}{c}$ and $V_l:=\frac{-N+v_l}{c}$. Then $U_l, V_l \in \frac{1}{c}\Z$ but $U_l+V_l \in \Z$. We define $$\begin{array}{rclrcll}
b_l&:=&t^{U_l}, &c_l&:=&t^{V_l} & \text{for }\; l=1,\ldots, a, \\
b_{a+i}&:=&\omega^it^{-m},
&c_{a+i}&:=&\omega^{-i}t^{-m}
& \text{for }\;i=1,\ldots,d, \\
b_{a+ld+i}&:=&\omega^it^{U_l},
&c_{a+ld+i}&:=&\omega^{-i}t^{V_l}
& \text{for }\;i=1,\ldots,d \text{ and } l=1,\ldots, c-1, \\
b_{g+i}&:=&-\omega^{i}t, &c_{g+i}&:=&-\omega^{-i}t &\text{for }\;i=1,\ldots,d,\\
b_{s-1}&:=&t^{-m}, &c_{s-1}&:=&t^{N+1}, &\\
b_s&\rightarrow&\infty,&c_s&\rightarrow&\infty.&\\
\end{array}$$ where $g=a+cd$ and $s=(c+1)\frac{b'}{2}+1$.

We now calculate the LHS of Andrew’s identity. Using the notation $$(\omega^{\pm 1} t^x)_n=(\omega t^x)_n(\omega^{-1} t^x)_n$$ and the identities $$\lim_{c\to \infty} \frac{(c)_n}{c^n}=(-1)^n t^{\frac{n(n-1)}{2}} \;\;\text{ and }\;\; \lim_{c\to\infty} \left(\frac{t}{c}\right)_n=1$$ we get $$\begin{aligned}
LHS&=&1+
\sum_{n\geq 1}
t^{n(n-1+s+N-y)}\;
(1+t^n)\frac{(t^{-N})_n}{(t^{N+1})_n}
\cdot
\prod_{l=1}^{a}
    \frac{
        (t^{U_l})_n (t^{V_l})_n
    }{
        (t^{1-U_l})_n (t^{1-V_l})_n
    }   
\cdot
\prod_{i=1}^{d}
    \frac{
        (\omega^{\pm i} t^{-m})_n
    }{
        (\omega^{\pm i} t^{1+m})_n
    }
\\  && \hspace{3cm}
\cdot\prod_{i=1}^{d}\prod_{l=1}^{c-1}
    \frac{
        (\omega^i t^{U_l})_n
        (\omega^{-i}t^{V_l})_n
    }{
    (\omega^{-i} t^{1-U_l})_n
    (\omega^{i}t^{1-V_l})_n 
    }
\cdot
\prod_{i=1}^{d}
    \frac{
        (-\omega^{\pm i}t)_n
    }{
        (-\omega^{\pm i})_n
    }
\cdot
\frac{
    (t^{-m})_n (t^{N+1})_n
}{
    (t^{1+m})_n (t^{-N})_n
}\end{aligned}$$ where $$y:=
\sum_{l=1}^{a} (U_l+V_l)+\sum_{i=1}^{d}\sum_{l=1}^{c-1}(U_l+V_l)-m(2d+1)+2d+1+N.$$ Since $\sum_{l=1}^{c-1}(U_l+V_l)=2\sum_{l=1}^{a}(U_l+V_l)=2(-N+m+\frac{c-1}{2})$ and $2d+1=b'$, we have $$n-1+s+N-y=n+Nb'.$$ Further, $$\prod_{i=1}^{d} \frac{(-\omega^{\pm i}t)_n}{(-\omega^{\pm i})_n}
=
\prod_{i=1}^{b'-1}\frac{1+\omega^i t^n}{1+\omega^i}=\frac{1+t^{b'n}}{1+t^n}$$ and $$\begin{aligned}
&&
\prod_{l=1}^{a}
    \frac{
        (t^{U_l})_n (t^{V_l})_n
    }{
        (t^{1-U_l})_n (t^{1-V_l})_n
    }   
\cdot
\prod_{i=1}^{d}
    \frac{
        (\omega^{\pm i} t^{-m})_n
    }{
        (\omega^{\pm i} t^{1+m})_n
    }
\cdot
\prod_{i=1}^{d}\prod_{l=1}^{c-1}
    \frac{
        (\omega^i t^{U_l})_n
        (\omega^{-i}t^{V_l})_n
    }{
    (\omega^{-i} t^{1-U_l})_n
    (\omega^{i}t^{1-V_l})_n 
    }
\cdot
\frac{
    (t^{-m})_n
}{
    (t^{1+m})_n
}
\\
&&\hspace{105mm}=
\prod_{i=0}^{b'-1}\prod_{l=0}^{c-1}
\frac{
(\omega^i t^{\frac{-N+l}{c}})_{n}
}{
(\omega^i t^{\frac{N+1+l}{c}})_{n}
}.\end{aligned}$$ 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 $\delta_j:=n_{j+1}-n_j$. Then the RHS is given by $$\begin{aligned}
RHS
&=&
(t)_N\sum_{n_s\geq \cdots\geq n_2 \geq n_1=0}
\frac{
t^{x}\cdot (t^{-N})_{n_s}(b_s)_{n_s}(c_s)_{n_s}
}{
\prod_{i=1}^{s-1} (t)_{\delta_{i}}(t^{-N}b_sc_s)_{n_s}
}
\cdot
\frac{
(t^{-m})_{n_{s-1}}(t^{N+1})_{n_{s-1}}
(t^{m-N})_{\delta_{s-1}}
}{
(t^{m+1})_{n_s}(t^{-N})_{n_s}
}
\\
&&\hspace{7mm}
\cdot
\prod_{l=1}^{a}
\frac{
(t^{U_l})_{n_l}(t^{V_l})_{n_l}
(t^{1-U_l-V_l})_{\delta_{l}}
}{
(t^{1-U_l})_{n_{l+1}}(t^{1-V_l})_{n_{l+1}}
}
\cdot
\prod_{i=1}^{d}
\frac{
(\omega^{\pm i}t^{-m})_{n_{a+i}}
(t^{2m+1})_{\delta_{a+i}}
}{
(\omega^{\pm i}t^{m+1})_{n_{a+i+1}}
}
\frac{
(-\omega^{\pm i}t)_{n_{g+i}}
(t^{-1})_{\delta_{g+i}}
}{
(-\omega^{\pm i})_{n_{g+i+1}}
}
\\
&&\hspace{7mm}
\cdot
\prod_{i=1}^{d}\prod_{l=1}^{c-1}
\frac{
(\omega^it^{U_l})_{n_{a+ld+i}}
(\omega^{-i}t^{V_l})_{n_{a+ld+i}}
(t^{1-U_l-V_l})_{\delta_{a+ld+i}}
}{
(\omega^{-i}t^{1-U_l})_{n_{a+ld+i+1}}
(\omega^{i}t^{1-V_l})_{n_{a+ld+i+1}}
}\end{aligned}$$ where $$\begin{aligned}
x
&=&\sum_{l=1}^{a}(1-U_l-V_l)\,n_l
+\sum_{i=1}^{d}(2m+1)\,n_{a+i}
\\&&\hspace{2cm}
+\sum_{i=1}^{d}\sum_{l=1}^{c-1}(1-U_l-V_l)\, n_{a+ld+i}
-\sum_{i=1}^{d}n_{g+i}+(m-N)\,n_{s-1}+n_s.\end{aligned}$$ 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 $$\lim_{b_s,c_s\to\infty}\frac{
(b_s)_{n_s}(c_s)_{n_s}
}{
(t^{-N}b_sc_s)_{n_s}
}
=
(-1)^{n_s}t^{\frac{n_s(n_s-1)}{2}}t^{Nn_s}.$$ The term $(t^{-1})_{\delta_{g+i}}$ is zero unless $\delta_{g+i}\in \{0,1\}$. Therefore, we get $$\prod_{i=1}^{d}\frac{
(-\omega^{\pm i}t)_{n_{g+i}}
}{
(-\omega^{\pm i})_{n_{g+i+1}}
}
=
\prod_{i=1}^{d}
(1+\omega^{\pm i}t^{n_{g+i}})^{1-\delta_{g+i}}.$$ Due to the term $(t^{-m})_{n_s}$, we have $n_s\leq m$ and therefore $n_i\leq m$ for all $i$. Multiplying the numerator and denominator of each term of the RHS by $$\begin{aligned}
&&
\prod_{l=1}^{a} (t^{1-U_l+n_{l+1}})_{m-n_{l+1}}(t^{1-V_l+n_{l+1}})_{m-n_{l+1}}
\prod_{i=1}^{d} (\omega^{\pm i}t^{m+1+n_{a+i+1}})_{m-n_{a+i+1}}
\\
&&\hspace{35mm}\cdot\prod_{i=1}^{d}\prod_{l=1}^{c-1}
(\omega^{-i}t^{1-U_l+n_{a+ld+i+1}})_{m-n_{a+ld+i+1}}
(\omega^{i}t^{1-V_l+n_{a+ld+i+1}})_{m-n_{a+ld+i+1}}\end{aligned}$$ gives in the denominator $\prod_{i=0}^{b'-1}\prod_{l=1}^{c-1}(\omega^it^{1-U_l})_m
\cdot
\prod_{i=1}^{b'-1}(\omega^it^{m+1})_m$. This is equal to $$\prod_{l=1}^{c-1}(t^{b'(1-U_l)};t^{b'})_m
\cdot \frac{(t^{b'(m+1)};t^{b'})_m}{(t^{m+1};t)_m}
=
\frac{(q^{N+1};q)_{cm}}{(t^{m+1};t)_m}.$$ Further, $$(t)_N(t^{N+1})_{n_{s-1}}=(t)_{N+n_{s-1}}=(t)_{m}(t^{m+1})_{N-m+n_{s-1}}.$$ The term $(t^{-N+m})_{\delta_{s-1}}$ is zero unless $\delta_{s-1}\leq N-m$ and therefore $$\frac{(t^{m+1})_{N-m+n_{s-1}}}{(t^{m+1})_{n_s}}
=
(t^{m+1+n_s})_{N-m-\delta_{s-1}}.$$ Taking the above calculations into account, we get $$\label{RHS}
RHS=
\frac{(t;t)_{2m}}{(q^{N+1};q)_{cm}}
\cdot T_k(q,t)$$ where $$\begin{aligned}
T_k(q,t)&:=&
\sum_{n_s\geq \cdots\geq n_2 \geq n_1=0}
(-1)^{n_s}t^{x'}\cdot
(t^{-m})_{n_{s-1}}\cdot
(t^{m+1+n_s})_{N-m-\delta_{s-1}}
\cdot\frac{
(t^{-N+m})_{\delta_{s-1}}
}{
\prod_{i=1}^{s-1} (t)_{\delta_i}
}
\\
&&\hspace{3mm}
\cdot
\prod_{l=1}^{a}(t^{1-U_l-V_l})_{\delta_l}
\cdot
\prod_{i=1}^{d} (t^{2m+1})_{\delta_{a+i}}(t^{-1})_{\delta_{g+i}}
\cdot
\prod_{i=1}^{d}\prod_{l=1}^{c-1}(t^{1-U_l-V_l})_{\delta_{a+ld+i}}
\\
&&\hspace{3mm}
\cdot\prod_{l=1}^{a}
(t^{U_l})_{n_l}(t^{V_l})_{n_l}
(t^{1-U_l+n_{l+1}})_{m-n_{l+1}}
(t^{1-V_l+n_{l+1}})_{m-n_{l+1}}
\cdot
\prod_{i=1}^{d}
(1+\omega^{\pm i}t^{n_{g+i}})^{1-\delta_{g+i}}
\\
&&\hspace{3mm}
\cdot
\prod_{i=1}^{d}
(\omega^{\pm i}t^{-m})_{n_{a+i}}
(\omega^{\pm i}t^{m+1+n_{a+i+1}})_{m-n_{a+i+1}}
\cdot
\prod_{i=1}^{d}
\prod_{l=1}^{c-1}
(\omega^{i}t^{U_l})_{n_{a+ld+i}}
(\omega^{-i}t^{V_l})_{n_{a+ld+i}}
\\
&&\hspace{3mm}
\cdot
\prod_{i=1}^{d}
\prod_{l=1}^{c-1}
(\omega^{-i}t^{1-U_l+n_{a+ld+i+1}})_{m-n_{a+ld+i+1}}
(\omega^{i}t^{1-V_l+n_{a+ld+i+1}})_{m-n_{a+ld+i+1}}\end{aligned}$$ and $x':=x+\frac{n_s(n_s-1)}{2}+Nn_s$.

We now define the element $Q_{k}(q,x_b,j)$ by $$Q_{k}(q,x_b,j):=
\left((-1)^{k+1}q^{-\frac{k(k+1)}{2}}\right)
^{\frac{1+{\operatorname{sn}}(b)}{2}}
\left(q^{(k+1)^2}\right)^{\frac{1-{\operatorname{sn}}(b)}{2}}
\frac{(x_{b;j};x_{b;j})_{2m}}{(q;q)_{N+cm}}
\;T_k(q,x_{b;j}).$$

By Lemmas \[S\_[b;j]{}(k,q)\] and \[LHS\], Equation (\[RHS\]) and the following Lemma \[bPosbNeg\], we see that this element satisfies Equation (\[imp\]).

\[bPosbNeg\] The following formula holds. $$(-1)^{k+1}{\text{$\left[\begin{array}{c}2k+1\\ k\end{array}
\right]$}}(q^{k+1};q)_{k+1}^{-1}=
(-1)^{k+1}\frac{q^{-k(k+1)/2}}{(q;q)_{k+1}}
=
\frac{q^{-(k+1)^2}}{(q^{-1};q^{-1})_{k+1}}$$

This is an easy calculation using $$(q^{k+1};q)_{k+1}=(-1)^{k+1} q^{(3k^2+5k+2)/4}\frac{\{2k+1\}!}{\{k\}!}.$$

Part 1, $b$ even case. {#part-1-b-even-case. .unnumbered}
----------------------

Let $b=\pm 2^l$. We have to prove Equality (\[imp\]) only for $j=0$, i.e. we have to show $$\frac{1}{(q^{k+1};q)_{k+1}}
\,\cL_{b;0}\left( \prod_{i=0}^k (z+z^{-1} - q^i -q^{-i})   \right)
=
2\, Q_k(q^{{\operatorname{sn}}(b)}, x_{b},0).$$ The calculation works similar to the odd case. Note that we have $c=1$ here. This case was already done in [@BL] and [@Le]. 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 $t:=x_{b;0}$, $d:=\frac{b}{2}-1$, $\omega$ a $b$th root of unity and choose a primitive square root $\nu$ of $\omega$. Define the parameters of Andrew’s identity by $$\begin{array}{rclrcll}
b_{i}&:=&\omega^it^{-N},
&c_{i}&:=&\omega^{-i}t^{-N}
&  \;\text{for }\;i=1,\ldots,d, \\
b_{d+i}&:=&-\nu^{2i-1}t, &c_{d+i}&:=&-\nu^{-(2i-1)}t &\;\text{for }\;i=1,\ldots,d+1,\\
b_{b}&:=&-t^{-N},
&c_{b}&:=&-t^0=-1, &\\
b_{s-1}&:=&t^{-N}, &c_{s-1}&:=&t^{N+1}, &\\
b_s&\rightarrow&\infty,&c_s&\rightarrow&\infty,&
\end{array}$$ where $s=b+2$. Now we can define the element $$Q_{k}(q,x_b,0):=
\left((-1)^{k+1}q^{-\frac{k(k+1)}{2}}\right)
^{\frac{1+{\operatorname{sn}}(b)}{2}}
\left(q^{(k+1)^2}\right)^{\frac{1-{\operatorname{sn}}(b)}{2}}
\frac{(x_{b;0};x_{b;0})_{2N}}{(q;q)_{2N}}
\frac{1}{(-x_{b;0};x_{b;0})_N} T_k(q,x_{b;0})$$ where $$\begin{aligned}
T_k(q,t)
&:=&
\sum_{n_{s-1}\geq \cdots \geq n_1=0}
(-1)^{n_{s-1}}t^{x''}
\cdot
\frac{
\prod_{i=1}^{d}
(t^{2N+1})_{\delta_{i}}
\cdot\prod_{i=1}^{d+1}(t^{-1})_{\delta_{d+i}}
\cdot
(t^{N+1})_{\delta_b}
}{
\prod_{i=1}^{s-2}(t)_{\delta_i}
}
\\
&&\hspace{13mm}
\cdot
(t^{-N})_{n_{s-1}}
\cdot (-t^{N+1+n_{s-1}})_{N-n_{s-1}}
\cdot(-t^{-N})_{n_{b}}
\cdot (-t)_{n_{b}-1}
\cdot (-t^{n_{s-1}+1})_{N-n_{s-1}}
\\
&&\hspace{13mm}
\cdot
\prod_{i=1}^{d}
(\omega^{\pm i}t^{-N})_{n_{i}}
(\omega^{\pm i}t^{N+1+n_{i+1}})_{N-n_{i+1}}
\cdot
\prod_{i=1}^{d+1}(1+\nu^{\pm(2i-1)}t^{n_{d+i}})^{1-\delta_{d+i}}\end{aligned}$$ and $
x'':=\sum_{i=1}^{d}(2N+1)n_i-\sum_{i=1}^{d+1}n_{d+i}+\frac{n_{s-1}(n_{s-1}-1)}{2}+(N+1)(n_b+n_{s-1})$. We use the notation $(a;b)_{-1}=\frac{1}{1-ab^{-1}}$.

Part 2 {#part-2 .unnumbered}
------

We have to show that $Q_{k}(q,x_b,j) \in\calS_{p,j}$, where $j \in \N\cup\{0\}$ if $p$ is odd, and $j=0$ for $p=2$. The following two lemmas do the proof.

For $t=x_{b;j}$, $$T_k(q,t)\in \Z[q^{\pm 1}, t^{\pm1}].$$

Let us first look at the case $b$ odd and positive. Since for $a\not=0$, $(t^a)_n$ 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 $T_k(q,t)\in\Z[t^{\pm 1/c},\omega]$. Since \[Sbj\] S\_[b;j]{}(k,q)= T\_k(q,t), there are $f_0,g_0\in\Z[q^{\pm 1},t^{\pm 1}]$ such that $T_k(q,t)= \frac{f_0}{g_0}$. This implies that $T_k(q,t)\in \Z[q^{\pm 1}, t^{\pm 1}]$ since $f_0$ and $g_0$ do not depend on $\omega$ and the $c$th root of $t$.

The proofs for the even and the negative case work similar.

For $t=x_{b;j}$, $$\frac{(t;t)_{2m}}{(q;q)_{N+cm}}
\frac{1}{((-t;t)_{N})^{\lambda}}
\in\calS_{p,j}$$ where $\lambda=1$ and $j=0$ if $p=2$, and $\lambda=0$ and $j\in \N\cup \{0\}$ otherwise.

Notice that $$(q;q)_{N+cm}=\widetilde{(q;q)}_{N+cm}(q^c;q^c)_{2m},$$ where we use the notation $$\widetilde{(q^a;q)}_{n}:=
\prod_{\substack{j=0\\c\nmid (a+j)}}^{n-1}(1-q^{a+j}).$$ We have to show that $$\frac{(q^c;q^c)_{2m}}{(t;t)_{2m}}
\cdot
\widetilde{(q;q)}_{N+cm}\cdot
((-t;t)_{N})^{\lambda}$$ is invertible in $\Z[1/p][q]$ modulo any ideal $(f)=(\prod_{n} \Phi^{k_n}_n (q))$ where $n$ runs through a subset of $p^j\N_p$. Recall that in a commutative ring $A$, an element $a$ is invertible in $A/(d)$ if and only if $(a)+(d)=(1)$. If $(a)+(d)=(1)$ and $(a)+(e)=(1)$, multiplying together we get $(a)+(de)=(1)$. Hence, it is enough to consider $f=\Phi_{p^j n}(q)$ with $(n,p)=1$. For any $X\in\N$, we have $$\begin{aligned}
\label{denomin1}
{\widetilde{(q;q)}_X}&=&\prod_{\text{\scriptsize{$\begin{array}{c}i=1 \\ c\nmid i \end{array}$}}}^{X} \prod_{d\mid i} \Phi_d(q),
\\\label{denomin2}
(-t;t)_X&=&\frac{(t^2;t^2)_X}{(t;t)_X}=\prod_{i=1}^{X}\prod_{d\mid i}\Phi_{2d}(t)
\\\label{denomin3}
\frac{(q^c;q^c)_X}{(t;t)_X}
&=&\frac{(t^{b'};t^{b'})_X}{(t;t)_X}=
\frac{
\prod_{i=1}^{X}\prod_{d\mid ib'}\Phi_d(t)
}{
\prod_{i=1}^{X}\prod_{d\mid i}\Phi_d(t)
}\end{aligned}$$ for $b'=b/c$. Recall that $(\Phi_r(q),\Phi_a(q))=(1)$ in $\Z[1/p][q]$ if either $r/a$ is not a power of a prime or a power of $p$. For $r=p^j n$ odd and $a$ such that $c\nmid a$, one of the conditions is always satisfied. Hence is invertible in $\calS_{p,j}$. If $b=c$ or $b'=1$, and do not contribute. For $c<b$, notice that $q$ is a $cn$th primitive root of unity in $\Z[1/p][q]/(\Phi_{cn}(q)) = \Z[1/p][e_{cn}]$. Therefore $t^{b'}=q^{c}$ is an $n$th primitive root of unity. Since $(n,b')=1$, $t$ must be a primitive $n$th root of unity in $\Z[1/p][e_{cn}]$, too, and hence $\Phi_n(t) = 0$ in that ring. Since for $j$ with $(j, p) > 1$, $(\Phi_j(t),\Phi_n(t)) = (1)$ in $\Z[1/p][t]$, we have $\Phi_j(t)$ is invertible in $\Z[1/p][e_{cn}]$, and therefore and are invertible, too.

Proof of Theorem \[GeneralizedHabiro\]
======================================

The appendix is devoted to the proof of Theorem \[GeneralizedHabiro\], a generalization of the deep integrality result of Habiro, namely Theorem 8.2 of [@Ha]. The existence of this generalization and some ideas of the proof were kindly communicated to us by Habiro.

Reduction to a result on values of the colored Jones polynomial {#reduction-to-a-result-on-values-of-the-colored-jones-polynomial .unnumbered}
---------------------------------------------------------------

We will use the notations of [@Ha]. We put $q=e^{h}$, and $v=e^{h/2}$, where $h$ is a free parameter. The quantum algebra $U_h=U_h(sl_2)$, generated by $E, F$ and $H$, subject to some relations, is the quantum deformation of the universal enveloping algebra $U(sl_2)$.

Let $V_n$ be the unique $(n+1)$–dimensional irreducible $U_h$–module. In [@Ha], Habiro defined a new basis $\tilde{P}'_k$, $k=0,1,2,\dots$, for the Grothendieck ring of finite–dimensional $U_h(sl_2)$–modules with $$\tilde{P}_k':=\frac{v^{\frac{1}{2}k(1-k)}}{\{k\}!} \, \prod_{i=0}^{k-1}
(V_1-v^{2i+1}-v^{-2i-1}).$$

Put $\tilde{P}'_{\bk}=\{\tilde{P}'_{k_1},\ldots,\tilde{P}'_{k_m}\}$. It follows from Lemma 6.1 of [@Ha] that we will have identity of Theorem \[GeneralizedHabiro\] if we put

$$C_{L\sqcup L'}(\bk,\bj)=
J_{L\sqcup L'}\,(\tilde{P}'_{\bk},\bj)\,
\prod_{i}(-1)^{k_i}q^{k^2_i+k_i+1}\, .$$

Hence to prove Theorem \[GeneralizedHabiro\] it is enough to show the following.

\[main-integrality\] Suppose $L\sqcup L'$ is a colored framed link in $S^3$ such that $L$ has zero linking matrix and $L'$ has odd colors. Then for $k=\max\{k_1,\dots, k_m\}$ we have $$J_{L\sqcup L'}(\tilde{P}'_{\bk},\bj) \in \frac{(q^{k+1};q)_{k+1}}{1-q}
\,\,\BZ[q^{\pm 1}].$$

In the case $L'=\emptyset$, this statement was proved in [@Ha 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.

Sketch of the proof of Habiro’s integrality theorem
---------------------------------------------------

Geometric part {#geometric-part .unnumbered}
--------------

Let us first recall the notion of a bottom tangle, introduced by Habiro in [@Ha-b].

An $n$–component bottom tangle $T=T_1\sqcup \dots \sqcup
T_n$ is a framed tangle consisting of $n$ arcs $T_1,\ldots ,T_n$ in a cube such that all the endpoints of $T$ are on a line at the bottom square of the cube, and for each $i=1,\ldots ,n$ the component $T_i$ runs from the $2i$th endpoint on the bottom to the $(2i-1)$st endpoint on the bottom, where the endpoints are counted from the left. An example, the Borromean bottom tangle $B$, is given in Figure \[borro\].

In [@Ha-b], Habiro defined a braided subcategory $\modB$ of the category of framed, oriented tangles which acts on the bottom tangles by composition (vertical pasting). The objects of $\modB $ are the symbols $\modb ^{\otimes n}$, $n\ge 0$, where $\modb:=\downarrow \uparrow $. For $m,n\ge 0$, a morphism $X$ of $\modB$ from $\modb ^{\otimes m}$ to $\modb ^{\otimes n}$ is the isotopy class of a framed, oriented tangle $X$ which we can compose with $m$–component bottom tangles to get $n$–component bottom tangles. Let $\modB (m,n)$ be the set of morphisms from $\modb ^{\otimes m}$ to $\modb ^{\otimes n}$. The composite $YX$ of two morphisms is the gluing of $Y$ to the bottom of $X$, and the identity morphism $1_{\modb ^{\otimes m}}=\downarrow \uparrow \cdots\downarrow \uparrow $ is a tangle consisting of $2m$ vertical arcs. The monoidal structure is given by pasting tangles side by side. The braiding for the generating object $\modb$ with itself is given by $\psi _{\modb ,\modb }=\incl{1.5em}{phibb+2}$.

Corollary 9.13 in [@Ha-b] states the following.

\[thmASL\][*(Habiro)*]{} If the linking matrix of a bottom tangle $T$ is zero then $T$ can be presented as $T=W B^{\otimes k}$, where $k\geq 0$ and $W\in \modB(3k,n)$ is obtained by horizontal and vertical pasting of finitely many copies of $1_\modb$, $\psi_{\modb,\modb}$, $\psi^{-1}_{\modb,\modb}$, and

$$\def\sss{1.5em}
    \eta _\modb =\incl{\sss}{bot0},\quad
    \def\sss{2em}
    \mu _\modb =\incl{\sss}{bomu},\quad
    \gamma_+=\incl{2.5em}{gamma+},\quad
    \gamma_-=\incl{2.5em}{gamma-}.
    $$

Algebraic part {#algebraic-part .unnumbered}
--------------

Let $K=v^{H}=e^\frac{hH}{2}$. Habiro introduced the integral version ${\mathcal{U}}_q$, which is the $\Z[q, q^{- 1}]$–subalgebra of $U_h$ freely spanned over $\Z[q, q^{- 1}]$ by $\tilde{F}^{(i)}K^{j}e^k$ for $i,k\geq 0, j\in \Z$, where $$\tilde{F}^{(n)}=\frac{F^nK^n}{v^{\frac{n(n-1)}{2}}[n]!}\;\;\;\;\text{ and
 }\;\;\;\; e=(v-v^{-1})E.$$

There is $\Z/2\Z$–grading, ${\mathcal{U}}_q={\mathcal{U}}^0_q\oplus {\mathcal{U}}^1_q$, where ${\mathcal{U}}^{0}_q $ (resp. ${\mathcal{U}}_q^{1}$) is spanned by $\tilde{F}^{(i)}K^{2j}e^k$ (resp. $\tilde{F}^{(i)}K^{2j+1}e^k$). We call this the $\ve$–grading, and ${\mathcal{U}}^{0}_q $ (resp. ${\mathcal{U}}_q^{1}$) the even (resp. odd) part.

The two–sided ideal $\F_p$ in ${\mathcal{U}}_q$ generated by $e^p$ induces a filtration on $({\mathcal{U}}_q)^{\otimes n}$, $n\geq 1$, by $$\F_p(({\mathcal{U}}_q^{})^{\otimes n})=
\sum^n_{i=1} ({\mathcal{U}}_q^{})^{\otimes i-1}\otimes
\F_p({\mathcal{U}}_q^{})\otimes ({\mathcal{U}}_q^{})^{\otimes n-i}
\subset ({\mathcal{U}}_q^{})^{\otimes n}\, .$$ Let $(\tilde{{\mathcal{U}}}_q)^{\tilde\otimes n}$ be the image of the homomorphism $$\lim_{\overleftarrow{\hspace{1mm}{p\geq 0}\hspace{1mm}}} \;\;
\frac{({\mathcal{U}}_q)^{\otimes n}}{\F_p(({\mathcal{U}}_q)^{\otimes n})} \to
U_h^{\hat \otimes n}$$ where $\hat\otimes$ is the $h$–adically completed tensor product. By using $\F_p({\mathcal{U}}_q^\ve):= \F_p({\mathcal{U}}_q)\cap {\mathcal{U}}^{\ve}_q$ one defines $(\tilde{{\mathcal{U}}}^{\ve}_q)^{\tilde\otimes n}$ for $\ve\in\{0,1\}$ in a similar fashion.

By definition (Section 4.2 of [@Ha]), the universal $sl_2$ invariant $J_T$ of an $n$–component bottom tangle $T$ is an element of $U_h^{\hat\otimes n}$. Theorem 4.1 in [@Ha] states that, in fact, for any bottom tangle $T$ with zero linking matrix, $J_T$ is even, i.e. \[even\] J\_T(\^[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 of $\tilde {\mathcal{U}}^0_q$, Habiro showed that $$J_K=\sum_{n\geq 0}  (-1)^n q^{n(n+1)}\frac{(1-q)}{(q^{n+1};q)_{n+1}}\,  J_K(\tilde P'_n)\, \sigma_n$$ where $$\sigma_n=\prod^n_{i=0}(C^2-(q^i+2+ q^{-i}))\, \quad {\rm with}\quad
C=(v-v^{-1}){\tilde{F}}^{(1)}K^{-1}e+vK+v^{-1}K^{-1}\, ,$$ the quantum Casimir operator. The $\sigma_n$ provide a basis for the even part of the center. From this, Habiro deduced that $J_K(\tilde P'_n)\in \frac{  (q^{n+1};q)_{n+1}}{(1-q)} \Z[q,q^{-1}]$.

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 [@Ha]) and the fact that $J_T$ is invariant under the adjoint action.

Algebro–geometric part {#algebrogeometric-part .unnumbered}
----------------------

The proof of uses Proposition \[thmASL\], which allows to build any bottom tangle $T$ with zero linking matrix from simple parts, i.e. $T=W(B^{\otimes k})$.

On the other hand, the construction of the universal invariant $J_T$ extends to the braided functor $J:\modB\to \operatorname{ Mod}_{U_h}$ from $\modB$ to the category of $U_h$–modules. This means that $J_{W(B^{\otimes k})}=J_W(J_{B^{\otimes k}})$. Therefore, in order to show , we need to check that $J_B \in (\tilde {\mathcal{U}}^0_q)^{\tilde \otimes 3} $, and then verify that $J_W$ maps the even part to itself. The first check can be done by a direct computation [@Ha Section 4.3]. The last verification is the content of Corollary 3.2 in [@Ha].

Proof of Theorem \[main-integrality\] 
--------------------------------------

Generalization of Equation  {#generalization-of-equation .unnumbered}
---------------------------

To prove Theorem \[main-integrality\] we need a generalization of Equation or Theorem 4.1 in [@Ha] to tangles with closed components. To state the result let us first introduce two new gradings.

Suppose $T$ is an $n$–component bottom tangle in a cube, homeomorphic to the 3–ball $D^3$. Let ${\tilde {\mathcal S}}(D^3\setminus T)$ be the $\Z[q^{\pm 1/4}]$–module freely generated by the isotopy classes of framed unoriented colored links in $D^3\setminus T$, including the empty link. For such a link $L \subset D^3\setminus T$ with $m$–components colored by $n_1,\dots, n_m$, we define our new gradings as follows. First provide the components of $L$ with arbitrary orientations. Let $l_{ij}$ be the linking number between the $i$th component of $T$ and the $j$th component of $L$, and $p_{ij}$ be the linking number between the $i$th and the $j$th components of $L$. For $X=T\sqcup L$ we put $${\operatorname{gr}}_\ve(X):= (\ve_1,\dots, \ve_n)\in (\Z/2\Z)^n, \quad  \text{where}\quad
 \ve_i := \sum_{j} l_{ij} n'_{j} \pmod 2, \quad \text{ and}
\label{3306}$$

$${\operatorname{gr}}_q(L) := \sum_{1\le i,j\le m }p_{ij} n_i'n_j' + 2\sum_{1\le j\le m}(p_{jj}+1) n_j' \pmod 4,
\quad \text{where} \quad n_i':= n_i-1.$$ It is easy to see that the definitions do not depend on the orientation of $L$.

The meaning of ${\operatorname{gr}}_q(L)$ 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 $q^{{\operatorname{gr}}_q(L)/4}$; see [@LeDuke] for this result and its generalization to other Lie algebras.

We further extend both gradings to ${\tilde {\mathcal S}}(D^3\setminus T)$ by $${\operatorname{gr}}_\ve(q^{1/4})=0, \quad {\operatorname{gr}}_q(q^{1/4}) =1 \pmod 4.$$

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 [@Ha-b Section 7.3]). In [@Ha-b], it is proved that $J_X$ is adjoint invariant. The generalization of Theorem 4.1 of [@Ha] is the following.

\[MMM\] Suppose $X=T \sqcup L$, where $T$ is a $n$–component bottom tangle with zero linking matrix and $L$ is a framed unoriented colored link with ${\operatorname{gr}}_\ve(X)= (\ve_1,\dots,\ve_n)$. Then $$J_X \in  q^{{{\operatorname{gr}}_q(L)/4}}\, \left ( \tilde {\mathcal{U}}_q^{\ve_1}\, \tilde \otimes
\dots \tilde \otimes\; \tilde {\mathcal{U}}_q^{\ve_n}\right ).$$ \[3310\]

\[MMM1\] Suppose $L$ is colored by a tuple of odd numbers, then $$J_X\in (\tilde {\mathcal{U}}_q^{0})^{\tilde\otimes n}\, .$$

Since $J_X$ is invariant under the adjoint action, Theorem \[main-integrality\] follows from Corollary \[MMM1\] by repeating Habiro’s arguments.

Hence it remains to prove Theorem \[MMM\]. In the proof we will need a notion of a [*good morphism*]{}.

### Good morphisms {#good-morphisms .unnumbered}

Let $I_m:=1_{\modb^{\otimes m}}\in \modB(m,m)$ be the identity morphism of $\modb ^{\otimes m}$ in the cube $C$. A framed link $L$ in the complement $C \setminus I_m$ is [*good*]{} if $L$ is geometrically disjoint from all the up arrows of $\modb ^{\otimes m}$, 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 $X =WY$, then it is easy to see that ${\operatorname{gr}}_\ve(X)$ does not depend on $Y$, and we define ${\operatorname{gr}}_\ve(W):={\operatorname{gr}}_\ve(X)$. Also define ${\operatorname{gr}}_q(W):= {\operatorname{gr}}_q(L)$.

As in the case with $L=\emptyset$, the universal invariant extends to a map $J_W: {\mathcal{U}}_h^{\otimes m} \to {\mathcal{U}}_h^{\otimes m}$.

Proof of Theorem \[MMM\] {#proof-of-theorem-mmm .unnumbered}
------------------------

The strategy here is again analogous to the Habiro case: In Proposition \[TWWB\] 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 $\mu_\modb$. Since, any bottom tangle with zero linking matrix satisfies Theorem \[MMM\] and $\mu_\modb$ is the product in ${\mathcal{U}}_q$, which preserves the gradings, it remains to show that any good morphism preserves the gradings. This is done in Proposition \[MM2\] below.

\[TWWB\] Assume $X=T\sqcup L$ 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 $\mu_\modb$.

Let us first define $\tilde \gamma_\pm\in \modB(i, i+1)$ for any $i\in {\mathbb N}$ as follows. $$\def\sss{3em}
  \tilde\gamma_+  =\incl{\sss}{tildegamma+}\quad
    \def\sss{3em}
    \tilde\gamma_- =\incl{\sss}{tildegamma-}.$$

If a copy of $\mu_\modb$ is directly above $\psi^{\pm 1}_{\modb, \modb}$ or $ \gamma_\pm$, one can move $\mu_\modb$ down by isotopy and represent the result by pasting copies of $\psi^{\pm 1}_{\modb, \modb}$ and $\tilde \gamma_\pm$. It is easy to see that after the isotopy $\gamma_\pm$ gets replaced by $\tilde\gamma_\pm$ and $\psi^{\pm 1}_{\modb, \modb}$ by two copies of $\psi^{\pm 1}_{\modb, \modb}$.

Using Proposition \[thmASL\] and reordering the basic morphisms so that the $\mu$’s are at the bottom, one can see that $T$ admits the following presentation: $$T= W_2 \tilde W_1 (B^{\otimes k})$$ where $B$ is the Borromean tangle, $W_2$ is obtained by pasting copies of $\mu_\modb$ and $\tilde{W}_1$ is obtained by pasting copies of $\psi^{\pm 1}_{\modb,\modb}$, $\tilde\gamma_{\pm}$ and $\eta_\modb$.

Let $P$ be the horizontal plane separating $\tilde W_1$ from $W_2$. Let $P_+$ ($P_-$) be the upper (lower, respectively) half–space. Note that $W_0=\tilde W_1(B^{\otimes k})$ 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 $(P_+,\; l \text{ trivial arcs})$. 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 $(P_+,\; l\text{ trivial arcs})$ one can isotope $L$ in $P_+$ to the bottom end points of down arrows. We then obtain the desired presentation.

\[MM2\] For every good morphism $W$, the operator $J_{W}$ preserves gradings in the following sense. If $x\in {\mathcal{U}}_q^{\ve_1} \otimes \dots \otimes {\mathcal{U}}_q^{\ve_m}$, then $$J_{W}(x) \in q^{{\operatorname{gr}}_q(W)/4} \left( {\mathcal{U}}_q^{\ve_1'} \otimes \dots \otimes
 {\mathcal{U}}_q^{\ve_m'}\right), \quad \text{where} \quad (\ve'_1,\dots, \ve'_m) =
 (\ve_1,\dots, \ve_m) + {\operatorname{gr}}_{\ve}(W).$$

The rest of the appendix is devoted to the proof of Proposition \[MM2\].

Proof of Proposition \[MM2\] {#ProofMM2 .unnumbered}
----------------------------

We proceed as follows. Since $J_X$ is invariant under cabling and skein relations, and by Lemma \[A8\] below, both relations preserve ${\operatorname{gr}}_\ve$ and ${\operatorname{gr}}_q$, we consider the quotient of $\tilde S(D^3\setminus T)$ by these relations known as a skein module of $D^3\setminus T$. For $T=I_n$, this module has a natural algebra structure, with good morphisms forming a subalgebra. By Lemma \[A9\] (see also Figure \[Wgamma\]), the basis elements $W_\gamma$ of this subalgebra are labeled by $n$–tuples $\gamma=(\gamma_1,\dots, \gamma_n)\in (\Z/2\Z)^n$. It’s clear that if the proposition holds for $W_{\gamma_1}$ and $W_{\gamma_2}$, then it holds for $W_{\gamma_1} W_{\gamma_2}$. Hence it remains to check the claim for $W_\gamma$’s. This is done in Corollary \[corA10\] for basic good morphisms corresponding to $\gamma$ whose non–zero $\gamma_j$’s are consecutive. Finally, any $W_\gamma$ can be obtained by pasting a basic good morphism with few copies of $\psi^{\pm}_{\modb, \modb}$. Since $J_{\psi^\pm}$ preserves gradings (compare (3.15), (3.16) in [@Ha]), the claim follows from Lemmas \[A9\], \[A8\] and Proposition \[3308\] below.

### Cabling and skein relations {#cabling-and-skein-relations .unnumbered}

Let us introduce the following relations in ${\tilde {\mathcal S}}(D^3\setminus T)$.

Cabling relations:

-   Suppose $n_i=1$ for some $i$. The first cabling relation is $L=\tilde L$, where $\tilde L$ is obtained from $L$ by removing the $i$th component.

-   Suppose $n_i \ge 3$ for some $i$. The second cabling relation is $L= L'' -L'$, where $L'$ is the link $L$ with the color of the $i$th component switched to $n_i-2$, and $L''$ is obtained from $L$ by replacing the $i$th component with two of its parallels, which are colored with $n_i-1$ and $2$.

Skein relations:

-   The first skein relation is $U=q^{\frac 12}+q^{-\frac 12}$, where $U$ denotes the unknot with framing zero and color 2.

-   Let $L_R$, $L_V$ and $L_H$ be unoriented framed links with colors 2 which are identical except in a disc where they are as shown in Figure \[Skein\]. Then the second skein relation is $L_R = q^{\frac 14}L_V + q^{-\frac 14}L_H$ if the two strands in the crossing come from different components of $L_R$, and $L_R = \epsilon(q^{\frac 14}L_V - q^{-\frac 14}L_H)$ if the two strands come from the same component of $L_R$, producing a crossing of sign $\epsilon=\pm 1$ (i.e. appearing as in $L_{\epsilon}$ of Figure \[Skein\] if $L_R$ is oriented).

We denote by $S(D^3\setminus T)$ the quotient of ${\tilde {\mathcal S}}(D^3\setminus T)$ by these relations. It is known as the [*skein module*]{} of $D^3\setminus T$ (compare [@Pr], [@SikPr] and [@Bullock]). Recall that the ground ring is $\Z[q^{\pm 1/4}]$.

Using the cabling relations, we can reduce all colors of $L$ in $S(D^3\setminus T)$ to be 2. Note that the skein module $S(C\setminus I_n)$ 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 $\gamma=(\gamma_1,\dots, \gamma_n)\in (\Z/2\Z)^n$ let $W_\gamma$ be a simple closed curve encircling the end points of those downward arrows with $\gamma_i=1$. See Figure \[Wgamma\] for an example.

Similarly to the case of Kauffman bracket skein module [@Bullock], one can easily prove the following.

The algebra $\mathcal A_n$ is generated by $2^n$ curves $W_\gamma$. \[A9\]

Using linearity, we can extend the definition of $J_X$ to $X =T \sqcup L$, where $L$ is [*any element*]{} of ${\tilde {\mathcal S}}(D^3\setminus T)$. It is known that $J_X$ is invariant under the cablings and skein relations (Theorem 4.3 of [@KM]), hence $J_X$ is defined for $L \in S(D^3\setminus T)$. Moreover, we have

\[A8\] Both gradings ${\operatorname{gr}}_\ve$ and ${\operatorname{gr}}_q$ are preserved under the cabling and skein relations.

The statement is obvious for the $\ve$–grading. For the $q$–grading, notice that $${\operatorname{gr}}_q(L)=2\sum_{1\leq i<j\leq m}p_{ij}n_i' n_j' +
\sum_{1\leq j \leq m}p_{jj}n_j'^2+2\sum_{1\leq j\leq m}(p_{jj}+1)n_j',$$ and therefore ${\operatorname{gr}}_q(L'')\equiv {\operatorname{gr}}_q(L')\equiv{\operatorname{gr}}_q(L)\pmod{4}$. This takes care of the cabling relations.

Let us now assume that all colors of $L$ are equal to 2 and therefore $${\operatorname{gr}}_q(L)=2\sum_{1\leq i< j\leq m} p_{ij}+3\sum_{i=1}^{m} p_{ii}+2m.$$ 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 \[Skein\] come from the same component of $L_R$ and that the crossing is positive. Then, $L_V$ and $L_H$ have one positive self–crossing less, and $L_V$ has one link component more than $L_R$. Therefore $$\begin{aligned}
{\operatorname{gr}}_q(q^{\frac 14}L_V)&=&{\operatorname{gr}}_q(L_R)-3+2+1 \equiv {\operatorname{gr}}_q{L_R} \pmod{4}, \\
{\operatorname{gr}}_q(q^{-\frac 14}L_H)&=&{\operatorname{gr}}_q(L_R)-3-1 \equiv {\operatorname{gr}}_q{L_R} \pmod{4}.\end{aligned}$$ 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.

### Basic good morphisms {#basic-good-morphisms .unnumbered}

Let $ Z_n$ be $W_\gamma$ for $\gamma=(1,1,\dots, 1)\in (\Z/2\Z)^n$. $$\label{Zn}
    \def\sss{3em}
      Z_n  =\incl{\sss}{ZnVers2}\quad\quad
$$

One has a presentation $$J_{ Z_n} = 
\sum z^{(n)}_{i_1} \otimes \sum z^{(n)}_{i_2} \otimes \dots \otimes
\sum z^{(n)}_{i_{2n}},$$ such that $z^{(n)}_{i_{2j-1}}\, z^{(n)}_{i_{2j}}\in v\; {\mathcal{U}}^1_q$ for every $j=1,\dots, n$. \[3308\]

\[corA10\] $J_{ Z_n}$ satisfies Proposition \[MM2\].

Assume $x\in {\mathcal{U}}_q^{\ve_1} \otimes \dots \otimes {\mathcal{U}}_q^{\ve_n}$, then we have $$J_{ Z_n}(x) = 
\sum z^{(n)}_{i_1} x_1  z^{(n)}_{i_2} \otimes \dots \otimes
\sum z^{(n)}_{i_{2n-1}} x_n z^{(n)}_{i_{2n}}.$$ Hence, by Proposition \[3308\], we get $$J_{ Z_n}(x) \in q^{1/2} \left( {\mathcal{U}}_q^{\ve_1'} \otimes \dots \otimes
 {\mathcal{U}}_q^{\ve_n'}\right), \quad \text{where} \quad (\ve'_1,\dots, \ve'_m) =
 (\ve_1,\dots, \ve_n) + (1,1,\dots, 1).$$ The claim follows now from the fact that ${\operatorname{gr}}_\ve( Z_n)=(1,1,\dots,1)$ and ${\operatorname{gr}}_q(L)=2$.

Proof of Proposition \[3308\]
-----------------------------

The statement holds true for $J_{ Z_1}= C\otimes {\operatorname{id}}_\uparrow$. Now Lemma 7.4 in [@Ha-b] states that applying $\Delta$ to the $i$th component of the universal quantum invariant of a tangle is the same as duplicating the $i$th component. Using this fact we represent \[Jn\] J\_[Z\_[n+1]{}]{} = (1\_[\^[n-1]{}]{} \_) ( J\_[Z\_[n]{}]{} ),where $\Phi$ is defined as follows. For $x \in {\mathcal{U}}_q$ with $\Delta (x)=\sum x_{(1)}\otimes x_{(2)}$, we put $$\Phi(x) :=   \sum_{(x), m,n} x_{(1)} \otimes \beta_m S(\beta_n)
\otimes \alpha_n \, x_{(2)} \alpha_m$$ where the $R$–matrix is given by $R =\sum_l {\alpha_l \otimes \beta_l}$. See Figure below for a picture.

We are left with the computation of the $\ve$–grading of each component of $\Phi(x)$.

In ${\mathcal{U}}_q$, in addition to the $\ve$–grading, there is also the $K$–grading, defined by $|K|=|K^{-1}|=0, |e|=1, |F|=-1$. In general, the co–product $\Delta$ does not preserve the $\ve$–grading. However, we have the following.

Suppose $x\in {\mathcal{U}}_q$ is homogeneous in both $\ve$–grading and $K$–grading. Then we have a presentation $$\Delta(x) = \sum_{(x)} x_{(1)} \otimes x_{(2)},$$ where each $x_{(1)}$, $x_{(2)}$ are homogeneous with respect to the $\ve$–grading and $K$–grading. In addition, for $x\in {\mathcal{U}}_q^\ve$, we have $x_{(2)}\in {\mathcal{U}}_q^\ve$ and $x_{(1)} \, K^{-|x_{(2)}|}\in {\mathcal{U}}_q^\ve$. \[3304\]

If the statements hold true for $x, y \in {\mathcal{U}}_q$, then they hold true for $xy$. Therefore, it is enough to check the statements for the generators $e, {\tilde{F}}^{(1)},$ and $K$, for which they follow from explicit formulas of the co–product.

Suppose $x \in {\mathcal{U}}_q$ is homogeneous in both $\ve$–grading and $K$–grading. There is a presentation $$\Phi(x) =\sum x_{i_0}\otimes x_{i_1}\otimes  x_{i_2}$$ such that each $x_{i_j}$ is homogeneous in both $\ve$–grading and $K$–grading, and for $x \in {\mathcal{U}}^\ve_q$, $x_{i_2}$ and $ x_{i_0}\, x_{i_1}$ belong to ${\mathcal{U}}^\ve_q$. \[3303\]

We put $D=\sum D' \otimes D'':=v^{\frac 12 H\otimes H}$. Using (see e.g. [@Ha]) $$R=D\left(\sum_{n} q^{\frac 12 n(n-1)}{\tilde{F}}^{(n)}K^{-n}\otimes e^n\right),$$ we get $$\begin{aligned}
\Phi(x)&=&\sum_{(x),n,m}
q^{\frac{1}{2}\left(m(m-1)+n(n-1)\right)}
x_{(1)}\otimes
D''_2 e^m S(D''_1e^{n})\otimes
D'_1{\tilde{F}}^{(n)}K^{-n}x_{(2)}D'_2{\tilde{F}}^{(m)}K^{-m}\\
&=&
\sum_{(x),n,m}(-1)^n q^{-\frac 12 m(m+1)-n(|x_{(2)}|+1)}
x_{(1)}\otimes e^{m}e^n K^{-|x_{(2)}|}\otimes {\tilde{F}}^{(n)}x_{(2)}{\tilde{F}}^{(m)},\end{aligned}$$ where we used $({\operatorname{id}}\otimes S)D=D^{-1}$ and $D^{\pm1}(1\otimes x)=(K^{\pm|x|}\otimes x)D^{\pm 1}$ for homogeneous $x\in {\mathcal{U}}_q$ with respect to the $K$–grading. Now, the claim follows from Lemma \[3304\].

By induction on $n$ in , given that $C\in v \;{\mathcal{U}}^1_q$, Lemma \[3303\] implies Proposition \[3308\].

[\[EMSS\]]{}

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[^1]: There are misprints in Lemma 4.21: $q^*\pm n$ should be replaced by $q^*\mp n$ for $n=1,2$.
