# Cyclotomic expansions for $\mathfrak{gl}_{N}$ link invariants
via interpolation Macdonald polynomials

Anna Beliakova  Institut für Mathematik
Universität Zürich
Winterthurerstrasse 190
CH-8057 Zürich
Switzerland<EMAIL_ADDRESS>and  Eugene Gorsky  Department of Matematics
University of California, Davis
One Shields Avenue, Davis CA 95616, USA<EMAIL_ADDRESS>
###### Abstract.

In this paper we construct a new basis for the cyclotomic completion of the
center of the quantum $\mathfrak{gl}_{N}$ in terms of the interpolation
Macdonald polynomials. Then we use a result of Okounkov to provide a dual
basis with respect to the quantum Killing form (or Hopf pairing). The main
applications are: 1) cyclotomic expansions for the $\mathfrak{gl}_{N}$
Reshetikhin–Turaev link invariants and the universal $\mathfrak{gl}_{N}$ knot
invariant; 2) an explicit construction of the unified $\mathfrak{gl}_{N}$
invariants for integral homology 3-spheres using universal Kirby colors. These
results generalize those of Habiro for $\mathfrak{sl}_{2}$. In addition, we
give a simple proof of the fact that the universal $\mathfrak{gl}_{N}$
invariant of any evenly framed link and the universal $\mathfrak{sl}_{N}$
invariant of any $0$-framed algebraically split link are $\Gamma$-invariant,
where $\Gamma=Y/2Y$ with the root lattice $Y$.

##  1\. Introduction

In a series of papers [1, 2, 15] Habiro, the first author et al. defined
unified invariants of homology 3-spheres that belong to the Habiro ring and
dominate Witten–Reshetikhin–Turaev (WRT) invariants. Unified invariants
provide an important tool to study structural properties of the WRT
invariants. In [3, 5] they were used to prove integrality of the
$\mathfrak{sl}_{2}$ WRT invariants for all 3-manifolds at all roots of unity.

The theory of unified invariants for $\mathfrak{sl}_{2}$ is based on
cyclotomic expansions for the colored Jones polynomial and for the universal
knot invariant constructed as follows. Given a framed oriented link $L$ in the
3-sphere, we open its components to obtain a bottom tangle $T$, presented by a
diagram $D$ (see Figure 1). For a ribbon Hopf algebra $U_{q}\mathfrak{g}$, the
universal link invariant $J_{L}(\mathfrak{g};q)$ is obtained by spliting $D$
intro elementary pieces: crossings, caps and cups and then by associating to
them $R^{\pm 1}$-matrices, and pivotal elements, respectively.

Figure 1. An example of the clasp bottom tangle

For a knot $K$, $J_{K}(\mathfrak{g};q)$ belongs to (some completion of) the
center $\mathcal{Z}(U_{q}\mathfrak{g})$. In the easiest case
$\mathfrak{g}=\mathfrak{sl}_{2}$, the center is generated by the Casimir $C$.
For a $0$-framed knot $K$, Habiro showed that there are coefficients
$a_{m}(K)\in\mathbb{Z}[q^{\pm 1}]$ such that

(1) $\displaystyle
J_{K}(\mathfrak{sl}_{2};q)=\sum^{\infty}_{m=0}a_{m}(K)\,\sigma_{m}\;\quad\text{with}\;\quad\sigma_{m}=\prod_{i=1}^{m}\left(C^{2}-(q^{i}+q^{-i}+2)\right).$

Replacing $C^{2}$ in (1) by its value $q^{n}+q^{-n}+2$ on the $n$-dimensional
irreducible representation $V_{n-1}$, we get the $n$-colored Jones polynomial
of $K$ (normalized to 1 for the unknot)

(2)
$J_{K}(V_{n-1},q)=\sum_{m=0}^{\infty}(-1)^{m}q^{-\frac{m(m+1)}{2}}a_{m}(K)\,(q^{1+n};q)_{m}(q^{1-n};q)_{m}$

where $(a;q)_{m}=(1-a)(1-aq)\dots(1-aq^{m-1})$. Equation (2) is known as a
cyclotomic expansion of the colored Jones polynomial. Thus, Habiro’s series
(1) dominates all colored Jones polynomials of $K$. To prove the fact that
$J_{K}(\mathfrak{sl}_{2};q)$ belongs to the even part of
$\mathcal{Z}(U_{q}\mathfrak{sl}_{2})$, generated by $C^{2}$, Habiro used the
whole power of the theory of bottom tangles developed in [16].

In this paper we give a simple proof for the “evenness” of the universal
invariant of algebraically split links for all quantum groups of type $A$.
Recall that $U_{q}\mathfrak{g}$ has a natural action of a finite group
$\Gamma=Y/2Y$ where $Y$ is the root lattice of $\mathfrak{g}$. For
$\mathfrak{g}=\mathfrak{gl}_{N}$, $\Gamma=\mathbb{Z}_{2}^{N}$ and for
$\mathfrak{g}=\mathfrak{sl}_{N}$, $\Gamma=\mathbb{Z}_{2}^{N-1}$.

######  Theorem 1.1.

The universal $\mathfrak{gl}_{N}$ invariant of any evenly framed link is
$\Gamma$-invariant. The universal $\mathfrak{sl}_{N}$ invariant of any
0-framed algebraically split link is $\Gamma$-invariant.

The quantum group $U_{q}\mathfrak{gl}_{N}$ admits a finite dimensional
irreducible representation $V(\lambda)$ with highest weight $v^{\lambda}$ for
any partition $\lambda=(\lambda_{1}\geq\dots\geq\lambda_{N})$ with $N$ parts
and $v^{2}=q$. To prove Theorem 1.1 we extend the Reshetikhin-Turaev
invariants to tangles colored with representations $L(\zeta)\otimes
V(\lambda)$ where $L(\zeta)$ is a one-dimensional representation of
$U_{q}\mathfrak{gl}_{N}$ for $\zeta\in\Gamma$. Then the claim follows from the
comparison of the $\mathfrak{gl}_{N}$ Reshetikhin-Turaev link invariants
colored with $L(\zeta)\otimes V(\lambda)$ and $V(\lambda)$.

The next main result of the paper establishes an explicit basis in the
$\Gamma$-invariant part of the center $\mathcal{Z}$ of
$U_{q}\mathfrak{gl}_{N}$. It generalizes Habiro’s basis
$\\{\sigma_{m}\,|\,m\in\mathbb{N}\\}$ for the even part of
$\mathcal{Z}(U_{q}\mathfrak{sl}_{2})$.

######  Theorem 1.2.

There exists a family of central elements $\sigma_{\lambda}\in\mathcal{Z}$
labeled by partitions $\lambda$ with at most $N$ parts with the following
properties:

  * (a)

$\sigma_{\lambda}$ is $\Gamma$-invariant and annihilates $L(\zeta)\otimes
V(\mu)$ for all $\zeta\in\Gamma$ and partitions $\mu$ with at most $N$ parts
not containing $\lambda$;

  * (b)

$\sigma_{\lambda}$ does not annihilate $V(\lambda)$ and acts on it by an
explicit scalar (see Theorem 8.2).

The proof uses the theory of interpolation Macdonald polynomials developed in
[23, 24, 29, 30, 31, 32, 36]. This theory allows one to reconstruct a
symmetric function $f(x_{1},\ldots,x_{N})$ from its values at special points
$x_{i}=q^{-\mu_{i}-N+i}$ where $\mu$ is an arbitrary partition with at most
$N$ parts. The connection between the center of $U_{q}\mathfrak{gl}_{N}$ and
symmetric functions goes through the quantum Harish-Chandra isomorphism, and
we interpret $f(q^{-\mu_{1}-N+1},\dots,q^{-\mu_{N}})$ as the scalar by which
the element of the center $f$ acts on the irreducible representation
$V({\mu})$. Interpolation Macdonald polynomials then correspond to a natural
basis in the center of $U_{q}\mathfrak{gl}_{N}$.

The polynomials $\sigma_{\lambda}$ yield a basis in the $\Gamma$-invariant
parts of both the center $\mathcal{Z}$ and its completion (a function in the
completion is determined by its values on all finite-dimensional
representations). We use a formula of Okounkov [29] to give explicit expansion
of a given central element $z$ in the basis $\sigma_{\lambda}$ in terms of the
scalars by which $z$ acts on all finite-dimensional representations
$V({\lambda})$. This leads to an expansion of the universal knot invariant in
the basis $\sigma_{\lambda}$, where the coefficients are related to
Reshetikhin-Turaev invariants of the same knot colored by $V({\mu})$ via an
explicit triangular matrix $(d_{\lambda,\mu})$ which does not depend on the
knot.

######  Theorem 1.3.

For any evenly framed knot $K$, there exist Laurent polynomials
$a_{\lambda}(K)\in\mathbb{Z}[q,q^{-1}]$ such that the universal invariant of
$K$ has the following expansion:

(3)
$J_{K}(\mathfrak{gl}_{N};q)=\sum_{\lambda}a_{\lambda}(K)\,\sigma_{\lambda}\ .$

Moreover, the coefficients $a_{\lambda}(K)$ can be computed in terms of the
Reshetikhin-Turaev invariants as follows:

$a_{\lambda}(K)=\sum_{\mu\subset\lambda}{d_{\lambda,\mu}(q^{-1})}\;J_{K}(V(\mu),q)$

where the coefficients $d_{\lambda,\mu}(q)$ are defined in Theorem 10.17.

We prove Theorem 1.3 as Proposition 8.7. We would like to emphasize that the
fact that $a_{\lambda}(K)$ are Laurent polynomials in $q$ is highly
nontrivial. Indeed, we have computed the tables of coefficients
$d_{\lambda,\mu}(q)$ for $\mathfrak{gl}_{2}$ in Section 11.4 and these are
complicated rational functions, so a priori $a_{\lambda}(K)$ are rational
functions as well. Theorem 1.3 thus encodes certain divisibility properties
for the linear combinations of colored invariants of $K$. We refer to Section
11.5 for the explicit computation of the coefficients $a_{\lambda}(K)$ for the
figure eight knot.

We call (3) a cyclotomic expansion of the universal $\mathfrak{gl}_{N}$ knot
invariant. The name cyclotomic is justified by the fact that (3) has well-
defined evaluations at any root of unity by Lemma 10.29 below. Note that for
$N=2$ and a $0$-framed knot, our expansion does not coincide with that of
Habiro, simply because if an element $z\in U_{q}\mathfrak{gl}_{2}$ is central
and $\Gamma$-invariant, it does not imply $z$ has a decomposition in even
powers of the Casimir. Therefore, our cyclotomic expansion is rather a
generalization of $F_{\infty}$ in [37] or [4, eq.(3.14)], both having
interesting application in the theory of non semisimple invariants of links
and 3-manifolds.

For our next application, assume $M$ is an integral homology 3-sphere obtained
by $\mathbf{\varepsilon}$-surgery on an $\ell$-component algebraically split
$0$-framed link $L$ with $\mathbf{\varepsilon}\in\\{\pm 1\\}^{\ell}$.
Following Habiro–Le, we define an $\mathfrak{gl}_{N}$ unified invariant $I(M)$
as

$I(M)=\langle\,r^{\otimes\mathbf{\varepsilon}},J_{L}(\mathfrak{gl}_{N};q)\,\rangle\
$

where $r$ is the $\mathfrak{gl}_{N}$ ribbon element and
$\langle\cdot,\cdot\rangle$ is the Hopf pairing. In the case of
$\mathfrak{sl}_{N}$ Habiro–Le proved [19] that the unified invariant belongs
to a cyclotomic completition of the polynomial ring

$\widehat{\mathbb{Z}[q]}:={\lim\limits_{\overleftarrow{\hskip 5.69054ptn\hskip
5.69054pt}}}\;\frac{\mathbb{Z}[q]}{((q;q)_{n})}\ $

known as Habiro ring. Using interpolation, we are able to express $I(M)$ in
terms of special linear combinations of Reshetikhin–Turaev invariants of $L$,
called Kirby colors. For this we diagonalize the Hopf pairing, i.e. find a
basis $P_{\mu}$ that is orthonormal to $\sigma_{\lambda}$ and orthogonal to
$V(\lambda)$ with respect to the Hopf pairing. This allows us to give explicit
formulas for the universal Kirby colors $\omega_{\pm}$ (see (25)) in the basis
$P_{\mu}$ and to prove the following result.

######  Theorem 1.4.

The unified invariant

$I(M)=J_{L}(\omega_{\epsilon_{1}},\dots,\omega_{\epsilon_{\ell}})\;\in\;\widehat{\mathbb{Z}[q]}$

belongs to the Habiro ring and dominates $\mathfrak{gl}_{N}$ WRT invariants of
$M_{\pm}$ at all roots of unity. Moreover, $I(M)$ is equal to the
$\mathfrak{sl}_{N}$ Habiro–Le invariant of $M_{\pm}$.

To prove that $I(M)$ is equal to the $\mathfrak{sl}_{N}$ Habiro–Le invariant
we show the equality of the universal $\mathfrak{gl}_{N}$ and
$\mathfrak{sl}_{N}$ invariants for $0$-framed algebraically split links, and
the fact that the $\mathfrak{gl}_{N}$ and $\mathfrak{sl}_{N}$ twist forms
$x\mapsto\langle r^{\pm 1},x\rangle$ on them coincide. It follows that $I(M)$
belongs to the Habiro ring. Then we establish invariance of Kirby colors
$\omega_{\pm}$ under Hoste moves (a version of Fenn–Rourke moves between
algebraically split links) in Lemma 9.1, and finally deduce the equality
$I(M)=J_{L}(\omega_{\epsilon_{1}},\dots,\omega_{\epsilon_{\ell}})$.

The main advantage of Theorem 1.4 compared to Habiro–Le approach is the
interpretation of $I(M)$ as the Reshetikhin–Turaev invariant of $L$ colored by
$\omega_{\mathbf{\varepsilon}}$. This leads to various striking divisibility
results and allows us to extend our cyclotomic expansion to links.

######  Corollary 1.5.

Given an $\ell$ component algebraically split $0$-framed link $L$, then for
all but finitely many partitions $\lambda_{i}$ with $1\leq i\leq\ell$, there
exist positive integers $n=n(\lambda_{i},N)$, such that

$J_{L}(P^{\prime}_{\lambda_{1}},\dots,P^{\prime}_{\lambda_{\ell}})\in(q;q)_{n}\,\mathbb{Z}[q,q^{-1}]\
$

where $P^{\prime}_{\lambda}=v^{|\lambda|}\,\dim_{q}V(\lambda)\,P_{\lambda}$ is
a scalar multiple of $P_{\lambda}$.

This is a generalization of the famous integrability theorem in [15, Thm.
8.2]. The authors do not know any direct proof of Corollary 1.5 without using
the theory of unified invariants. Based on Corollary 9.4 we obtain a
cyclotomic expansion for the Reshetikhin-Turaev invariants of $L$:

(4)
$J_{L}(\lambda_{1},\dots,\lambda_{\ell})=v^{\sum_{i}|\lambda_{i}|}\sum_{\mu_{i}\subset\lambda_{i}}\prod^{l}_{j=1}c_{\lambda_{j},\mu_{j}}(q^{-1})\,J_{L}(P^{\prime}_{\mu_{1}},\dots,P^{\prime}_{\mu_{\ell}})$

where the matrix
$\left[c_{\lambda,\mu}(q)\right]_{\lambda,\mu}:=\left[F_{\lambda}(q^{-\mu_{i}-N+i})\right]_{\lambda,\mu}$
is the inverse of $\left[d_{\lambda,\mu}(q)\right]_{\lambda,\mu}$. This
generalizes equation $(8.2)$ in [15].

In addition, in the case of knot surgeries we give a direct proof of the fact
that

$I(M_{\pm})=J_{L}(\omega_{\pm})\;\in\;\widehat{\mathbb{Z}[v]}$

by using our cyclotomic expansion and the interpolation theory.

Finally, we would like to comment on potential ideas for categorification of
these results. The ring of symmetric polynomials in $N$ variables is naturally
categorified by the category of annular $\mathfrak{gl}_{N}$-webs, with
morphisms given by annular foams [6, 33, 34, 13, 11]. By the work of the
second author and Wedrich [13], one can interpret it as a symmetric monoidal
Karoubian category generated by one object $E$ corresponding to a single
essential circle. The symmetric polynomials are then categorified by the Schur
functors of $E$.

We expect the categorified interpolation polynomials to correspond to
interpolation Macdonald polynomials where $q$ plays the role of quantum
grading and $t$ of the homological grading (after some change of variables).
We recall the general definitions and properties of these polynomials from
[29] in Appendix. The key obstacle for categorification of interpolation
polynomials is that they are not homogeneous. Therefore one needs to enrich
the category and allow additional morphisms between $E$ and identity.

On the other hand, the conjectures of the second author, Negut and Rasmussen
([12], see [10, 11] for further discussions) relate a version of the annular
category to the derived category of the Hilbert scheme of points on the plane.
The interpolation Macdonald polynomials appear in that context as well [7].

The paper is organized as follows. After recalling the definitions, we compare
the Reshetikhin–Turaev invariants of tangles colored by $V(\lambda)$ and
$L(\zeta)\otimes V(\lambda)$ in Section 4. In the next two sections we
summarize known results about the center of $U_{q}\mathfrak{gl}_{N}$, define
its completion and prove Theorem 1.1 in Section 6.2. The remaining results are
proven in Sections 8, 9 assuming some facts about interpolation.

In the last sections we develop the theory of the interpolation Macdonald
polynomials, starting from the one variable case. We define multi-variable
interpolation polynomials, state and prove their properties in Section 10.2.
Next, we solve the interpolation problem in two ways, one using the approach
of Okounkov (Theorem 10.17), and another using Hopf pairing (see (38)). We
study divisibility of $F_{\lambda}(q^{a_{1}},\ldots,q^{a_{n}})$ by quantum
factorials in Section 10.5 (see Lemma 10.29). Section 11 is focused on various
stability properties of the interpolation polynomials such as adding a column
to a partition $\lambda$ (Proposition 11.8) and changing $N$ for a fixed Young
diagram $\lambda$. In particular, in Proposition 11.5 we describe a HOMFLY-PT
analogue of the interpolation polynomials depending on an additional parameter
$A=q^{N}$. We provide lots of examples and tables of interpolation
polynomials, especially for $\mathfrak{gl}_{2}$. In Appendix A we collect some
additional known facts about the interpolation Macdonald polynomials and the
Habiro ring.

## Acknowledgments

The authors would like to thank Pavel Etingof, Kazuo Habiro, Thang Le, Matt
Hogancamp, Andrei Okounkov and Paul Wedrich for the useful discussions. We
thank Satoshi Nawata for his comments on the first version of the paper and
for bringing the reference [22] to our attention. Our work was partially
supported by the NSF grants DMS-1700814 (E.G.), DMS-1760329 (E.G.) and the
NCCR SwissMAP (A.B.).

##  2\. Notations and conventions

###  2.1. $q$-binomial formulas

Throughout the paper we will use the following notations for the $q$-series.
The $q$-Pochhammer symbols are defined as

$(a;q)_{m}=\prod_{i=0}^{m-1}(1-aq^{i}),\
(a;q)_{\infty}=\prod_{i=0}^{\infty}(1-aq^{i}),\ m\geq 0.$

It is easy to see that

$(a;q)_{m+k}=(a;q)_{m}(aq^{m};q)_{k},\
(a;q)_{m}=\frac{(a;q)_{\infty}}{(aq^{m};q)_{\infty}}.$

We will use two normalizations for $q$-binomial coefficients defined as
follows:

$\\{a\\}_{q}=1-q^{a},\ [a]_{q}=\frac{\\{a\\}_{q}}{\\{1\\}_{q}},\
[a]_{q}!=[1]_{q}\cdots[a]_{q},\
\binom{a}{b}_{q}=\frac{[a]_{q}!}{[b]_{q}![a-b]_{q}!}\ .$

Note that

$[a]_{q}=\frac{(q;q)_{a}}{(1-q)^{a}},\
\binom{a}{b}_{q}=\frac{(q;q)_{a}}{(q;q)_{b}(q;q)_{a-b}}.$

Finally, the $q$-binomial formula gives

$(a;q)_{m}=\sum_{j=0}^{m}(-1)^{j}q^{\frac{j(j-1)}{2}}\binom{m}{j}_{q}a^{j}.$

Let us also define symmetric $q$-numbers. For this we chose $v$ such that
$v^{2}=q$ and set

$\\{a\\}=v^{a}-v^{-a},\quad[a]:=\frac{\\{a\\}}{\\{1\\}},\quad\left[\begin{array}[]{cc}\\!a\\!\\\
\\!b\\!\end{array}\right]=\frac{\\{a\\}!}{\\{b\\}!\\{a-b\\}!}\ .$

We will use all these formulas throughout the paper without a reference.

###  2.2. Partitions

We will work with partitions
$\lambda=(\lambda_{1}\geq\lambda_{2}\geq\ldots\lambda_{N})$ which we will
identify with the corresponding Young diagrams in French notation, where the
rows have length $\lambda_{i}$. Transpose diagram to $\lambda$ is denoted by
$\lambda^{\prime}$, and $|\lambda|=\sum\lambda_{i}$. Given a box in a Young
diagram, we define its arm, co-arm, leg and co-leg as in Figure 2.

$l^{\prime}$$l$$a^{\prime}$$a$ Figure 2. Arm, co-arm, leg and co-leg

We define the hook length as $h(\square)=a(\square)+l(\square)+1$, and the
content $c(\square)=a^{\prime}-l^{\prime}$.

Let

$n(\lambda)=\sum(i-1)\lambda_{i}=\sum_{\square}l^{\prime}(\square)=\sum_{\square}l(\square),$

then

$n(\lambda^{\prime})=\sum\frac{\lambda_{i}(\lambda_{i}-1)}{2}=\sum_{\square}a^{\prime}(\square)=\sum_{\square}a(\square).$

The content of $\lambda$ is defined as

$c(\lambda)=\sum_{\square}c(\square)=n(\lambda^{\prime})-n(\lambda).$

Let $\bar{\lambda}_{i}=\lambda_{i}+N-i$ for $1\leq i\leq N$, then we have the
following identity

(5) $\prod_{\Box\in\lambda}(1-t^{h(\Box)})=\frac{\prod_{i\geq
1}\prod^{\bar{\lambda}_{i}}_{j=1}\left(1-t^{j}\right)}{\prod_{i<j}\left(1-t^{\bar{\lambda}_{i}-\bar{\lambda}_{j}}\right)}$

and we define

(6) $\displaystyle D_{N}(\lambda)$
$\displaystyle=\sum_{i=1}^{N}\frac{(\bar{\lambda}_{i})(\bar{\lambda}_{i}-1)}{2}=\sum_{i}\frac{\lambda_{i}(\lambda_{i}-1)}{2}+\sum_{i}(N-i)\lambda_{i}+\sum_{i=1}^{N}\binom{N-i}{2}$
(7)
$\displaystyle=n(\lambda^{\prime})+(N-1)|\lambda|-n(\lambda)+\binom{N}{3}=c(\lambda)+(N-1)|\lambda|+\binom{N}{3}.$

##  3\. Quantum groups

###  3.1. Quantum $\mathfrak{gl}_{N}$

The quantum group $\mathcal{U}=U_{q}\mathfrak{gl}_{N}$ is a
$\mathbb{C}(v)$-algebra generated by $E_{1},\ldots,E_{N-1}$,
$F_{1},\ldots,F_{N-1}$, $K^{\pm 1}_{1},\ldots,K_{N}^{\pm 1}$ satisfying the
following relations:

(8) $K_{i}E_{i}=vE_{i}K_{i},\ K_{i}F_{i}=v^{-1}F_{i}K_{i},\
K_{i+1}E_{i}=v^{-1}E_{i}K_{i+1},\ K_{i+1}F_{i}=vF_{i}K_{i+1}$ (9)
$[E_{i},F_{j}]=\delta_{ij}\frac{K_{i}K_{i+1}^{-1}-K_{i+1}K_{i}^{-1}}{v-v^{-1}},\
[K_{i},K_{j}]=0,$ (10) $E_{i}^{2}E_{j}-[2]E_{i}E_{j}E_{i}+E_{j}E_{i}^{2}=0\
\text{if}\ |i-j|=1\ \text{and}\ [E_{i},E_{j}]=0\ \text{otherwise}$

and analogously for $F_{i}$, where $v^{2}=q$. To simplify the notation we set
$\mathcal{K}_{i}:=K_{i}K^{-1}_{i+1}$. Then the Hopf algebra structure on
$\mathcal{U}$ (i.e. coproduct, antipode and counit) can be defined as follows:

$\Delta(E_{i})=E_{i}\otimes 1+\mathcal{K}_{i}\otimes E_{i},\
\Delta(F_{i})=1\otimes F_{i}+F_{i}\otimes\mathcal{K}_{i}^{-1},\
\Delta(K_{i}^{\pm 1})=K_{i}^{\pm 1}\otimes K_{i}^{\pm 1},$ $S(K_{i}^{\pm
1})=K_{i}^{\mp 1},\ S(E_{i})=-E_{i}\mathcal{K}_{i}^{-1},\
S(F_{i})=-\mathcal{K}_{i}F_{i}$ $\varepsilon(K_{i}^{\pm
1})=1,\varepsilon(E_{i})=\varepsilon(F_{i})=0.$

Usually $\mathcal{U}$ is considered as a subalgebra of $\mathcal{U}_{h}$ that
is an $h$-adically complete $\mathbb{C}[[h]]$-algebra topologically generated
by $E_{i}$, $F_{i}$ and $H_{j}$ for $1\leq i\leq N-1$ and $1\leq j\leq N$ with

$v=\exp{h/2},\quad K_{i}=v^{H_{i}}=\exp{hH_{i}/2}$

satisfying (9), (10) and

$H_{i}E_{i}-E_{i}H_{i}=E_{i},\;\;H_{i}F_{i}-F_{i}H_{i}=-F_{i},\;\;H_{i+1}E_{i}-E_{i}H_{i+1}=-E_{i},\;\;H_{i+1}F_{i}-F_{i}H_{i+1}=F_{i}$

replacing (8). Rewriting the defining relations in terms of the generators

$e_{i}={E_{i}}(v-v^{-1}),\quad
F_{i}^{(n)}=\frac{F_{i}^{n}}{[n]!}\quad\text{and}\quad
K_{j}\quad\text{for}\quad 1\leq i\leq N-1,\quad 1\leq j\leq N$

we obtain an integral version $\mathcal{U}_{\mathbb{Z}}$ as a Hopf algebra
over $\mathbb{Z}[v,v^{-1}]\subset\mathbb{C}(v)\subset\mathbb{C}[[h]]$.

The quantum group $\mathfrak{gl}_{N}$ has a fundamental representation
$\mathbb{C}^{N}$ with basis $v_{1},\ldots,v_{N}$ such that

$K_{i}v_{j}=v^{\delta_{ij}}v_{j},\ E_{i}v_{j}=\begin{cases}v_{i}&\text{if}\
j=i+1\\\ 0&\text{otherwise}\\\
\end{cases},F_{i}v_{j}=\begin{cases}v_{i+1}&\text{if}\ j=i\\\
0&\text{otherwise}.\\\ \end{cases}$

It generates a braided monoidal category with simple objects $V(\lambda)$,
where $\lambda$ is a partition with at most $N$ parts. These are highest
weight modules where $K_{i}$ act on the highest weight vector by
$v^{\lambda_{i}}$. The fundamental representation corresponds to
$\lambda=(1)$. The representations $V(\lambda)$ have integral basis where
$\mathcal{U}_{\mathbb{Z}}$ acts by $\mathbb{Z}[v,v^{-1}]$-valued matrices.

###  3.2. Ribbon structure

The Hopf algebra $\mathcal{U}_{h}$ admits a ribbon Hopf algebra structure (see
e.g. [8, Cor. 8.3.16]). The universal $R$-matrix has the form
$\mathcal{R}=D\Theta$ where the diagonal part $D$ and the quasi-$R$-matrix are
defined as follows

$D=v^{\sum^{N}_{i=1}H_{i}\otimes
H_{i}}\quad\text{and}\quad\Theta=\sum_{\mathbf{n}\in\mathbb{N}^{N-1}}F_{\mathbf{n}}\otimes
e_{\mathbf{n}}$

where for any sequence of non-negative integers
$\mathbf{n}=(n_{1},\dots,n_{N-1})$, the elements $e_{\mathbf{n}}$ and
$F_{\mathbf{n}}$ are defined by equations (66) and (67) in [19] and form
topological bases of the positive and negative parts in the triangular
decomposition of $\mathcal{U}_{\mathbb{Z}}$. The inverse matrix
$\mathcal{R}^{-1}=\iota(\Theta)D^{-1}$ is obtained by applying the involution
$\iota:v\to v^{-1}$.

The ribbon element and its inverse have the form

(11)
$r=\sum_{\mathbf{n}}F_{\mathbf{n}}\;\mathcal{K}_{\mathbf{n}}\;r_{0}\;e_{\mathbf{n}}\quad\text{and}\quad
r^{-1}=\sum_{\mathbf{n}}\iota(F_{\mathbf{n}})\;\mathcal{K}_{-\mathbf{n}}\;r^{-1}_{0}\;\iota(e_{\mathbf{n}})$

where $r_{0}=K_{-2\rho}v^{-\sum_{i}H^{2}_{i}}$ and
$K_{-2\rho}=\prod^{N}_{i=1}K_{i}^{2i-N-1}$ is the pivotal element. Here for
any sequence of integers $\mathbf{n}\in\mathbb{Z}^{N-1}$ we set
$\mathcal{K}_{\mathbf{n}}=\prod_{i}\mathcal{K}^{n_{i}}_{i}$, and denote by

$\rho=\left(\frac{N-1}{2},\frac{N-3}{2},\ldots,\frac{1-N}{2}\right)=\frac{1-N}{2}(1,\ldots,1)+(N-1,N-2,\ldots,0)$

the half sum of all positive roots. Using the central element
$K=\prod^{N}_{i=1}K_{i}$, we can write the previous definitions as follows:

$r^{-1}_{0}=K^{N}\;\prod^{N}_{i=1}K^{-2i}_{i}\;v^{\sum_{i}H_{i}(H_{i}+1)},\quad
K_{-2\rho}=K^{-N-1}\prod^{N}_{i=1}K^{2i}_{i}\ .$

The central element $r^{-1}$ acts on $V(\lambda)$ by the multiplication with

$\theta_{V(\lambda)}=v^{(\lambda,\lambda+2\rho)}=v^{N|\lambda|}q^{c(\lambda)}$

where $(\lambda,\mu)=\sum^{N}_{i=1}\lambda_{i}\mu_{i}$, $c(\lambda)$ is the
content of $\lambda$ and $v^{2}=q$.

###  3.3. Even part of $\mathcal{U}$

The algebra $\mathcal{U}$ has a natural grading by
$\Gamma=\mathbb{Z}_{2}^{N}=\\{\pm 1\\}^{N}$ where
$\zeta=(\zeta_{1},\ldots,\zeta_{N})\in\Gamma$ acts on $K_{i}$ by $\zeta_{i}$,
on $E_{i}$ by 1 and on $F_{i}$ by $\zeta_{i}\zeta_{i+1}$. It is easy to see
that the defining relations are preserved under this action. Following [19],
we call an element of $\mathcal{U}_{N}$ even or $\Gamma$-invariant if it is
preserved under the action of $\Gamma$.

Let us denote by $\mathcal{U}^{\text{ev}}_{\mathbb{Z}}$ a
$\mathbb{Z}[q,q^{-1}]$-subalgebra of $\mathcal{U}_{\mathbb{Z}}$ generated by
$e_{i}$, $F^{(n)}_{i}\mathcal{K}_{i}$ and $K^{2}_{j}$ for $1\leq i\leq N-1$
and $1\leq j\leq N$. It is easy to check that
$\mathcal{U}^{\text{ev}}_{\mathbb{Z}}$ is $\Gamma$-invariant.

The action of $\Gamma$ descends on the category $\mathit{Rep}(\mathcal{U})$ of
all finite-dimensional representations. Given
$\zeta=(\zeta_{1},\ldots,\zeta_{N})\in\Gamma$, we can define a one-dimensional
representation $L(\zeta)$ where $E_{i}$ and $F_{i}$ act by zero, and $K_{i}$
act by $\zeta_{i}$. We can also define representation $V(\lambda)\otimes
L(\zeta)$ where $K_{i}$ act on the highest weight vector by
$\zeta_{i}v^{\lambda_{i}}$.

######  Lemma 3.1.

The action of $\mathcal{U}$ on $V(\lambda)\otimes L(\zeta)$ agrees with the
$\Gamma$-twisted action of $\mathcal{U}$ on $V(\lambda)$.

###### Proof.

Indeed, $\Delta(F_{i})=1\otimes F_{i}+F_{i}\otimes\mathcal{K}_{i}^{-1}$, so
$F_{i}$ acts on $V(\lambda)\otimes L(\zeta)$ via
$F_{i}\otimes\mathcal{K}_{i}^{-1}=F_{i}\zeta_{i}\zeta_{i+1}$. Similarly,
$E_{i}$ acts on $V(\lambda)\otimes L(\zeta)$ via $E_{i}\otimes 1=E_{i}$ and
$K_{i}$ acts via $K_{i}\otimes K_{i}=K_{i}\zeta_{i}.$ ∎

###  3.4. The subalgebra $U_{q}\mathfrak{sl}_{N}$

We define $U_{q}\mathfrak{sl}_{N}$ as a subalgebra of $\mathcal{U}$ generated
by $E_{i},F_{i}$ and $\mathcal{K}^{\pm 1}_{i}:=K^{\pm 1}_{i}K^{\mp 1}_{i+1}$
for $1\leq i\leq N-1$. The Hopf algebra $U_{q}\mathfrak{sl}_{N}$ also admits
an integral version $\mathcal{U}_{\mathbb{Z}}\mathfrak{sl}_{N}$ generated by

$e_{i},\;\;F^{(n)}_{i}\quad\text{and}\quad\mathcal{K}^{\pm 1}_{i}$

over $\mathbb{Z}[q,q^{-1}]$. The braiding $\mathcal{R}=D^{\prime}\Theta$ with
$\Theta$ as for $\mathfrak{gl}_{N}$, but different diagonal part

$D^{\prime}=v^{\sum^{N-1}_{i=1}\frac{{\mathcal{H}}_{i}\otimes{\mathcal{H}}_{i}}{2}}\quad\text{where}\quad{\mathcal{H}}_{i}=H_{i}-H_{i+1}\
.$

The ribbon element is defined by (11) with
$r_{0}=K_{-2\rho}\prod^{N-1}_{i=1}v^{{-\mathcal{H}}_{i}^{2}/2}$. The pivotal
element $K_{-2\rho}$ does not change. Note that the $\Gamma$-invariant part of
$U_{q}\mathfrak{sl}_{N}$ generated by $e_{i}$, $F^{(n)}_{i}\mathcal{K}_{i}$
and $\mathcal{K}^{2}_{j}$ for $1\leq i,j\leq N-1$ has a smaller Cartan part
than its $\mathfrak{gl}_{N}$ analogue.

######  Example 3.2.

For $N=2$ the product $K_{1}K_{2}$ is central. By denoting
$\mathcal{K}=K_{1}K_{2}^{-1},E=E_{1},F=F_{1}$ we get the standard presentation
for $U_{q}(\mathfrak{sl}_{2})$:

$\mathcal{K}E=v^{2}E\mathcal{K},\;\mathcal{K}F=v^{-2}F\mathcal{K},\;[E,F]=\frac{\mathcal{K}-\mathcal{K}^{-1}}{v-v^{-1}}.$

###  3.5. Universal invariant

Lawrence, Reshetikhin, Ohtsuki and Kauffman constructed quantum group valued
universal link invariants. As it was already mentioned in the introduction,
the universal invariant of a link is defined by splitting a diagram of its
bottom tangle into elementary pieces and by associating $R$-matrices and
pivotal elements to them. For more details and references we recommend to
consult [16, Sec. 7.3]. However, we admit here the convention from [19, Sec.
2.7] and write the contributions from left to right along the orientation of
each component.

##  4\. Ribbon structure on $\mathit{Rep}(\mathcal{U})$

The aim of this section is to compare the Reshetikhin-Turaev invariants of a
bottom tangle whose components are colored with $V(\lambda)$ and
$V(\lambda)\otimes L(\zeta)$. This will be later used to prove Theorem 1.1.

Let us denote by $\mathcal{R}_{\mathbb{Q}}$ the representation ring of
$\mathit{Rep}(\mathcal{U})$ over $\mathbb{Q}(v)$. Given an $l$ component link
$L$, Reshetikhin–Turaev functor associated with Lie algebra $\mathfrak{g}$
provides a $\mathbb{Q}(v)$-multilinear map

$\displaystyle
J_{L}:\mathcal{R}_{\mathbb{Q}}\times\dots\times\mathcal{R}_{\mathbb{Q}}$
$\displaystyle\to\mathbb{Q}(v)$ $\displaystyle(\mu_{1},\dots,\mu_{l})$
$\displaystyle\mapsto\bigotimes_{i}\mathrm{Tr}^{V(\mu_{i})}_{q}\left(J_{L}(\mathfrak{g};q)\right)=:J_{L}(\mathfrak{g};\mu_{1},...,\mu_{l})$

normalized to $\prod_{i}\dim_{q}(V(\mu_{i}))$ for the $0$-framed
$(\mu_{1},\dots,\mu_{l})$-colored unlink. In cases when $\mathfrak{g}$ is
fixed in the context, we will remove it from the notation for simplicity.

Note that in the case of a knot, we have
$J_{K}(\lambda)=\dim_{q}(V(\lambda))J_{K}(V(\lambda),q)$ where the last
invariant is the colored Jones polynomial used in Introduction and normalized
to be 1 for the unknot.

The universal $R$-matrix defines a braiding between the representations
$V(\lambda)$. We can extend this braiding to $\mathit{Rep}(\mathcal{U})$ as
follows. Clearly, $L(\zeta)\otimes L(\zeta^{\prime})\simeq
L(\zeta\zeta^{\prime})$ and we define the braiding between $L(\zeta)$ and
$L(\zeta^{\prime})$ to be trivial. Let $V$ be a finite-dimensional
representation of $\mathcal{U}$ where the eigenvalues of $K_{i}$ are integral
powers of $v$. Given $\zeta\in\Gamma$ we consider a $\mathbb{C}$-linear map
$T_{V}(\zeta):V\to V$ which acts by $\prod\zeta_{i}^{a_{i}}$ on the weight
subspace of $V$ where $K_{i}$ acts as $v^{a_{i}}$.

######  Lemma 4.1.

The maps

$c_{\zeta,V}:=\mathrm{swap}\circ(\mathrm{Id}\otimes
T_{V}(\zeta)):L(\zeta)\otimes V\to V\otimes L(\zeta)$

with inverses

$c_{V,\zeta}:=\mathrm{swap}\circ(T_{V}(\zeta)\otimes\mathrm{Id}):V\otimes
L(\zeta)\to L(\zeta)\otimes V$

define a braiding on $\mathit{Rep}(\mathcal{U})$.

###### Proof.

First, let us check that $\text{swap}\circ(\mathrm{Id}\otimes T_{V}(\zeta))$
intertwines the actions of $\mathcal{U}$ on both sides. Indeed, let $v\in V$
be a vector with weight $(v^{a_{1}},\ldots,v^{a_{N}})$, then $E_{i}v$ has
weight $(v^{a_{1}},\ldots,v^{a_{i}+1},v^{a_{i+1}-1},\ldots,v^{a_{N}})$ while
$F_{i}v$ has weight
$(v^{a_{1}},\ldots,v^{a_{i}-1},v^{a_{i+1}+1},\ldots,v^{a_{N}})$.

Let $\bullet$ denote the basis vector in $L(\zeta)$, then

$c_{\zeta,V}E_{i}(\bullet\otimes
v)=c_{\zeta,V}\left(\zeta_{i}\zeta_{i+1}\bullet\otimes
E_{i}(v)\right)=\zeta_{1}^{a_{1}}\cdots\zeta_{i}^{a_{i}}\zeta_{i+1}^{a_{i+1}}\cdots\zeta^{a_{N}}_{N}E_{i}(v)\otimes\bullet,\\\
c_{\zeta,V}F_{i}(\bullet\otimes v)=c_{\zeta,V}\left(\bullet\otimes
F_{i}(v)\right)=\zeta_{1}^{a_{1}}\cdots\zeta_{i}^{a_{i}-1}\zeta_{i+1}^{a_{i+1}+1}\cdots\zeta^{a_{N}}_{N}F_{i}(v)\otimes\bullet,\\\
c_{\zeta,V}K_{i}(\bullet\otimes v)=c_{\zeta,V}\left(\zeta_{i}\bullet\otimes
K_{i}(v)\right)=\zeta_{1}^{a_{1}}\cdots\zeta_{i}^{a_{i}+1}\cdots\zeta^{a_{N}}_{N}K_{i}(v)\otimes\bullet,\\\
$

while

$E_{i}c_{\zeta,V}(\bullet\otimes
v)=E_{i}(\zeta_{1}^{a_{1}}\cdots\zeta_{N}^{a_{N}}v\otimes\bullet)=\zeta_{1}^{a_{1}}\cdots\zeta^{a_{N}}_{N}E_{i}(v)\otimes\bullet,\\\
F_{i}c_{\zeta,V}(\bullet\otimes
v)=F_{i}(\zeta_{1}^{a_{1}}\cdots\zeta_{N}^{a_{N}}v\otimes\bullet)=\zeta_{1}^{a_{1}}\cdots\zeta_{i}^{a_{i}-1}\zeta_{i+1}^{a_{i+1}+1}\cdots\zeta^{a_{N}}_{N}F_{i}(v)\otimes\bullet,\\\
K_{i}c_{\zeta,V}(\bullet\otimes
v)=K_{i}(\zeta_{1}^{a_{1}}\cdots\zeta_{N}^{a_{N}}v\otimes\bullet)=\zeta_{1}^{a_{1}}\cdots\zeta_{i}^{a_{i}+1}\cdots\zeta^{a_{N}}_{N}K_{i}(v)\otimes\bullet.\\\
$

Next, we observe that
$T_{V}(\zeta)T_{V}(\zeta^{\prime})=T_{V}(\zeta\zeta^{\prime})$ and
$T_{U\otimes V}(\zeta)=T_{U}(\zeta)\otimes T_{V}(\zeta)$, so $c_{\zeta,V}$
indeed defines a braiding. Even more concretely, we get the braiding as the
composition

(12) $c_{L(\zeta)\otimes V,L(\zeta^{\prime})\otimes U}:L(\zeta)\otimes
V\otimes L(\zeta^{\prime})\otimes
U\xrightarrow{c_{V,\zeta^{\prime}}}L(\zeta)\otimes L(\zeta^{\prime})\otimes
V\otimes U=L(\zeta^{\prime})\otimes L(\zeta)\otimes V\otimes
U\xrightarrow{c_{V,U}}\\\ L(\zeta^{\prime})\otimes L(\zeta)\otimes U\otimes
V\xrightarrow{c_{\zeta,U}}L(\zeta^{\prime})\otimes U\otimes L(\zeta)\otimes
V.$

∎

The representations $L(\zeta)$ are self-dual, and it is easy to see that the
braiding $c_{\zeta,V}$ is compatible with changing $V$ to $V^{*}$. Therefore,
$\mathit{Rep}(\mathcal{U})$ with objects $L(\zeta)\otimes V$ form a pivotal
braided monoidal category.

The quantum dimension of $L(\zeta)$ equals to the trace of the action of the
pivotal element, which is $(\prod_{i}\zeta_{i})^{N+1}$. The twist coefficient
$\theta_{L(\zeta)}$ is defined as the action of the ribbon element on
$L(\zeta)$, and is given by $(\prod_{i}\zeta_{i})^{N}$.

######  Lemma 4.2.

$\mathit{Rep}(\mathcal{U})$ is a ribbon category with twist
$\theta_{L(\zeta)\otimes V}=\theta_{L(\zeta)}\theta_{V}$.

###### Proof.

By definition $\theta_{L(\zeta)\otimes
V}=c_{\zeta,V}\theta_{L(\zeta)}\theta_{V}c_{V,\zeta}=\theta_{L(\zeta)}\theta_{V}$.
∎

###  4.1. Braiding in $\mathit{Rep}(U_{q}\mathfrak{sl}_{N})$

In this section, we study the action of $\Gamma$ and the corresponding
braiding for $U_{q}\mathfrak{sl}_{N}$, starting from $N=2$. Similarly to the
previous section, $U_{q}\mathfrak{sl}_{2}$ has a one dimensional
representation $L(-1)$ where $E$ and $F$ act by 0 and $\mathcal{K}$ acts by
$-1$. The action of $U_{q}\mathfrak{sl}_{2}$ on $L(-1)\otimes V$ is equivalent
to $\mathbb{Z}_{2}$-twisted action on $V$ where $\mathbb{Z}_{2}$ scales $E$ by
1 and $F,\mathcal{K}$ by $-1$.

One can attempt to define a braiding for $U_{q}\mathfrak{sl}_{2}$. Since $E$
and $F$ shift the weights by $2$, it is easy to see that the analogue of
$T_{V}$ should act by $(\sqrt{-1})^{a}$ on a subspace with weight $v^{a}$, and
it does not square to identity. Nevertheless, it squares to $\pm\text{id}$ on
each irreducible representation. This means that braiding relations on
$\mathit{Rep}(U_{q}\mathfrak{sl}_{2})$ hold up to sign.

To pin down this sign, we define the sign automorphism $\Sigma_{V}$ which acts
by $(-1)^{a}$ on a subspace with weight $v^{a}$. Since $E,F$ shift the weight
by $\pm v^{2}$, $\Sigma_{V}$ commutes with the action of
$U_{q}\mathfrak{sl}_{2}$ on $V$. The operator $\Sigma_{V}$ acts on the
irreducible representation $V(n)$ by a scalar $(-1)^{n}$. Also, it is easy to
see that $\Sigma_{V\oplus W}=\Sigma_{V}\oplus\Sigma_{W}$ and $\Sigma_{V\otimes
W}=\Sigma_{V}\otimes\Sigma_{W}$.

######  Lemma 4.3.

The operators $T_{V}$ and $\Sigma_{V}$ satisfy the following properties:

  * (a)

We have

$T_{V}^{2}=\Sigma_{V},\
c_{L(-1),V}=c^{-1}_{L(-1),V}(1\otimes\Sigma_{V})=(\Sigma_{V}\otimes
1)c^{-1}_{L(-1),V}$

  * (b)

Let $c_{V,W}:V\otimes W\to W\otimes V$ be the braiding, then

$c_{V,W}(\Sigma_{V}\otimes 1)=(1\otimes\Sigma_{V})c_{V,W},\
c_{V,W}(1\otimes\Sigma_{W})=(\Sigma_{W}\otimes 1)c_{V,W}$

  * (c)

We have $c_{L(-1),V\otimes W}=c_{L(-1),V}\circ c_{L(-1),W}.$

  * (d)

The braiding with $L(-1)$ satisfies Yang-Baxter equation, that is, the
following diagram commutes:

${L(-1)\otimes V\otimes W}$${V\otimes L(-1)\otimes W}$${V\otimes W\otimes
L(-1)}$${L(-1)\otimes W\otimes V}$${W\otimes L(-1)\otimes V}$${W\otimes
V\otimes
L(-1)}$$\scriptstyle{c_{L(-1),V}}$$\scriptstyle{c_{V,W}}$$\scriptstyle{c_{L(-1),W}}$$\scriptstyle{c_{V,W}}$$\scriptstyle{c_{L(-1),W}}$$\scriptstyle{c_{L(-1),V}}$

###### Proof.

Part (a) is clear. To prove (b), observe that the action of
$U_{q}\mathfrak{sl}_{2}\otimes U_{q}\mathfrak{sl}_{2}$ on $V\otimes W$
commutes with both $\Sigma_{V}\otimes 1$ and $1\otimes\Sigma_{V}$, and the
$R$-matrix is an element of the completion of $U_{q}\mathfrak{sl}_{2}\otimes
U_{q}\mathfrak{sl}_{2}$.

Given a pair of vectors $u\in V,w\in W$ such that $Ku=v^{i}u$ and $Kw=v^{j}w$,
we get $K(u\otimes w)=v^{i+j}u\otimes w$, so $T_{V\otimes W}=T_{V}\otimes
T_{W}$. Since $c_{L(-1),V}=\mathrm{swap}\circ(\mathrm{Id}\otimes T_{V})$, we
get the desired relation. Finally, (d) follows from (c). ∎

We can generalize the above results to representations of
$U_{q}\mathfrak{sl}_{N}$ as follows. For $\zeta\in\mathbb{Z}_{2}^{N-1}$ there
is a one-dimensional representation $L(\zeta)$ of $U_{q}(\mathfrak{sl}_{N})$
where $E_{i},F_{i}$ act by 0 and $\mathcal{K}_{i}=K_{i}K_{i+1}^{-1}$ act by
$\zeta_{i}$ ($1\leq i\leq N-1$). Given a representation $V$ where all weights
of $\mathcal{K}_{i}$ are integral powers of $v$, we can define an operator
$T_{\zeta,V}:V\to V$ which acts by $\zeta^{A^{-1}\mathbf{a}}$ on a subspace
where $\mathcal{K}_{i}$ acts by $v^{a_{i}}$. Here $A$ is the Cartan matrix for
$\mathfrak{sl}_{N}$ given by

(13) $A=\left(\begin{matrix}2&-1&0&\ldots&0\\\ -1&2&-1&\ldots&0\\\
0&-1&2&\ldots&0\\\ \vdots&\vdots&\vdots&\ddots&\vdots\\\ 0&0&0&\ldots&2\\\
\end{matrix}\right)$

and $\mathbf{a}=(a_{1},\ldots,a_{N-1})$. Note that $\det(A)=N$, so $A^{-1}$
has rational entries with denominator $N$ and one needs to choose an $N$-th
root of $(-1)$ to define $\zeta^{A^{-1}\mathbf{a}}$. Define
$\Sigma_{\zeta,V}=T^{2}_{\zeta,V}$.

######  Lemma 4.4.

The operators $T_{\zeta,V}$ and $\Sigma_{\zeta,V}$ satisfy the following
properties:

  * (a)

$T_{\zeta,V}E_{i}=\zeta_{i}E_{i}T_{\zeta,V},T_{\zeta,V}F_{i}=\zeta_{i}F_{i}T_{\zeta,V}$

  * (b)

$\Sigma_{\zeta,V}$ commutes with the action of $U_{q}\mathfrak{sl}_{N}$ on $V$

  * (c)

The map $c_{L(\zeta),V}=\mathrm{swap}\circ(\mathrm{Id}\otimes
T_{\zeta,V}):L(\zeta)\otimes V\to V\otimes L(\zeta)$ is a morphism of
$U_{q}\mathfrak{sl}_{N}$-representations

  * (d)

The maps $T_{\zeta,V}$ and $\Sigma_{\zeta,V}$ satisfy all equations in Lemma
4.3 with $L(-1)$ changed to $L(\zeta)$.

###### Proof.

(a) The operator $F_{i}$ changes the weight
$\mathbf{a}=(a_{1},\ldots,a_{N-1})$ by $Ae_{i}$, so if
$\mathcal{K}_{i}v=v^{a_{i}}v$ then

$T_{\zeta,V}F_{i}(v)=\zeta^{A^{-1}(\mathbf{a}+Ae_{i})}F_{i}v=\zeta^{A^{-1}\mathbf{a}+e_{i}}F_{i}v=\zeta_{i}F_{i}T_{\zeta,V}(v).$

The proof for $E_{i}$ is similar. Part (b) immediately follows from (a).

For (c), we observe that the action of $E_{i}$ on $L(\zeta)\otimes V$ is the
same as the action on $V$, while the actions of $F_{i},\mathcal{K}_{i}$ are
twisted by $\zeta_{i}$. On the other hand, the action of $F_{i}$ on $V\otimes
L(\zeta)$ is the same as the action on $V$, while the actions of
$E_{i},\mathcal{K}_{i}$ are twisted by $\zeta_{i}$. Therefore by (a) the
operator $c_{L(\zeta),V}$ intertwines the actions of $U_{q}\mathfrak{sl}_{N}$
on $L(\zeta)\otimes V$ and $V\otimes L(\zeta)$.

Finally, the proof of the rest of Lemma 4.3 extends to
$U_{q}\mathfrak{sl}_{N}$ verbatim. ∎

######  Remark 4.5.

The above construction of $T_{\zeta,V}$ and $\Sigma_{\zeta,V}$ can be extended
to an arbitrary semisimple Lie algebra with Cartan matrix $A$. The action of
$\Sigma_{\zeta,V}$ can be interpreted in terms of projection of the weight
lattice to its quotient by the root lattice.

We draw a tangle colored by a representation $V=V(\lambda)$ using solid lines,
and a tangle colored by $L(\zeta)$ by dotted lines. If a component is colored
by $L(\zeta)\otimes V$, we draw a dotted line on the left of a solid line and
parallel to it. The crossings between solid and dotted lines correspond to
$c^{\pm}_{L(\zeta),V}$ depicted in Figure 3. Note that unlike
$\mathfrak{gl}_{N}$ case, $c_{L(\zeta),V}$ does not square to identity and we
have to distinguish under- and over-crossings between solid and dotted lines.
This allows us to define Reshetikhin-Turaev invariants for framed tangles
colored by representations of $U_{q}\mathfrak{sl}_{N}$ of the form
$L(\zeta)\otimes V(\lambda)$.

Using the notations as in Figure 3, we can visualize the statement of Lemma
4.4 in Figure 4.

$\Sigma$ Figure 3. The operators $c_{L(\zeta),V}$, $c^{-1}_{L(\zeta),V}$ and
$\Sigma_{\zeta}$.

(a)$=$$\Sigma$$=$$\Sigma$$=$$\Sigma$

(b)$\Sigma$$=$$\Sigma$

(d)

Figure 4. Diagrammatics for Lemma 4.4

######  Theorem 4.6.

(a) Let $L$ be an algebraically split $0$-framed link with $\ell$ components.
Then for arbitrary partitions $\lambda_{1},\ldots,\lambda_{\ell}$ and
$\zeta_{1},\ldots,\zeta_{\ell}\in\Gamma$ the following identity of
Reshetikhin–Turaev invariants holds:

(14) $J_{L}\left({\mathfrak{sl}_{N}};V(\lambda_{1})\otimes
L(\zeta_{1}),\ldots,V(\lambda_{\ell})\otimes L(\zeta_{\ell})\right)=\\\
J_{L}\left({\mathfrak{sl}_{N}};V(\lambda_{1}),\ldots,V(\lambda_{\ell})\right)\cdot\dim_{q}L(\zeta_{1})\cdots\dim_{q}L(\zeta_{\ell}),$

where $\dim_{q}L(\zeta_{i})=\mathrm{Tr}^{L(\zeta_{i})}_{q}(1)=\pm 1$.

(b) Let $L$ be an arbitrary link with evenly framed components, if $N$ is odd.
Then (14) holds for $\mathfrak{gl}_{N}$ Reshetikhin–Turaev invariants.

###### Proof.

(a) We use the results of Lemmas 4.3 and 4.4 and the above diagrammatic
notation. By Lemma 4.3(a), we can change crossings between dotted and solid
lines at a cost of placing $\Sigma_{\zeta}$ on solid lines. By doing this
iteratively, we can make all dotted lines to be above solid lines. At this
stage, each solid component of $L$ acquires several copies of $\Sigma_{\zeta}$
and $\Sigma_{\zeta}^{-1}$ at various places of the link diagram. The number of
these copies (with signs) equals the linking number between this component and
the dotted part which is even by our assumption. By Lemma 4.3(b) we can
combine all these copies of $\Sigma_{\zeta}$ together and cancel out. Finally,
using Lemma 4.3(d), we can separate the dotted and solid links. By changing
the crossings in the dotted link, we transform it to the $0$-framed unlink.
Therefore the invariant of the solid link equals
$J_{L}\left({\mathfrak{sl}_{N}};V(\lambda_{1}),\ldots,V(\lambda_{\ell})\right)$
while the invariant of the dotted link equals
$\dim_{q}L(\zeta_{1})\cdots\dim_{q}L(\zeta_{\ell})$.

The proof of (b) is similar, except that $\Sigma_{V}$ is trivial for all $V$.
As before we can unknot dotted components. Now the ribbon element acts on
$L(\zeta)$ by $\theta_{L(\zeta)}=(\prod_{i}\zeta_{i})^{N}$, and hence, any
(even if $N$ is odd) number of them acts by $1$. The result follows.

∎

##  5\. Center of $\mathcal{U}$

Let $\mathcal{Z}$ be the center of $\mathcal{U}_{\mathbb{Z}}$. In this section
we recall the main facts known about $\mathcal{Z}$.

###  5.1. Harish-Chandra isomorphism

Let $(\mathcal{U}^{0}_{\mathbb{Z}})^{S_{N}}:=\mathbb{Z}[v,v^{-1}][{K^{\pm
1}_{1}},\ldots,K_{N}^{\pm 1}]^{S_{N}}$ be the Cartan part of
$\mathcal{U}_{\mathbb{Z}}$ invariant under the Weyl group action. After a
multiplication by an appropriate power of the central element
$K:=\prod^{N}_{i=1}K_{i}$, each element of $(\mathcal{U}^{0})^{S_{N}}$ can be
viewed as a symmetric function in $N$ variables. This allows to identify
$(\mathcal{U}^{0}_{\mathbb{Z}})^{S_{N}}$ with the ring of symmetric functions
divided by powers of the elementary symmetric polynomial $e_{N}=K$. In the
classical case, this ring can be identified with the center using the Harish-
Chandra isomorphism. After quantization, the image of the Harish-Chandra
homomorphism belongs to

$\mathit{Sym}=\mathbb{Z}[v^{\pm 1},e_{N}^{-1}][x_{1},\dots,x_{N}]^{S_{N}}$

where $x_{i}=K_{i}^{2}$ (compare e.g. [20, Ch. 6]). In this section we will
furthermore identify $\mathit{Sym}$ with the Grothendieck ring $\mathcal{R}$
of $\mathit{Rep}(\mathcal{U})$ with coefficients $\mathbb{Z}[v^{\pm 1}]$.

First, the character map

$\mathit{ch}:\mathcal{R}\to\mathit{Sym}$

sends a representation $U$ to its character $\mathit{ch}(U)$. Clearly,
$\mathit{ch}(U\oplus V)=\mathit{ch}(U)+\mathit{ch}(V)$ and
$\mathit{ch}(U\otimes V)=\mathit{ch}(U)\mathit{ch}(V)$, so $\mathit{ch}$ is a
ring homomorphism. The character of $V(\lambda)$ equals the Schur function
$s_{\lambda}(x_{1},\dots,x_{N})$, while the character of $L(\zeta)$ equals
$\zeta_{1}\cdots\zeta_{N}$.

The Harish-Chandra map

$\mathit{hc}:\mathcal{Z}\to\mathit{Sym}$

is defined as follows. Let $\phi$ be a central element in $\mathcal{U}$, it
acts in the Verma module $\Delta(\lambda)$ by some scalar
$\phi|_{\Delta(\lambda)}$. We define $\mathit{hc}(\phi)$ to be the polynomial
in $\mathit{Sym}$ defined by the condition

$\mathit{hc}(\phi)(q^{\rho+\lambda})=\phi|_{\Delta(\lambda)}\quad\text{for
all}\quad\lambda$

where $\rho=\left(\frac{N-1}{2},\frac{N-3}{2},\ldots,\frac{1-N}{2}\right)$.
Note that the product $\phi\phi^{\prime}$ acts on $\Delta(\lambda)$ by the
product of the corresponding scalars, so $\mathit{hc}$ is also a ring
homomorphism. It is known to be an isomorphism (see e.g. [20, Ch. 6]).

Finally, the map $\xi:\mathcal{R}\to\mathcal{Z}$ is defined by
$\xi=\mathit{hc}^{-1}\circ\mathit{ch}$. It is a composition of two ring
homomorphisms and hence a ring homomorphism too. Hence, we get the commutative
diagram:

${\mathcal{R}}$${\mathcal{Z}}$${\mathit{Sym}}$$\scriptstyle{\mathit{ch}}$$\scriptstyle{\xi}$$\scriptstyle{\mathit{hc}}$

In Lemma 5.3 we will show that $\xi$ actually coincides with the Drinfeld map.

######  Example 5.1.

The central element $K=K_{1}\cdots K_{N}$ acts on $V(\lambda)$ by a scalar
$v^{\sum\lambda_{i}}$. Since $\sum\rho_{i}=0$, we get $\mathit{hc}(K_{1}\cdots
K_{N})=y_{1}\cdots y_{N}$.

######  Example 5.2.

The center of $U_{q}\mathfrak{sl}_{2}$ is generated by the Casimir element:

$C=(v-v^{-1})^{2}FE+v\mathcal{K}+v^{-1}\mathcal{K}^{-1}$

It acts on a representation $V_{m}$ by $v^{m+1}+v^{-m-1}$, so
$\mathit{hc}(C)=y+y^{-1}$ (note that $v^{\rho}=v$ in this case). On the other
hand, $\mathit{ch}(V_{1})=y+y^{-1}$, so $\xi(V_{1})=C$, where $V_{1}$ the
2-dimensional representation.

Similarly, we can consider the corresponding central element in
$U_{q}\mathfrak{gl}_{2}$ defined by

$C_{\mathfrak{gl}_{2}}=(v-v^{-1})^{2}FE+vK_{1}K_{2}^{-1}+v^{-1}K_{1}^{-1}K_{2}.$

It acts on a representation $V(\lambda)$ by a scalar

$v^{1+\lambda_{1}-\lambda_{2}}+v^{-1-\lambda_{1}+\lambda_{2}}=\frac{y_{1}}{y_{2}}+\frac{y_{2}}{y_{1}},\
\quad y_{1}=v^{1/2+\lambda_{1}},y_{2}=v^{-1/2+\lambda_{2}},$

so
$\mathit{hc}(C_{\mathfrak{gl}_{2}})=\frac{y_{1}}{y_{2}}+\frac{y_{2}}{y_{1}}=\frac{y^{2}_{1}+y^{2}_{2}}{y_{1}y_{2}}=e^{-1}_{2}(y_{1},y_{2})(x_{1}+x_{2})$.

###  5.2. Hopf pairing

The Hopf pairing $\langle U,V\rangle$ of two representations
$U,V\in\mathcal{R}$ is defined as the Reshetikhin–Turaev invariant of the Hopf
link with components labeled by $U$ and $V$. This is a symmetric bilinear
pairing on $\mathcal{R}$. The map $\xi$ is related to the Hopf pairing as
follows:

######  Lemma 5.3.

The Hopf pairing on $\mathcal{R}$ can be computed as

$\langle U,V\rangle=\mathrm{Tr}_{q}^{U}(\xi(V)).$

###### Proof.

Consider the Drinfeld map $D$ [9] which sends a representation $V$ to a
central element corresponding to the universal invariant of the following
tangle:

$D(V):=$$V$

By e.g. [14, eq. (20)] (see also [19, Proposition 8.19] and references
therein) the eigenvalue of $D(V)$ on the irreducible representation
$V(\lambda)$ equals $\mathit{ch}(q^{\lambda+\rho})$ where $\mathit{ch}$ is the
character of $V$. By the definition of the Harish-Chandra map, this means that
$\mathit{hc}(D(V))=\mathit{ch}(V)$, and

$D(V)=\mathit{hc}^{-1}(ch(V))=\xi(V),$

so $\xi$ agrees with the Drinfeld map. Now $\langle
U,V\rangle=\mathrm{Tr}_{q}^{U}(D(V))=\mathrm{Tr}_{q}^{U}(\xi(V))$ or more
precisely,

$\langle
V(\lambda),V(\mu)\rangle=s_{\lambda}(q^{\mu+\rho})s_{\mu}(q^{\rho})\quad\text{where}\quad\dim_{q}V(\mu)=s_{\mu}(q^{\rho}).$

∎

Using the Drinfeld isomorphism $\xi$ we can extend the Hopf pairing to the
center by setting

$\langle
z_{1},z_{2}\rangle:=\langle\xi^{-1}(z_{1}),\xi^{-1}(z_{2})\rangle\quad\text{for
any}\quad z_{1},z_{2}\in\mathcal{Z}\ .$

##  6\. Cyclotomic completion and the universal invariant

The universal invariant of a link belongs a priori to a (completed) tensor
product of copies of $\mathcal{U}_{h}$, rather than $\mathcal{U}$ or
$\mathcal{U}_{\mathbb{Z}}$, due to the diagonal part of the $R$-matrix. The
aim of this section is to define a certain completion of
$\mathcal{U}_{\mathbb{Z}}$ and its tensor powers, such that the universal
$\mathfrak{gl}_{N}$ invariant of evenly framed links belongs to it. Since the
action of $\Gamma$ extends to the completion, this will allow us to speak
about $\Gamma$-invariance of $J_{L}(\mathfrak{gl}_{N};q)$.

###  6.1. Cyclotomic completion of $\mathcal{U}_{\mathbb{Z}}$

Given $n\in\mathbb{N}$, we define a family of two-sided ideals
$\mathcal{U}^{(n)}_{\mathbb{Z}}$ as the minimal filtration such that
$\mathcal{U}^{(n)}_{\mathbb{Z}}\mathcal{U}^{(m)}_{\mathbb{Z}}\subset\mathcal{U}^{(m+n)}_{\mathbb{Z}}$
and

$(q;q)_{n},\;\;e^{n}_{i},\;\;f_{n}(K^{2}_{j})\in\mathcal{U}^{(n)}_{\mathbb{Z}}$

for any $1\leq i\leq N-1$ and $1\leq j\leq N$ where $f_{n}(x)=(x;q)_{n}$. In
other words, $\mathcal{U}^{(n)}_{\mathbb{Z}}$ is the two-sided ideal generated
by the products

(15) $(q;q)_{a}\;e_{\mathbf{m}}\;f_{c_{1}}(K^{2}_{1})\cdots
f_{c_{N}}(K^{2}_{N}),\quad\text{with}\quad a+\sum_{i}m_{i}+\sum_{i}c_{i}=n\ .$

######  Lemma 6.1.

We have

$\Delta\left(f_{n}\left(K^{2}_{i}\right)\right)=\sum_{a=0}^{n}\binom{n}{a}_{q}f_{a}(K^{2}_{i})\otimes
K_{i}^{2(n-a)}f_{n-a}(K^{2}_{i}).$

###### Proof.

We prove Lemma by induction in $n$. For $n=0$ it is clear. The induction step
follows from the identities

$f_{n+1}(K^{2}_{i})=f_{n}(K_{i})(1-q^{n}K^{2}_{i})$

and

$\Delta(1-q^{n}K^{2}_{i})=1\otimes 1-q^{n}K^{2}_{i}\otimes
K^{2}_{i}=(1-q^{a}K^{2}_{i})\otimes
q^{n-a}K^{2}_{i}+1\otimes(1-q^{n-a}K^{2}_{i}).$

∎

######  Proposition 6.2.

a) $\mathcal{U}^{(n)}_{\mathbb{Z}}$ is the left ideal generated by (15).

b) $\mathcal{U}^{(n)}_{\mathbb{Z}}$ form a Hopf algebra filtration, that is
$\Delta\,\mathcal{U}^{(n)}_{\mathbb{Z}}\subset\,\sum_{i+j=n}\,\mathcal{U}^{(i)}_{\mathbb{Z}}\otimes\mathcal{U}^{(j)}_{\mathbb{Z}}$.

c) Assume that $\lambda_{i}\leq k$ for all $i$. Given arbitrary $m$, there
exists $n=n(k,m)$ such that the elements of $\mathcal{U}^{(n)}_{\mathbb{Z}}$
act on the integral basis of $V(\lambda)$ by matrices divisible by
$(q;q)_{m}$.

###### Proof.

a) Observe that by Lemma 10.5 we get
$f_{n}(q^{s}K^{2}_{i})\in\mathcal{U}^{(n)}_{\mathbb{Z}}$ for all integer $s$.
Now the statement follows from the identities

$f_{n}(K^{2}_{i})F_{i}^{(s)}=F_{i}^{(s)}f_{n}(q^{-s}K^{2}_{i}),\
f_{n}(K^{2}_{i+1})F_{i}^{(s)}=F_{i}^{(s)}f_{n}(q^{s}K^{2}_{i+1})$

and

$f_{n}(K^{2}_{i})e^{s}_{i}=e^{s}_{i}f_{n}(q^{s}K^{2}_{i}),\
f_{n}(K^{2}_{i+1})e^{s}_{i}=e^{s}_{i}f_{n}(q^{-s}K^{2}_{i+1}).$

b) Follows from the identity

$\Delta(e_{j}^{m})=\sum_{i=0}^{m}\binom{m}{i}_{q}e_{j}^{m-i}\mathcal{K}^{i}\otimes
e_{j}^{i}$

and Lemma 6.1.

c) By (a), it is sufficient to check the statement for $e_{i}^{n}$ and
$f_{n}(K^{2}_{i})$. If $\lambda_{i}\leq k$ then for $n>k$ $e_{i}^{n}$
annihilates $V(\lambda)$, while $f_{n}(K^{2}_{i})$ acts on a vector with
weight $(v^{\lambda_{1}},\ldots,v^{\lambda_{N}})$ by
$f_{n}(q^{\lambda_{i}})=(q^{\lambda_{i}};q)_{n}$ which is divisible by
$(q;q)_{n}$. ∎

By Proposition 6.2(b), the filtration

$\mathcal{U}_{\mathbb{Z}}=\mathcal{U}^{(0)}_{\mathbb{Z}}\supset\mathcal{U}^{(1)}_{\mathbb{Z}}\supset\dots\mathcal{U}^{(n)}_{\mathbb{Z}}\supset\dots$

is a Hopf algebra filtration of $\mathcal{U}_{\mathbb{Z}}$ with respect to a
descending filtration of ideals $I_{n}=((q;q)_{n})$ in $\mathbb{Z}[v,v^{-1}]$
in the sense of [18, Sec. 4]. Hence, the completion

$\widehat{\mathcal{U}}:={\lim\limits_{\overleftarrow{\hskip 5.69054ptn\hskip
5.69054pt}}}\;\;\frac{\mathcal{U}_{\mathbb{Z}}}{\mathcal{U}^{(n)}_{\mathbb{Z}}}$

is a complete Hopf algebra over the ring

$\widehat{\mathbb{Z}[v]}:={\lim\limits_{\overleftarrow{\hskip 5.69054ptn\hskip
5.69054pt}}}\;\;\frac{\mathbb{Z}[v]}{((q;q)_{n})}.$

We refer to [18, Section 4] for details. Analogously, we define the
$\Gamma$-invariant subalgebra

$\widehat{\mathcal{U}^{\text{ev}}}:={\lim\limits_{\overleftarrow{\hskip
5.69054ptn\hskip
5.69054pt}}}\;\;\frac{\mathcal{U}^{\text{ev}}_{\mathbb{Z}}}{\mathcal{U}^{(n)}_{\mathbb{Z}}}$

as a complete Hopf algebra over the Habiro ring $\widehat{\mathbb{Z}[q]}$. Let
us now extend the completion to the tensor powers of
$\mathcal{U}_{\mathbb{Z}}$. For this we define the filtration for
$\mathcal{U}_{\mathbb{Z}}^{\otimes l}$ for $l\geq 1$ as follows

${\mathcal{F}}_{n}(\mathcal{U}_{\mathbb{Z}}^{\otimes
l})=\sum^{l}_{i=1}\mathcal{U}_{\mathbb{Z}}^{\otimes
i-1}\otimes\mathcal{U}^{(n)}_{\mathbb{Z}}\otimes\mathcal{U}_{\mathbb{Z}}^{\otimes
l-i}$

and the completed tensor product $\mathcal{U}_{\mathbb{Z}}^{\hat{\otimes}l}$
with respect to this filtration will be the image of the homomorphism

${\lim\limits_{\overleftarrow{\hskip 5.69054ptn\hskip
5.69054pt}}}\;\;\frac{\mathcal{U}_{\mathbb{Z}}^{\otimes
l}}{{\mathcal{F}}_{n}(\mathcal{U}_{\mathbb{Z}}^{\otimes
l})}\;\;\to\;\;\mathcal{U}_{h}^{{\otimes}k}$

where on the right hand side we use the $h$-adically completed tensor product.

###  6.2. Hopf pairing and universal invariants

Let us denote by $c\in\mathcal{U}_{h}{\otimes}\,\mathcal{U}_{h}$ the double
braiding or the universal invariant of the clasp tangle in Figure 1, given by

$c=(S\,\otimes\,\text{id})\,\mathcal{R}_{21}\mathcal{R}\ .$

The main point about this element is that it is dual to the Hopf pairing or
the quantum Killing form (compare [19, Sec. 4]). Hence, after writing
$c=\sum_{i}c(i)\otimes c^{\prime}(i)$ the Hopf pairing is defined by setting

(16) $\langle c(i),c^{\prime}(j)\rangle:=\delta_{ij}$

Restricting to the Cartan part this gives us (compare [19, Lemma 3.12])

(17) $\displaystyle D^{-2}=\prod^{N}_{i=1}q^{-H_{i}\otimes
H_{i}}=\prod^{N}_{i=1}\sum_{n_{i}}(-1)^{n_{i}}\frac{h^{n_{i}}}{n_{i}!}H_{i}^{n_{i}}\,\otimes\,H_{i}^{n_{i}}$

and hence, $\langle
H_{i}^{n},H_{j}^{m}\rangle=\delta_{ij}\delta_{nm}(-1)^{n}\frac{n!}{h^{n}}$. We
deduce that $\langle K_{i}^{2},K_{j}^{2}\rangle=q^{-1}$ or, more generally,

$\langle K_{i}^{2a},K_{j}^{2b}\rangle=\delta_{ij}q^{-ab}$

defines the Hopf pairing on the $\Gamma$-invariant part of the Cartan. In
Section 10 we construct another basis for the Cartan given by
$\prod^{N}_{i=1}f_{n_{i}}(K^{2}_{i})$ such that $\langle
f_{n},f_{m}\rangle=\delta_{nm}(-1)^{n}q^{-n}(q;q)_{n}$. In this new basis, we
can rewrite the Cartan part of the clasp element as follows:

(18)
$D^{-2}=\sum_{\mathbf{n}\in\mathbb{N}^{N}}\prod^{N}_{i=1}\frac{(-1)^{n_{i}}q^{n_{i}}}{(q;q)_{n_{i}}}f_{n_{i}}(K^{2}_{i})\otimes
f_{n_{i}}(K^{2}_{i})$

For $\mathfrak{sl}_{N}$ similar computations will give

$(D^{\prime})^{-2}=\sum_{\mathbf{n}\in\mathbb{N}^{N-1}}\prod^{N-1}_{i=1}\frac{(-1)^{n_{i}}q^{n_{i}}}{(q;q)_{n_{i}}}f_{n_{i}}(\mathcal{K}_{i})\otimes
f_{n_{i}}(\mathcal{K}^{2}_{i})\ $

(compare Section B.1 in [19]).

Let us denote by

$\text{Inv}\,(\mathcal{U})=\\{u\in\mathcal{U}\,|\,x\vartriangleright
u=\epsilon(x)u\quad\forall x\in\mathcal{U}\\}$

the invariant part of $\mathcal{U}$ under the adjoint action
$x\vartriangleright u:=x_{(1)}uS(x_{(2)})$ in Sweedler notation. The main
advantage of the usage of bottom tangles in the definition of
$J_{L}(\mathfrak{gl}_{N};q)$ is that in this case
$J_{L}(\mathfrak{gl}_{N};q)\in\text{Inv}\,(\mathcal{U})$ (compare [15,
Sec.4.3]). As a corollary, we get the following:

######  Proposition 6.3.

Given an $l$-component evenly framed link $L$, the universal invariant
$J_{L}(\mathfrak{gl}_{N};q)$ is a well defined element of $\text{\rm
Inv}\,\left(\widehat{\mathcal{U}}^{\hat{\otimes}l}\right)$.

###### Proof.

By definition, $J_{K}$ is obtained by multiplying together elementary pieces,
such as $F_{\mathbf{n}}$, $e_{\mathbf{n}}$, $K^{\pm 1}_{2\rho}$, $D^{\pm 1}$,
and by then taking a sum over all indices. The linking between different
components and framing will make appear powers of $D^{\pm 2}$ that we can
decompose using the basis elements $f_{n}(K^{2}_{i})$ of the completion by
(18). Note that we can collect all diagonal contributions of each component by
using formulas like

$D(E_{i}\otimes 1)D^{-1}=E_{i}\otimes\mathcal{K}_{i}\quad\text{and}\quad
D(1\otimes F_{j})D^{-1}=\mathcal{K}^{-1}_{j}\otimes F_{j}\ .$

Since framing is assumed to be even, we will have an even number of $D$-parts.
Hence using (18) and the explicit form of the quasi $R$-matrix $\Theta$, we
get the claim. ∎

######  Remark 6.4.

For $\mathfrak{sl}_{N}$ we can build the same completion after replacing
$K_{i}$ with $\mathcal{K}_{i}$. Then the arguments in the proof of Proposition
6.3 will show us that for any algebraically split link the universal
invariants belongs to this completion.

Proof of Theorem 1.1.

Using Proposition 6.3 and remark above, we can define the action of $\Gamma$
on each component of $J_{L}(\mathfrak{g};q)$ separately. We will denote by
$J_{L}^{\zeta_{1},\ldots,\zeta_{\ell}}(\mathfrak{g};q)$ the result of this
action. Then we have

$J_{L}\left(V(\lambda_{1})\otimes L(\zeta_{1}),\ldots,V(\lambda_{\ell})\otimes
L(\zeta_{\ell})\right)=$
$\bigotimes^{l}_{i=1}\mathrm{Tr}^{V(\lambda_{i})\otimes
L(\zeta_{i})}_{q}\left(J_{L}(\mathfrak{g};q)\right)=\bigotimes^{l}_{i=1}\mathrm{Tr}^{V(\lambda_{i})}_{q}\left(J_{T}^{\zeta_{1},\ldots,\zeta_{\ell}}(\mathfrak{g};q)\right)\cdot\dim_{q}L(\zeta_{1})\cdots\dim_{q}L(\zeta_{\ell}).$

The second equation follows from Lemma 3.1. By Theorem 4.6 we conclude that

$J_{L}^{\zeta_{1},\ldots,\zeta_{\ell}}\left(\lambda_{1},\ldots,\lambda_{\ell}\right)=J_{L}\left(\lambda_{1},\ldots,\lambda_{\ell}\right)$

for all $\lambda_{1},\ldots,\lambda_{\ell}$ under the assumptions of Theorem
1.1, therefore
$J_{L}(\mathfrak{g};q)=J_{L}^{\zeta_{1},\ldots,\zeta_{\ell}}(\mathfrak{g};q)$
and hence, $J_{L}(\mathfrak{g};q)$ is $\Gamma$-invariant under the same
assumptions.

$\hfill\Box$

######  Corollary 6.5.

For any $\ell$-component evenly framed link $L$, $J_{L}(\mathfrak{gl}_{N};q)$
belongs to the $\Gamma$-invariant part of $\text{\rm
Inv}\left(\widehat{\mathcal{U}}^{\hat{\otimes}\ell}\right)$. Moreover, for
every $0$-framed algebraically split link $L$,

$J_{L}(\mathfrak{gl}_{N};q)=J_{L}(\mathfrak{sl}_{N};q)\ .$

###### Proof.

The first statement is the direct consequence of Theorem 1.1. The second one
follows from the fact that the only difference in the definitions of both
invariants is in the diagonal part of the $R$-matrix, that does not contribute
since the linking matrix vanishes and the rules for moving of $D$ and
$D^{\prime}$ along a component of the link coincide. ∎

###  6.3. Twist forms

Let us denote by $\widehat{\mathcal{Z}}$ the center of
$\widehat{\mathcal{U}}$. In what follows, we will be particularly interested
in the following twist forms

${\mathcal{T}}_{\pm}:\widehat{\mathcal{Z}}\to\widehat{\mathcal{Z}}\quad\text{
given by}\quad{\mathcal{T}}_{\pm}(z):=\langle r^{\pm 1},z\rangle$

the Hopf pairing with the ribbon element. On the $\Gamma$-invariant Cartan
part they are easy to compute, given the Hopf pairing between the generators
$H_{i}$ in Section 6.2 . We have

(19) ${\mathcal{T}}_{\pm}(K_{2\mathbf{a}})=\langle r^{\pm
1}_{0},K_{2\mathbf{a}}\rangle=v^{\pm(\mathbf{a},2\rho-\mathbf{a})}\in\mathbb{Z}[v,v^{-1}]\
$

for any $\mathbf{a}\in\mathbb{Z}^{N}$. Now equation (16) allows to extend the
twists form to $\widehat{\mathcal{U}}^{\text{ev}}$ as follows:

${\mathcal{T}}_{\pm}(F_{\mathbf{m}}\mathcal{K}_{\mathbf{m}}K_{2\mathbf{a}}e_{\mathbf{n}})=\delta_{\mathbf{m},\mathbf{n}}q^{(\rho,\sum_{i}n_{i}\alpha_{i})}v^{\pm(\mathbf{a},2\rho-\mathbf{a})}\in\mathbb{Z}[v,v^{-1}]\
$

where $\alpha_{i}=e_{i}-e_{i+1}$ are the simple roots. Observe that after
restriction to $\mathcal{U}_{q}^{\text{ev}}\mathfrak{sl}_{N}$, i.e. replacing
$K_{2\mathbf{a}}$ with $\mathcal{K}_{2\mathbf{b}}$ in the above formula, the
result belong to $\mathbb{Z}[q,q^{-1}]$ and coincide with [19, eq. (102)] for
any $\mathbf{b}\in\mathbb{Z}^{N-1}$.

##  7\. Habiro’s basis for $\mathcal{Z}(U_{q}\mathfrak{sl}_{2})$

In this section we summarize Habiro’s results for $\mathfrak{sl}_{2}$ in the
way suitable for our generalization.

Habiro [15] defined a remarkable family of central elements in
$\mathcal{Z}(U_{q}\mathfrak{sl}_{2})$:

(20)
$\sigma_{m}:=\prod^{m}_{i=1}\left(C^{2}-(v^{i}+v^{-i})^{2}\right)=\prod_{i=1}^{m}(C-v^{i}-v^{-i})(C+v^{i}+v^{-i}).$

Since $C$ acts on the $(j+1)$-dimensional representation $V_{j}$ by a scalar
$v^{j+1}+v^{-j-1}$, the polynomial $\sigma_{m}$ is completely characterized by
the following properties:

  * (a)

(Parity) $\sigma_{m}$ is $\Gamma=\mathbb{Z}_{2}$-invariant.

  * (b)

(Vanishing) $\sigma_{m}$ annihilates the representations $V_{j}$ for $j<m$.

  * (c)

(Normalization) $\sigma_{m}$ acts on the representation $V_{m}$ by a scalar

(21) $\prod^{m}_{i=1}\left((v^{m+1}+v^{-m-1})^{2}-(v^{i}+v^{-i})^{2}\right)\
.$

Note that parity implies that $\sigma_{m}$ also annihilates the
representations $L(-1)\otimes V_{j}$ for $j<m$. By using the Harish-Chandra
isomorphism, we can alternatively consider the polynomials

$T_{m}(y):=\mathit{hc}(\sigma_{m}):=\prod^{m}_{i=1}(yv^{i}-y^{-1}v^{-i})(yv^{-i}-y^{-1}v^{i})=(-1)^{m}\prod^{m}_{i=1}q^{-i}(1-y^{2}q^{i})(1-y^{-2}q^{i})$

which are characterized by the following properties:

  * (a)

(Parity) $T_{m}$ is $\mathbb{Z}_{2}$-invariant, that is, $T_{m}(-y)=T_{m}(y)$

  * (b)

(Vanishing) $T_{m}(\pm v^{j+1})=0$ for $j<m$

  * (c)

(Normalization) $T_{m}(v^{m+1})$ is given in (21).

Habiro proved that $\\{\sigma_{m}\\}_{m\geq 0}$ form a basis in (a certain
completion of) the $\Gamma$-invariant part of the center. Hence, the elements
$S_{m}=\xi^{-1}(\sigma_{m})$, given by

$S_{m}:=\prod_{i=1}^{m}(V_{1}-v^{i}-v^{-i})(V_{1}+v^{i}+v^{-i})$

form a basis of $\mathcal{R}$. We will show that

$P_{n}=\prod_{i=0}^{n-1}(V_{1}-v^{2i+1}-v^{-2i-1})\in\mathcal{R}$

is a dual basis to $\\{S_{m}\\}_{m\geq 0}$ with respect to the Hopf pairing.
The following is a slight reformulation of [15, Prop. 6.3].

######  Lemma 7.1.

We have

$\langle P_{n},S_{m}\rangle=\frac{\\{2n+1\\}!}{\\{1\\}}\delta_{n,m}\ .$

###### Proof.

Clearly, one has

$\xi(P_{n})=\prod_{i=0}^{n-1}(C-v^{2i+1}-v^{-2i-1})$

which annihilates $V_{2i}$ for $i<n$. We have the following cases:

1) For $n<m$ we have $\langle
P_{n},S_{m}\rangle=\mathrm{Tr}_{q}^{P_{n}}(\sigma_{m})$. Since $P_{n}$ is in
span of $V_{i}$ for $i\leq n$ and $\sigma_{m}$ annihilates all these, we get
$\langle P_{n},S_{m}\rangle=0$.

2) For $m<n$ we have $\langle
P_{n},S_{m}\rangle=\mathrm{Tr}_{q}^{S_{m}}(\xi(P_{n}))$. Since $S_{m}$ is in
span of $V_{2i}$ for $i\leq n$ and

$\langle P_{n},V_{2i}\rangle=\\{i+n\\}\dots\\{i-n+1\\}[2i+1]\ .$

Hence $P_{n}$ annihilates all these, we get $\langle P_{n},S_{m}\rangle=0$.

3) Finally, for $n=m$ we observe that $P_{n}$ has a unique copy of $V_{n}$ and

$\langle P_{n},S_{n}\rangle=\langle
V_{n},S_{n}\rangle=\mathrm{Tr}_{q}^{V_{n}}(\sigma_{n})$

which is easy to compute. ∎

We can use the above results to compute the coefficients in the decomposition
of any central element into $\\{\sigma_{m}\\}_{m\geq 0}$.

######  Lemma 7.2.

Let $\phi$ be a $\mathbb{Z}_{2}$-invariant element in
$\mathcal{Z}(U_{q}\mathfrak{sl}_{2})$ which acts on $V_{j}$ by a scalar
$\phi_{j}$. Then

$\phi=\sum
a_{n}\sigma_{n},\;\text{where}\;\;a_{n}=\sum_{i=0}^{n}(-1)^{n-i}\frac{\\{2i+2\\}\\{i+1\\}}{\\{n+i+2\\}!\\{n-i\\}!}\;\phi_{i}\
.$

###### Proof.

We have ([15, Lemma 6.1])

$P_{n}=\sum_{i=0}^{n}(-1)^{n-i}\frac{[2i+2]}{[n+i+2]}\left[\begin{array}[]{cc}\\!2n+1\\!\\\
\\!n+1+i\\!\end{array}\right]V_{i}.$

If $\phi=\sum a_{m}\sigma_{m}$ then

$a_{n}=\frac{\\{1\\}}{\\{2n+1\\}!}\mathrm{Tr}_{q}^{P_{n}}(\phi)=\frac{\\{1\\}}{\\{2n+1\\}!}\sum_{i=0}^{n}(-1)^{n-i}\frac{[2i+2]}{[n+i+2]}\left[\begin{array}[]{cc}\\!2n+1\\!\\\
\\!n+1+i\\!\end{array}\right]\mathrm{Tr}_{q}^{V_{i}}(\phi)=$
$\sum_{i=0}^{n}(-1)^{n-i}\frac{\\{2i+2\\}\\{1\\}}{\\{n+i+2\\}!\\{n-i\\}!}\dim_{q}(V_{i})\phi_{i}.$

Using $\dim_{q}(V_{i})=[i+1]$ we obtain the result. ∎

Habiro proved that for any $0$-framed knot $K$, there exist
$a_{n}(K)\in\mathbb{Z}[q,q^{-1}]$ such that

$J_{K}(\mathfrak{sl}_{2};q)=\sum_{n\geq 0}a_{n}(K)\,\sigma_{n}$

known as a cyclotomic expansion of the colored Jones polynomial of the knot
$K$.

##  8\. New basis for the center of $\widehat{\mathcal{U}}$

Recall that $\widehat{\mathcal{Z}}$ is the center of the completion
$\widehat{\mathcal{U}}$. In this section we construct the basis
$\\{\sigma_{\lambda}\\}_{\lambda}$ of the $\Gamma$-invariant part of
$\widehat{\mathcal{Z}}$. Furthermore, we explicitly define its dual
$\\{P_{\lambda}\\}_{\lambda}$ with respect to the Hopf pairing. This allows us
to construct the cyclotomic expansion of $J_{K}(\mathfrak{gl}_{N};q)$ for any
$0$-framed knot $K$.

The proof uses the existence and properties of interpolation Macdonald
polynomials [29] which are summarized in the following theorem.

######  Theorem 8.1.

There is a family of symmetric polynomials $F_{\lambda}(x_{1},\ldots,x_{N};q)$
such that:

  * (a)

$F_{\lambda}$ is in the span of Schur functions $s_{\mu}$ for $\mu\leq\lambda$
with the leading term

$F_{\lambda}=(-1)^{|\lambda|+\binom{N}{2}}q^{D_{N}(\lambda)}s_{\lambda}+\ldots\
.$

  * (b)

$F_{\lambda}(q^{-\mu_{1}-N+1},\ldots,q^{-\mu_{N}})=0$ unless $\mu$ contains
$\lambda$.

  * (c)

$F_{\lambda}(q^{-\lambda_{1}-N+1},\ldots,q^{-\lambda_{N}})=(-1)^{\binom{N}{2}}q^{n(\lambda)+\binom{N}{3}}\prod_{\square\in\lambda}(1-q^{-h(\square)})$.

  * (d)

Any function $F$ in the completion can be written as

(22)
$F(x_{1},\ldots,x_{N})=\sum_{\lambda,\;\mu\subset\lambda}d_{\mu,\lambda}(q)F(q^{-\mu_{1}-N+1},\ldots,q^{-\mu_{N}})F_{\lambda}(x_{1},\ldots,x_{N};q)$

where $d_{\lambda,\mu}$ are explicit coefficients prescribed by Theorem 10.17.

We discuss the definition and give more details on interpolation Macdonald
polynomials in Section 10.

######  Theorem 8.2.

There exists a family of central elements $\sigma_{\lambda}\in\mathcal{Z}$
with the following properties:

  * (a)

$\sigma_{\lambda}$ is $\Gamma$-invariant and annihilates $L(\zeta)\otimes
V(\mu)$ for all $\mu$ not containing $\lambda$ and $\zeta\in\Gamma$.

  * (b)

$\mathit{hc}(\sigma_{\lambda})$ is in the span of
$s_{\mu}(x_{1},\ldots,x_{N})$ for $\mu\leq\lambda$, with the leading term

$\mathit{hc}(\sigma_{\lambda})=(-1)^{|\lambda|+\binom{N}{2}}v^{(N-1)|\lambda|}q^{D_{N}(\lambda)}s_{\lambda}+\ldots\
.$

  * (c)

$\sigma_{\lambda}$ acts on $V(\lambda)$ by a scalar

$\sigma_{\lambda}|_{V(\lambda)}=(-1)^{\binom{N}{2}}q^{-n(\lambda)-\binom{N}{3}}\prod_{\square\in\lambda}(1-q^{h(\square)})\
.$

###### Proof.

Define $\sigma_{\lambda}=\mathit{hc}^{-1}(g_{\lambda})$, where
$g_{\lambda}(x_{1},\dots,x_{N})=F_{\lambda}(v^{N-1}x_{1},\ldots,v^{N-1}x_{N};q^{-1})$.
Then $\sigma_{\lambda}$ is clearly $\Gamma$-invariant and

$\sigma_{\lambda}|_{L(\zeta)\otimes V({\mu})}=g_{\lambda}(\zeta_{i}\cdot
v^{\mu_{i}+\rho_{i}})=F_{\lambda}(q^{(\mu_{1}+N-1)},\ldots,q^{\mu_{N}};q^{-1}).$

Indeed, if $y_{i}=\zeta_{i}\cdot
v^{\mu_{i}+\rho_{i}}=\zeta_{i}v^{(\mu_{i}-\frac{N-1}{2}+N-i)}$ then
$v^{N-1}y_{i}^{2}=q^{(\mu_{i}+N-i)}$.

Now $F_{\lambda}(q^{(\mu_{1}+N-1)},\ldots,q^{\mu_{N}};q^{-1})$ vanishes unless
$\mu$ contains $\lambda$, and has the nonzero value prescribed by the previous
theorem for $\mu=\lambda$. ∎

Let us define $\mathcal{R}_{\mathbb{Q}}:=\mathcal{R}\otimes\mathbb{Q}(v)$ by
extending the coefficient ring $\mathbb{Z}[v^{\pm 1}]$ of $\mathcal{R}$ to the
rational functions in $v$.

######  Theorem 8.3.

Define the following formal elements of $\mathcal{R}_{\mathbb{Q}}$

$P_{\lambda}=\sum_{\mu\subset\lambda}\frac{d_{\lambda,\mu}(q^{-1})}{\dim_{q}V(\mu)}\;V(\mu)\in\mathcal{R}_{\mathbb{Q}},$

then one has

(23) $\langle
P_{\lambda},\sigma_{\nu}\rangle:=\mathrm{Tr}^{P_{\lambda}}_{q}(\sigma_{\nu})=\delta_{\lambda,\nu}\
.$

###### Proof.

First, let us write the interpolation formula (22) for $F=F_{\nu}$:

$F_{\nu}(x_{1},\ldots,x_{N};q)=\sum_{\mu\subset\lambda}d_{\lambda,\mu}(q)F_{\nu}(q^{-\mu_{1}-N+1},\ldots,q^{-\mu_{N}};q)F_{\lambda}(x_{1},\ldots,x_{N};q),$

so

$\sum_{\mu\subset\lambda}d_{\lambda,\mu}(q)F_{\nu}(q^{-\mu_{1}-N+1},\ldots,q^{-\mu_{N}};q)=\delta_{\lambda,\nu}.$

By changing $q$ to $q^{-1}$ we get

(24)
$\sum_{\mu\subset\lambda}d_{\lambda,\mu}(q^{-1})F_{\nu}(q^{\mu_{1}+N-1},\ldots,q^{\mu_{N}};q^{-1})=\delta_{\lambda,\nu}.$

Now
$\mathrm{Tr}^{V(\mu)}_{q}(\sigma_{\nu})=\dim_{q}(V(\mu))\;g_{\nu}(q^{\mu_{1}+N-1},\ldots,q^{\mu_{N}})$,
hence

$\mathrm{Tr}^{P_{\lambda}}_{q}(\sigma_{\nu})=\sum_{\mu\subset\lambda}\frac{d_{\lambda,\mu}(q^{-1})}{\dim_{q}V({\mu})}\mathrm{Tr}^{V({\mu})}_{q}(\sigma_{\nu})=\;\delta_{\lambda,\nu}\
.$

∎

Next, we would like to study the integrality properties of the universal knot
invariant.

######  Lemma 8.4.

(a) Let $\sigma\in\mathcal{U}^{\text{ev}}_{\mathbb{Z}}$. Then
$\sigma=(K_{1}\cdots K_{N})^{-2s}\sum a_{\lambda}\sigma_{\lambda}$ with
$a_{\lambda}\in\mathbb{Z}[q,q^{-1}]$.

(b) Given $k$ and $m$, there exists $n=n(k,m)$ such that for all
$\Gamma$-invariant central elements $\sigma$ in the ideal
$\mathcal{U}^{(n)}_{\mathbb{Z}}$ the coefficients $a_{\lambda}$ are divisible
by $(q;q)_{m}$ for $|\lambda|\leq k$.

###### Proof.

(a) Recall that Harish-Chandra transform $\mathit{hc}$ identifies the
$\Gamma$-invariant part of the center of $\mathcal{U}_{\mathbb{Z}}$ with the
space of symmetric functions in $x_{1},\ldots,x_{N}$ with coefficients in
$\mathbb{Z}[q,q^{-1}]$. Since $F_{\lambda}$ is a polynomial with top degree
part equal to the Schur polynomial (up to a monomial in $q$), we can write
$(x_{1}\cdots
x_{N})^{s}f(x_{1},\ldots,x_{N})=\sum_{\lambda}a_{\lambda}F_{\lambda}(x_{1},\ldots,x_{N};q^{-1})$
and the result follows.

(b) If $\sigma$ is in the ideal $\mathcal{U}^{(n)}_{\mathbb{Z}}$ for
sufficiently large $n$, then by Proposition 6.2 its matrix elements in the
integral basis of $V(\lambda)$ are divisible by $(q;q)_{m}$. By definition of
Harish-Chandra transform, this implies that the values
$f(q^{-\lambda_{1}-N+1},\ldots,q^{-\lambda_{N}})$ are divisible by $(q;q)_{m}$
and hence by the interpolation formula (22) the coefficients $a_{\lambda}$ are
divisible by $(q;q)_{m}$ as well. ∎

######  Corollary 8.5.

The center of the completion $\widehat{\mathcal{U}}$ is isomorphic to the
completion of the space of symmetric polynomials with coefficients in
$\widehat{\mathbb{Z}[v]}$ with respect to the basis $F_{\lambda}$.

###### Proof.

By Lemma 8.4 any element of the center of $\widehat{\mathcal{U}}$ can be
written as an infinite series $\sum a_{\lambda}F_{\lambda}$ with coefficients
in $\widehat{\mathbb{Z}[v]}$, up to a factor $(x_{1}\cdots x_{N})^{-s}$. By
Corollary 11.10 the multiplication by $(x_{1}\cdots x_{N})^{-s}$ preserves the
space of such series. ∎

######  Corollary 8.6.

Any $\sigma\in\widehat{\mathcal{U}^{\text{ev}}}$ can be written as an infinite
sum $\sigma=\sum a_{\lambda}\sigma_{\lambda}$ with coefficients
$a_{\lambda}=\mathrm{Tr}^{P_{\mu}}_{q}(\sigma)\in\widehat{\mathbb{Z}[q]}$.

######  Proposition 8.7.

The universal knot invariant admits an expansion

$J_{K}(\mathfrak{gl}_{N};q)=\sum_{\lambda}a_{\lambda}(K)\sigma_{\lambda}\quad\text{with}\quad
a_{\lambda}(K)=\sum_{\mu\subset\lambda}{d_{\lambda,\mu}(q^{-1})}\,J_{K}(V(\mu),q)\in\mathbb{Z}[q,q^{-1}]\
$

called a cyclotomic expansion of the universal $\mathfrak{gl}_{N}$ knot
invariant.

Proposition 8.7 implies Theorem 1.3 in Introduction. Note that the knot
invariant $J_{K}(V(\mu),q)$ is normalized to be 1 for the unknot.

###### Proof.

By Corollary 6.5, $J_{K}(\mathfrak{gl}_{N};q)$ is a central element in
$\widehat{\mathcal{U}^{\text{ev}}}$, so it can be written as $\sigma=\sum
a_{\lambda}\sigma_{\lambda}$ with coefficients
$a_{\lambda}\in\widehat{\mathbb{Z}[q]}$. On the other hand, the value of
$J_{K}$ on any representation $V_{\lambda}$ is in $\mathbb{Z}[q,q^{-1}]$, so
by the interpolation formula (22) the coefficients $a_{\lambda}$ can be
written as rational functions with numerators in $\mathbb{Z}[q,q^{-1}]$ and
cyclotomic denominators. By Proposition 12.1 this implies that
$a_{\lambda}\in\mathbb{Z}[q,q^{-1}]$. The explicit formula for $a_{\lambda}$
is obtained by taking Hopf pairing with $P_{\mu}$ and observing that
$\mathrm{Tr}^{V({\mu})}_{q}\left(J_{K}(\mathfrak{gl}_{N};q)\right)=\dim_{q}(V(\mu))J_{K}(V(\mu),q)$
according to our normalization. ∎

The last result shows that
$a_{\lambda}(K)=\mathrm{Tr}_{q}^{P_{\lambda}}(J_{K}(\mathfrak{gl}_{N};q))\in\mathbb{Z}[q^{\pm
1}]$, even through the coefficients $d_{\lambda,\mu}(q)$ are rational
functions in $q$ (compare Example 10.23).

##  9\. Unified invariants of integral homology 3-spheres

This section is devoted to our main application of the previous results — a
construction of the unified invariants for integral homology 3-spheres. We
start with few auxiliary results.

Let us denote by

$P^{\prime}_{\lambda}=v^{-|\lambda|}\dim_{q}V(\lambda)\sum_{\mu\subset\lambda}\frac{d_{\lambda,\mu}(q^{-1})}{\dim_{q}V(\mu)}\;V(\mu)\in\mathcal{R}_{\mathbb{Q}}$

and define

(25) $\omega_{\pm}=\sum_{\lambda}(-1)^{|\lambda|+\binom{N}{2}}q^{\mp
c(\lambda)}q^{w_{\pm}(\lambda)}P^{\prime}_{\lambda}\in\widehat{\mathcal{R}}_{\mathbb{Q}}\quad\text{with}\quad\begin{array}[]{ll}w_{+}(\lambda)&=D_{N}(\lambda)\\\
w_{-}(\lambda)&=D_{N}(\lambda)+N|\lambda|\end{array}$

where $c(\lambda)$ is the content of $\lambda$. The next Lemma implies that
$\omega_{\pm}$ is the universal Kirby color for $(\pm 1)$-surgery.

######  Lemma 9.1.

For any $x\in\widehat{\mathcal{R}}_{\mathbb{Q}}$, we have

(26) $\langle\omega_{\pm},x\rangle=J_{U_{\mp}}(x)=\langle r^{\pm
1},\xi(x)\rangle$

where $J_{U_{\pm}}(x)$ is the Reshetikhin–Turaev invariant of the $(\pm
1)$-framed unknot colored by $x$.

###### Proof.

It is enough to check (26) for the basis elements $x=V(\nu)$. We compute

$\displaystyle\langle P^{\prime}_{\lambda},V(\nu)\rangle$
$\displaystyle=v^{-|\lambda|}\dim_{q}V(\lambda)\sum_{\mu\subset\lambda}\frac{{d_{\lambda,\mu}(q^{-1})}}{\dim_{q}V(\mu)}\;\langle
V(\mu),V(\nu)\rangle=v^{-|\lambda|}\dim_{q}V(\lambda)\sum_{\mu\subset\lambda}{d_{\lambda,\mu}(q^{-1})}s_{\nu}(q^{\mu_{i}+N-i})$
$\displaystyle=\dim_{q}V(\lambda)\sum_{\mu\subset\lambda}{d_{\lambda,\mu}(q^{-1})}C_{\nu}F_{\nu}(q^{\mu_{i}+N-i})={C_{\lambda}}\delta_{\lambda,\nu}\dim_{q}V(\lambda)$

where we used Lemma 5.3, equation (24) and the expansion
$s_{\nu}=(-1)^{|\lambda|+\binom{N}{2}}q^{-D_{N}(\lambda)}v^{(1-N)|\lambda|}F_{\nu}+\text{lower
terms}$ and hence,

$C_{\lambda}=(-1)^{|\lambda|+\binom{N}{2}}q^{-D_{N}(\lambda)}v^{-N|\lambda|}\
.$

Using this computation it is easy to check that

(27) $\langle\omega_{\pm},V(\nu)\rangle=v^{\mp N|\nu|}q^{\mp
c(\nu)}\dim_{q}V(\nu)=v^{\mp(\nu,\nu+2\rho)}\dim_{q}V(\nu)=\mathrm{Tr}^{V(\nu)}_{q}(r^{\pm
1})$

is equal to $J_{U_{\mp}}(V(\nu))$. ∎

From the following computation for
$V^{\prime}(\nu)=\frac{V(\nu)}{\dim_{q}(V(\nu))}$

$\langle\omega_{+}\omega_{-},V^{\prime}(\nu)\rangle=\langle\omega_{+},V^{\prime}(\nu)\rangle\langle\omega_{-},V^{\prime}(\nu)\rangle=v^{(\nu,\nu+2\rho)}v^{-(\nu,\nu+2\rho)}=\langle
1,V^{\prime}(\nu)\rangle$

we see that $\omega_{+}$ and $\omega_{-}$ are inverse to each other in the
algebra $\widehat{\mathcal{R}}_{\mathbb{Q}}$ isomorphic to
$\widehat{\mathcal{Z}}_{\mathbb{Q}}$.

A direct consequence of the above Lemma and the fusion rules is the following
result.

######  Theorem 9.2.

Let $L\cup K=L_{1}\cup L_{2}\dots\cup L_{l}\cup K$ be an $(l+1)$ component
algebraically split $0$-framed link such that $K$ is the unknot. We denote by
$L_{(K,\pm 1)}$ the framed link in $S^{3}$ obtained from $L$ by $\pm
1$-surgery along $K$, then for any $p_{1},\dots,p_{l}\in\mathcal{R}$

$J_{L\cup K}(p_{1},\dots,p_{l},\omega^{\pm 1})=J_{L_{(K,\pm
1)}}(p_{1},\dots,p_{l}).$

###### Proof.

The proof is given in [15, Thm. 9.4]. ∎

###  9.1. Construction of the unified invariants

Without loss of generality, we can assume that an integral homology 3-sphere
$M$ is obtained by $\mathbf{\varepsilon}$-surgery on an $\ell$ component
algebraically split link $L$, where $\mathbf{\varepsilon}\in\\{\pm
1\\}^{\ell}$. For $\mathfrak{sl}_{N}$ Habiro and Le defined a unified
invariant of $M$ as follows

$I^{\text{HL}}(M):=\mathcal{T}^{\prime}_{\mathbf{\varepsilon}}(J_{L}(\mathfrak{sl}_{N};q))\;\in\;\widehat{\mathbb{Z}[q]}\quad\text{
where}\quad\mathcal{T}^{\prime}_{\varepsilon}=\bigotimes^{\ell}_{i=1}\mathcal{T}^{\prime}_{\varepsilon_{i}}\
$

is the $\mathfrak{sl}_{N}$ twist form. They also proved that
$I^{\text{HL}}(M)$ belongs to the Habiro ring [19].

For $\mathfrak{gl}_{N}$ we define the unified invariant of $M$ similarly

$I(M):=\mathcal{T}_{\mathbf{\varepsilon}}(J_{L}(\mathfrak{gl}_{N};q))\;\in\;\widehat{\mathbb{Z}[q]}\quad\text{
where}\quad\mathcal{T}_{\varepsilon}=\bigotimes^{\ell}_{i=1}\mathcal{T}_{\varepsilon_{i}}\
$

by using the $\mathfrak{gl}_{N}$ twist forms.

######  Theorem 9.3.

For any integral homology $3$-sphere $M$,

$I(M)=J_{L}(\omega_{\varepsilon_{1}},\dots,\omega_{\varepsilon_{l}})\in\;\widehat{\mathbb{Z}[q]}$

Moreover, its evaluation at any root of unity coincides with the
$\mathfrak{sl}_{N}$ Witten–Reshetikhin–Turaev invariant of $M$.

This implies Theorem 1.4 from Introduction.

###### Proof.

By Corollary 6.5, for any algebraically split $0$-framed link $L$ we have

$J_{L}(\mathfrak{gl}_{N};q)=J_{L}(\mathfrak{sl}_{N};q).$

Hence, as explained in Section 6.3, the $\mathfrak{gl}_{N}$ and
$\mathfrak{sl}_{N}$ twist forms on $J_{L}$ do coincide. This implies
$I(M)=I^{\text{HL}}(M)$. Since the Habiro–Le invariant is known to belong to
the Habiro ring and to evaluate at a root of unity to the Witten-Reshetikhin-
Turaev (WRT) one, it remain to show
$I(M)=J_{L}(\omega_{\epsilon_{1}},\dots,\omega_{\epsilon_{\ell}})$. We prove
this claim in two steps.

Step 1: Assume $\ell=1$, then $J_{L}(\omega_{\pm})=I(M)$ by Lemma 9.1.

Step 2: For any $x=x_{1}\otimes\dots\otimes
x_{\ell}\in\text{Inv}(\widehat{\mathcal{U}}_{\mathbb{Z}}^{\hat{\otimes}\ell})$
we define $a_{k}$ for $k=0,1,\dots,\ell$ and $b_{k}=1,\dots,\ell$ as follows:

$a_{k}=\prod^{k}_{i=1}\langle
r^{\epsilon_{i}},x_{i}\rangle\prod^{\ell}_{j=k+1}\langle\omega_{\epsilon_{j}},x_{j}\rangle,\quad
b_{k}=x_{k}\prod^{k-1}_{i=1}\langle
r^{\epsilon_{i}},x_{i}\rangle\prod^{\ell}_{j=k+1}\langle\omega_{\epsilon_{j}},x_{j}\rangle.$

Then

$a_{k-1}=\langle\omega_{\epsilon_{k}},b_{k}\rangle\quad\text{and}\quad
a_{k}=\langle r^{\epsilon_{k}},b_{k}\rangle.$

where we identify $\omega_{\pm}$ with their image under $\xi$ for simplicity.
Since $b_{k}\in\mathcal{Z}_{\mathbb{Q}}$, we have $a_{k}=a_{k-1}$ by Step 1
for $k=1,2,\dots,\ell$. Hence, we have $a_{0}=a_{\ell}$ which is our claim. ∎

Theorem 9.3 has striking consequences. Indeed, for any $\lambda\in\mathcal{R}$
let us denote

$\mathcal{P}_{\lambda}=\text{\rm Span}_{\mathbb{Z}[q^{\pm
1}]}\\{P^{\prime}_{\mu}\,|\,\lambda\subset\mu\\},\quad\mathcal{P}=\mathcal{P}_{\emptyset}\quad\text{and}\quad\widehat{\mathcal{P}}:={\lim\limits_{\overleftarrow{\hskip
5.69054pt\lambda\hskip
5.69054pt}}}\;\;\frac{\mathcal{P}}{\mathcal{P}_{\lambda}}.$

For any framed link $L$, the Reshetikhin–Turaev functor provides a
$\mathbb{Q}(v)$-multilinear map
$J_{L}:\mathcal{R}_{\mathbb{Q}}\times\dots\times\mathcal{R}_{\mathbb{Q}}\to\mathbb{Q}(v)$.
For any algebraically split $0$-framed link $L$, Theorem 9.3 implies that its
restriction to $\mathcal{P}$ provides a $\mathbb{Z}[q^{\pm 1}]$-multilinear
map

$J_{L}:\mathcal{P}\times\dots\times\mathcal{P}\to\mathbb{Z}[q,q^{-1}]$

inducing

$J_{L}:\widehat{\mathcal{P}}\times\dots\times\widehat{\mathcal{P}}\to\widehat{\mathbb{Z}[q]}\
.$

This leads to a generalization of the famous integrability theorem in [15,
Thm. 8.2].

######  Corollary 9.4.

Given an $\ell$ component algebraically split $0$-framed link $L$, then for
all but finitely many partitions $\lambda_{i}$ with $1\leq i\leq\ell$, there
exist positive integers $n=n(\lambda_{i},N)$, such that

$J_{L}(P^{\prime}_{\lambda_{1}},\dots,P^{\prime}_{\lambda_{\ell}})\in(q;q)_{n}\mathbb{Z}[q,q^{-1}]\
.$

It would be interesting to have a direct proof of Corollary 9.4 without using
the theory of unified invariants.

Based on Corollary 9.4 we can give a cyclotomic expansion of the
Reshetikhin–Turaev invariant of $L$ as follows:

(28)
$J_{L}(\lambda_{1},\dots,\lambda_{\ell})=v^{\sum_{i}|\lambda_{i}|}\sum_{\mu_{i}\subset\lambda_{i}}\prod^{\ell}_{j=1}c_{\lambda_{j},\mu_{j}}(q^{-1})\,J_{L}(P^{\prime}_{\mu_{1}},\dots,P^{\prime}_{\mu_{\ell}})$

where the matrix
$\left[c_{\lambda,\mu}(q)\right]_{\lambda,\mu}:=\left[F_{\lambda}(q^{-\mu_{i}-N+i})\right]_{\lambda,\mu}$
is the inverse of $\left[d_{\lambda,\mu}(q)\right]_{\lambda,\mu}$ by Theorem
10.17 below. This generalizes equation $(8.2)$ in [15].

###  9.2. Few direct arguments

Our proof of the fact that $I(M)$ belongs to the Habiro ring is based on the
result that $I^{\text{HL}}(M)\in\widehat{\mathbb{Z}[q]}$ proven in [19] on
more than 100 pages. Given the complexity of their argument, we decided to
collect here different facts that can be shown without reference to [19].

######  Theorem 9.5.

Assume $M_{\pm}$ is obtained by $(\pm 1)$-surgery on the knot $K$, then

$I(M_{\pm})=J_{K}(\omega_{\pm})\;\in\;\widehat{\mathbb{Z}[v]}$

belongs to the Habiro ring.

###### Proof.

By Theorem 1.3 we know

$J_{K}(\mathfrak{gl}_{N};q)=\sum_{\mu}a_{\mu}(K)\sigma_{\mu}\quad\text{
with}\quad a_{\mu}(K)\in\mathbb{Z}[q,q^{-1}]$

The fact that $I(M_{\pm})$ belongs to the Habiro ring easily follows from the
claim that $\mathcal{T}_{\pm}(\sigma_{\mu})$ is divisible by $(v;v)_{m}$ for
some $m$ depending on $\mu$ and $N$. Let us prove this claim. By (19) the Hopf
pairing with $r^{\pm 1}$ replaces an element $x^{k_{i}}_{i}$ with
$v^{Q_{\pm}(k_{i})}$ where $Q_{\pm}$ is a quadratic form. By Lemma 10.27 we
can rewrite $\sigma_{\mu}$ as a linear combination of
$\prod^{d}_{i=1}f_{n_{i}}(q^{s_{i}}x_{i})$ such that $\sum_{i}n_{i}=|\mu|$,
$d=N(N+1)/2$ and $s_{i}\in\mathbb{Z}$. Moreover, each $f_{n}(q^{a}x_{i})$ is
divisible by $f_{n}(v^{a}y_{i})$ where $y_{i}^{2}=x_{i}$ and hence belongs to
the ideal $I_{n}$ of $\mathbb{Z}[v^{\pm 1},y^{\pm 1}_{i}]$ characterized in
Proposition 2.1 of [3]. The result follows now from [3, Theorem 2.2]. The
number $m$ we are looking for is
$\left\lfloor\frac{|\mu|}{N(N+1)}\right\rfloor$.

∎

Combining previous results we obtain an explicit expression for the unified
invariant for knot surgeries:

(29)
$I_{M_{\pm}}=J_{K}(\omega_{\pm})=\mathrm{Tr}_{q}^{\omega_{\pm}}J_{K}(\mathfrak{gl}_{N};q)=\sum_{\lambda}(-1)^{|\lambda|+\binom{N}{2}}q^{\mp
c(\lambda)}q^{w_{\pm}(\lambda)}\,J_{K}(P^{\prime}_{\lambda})$

Assuming that
$I(M)=J_{L}(\omega_{\varepsilon_{1}},\dots,\omega_{\varepsilon_{l}})$ is well
defined, its topological invariance can be shown directly as follows. Since
$I(M)$ depends only on the isotopy class of $L$, it remains to check its
invariance under Hoste moves (a version of Fenn-Rourke moves between
algebraically split links). Without loss of generality, we can assume that the
last component is an unknot, then the statement follows from Theorem 9.2.

Assuming $I(M)$ belongs to the Habiro ring, and as such has well defined
evaluations at roots of unity [17, Thm. 6.3] we can use the same trick as
above to show that for any root of unity $\zeta$

$\text{ev}_{\zeta}I(M)=\text{WRT}(M,\zeta).$

Let us recall that the WRT invariant is obtained from
$J_{L}(\mathfrak{sl}_{N};q)$ by taking trace along each component with the
Kirby color

$\Omega_{\pm}=\frac{\sum_{\lambda}v^{\pm(\lambda,\lambda+2\rho)}\dim_{q}V(\lambda)}{\sum_{\mu}v^{\pm(\mu,2\rho+\mu)}\dim^{2}_{q}V(\mu)}\,V(\lambda)$

where the sums are taken over all
$\lambda,\mu\in\mathcal{R}^{\text{fin}}=\\{\lambda|\dim_{\zeta}V(\lambda)\neq
0\\}$ and $v^{2}$ is evaluated to $\zeta$. Hence, we need to show that for any
$x$ in the ad-invariant part of the completed $\ell$th tensor power of
$\widehat{\mathcal{U}}_{\mathbb{Z}}$, we have

$\mathrm{Tr}_{q}^{\Omega_{\varepsilon}}(x)\stackrel{{\scriptstyle\zeta}}{{=}}\mathrm{Tr}_{q}^{\omega_{\varepsilon}}(x)\quad\quad\forall
x\in\text{Inv}(\widehat{\mathcal{U}}_{\mathbb{Z}}^{\hat{\otimes}\ell})$

where $\stackrel{{\scriptstyle\zeta}}{{=}}$ means the equality after
evaluation $v^{2}=\zeta$. We will prove this fact in two steps.

Step 1: Assume $\ell=1$, in this case
$\text{Inv}\,\widehat{\mathcal{U}}_{\mathbb{Z}}=\widehat{\mathcal{Z}}$ with
basis given by $z_{\lambda}=\xi(V(\lambda))$. Since $\Omega_{\pm}$ is
invariant under Hoste moves, we have

$\mathrm{Tr}_{q}^{\Omega_{\pm}}\left(z_{\nu}\right)=\langle\Omega_{\pm},V(\nu)\rangle=\text{ev}_{\zeta}\left(v^{\mp(\nu,\nu+2\rho)}\dim_{q}V(\nu)\right)$

where we interpret the left hand side as a Hopf link with components colored
by $\Omega_{\pm}$ and $V(\nu)$, and the right hand side is the result of the
sliding. Comparing this computation with (27), we deduce that at roots of
unity the actions of $\Omega_{\pm}$ and $\omega_{\pm}$ do coincide on
$\xi(\mathcal{R}^{\text{fin}})$, and they vanish on
$x\in\widehat{\mathcal{Z}}\setminus\xi(\mathcal{R}^{\text{fin}})$ after
evaluation.

Step 2: Define $a_{k}$ for $k=0,1,\dots,\ell$ and $b_{k}=1,\dots,\ell$ as
follows:

$a_{k}=\bigotimes^{k}_{j=1}\mathrm{Tr}_{q}^{\Omega_{\varepsilon_{j}}}\otimes\bigotimes^{\ell}_{j=k+1}\mathrm{Tr}_{q}^{\omega_{\varepsilon_{j}}}(x),\quad
b_{k}=\bigotimes^{k-1}_{j=1}\mathrm{Tr}_{q}^{\Omega_{\varepsilon_{j}}}\otimes
1\otimes\bigotimes^{\ell}_{j=k+1}\mathrm{Tr}_{q}^{\omega_{\varepsilon_{j}}}(x).$

Then

$a_{k-1}=\mathrm{Tr}_{q}^{\omega_{\varepsilon_{k}}}(b_{k})\quad\text{and}\quad
a_{k}=\mathrm{Tr}_{q}^{\Omega_{\varepsilon_{k}}}(b_{k}).$

Since $b_{k}\in\mathcal{Z}_{\mathbb{Q}}$, we have
$a_{k}\stackrel{{\scriptstyle\zeta}}{{=}}a_{k-1}$ by Step 1 and Lemma 9.1 for
$k=1,2,\dots,\ell$. Hence, we have
$a_{0}\stackrel{{\scriptstyle\zeta}}{{=}}a_{\ell}$ which is our claim.

##  10\. Interpolation polynomials

In this section we summarize the theory of interpolation Macdonald
polynomials.

###  10.1. One variable case

Consider the space of polynomials in one variable $x$ over $\mathbb{C}(q)$
with the following bilinear form

$(x^{k},x^{m})=q^{-km}.$

Let us define polynomials $f_{m}(x),m=0,1,\ldots$ by the equation $f_{0}(x)=1$
and

(30) $f_{m}(x)=(x;q)_{m}=(1-x)\cdots(1-xq^{m-1})\quad\text{for}\quad m\geq 1.$

Clearly, $f_{m}(x)$ is a degree $m$ polynomial with leading term
$(-1)^{m}q^{\frac{m(m-1)}{2}}x^{m}$, so $\\{f_{m}\\}_{m\geq 0}$ form a basis
in $\mathbb{Z}_{q,q^{-1}}[x]$. Our next aim is to show that this basis is
orthogonal. Observe that $f_{m}(q^{-k})=0$ for $k<m$.

######  Lemma 10.1.

We have $(f_{m}(x),f_{k}(x))=\delta_{km}q^{-m}(q;q)_{m}$

###### Proof.

First, observe that $(g(x),x^{k})=g(q^{-k})$ for any polynomial $g(x)$.
Therefore for $m>k$ we have $(f_{m}(x),x^{k})=f_{m}(q^{-k})=0$, so
$(f_{m}(x),g(x))=0$ for any polynomial $g(x)$ of degree strictly less than
$m$. In particular, $(f_{m}(x),f_{k}(x))=0$ for $m>k$ and

$(f_{m}(x),f_{m}(x))=(-1)^{m}q^{\frac{m(m-1)}{2}}(f_{m}(x),x^{m})=(-1)^{m}q^{\frac{m(m-1)}{2}}f_{m}(q^{-m})=$
$(-1)^{m}q^{\frac{m(m-1)}{2}}(1-q^{-m})\cdots(1-q^{-1})=q^{-m}(1-q^{m})\cdots(1-q).$

∎

######  Lemma 10.2.

The transition matrix between the monomial basis $x^{a}$ and the basis
$f_{b}(x)$ has the following form:

(31) $x^{a}=\sum_{b\leq a}k_{a,b}f_{b}(x),\quad
k_{a,b}=(-1)^{b}q^{-ab+\frac{b(b+1)}{2}}\binom{a}{b}_{q}.$

###### Proof.

To find the coefficients we compute the pairing $(f_{b}(x),x^{a})$, then using
orthogonality we obtain

$k_{a,b}=\frac{(f_{b}(x),x^{a})}{(f_{b}(x),f_{b}(x))}=\frac{f_{b}(q^{-a})}{(f_{b}(x),f_{b}(x))}.$

For $a\geq b$ from Lemma 10.1 we get

$(f_{b}(x),f_{b}(x))=q^{-b}(q;q)_{b},$

while

$\displaystyle f_{b}(q^{-a})$
$\displaystyle=(1-q^{-a})\cdots(1-q^{-a+b-1})=(-1)^{b}q^{-ab+\frac{b(b-1)}{2}}(1-q^{a})\cdots(1-q^{a-b+1})$
$\displaystyle=(-1)^{b}q^{-ab+\frac{b(b-1)}{2}}\frac{(q;q)_{a}}{(q;q)_{a-b}}.$

and the equation follows. ∎

Our next goal is to expand arbitrary polynomial $f(x)$ in the basis
$f_{m}(x)$. This can be done in two different ways. First, we can expand
$f(x)$ in the monomial basis and apply (31). Alternatively, we can apply
Newton interpolation method: if $f(x)=\sum a_{m}f_{m}(x)$ then

$f(q^{-j})=\sum_{m\geq j}a_{m}f_{m}(q^{-j}),$

which is a triangular system of equations for the unknown coefficients
$a_{m}$. Thus knowing $f(q^{-j})$ one can at least theoretically reconstruct
the coefficients $a_{m}$. This can be made explicit by the following:

######  Lemma 10.3.

We have

(32) $f(x)=\sum_{m=0}^{\infty}a_{m}f_{m}(x),\
a_{m}=\frac{1}{(f_{m},f_{m})}\sum_{j=0}^{m}(-1)^{j}q^{\frac{j(j-1)}{2}}\binom{m}{j}_{q}f(q^{-j}).$

###### Proof.

By $q$-binomial theorem we have

(33)
$f_{m}(x)=\sum_{j=0}^{m}(-1)^{j}q^{\frac{j(j-1)}{2}}\binom{m}{j}_{q}x^{j}.$

Now

$a_{m}=\frac{(f,f_{m})}{(f_{m},f_{m})}=\frac{1}{(f_{m},f_{m})}\sum_{j=0}^{m}(-1)^{j}q^{\frac{j(j-1)}{2}}\binom{m}{j}_{q}(f,x^{j}).$

Finally, $(f,x^{j})=f(q^{-j})$. ∎

######  Remark 10.4.

Equation (33) can be interpreted as an explicit inverse of the matrix in (31).

One can consider completion $\widehat{\mathbb{Z}_{q}[x]}$ of the space of
polynomials with respect to the basis $f_{m}(x)$. In this completion, infinite
sums $\sum_{m=0}^{\infty}a_{m}f_{m}(x)$ are allowed. Newton interpolation
method and (32) identify this completion with the space of distributions on
the interpolation nodes $1,q^{-1},\ldots$.

We will need the following lemma.

######  Lemma 10.5.

We have

$(x-q^{s})(x-q^{s+1})\cdots(x-q^{s+m-1})=\\\
\sum_{j=0}^{m}(-1)^{j}q^{-jm+\binom{j+1}{2}}\binom{m}{j}_{q}(1-q^{s+j})\cdots(1-q^{s+m-1})f_{j}(x).$

###### Proof.

We prove it by induction in $m$. For $m=1$ we get

$x-q^{s}=-(1-x)+(1-q^{s})=-f_{1}+(1-q^{s})f_{0}.$

For the step of induction we observe

(34) $\displaystyle(x-q^{s+m})f_{j}(x)$
$\displaystyle=-q^{-j}(1-q^{j}x)f_{j}(x)+(q^{-j}-q^{s+m})f_{j}(x)$
$\displaystyle=-q^{-j}f_{j+1}(x)+q^{-j}(1-q^{s+m+j})f_{j}(x).$

Using (34), it is easy to identify the coefficient at $f_{j}(x)$ in

$(x-q^{s+m})\sum(-1)^{j}q^{-jm+\binom{j+1}{2}}\binom{m}{j}_{q}(1-q^{s+j})\cdots(1-q^{s+m-1})f_{j}(x)$

as

$-q^{-j+1}(-1)^{j-1}q^{-(j-1)m+\binom{j}{2}}\binom{m}{j-1}_{q}(1-q^{s+j-1})\cdots(1-q^{s+m-1})\\\
+q^{-j}q^{-jm+\binom{j+1}{2}}\binom{m}{j}_{q}(1-q^{s+j})\cdots(1-q^{s+m-1})(1-q^{s+m+j})\\\
=-q^{-j(m+1)+\binom{j+1}{2}}(1-q^{s+j})\cdots(1-q^{s+m-1})\\\
\times\left[q^{m-j+1}\binom{m}{j-1}_{q}(1-q^{s+j-1})+\binom{m}{j}_{q}(1-q^{s+m+j})\right].$

It remains to notice that

$q^{m-j+1}\binom{m}{j-1}_{q}(1-q^{s+j-1})+\binom{m}{j}_{q}(1-q^{s+m+j})$
$=\left[q^{m-j+1}\binom{m}{j-1}_{q}+\binom{m}{j}_{q}\right]-q^{s+m}\left[\binom{m}{j-1}_{q}+q^{j}\binom{m}{j}_{q}\right]$
$=\binom{m+1}{j}_{q}-q^{s+m}\binom{m+1}{j}_{q}=(1-q^{s+m})\binom{m+1}{j}_{q}.$

∎

######  Remark 10.6.

If we set a formal variable $y=q^{s}$ in Lemma 10.5, then we get the identity

$(x-y)(x-qy)\cdots(x-yq^{m-1})=\sum_{j=0}^{m}(-1)^{j}q^{-jm+\binom{j+1}{2}}\binom{m}{j}_{q}f_{m-j}(yq^{j})f_{j}(x).$

This is a $q$-analogue of the binomial identity

$(x-y)^{m}=\sum_{j=0}^{m}(-1)^{j}\binom{m}{j}(1-y)^{m-j}(1-x)^{j}.$

###  10.2. Multi-variable case: polynomials

Let us generalize the above results to the case of $N$ variables. The pairing
has the form

$(x_{1}^{a_{1}}\cdots x_{N}^{a_{N}},x_{1}^{b_{1}}\cdots
x_{N}^{b_{N}})=q^{-\sum
a_{i}b_{i}}=(x_{1}^{a_{1}},x_{1}^{b_{1}})\cdots(x_{N}^{a_{N}},x_{N}^{b_{N}}).$

Note that for $\mathbf{x}=(x_{1},\dots,x_{N})$

$(g(\mathbf{x}),x_{1}^{b_{1}}\cdots
x_{N}^{b_{N}})=g(q^{-b_{1}},\ldots,q^{-b_{N}}).$

Consider the products

$f_{k_{1},\ldots,k_{N}}(\mathbf{x})=f_{k_{1}}(x_{1})\cdots f_{k_{N}}(x_{N}).$

Since $f_{k}(x)$ give a basis in $\mathbb{C}(q)[x]$, the polynomials
$f_{k_{1},\ldots,k_{N}}$ give a basis in $\mathbb{C}(q)[x_{1},\ldots,x_{N}]$.
Clearly,

$(f_{k_{1},\ldots,k_{N}},x_{1}^{b_{1}}\cdots x_{N}^{b_{N}})=0\
\textrm{unless}\ b_{i}\geq k_{i}\ \textrm{for all}\ i.$

######  Lemma 10.7.

We have $(f_{k_{1},\ldots,k_{N}},f_{m_{1},\ldots,m_{N}})=0$ unless
$k_{i}=m_{i}$ for all $i$.

###### Proof.

Suppose that $k_{i}>m_{i}$ for some $i$. Since $f_{m_{1},\ldots,m_{N}}$
contains only monomials of the form $x_{1}^{b_{1}}\cdots x_{N}^{b_{N}}$ with
$b_{i}\leq m_{i}$, we have $(f_{k_{1},\ldots,k_{N}},x_{1}^{b_{1}}\cdots
x_{N}^{b_{N}})=0$ for all such monomials and hence
$(f_{k_{1},\ldots,k_{N}},f_{m_{1},\ldots,m_{N}})=0$. ∎

Next, we would like to describe the basis in symmetric polynomials. It will be
labeled by partitions
$\lambda=(\lambda_{1}\geq\lambda_{2}\geq\ldots\geq\lambda_{N})$ with at most
$N$ parts. We define

(35)
$F_{\lambda}(\mathbf{x})=\frac{\det(f_{\lambda_{i}+N-i}(x_{j}))}{\prod_{i<j}(x_{i}-x_{j})}.$

Clearly, the numerator in (35) is antisymmetric in $x_{i}$, so it is divisible
by $\prod_{i<j}(x_{i}-x_{j})$ and the ratio is a symmetric function. It is
easy to see that $F_{\lambda}(\mathbf{x})$ is a non-homogeneous polynomial of
degree $|\lambda|$, and the top degree component equals
$(-1)^{|\lambda|+\binom{N}{2}}q^{D_{N}(\lambda)}s_{\lambda}$ where
$s_{\lambda}$ is the Schur function and $D_{N}(\lambda)$ is defined by (6).
The function $F_{\lambda}(\mathbf{x})$ is known as a special case of a
factorial Schur function [26, 27, 28], it is also a specialization of
nonsymmetric Macdonald polynomials described below.

######  Lemma 10.8.

Suppose that $b_{1}>\ldots>b_{N}$. Then
$F_{\lambda}(q^{-b_{1}},\ldots,q^{-b_{N}})=0$ unless
$b_{i}\geq\lambda_{i}+N-i$ for all $i$.

###### Proof.

Suppose that $b_{j}<\lambda_{j}+N-j$ for some $j$, then for all $i\leq j$ and
$\ell>j$ one has $\lambda_{i}+N-i\geq\lambda_{j}+N-j>b_{j}\geq b_{\ell}$, so
$f_{\lambda_{i}+N-i}(q^{-b_{\ell}})=0$. This implies
$\det[f_{\lambda_{i}+N-i}(q^{-b_{\ell}})]_{i,\ell=1}^{N}=0$. On the other
hand, since $b_{i}\neq b_{j}$ the denominator
$\prod_{i<j}(q^{-b_{i}}-q^{-b_{j}})$ does not vanish. ∎

######  Corollary 10.9.

If $\mu$ is another partition then we can define $b_{i}=\mu_{i}+N-i$, and
conclude that $F_{\lambda}(q^{-\mu_{i}-N+i})=0$ unless
$\mu_{i}\geq\lambda_{i}$ for all $i$, that is, partition $\mu$ contains
$\lambda$.

######  Example 10.10.

Suppose that $\lambda=(1)$, then $F_{(1)}$ is a symmetric function of degree 1
with leading term $(-1)^{1+\binom{N}{2}}q^{D_{N}(1)}s_{(1)}=q^{D_{N}(1)}\sum
x_{i}$. We have $D_{N}(1)=N-1+\binom{N}{3}$, so
$F_{(1)}(x_{1},\ldots,x_{N})=(-1)^{1+\binom{N}{2}}q^{N-1+\binom{N}{3}}\sum
x_{i}+c.$ To find the constant $c$, we observe that by Corollary 10.9 we get
$F_{(1)}(q^{-N+1},q^{-N+2},\ldots,1)=0$, so

$c=(-1)^{\binom{N}{2}}q^{N-1+\binom{N}{3}}(q^{-N+1}+q^{-N+2}+\ldots+1)=(-1)^{\binom{N}{2}}q^{\binom{N}{3}}[N]_{q}.$

######  Lemma 10.11.

We have

$F_{\lambda}(q^{-\lambda_{i}-N+i})=(-1)^{\binom{N}{2}}q^{n(\lambda)+\binom{N}{3}}\prod_{\square\in\lambda}(1-q^{-h(\square)}),$

where $h(\square)$ is the hook length of a box $\square$ in the Young diagram
corresponding to $\lambda$.

###### Proof.

Since the sequence $\lambda_{i}+N-i$ is strictly decreasing, we have
$f_{\lambda_{j}+N-j}(q^{-\lambda_{i}-N+i})=0$ for $j>i$ and

$f_{\lambda_{i}+N-i}(q^{-\lambda_{i}-N+i})=\\{\lambda_{i}+N-i\\}_{q^{-1}}!$

and

$\det(f_{\lambda_{j}+N-j}(q^{-\lambda_{i}-N+i}))=\prod_{i}\\{\lambda_{i}+N-i\\}_{q^{-1}}!\
.$

On the other hand,

$\prod_{i<j}(q^{-\lambda_{i}-N+i}-q^{-\lambda_{j}-N+j})=(-1)^{\binom{N}{2}}q^{-\sum(\lambda_{j}+N-j)(j-1)}\prod_{i<j}(1-q^{-\lambda_{i}+i+\lambda_{j}-j}).$

and the statement follows now from formula (5) and the identity

$\sum(\lambda_{j}+N-j)(j-1)=n(\lambda)+\binom{N}{3}.$

∎

######  Example 10.12.

For arbitrary $N$ and $\lambda=(1)$ we computed in Example 10.10 that

$F_{(1)}=(-1)^{1+\binom{N}{2}}q^{\binom{N}{3}}(q^{N-1}(x_{1}+\ldots+x_{N})-[N]_{q}).$

Hence,

$F_{(1)}(q^{-N},q^{-N+2},\ldots,1)=q^{\binom{N}{3}}(q^{N-1}(q^{-N}+q^{-N+2}+\ldots+1)-[N]_{q})=$
$(-1)^{1+\binom{N}{2}}q^{\binom{N}{3}}(q^{-1}-1)=(-1)^{\binom{N}{2}}q^{\binom{N}{3}}(1-q^{-1}).$

We summarize the above results in the following proposition:

######  Proposition 10.13.

[29] There exists a unique collection of nonhomogeneous symmetric polynomials
$F_{\lambda}(x_{1},\ldots,x_{N})$ with the following properties:

  * •

$F_{\lambda}(x_{1},\ldots,x_{N})$ has degree $|\lambda|$.

  * •

$F_{\lambda}(q^{-\mu_{i}-N+i})=0$ for all partitions $\mu$ not containing
$\lambda$.

  * •

$F_{\lambda}(q^{-\lambda_{i}-N+i})=(-1)^{\binom{N}{2}}q^{n(\lambda)+\binom{N}{3}}\prod_{\square\in\lambda}(1-q^{-h(\square)}).$

We will denote the value
$F_{\lambda}(q^{-\lambda_{i}-N+i})=(-1)^{\binom{N}{2}}q^{n(\lambda)+\binom{N}{3}}\prod_{\square\in\lambda}(1-q^{-h(\square)})$
by $c_{\lambda,\lambda}$.

######  Lemma 10.14.

Suppose that $q$ is a root of unity. Then $c_{\lambda,\lambda}$ vanishes for
all but finitely many partitions $\lambda$.

###### Proof.

Observe that $\prod_{\square\in\lambda}(1-q^{-h(\square)})$ is divisible by
$\prod_{i}[\lambda_{i}-\lambda_{i+1}]_{q}!$ and

$\sum_{i=1}^{N}i(\lambda_{i}-\lambda_{i+1})=|\lambda|.$

This means that for some $i$ we must have

$i(\lambda_{i}-\lambda_{i+1})\geq\frac{|\lambda|}{N},\
\lambda_{i}-\lambda_{i+1}\geq\frac{|\lambda|}{iN}\geq\frac{|\lambda|}{N^{2}},$

and $c_{\lambda,\lambda}$ is divisible by
$(1-q)\cdots(1-q^{\lfloor\frac{|\lambda|}{N^{2}}\rfloor})$. If $q^{s}=1$ then
it vanishes for $|\lambda|\geq sN^{2}$. ∎

######  Remark 10.15.

A partition is called an $s$-core if none of its hook lengths is divisible by
$s$. The $s$-core partitions play an important role in representation theory
of symmetric groups in finite characteristic, and of Hecke algebras at roots
of unity [21]. If $q^{s}=1$ then clearly $c_{\lambda,\lambda}(q)\neq 0$ if and
only if $\lambda$ is an $s$-core. Although there are infinitely many
$s$-cores, Lemma 10.14 shows that there are finitely many $s$-cores with at
most $N$ rows.

For example, for $s=2$ the $2$-cores are “staircase partitions”
$\lambda=(k,k-1,\ldots,1)$, and the maximal $2$-core with at most $N$ rows has
size $N+(N-1)+\ldots+1=\binom{N+1}{2}$.

###  10.3. Multi-variable case: interpolation

One can use the polynomials $F_{\lambda}$ to solve the following interpolation
problem.

######  Problem 10.16.

Find a symmetric function $f=\sum a_{\lambda}F_{\lambda}$ given its values
$f(q^{-\mu_{i}-N+i})$ for all $\mu$.

We have

$f(q^{-\mu_{i}-N+i})=\sum a_{\lambda}F_{\lambda}(q^{-\mu_{i}-N+i})$

This is a linear system on $a_{\lambda}$ with the triangular matrix

(36) ${\sf C}=\left[c_{\lambda,\mu}\right]_{\lambda,\mu},\
c_{\lambda,\mu}(q):=F_{\lambda}(q^{-\mu_{i}-N+i})$

It is clear from Proposition 10.13 that to find $a_{\lambda}$ for a given
$\lambda$ it is sufficient to know all coefficients $c_{\mu,\nu}$ for
$\mu\subset\nu\subset\lambda$.

In [29] Okounkov computed the inverse matrix $D={\sf C}^{-1}$ which allows one
to explicitly compute the coefficients $a_{\lambda}$.

######  Theorem 10.17.

[29] Define $c^{*}_{\lambda,\mu}(q)=c_{\lambda,\mu}(q^{-1})$ and
$\operatorname{cont}(\lambda)=n(\lambda)-n(\lambda^{\prime})$. Then

$D=\left[d_{\lambda,\mu}\right]_{\lambda,\mu},\
d_{\lambda,\mu}=(-1)^{|\mu|-|\lambda|}q^{\operatorname{cont}(\lambda)-\operatorname{cont}(\mu)}\frac{c^{*}_{\lambda,\mu}}{c_{\mu,\mu}c^{*}_{\lambda,\lambda}}$

and

$a_{\mu}=\sum_{\lambda\subset\mu}d_{\lambda,\mu}f(q^{-\lambda_{i}-N+i})=\frac{1}{c_{\mu,\mu}}\sum_{\lambda\subset\mu}(-1)^{|\mu|-|\lambda|}q^{\operatorname{cont}(\lambda)-\operatorname{cont}(\mu)}\frac{c^{*}_{\lambda,\mu}}{c^{*}_{\lambda,\lambda}}f(q^{-\lambda_{i}-N+i}).$

######  Example 10.18.

If $\lambda=\mu$ then clearly $d_{\lambda,\mu}=\frac{1}{c_{\lambda,\lambda}}$.

######  Example 10.19.

We have $F_{(\emptyset)}=(-1)^{\binom{N}{2}}q^{\binom{N}{3}}$, so

$c_{(\emptyset),(\emptyset)}=c_{(\emptyset),(1)}=(-1)^{\binom{N}{2}}q^{\binom{N}{3}},\quad
c^{*}_{(\emptyset),(\emptyset)}=c^{*}_{(\emptyset),(1)}=(-1)^{\binom{N}{2}}q^{-\binom{N}{3}}.$

Since

$c_{(1),(1)}=(-1)^{\binom{N}{2}}q^{\binom{N}{3}}(1-q^{-1})=(-1)^{\binom{N}{2}+1}q^{\binom{N}{3}-1}(1-q),$

we get

$d_{(\emptyset),(1)}=\frac{(-1)^{\binom{N}{2}}q^{-\binom{N}{3}+1}}{(1-q)}.$

So the first two terms of interpolation series have the following form:

$f(x_{1},\ldots,x_{N})=(-1)^{\binom{N}{2}}q^{-\binom{N}{3}}f(q^{1-N},q^{2-N},\ldots,1)F_{(\emptyset)}(\mathbf{x})+\\\
\frac{(-1)^{\binom{N}{2}+1}q^{-\binom{N}{3}+1}}{1-q}\left[-f(q^{1-N},q^{2-N},\ldots,1)+f(q^{-N},q^{2-N},\ldots,1)\right]F_{(1)}(\mathbf{x})+\ldots$

######  Example 10.20.

For $N=1$ and $a\geq b$ we have

$c_{(b),(a)}=f_{b}(q^{-a})=(1-q^{-a})\cdots(1-q^{-a+b-1})$

hence

$c^{*}_{(b),(a)}=(1-q^{a})\cdots(1-q^{a-b+1})$

Now

$\frac{c^{*}_{(b),(a)}}{c^{*}_{(b),(b)}}=\frac{(1-q^{a})\cdots(1-q^{a-b+1})}{(1-q^{b})\cdots(1-q)}=\binom{a}{b}_{q},$

and

$d_{(b),(a)}=(-1)^{a-b}q^{\frac{b(b-1)}{2}-\frac{a(a-1)}{2}}\frac{c^{*}_{(b),(a)}}{c_{(a),(a)}c^{*}_{(b),(b)}}=\frac{(-1)^{a-b}}{c_{(a),(a)}}q^{\frac{b(b-1)}{2}-\frac{a(a-1)}{2}}\binom{a}{b}_{q},$

which matches (32).

######  Example 10.21.

Let $N=2$, $\lambda=(1)$ and $\mu=(3,2)$. We have
$F_{\lambda}=q(x_{1}+x_{2})-(1+q)$, so

$c_{\lambda,\mu}=F_{\lambda}(q^{-4},q^{-2})=(-q-1+q^{-1}+q^{-3}),\
c^{*}_{\lambda,\mu}=q^{3}+q-1-q^{-1},\\\ $

and using Lemma 10.11

$c_{\lambda,\lambda}=-(1-q^{-1}),\ c^{*}_{\lambda,\lambda}=-(1-q),$
$c_{\mu,\mu}=-q^{2}(1-q^{-1})^{2}(1-q^{-2})(1-q^{-3})(1-q^{-4})=q^{-9}(1-q)^{2}(1-q^{2})(1-q^{3})(1-q^{4}).$

Now

$d_{\lambda,\mu}=q^{-2}\frac{c^{*}_{\lambda,\mu}}{c_{\mu,\mu}c^{*}_{\lambda,\lambda}}=-q^{6}\frac{q^{4}+q^{2}-q-1}{(1-q)^{3}(1-q^{2})(1-q^{3})(1-q^{4})}.$

###  10.4. Hopf pairing

We have a symmetric bilinear form $(\cdot,\cdot)$ on
$\mathbb{Z}[x_{1},\ldots,x_{N}]^{S_{N}}$ defined by its values on Schur
polynomials

$(s_{\lambda},s_{\mu})=s_{\lambda}(q^{-\mu_{1}-N+1},\ldots,q^{-\mu_{N}})s_{\mu}(q^{-N+1},\ldots,1).$

It is closely related to the Hopf pairing $\langle\cdot,\cdot\rangle$ for
$\mathcal{R}=\mathit{Rep}(\mathcal{U})$ defined in Section 5.2. Note that

$(f,s_{\mu})=f(q^{-\mu_{1}-N+1},\ldots,q^{-\mu_{N}})s_{\mu}(q^{-N+1},\ldots,1).$

for any symmetric function $f$.

######  Proposition 10.22.

We have

(37)
$(F_{\lambda},F_{\nu})=\delta_{\lambda,\nu}q^{-|\lambda|+2\binom{N}{3}}\prod_{\square\in\lambda}(1-q^{N+c(\square)}),$

so the Hopf pairing is diagonal in the basis $\\{F_{\lambda}\\}_{\lambda}$.

###### Proof.

We have

$(F_{\lambda},s_{\mu})=F_{\lambda}(q^{-\mu_{1}-N+1},\ldots,q^{-\mu_{N}})s_{\mu}(q^{-N+1},\ldots,1)=0$

unless $\lambda\subset\mu$. On the other hand, $F_{\nu}$ can be expanded in
$s_{\mu}$ for $\mu\preceq\nu$, so $(F_{\lambda},F_{\nu})$ vanishes unless
there exists $\mu\preceq\nu$ such that $\lambda\subset\mu$, in particular,
$\lambda\preceq\nu$.

Since the Hopf pairing is symmetric, $(F_{\lambda},F_{\nu})$ vanishes unless
$\lambda\preceq\nu$ and $\nu\preceq\lambda$, so $\lambda=\nu$. Finally,

$(F_{\lambda},F_{\lambda})=(-1)^{|\lambda|+\binom{N}{2}}q^{D_{N}(\lambda)}(F_{\lambda},s_{\lambda})=(-1)^{|\lambda|+\binom{N}{2}}q^{D_{N}(\lambda)}F_{\lambda}(q^{-\lambda_{1}-N+1},\ldots,q^{-\lambda_{N}})s_{\lambda}(q^{-N+1},\ldots,1).$

Now

$F_{\lambda}(q^{-\lambda_{1}-N+1},\ldots,q^{-\lambda_{N}})=(-1)^{\binom{N}{2}}q^{n(\lambda)+\binom{N}{3}}\prod_{\square\in\lambda}(1-q^{-h(\square)})$

while

$s_{\lambda}(q^{-N+1},\ldots,1)=q^{-n(\lambda)}\prod_{\square\in\lambda}\frac{(1-q^{-N-c(\square)})}{(1-q^{-h(\square)})}.$

hence

$F_{\lambda}(q^{-\lambda_{1}-N+1},\ldots,q^{-\lambda_{N}})s_{\lambda}(q^{-N+1},\ldots,1)=(-1)^{\binom{N}{2}}q^{\binom{N}{3}}\prod_{\square\in\lambda}(1-q^{-N-c(\square)})=$
$(-1)^{|\lambda|+\binom{N}{2}}q^{-N|\lambda|-c(\lambda)+\binom{N}{3}}(1-q^{N+c(\square)}).$

On the other hand, $D_{N}(\lambda)=c(\lambda)+(N-1)|\lambda|+\binom{N}{3}$. ∎

This provides us with a different perspective for the interpolation problem.
Suppose that we have a Schur expansion for $F_{\lambda}$:

$F_{\lambda}=\sum_{\mu\preceq\lambda}b_{\lambda,\mu}s_{\mu}.$

Then for an arbitrary symmetric function $f(x_{1},\ldots,x_{N})$ we can write

$f=\sum_{\lambda}\frac{(f,F_{\lambda})}{(F_{\lambda},F_{\lambda})}F_{\lambda}=\sum_{\lambda}\sum_{\mu\preceq\lambda}b_{\lambda,\mu}\frac{(f,s_{\mu})}{(F_{\lambda},F_{\lambda})}F_{\lambda}=\sum_{\lambda}\sum_{\mu\preceq\lambda}\frac{b_{\lambda,\mu}s_{\mu}(q^{-N-i})}{(F_{\lambda},F_{\lambda})}f(q^{-\mu_{i}-N+i})F_{\lambda},$

and the interpolation coefficient is equal to

(38)
$d_{\lambda,\mu}=\frac{b_{\lambda,\mu}s_{\mu}(q^{-N+i})}{(F_{\lambda},F_{\lambda})}.$

######  Example 10.23.

For $N=2$ and $\lambda=(3,2)$ we have

$F_{(3,2)}=q^{2}(1-x_{1})(1-qx_{1})(1-x_{2})(1-qx_{2})(q^{3}(x_{1}+x_{2})-(1+q))=\\\
q^{7}s_{3,2}-q^{6}(1+q)s_{3,1}-q^{4}(1+q+q^{2}+q^{3})s_{2,2}+q^{6}s_{3,0}+q^{3}(1+q+q^{2}+q^{3})(1+q)s_{2,1}-\\\
q^{3}(1+q+q^{2}+q^{3})s_{2,0}-q^{2}(1+q+q^{2}+q^{3})(1+q)s_{1,1}+(q^{5}+q^{4}+2q^{3}+q^{2})s_{1,0}-(q^{3}+q^{2}).$

Also

$(F_{3,2},F_{3,2})=-q^{-5}(1-q^{4})(1-q^{3})(1-q^{2})^{2}(1-q)$

Therefore the interpolation coefficient for $\lambda=(3,2)$ and $\mu=(1,0)$
equals

$d_{(3,2),(1,0)}=(q^{5}+q^{4}+2q^{3}+q^{2})\frac{s_{1,0}(q^{-1},1)}{(F_{3,2},F_{3,2})}=$
$-\frac{(q^{5}+q^{4}+2q^{3}+q^{2})(1+q^{-1})}{q^{-5}(1-q^{4})(1-q^{3})(1-q^{2})^{2}(1-q)}=-\frac{q^{6}(q^{4}+q^{2}-q-1)}{(1-q^{4})(1-q^{3})(1-q^{2})(1-q)^{3}}.$

This agrees with Example 10.21.

###  10.5. Divisibility

Given a polynomial $f(x)$, define

$\partial_{xy}(f):=\frac{f(x)-f(y)}{x-y}.$

Observe that

$\partial_{xy}(fg)=\frac{f(x)-f(y)}{x-y}g(x)+f(y)\frac{g(x)-g(y)}{x-y}=\partial_{xy}f\cdot
g(x)+f(y)\cdot\partial_{xy}(g).$

More generally, we have

(39) $\displaystyle\partial_{xy}(f_{1}\cdots f_{k})$
$\displaystyle=\partial_{xy}(f_{1})f_{2}(x)\cdots
f_{k}(x)+f_{1}(y)\partial_{xy}(f_{2})f_{3}(x)\cdots f_{k}(x)+\ldots$
$\displaystyle+f_{1}(y)f_{2}(y)\cdots\partial_{xy}(f_{k}).$

######  Example 10.24.

For $f_{n}(x)=(1-x)\cdots(1-q^{n-1}x)$, note that
$\partial_{xy}(1-q^{i}x)=-q^{i}$, so we get

$\displaystyle F_{n,0}(x,y)$
$\displaystyle=\partial_{x,y}f_{n+1}(x)=\sum_{i=0}^{n}(1-y)\cdots(1-q^{i-1}y)[\partial_{x,y}(1-q^{i}x)](1-q^{i+1}x)\cdots(1-q^{n}x)$
$\displaystyle=\sum_{i=0}^{n}f_{i}(y)\cdot(-q^{i})f_{n-i}(q^{i+1}x).$

######  Example 10.25.

For example,

$F_{1,0}(x,y)=q(x+y)-(1+q)=q(y-1)+(qx-1)=-[qf_{1}(y)+f_{1}(qx)].$

Similarly,

$\displaystyle F_{2,0}(x,y)$
$\displaystyle=-q^{3}(x^{2}+xy+y^{2})+(q+q^{2}+q^{3})(x+y)-(1+q+q^{2})$
$\displaystyle=-[(1-qx)(1-q^{2}x)+q(1-q^{2}x)(1-y)+q^{2}(1-y)(1-qy)]$
$\displaystyle=-[f_{2}(qx)+qf_{1}(x)f_{2}(y)+q^{2}f_{2}(y)].$

######  Corollary 10.26.

For all integers $a$ and $b$ the value $F_{n,0}(q^{a},q^{b})$ is divisible by
$\left(\left\lfloor\frac{n}{2}\right\rfloor\right)_{q}!$

###### Proof.

Let $k=\left\lfloor\frac{n}{2}\right\rfloor$. In the above equation either
$i\geq k$ or $n-i\geq k$, so each term in the sum is either divisible by
$f_{k}(q^{i+1+a})$ or by $f_{k}(q^{b})$, so by $q$-binomial theorem it is
divisible by $(k)_{q}!$ ∎

More generally, let $\partial_{i}=\partial_{x_{i},x_{i+1}}$ then it is well
known that $\partial_{i}$ satisfy braid relations, so one can define
$\partial_{w}$ for any permutation $w$. Furthermore,

$F_{\lambda}(x_{1},\ldots,x_{N})=\partial_{w_{0}}[f_{\lambda_{1}+N-1}(x_{1})\cdots
f_{\lambda_{N}}(x_{N})],$

where $w_{0}=(N\ N-1\ \ldots 1)$ is the longest element in $S_{N}$.

######  Lemma 10.27.

For all $\lambda$ one can write $F_{\lambda}(x_{1},\ldots,x_{N})$ as the sum
where each term has the form

(40) $f_{j_{1}}(q^{s_{1}}x_{m_{1}})\cdots f_{j_{d}}(q^{s_{d}}x_{m_{d}}),\
\text{where}\ j_{1}+\ldots+j_{d}=|\lambda|\ \text{and}\ d=\binom{N+1}{2}.$

Here the indices $m_{i}$ might repeat arbitrarily.

###### Proof.

From (39) and Example 10.24 it is clear that $\partial_{i}$ applied to a
product (40) with $\ell$ factors produces a sum of similar products with
$\ell+1$ factors. We start from a product of $N$ factors, and $\partial_{w}$
is a composition of $\binom{N}{2}$ operators $\partial_{i}$, so the terms in
the resulting sum have $N+\binom{N}{2}=\binom{N+1}{2}$ factors. Also, each
$\partial_{i}$ decreases the degree by 1, so

$j_{1}+\ldots+j_{d}=\sum(\lambda_{i}+N-i)-\binom{N}{2}=|\lambda|.$

∎

######  Remark 10.28.

A more careful analysis of this proof leads to a combinatorial formula for
$F_{\lambda}$ where the terms are labeled by semistandard tableaux, but we do
not need it here. This is a $q$-analogue of the expansion of a Schur function
in the monomial basis.

######  Lemma 10.29.

For any sequence of integers $a_{1},\ldots,a_{N}$ the value
$F_{\lambda}(q^{a_{1}},\ldots,q^{a_{N}})$ is divisible by $(k)_{q}!$ where
$k=\left\lfloor\frac{|\lambda|}{\binom{N+1}{2}}\right\rfloor$.

###### Proof.

In each term (40) there are $d=\binom{N+1}{2}$ indices $j_{1},\cdots,j_{d}$
which add up to $|\lambda|$, so at least one of these indices is greater than
$|\lambda|/d$. It remains to notice that $f_{j}(q^{a})$ is divisible by
$(q)_{j}!$ for all integers $a$. ∎

The following lemma gives a rough description of the expansion

(41)
$F_{\lambda}(x_{1},\ldots,x_{N})=\sum_{m_{1},\ldots,m_{N}}b_{m_{1},\ldots,m_{k}}f_{m_{1}}(x_{1})\cdots
f_{m_{N}}(x_{N}).$

of the symmetric interpolation polynomial $F_{\lambda}$ in terms of
nonsymmetric ones.

######  Lemma 10.30.

Given $k$, for sufficiently large $|\lambda|$ for all terms of the expansion
(41) either the coefficient $b_{m_{1},\ldots,m_{k}}$ is divisible by
$(k)_{q}!$ or there exists $m_{i}\geq k$ for some $1\leq i\leq N$.

###### Proof.

We follow the same logic as in Lemma 10.29. For $|\lambda|>2k\binom{N+1}{2}$
every term (40) is divisible by $f_{2k}(q^{s}x_{i})$ for some $s$ and $i$. By
Lemma 10.5 this can be further decomposed into terms which are divisible by
$(j)_{q}!f_{2k-j}(x_{i})$, and either $j$ or $2k-j$ is greater than or equal
to $k$. Overall, we presented

$F_{\lambda}(x_{1},\ldots,x_{N})=A(k)_{q}!+\sum B_{i}f_{k}(x_{i})$

for some polynomials $A$ and $B_{i}$. It remains to notice that the polynomial
$B_{i}f_{k}(x_{i})$ can be presented as the sum of $f_{m_{1}}(x_{1})\cdots
f_{m_{N}}(x_{N})$ where $m_{i}\geq k$. ∎

##  11\. Stability of interpolation and the case $N=2$

###  11.1. Stability of interpolation matrices

In this section study the dependence of the interpolation polynomials on $N$.

As above, if partition $\lambda$ has less than $N$ parts we can complete it
with zeroes. We denote by $F_{\lambda;N}(x_{1},\ldots,x_{N})$ the
corresponding polynomial in $N$ variables.

######  Lemma 11.1.

Let $\lambda$ be a partition with at most $N$ parts. Then

$F_{\lambda;N}(x_{1},\ldots,x_{N-1},1)=\begin{cases}(-1)^{N-1}q^{\binom{N-1}{2}}F_{\lambda;N-1}(qx_{1},\ldots,qx_{N-1})&\text{if}\
\lambda_{N}=0\\\ 0&\text{otherwise}.\end{cases}$

###### Proof.

Let $\mu$ be a partition with at most $N-1$ parts. Then by Proposition 10.13

$F_{\lambda;N}(q^{-\mu_{1}-N+1},\ldots,q^{-\mu_{N-1}-1},1)=0$

unless $\mu$ contains $\lambda$. If $\lambda_{N}>0$ then this never happens
and $F_{\lambda;N}(x_{1},\ldots,x_{N-1},1)=0$. If $\lambda_{N}=0$ we write
$L(x_{1},\ldots,x_{N-1})=F_{\lambda;N-1}(qx_{1},\ldots,qx_{N-1})$. We have

$L(q^{-\mu_{1}-N+1},\ldots,q^{-\mu_{N-1}-1})=F_{\lambda;N-1}(q^{-\mu_{1}-(N-1)+1},\ldots,q^{-\mu_{N-1}})$

which vanishes unless $\mu$ contains $\lambda$, so by Proposition 10.13
$F_{\lambda;N}(x_{1},\ldots,x_{N-1},1)$ is proportional to
$L(x_{1},\ldots,x_{N-1})$. Finally, at $\mu=\lambda$ we can use Lemma 10.11 to
determine the coefficient. ∎

######  Remark 11.2.

We can also prove the lemma using the explicit determinantal formula. Indeed,
$f_{\lambda_{i}+N-i}(1)=0$ unless $f_{\lambda_{i}+N-i}=0$ which is equivalent
to $i=N$ and $\lambda_{N}=0$. Therefore for $\lambda_{N}\neq 0$ the last row
in the matrix $f_{\lambda_{i}+N-i}(x_{j})$ vanishes (where $x_{N}=1$), and
$F_{\lambda;N}(x_{1},\ldots,x_{N-1},1)=0$. For $\lambda_{N}=0$ we have

$F_{\lambda;N}(x_{1},\ldots,x_{N-1},1)=\frac{\det\left[f_{\lambda_{i}+N-i}(x_{j})\right]_{i,j=1}^{N-1}}{\prod_{i<j\leq
N-1}(x_{i}-x_{j})\prod_{i\leq N-1}(x_{i}-1)}.$

Note that $f_{k+1}(x)=(1-x)f_{k}(qx)$, so

$f_{\lambda_{i}+N-i}(x_{j})=(1-x_{j})f_{\lambda_{i}+(N-1)-i}(qx_{j})$

Therefore

$F_{\lambda;N}(x_{1},\ldots,x_{N-1},1)=\frac{\prod_{i=1}^{n}(1-x_{i})\det\left[f_{\lambda_{i}+(N-1)-i}(qx_{j})\right]_{i,j=1}^{N-1}}{\prod_{i<j\leq
N-1}(x_{i}-x_{j})\prod_{i\leq N-1}(x_{i}-1)}=$
$(-1)^{N-1}q^{\binom{N-1}{2}}F_{\lambda;N-1}(qx_{1},\ldots,qx_{N-1}).$

######  Corollary 11.3.

Let $c_{\lambda,\mu}^{(N)}$ be the coefficient defined in previous section for
symmetric functions in $N$ variables. Then the expressions

$(-1)^{\binom{N}{2}}q^{-\binom{N}{3}}c^{(N)}_{\lambda,\mu},(-1)^{\binom{N}{2}}q^{\binom{N}{3}}c^{(N)*}_{\lambda,\mu},(-1)^{\binom{N}{2}}q^{\binom{N}{3}}d^{(N)}_{\lambda,\mu}$

are independent of $N$ (provided that $\lambda$ and $\mu$ have at most $N$
parts).

######  Example 11.4.

For one-row partitions $\lambda=(b)$ and $\mu=(a)$ the interpolation
coefficients are given by the formulas in Example 10.20 up to a monomial
factor.

The above results allow us to describe Schur expansion of interpolation
polynomials:

######  Proposition 11.5.

We have

(42)
$F_{\lambda}^{(N)}=(-1)^{\binom{N}{2}}q^{\binom{N}{3}}\sum_{\mu\subset\lambda}\overline{b_{\lambda,\mu}}A^{|\mu|}\prod_{\square\in\lambda\setminus\mu}(1-Aq^{c(\square)})s_{\lambda}^{(N)}$

where $A=q^{N}$ and the coefficients

$\overline{b_{\lambda,\mu}}=(-1)^{\binom{N}{2}}q^{\binom{N}{3}}d^{(N)}_{\lambda,\mu}q^{-|\lambda|-|\mu|-n(\mu)}\prod_{\square\in\mu}(1-q^{h(\square)})$

do not depend on $N$.

###### Proof.

It follows from (38) that

$F_{\lambda}=\sum b_{\lambda,\mu}s_{\mu},\
b_{\lambda,\mu}=\frac{d_{\lambda,\mu}(F_{\lambda},F_{\lambda})}{s_{\mu}(q^{-N+i})}.$

Since $d_{\lambda,\mu}$ vanishes unless $\mu\subset\lambda$, the same is true
for $b_{\lambda,\mu}$. By Corollary 11.3 the product
$\overline{d_{\lambda,\mu}}=(-1)^{\binom{N}{2}}q^{\binom{N}{3}}d_{\lambda,\mu}$
does not depend on $N$, and we can use the formulas

$(F_{\lambda},F_{\lambda})=q^{-|\lambda|+2\binom{N}{3}}\prod_{\square\in\lambda}(1-Aq^{c(\square)}),$
$s_{\mu}(q^{-N+i})=q^{n(\mu)-(N-1)|\mu|}\prod_{\square\in\mu}\frac{(1-Aq^{c(\square)})}{(1-q^{h(\square)})}.$

to write

$b_{\lambda,\mu}=(-1)^{\binom{N}{2}}q^{-\binom{N}{3}}\overline{d_{\lambda,\mu}}\cdot
q^{-|\lambda|+2\binom{N}{3}-n(\mu)+N|\mu|-|\mu|}\prod_{\square\in\lambda}(1-Aq^{c(\square)})\prod_{\square\in\mu}\frac{(1-q^{h(\square)})}{(1-Aq^{c(\square)})}.$

The result follows. ∎

######  Corollary 11.6.

The one-row interpolation polynomials have the following Schur expansion:

$F_{(m)}^{(N)}=(-1)^{\binom{N}{2}}q^{\binom{N}{3}}\sum_{\mu\subset\lambda}(-1)^{j}q^{\frac{j(j-3)}{2}}A^{j}\frac{(1-Aq^{j})\cdots(1-Aq^{m-1})}{(1-q)\cdots(1-q^{m-j})}h_{j}^{(N)}=$
$(-1)^{\binom{N}{2}}q^{\binom{N}{3}}\sum_{\mu\subset\lambda}(-1)^{j}q^{\frac{j(j-3+2N)}{2}}\binom{N+m-1}{m-j}_{q}h_{j}^{(N)}.$

Here $h_{j}^{(N)}=s_{(j)}^{(N)}$ are complete symmetric functions in $N$
variables.

###### Proof.

For $\lambda=(m)$ and $N=1$ we have

$f_{m}(x)=\sum_{j=0}^{m}(-1)^{j}q^{\frac{j(j-1)}{2}}\binom{m}{j}_{q}x^{j}.$

By writing $A=q$ and $\mu=(j)$ we get

$A^{|\mu|}\prod_{\square\in\lambda\setminus\mu}(1-Aq^{c(\square)})=q^{j}(1-q^{j+1})\cdots(1-q^{m}),$

so

$\overline{b_{(m),(j)}}=(-1)^{j}\frac{q^{\frac{j(j-3)}{2}}}{(1-q)\cdots(1-q^{m-j})}.$

∎

######  Remark 11.7.

The HOMFLY-PT limit of interpolation polynomials in Proposition 11.5 appears
to be related to the results and conjectures in [22], it would be interesting
to find a precise connection.

###  11.2. Adding a column

It is well known that in symmetric functions in $N$ variables one has the
identity

$s_{\lambda+1^{N}}=x_{1}\cdots x_{N}\cdot s_{\lambda}.$

Here $\lambda+1^{N}=(\lambda_{1}+1,\ldots,\lambda_{N}+1)$ and the
corresponding Young diagram is obtained from the Young diagram for $\lambda$
by adding a vertical column.

For interpolation polynomials we have two different generalizations of this
identity: the first relates $F_{\lambda+1^{N}}$ to $F_{\lambda}$ and the
second describe the action of the multiplication by $x_{1}\cdots x_{N}$.

######  Proposition 11.8.

We have
$F_{\lambda+1^{N}}(x_{1},\ldots,x_{N})=q^{\binom{N}{2}}\prod_{i=1}^{N}(1-x_{i})F_{\lambda}(qx_{1},\ldots,qx_{N})$.
More generally,

(43)
$F_{\lambda+k^{N}}(x_{1},\ldots,x_{N})=q^{k\binom{N}{2}}\prod_{i=1}^{N}f_{k}(x_{i})F_{\lambda}(q^{k}x_{1},\ldots,q^{k}x_{N}).$

###### Proof.

We have $f_{m+1}(x)=(1-x)f_{m}(qx)$, therefore

$\det\left[f_{\lambda_{i}+1+N-i}(x_{j})\right]=\det\left[(1-x_{j})f_{\lambda_{i}+N-i}(qx_{j})\right]=\prod_{j=1}^{N}(1-x_{j})\det\left[f_{\lambda_{i}+N-i}(x_{j})\right].$

Since each factor $(x_{i}-x_{j})$ in the denominator gets multiplied by $q$
after changing $x_{i}\to qx_{i}$, this implies the first equation. Now (43)
can be obtained by applying it $k$ times. ∎

Let $e_{i}$ denote the $i$-th basic vector in $\mathbb{Z}^{N}$ with $1$ at
$i$-th position and $0$ at other positions. Given $I\subset\\{1,\ldots,n\\}$,
we define $e_{I}=\sum_{i\in I}e_{i}$.

######  Proposition 11.9.

We have

$x_{1}\cdots
x_{N}F_{\lambda}(x_{1},\ldots,x_{N})=q^{-|\lambda|-\binom{N}{2}}\sum_{I\subset\\{1,\ldots,n\\}}(-1)^{|I|}F_{\lambda+e_{I}}(x_{1},\ldots,x_{N}).$

Here we use the convention that $F_{\lambda+e_{I}}=0$ unless the entries of
$\lambda+e_{I}$ are non-increasing (that is, $\lambda+e_{I}$ is a partition).

###### Proof.

We have $f_{m+1}(x)=f_{m}(x)(1-q^{m}x)$, so

$xf_{m}(x)=q^{-m}(f_{m}(x)-f_{m+1}(x)).$

Therefore

$x_{1}\cdots
x_{N}\det\left[f_{\lambda_{i}+N-i}(x_{j})\right]=\det\left[x_{j}f_{\lambda_{i}+N-i}(x_{j})\right]=$
$\det\left[q^{-\lambda_{i}-N+i}(f_{\lambda_{i}+N-i}(x_{j})-f_{\lambda_{i}+1+N-i}(x_{j}))\right].$

∎

######  Corollary 11.10.

Consider the completion of the space of symmetric functions with coefficients
in $\mathbb{Z}[q,q^{-1}]$ with respect to the basis $F_{\lambda}$. Then the
operator of multiplication by $x_{1}\cdots x_{N}$ is invertible in this
completion and its inverse is given by the equation

$(x_{1}\cdots
x_{N})^{-1}F_{\lambda}(x_{1},\ldots,x_{N})=q^{\binom{N}{2}}\sum_{v\in\mathbb{Z}_{\geq
0}^{N}}q^{|\lambda|+v}F_{\lambda+v}(x_{1},\ldots,x_{N}).$

###### Proof.

Define the operators $A_{i}$ by $A_{i}(F_{\lambda})=F_{\lambda+e_{i}}$, and
$p_{i}(F_{\lambda})=q^{\lambda_{i}}F_{\lambda}$ for $i=1,\ldots,N$. Clearly,
$[A_{i},A_{j}]=[p_{i},p_{j}]=[A_{i},p_{j}]$ for $i\neq j$ and by Proposition
11.9 we have

$x_{1}\cdots x_{N}=q^{-\binom{N}{2}}\prod_{i}(1-A_{i})p^{-1}_{i},$

hence

$(x_{1}\cdots
x_{N})^{-1}=q^{\binom{N}{2}}\prod_{i}p_{i}(1+A_{i}+A_{i}^{2}+\ldots).$

∎

######  Example 11.11.

For $N=1$ and $\lambda=(0)$ we get a curious identity

$x^{-1}=\sum_{m=0}^{\infty}f_{m}(x)q^{m}$

We can check this identity directly, by computing the values of both sides at
$q^{-j}$ for all $j$. Denote

$u_{j}=\sum_{m=0}^{\infty}f_{m}(q^{-j})q^{m}=\sum_{m=0}^{j}f_{m}(q^{-j})q^{m}.$

Then $u_{j+1}=1+q(1-q^{-j-1})u_{j}$ and $u_{0}=1$, so it is easy to see that
$u_{j}=q^{j}$.

###  11.3. Interpolation polynomials for $\mathfrak{gl}_{2}$

In this subsection we describe the interpolation polynomials for
$\mathfrak{gl}_{2}$ explicitly. By definition, we have polynomials
$F_{\lambda}(x_{1},x_{2})$ where $\lambda_{1}\geq\lambda_{2}$:

$F_{\lambda_{1},\lambda_{2}}(x_{1},x_{2})=\frac{1}{x_{1}-x_{2}}\left|\begin{matrix}f_{\lambda_{1}+1}(x_{1})&f_{\lambda_{1}+1}(x_{2})\\\
f_{\lambda_{2}}(x_{1})&f_{\lambda_{2}}(x_{2})\end{matrix}\right|$

Let us consider the case $\lambda_{2}=0$ first, and write $\lambda_{1}=k$.
Then

$F_{k,0}(x_{1},x_{2})=\frac{1}{x_{1}-x_{2}}\det\left|\begin{matrix}f_{k+1}(x_{1})&f_{k+1}(x_{2})\\\
1&1\end{matrix}\right|=\frac{f_{k+1}(x_{1})-f_{k+1}(x_{2})}{x_{1}-x_{2}}.$

Let

$h_{i}(x_{1},x_{2})=\frac{x_{1}^{i+1}-x_{2}^{i+1}}{x_{1}-x_{2}}.$

Recall that
$f_{k+1}(x)=\sum_{j=0}^{k+1}(-1)^{j}q^{\frac{j(j-1)}{2}}\binom{k+1}{j}_{q}x^{j}$,
so

$F_{k,0}(x_{1},x_{2})=\sum_{j=1}^{k+1}(-1)^{j}q^{\frac{j(j-1)}{2}}\binom{k+1}{j}_{q}h_{j-1}(x_{1},x_{2}),$

compare with Corollary 11.6. We just replace each $x^{j}$ in the expression
for $f_{k+1}(x)$ by $h_{j-1}(x_{1},x_{2})$.

######  Example 11.12.

We have

$f_{1}(x)=1-x,\ f_{2}(x)=(1-x)(1-qx)=1-(1+q)x+qx^{2},$
$f_{3}(x)=(1-x)(1-qx)(1-q^{2}x)=1-(1+q+q^{2})x+(q+q^{2}+q^{3})x^{2}-q^{3}x^{3}$

so

$F_{0,0}(x_{1},x_{2})=-1,\ F_{1,0}(x_{1},x_{2})=q(x_{1}+x_{2})-(1+q),\ $
$F_{2,0}=-q^{3}(x_{1}^{2}+x_{1}x_{2}+x_{2}^{2})+(q+q^{2}+q^{3})(x_{1}+x_{2})-(1+q+q^{2}).$

By Proposition 11.8 we have

$F_{\lambda_{1},\lambda_{2}}(x_{1},x_{2})=q^{\lambda_{2}}f_{\lambda_{2}}(x_{1})f_{\lambda_{2}}(x_{2})F_{\lambda_{1}-\lambda_{2},0}(q^{\lambda_{2}}x_{1},q^{\lambda_{2}}x_{2}).$

In particular, for $(\lambda_{1},\lambda_{2})=(k,k)$ we have

$F_{k,k}(x_{1},x_{2})=q^{k}f_{k}(x_{1})f_{k}(x_{2}).$

Also, by Lemma 11.1 we get

(44)
$F_{\lambda_{1},\lambda_{2}}(x_{1},1)=\begin{cases}-f_{\lambda_{1}}(qx_{1})&\text{if}\
\lambda_{2}=0\\\ 0&\text{otherwise}.\end{cases}$

###  11.4. Interpolation tables for $\mathfrak{gl}_{2}$

For the reader’s convenience, we have computed the polynomials
$F_{\lambda}(x_{1},x_{2})$ and the corresponding interpolation matrices using
Sage [35]. First, we present $F_{\lambda}$ in Schur basis:

$F_{0}=-1,\quad F_{1}=qs_{1}-(q+1),\quad
F_{2}=-q^{3}s_{2}+(q^{3}+q^{2}+q)s_{1}-(q^{2}+q+1),\\\
F_{1,1}=-qs_{1,1}+qs_{1}-q=-q(1-x_{1})(1-x_{2})\\\
F_{3}=q^{6}s_{3}-(q^{6}+q^{5}+q^{4}+q^{3})s_{2}+(q^{5}+q^{4}+2q^{3}+q^{2}+q)s_{1}-(q^{3}+q^{2}+q+1)\\\
F_{2,1}=q^{3}s_{2,1}-q^{3}s_{2}-(q^{3}+q^{2}+q)s_{1,1}+(q^{3}+q^{2}+q)s_{1}-(q^{2}+q)\\\
F_{3,1}=-q^{6}s_{3,1}+q^{6}s_{3}+(q^{6}+q^{5}+q^{4}+q^{3})s_{2,1}-(q^{6}+q^{5}+q^{4}+q^{3})s_{2}-\\\
(q^{5}+q^{4}+2q^{3}+q^{2}+q)s_{1,1}+(q^{5}+q^{4}+2q^{3}+q^{2}+q)s_{1}-(q^{3}+q^{2}+q)\\\
F_{2,2}=-q^{4}s_{2,2}+(q^{4}+q^{3})s_{2,1}-q^{3}s_{2}-(q^{4}+q^{3}+q^{2})s_{1,1}+(q^{3}+q^{2})s_{1}-q^{2}\\\
F_{3,2}=q^{7}s_{3,2}-(q^{7}+q^{6})s_{3,1}-(q^{7}+q^{6}+q^{5}+q^{4})s_{2,2}+q^{6}s_{3}+(q^{7}+2q^{6}+2q^{5}+2q^{4}+q^{3})s_{2,1}-\\\
(q^{6}+q^{5}+q^{4}+q^{3})s_{2}-(q^{6}+2q^{5}+2q^{4}+2q^{3}+q^{2})s_{1,1}+(q^{5}+q^{4}+2q^{3}+q^{2})s_{1}-(q^{3}+q^{2})\\\
F_{3,3}=-q^{9}s_{3,3}+(q^{9}+q^{8}+q^{7})s_{3,2}-(q^{8}+q^{7}+q^{6})s_{3,1}-(q^{9}+q^{8}+2q^{7}+q^{6}+q^{5})s_{2,2}+q^{6}s_{3}+\\\
(q^{8}+2q^{7}+2q^{6}+2q^{5}+q^{4})s_{2,1}-(q^{6}+q^{5}+q^{4})s_{2}-(q^{7}+q^{6}+2q^{5}+q^{4}+q^{3})s_{1,1}+(q^{5}+q^{4}+q^{3})s_{1}-q^{3}$

Next, we list the values of the evaluations
$c_{\lambda,\mu}=F_{\lambda}(q^{-\mu_{1}-1},q^{-\mu_{2}})$ for various
$\lambda$ and $\mu$ in Tables 1, 2, 3 below. The resulting matrix ${\sf
C}=(c_{\lambda,\mu})$ is upper-triangular, with diagonal entries prescribed by
Lemma 10.11. Zero entries correspond to pairs $(\lambda,\mu)$ where $\mu$ does
not contain $\lambda$. The entry corresponding to $(\lambda,\mu)=((1),(3,2))$
is marked in bold, it is divisible by $1-q$ but does not factor any further.

Using either Theorem 10.17 or equation (38), one can easily reconstruct the
inverse matrix $D={\sf C}^{-1}$, and we list part of it in Table 4 (see
Examples 10.21 and 10.23 for more computations).

Note that by Corollary 11.3 this determines the coefficients $c_{\lambda,\mu}$
and $d_{\lambda,\mu}$ for $\lambda\subset\mu\subset(3,3)$ and arbitrary $N$.

###  11.5. Link invariants for $\mathfrak{gl}_{2}$

We can use the interpolation tables to expand the invariants of simple knots
in the basis $F_{\lambda}$. Indeed, the colored $\mathfrak{gl}_{2}$ invariants
are determined by the colored $\mathfrak{sl}_{2}$ invariants (that is, colored
Jones polynomial) by the formula

$J_{K}(V(\lambda_{1},\lambda_{2}),q)=J_{K}(V_{\lambda_{1}-\lambda_{2}},q).$

The coefficients $a_{\lambda}(K)$ are then determined by Theorem 1.3

$a_{\lambda}(K)=\sum_{\mu\subset\lambda}d_{\lambda,\mu}(q^{-1})J_{K}(V(\mu),q).$

For example, for the figure eight knot we have the following values of the
colored Jones polynomial:

$J_{K}(V_{0},q)=1=J_{K}(V({1,1}),q),\
J_{K}(V_{1},q)=J_{K}(V({2,1}),q)=1+q^{2}+q^{-2}-q-q^{-1},$
$J_{K}(V_{2},q)=1+q^{3}+q^{-3}-q-q^{-1}+(q^{3}+q^{-3}-q-q^{-1})(q^{3}+q^{-3}-q^{2}-q^{-2}).$

Using the values of $d_{\lambda,\mu}$ from Table 4 (and changing $q$ to
$q^{-1}$) we obtain

$a_{0}(K)=-J_{K}(V_{0},q)=-1,\
a_{1}(K)=-\frac{q^{-1}}{1-q^{-1}}J_{K}(V_{0},q)+\frac{q^{-1}}{1-q^{-1}}J_{K}(V_{1},q)=q^{-2}(q^{3}-1),$
$a_{2}(K)=-\frac{q^{-2}}{(1-q^{-1})(1-q^{-2})}J_{K}(V_{0},q)+\frac{q^{-2}}{(1-q^{-1})^{2}}J_{K}(V_{1},q)-\\\
\frac{q^{-3}}{(1-q^{-1})(1-q^{-2})}J_{K}(V_{2},q)=q^{-6}(-q^{9}+q^{5}+q^{4}-q^{3}-1),$
$a_{1,1}(K)=-\frac{q^{-3}}{(1-q^{-1})(1-q^{-2})}J_{K}(V_{0},q)+\frac{q^{-2}}{(1-q^{-1})^{2}}J_{K}(V_{1},q)-\\\
\frac{q^{-2}}{(1-q^{-1})(1-q^{-2})}J_{K}(V({1,1}),q)=q^{-2}(q^{2}+q+1),$
$a_{2,1}(K)=-\frac{q^{-4}}{(1-q^{-1})^{2}(1-q^{-3})}J_{K}(V_{0},q)+\frac{q^{-3}}{(1-q^{-1})^{3}}J_{K}(V_{1},q)-\\\
\frac{q^{-4}}{(1-q^{-1})^{2}(1-q^{-2})}J_{K}(V_{2},q)-\frac{q^{-3}}{(1-q^{-1})^{2}(1-q^{-2})}J_{K}(V({1,1}),q)+\\\
\frac{q^{-4}}{(1-q^{-1})^{2}(1-q^{-3})}J_{K}(V({2,1}),q)=q^{-6}(-q^{8}-q^{7}-q^{6}-q^{5}+q^{4}+2q^{3}+q^{2}+q+1).$

Using Tables 1, 2, 3, one can similarly compute the values of $a_{\lambda}(K)$
for all $\lambda\subset(3,3)$ and verify that these are indeed Laurent
polynomials in $q$.

$\lambda\backslash\mu$ | (0) | (1) | (2) | (1,1)
---|---|---|---|---
(0) | $-1$ | $-1$ | $-1$ | $-1$
(1) | $0$ | $q^{-1}(1-q)$ | $q^{-2}(1-q^{2})$ | $q^{-1}(1-q^{2})$
(2) | $0$ | $0$ | $-q^{-3}(1-q)(1-q^{2})$ | $0$
(1,1) | $0$ | $0$ | $0$ | $-q^{-2}(1-q)(1-q^{2})$
(3) | $0$ | $0$ | $0$ | $0$
(2,1) | $0$ | $0$ | $0$ | $0$
(3,1) | $0$ | $0$ | $0$ | $0$
(2,2) | $0$ | $0$ | $0$ | $0$
(3,2) | $0$ | $0$ | $0$ | $0$
(3,3) | $0$ | $0$ | $0$ | $0$
Table 1. Evaluations of interpolation polynomials: matrix
$C=(c_{\lambda,\mu})$

$\lambda\backslash\mu$ | (3) | (2,1) | (3,1)
---|---|---|---
(0) | $-1$ | $-1$ | $-1$
(1) | $q^{-3}(1-q^{3})$ | $q^{-2}(1-q^{3})$ | $q^{-3}(1-q^{4})$
(2) | $-q^{-5}(1-q^{2})(1-q^{3})$ | $-q^{-3}(1-q)(1-q^{3})$ | $-q^{-5}(1-q^{2})(1-q^{4})$
(1,1) | $0$ | $-q^{-3}(1-q)(1-q^{3})$ | $-q^{-4}(1-q)(1-q^{4})$
(3) | $q^{-6}(1-q)(1-q^{2})(1-q^{3})$ | $0$ | $q^{-6}(1-q)(1-q^{2})(1-q^{4})$
(2,1) | $0$ | $q^{-4}(1-q)^{2}(1-q^{3})$ | $q^{-6}(1-q)(1-q^{2})(1-q^{4})$
(3,1) | $0$ | $0$ | $-q^{-7}(1-q)^{2}(1-q^{2})(1-q^{4})$
(2,2) | $0$ | $0$ | $0$
(3,2) | $0$ | $0$ | $0$
(3,3) | $0$ | $0$ | $0$

Table 2.  Matrix $C=(c_{\lambda,\mu})$, continued

$\lambda\backslash\mu$ | (2,2) | (3,2) | (3,3)
---|---|---|---
(0) | $-1$ | $-1$ | $-1$
(1) | $q^{-2}(1+q)(1-q^{2})$ | $\mathbf{q^{-3}(-q^{4}-q^{3}+q^{2}+1)}$ | $q^{-3}(1+q)(1-q^{3})$
(2) | $-q^{3}(1-q^{2})(1-q^{3})$ | $-q^{-5}(1-q^{3})(1-q^{4})$ | $-q^{-5}(1+q)(1-q^{3})^{2}$
(1,1) | $-q^{-4}(1-q^{2})(1-q^{3})$ | $-q^{-5}(1-q^{2})(1-q^{4})$ | $-q^{-6}(1-q^{3})(1-q^{4})$
(3) | $0$ | $q^{-6}(1-q)(1-q^{3})(1-q^{4})$ | $q^{-6}(1-q^{2})(1-q^{3})(1-q^{4})$
(2,1) | $q^{-5}(1-q^{2})^{2}(1-q^{3})$ | $q^{-7}(1-q^{2})(1-q^{3})(1-q^{4})$ | $q^{-8}(1+q)(1-q^{2})(1-q^{3})(1-q^{4})$
(3,1) | $0$ | $-q^{-8}(1-q)(1-q^{2})(1-q^{3})(1-q^{4})$ | $-q^{-9}(1-q^{2})(1-q^{3})^{2}(1-q^{4})$
(2,2) | $-q^{-6}(1-q)(1-q^{2})^{2}(1-q^{3})$ | $-q^{-8}(1-q)(1-q^{2})(1-q^{3})(1-q^{4})$ | $-q^{-10}(1-q^{2})(1-q^{3})^{2}(1-q^{4})$
(3,2) | $0$ | $q^{-9}(1-q)^{2}(1-q^{2})(1-q^{3})(1-q^{4})$ | $q^{-11}(1-q^{2})^{2}(1-q^{3})^{2}(1-q^{4})$
(3,3) | $0$ | $0$ | $-q^{-12}(1-q)(1-q^{2})^{2}(1-q^{3})^{2}(1-q^{4})$

Table 3. Matrix $C=(c_{\lambda,\mu})$, continued $\lambda\backslash\mu$ | (0) | (1) | (2) | (1,1) | (2,1)
---|---|---|---|---|---
(0) | $-1$ | $-\frac{q}{1-q}$ | $-\frac{q^{2}}{(1-q)(1-q^{2})}$ | $-\frac{q^{3}}{(1-q)(1-q^{2})}$ | $-\frac{q^{4}}{(1-q)^{2}(1-q^{3})}$
(1) | $0$ | $\frac{q}{1-q}$ | $\frac{q^{2}}{(1-q)^{2}}$ | $\frac{q^{2}}{(1-q)^{2}}$ | $\frac{q^{3}}{(1-q)^{3}}$
(2) | $0$ | $0$ | $-\frac{q^{3}}{(1-q)(1-q^{2})}$ | $0$ | $-\frac{q^{4}}{(1-q)^{2}(1-q^{2})}$
(1,1) | $0$ | $0$ | $0$ | $-\frac{q^{2}}{(1-q)(1-q^{2})}$ | $-\frac{q^{3}}{(1-q)^{2}(1-q^{2})}$
(2,1) | $0$ | $0$ | $0$ | $0$ | $\frac{q^{4}}{(1-q)^{2}(1-q^{3})}$
Table 4. Interpolation matrix $D=(d_{\lambda,\mu})=C^{-1}$

##  12\. Appendix

Here we collect some useful definitions and facts about Habiro’s ring and
interpolation Macdonald polynomials.

###  12.1. Habiro’s ring

The Habiro ring [17] is defined as

$\widehat{\mathbb{Z}[q]}:={\lim\limits_{\overleftarrow{\hskip 5.69054ptn\hskip
5.69054pt}}}\;\frac{\mathbb{Z}[q]}{((q;q)_{n})}$