A $q$ -analogue of symmetric multiple zeta value

Yoshihiro Takeyama Department of Mathematics, Institute of Pure and Applied Sciences, University of Tsukuba, Tsukuba, Ibaraki 305-8571, Japan [email protected]

Abstract.

We construct a $q$ -analogue of truncated version of symmetric multiple zeta values which satisfies the double shuffle relation. Using it, we define a $q$ -analogue of symmetric multiple zeta values and see that it satisfies many of the same relations as symmetric multiple zeta values, which are the inverse relation and a part of the double shuffle relation and the Ohno-type relation.

This work was supported by JSPS KAKENHI Grant Number 22K03243.
This is a pre-print of an article published in The Ramanujan Journal. The final authenticated version is available online at: https://doi.org/10.1007/s11139-023-00755-9.

1. Introduction

The multiple zeta value (MZV) is the real value defined by

(1.1)

\displaystyle\zeta(\mathbf{k})=\sum_{0<m_{1}<\cdots<m_{r}}\frac{1}{m_{1}^{k_{1}}\cdots m_{r}^{k_{r}}}

for a tuple of positive integers $\mathbf{k}=(k_{1},\ldots,k_{r})$ with $k_{r}\geq 2$ . We set $\zeta(\varnothing)=1$ and regard it as a MZV. We denote by $\mathcal{Z}$ the $\mathbb{Q}$ -linear subspace of $\mathbb{R}$ spanned by the MZVs. It is known that $\mathcal{Z}$ forms a $\mathbb{Q}$ -algebra with respect to the usual multiplication on $\mathbb{R}$ .

Kaneko and Zagier introduced two kinds of variants of MZVs called finite multiple zeta values and symmetric multiple zeta values (see, e.g., [5]). The symmetric multiple zeta value (SMZV) is defined as an element of the quotient $\mathcal{Z}/\zeta(2)\mathcal{Z}$ . The purpose of this paper is to construct a $q$ -analogue of the SMZVs which shares many of the relations among them.

The SMZV is defined as follows. Although the infinite sum (1.1) diverges if $k_{r}=1$ , we have two kinds of its regularization, which are called the harmonic regularized MZV $\zeta^{\ast}(\mathbf{k})$ and the shuffle regularized MZV $\zeta^{\mathcyr{sh}}(\mathbf{k})$ (see Section 2.2 for the details). If $k_{r}\geq 2$ , they are equal to the MZV $\zeta(\mathbf{k})$ . Using them we set

\displaystyle\zeta^{\mathcal{S},\bullet}(k_{1},\ldots,k_{r})=\sum_{i=0}^{r}(-1)^{k_{i+1}+\cdots+k_{r}}\zeta^{\bullet}(k_{1},k_{2},\ldots,k_{i})\zeta^{\bullet}(k_{r},k_{r-1},\ldots,k_{i+1})

for $\bullet\in\{\ast,\mathcyr{sh}\}$ . It is known that the difference $\zeta^{\ast}(\mathbf{k})-\zeta^{\mathcyr{sh}}(\mathbf{k})$ belongs to the ideal $\zeta(2)\mathcal{Z}$ for any tuple $\mathbf{k}$ of positive integers. The SMZV $\zeta^{\mathcal{S}}(\mathbf{k})$ is defined by

\displaystyle\zeta^{\mathcal{S}}(\mathbf{k})=\zeta^{\mathcal{S},\bullet}(\mathbf{k})\qquad\hbox{mod}\,\,\zeta(2)\mathcal{Z}

as an element of the quotient $\mathcal{Z}/\zeta(2)\mathcal{Z}$ .

Now we fix a complex parameter $q$ satisfying $0<|q|<1$ and define the $q$ -integer $[n]$ for $n\geq 1$ by $[n]=(1-q^{n})/(1-q)$ . There are various models of a $q$ -analogue of the MZV (see, e.g., [11, Chapter 12]). Most of them are of the following form:

(1.2)

\displaystyle\sum_{0<m_{1}<\cdots<m_{r}}\frac{P_{1}(q^{m_{1}})\cdots P_{r}(q^{m_{r}})}{[m_{1}]^{k_{1}}\cdots[m_{r}]^{k_{r}}},

where $P_{j}(x)$ is a polynomial whose degree is less than or equal to $k_{j}$ for $1\leq j\leq r$ . If $P_{r}(0)=0$ , the infinite sum (1.2) is absolutely convergent. In this paper we call a value of the form (1.2) a $q$ -analogue of the MZV ( $q$ MZV) without specifying the model.

To construct a $q$ -analogue of SMZVs, it would be natural to consider two kinds of regularizations of the sum (1.2) with $P_{r}(0)\not=0$ which turn into the harmonic regularized MZV and the shuffle regularized MZV in the limit as $q\to 1$ . However, there does not seem to exist any standard definition of a shuffle regularization of the sum (1.2).

To avoid the difficulty, we construct a $q$ -analogue of SMZVs in a different approach. In [6], Ono, Seki and Yamamoto introduced two kinds of truncations of the $t$ -adic symmetric multiple zeta value ( $t$ -adic SMZV), which is an element of the formal power series ring $(\mathcal{Z}/\zeta(2)\mathcal{Z})[[t]]$ whose constant term is equal to the SMZV. As a corollary, we have the following expression for $\zeta^{\mathcal{S},\bullet}(\mathbf{k})$ . Set

(1.3)		$\displaystyle\zeta_{M}^{\mathcal{S},\ast}(k_{1},\ldots,k_{r})$	$\displaystyle=\sum_{i=0}^{r}(-1)^{k_{i+1}+\cdots+k_{r}}\sum_{\begin{subarray}{c}0<m_{1}<\cdots<m_{i}<M\\ 0<m_{r}<\cdots<m_{i+1}<M\end{subarray}}\frac{1}{m_{1}^{k_{1}}\cdots m_{r}^{k_{r}}},$
(1.4)		$\displaystyle\zeta_{M}^{\mathcal{S},\mathcyr{sh}}(k_{1},\ldots,k_{r})$	$\displaystyle=\sum_{i=0}^{r}(-1)^{k_{i+1}+\cdots+k_{r}}\sum_{\begin{subarray}{c}0<m_{1}<\cdots<m_{i}\\ 0<m_{r}<\cdots<m_{i+1}\\ m_{i}+m_{i+1}<M\end{subarray}}\frac{1}{m_{1}^{k_{1}}\cdots m_{r}^{k_{r}}}.$

Then we have $\zeta^{\mathcal{S},\bullet}(\mathbf{k})=\lim_{M\to\infty}\zeta_{M}^{\mathcal{S},\bullet}(\mathbf{k})$ for $\bullet\in\{\ast,\mathcyr{sh}\}$ [6, Corollary 2.7]. Note that we do not need any regularization here. Hence we could follow the above procedure to construct a $q$ -analogue of SMZVs. It is the main theme of this paper.

One important example of the relations among SMZVs is the double shuffle relation (see Theorem 2.4 below) due to Kaneko and Zagier. Jarossay proved that the $t$ -adic SMZVs satisfy a generalization of the double shuffle relation [4]. The point is that the truncated version of the $t$ -adic SMZVs also satisfies them [6, Theorem 1.6]. As a corollary we see that the truncated version of SMZVs (1.3) and (1.4) satisfies the same double shuffle relation as SMZVs.

In this paper we first construct a $q$ -analogue of the truncated SMZVs (1.3) and (1.4) which satisfies the double shuffle relation (Proposition 4.1 and Proposition 4.2). Second, we define a $q$ -analogue of SMZVs as a limit of the above truncated version. As SMZVs are defined to be the elements of the quotient $\mathcal{Z}/\zeta(2)\mathcal{Z}$ , our $q$ -analogue of SMZVs is an element of the quotient of the space of the $q$ MZVs by a sum of two ideals $\mathcal{N}_{q}+\mathcal{P}_{q}$ . Roughly speaking, $\mathcal{N}_{q}$ is the ideal generated by the $q$ MZVs which turn into zero in the limit as $q\to 1-0$ , and $\mathcal{P}_{q}$ is the ideal generated by the values $\sum_{m>0}q^{2km}/[m]^{2k}$ with $k\geq 1$ , which turns into $\zeta(2k)$ in the limit as $q\to 1-0$ . Hence the ideal $\mathcal{N}_{q}+\mathcal{P}_{q}$ could be regarded as a $q$ -analogue of to the ideal $\zeta(2)\mathcal{Z}$ .

Unlike the truncated SMZVs, the double shuffle relation of the truncated $q$ SMZVs does not imply that of the $q$ SMZVs. It is because not all of the truncated $q$ SMZVs converge in the limit as $M\to\infty$ and the space spanned by them which converge is not closed under the shuffle product (see Example 6.3 below). Hence our $q$ -analogue of SMZVs satisfies only a part of the double shuffle relation of SMZVs. At this stage the author does not know how to overcome this point.

The paper is organized as follows. In Section 2 we review on SMZV and its truncated version and the relations of them including the double shuffle relation. Section 3 gives some preliminaries on $q$ MZVs. In Section 4 we define a $q$ -analogue of the truncated SMZVs and prove the double shuffle relation. The proof is quite similar to that for the truncated SMZVs given in [6]. In Section 5 we construct a $q$ -analogue of SMZVs. In Section 6 we see that our $q$ -analogue shares many of the relations with SMZVs, which are the reversal relation and a part of the double shuffle relation and the Ohno-type relation due to Oyama [7]. Additionally two appendices follow. In Appendix A we discuss the asymptotic behavior of $q$ MZVs in the limit as $q\to 1-0$ to prove Proposition 3.4 below. Appendix B provides proofs of technical propositions.

Here we give notation used throughout. For a tuple of non-negative integers $\mathbf{k}=(k_{1},\ldots,k_{r})$ , we define its weight $\mathrm{wt}(\mathbf{k})$ and depth $\mathrm{dep}(\mathbf{k})$ by $\mathrm{wt}(\mathbf{k})=\sum_{i=1}^{r}k_{i}$ and $\mathrm{dep}(\mathbf{k})=r$ , respectively. We call a tuple of positive integers an index. We regard the empty set $\varnothing$ as an index whose weight and depth are zero. An index $\mathbf{k}=(k_{1},\ldots,k_{r})$ is said to be admissible if $\mathbf{k}=\varnothing$ , or $r\geq 1$ and $k_{r}\geq 2$ . We denote the set of indices (resp. admissible indices) by $I$ (resp. $I_{0}$ ). The reversal $\overline{\mathbf{k}}$ of an index $\mathbf{k}=(k_{1},k_{2},\ldots,k_{r})$ is defined by $\overline{\mathbf{k}}=(k_{r},\ldots,k_{2},k_{1})$ . For the empty index we set $\overline{\varnothing}=\varnothing$ .

In this paper we often make use of generating functions in proofs. Then we use the operations defined below without mention. Suppose that $\mathfrak{A}$ is a unital algebra over a commutative ring $\mathcal{C}$ . Then we extend the addition $+$ and the multiplication $\cdot$ on $\mathfrak{A}$ to the formal power series ring $\mathfrak{A}[[X]]$ by $f(X)+g(X)=\sum_{j\geq 0}(a_{j}+b_{j})X^{j}$ and $f(X)\cdot g(X)=\sum_{j,l\geq 0}(a_{j}\cdot b_{l})X^{j+l}$ for $f(X)=\sum_{j\geq 0}a_{j}X^{j}$ and $g(X)=\sum_{l\geq 0}b_{l}X^{l}$ with $a_{j},b_{l}\in\mathfrak{A}$ , respectively. Similarly, we extend a $\mathcal{C}$ -linear map $\varphi:\mathfrak{A}\rightarrow\mathfrak{B}$ to the $\mathcal{C}$ -linear map $\varphi:\mathfrak{A}[[X]]\rightarrow\mathfrak{B}[[X]]$ by $\varphi(f(X))=\sum_{j\geq 0}\varphi(a_{j})X^{j}$ for $f(X)=\sum_{j\geq 0}a_{j}X^{j}$ . We adopt the above convention for the formal power series ring of several variables.

2. Symmetric multiple zeta value

2.1. Multiple zeta value

For an index $\mathbf{k}=(k_{1},\ldots,k_{r})$ and a positive integer $M$ , we define the truncated multiple zeta value $\zeta_{M}(\mathbf{k})$ by

\displaystyle\zeta_{M}(\mathbf{k})=\sum_{0<m_{1}<\cdots<m_{r}<M}\frac{1}{m_{1}^{k_{1}}\cdots m_{r}^{k_{r}}}

if $\mathbf{k}$ is not empty, and $\zeta_{M}(\varnothing)=1$ for the empty index. If $1\leq M\leq\mathrm{dep}(\mathbf{k})$ , we set $\zeta_{M}(\mathbf{k})=0$ .

For an admissible index $\mathbf{k}$ , we define the multiple zeta value (MZV) $\zeta(\mathbf{k})$ by $\zeta(\mathbf{k})=\lim_{M\to\infty}\zeta_{M}(\mathbf{k})$ . If $\mathbf{k}=(k_{1},\ldots,k_{r})$ is a non-empty admissible index, we have

\displaystyle\zeta(\mathbf{k})=\sum_{0<m_{1}<\cdots<m_{r}}\frac{1}{m_{1}^{k_{1}}\cdots m_{r}^{k_{r}}},

which converges since $k_{r}\geq 2$ .

Let $\mathfrak{h}=\mathbb{Q}\langle x,y\rangle$ be the non-commutative polynomial ring of two variables $x$ and $y$ over $\mathbb{Q}$ . We set $z_{k}=yx^{k-1}$ for $k\geq 1$ , and $z_{\mathbf{k}}=z_{k_{1}}\cdots z_{k_{r}}$ for a non-empty index $\mathbf{k}=(k_{1},\ldots,k_{r})$ . For the empty index we set $z_{\varnothing}=1$ .

Set $\mathfrak{h}^{1}=\mathbb{Q}+y\mathfrak{h}$ . It is a $\mathbb{Q}$ -subalgebra of $\mathfrak{h}$ , which can be identified with the non-commutative polynomial ring over $\mathbb{Q}$ with the set of variables $\{z_{k}\mid k\geq 1\}$ . We also set $\mathfrak{h}^{0}=\mathbb{Q}+y\mathfrak{h}x$ , which is a $\mathbb{Q}$ -submodule of $\mathfrak{h}^{1}$ with a basis $\{z_{\mathbf{k}}\mid\mathbf{k}\in I_{0}\}$ .

The harmonic product $\ast$ on $\mathfrak{h}^{1}$ is the $\mathbb{Q}$ -bilinear map $\ast:\mathfrak{h}^{1}\times\mathfrak{h}^{1}\longrightarrow\mathfrak{h}^{1}$ defined by the following properties:

(i)

For any $w\in\mathfrak{h}^{1}$ , it holds that $w\ast 1=w$ and $1\ast w=w$ .
(ii)

For any $w,w^{\prime}\in\mathfrak{h}^{1}$ and $k,l\geq 1$ , it holds that $(wz_{k})\ast(w^{\prime}z_{l})=(w\ast w^{\prime}z_{l})z_{k}+(wz_{k}\ast w^{\prime})z_{l}+(w\ast w^{\prime})z_{k+l}$ .

The shuffle product $\mathcyr{sh}:\mathfrak{h}\times\mathfrak{h}\longrightarrow\mathfrak{h}$ is similarly defined by the following properties and $\mathbb{Q}$ -linearity:

(i)

For any $w\in\mathfrak{h}$ , it holds that $w\,\mathcyr{sh}\,1=w$ and $1\,\mathcyr{sh}\,w=w$ .
(ii)

For any $w,w^{\prime}\in\mathfrak{h}$ and $u,v\in\{x,y\}$ , it holds that $(wu)\,\mathcyr{sh}\,(w^{\prime}v)=(w\,\mathcyr{sh}\,w^{\prime}v)u+(wu\,\mathcyr{sh}\,w^{\prime})v$ .

Then $\mathfrak{h}^{1}$ (resp. $\mathfrak{h}$ ) becomes a commutative $\mathbb{Q}$ -algebra with respect to the harmonic (resp. shuffle) product, which we denote by $\mathfrak{h}^{1}_{\ast}$ (resp. $\mathfrak{h}_{\mathcyr{sh}}$ ). We see that $\mathfrak{h}^{0}$ is a $\mathbb{Q}$ -subalgebra of $\mathfrak{h}^{1}_{\ast}$ with respect to the harmonic product. We denote it by $\mathfrak{h}^{0}_{\ast}$ . We also see that $\mathfrak{h}^{1}$ and $\mathfrak{h}^{0}$ are $\mathbb{Q}$ -subalgebras of $\mathfrak{h}_{\mathcyr{sh}}$ with respect to the shuffle product. We denote them by $\mathfrak{h}^{1}_{\mathcyr{sh}}$ and $\mathfrak{h}^{0}_{\mathcyr{sh}}$ , respectively.

For a positive integer $M$ , we define the $\mathbb{Q}$ -linear map $Z_{M}:\mathfrak{h}^{1}\to\mathbb{Q}$ by $Z_{M}(z_{\mathbf{k}})=\zeta_{M}(\mathbf{k})$ for an index $\mathbf{k}$ . Similarly, we define the $\mathbb{Q}$ -linear map $Z:\mathfrak{h}^{0}\to\mathbb{R}$ by $Z(z_{\mathbf{k}})=\zeta(\mathbf{k})$ , for an admissible index $\mathbf{k}$ .

Proposition 2.1.

(i)

For any $w,w^{\prime}\in\mathfrak{h}^{1}$ and $M\geq 1$ , it holds that

$\displaystyle Z_{M}(w\ast w^{\prime})=Z_{M}(w)Z_{M}(w^{\prime}).$

Hence, if $w$ and $w^{\prime}$ belong to $\mathfrak{h}^{0}$ , we have

(2.1) $\displaystyle Z(w\ast w^{\prime})=Z(w)Z(w^{\prime}).$
(ii)

For any $w,w^{\prime}\in\mathfrak{h}^{0}$ , it holds that

(2.2) $\displaystyle Z(w\,\mathcyr{sh}\,w^{\prime})=Z(w)Z(w^{\prime}).$

The relations (2.1) and (2.2) are called the finite double shuffle relation. They imply that the map $Z$ is a $\mathbb{Q}$ -algebra homomorphism to $\mathbb{R}$ with respect to both the harmonic and shuffle product. We denote the image of $Z$ by $\mathcal{Z}$ , which is a $\mathbb{Q}$ -subalgebra of $\mathbb{R}$ .

2.2. Symmetric multiple zeta value

It is known that $\mathfrak{h}^{1}_{\bullet}\simeq\mathfrak{h}^{0}_{\bullet}[y]$ for $\bullet\in\{\ast,\mathcyr{sh}\}$ (see [2, 8]). Hence, for $\bullet\in\{\ast,\mathcyr{sh}\}$ , there uniquely exists the $\mathbb{Q}$ -algebra homomorphism $Z^{\bullet}$ from $\mathfrak{h}^{1}_{\bullet}$ to the polynomial ring $\mathcal{Z}[T]$ such that $Z^{\bullet}(y)=T$ . For an index $\mathbf{k}$ , we define $\zeta^{\ast}(\mathbf{k})=Z^{\ast}(z_{\mathbf{k}})|_{T=0}$ and $\zeta^{\mathcyr{sh}}(\mathbf{k})=Z^{\mathcyr{sh}}(z_{\mathbf{k}})|_{T=0}$ . It is known that

(2.3)

\displaystyle\sum_{s\geq 0}X^{s}\zeta^{\ast}(\mathbf{k},\underbrace{1,\ldots,1}_{s})=\exp{(\sum_{n\geq 2}\frac{(-1)^{n-1}}{n}\zeta(n)X^{n})}\sum_{s\geq 0}X^{s}\zeta^{\mathcyr{sh}}(\mathbf{k},\underbrace{1,\ldots,1}_{s})

in the formal power series ring $\mathcal{Z}[[X]]$ for any index $\mathbf{k}$ [3, Theorem 1 and Proposition 10].

For an index $\mathbf{k}=(k_{1},\ldots,k_{r})$ and $\bullet\in\{\ast,\mathcyr{sh}\}$ , we set

\displaystyle\zeta^{\mathcal{S},\bullet}(\mathbf{k})=\sum_{i=0}^{r}(-1)^{k_{i+1}+\cdots+k_{r}}\zeta^{\bullet}(k_{1},k_{2},\ldots,k_{i})\zeta^{\bullet}(k_{r},k_{r-1},\ldots,k_{i+1}).

From (2.3), we see that the difference $\zeta^{\mathcal{S},\ast}(\mathbf{k})-\zeta^{\mathcal{S},\mathcyr{sh}}(\mathbf{k})$ belongs to the ideal $\zeta(2)\mathcal{Z}=\pi^{2}\mathcal{Z}$ of $\mathcal{Z}$ . The symmetric MZV (SMZV) $\zeta_{\mathcal{S}}(\mathbf{k})$ is defined by

\displaystyle\zeta^{\mathcal{S}}(\mathbf{k})=\zeta^{\mathcal{S},\bullet}(\mathbf{k})\,\,\hbox{mod}\,\,\pi^{2}\mathcal{Z}

as an element of $\mathcal{Z}/\zeta(2)\mathcal{Z}$ .

2.3. Truncated symmetric multiple zeta value

We define the $\mathbb{Q}$ -algebra anti-automorphism $\psi$ on $\mathfrak{h}^{1}\simeq\mathbb{Q}\langle z_{1},z_{2},\ldots\rangle$ by $\psi(z_{k})=(-1)^{k}z_{k}$ for $k\geq 1$ . For $\bullet\in\{\ast,\mathcyr{sh}\}$ , we define the $\mathbb{Q}$ -linear map $w^{\mathcal{S},\bullet}:\mathfrak{h}^{1}\to\mathfrak{h}^{1}$ by $w^{\mathcal{S},\bullet}(1)=1$ and

\displaystyle w^{\mathcal{S},\bullet}(u_{1}\cdots u_{r})=\sum_{i=0}^{r}u_{1}\cdots u_{i}\bullet\psi(u_{i+1}\cdots u_{r})

for $r\geq 1$ and $u_{1},\ldots,u_{r}\in\{z_{k}\}_{k\geq 1}$ .

For a positive integer $M$ and $\bullet\in\{\ast,\mathcyr{sh}\}$ , we define the $\mathbb{Q}$ -linear map $Z^{\mathcal{S},\bullet}_{M}:\mathfrak{h}^{1}\to\mathbb{Q}$ by $Z^{\mathcal{S},\bullet}_{M}=Z_{M}\circ w_{\mathcal{S}}^{\bullet}$ . For an index $\mathbf{k}=(k_{1},\ldots,k_{r})$ and a positive integer $M$ , the value $Z^{\mathcal{S},\bullet}_{M}(z_{\mathbf{k}})$ $(\bullet\in\{\ast,\mathcyr{sh}\})$ is expressed as follows:

	$\displaystyle Z_{M}^{\mathcal{S},\ast}(z_{\mathbf{k}})$	$\displaystyle=\sum_{i=0}^{r}(-1)^{k_{i+1}+\cdots+k_{r}}\sum_{\begin{subarray}{c}0<m_{1}<\cdots<m_{i}<M\\ 0<m_{r}<\cdots<m_{i+1}<M\end{subarray}}\frac{1}{m_{1}^{k_{1}}\cdots m_{r}^{k_{r}}},$
	$\displaystyle Z_{M}^{\mathcal{S},\mathcyr{sh}}(z_{\mathbf{k}})$	$\displaystyle=\sum_{i=0}^{r}(-1)^{k_{i+1}+\cdots+k_{r}}\sum_{\begin{subarray}{c}0<m_{1}<\cdots<m_{i}\\ 0<m_{r}<\cdots<m_{i+1}\\ m_{i}+m_{i+1}<M\end{subarray}}\frac{1}{m_{1}^{k_{1}}\cdots m_{r}^{k_{r}}},$

where $m_{0}$ and $m_{r+1}$ in the condition $m_{i}+m_{i+1}<M$ are set equal to zero in the latter formula. We call the above values the truncated symmetric multiple zeta value.

Theorem 2.2 ([4, 6]).

(i)

For any $w,w^{\prime}\in\mathfrak{h}^{1}$ and $M\geq 1$ , it holds that

\displaystyle Z_{M}^{\mathcal{S},\ast}(w\ast w^{\prime})=Z_{M}^{\mathcal{S},\ast}(w)Z_{M}^{\mathcal{S},\ast}(w^{\prime}),\qquad Z_{M}^{\mathcal{S},\mathcyr{sh}}(w\,\mathcyr{sh}\,w^{\prime})=Z_{M}^{\mathcal{S},\mathcyr{sh}}(w\psi(w^{\prime})).

(ii)

It holds that $w^{\mathcal{S},\bullet}(\mathfrak{h}^{1})\subset\mathfrak{h}^{0}$ for $\bullet\in\{\ast,\mathcyr{sh}\}$ . Hence the limit $\lim_{M\to\infty}Z_{M}^{\mathcal{S},\bullet}(w)$ converges for any $w\in\mathfrak{h}^{1}$ and $\bullet\in\{\ast,\mathcyr{sh}\}$ .
(iii)

For any index $\mathbf{k}$ and $\bullet\in\{\ast,\mathcyr{sh}\}$ , it holds that

$\displaystyle\lim_{M\to\infty}Z_{M}^{\mathcal{S},\bullet}(z_{\mathbf{k}})=\zeta^{\mathcal{S},\bullet}(\mathbf{k}).$

2.4. Relations of SMZVs

From the definition of $\zeta^{\mathcal{S},\bullet}$ , we obtain the following equality, which is called the reversal relation:

Theorem 2.3.

For any index $\mathbf{k}$ , it holds that

\displaystyle\zeta^{\mathcal{S}}(\overline{\mathbf{k}})=(-1)^{\mathrm{wt}(\mathbf{k})}\zeta^{\mathcal{S}}(\mathbf{k}).

For indices $\mathbf{k},\mathbf{l}$ and $\mathbf{m}$ , we define the non-negative integer $d_{\mathbf{k},\mathbf{l}}^{\,\bullet,\mathbf{m}}\,(\bullet\in\{\ast,\mathcyr{sh}\})$ by

(2.4)

\displaystyle z_{\mathbf{k}}\bullet z_{\mathbf{l}}=\sum_{\mathbf{m}}d_{\mathbf{k},\mathbf{l}}^{\,\bullet,\mathbf{m}}z_{\mathbf{m}}.

The following relation is called the double shuffle relation of SMZVs, which can be obtained from Theorem 2.2.

Theorem 2.4.

For any index $\mathbf{k}$ and $\mathbf{l}$ , it holds that

	$\displaystyle\zeta^{\mathcal{S}}(\mathbf{k})\zeta^{\mathcal{S}}(\mathbf{l})=\sum_{\mathbf{m}}d_{\mathbf{k},\mathbf{l}}^{\,\ast,\mathbf{m}}\zeta^{\mathcal{S}}(\mathbf{m}),$
	$\displaystyle(-1)^{\mathrm{wt}(\mathbf{l})}\zeta^{\mathcal{S}}(\mathbf{k},\overline{\mathbf{l}})=\sum_{\mathbf{m}}d_{\mathbf{k},\mathbf{l}}^{\,\mathcyr{sh},\mathbf{m}}\zeta^{\mathcal{S}}(\mathbf{m}).$

In [7], Oyama proved that Theorem 2.4 and the identity $\zeta^{\mathcal{S}}(k,\ldots,k)=0$ for any $k\geq 1$ imply the Ohno-type relation. To write down it, we define the Hoffman dual $\mathbf{k}^{\vee}$ of an index $\mathbf{k}$ . We define the $\mathbb{Q}$ -algebra automorphism $\tau$ on $\mathfrak{h}$ by $\tau(x)=y$ and $\tau(y)=x$ . For a non-empty index $\mathbf{k}$ , the monomial $z_{\mathbf{k}}$ is written in the form $z_{\mathbf{k}}=yw$ with a monomial $w\in\mathfrak{h}$ . Then the Hoffman dual index of $\mathbf{k}$ is defined to be the index $\mathbf{k}^{\vee}$ satisfying $z_{\mathbf{k}^{\vee}}=y\tau(w)$ . For example, if $\mathbf{k}=(3,1,2,1)$ , we have $z_{\mathbf{k}}=yx^{2}y^{2}xy$ and $z_{\mathbf{k}^{\vee}}=y\tau(x^{2}y^{2}xy)=y^{3}x^{2}yx=z_{1}^{2}z_{3}z_{2}$ , hence $\mathbf{k}^{\vee}=(1,1,3,2)$ . For the empty index we set $\varnothing^{\vee}=\varnothing$ .

Theorem 2.5 (Ohno-type relation [7]).

For any index $\mathbf{k}$ and any non-negative integer $m$ , it holds that

\displaystyle\sum_{\begin{subarray}{c}\mathbf{e}\in(\mathbb{Z}_{\geq 0})^{r}\\ \mathrm{wt}(\mathbf{e})=m\end{subarray}}\zeta^{\mathcal{S}}(\mathbf{k}+\mathbf{e})=\sum_{\begin{subarray}{c}\mathbf{e}\in(\mathbb{Z}_{\geq 0})^{s}\\ \mathrm{wt}(\mathbf{e})=m\end{subarray}}\zeta^{\mathcal{S}}((\mathbf{k}^{\vee}+\mathbf{e})^{\vee}),

where $r=\mathrm{dep}(\mathbf{k})$ and $s=\mathrm{dep}(\mathbf{k}^{\vee})$ .

3. A $q$ -analogue of multiple zeta value

3.1. Algebraic formulation

Set $\mathcal{C}=\mathbb{Q}[\hbar]$ , where $\hbar$ is a formal variable. Let $\mathfrak{H}=\mathcal{C}\langle a,b\rangle$ be the non-commutative polynomial ring of two variables $a$ and $b$ over $\mathcal{C}$ . For $k\geq 1$ we set

\displaystyle g_{k}=ba^{k},\qquad e_{k}=b(a+\hbar)a^{k-1}.

They are related to each other as

\displaystyle g_{k}=(-\hbar)^{k-1}g_{1}+\sum_{j=2}^{k}(-\hbar)^{k-j}e_{j}

for $k\geq 1$ , and

(3.1)

\displaystyle e_{k}=g_{k}+\hbar g_{k-1}

for $k\geq 2$ . Note that $e_{1}-g_{1}=\hbar b$ .

We set

\displaystyle A=\{\hbar b\}\cup\{ba^{k}\mid k\geq 1\}=\{e_{1}-g_{1}\}\cup\{g_{k}\mid k\geq 1\},

which is an algebraically independent set, and denote by $\mathcal{C}\langle A\rangle$ the $\mathcal{C}$ -subalgebra of $\mathfrak{H}$ generated by $1$ and $A$ . The depth of a monomial $u_{1}\cdots u_{r}\,(u_{1},\ldots,u_{r}\in A)$ is defined to be $r$ .

We define the $\mathcal{C}$ -submodule $\widehat{\mathfrak{H}^{0}}$ of $\mathcal{C}\langle A\rangle$ by

\displaystyle\widehat{\mathfrak{H}^{0}}=\mathcal{C}+\sum_{k\geq 1}\mathcal{C}\langle A\rangle g_{k}.

Then the set consisting of the elements

(3.2)

\displaystyle(e_{1}-g_{1})^{\alpha_{1}}g_{\beta_{1}+1}\cdots(e_{1}-g_{1})^{\alpha_{r}}g_{\beta_{r}+1}

with $r\geq 0$ and $\alpha_{1},\ldots,\alpha_{r},\beta_{1},\ldots,\beta_{r}\geq 0$ forms a free basis of $\widehat{\mathfrak{H}^{0}}$ .

Let $q$ be a complex parameter satisfying $0<|q|<1$ . We endow the complex number field $\mathbb{C}$ with $\mathcal{C}$ -module structure such that $\hbar$ acts as multiplication by $1-q$ .

We denote by $\mathfrak{z}$ the $\mathcal{C}$ -submodule of $\mathcal{C}\langle A\rangle$ spanned by the set $A$ . For a positive integer $m$ , we define the $\mathcal{C}$ -linear map $F_{q}(m;\cdot):\mathfrak{z}\longrightarrow\mathbb{C}$ by

(3.3)

\displaystyle F_{q}(m;e_{1}-g_{1})=1-q,\qquad F_{q}(m;g_{k})=\frac{q^{km}}{[m]^{k}}

for $k\geq 1$ , where $[m]$ is the $q$ -integer

\displaystyle[m]=\frac{1-q^{m}}{1-q}=1+q+\cdots+q^{m-1}.

Then, from $e_{1}=(e_{1}-g_{1})+g_{1}$ and (3.1) for $k\geq 2$ , we have

(3.4)

\displaystyle F_{q}(m;e_{k})=\frac{q^{(k-1)m}}{[m]^{k}}

for any $k\geq 1$ because $(1-q)[m]+q^{m}=1$ .

For a positive integer $M$ , we define the $\mathcal{C}$ -linear map $Z_{q,M}:\mathcal{C}\langle A\rangle\longrightarrow\mathbb{C}$ by $Z_{q,M}(1)=1$ and

\displaystyle Z_{q,M}(u_{1}\cdots u_{r})=\sum_{0<m_{1}<\cdots<m_{r}<M}\prod_{i=1}^{r}F_{q}(m_{i};u_{i})

for $u_{1},\ldots,u_{r}\in A$ . Then we see that, if $w\in\widehat{\mathfrak{H}^{0}}$ , $Z_{q,M}(w)$ converges in the limit as $M\to\infty$ . Hence we can define the $\mathcal{C}$ -linear map $Z_{q}:\widehat{\mathfrak{H}^{0}}\longrightarrow\mathbb{C}$ by

\displaystyle Z_{q}(w)=\lim_{M\to\infty}Z_{q,M}(w)

for $w\in\widehat{\mathfrak{H}^{0}}$ . In this paper we call the value $Z_{q}(w)$ with $w\in\widehat{\mathfrak{H}^{0}}$ a $q$ -analogue of MZV ( $q$ MZV).

Remark 3.1.

There are various models of $q$ -analogue of MZV (see [11, Chapter 12]). We can represent them in the form of $Z_{q}(w)$ with some $w\in\widehat{\mathfrak{H}^{0}}$ . For example, from (3.3) and (3.4), we see that

	$\displaystyle Z_{q}(g_{k_{1}}\cdots g_{k_{r}})=\sum_{0<m_{1}<\cdots<m_{r}}\frac{q^{k_{1}m_{1}+\cdots+k_{r}m_{r}}}{[m_{1}]^{k_{1}}\cdots[m_{r}]^{k_{r}}},$
	$\displaystyle Z_{q}(e_{k_{1}}\cdots e_{k_{r}})=\sum_{0<m_{1}<\cdots<m_{r}}\frac{q^{(k_{1}-1)m_{1}+\cdots+(k_{r}-1)m_{r}}}{[m_{1}]^{k_{1}}\cdots[m_{r}]^{k_{r}}},$

which are called the Schlesinger-Zudilin model and the Bradley-Zhao model, respectively.

3.2. Double shuffle relation of $q$ MZVs

The harmonic product and the shuffle product associated with the $q$ MZV are defined as follows.

First, we define the symmetric $\mathcal{C}$ -bilinear map $\circ_{\hbar}:\mathfrak{z}\times\mathfrak{z}\to\mathfrak{z}$ by

\displaystyle(e_{1}-g_{1})\circ_{\hbar}(e_{1}-g_{1})=\hbar(e_{1}-g_{1}),\quad(e_{1}-g_{1})\circ_{\hbar}g_{k}=\hbar g_{k},\quad g_{k}\circ_{\hbar}g_{l}=g_{k+l}

for $k,l\geq 1$ . Then we see that

(3.5)

\displaystyle F_{q}(m;u\circ_{\hbar}v)=F_{q}(m;u)F_{q}(m;v)

for any $u,v\in\mathfrak{z}$ and $m\geq 1$ . The harmonic product $\ast_{\hbar}$ is the $\mathcal{C}$ -bilinear binary operation on $\mathcal{C}\langle A\rangle$ uniquely defined by the following properties:

(i)

For any $w\in\mathcal{C}\langle A\rangle$ , it holds that $w\ast_{\hbar}1=w$ and $1\ast_{\hbar}w=w$ .
(ii)

For any $w,w^{\prime}\in\mathcal{C}\langle A\rangle$ and $u,v\in A$ , it holds that $(wu)\ast_{\hbar}(w^{\prime}v)=(w\ast_{\hbar}w^{\prime}v)u+(wu\ast_{\hbar}w^{\prime})v+(w\ast_{\hbar}w^{\prime})(u\circ_{\hbar}v)$ .

The harmonic product $\ast_{\hbar}$ on $\mathcal{C}\langle A\rangle$ is commutative and associative.

Proposition 3.2.

The $\mathcal{C}$ -submodule $\widehat{\mathfrak{H}^{0}}$ of $\mathcal{C}\langle A\rangle$ is closed under the harmonic product $\ast_{\hbar}$ , and it holds that $Z_{q,M}(w\ast_{\hbar}w^{\prime})=Z_{q,M}(w)Z_{q,M}(w^{\prime})$ for any $w,w^{\prime}\in\mathcal{C}\langle A\rangle$ and $M\geq 1$ . Therefore, we have $Z_{q}(w\ast_{\hbar}w^{\prime})=Z_{q}(w)Z_{q}(w^{\prime})$ for any $w,w^{\prime}\in\widehat{\mathfrak{H}^{0}}$ .

Next, we define the shuffle product. Consider the $\mathbb{Q}$ -linear right action of $\mathfrak{H}$ on a $\mathbb{C}$ -valued function $f(t)$ defined by

\displaystyle(f\hbar)(t)=(1-q)f(t),\qquad(fa)(t)=(1-q)\sum_{j=1}^{\infty}f(q^{j}t),\qquad(fb)(t)=\frac{t}{1-t}f(t).

Then, for any function $f$ and $g$ , it holds that

(3.6)

\displaystyle fa\cdot ga=(fa\cdot g+f\cdot ga+(1-q)f\cdot g)a,\quad fb\cdot g=f\cdot gb=(f\cdot g)b

if all terms are well-defined, where $\cdot$ denotes the usual multiplication of functions. Motivated by the above relations we define the shuffle product on $\mathfrak{H}$ as the $\mathcal{C}$ -bilinear binary operation uniquely defined by the following properties:

(i)

For any $w\in\mathfrak{H}$ , it holds that $w\,\mathcyr{sh}_{\hbar}\,1=w$ and $1\,\mathcyr{sh}_{\hbar}\,w=w$ .

(ii)

For any $w,w^{\prime}\in\mathfrak{H}$ , it holds that

	$\displaystyle wa\,\mathcyr{sh}_{\hbar}\,w^{\prime}a=(wa\,\mathcyr{sh}_{\hbar}\,w^{\prime}+w\,\mathcyr{sh}_{\hbar}\,w^{\prime}a+\hbar w\,\mathcyr{sh}_{\hbar}\,w^{\prime})a,$
	$\displaystyle wb\,\mathcyr{sh}_{\hbar}\,w^{\prime}=w\,\mathcyr{sh}_{\hbar}\,w^{\prime}b=(w\,\mathcyr{sh}_{\hbar}\,w^{\prime})b.$

The shuffle product is commutative and associative. Note that

(3.7)

\displaystyle(w(e_{1}-g_{1}))\,\mathcyr{sh}_{\hbar}\,w^{\prime}=w\,\mathcyr{sh}_{\hbar}\,(w^{\prime}(e_{1}-g_{1}))=(w\,\mathcyr{sh}_{\hbar}\,w^{\prime})(e_{1}-g_{1})

for any $w,w^{\prime}\in\mathfrak{H}$ since $e_{1}-g_{1}=\hbar b$ .

Proposition 3.3.

The $\mathcal{C}$ -submodules $\mathcal{C}\langle A\rangle$ and $\widehat{\mathfrak{H}^{0}}$ of $\mathfrak{H}$ are closed under the shuffle product $\mathcyr{sh}_{\hbar}$ . For any $u_{1},\ldots,u_{r},v_{1},\ldots,v_{s}\in\mathfrak{z}$ and $M\geq 1$ , it holds that

(3.8)

\displaystyle Z_{q,M}(u_{1}\cdots u_{r}\,\mathcyr{sh}_{\hbar}\,v_{1}\cdots v_{s})=\sum_{\begin{subarray}{c}0<m_{1}<\cdots<m_{r}\\ 0<n_{1}<\cdots<n_{s}\\ m_{r}+n_{s}<M\end{subarray}}\prod_{i=1}^{r}F_{q}(m_{i};u_{i})\prod_{j=1}^{s}F_{q}(n_{j};v_{j}).

Therefore, we have $Z_{q}(w\,\mathcyr{sh}_{\hbar}\,w^{\prime})=Z_{q}(w)Z_{q}(w^{\prime})$ for any $w,w^{\prime}\in\widehat{\mathfrak{H}^{0}}$ .

Proof.

Note that $g_{k+1}=g_{k}a$ for $k\geq 1$ . For $w,w^{\prime}\in\mathcal{C}\langle A\rangle$ and $k,l\geq 1$ , it holds that

(3.9)	$\displaystyle wg_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}g_{1}$	$\displaystyle=(w\,\mathcyr{sh}_{\hbar}\,w^{\prime}g_{1}+wg_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}+(w\,\mathcyr{sh}_{\hbar}\,w^{\prime})(e_{1}-g_{1}))g_{1},$
(3.10)	$\displaystyle wg_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}g_{k+1}$	$\displaystyle=(w\,\mathcyr{sh}_{\hbar}\,w^{\prime}g_{k+1})g_{1}+(wg_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}g_{k})a+\hbar(w\,\mathcyr{sh}_{\hbar}\,w^{\prime}g_{k})g_{1},$
(3.11)	$\displaystyle wg_{k+1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}g_{l+1}$	$\displaystyle=(wg_{k}\,\mathcyr{sh}_{\hbar}\,w^{\prime}g_{l+1}+wg_{k+1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}g_{l}+\hbar wg_{k}\,\mathcyr{sh}_{\hbar}\,w^{\prime}g_{l})a.$

These formulas imply that

(3.12)

\displaystyle\mathcal{C}\langle A\rangle g_{k}\,\mathcyr{sh}_{\hbar}\,\mathcal{C}\langle A\rangle g_{l}\subset\sum_{\min{(k,l)}\leq j\leq k+l-1}\mathcal{C}\langle A\rangle g_{j}

for $k,l\geq 1$ . From (3.7) and (3.12), we see that $\mathcal{C}\langle A\rangle$ and $\widehat{\mathfrak{H}^{0}}$ are closed under the shuffle product.

Denote by $\mathcal{F}$ the $\mathbb{C}$ -vector space of holomorphic functions on the unit disc. We endow $\mathcal{F}$ with $\mathcal{C}$ -module structure such that $\hbar$ acts as multiplication by $1-q$ . Let $\mathbf{1}(t)=1$ be the constant function. We define the $\mathcal{C}$ -linear map $L_{q}(\cdot):\mathcal{C}\langle A\rangle\longrightarrow\mathcal{F}$ by

\displaystyle L_{q}(u_{1}\cdots u_{r})(t)=((\mathbf{1})u_{1}\cdots u_{r})(t),

where $u_{1},\ldots,u_{r}\in A$ . We see that

\displaystyle L_{q}(u_{1}\cdots u_{r})(t)=\sum_{0<m_{1}<\cdots<m_{r}}t^{m_{r}}\prod_{i=1}^{r}F_{q}(m_{i};u_{i})

for $u_{1},\ldots,u_{r}\in\mathfrak{z}$ . Hence, $Z_{q,M}(w)$ is equal to the sum of the coefficients of $t^{m}$ in $L_{q}(w)(t)$ over $0\leq m<M$ for $w\in\mathcal{C}\langle A\rangle$ . By (3.6), we have $L_{q}(w\,\mathcyr{sh}_{\hbar}\,w^{\prime})=L_{q}(w)L_{q}(w^{\prime})$ for any $w,w^{\prime}\in\mathcal{C}\langle A\rangle$ . Thus we obtain (3.8). ∎

As stated in Proposition 3.2 and Proposition 3.3, we have

(3.13)

\displaystyle Z_{q}(w\ast w^{\prime})=Z_{q}(w)Z_{q}(w^{\prime}),\qquad Z_{q}(w\,\mathcyr{sh}_{\hbar}\,w^{\prime})=Z_{q}(w)Z_{q}(w^{\prime})

for any $w,w^{\prime}\in\widehat{\mathfrak{H}^{0}}$ . We call them the double shuffle relation for the $q$ MZVs. We denote by $\widehat{\mathcal{Z}_{q}}$ the image of the map $Z_{q}:\widehat{\mathfrak{H}^{0}}\rightarrow\mathbb{C}$ . Then $\widehat{\mathcal{Z}_{q}}$ is a $\mathcal{C}$ -subalgebra of $\mathbb{C}$ .

3.3. Limit of $q$ MZVs as $q\to 1-0$

We denote by $\mathfrak{n}_{0}$ the $\mathcal{C}$ -submodule of $\widehat{\mathfrak{H}^{0}}$ spanned by the elements of the form (3.2) with $r\geq 1$ and $\alpha_{s},\beta_{t}\geq 1$ for some $1\leq s\leq t\leq r$ . We define the $\mathcal{C}$ -modules $\mathfrak{n}$ and $\mathfrak{H}^{0}$ by

\displaystyle\mathfrak{n}=\mathfrak{n}_{0}+\hbar\,\widehat{\mathfrak{H}^{0}},\qquad\mathfrak{H}^{0}=\mathcal{C}+\sum_{k\geq 2}\mathcal{C}\langle A\rangle g_{k}+\mathfrak{n}.

For a non-empty index $\mathbf{k}=(k_{1},\ldots,k_{r})$ , we set

\displaystyle g_{\mathbf{k}}=g_{k_{1}}\cdots g_{k_{r}}.

For the empty index, we set $g_{\varnothing}=1$ . Note that the quotient $\mathfrak{H}^{0}/\mathfrak{n}$ is a free $\mathbb{Q}$ -module which has a basis $\{g_{\mathbf{k}}\mid\mathbf{k}\in I_{0}\}$ .

Proposition 3.4.

Here we consider a limit as $q\to 1-0$ with $q$ being real.

(i)

If $\mathbf{k}$ is an admissible index, it holds that $\lim_{q\to 1-0}Z_{q}(g_{\mathbf{k}})=\zeta(\mathbf{k})$ .
(ii)

For any $w\in\mathfrak{n}$ , it holds that $\lim_{q\to 1-0}Z_{q}(w)=0$ .

See Appendix A for the proof.

We define the unital $\mathbb{Q}$ -algebra homomorphism $\iota:\mathcal{C}\langle A\rangle\rightarrow\mathfrak{h}^{1}$ by

(3.14)

\displaystyle\iota(\hbar)=0,\quad\iota(e_{1}-g_{1})=0,\quad\iota(g_{k})=z_{k}

for $k\geq 1$ . Then we see that the restriction of $\iota$ to $\mathfrak{H}^{0}$ is a surjection onto $\mathfrak{h}^{0}$ and its kernel is equal to $\mathfrak{n}$ . Therefore, from Proposition 3.4, we obtain the following corollary.

Corollary 3.5.

For any $w\in\mathfrak{H}^{0}$ , it holds that

(3.15)

\displaystyle\lim_{q\to 1-0}Z_{q}(w)=Z(\iota(w)).

Remark 3.6.

The equality (3.15) does not necessarily hold for $w\in\widehat{\mathfrak{H}^{0}}\setminus\mathfrak{H}^{0}$ such that the limit of $Z_{q}(w)$ as $q\to 1-0$ converges. For example, we see that

	$\displaystyle Z_{q}((e_{1}-g_{1})g_{1})$	$\displaystyle=(1-q)^{2}\sum_{0<m<n}\frac{q^{n}}{1-q^{n}}=(1-q)^{2}\sum_{0<n,l}(n-1)q^{nl}$
		$\displaystyle=(1-q)^{2}\sum_{0<l}\frac{q^{2l}}{(1-q^{l})^{2}}=Z_{q}(g_{2}),$

which is a special case of the resummation identity

\displaystyle Z_{q}((e_{1}-g_{1})^{\alpha_{1}}g_{\beta_{1}+1}\cdots(e_{1}-g_{1})^{\alpha_{r}}g_{\beta_{r}+1})=Z_{q}((e_{1}-g_{1})^{\beta_{r}}g_{\alpha_{r}+1}\cdots(e_{1}-g_{1})^{\beta_{1}}g_{\alpha_{1}+1})

for $\alpha_{1},\ldots,\alpha_{r},\beta_{1},\ldots,\beta_{r}\geq 0$ [9, Theorem 4]. Hence, the limit of $Z_{q}((e_{1}-g_{1})g_{1})$ as $q\to 1-0$ is equal to $\zeta(2)$ . However, we have $\iota((e_{1}-g_{1})g_{1})=0$ .

3.4. Restoration of finite double shuffle relation of MZVs

Proposition 3.7.

The $\mathcal{C}$ -module $\mathfrak{n}$ is an ideal of $\widehat{\mathfrak{H}^{0}}$ with respect to both the harmonic product and the shuffle product.

Proof.

It suffices to show that $\mathfrak{n}_{0}\ast_{\hbar}\widehat{\mathfrak{H}^{0}}\subset\mathfrak{n}$ and $\mathfrak{n}_{0}\,\mathcyr{sh}_{\hbar}\,\widehat{\mathfrak{H}^{0}}\subset\mathfrak{n}$ . The former follows from the definition of the harmonic product. The latter follows from (3.7) and (3.12). ∎

Corollary 3.8.

The $\mathcal{C}$ -module $\mathfrak{H}^{0}$ is closed under both the harmonic product and the shuffle product.

Proof.

From $g_{k}\circ_{\hbar}g_{l}=g_{k+l}$ , we see that $\sum_{k\geq 2}\mathcal{C}\langle A\rangle g_{k}$ is closed under the harmonic product. It is also closed under the shuffle product because of (3.12). Hence Proposition 3.7 implies that $\mathfrak{H}^{0}$ is closed. ∎

We set

\displaystyle\mathcal{Z}_{q}=Z_{q}(\mathfrak{H}^{0}).

Corollary 3.8 implies that $\mathcal{Z}_{q}$ is a $\mathcal{C}$ -subalgebra of $\widehat{\mathcal{Z}_{q}}$ .

Proposition 3.9.

For any $w,w^{\prime}\in\mathcal{C}\langle A\rangle$ and $\bullet\in\{\ast,\mathcyr{sh}\}$ , it holds that

(3.16)

\displaystyle\iota(w\bullet_{\hbar}w^{\prime})=\iota(w)\bullet\iota(w^{\prime}).

Proof.

We may assume that $w$ and $w^{\prime}$ are monomials in $A$ . If $w=1$ or $w^{\prime}=1$ , the desired equality (3.16) is trivial. Now we proceed the proof by induction on the sum of the depth of $w$ and $w^{\prime}$ .

Since $\iota((e_{1}-g_{1})\circ_{\hbar}u)=0$ for any $u\in\mathfrak{z}$ and $\iota(g_{k}\circ_{\hbar}g_{l})=z_{k+l}$ for $k,l\geq 1$ , we see that (3.16) holds for $\bullet=\ast$ .

We consider the case of $\bullet=\mathcyr{sh}$ . If $w$ or $w^{\prime}$ is the form of $u(e_{1}-g_{1})$ with a monomial $u$ of $A$ , we see that (3.16) from $\iota(e_{1}-g_{1})=0$ and (3.7). Hence it suffices to prove the case where $w=ug_{k}$ and $w^{\prime}=vg_{l}$ with monomials $u,v$ in $A$ and $k,l\geq 1$ . For that purpose we use the following formula. Note that, in the formal power series ring $\mathfrak{H}[[X]]$ , we have

\displaystyle\sum_{k=1}^{\infty}g_{k}X^{k-1}=g_{1}\frac{1}{1-aX}.

Lemma 3.10.

For any $u,v\in\mathfrak{H}$ , we have

	$\displaystyle ug_{1}\frac{1}{1-aX}\,\mathcyr{sh}_{\hbar}\,vg_{1}\frac{1}{1-aY}$
	$\displaystyle=\left\{(1+\hbar X)\left(ug_{1}\frac{1}{1-aX}\,\mathcyr{sh}_{\hbar}\,v\right)+(1+\hbar Y)\left(u\,\mathcyr{sh}_{\hbar}\,vg_{1}\frac{1}{1-aY}\right)+(u\,\mathcyr{sh}_{\hbar}\,v)(e_{1}-g_{1})\right\}$
	$\displaystyle\times g_{1}\frac{1}{1-a(X+Y+\hbar XY)}.$

See Appendix B.1 for the proof. From the above equality, we see that

\displaystyle\iota(ug_{1}\frac{1}{1-aX}\,\mathcyr{sh}_{\hbar}\,vg_{1}\frac{1}{1-aY})=\iota(ug_{1}\frac{1}{1-aX}\,\mathcyr{sh}_{\hbar}\,v+u\,\mathcyr{sh}_{\hbar}\,vg_{1}\frac{1}{1-aY})z_{1}\frac{1}{1-x(X+Y)}

in the formal power series ring $\mathfrak{h}[[X,Y]]$ . The induction hypothesis implies that the right hand side is equal to

\displaystyle\left\{\iota(u)z_{1}\frac{1}{1-xX}\,\mathcyr{sh}\,\iota(v)+\iota(u)\,\mathcyr{sh}\,\iota(v)z_{1}\frac{1}{1-xY}\right\}z_{1}\frac{1}{1-(X+Y)x}.

We see that it is equal to

\displaystyle\iota(u)z_{1}\frac{1}{1-xX}\,\mathcyr{sh}\,\iota(v)z_{1}\frac{1}{1-xY}=\iota(ug_{1}\frac{1}{1-aX})\,\mathcyr{sh}\,\iota(vg_{1}\frac{1}{1-aY})

in the same way as the proof of Lemma 3.10. This completes the proof of Proposition 3.9 for $\bullet=\mathcyr{sh}$ . ∎

Using Corollary 3.8 and Proposition 3.9, we restore the finite double shuffle relation of MZVs from (3.13) as follows. Let $\mathbf{k}$ and $\mathbf{l}$ be admissible indices. Then, for $\bullet\in\{\ast,\mathcyr{sh}\}$ , we see that

	$\displaystyle Z(z_{\mathbf{k}})Z(z_{\mathbf{l}})$	$\displaystyle=\lim_{q\to 1-0}Z_{q}(g_{\mathbf{k}})Z_{q}(g_{\mathbf{l}})=\lim_{q\to 1-0}Z_{q}(g_{\mathbf{k}}\bullet g_{\mathbf{l}})$
		$\displaystyle=Z(\iota(g_{\mathbf{k}}\bullet g_{\mathbf{l}}))=Z(\iota(g_{\mathbf{k}})\bullet\iota(g_{\mathbf{l}}))=Z(z_{\mathbf{k}}\bullet z_{\mathbf{l}}).$

4. A $q$ -analogue of truncated SMZV

Let $\psi^{\bullet}\,(\bullet\in\{\ast,\mathcyr{sh}\})$ be the $\mathcal{C}$ -algebra anti-involution on $\mathcal{C}\langle A\rangle=\mathcal{C}\langle\hbar b,ba,ba^{2},\ldots\rangle$ defined by

	$\displaystyle\psi^{\ast}(\hbar b)=\hbar b,\quad\psi^{\ast}(ba^{k})=b(-a)^{k},$
	$\displaystyle\psi^{\mathcyr{sh}}(\hbar b)=\hbar b,\quad\psi^{\mathcyr{sh}}(ba^{k})=b(-a-\hbar)^{k}$

for $k\geq 1$ . For $\bullet\in\{\ast,\mathcyr{sh}\}$ , we define the $\mathcal{C}$ -linear map $w^{\mathcal{S},\bullet}_{\hbar}:\mathcal{C}\langle A\rangle\to\mathcal{C}\langle A\rangle$ by $w^{\mathcal{S},\bullet}_{\hbar}(1)=1$ and

\displaystyle w^{\mathcal{S},\bullet}_{\hbar}(u_{1}\cdots u_{r})=\sum_{i=0}^{r}u_{1}\cdots u_{i}\bullet_{\hbar}\psi^{\bullet}(u_{i+1}\cdots u_{r})

for $r\geq 1$ and $u_{1},\ldots,u_{r}\in A$ .

Now we define the $\mathcal{C}$ -linear map $Z^{\mathcal{S},\bullet}_{q,M}:\mathcal{C}\langle A\rangle\to\mathbb{C}$ for $\bullet\in\{\ast,\mathcyr{sh}\}$ and $M\geq 1$ by $Z^{\mathcal{S},\bullet}_{q,M}=Z_{q,M}\circ w^{\mathcal{S},\bullet}_{\hbar}$ . We call the value $Z^{\mathcal{S},\bullet}_{q,M}(w)$ with $w\in\mathcal{C}\langle A\rangle$ a $q$ -analogue of truncated symmetric multiple zeta value.

From Proposition 3.2 and Proposition 3.3, we see that

(4.1)		$\displaystyle Z_{q,M}^{\mathcal{S},\ast}(u_{1}\cdots u_{r})$	$\displaystyle=\sum_{i=0}^{r}\sum_{\begin{subarray}{c}0<m_{1}<\cdots<m_{i}<M\\ 0<m_{r}<\cdots<m_{i+1}<M\end{subarray}}\prod_{j=1}^{i}F_{q}(m_{j};u_{j})\prod_{j=i+1}^{r}F_{q}(m_{j};\psi^{\ast}(u_{j})),$
(4.2)		$\displaystyle Z_{q,M}^{\mathcal{S},\mathcyr{sh}}(u_{1}\cdots u_{r})$	$\displaystyle=\sum_{i=0}^{r}\sum_{\begin{subarray}{c}0<m_{1}<\cdots<m_{i}\\ 0<m_{r}<\cdots<m_{i+1}\\ m_{i}+m_{i+1}<M\end{subarray}}\prod_{j=1}^{i}F_{q}(m_{j};u_{j})\prod_{j=i+1}^{r}F_{q}(m_{j};\psi^{\mathcyr{sh}}(u_{j}))$

for $u_{1},\ldots,u_{r}\in A$ .

We prove the double shuffle relation of the $q$ -analogue of truncated SMZVs. The proof is similar to that of the truncated SMZVs in [6].

Proposition 4.1.

For any $w,w^{\prime}\in\mathcal{C}\langle A\rangle$ and $M\geq 1$ , it holds that

(4.3)

\displaystyle Z^{\mathcal{S},\ast}_{q,M}(w\ast_{\hbar}w^{\prime})=Z^{\mathcal{S},\ast}_{q,M}(w)Z^{\mathcal{S},\ast}_{q,M}(w^{\prime}).

Proof.

Note that $\psi^{\ast}$ is an anti-automorphism. We extend the $\mathcal{C}$ -linear map $F_{q}(m;\cdot):\mathfrak{z}\to\mathbb{C}$ for $m<0$ by $F_{q}(m;u)=F_{q}(-m;\psi^{\ast}(u))$ for $u\in\mathfrak{z}$ . Then, from (4.1), we see that

(4.4)

\displaystyle Z^{\mathcal{S},\ast}_{q,M}(u_{1}\cdots u_{r})=\sum_{\begin{subarray}{c}m_{1}\prec\cdots\prec m_{r}\\ 0<|m_{1}|,\ldots,|m_{r}|<M\end{subarray}}\prod_{i=1}^{r}F_{q}(m_{i};u_{i})

for $u_{1},\ldots,u_{r}\in A$ , where $\prec$ is Kontsevich’s order on the set $(\mathbb{Z}\setminus\{0\})\sqcup\{\infty=-\infty\}$ defined by $1\prec 2\prec\cdots\prec\infty=-\infty\prec\cdots\prec-2\prec-1$ .

From the definition of $\circ_{\hbar}$ and $\psi^{\ast}$ , we see that

\displaystyle\psi^{\ast}(u\circ_{\hbar}v)=\psi^{\ast}(u)\circ_{\hbar}\psi^{\ast}(v)

for any $u,v\in A$ . Hence the relation (3.5) holds for any non-zero integer $m$ and $u,v\in A$ , and we obtain (4.3) by using the expression (4.4) in the same way as the proof for the harmonic product relation of the truncated MZVs $\zeta_{M}(\mathbf{k})$ . ∎

Proposition 4.2.

For any $w,w^{\prime}\in\mathcal{C}\langle A\rangle$ and $M\geq 1$ , it holds that

(4.5)

\displaystyle Z^{\mathcal{S},\mathcyr{sh}}_{q,M}(w\,\mathcyr{sh}_{\hbar}\,w^{\prime})=Z^{\mathcal{S},\mathcyr{sh}}_{q,M}(w\,\psi^{\mathcyr{sh}}(w^{\prime})).

Proof.

We extend the $\mathcal{C}$ -linear map $F_{q}(m;\cdot):\mathfrak{z}\to\mathbb{C}$ for $m<0$ by $F_{q}(m;u)=F_{q}(-m;\psi^{\mathcyr{sh}}(u))$ for $u\in A$ . Then we see that (3.3) holds for any nonzero integer $m$ and $k\geq 1$ .

Let $T_{q,M}:\mathcal{C}\langle A\rangle\times\mathcal{C}\langle A\rangle\times\mathcal{C}\langle A\rangle\to\mathbb{C}$ be the $\mathcal{C}$ -trilinear map defined by $T_{q,M}(1,1,1)=1$ and

	$\displaystyle T_{q,M}(u_{1}\cdots u_{r_{1}},v_{1}\cdots v_{r_{2}},w_{1}\cdots w_{r_{3}-1})$
	$\displaystyle=\sum_{t=1}^{3}\sum_{j=1}^{r_{t}}\sum_{(l_{i}^{(s)})\in D_{j}^{(t)}}\prod_{i=1}^{r_{1}}F_{q}(l_{1}^{(1)}+\cdots+l_{i}^{(1)};u_{i})\prod_{i=1}^{r_{2}}F_{q}(l_{1}^{(2)}+\cdots+l_{i}^{(2)};v_{i})$
	$\displaystyle\qquad\qquad\qquad\qquad{}\times\prod_{i=1}^{r_{3}-1}F_{q}(\|l^{(1)}\|+\|l^{(2)}\|+l_{1}^{(3)}+\cdots+l_{i}^{(3)};w_{i}),$

where $r_{1},r_{2}\geq 0,r_{3}\geq 1$ , $u_{1},\ldots,u_{r_{1}},v_{1},\ldots,v_{r_{2}},w_{1},\ldots,w_{r_{3}-1}\in A$ and $|l^{(s)}|=\sum_{j=1}^{r_{s}}l_{j}^{(s)}\,(s=1,2)$ . The summation region $D_{j}^{(t)}$ is the subset of $\mathbb{Z}^{r_{1}+r_{2}+r_{3}}$ consisting of tuples $(l_{i}^{(s)})_{\begin{subarray}{c}1\leq s\leq 3\\ 1\leq i\leq r_{s}\end{subarray}}$ satisfying the following conditions:

\displaystyle\sum_{s=1}^{3}\sum_{i=1}^{r_{s}}l_{i}^{(s)}=0,\quad-M<l_{j}^{(t)}<0,\quad l_{i}^{(s)}>0\,\,\hbox{for any}\,(s,i)\not=(t,j).

Then we see that $T_{q,M}(u,v,1)=Z_{q,M}^{\mathcal{S},\mathcyr{sh}}(u\,\psi^{\mathcyr{sh}}(v))$ for $u,v\in\mathcal{C}\langle A\rangle$ . Hence it suffices to show that

(4.6)

\displaystyle T_{q,M}(u,v,w)=Z_{q,M}^{\mathcal{S},\mathcyr{sh}}((u\,\mathcyr{sh}_{\hbar}\,v)w)

for any $u,v,w\in\mathcal{C}\langle A\rangle$ .

From (4.2) and the definition of $T_{q,M}$ , we see that (4.6) holds if $u=1$ or $v=1$ . Set $u=u_{1}\cdots u_{r_{1}}$ and $v=v_{1}\cdots v_{r_{2}}$ . We prove (4.6) by induction on $r_{1}+r_{2}$ . Since $T_{q,M}(u,v,w)$ is symmetric with respect to $u$ and $v$ , it suffices to consider the two cases: (i) $u_{r_{1}}=e_{1}-g_{1}$ , (ii) $u_{r_{1}}=g_{k},\,v_{r_{2}}=g_{l}$ with $k,l\geq 1$ . The case (i) follows from (3.7) and $F_{q}(m;e_{1}-g_{1})=F_{q}(m+n;e_{1}-g_{1})=1-q$ for integers $m$ and $n$ satisfying $m\not=0$ and $m+n\not=0$ . To show the case (ii), we set

\displaystyle H_{q}(m;X)=\sum_{k\geq 1}X^{k-1}F_{q}(m;g_{k})=\frac{q^{m}}{[m]-q^{m}X}.

It holds that

	$\displaystyle H_{q}(m;X)H_{q}(n;Y)$
	$\displaystyle=\left\{(1+(1-q)X)H_{q}(m;X)+(1+(1-q)Y)H_{q}(n;Y)+1-q\right\}$
	$\displaystyle\times H_{q}(m+n;X+Y+(1-q)XY)$

for non-zero integers $m$ and $n$ satisfying $m+n\not=0$ . From the above relation and Lemma 3.10, we see that (4.6) holds in the case (ii) under the induction hypothesis. ∎

5. A $q$ -analogue of SMZV

In this section, we define a $q$ -analogue of the SMZV. To this aim, we use the map $Z_{q}^{\mathcal{S},\bullet}\,(\bullet\in\{\ast,\mathcyr{sh}\})$ , which corresponds to $\zeta^{\mathcal{S},\bullet}$ , defined as follows.

First we define the map $Z_{q}^{\mathcal{S},\ast}$ .

Proposition 5.1.

For any index $\mathbf{k}$ , the element $w_{\hbar}^{\mathcal{S},\ast}(g_{\mathbf{k}})$ belongs to the $\mathbb{Z}$ -submodule $\oplus_{\mathbf{l}\in I_{0}}\mathbb{Z}\,g_{\mathbf{l}}$ of $\mathfrak{H}^{0}$ .

Proof.

From the definition of the harmonic product $\ast_{\hbar}$ , we see that $g_{\mathbf{k}}\ast_{\hbar}g_{\mathbf{l}}=\sum_{\mathbf{m}}d_{\mathbf{k},\mathbf{l}}^{\,\ast,\mathbf{m}}g_{\mathbf{m}}$ , where $d_{\mathbf{k},\mathbf{l}}^{\ast,\mathbf{m}}$ is given by (2.4). Hence the proof of Proposition 5.4 in [6] with $\bullet=\ast$ and $n=0$ works also for our map $w_{\hbar}^{\mathcal{S},\ast}$ . ∎

Definition 5.2.

Set

\displaystyle\mathfrak{g}=\bigoplus_{\mathbf{k}\in I}\mathcal{C}g_{\mathbf{k}},

which is a $\mathcal{C}$ -submodule of $\widehat{\mathfrak{H}^{0}}$ . We define the $\mathcal{C}$ -linear map $Z_{q}^{\mathcal{S},\ast}:\mathfrak{g}\rightarrow\mathcal{Z}_{q}$ by

\displaystyle Z_{q}^{\mathcal{S},\ast}(g_{\mathbf{k}})=\lim_{M\to\infty}Z_{q,M}^{\mathcal{S},\ast}(g_{\mathbf{k}})=Z_{q}(w_{\hbar}^{\mathcal{S},\ast}(g_{\mathbf{k}}))

for an index $\mathbf{k}$ . More explicitly, we have

\displaystyle Z_{q}^{\mathcal{S},\ast}(g_{\mathbf{k}})=\sum_{i=0}^{r}(-1)^{k_{i+1}+\cdots+k_{r}}\sum_{\begin{subarray}{c}0<m_{1}<\cdots<m_{i}\\ 0<m_{r}<\cdots<m_{i+1}\end{subarray}}\frac{q^{k_{1}m_{1}+\cdots+k_{r}m_{r}}}{[m_{1}]^{k_{1}}\cdots[m_{r}]^{k_{r}}}

for a non-empty index $\mathbf{k}=(k_{1},\ldots,k_{r})$ .

We want to define the map $Z_{q}^{\mathcal{S},\mathcyr{sh}}$ similarly by taking the limit of $Z_{q,M}^{\mathcal{S},\mathcyr{sh}}$ as $M\to\infty$ . For that purpose, however, we should determine the domain carefully because it is not closed under the concatenation product unlike the $\mathcal{C}$ -module $\mathfrak{g}$ in Definition 5.2 as seen by the following example.

Example 5.3.

We have $w_{\hbar}^{\mathcal{S},\mathcyr{sh}}(e_{1})=-w_{\hbar}^{\mathcal{S},\mathcyr{sh}}(g_{1})=e_{1}-g_{1}$ and $Z_{q,M}^{\mathcal{S},\mathcyr{sh}}(e_{1})=-Z_{q,M}^{\mathcal{S},\mathcyr{sh}}(g_{1})=(1-q)(M-1)$ . Hence a $\mathcal{C}$ -linear combination of $e_{1}$ and $g_{1}$ whose image by $Z_{q,M}^{\mathcal{S},\mathcyr{sh}}$ converges in the limit as $M\to\infty$ should be proportional to $e_{1}+g_{1}$ . However, we see that

\displaystyle Z_{q,M}^{\mathcal{S},\mathcyr{sh}}((e_{1}+g_{1})^{2})=Z_{M}((e_{1}-g_{1})^{2})=(1-q)^{2}\binom{M-1}{2}\to\infty\qquad(M\to\infty).

Therefore, the concatenation product $(e_{1}+g_{1})^{2}$ of $e_{1}+g_{1}$ does not belong to the domain of the map $\lim_{M\to\infty}Z_{q,M}^{\mathcal{S},\mathcyr{sh}}$ . Here we note that, because $w_{\hbar}^{\mathcal{S},\mathcyr{sh}}(e_{1}g_{1}+g_{1}e_{1})=-(e_{1}-g_{1})^{2}$ , it holds that $Z_{q,M}^{\mathcal{S},\mathcyr{sh}}(e_{1}^{2}+2e_{1}g_{1}+2g_{1}e_{1}+g_{1}^{2})=0$ , which clearly converges as $M\to\infty$ .

To describe the domain of the map $w_{\hbar}^{\mathcal{S},\mathcyr{sh}}$ , we introduce the element $E_{1^{m}}\,(m\geq 0)$ of $\mathcal{C}\langle A\rangle$ defined by

\displaystyle E_{1^{m}}=\frac{1}{(m+1)!}\sum_{j=0}^{m}g_{1}^{\mathcyr{sh}_{\hbar}j}\mathcyr{sh}_{\hbar}e_{1}^{\mathcyr{sh}_{\hbar}(m-j)},

where $u^{\mathcyr{sh}_{\hbar}0}=1$ and $u^{\mathcyr{sh}_{\hbar}j}=u\mathcyr{sh}_{\hbar}u^{\mathcyr{sh}_{\hbar}(j-1)}$ for $u\in\mathfrak{H}$ and $j\geq 1$ . For example, we have

\displaystyle E_{1^{0}}=1,\qquad E_{1^{1}}=\frac{1}{2}(e_{1}+g_{1}),\qquad E_{1^{2}}=\frac{1}{6}(e_{1}^{2}+2e_{1}g_{1}+2g_{1}e_{1}+g_{1}^{2}).

Let $\mathbf{k}$ be a non-empty index. It can be written uniquely in the form

(5.1)

\displaystyle\mathbf{k}=(\underbrace{1,\ldots,1}_{s_{0}},t_{1}+2,\underbrace{1,\ldots,1}_{s_{1}},\ldots,t_{r}+2,\underbrace{1,\ldots,1}_{s_{r}})

with $r\geq 0$ and $s_{0},\ldots,s_{r},t_{1},\ldots,t_{r}\geq 0$ , where the right hand side reads $(\underbrace{1,\ldots,1}_{s_{0}})$ if $r=0$ . Then we set

\displaystyle E_{\mathbf{k}}=E_{1^{s_{0}}}e_{t_{1}+2}E_{1^{s_{1}}}\cdots e_{t_{r}+2}E_{1^{s_{r}}}.

For the empty index, we set $E_{\varnothing}=1$ .

Proposition 5.4.

For any index $\mathbf{k}$ , the element $w_{\hbar}^{\mathcal{S},\mathcyr{sh}}(E_{\mathbf{k}})$ belongs to $\mathfrak{H}^{0}$ .

See Appendix B.2 and Appendix B.3 for the proof.

Definition 5.5.

Set

\displaystyle\mathfrak{e}=\bigoplus_{\mathbf{k}\in I}\mathcal{C}E_{\mathbf{k}}.

We define the $\mathcal{C}$ -linear map $Z_{q}^{\mathcal{S},\mathcyr{sh}}:\mathfrak{e}\rightarrow\mathcal{Z}_{q}$ by

\displaystyle Z_{q}^{\mathcal{S},\mathcyr{sh}}(E_{\mathbf{k}})=\lim_{M\to\infty}Z_{q,M}^{\mathcal{S},\mathcyr{sh}}(E_{\mathbf{k}})=Z_{q}(w_{\hbar}^{\mathcal{S},\mathcyr{sh}}(E_{\mathbf{k}})).

for an index $\mathbf{k}$ .

Example 5.6.

If $k\geq 2$ , it holds that $F_{q}(m;\psi^{\mathcyr{sh}}(e_{k}))=F_{q}(-m;e_{k})=(-1)^{k}q^{m}/[m]^{k}$ for $m\geq 1$ . Hence, for a non-empty index $\mathbf{k}=(k_{1},\ldots,k_{r})$ whose all components are larger than one, we have

	$\displaystyle Z_{q}^{\mathcal{S},\mathcyr{sh}}(E_{\mathbf{k}})$	$\displaystyle=Z_{q}^{\mathcal{S},\mathcyr{sh}}(e_{k_{1}}\cdots e_{k_{r}})$
		$\displaystyle=\sum_{i=0}^{r}(-1)^{k_{i+1}+\cdots+k_{r}}\sum_{\begin{subarray}{c}0<m_{1}<\cdots<m_{i}\\ 0<m_{r}<\cdots<m_{i+1}\end{subarray}}\frac{q^{(k_{1}-1)m_{1}+\cdots+(k_{i}-1)m_{i}}}{[m_{1}]^{k_{1}}\cdots[m_{i}]^{k_{i}}}\frac{q^{m_{i+1}+\cdots+m_{r}}}{[m_{i+1}]^{k_{i+1}}\cdots[m_{r}]^{k_{r}}}.$

Example 5.7.

From Proposition B.4 and Lemma B.7, we see that $w_{\hbar}^{\mathcal{S},\mathcyr{sh}}(E_{1^{m}})=0$ for $m\geq 1$ . Hence we have $Z_{q}^{\mathcal{S},\mathcyr{sh}}(E_{1,\ldots,1})=Z_{q}^{\mathcal{S},\mathcyr{sh}}(E_{1^{m}})=0$ .

Using the two maps $Z_{q}^{\mathcal{S},\ast}$ and $Z_{q}^{\mathcal{S},\mathcyr{sh}}$ , we define a $q$ -analogue of the SMZV. Recall that we need the relation (2.3) to define the SMZV. We show the corresponding relation in the $q$ -analogue case.

Set

\displaystyle\mathcal{N}_{q}=Z_{q}(\mathfrak{n}).

From Proposition 3.7, we see that $\mathcal{N}_{q}$ is an ideal of $\widehat{\mathcal{Z}_{q}}$ . Note that $(1-q)\mathcal{Z}_{q}\subset\mathcal{N}_{q}$ since $\hbar\,\mathfrak{H}^{0}\subset\mathfrak{n}$ .

For $\bullet\in\{\ast,\mathcyr{sh}\}$ and $f(X)\in\mathcal{C}\langle A\rangle[[X]]$ satisfying $f(0)=0$ , we define the exponential $\exp_{\bullet_{\hbar}}(f(X))$ with respect to the product $\bullet_{\hbar}$ by

\displaystyle\exp_{\bullet_{\hbar}}{\left(f(X)\right)}=1+\sum_{n\geq 1}\frac{1}{n!}\underbrace{f(X)\bullet_{\hbar}\cdots\bullet_{\hbar}f(X)}_{\hbox{\tiny$n$ times}}.

Theorem 5.8.

For any admissible index $\mathbf{k}$ , it holds that

(5.2)

\displaystyle Z_{q}\left(g_{\mathbf{k}}\,\frac{1}{1-g_{1}X}\right)\equiv\exp{\left(\sum_{n\geq 2}\frac{(-1)^{n-1}}{n}Z_{q}(g_{n})X^{n}\right)}Z_{q}\left(E_{\mathbf{k}}\exp_{\mathcyr{sh}_{\hbar}}(g_{1}X)\right)

in $\widehat{\mathcal{Z}_{q}}[[X]]$ modulo $\mathcal{N}_{q}[[X]]$ .

Proof.

We start from the following equality. See Appendix B.4 for the proof.

Lemma 5.9.

Set

(5.3)

\displaystyle R(X)=\frac{e^{\hbar bX}-1}{\hbar b}=\sum_{n=1}^{\infty}\frac{X^{n}}{n!}(e_{1}-g_{1})^{n-1}.

For any admissible index $\mathbf{k}$ , it holds that

(5.4)

\displaystyle\frac{1}{1-R(X)g_{1}}\,\mathcyr{sh}_{\hbar}\,g_{\mathbf{k}}\frac{1}{1-g_{1}X}\equiv\frac{1}{1-g_{1}X}\,\mathcyr{sh}_{\hbar}\,E_{\mathbf{k}}\frac{1}{1-R(X)g_{1}}

modulo the $\mathcal{C}$ -submodule $\mathfrak{n}[[X]]$ of $\mathfrak{H}[[X]]$ .

We calculate the image by the map $Z_{q}$ of the both hand sides of (5.4) using Proposition 3.3 and (B.2). Then we obtain

\displaystyle\exp{(Z_{q}(g_{1})X)}Z_{q}\left(g_{\mathbf{k}}\,\frac{1}{1-g_{1}X}\right)\equiv Z_{q}\left(\frac{1}{1-g_{1}X}\right)Z_{q}\left(E_{\mathbf{k}}\exp_{\mathcyr{sh}_{\hbar}}(g_{1}X)\right)

modulo $\mathcal{N}_{q}[[X]]$ . It holds that

(5.5)

\displaystyle\frac{1}{1-g_{k}X}=\exp_{\ast_{\hbar}}{\left(\sum_{n\geq 1}\frac{(-1)^{n-1}}{n}g_{nk}X^{n}\right)}

for any $k\geq 1$ (see [3, Corollary 1]). Hence, from Proposition 3.2, we have

\displaystyle Z_{q}\left(\frac{1}{1-g_{1}X}\right)=\exp{\left(\sum_{n\geq 1}\frac{(-1)^{n-1}}{n}Z_{q}(g_{n})X^{n}\right)}.

It implies (5.2) since $\mathcal{N}_{q}$ is an ideal of $\widehat{\mathcal{Z}_{q}}$ . ∎

We denote by $\mathcal{P}_{q}$ the ideal of $\mathcal{Z}_{q}$ generated by the set $\{Z_{q}(g_{2k})\mid k\geq 1\}$ .

Corollary 5.10.

For any index $\mathbf{k}$ , the difference $Z_{q}^{\mathcal{S},\ast}(g_{\mathbf{k}})-Z_{q}^{\mathcal{S},\mathcyr{sh}}(E_{\mathbf{k}})$ belongs to the ideal $\mathcal{N}_{q}+\mathcal{P}_{q}$ .

Proof.

From the definition of $w_{\hbar}^{\mathcal{S},\bullet}\,(\bullet\in\{\ast,\mathcyr{sh}\})$ , Proposition B.4, Corollary B.6 and Lemma B.7, we see that the difference $Z_{q}^{\mathcal{S},\ast}(g_{\mathbf{k}})-Z_{q}^{\mathcal{S},\mathcyr{sh}}(E_{\mathbf{k}})$ is a signed sum of

(5.6)

\displaystyle\sum_{s=0}^{m}(-1)^{m-s}\left(Z_{q}(g_{\mathbf{m}}g_{1}^{s})Z_{q}(g_{\mathbf{m}^{\prime}}g_{1}^{m-s})-Z_{q}(E_{\mathbf{m}}\frac{g_{1}^{\mathcyr{sh}_{\hbar}s}}{s!})Z_{q}(E_{\mathbf{m}^{\prime}}\frac{g_{1}^{\mathcyr{sh}_{\hbar}(m-s)}}{(m-s)!})\right)

with admissible indices $\mathbf{m},\mathbf{m}^{\prime}$ and $m\geq 0$ modulo $Z_{q}(\hbar\widehat{\mathfrak{H}^{0}})$ . We denote (5.6) by $J_{m}$ and calculate the generating function $J(X)=\sum_{m\geq 0}J_{m}X^{m}$ . From Theorem 5.8, we see that

	$\displaystyle J(X)$	$\displaystyle=Z_{q}(g_{\mathbf{m}}\frac{1}{1-g_{1}X})Z_{q}(g_{\mathbf{m}^{\prime}}\frac{1}{1+g_{1}X})-Z_{q}(E_{\mathbf{m}}\exp_{\mathcyr{sh}_{\hbar}}{(g_{1}X)})Z_{q}(E_{\mathbf{m}^{\prime}}\exp_{\mathcyr{sh}_{\hbar}}{(-g_{1}X)})$
		$\displaystyle\equiv\left\{\exp{\left(\sum_{k\geq 1}\frac{Z_{q}(g_{2k})}{k}X^{2k}\right)}-1\right\}Z_{q}(g_{\mathbf{m}}\frac{1}{1-g_{1}X})Z_{q}(g_{\mathbf{m}^{\prime}}\frac{1}{1+g_{1}X})$

modulo $\mathcal{N}_{q}[[X]]$ . The right hand side belongs to $\mathcal{P}_{q}[[X]]$ . ∎

Now we are in a position to define a $q$ -analogue of the SMZV.

Definition 5.11.

For an index $\mathbf{k}$ , we define a $q$ -analogue of SMZV ( $q$ SMZV) $\zeta_{q}^{\mathcal{S}}(\mathbf{k})$ as an element of the quotient $\mathcal{Z}_{q}/(\mathcal{N}_{q}+\mathcal{P}_{q})$ by

\displaystyle\zeta_{q}^{\mathcal{S}}(\mathbf{k})=Z_{q}^{\mathcal{S},\ast}(g_{\mathbf{k}})=Z_{q}^{\mathcal{S},\mathcyr{sh}}(E_{\mathbf{k}})

modulo $\mathcal{N}_{q}+\mathcal{P}_{q}$ .

Example 5.12.

Let $k$ be a positive integer. From the definition of $w_{\hbar}^{\mathcal{S},\ast}$ and (5.5), we see that

\displaystyle\sum_{s\geq 0}X^{s}w_{\hbar}^{\mathcal{S},\ast}(g_{k}^{s})=\frac{1}{1-g_{k}X}\ast_{\hbar}\frac{1}{1-(-1)^{k}g_{k}X}=\exp_{\ast_{\hbar}}{\left(\sum_{n\geq 1}\frac{(-1)^{n-1}}{n}(1+(-1)^{kn})g_{kn}X^{n}\right)}.

Hence Proposition 3.2 implies that

\displaystyle\sum_{s\geq 0}X^{s}Z_{q}^{\mathcal{S},\ast}(g_{k}^{s})=\exp{\left(\sum_{n\geq 1}\frac{(-1)^{n-1}}{n}(1+(-1)^{kn})Z_{q}(g_{kn})X^{n}\right)}.

If $k$ is odd, then $1+(-1)^{kn}=0$ unless $n$ is even. Therefore, the right hand side belongs to $1+\mathcal{P}_{q}[[X]]$ and we have

(5.7)

\displaystyle\zeta_{q}^{\mathcal{S}}(\underbrace{k,\ldots,k}_{r})=0

for any $k,r\geq 1$ . In particular, the $q$ SMZV of depth one is always equal to zero.

Example 5.13.

We consider the $q$ SMZV of depth two. Set $\mathbf{k}=(k_{1},k_{2})$ . If $k_{1}$ and $k_{2}$ are even, then $w_{\hbar}^{\mathcal{S},\ast}(g_{\mathbf{k}})=2g_{k_{1}}\ast_{\hbar}g_{k_{2}}-g_{k_{1}+k_{2}}$ . If $k_{1}$ and $k_{2}$ are odd, we have $w_{\hbar}^{\mathcal{S},\ast}(g_{\mathbf{k}})=-g_{k_{1}+k_{2}}$ . In both cases we see that $Z_{q}(w_{\hbar}^{\mathcal{S},\ast}(g_{\mathbf{k}}))$ belongs to $\mathcal{P}_{q}$ from Proposition 3.2. Therefore $\zeta_{q}^{\mathcal{S}}(k_{1},k_{2})=0$ if $k_{1}+k_{2}$ is even.

We consider the case where $k_{1}+k_{2}$ is odd. To this aim we calculate the $q$ MZV $Z_{q}(g_{k_{1}}g_{k_{2}})$ whose weight is odd modulo $\mathcal{N}_{q}+\mathcal{P}_{q}$ . The calculation is similar to that for MZV in [1, 10].

Suppose that $k$ is odd and $k\geq 3$ . For $1\leq m<k$ , we have

\displaystyle g_{m}\ast_{\hbar}g_{k-m}=g_{m}g_{k-m}+g_{k-m}g_{m}+g_{k}.

From Lemma 3.10, we also see that

	$\displaystyle\sum_{m,l\geq 1}X^{m-1}Y^{l-1}g_{m}\,\mathcyr{sh}_{\hbar}\,g_{l}$	$\displaystyle=(1+\hbar X)\sum_{m,l\geq 1}X^{k-1}(X+Y+\hbar XY)^{l-1}g_{m}g_{l}$
		$\displaystyle+(1+\hbar Y)\sum_{m,l\geq 1}Y^{k-1}(X+Y+\hbar XY)^{l-1}g_{m}g_{l}$
		$\displaystyle+\sum_{l\geq 1}(X+Y+\hbar XY)^{l-1}(e_{1}-g_{1})g_{l}.$

Since $k\geq 3$ and $(e_{1}-g_{1})g_{l}$ belongs to $\mathfrak{n}_{0}$ if $l\geq 2$ , we have

\displaystyle g_{m}\,\mathcyr{sh}_{\hbar}\,g_{k-m}\equiv\sum_{j\geq 1}\left(\binom{k-j-1}{k-m-1}+\binom{k-j-1}{m-1}\right)g_{j}g_{k-j}

modulo $\mathfrak{n}$ . Note that $m$ or $k-m$ is even. Hence, from Proposition 3.2 and Proposition 3.3, we see that

(5.8)		$\displaystyle Z_{q}(g_{m}g_{k-m}+g_{k-m}g_{m}+g_{k})\equiv 0,$
(5.9)		$\displaystyle\sum_{j\geq 1}\left(\binom{k-j-1}{k-m-1}+\binom{k-j-1}{m-1}\right)Z_{q}(g_{j}g_{k-j})\equiv 0$

modulo $\mathcal{N}_{q}+\mathcal{P}_{q}$ . Set

\displaystyle\mathcal{D}(X,Y)=\sum_{m=1}^{k-1}X^{m-1}Y^{k-m-1}Z_{q}(g_{m}g_{k-m}),\qquad\mathcal{Z}(X,Y)=\frac{X^{k-1}-Y^{k-1}}{X-Y}Z_{q}(g_{k}).

Then (5.8) and (5.9) imply that

\displaystyle\mathcal{D}(X,Y)+\mathcal{D}(Y,X)+\mathcal{Z}(X,Y)\equiv 0,\qquad\mathcal{D}(X,X+Y)+\mathcal{D}(Y,X+Y)\equiv 0

modulo $(\mathcal{N}_{q}+\mathcal{P}_{q})[X,Y]$ , respectively. Using these relations and $\mathcal{D}(-X,-Y)=-\mathcal{D}(X,Y)$ , we obtain

\displaystyle\mathcal{D}(X,Y)\equiv-\frac{1}{2}\left(\mathcal{Z}(X,Y)+\mathcal{Z}(X-Y,X)+\mathcal{Z}(-Y,X-Y)\right).

Hence we find that

(5.10)

\displaystyle Z_{q}(g_{m}g_{k-m})\equiv-\frac{1}{2}\left(1+(-1)^{m}\binom{k}{m}\right)Z_{q}(g_{k})

modulo $\mathcal{N}_{q}+\mathcal{P}_{q}$ if $1\leq m<k$ and $k$ is odd.

Now suppose that $k_{1},k_{2}\geq 1$ and $k_{1}+k_{2}$ is odd. Since $k_{1}$ or $k_{2}$ is even, we see that

	$\displaystyle Z_{q}^{\mathcal{S},\ast}(g_{k_{1}}g_{k_{2}})$	$\displaystyle=Z_{q}(g_{k_{1}}g_{k_{2}}+(-1)^{k_{2}}g_{k_{1}}\ast_{\hbar}g_{k_{2}}-g_{k_{2}}g_{k_{1}})$
		$\displaystyle\equiv Z_{q}(g_{k_{1}}g_{k_{2}}-g_{k_{2}}g_{k_{1}})$
		$\displaystyle\equiv(-1)^{k_{2}}\binom{k_{1}+k_{2}}{k_{1}}Z_{q}(g_{k_{1}+k_{2}})$

modulo $\mathcal{N}_{q}+\mathcal{P}_{q}$ by using Proposition 3.2 and (5.10).

From the above arguments we see that

\displaystyle\zeta_{q}^{\mathcal{S}}(k_{1},k_{2})=(-1)^{k_{2}}\binom{k_{1}+k_{2}}{k_{1}}Z_{q}(g_{k_{1}+k_{2}})

in the quotient $\mathcal{Z}_{q}/(\mathcal{N}_{q}+\mathcal{P}_{q})$ for any $k_{1},k_{2}\geq 1$ . It is a $q$ -analogue of the formula for the SMZV of depth two

\displaystyle\zeta^{\mathcal{S}}(k_{1},k_{2})=(-1)^{k_{2}}\binom{k_{1}+k_{2}}{k_{1}}\zeta(k_{1}+k_{2})

modulo $\zeta(2)\mathcal{Z}$ (see, e.g., [5]).

We check that our $q$ SMZV is really a $q$ -analogue of the SMZV.

Theorem 5.14.

For any index $\mathbf{k}$ , it holds that

\displaystyle\lim_{q\to 1-0}Z_{q}^{\mathcal{S},\ast}(g_{\mathbf{k}})=\zeta^{\mathcal{S},\ast}(\mathbf{k}),\qquad\lim_{q\to 1-0}Z_{q}^{\mathcal{S},\mathcyr{sh}}(E_{\mathbf{k}})=\zeta^{\mathcal{S},\mathcyr{sh}}(\mathbf{k}).

Proof.

We see that $\iota(\psi^{\bullet}(w))=\psi(\iota(w))$ for any $w\in\mathcal{C}\langle A\rangle$ from the definition of $\iota,\psi^{\bullet}$ and $\psi$ . Hence, Proposition 3.9 implies that $\iota(w_{\hbar}^{\mathcal{S},\bullet}(w))=w^{\mathcal{S},\bullet}(\iota(w))$ for any $w\in\mathcal{C}\langle A\rangle$ and $\bullet\in\{\ast,\mathcyr{sh}\}$ . From the definition of $\iota$ , we have $\iota(g_{\mathbf{k}})=z_{\mathbf{k}}$ for any index $\mathbf{k}$ . Moreover, we have $\iota(e_{k})=z_{k}$ for $k\geq 1$ and

\displaystyle\iota(E_{1^{m}})=\frac{1}{(m+1)!}\sum_{j=0}^{m}\iota(g_{1})^{\mathcyr{sh}j}\,\mathcyr{sh}\,\iota(e_{1})^{\mathcyr{sh}(m-j)}=\frac{z_{1}^{\mathcyr{sh}m}}{m!}=z_{1}^{m}

for $m\geq 0$ . Hence $\iota(E_{\mathbf{k}})=z_{\mathbf{k}}$ for any index $\mathbf{k}$ . Now the desired formula follows from Corollary 3.5. ∎

The limit as $q\to 1-0$ of any element of $\mathcal{N}_{q}$ is zero and that of $\mathcal{P}_{q}$ is contained in $\zeta(2)\mathcal{Z}$ because $\lim_{q\to 1-0}Z_{q}(g_{2k})=\zeta(2k)\in\mathbb{Q}\,\zeta(2)^{k}$ for $k\geq 1$ . Therefore, we have the well-defined map $\mathcal{Z}_{q}/(\mathcal{N}_{q}+\mathcal{P}_{q})\to\mathcal{Z}/\zeta(2)\mathcal{Z}$ which sends the equivalent class of $f(q)\in\mathcal{Z}_{q}$ to that of $\lim_{q\to 1-0}f(q)$ . Theorem 5.14 implies that the map sends the $q$ SMZV $\zeta_{q}^{\mathcal{S}}(\mathbf{k})$ to the SMZV $\zeta^{\mathcal{S}}(\mathbf{k})$ . In this sense we may regard $\zeta_{q}^{\mathcal{S}}(\mathbf{k})$ as a $q$ -analogue of $\zeta^{\mathcal{S}}(\mathbf{k})$ .

6. Relations of the $q$ -analogue of symmetric multiple zeta value

6.1. Reversal relation

Theorem 6.1.

For any index $\mathbf{k}$ , we have

\displaystyle Z_{q}^{\mathcal{S},\ast}(g_{\overline{\mathbf{k}}})=(-1)^{\mathrm{wt}(\mathbf{k})}Z_{q}^{\mathcal{S},\ast}(g_{\mathbf{k}}),

and

\displaystyle Z_{q}^{\mathcal{S},\mathcyr{sh}}(E_{\overline{\mathbf{k}}})\equiv(-1)^{\mathrm{wt}(\mathbf{k})}Z_{q}^{\mathcal{S},\mathcyr{sh}}(E_{\mathbf{k}})

modulo $(1-q)\mathcal{Z}_{q}$ . Therefore, for any index $\mathbf{k}$ , it holds that

\displaystyle\zeta_{q}^{\mathcal{S}}(\overline{\mathbf{k}})=(-1)^{\mathrm{wt}(\mathbf{k})}\zeta_{q}^{\mathcal{S}}(\mathbf{k}).

Proof.

Since $\psi^{\bullet}$ is an anti-involution, we see that $w_{\hbar}^{\mathcal{S},\bullet}(\psi^{\bullet}(w))=w_{\hbar}^{\mathcal{S},\bullet}(w)$ for any $w\in\mathcal{C}\langle A\rangle$ and $\bullet\in\{\ast,\mathcyr{sh}\}$ . From the definition of $\psi^{\ast}$ , we have $\psi^{\ast}(g_{\mathbf{k}})=(-1)^{\mathrm{wt}(\mathbf{k})}g_{\overline{\mathbf{k}}}$ . We also have $\psi^{\mathcyr{sh}}(E_{\mathbf{k}})\equiv(-1)^{\mathrm{wt}(\mathbf{k})}E_{\overline{\mathbf{k}}}$ modulo $\hbar\,\mathfrak{e}$ (see Corollary B.6). Thus we obtain the desired equalities. ∎

6.2. Double shuffle relation

From Proposition 4.1, we obtain the following relation of $q$ SMZVs.

Proposition 6.2.

For any index $\mathbf{k}$ and $\mathbf{l}$ , it holds that

\displaystyle Z_{q}^{\mathcal{S},\ast}(g_{\mathbf{k}}\ast_{\hbar}g_{\mathbf{l}})=Z_{q}^{\mathcal{S},\ast}(g_{\mathbf{k}})Z_{q}^{\mathcal{S},\ast}(g_{\mathbf{l}}).

Next we consider the shuffle relation. Note that the product $E_{\mathbf{k}}\,\mathcyr{sh}_{\hbar}\,E_{\mathbf{l}}$ does not necessarily belong to $\mathfrak{e}$ as follows.

Example 6.3.

We have

\displaystyle E_{1}\,\mathcyr{sh}_{\hbar}\,E_{1}=\frac{1}{4}(g_{1}+e_{1})\,\mathcyr{sh}_{\hbar}\,(g_{1}+e_{1})=\frac{1}{4}(g_{1}^{2}+3g_{1}e_{1}+3e_{1}g_{1}+e_{1}^{2}),

which is not a $\mathcal{C}$ -linear combination of $E_{11}=E_{1^{2}}=(e_{1}^{2}+2e_{1}g_{1}+2g_{1}e_{1}+g_{1}^{2})/6,E_{2}=e_{2}$ and $E_{1}=(e_{1}+g_{1})/2$ .

However, we have the following proposition. We set

\displaystyle\mathfrak{e}^{0}=\sum_{\mathbf{k}\in I_{0}}\mathcal{C}E_{\mathbf{k}}.

Proposition 6.4.

Let $\mathbf{k}$ and $\mathbf{l}$ be an index. If at least one of $\mathbf{k}$ and $\mathbf{l}$ is admissible, then $E_{\mathbf{k}}\,\mathcyr{sh}_{\hbar}\,E_{\mathbf{l}}$ belongs to $\mathfrak{e}$ . If both $\mathbf{k}$ and $\mathbf{l}$ are admissible, then $E_{\mathbf{k}}\,\mathcyr{sh}_{\hbar}\,E_{\mathbf{l}}$ belongs to $\mathfrak{e}^{0}$ .

See Appendix B.5 for the proof.

Proposition 6.5.

Let $\mathbf{k}$ and $\mathbf{l}$ be an index and suppose that at least one of them is admissible. Then it holds that

\displaystyle Z_{q}^{\mathcal{S},\mathcyr{sh}}(E_{\mathbf{k}}\mathcyr{sh}_{\hbar}E_{\mathbf{l}})\equiv(-1)^{\mathrm{wt}(\mathbf{l})}Z_{q}^{\mathcal{S},\mathcyr{sh}}(E_{(\mathbf{k},\overline{\mathbf{l}})}),

modulo $(1-q)\mathcal{Z}_{q}$ , where $(\mathbf{k},\overline{\mathbf{l}})$ is the concatenation of $\mathbf{k}$ and $\overline{\mathbf{l}}$ .

Proof.

Proposition 4.2 implies that $Z_{q}^{\mathcal{S},\mathcyr{sh}}(E_{\mathbf{k}}\,\mathcyr{sh}_{\hbar}\,E_{\mathbf{l}})=Z_{q}^{\mathcal{S},\mathcyr{sh}}(E_{\mathbf{k}}\psi^{\mathcyr{sh}}(E_{\mathbf{l}}))$ . From Corollary B.6, we see that $E_{\mathbf{k}}\psi^{\mathcyr{sh}}(E_{\mathbf{l}})\equiv(-1)^{\mathrm{wt}(\mathbf{l})}E_{(\mathbf{k},\overline{\mathbf{l}})}$ modulo $\hbar\,\mathfrak{e}$ if $\mathbf{k}$ or $\mathbf{l}$ is admissible. ∎

From Proposition 6.2 and Proposition 6.5, we obtain the double shuffle relation of $q$ SMZVs as follows.

Theorem 6.6.

Let $d_{\mathbf{k},\mathbf{l}}^{\,\bullet,\mathbf{m}}$ be the non-negative integer defined by (2.4).

(i)

For any index $\mathbf{k}$ and $\mathbf{l}$ , it holds that

\displaystyle\zeta_{q}^{\mathcal{S}}(\mathbf{k})\zeta_{q}^{\mathcal{S}}(\mathbf{l})=\sum_{\mathbf{m}}d_{\mathbf{k},\mathbf{l}}^{\,\ast,\mathbf{m}}\zeta_{q}^{\mathcal{S}}(\mathbf{m}).

(ii)

Moreover, if at least one of $\mathbf{k}$ and $\mathbf{l}$ is admissible, it holds that

\displaystyle(-1)^{\mathrm{wt}(\mathbf{l})}\zeta_{q}^{\mathcal{S}}(\mathbf{k},\overline{\mathbf{l}})=\sum_{\mathbf{m}}d_{\mathbf{k},\mathbf{l}}^{\,\mathcyr{sh},\mathbf{m}}\zeta_{q}^{\mathcal{S}}(\mathbf{m}).

Proof.

We prove (ii) by using the map $\iota$ defined by (3.14). Proposition 6.4 implies that there exists $c_{\mathbf{k},\mathbf{l}}^{\mathbf{m}}(\hbar)\in\mathbb{Q}[\hbar]$ such that $E_{\mathbf{k}}\,\mathcyr{sh}_{\hbar}\,E_{\mathbf{l}}=\sum_{\mathbf{m}}c_{\mathbf{k},\mathbf{l}}^{\mathbf{m}}(\hbar)E_{\mathbf{m}}$ . As shown in the proof of Theorem 5.14, we have $\iota(E_{\mathbf{k}})=z_{\mathbf{k}}$ for any index $\mathbf{k}$ . Hence we see that

\displaystyle\iota(E_{\mathbf{k}}\,\mathcyr{sh}_{\hbar}\,E_{\mathbf{l}})=\sum_{\mathbf{m}}c_{\mathbf{k},\mathbf{l}}^{\mathbf{m}}(0)z_{\mathbf{m}}.

On the other hand, Proposition 3.9 implies that

\displaystyle\iota(E_{\mathbf{k}}\,\mathcyr{sh}_{\hbar}\,E_{\mathbf{l}})=z_{\mathbf{k}}\,\mathcyr{sh}\,z_{\mathbf{l}}=\sum_{\mathbf{m}}d_{\mathbf{k},\mathbf{l}}^{\,\mathcyr{sh},\mathbf{m}}z_{\mathbf{m}}.

Hence $c_{\mathbf{k},\mathbf{l}}^{\mathbf{m}}(0)=d_{\mathbf{k},\mathbf{l}}^{\,\mathcyr{sh},\mathbf{m}}$ and

\displaystyle E_{\mathbf{k}}\,\mathcyr{sh}_{\hbar}\,E_{\mathbf{l}}-\sum_{\mathbf{m}}d_{\mathbf{k},\mathbf{l}}^{\,\mathcyr{sh},\mathbf{m}}E_{\mathbf{m}}\in\hbar\,\mathfrak{e}.

Therefore we see that (ii) is true from Proposition 6.5 and $(1-q)\mathcal{Z}_{q}\subset\mathcal{N}_{q}$ . ∎

6.3. Ohno-type relation

From Theorem 6.6, we see that the $q$ SMZVs satisfy a part of the Ohno-type relation.

Theorem 6.7.

Let $\mathbf{k}$ be a non-empty admissible index. Then it holds that

(6.1)

\displaystyle\sum_{\begin{subarray}{c}\mathbf{e}\in(\mathbb{Z}_{\geq 0})^{r}\\ \mathrm{wt}(\mathbf{e})=m\end{subarray}}\zeta_{q}^{\mathcal{S}}(\mathbf{k}+\mathbf{e})=\sum_{\begin{subarray}{c}\mathbf{e}\in(\mathbb{Z}_{\geq 0})^{s}\\ \mathrm{wt}(\mathbf{e})=m\end{subarray}}\zeta_{q}^{\mathcal{S}}((\mathbf{k}^{\vee}+\mathbf{e})^{\vee})

for any $m\geq 1$ , where $r=\mathrm{dep}(\mathbf{k})$ and $s=\mathrm{dep}(\mathbf{k}^{\vee})$ .

Proof.

We define the $\mathbb{Q}$ -linear map $\eta:\mathfrak{h}^{1}\rightarrow\mathcal{Z}_{q}/(\mathcal{N}_{q}+\mathcal{P}_{q})$ by $\eta(z_{\mathbf{k}})=\zeta^{\mathcal{S}}_{q}(\mathbf{k})$ for an index $\mathbf{k}$ . Theorem 6.6 and (5.7) imply the following properties:

(i)

For any index $\mathbf{k}$ and $n\geq 1$ , it holds that $\eta(z_{\mathbf{k}}\ast z_{1}^{n})=0$ .
(ii)

For any admissible index $\mathbf{k}$ and $n\geq 0$ , it holds that $\eta(z_{\mathbf{k}}\,\mathcyr{sh}\,z_{1}^{n})=(-1)^{n}\eta(z_{\mathbf{k}}z_{1}^{n})$ .

For a non-empty index $\mathbf{k}=(k_{1},\ldots,k_{r})$ and $s\geq 0$ , we define the element $a_{s}(\mathbf{k})$ of $\mathfrak{h}^{1}$ by

\displaystyle a_{s}(\mathbf{k})=\sum\prod_{1\leq i\leq r}^{\curvearrowright}\left(y\prod_{1\leq l\leq k_{i}}^{\curvearrowright}(y^{j_{l}^{(i)}}x)\right),

where the sum is over the set

\displaystyle\left\{(j_{l}^{(i)})_{\begin{subarray}{c}1\leq i\leq r\\ 1\leq l\leq k_{i}\end{subarray}}\in(\mathbb{Z}_{\geq 0})^{\mathrm{wt}(\mathbf{k})}\mid\sum_{i=1}^{r}\sum_{l=1}^{k_{i}}j_{l}^{(i)}=s\right\}.

We also set

\displaystyle A_{\mathbf{k},s,p}=\sum_{\begin{subarray}{c}\lambda_{1},\ldots,\lambda_{r}\in\{0,1\}\\ \lambda_{1}+\cdots+\lambda_{r}=p\end{subarray}}a_{s}(k_{1}+\lambda_{1}-1,\ldots,k_{r}+\lambda_{r}-1)

for $s,p\geq 0$ . Then we have

\displaystyle\sum_{p=0}^{\min(n,r)}\sum_{\begin{subarray}{c}m+s=n-p\\ m,s\geq 0\end{subarray}}(-1)^{s}A_{\mathbf{k},s,p}\,\mathcyr{sh}\,z_{1}^{m}=z_{\mathbf{k}}\ast z_{1}^{n}

for any $n\geq 1$ (see equation (3) in [7]).

Now suppose that $\mathbf{k}$ is admissible. Then $A_{\mathbf{k},s,p}$ belongs to $\sum_{\mathbf{l}\in I_{0}}\mathbb{Q}z_{\mathbf{l}}$ for any $s,p\geq 0$ . Therefore we can use the properties (i) and (ii), and see that

\displaystyle\sum_{p=0}^{\min(n,r)}(-1)^{p}\sum_{\begin{subarray}{c}m+s=n-p\\ m,s\geq 0\end{subarray}}\eta(A_{\mathbf{k},s,p}z_{1}^{m})=0

for any $n\geq 1$ . From the above identity, we obtain (6.1) in the same way as the proof of Theorem 2.5 in [7]. ∎

Appendix A Proof of Proposition 3.4

Lemma A.1.

Suppose that $\alpha\geq 0$ and $\max{(1,\alpha)}\leq\beta\leq 2\alpha+1$ . Then the function $f(x)=x^{\alpha}(1-x)/(1-x^{\beta})$ is non-decreasing on the interval $[0,1)$ .

Proof.

If $\beta=1$ , the statement is trivial. We assume that $\beta>1$ . From the assumption we see that $\alpha\geq(\beta-1)/2>0$ . Set

\displaystyle g(x)=x^{1-\alpha}(1-x^{\beta})^{2}f^{\prime}(x)=\alpha-(\alpha+1)x+(\beta-\alpha)x^{\beta}+(\alpha-\beta+1)x^{\beta+1}.

Since $g(0)=\alpha>0$ and $g(1)=0$ , it suffices to show that $g^{\prime}(x)<0$ on the interval $(0,1)$ . Set

\displaystyle h(x)=x^{-\beta}g^{\prime}(x)=-(\alpha+1)x^{-\beta}+\beta(\beta-\alpha)x^{-1}+(\alpha-\beta+1)(\beta+1).

Then we see that $h(1)=0$ and, from the assumption,

\displaystyle h^{\prime}(x)=\beta x^{-\beta-1}(\alpha+1-(\beta-\alpha)x^{\beta-1})>\beta x^{-\beta-1}(2\alpha-\beta+1)\geq 0

for $0<x<1$ . Therefore we see that $h(x)<0$ , and hence $g^{\prime}(x)<0$ , on the interval $(0,1)$ . ∎

Corollary A.2.

Under the assumption of Lemma A.1, it holds that

\displaystyle x^{\alpha}\frac{1-x}{1-x^{\beta}}\leq\frac{1}{\beta}

for $0<x<1$ .

Proof.

It is because $f(x)\to 1/\beta$ as $x\to 1-0$ . ∎

Proposition A.3.

Suppose that $0<q<1$ . Set

\displaystyle B_{q}(\alpha_{1},\ldots,\alpha_{r};\beta_{1},\ldots,\beta_{r})=\sum_{l_{1},\ldots,l_{r}\geq 1}\prod_{j=1}^{r}\binom{l_{j}}{\alpha_{j}}\frac{(1-q)^{\alpha_{j}}q^{l_{j}/2}}{(l_{1}+\cdots+l_{j})^{\beta_{j}+1}}.

Then, for any non-negative integer $\alpha_{1},\ldots,\alpha_{r},\beta_{1},\ldots,\beta_{r}$ , it holds that

\displaystyle 0\leq Z_{q}((e_{1}-g_{1})^{\alpha_{1}}g_{\beta_{1}+1}\cdots(e_{1}-g_{1})^{\alpha_{r}}g_{\beta_{r}+1})\leq B_{q}(\alpha_{1},\ldots,\alpha_{r};\beta_{1},\ldots,\beta_{r}).

Proof.

From Corollary A.2, we see that

\displaystyle 0\leq\frac{q^{n}}{[n]}=q^{(n-1)/2}\frac{1-q}{1-q^{n}}q^{(n+1)/2}\leq\frac{q^{(n+1)/2}}{n}

for any $n\geq 1$ . Therefore, it holds that

	$\displaystyle 0$	$\displaystyle\leq Z_{q}((e_{1}-g_{1})^{\alpha_{1}}g_{\beta_{1}+1}\cdots(e_{1}-g_{1})^{\alpha_{r}}g_{\beta_{r}+1})$
		$\displaystyle=\sum_{0=n_{0}<n_{1}<\cdots<n_{r}}\prod_{j=1}^{r}(1-q)^{\alpha_{j}}\binom{n_{j}-n_{j-1}-1}{\alpha_{j}}\left(\frac{q^{n_{j}}}{[n_{j}]}\right)^{\beta_{j}+1}$
		$\displaystyle\leq\sum_{0=n_{0}<n_{1}<\cdots<n_{r}}\prod_{j=1}^{r}(1-q)^{\alpha_{j}}\binom{n_{j}-n_{j-1}}{\alpha_{j}}\left(\frac{q^{(n_{j}+1)/2}}{n_{j}}\right)^{\beta_{j}+1}.$

Set $l_{j}=n_{j}-n_{j-1}$ for $1\leq j\leq r$ . Then the right-hand side is dominated by $B_{q}(\alpha_{1},\ldots,\alpha_{r};\beta_{1},\ldots,\beta_{r})$ since $0<q<1$ . ∎

Proposition A.4.

Suppose that $0<q<1$ and $\alpha_{1},\ldots,\alpha_{r},\beta_{1},\ldots,\beta_{r}$ are non-negative integers. Then it holds that

(A.1)

\displaystyle B_{q}(\alpha_{1},\ldots,\alpha_{r};\beta_{1},\ldots,\beta_{r})=O((-\log{(1-q)})^{r})\qquad(q\to 1-0).

Moreover, if $\alpha_{s}\geq 1$ and $\beta_{t}\geq 1$ for some $1\leq s\leq t\leq r$ , then we have

(A.2)

\displaystyle B_{q}(\alpha_{1},\ldots,\alpha_{r};\beta_{1},\ldots,\beta_{r})=(1-q)\,O((-\log{(1-q)})^{r})\qquad(q\to 1-0).

Proof.

For a non-negative integer $\alpha$ , we define

\displaystyle\varphi_{\alpha}(x)=(1-x)^{\alpha}\sum_{l\geq 1}\binom{l}{\alpha}\frac{x^{l/2}}{l},\qquad\tilde{\varphi}_{\alpha}(x)=(1-x)^{\alpha}\sum_{l\geq 1}\binom{l}{\alpha}\frac{x^{l/2}}{l^{2}}.

We have

\displaystyle\varphi_{0}(x)=-\log{(1-x^{1/2})},\qquad\varphi_{\alpha}(x)=\frac{x^{\alpha/2}}{\alpha}(1+x^{1/2})^{\alpha}\quad(\alpha\geq 1).

Hence $\varphi_{\alpha}(x)=O(-\log{(1-x)})$ as $x\to 1-0$ for any $\alpha\geq 0$ . If $\alpha\geq 1$ , we see that

\displaystyle 0\leq\tilde{\varphi}_{\alpha}(x)=(1-x)^{\alpha}\sum_{l\geq 1}\frac{1}{\alpha}\binom{l-1}{\alpha-1}\frac{x^{l/2}}{l}\leq\frac{1-x}{\alpha}\varphi_{\alpha-1}(x)

for $0\leq x\leq 1$ . Hence $\tilde{\varphi}_{\alpha}(x)=(1-x)\,O(-\log{(1-x)})$ as $x\to 1-0$ for any $\alpha\geq 1$ .

Since $\beta_{j}\geq 0$ for $1\leq j\leq r$ , we see that

\displaystyle 0\leq B_{q}(\alpha_{1},\ldots,\alpha_{r};\beta_{1},\ldots,\beta_{r})\leq\prod_{j=1}^{r}\varphi_{\alpha_{j}}(q).

Hence we have (A.1). Now assume further that $\alpha_{s}\geq 1$ and $\beta_{t}\geq 1$ for some $1\leq s\leq t\leq r$ . If $s=t$ , it holds that

\displaystyle 0\leq B_{q}(\alpha_{1},\ldots,\alpha_{r};\beta_{1},\ldots,\beta_{r})\leq\tilde{\varphi}_{\alpha_{s}}(q)\prod_{\begin{subarray}{c}1\leq j\leq r\\ j\not=s\end{subarray}}\varphi_{\alpha_{j}}(q).

Hence we obtain (A.2) if $s=t$ . If $s<t$ , we see that

	$\displaystyle 0$	$\displaystyle\leq B_{q}(\alpha_{1},\ldots,\alpha_{r};\beta_{1},\ldots,\beta_{r})$
		$\displaystyle\leq\sum_{l_{1},\ldots,l_{r}\geq 1}\prod_{\begin{subarray}{c}1\leq j\leq r\\ j\not=s,t\end{subarray}}\left\{(1-q)^{\alpha_{j}}\binom{l_{j}}{\alpha_{j}}\frac{q^{l_{j}/2}}{l_{j}}\right\}(1-q)^{\alpha_{s}+\alpha_{t}}\binom{l_{s}}{\alpha_{s}}\binom{l_{t}}{\alpha_{t}}\frac{q^{l_{s}/2}}{(l_{1}+\cdots+l_{s})^{\beta_{s}+\beta_{t}+1}}\frac{q^{l_{t}/2}}{l_{1}+\cdots+l_{t}}$
		$\displaystyle\leq\sum_{l_{1},\ldots,l_{r}\geq 1}\prod_{\begin{subarray}{c}1\leq j\leq r\\ j\not=s\end{subarray}}\left\{(1-q)^{\alpha_{j}}\binom{l_{j}}{\alpha_{j}}\frac{q^{l_{j}/2}}{l_{j}}\right\}(1-q)^{\alpha_{s}}\binom{l_{s}}{\alpha_{s}}\frac{q^{l_{s}/2}}{l_{s}^{2}}$
		$\displaystyle=\tilde{\varphi}_{\alpha_{s}}(q)\prod_{\begin{subarray}{c}1\leq j\leq r\\ j\not=s\end{subarray}}\varphi_{\alpha_{j}}(q)$

since $\beta_{t}\geq 1$ . Thus we get (A.2) in the case where $s<t$ either. ∎

Proof of Proposition 3.4.

From Lemma A.1 and the monotone convergence theorem, we see that $Z_{q}(g_{\mathbf{k}})\to\zeta(\mathbf{k})$ as $q\to 1-0$ for any admissible index $\mathbf{k}$ . Suppose that $w$ is an element of $\widehat{\mathfrak{H}^{0}}$ of the form (3.2). Proposition A.3 and (A.1) imply that $Z_{q}(w)=O((-\log{(1-q)})^{r})$ as $q\to 1-0$ . Hence $Z_{q}(\hbar w)\to 0$ in the limit as $q\to 1-0$ . If $w\in\mathfrak{n}_{0}$ , then there exist $s$ and $t$ such that $\alpha_{s}\geq 1,\beta_{t}\geq 1$ and $1\leq s\leq t\leq r$ . Then Proposition A.3 and (A.2) imply that $Z_{q}(w)\to 0$ in the limit as $q\to 1-0$ . Thus we see that $\lim_{q\to 1-0}Z_{q}(w)=0$ for any $w\in\mathfrak{n}$ . ∎

Appendix B Proofs

B.1. Formulas of shuffle product

Proposition B.1.

Suppose that $U(X)$ and $V(X)$ belong to the ideal $X\mathcal{C}[b][[X]]$ of the formal power series ring $\mathcal{C}[b][[X]]$ generated by $X$ . Then it holds that

	$\displaystyle\left(w\frac{1}{1-U(X)a}\,\mathcyr{sh}_{\hbar}\,w^{\prime}\frac{1}{1-V(Y)a}\right)\left\{1-(U(X)+V(Y)+\hbar\,U(X)V(Y))a\right\}$
	$\displaystyle=-w\,\mathcyr{sh}_{\hbar}\,w^{\prime}+\left(w\frac{1}{1-U(X)a}\,\mathcyr{sh}_{\hbar}\,w^{\prime}\right)(1-U(X)a)+\left(w\,\mathcyr{sh}_{\hbar}\,w^{\prime}\frac{1}{1-V(Y)a}\right)(1-V(Y)a)$

for any $w,w^{\prime}\in\mathfrak{H}$ .

Proof.

Set

\displaystyle I(X,Y)=w\frac{1}{1-U(X)a}\,\mathcyr{sh}_{\hbar}\,w^{\prime}\frac{1}{1-V(Y)a}.

Using $(1-U(X)a)^{-1}=1+(1-U(X)a)^{-1}U(X)a$ , we see that

	$\displaystyle I(X,Y)$	$\displaystyle=-w\,\mathcyr{sh}_{\hbar}\,w^{\prime}+w\,\mathcyr{sh}_{\hbar}w^{\prime}\frac{1}{1-V(Y)a}+w\frac{1}{1-U(X)a}\,\mathcyr{sh}_{\hbar}w^{\prime}$
		$\displaystyle+w\frac{1}{1-U(X)a}U(X)a\,\mathcyr{sh}_{\hbar}w^{\prime}\frac{1}{1-V(Y)a}V(Y)a.$

Since all the coefficients of $U(X)$ and $V(Y)$ are polynomials in $b$ , the fourth term of the right hand side is equal to

	$\displaystyle\left(w\frac{1}{1-U(X)a}\,\mathcyr{sh}_{\hbar}w^{\prime}\frac{1}{1-V(Y)a}V(Y)a\right)U(X)a$
	$\displaystyle+\left(w\frac{1}{1-U(X)a}U(X)a\,\mathcyr{sh}_{\hbar}w^{\prime}\frac{1}{1-V(Y)a}\right)V(Y)a$
	$\displaystyle+\left(w\frac{1}{1-U(X)a}\,\mathcyr{sh}_{\hbar}w^{\prime}\frac{1}{1-V(Y)a}\right)\hbar\,U(X)V(Y)a.$

Using $(1-U(X)a)^{-1}=1+(1-U(X)a)^{-1}U(X)a$ again, we see that it is equal to

	$\displaystyle I(X,Y)(U(X)+V(Y)+\hbar\,U(X)V(Y))a$
	$\displaystyle-\left(w\frac{1}{1-U(X)a}\,\mathcyr{sh}_{\hbar}w^{\prime}\right)U(X)a-\left(w\,\mathcyr{sh}_{\hbar}w^{\prime}\frac{1}{1-V(Y)a}\right)V(Y)a.$

Thus we get the desired equality. ∎

Proposition B.2.

Suppose that $U(X)\in X\mathcal{C}[b][[X]]$ . Then it holds that

	$\displaystyle\left(w\frac{1}{1-U(X)a}\,\mathcyr{sh}_{\hbar}\,w^{\prime}a\right)(1-U(X)a)$
	$\displaystyle=w\,\mathcyr{sh}_{\hbar}\,w^{\prime}a-(w\,\mathcyr{sh}_{\hbar}\,w^{\prime})a+\left(w\frac{1}{1-U(X)a}\,\mathcyr{sh}_{\hbar}\,w^{\prime}\right)(1+\hbar\,U(X))a$

for any $w,w^{\prime}\in\mathfrak{H}$ .

Proof.

Set $V(Y)=Y$ in Proposition B.1, differentiate the both hand sides with respect to $Y$ and set $Y=0$ . Then we get the desired equality. ∎

Proof of Lemma 3.10.

From Proposition B.1 and Proposition B.2, we see that

	$\displaystyle\left(ug_{1}\frac{1}{1-aX}\,\mathcyr{sh}_{\hbar}\,vg_{1}\frac{1}{1-aY}\right)(1-(X+Y+\hbar XY)a)$
	$\displaystyle=-ug_{1}\,\mathcyr{sh}_{\hbar}\,vg_{1}+\left(ug_{1}\frac{1}{1-aX}\,\mathcyr{sh}_{\hbar}\,vg_{1}\right)(1-aX)+\left(ug_{1}\,\mathcyr{sh}_{\hbar}\,vg_{1}\frac{1}{1-aY}\right)(1-aY)$
	$\displaystyle=ug_{1}\,\mathcyr{sh}_{\hbar}\,vg_{1}-(ug_{1}\,\mathcyr{sh}_{\hbar}\,v+u\,\mathcyr{sh}_{\hbar}\,vg_{1})g_{1}$
	$\displaystyle+(1+\hbar X)\left(ug_{1}\frac{1}{1-aX}\,\mathcyr{sh}_{\hbar}\,v\right)g_{1}+(1+\hbar Y)\left(u\,\mathcyr{sh}_{\hbar}\,v\frac{1}{1-aX}\right)g_{1}.$

Using (3.9), we get the desired equality. ∎

Corollary B.3.

For $U(X)\in X\mathcal{C}[b][[X]]$ , we define the map $\rho_{U(X)}:\mathfrak{H}\rightarrow\mathfrak{H}[[X]]$ by

\displaystyle\rho_{U(X)}(w)=\left(\frac{1}{1-U(X)g_{1}}\,\mathcyr{sh}_{\hbar}\,w\right)\left(1-U(X)g_{1}\right).

Then the map $\rho_{U(X)}$ is a $\mathcal{C}$ -algebra homomorphism with respect to the concatenation product on $\mathfrak{H}$ , and we have

(B.1)

\displaystyle\rho_{U(X)}(a)=\frac{1}{1-U(X)g_{1}}\left(1+\hbar b\,U(X)\right)a,\quad\rho_{U(X)}(b)=\frac{1}{1-U(X)g_{1}}b\left(1-U(X)g_{1}\right).

Proof.

Set $w=1$ and $U(X)$ to $bU(X)$ in Proposition B.2. Then we see that

\displaystyle\rho_{U(X)}(w^{\prime}a)=\rho_{U(X)}(w^{\prime})\frac{1}{1-U(X)g_{1}}(1+\hbar bU(X))a.

From the definition of $\rho_{U(X)}$ , we also see that

\displaystyle\rho_{U(X)}(w^{\prime}b)=\rho_{U(X)}(w^{\prime})\frac{1}{1-U(X)g_{1}}b(1-U(X)g_{1}).

Since $\rho_{U(X)}(1)=1$ , we have (B.1) and see that $\rho_{U(X)}$ is an algebra homomorphism. ∎

B.2. Generating function of $E_{1^{m}}$

Here we calculate the generating function

\displaystyle E(X)=\sum_{m=0}^{\infty}E_{1^{m}}X^{m}.

Recall that

\displaystyle R(X)=\frac{e^{\hbar bX}-1}{\hbar b}=\sum_{n=1}^{\infty}\frac{X^{n}}{n!}(e_{1}-g_{1})^{n-1}.

Note that $R(X)$ belongs to $X\mathcal{C}[b][[X]]$ .

Proposition B.4.

It holds that

(B.2)

\displaystyle\exp_{\mathcyr{sh}_{\hbar}}{\left(g_{1}X\right)}=\frac{1}{1-R(X)g_{1}}.

Proof.

The power series $\Psi(X)=\exp_{\mathcyr{sh}_{\hbar}}{\left(g_{1}X\right)}$ is the unique solution of the differential equation $\Psi^{\prime}(X)=g_{1}\,\mathcyr{sh}_{\hbar}\,\Psi(X)$ satisfying $\Psi(0)=1$ . Since the right hand side with $X=0$ is equal to one, it suffices to show that

\displaystyle\frac{d}{dX}\frac{1}{1-R(X)g_{1}}=g_{1}\,\mathcyr{sh}_{\hbar}\,\frac{1}{1-R(X)g_{1}}.

The right hand side is equal to $\rho_{R(X)}(g_{1})(1-R(X)g_{1})^{-1}$ . Hence (B.1) implies that

	$\displaystyle g_{1}\,\mathcyr{sh}_{\hbar}\,\frac{1}{1-R(X)g_{1}}$	$\displaystyle=\rho_{R(X)}(b)\rho_{R(X)}(a)\frac{1}{1-R(X)g_{1}}$
		$\displaystyle=\frac{1}{1-R(X)g_{1}}e^{\hbar bX}g_{1}\frac{1}{1-R(X)g_{1}}=\frac{d}{dX}\frac{1}{1-R(X)g_{1}}$

since $R^{\prime}(X)=e^{\hbar bX}$ . ∎

Proposition B.5.

It holds that

\displaystyle E(X)=\frac{1}{1-R(X)g_{1}}\frac{R(X)}{X}.

Proof.

Note that $g_{1}$ and $\hbar b$ are commutative with respect to the shuffle product $\mathcyr{sh}_{\hbar}$ . From the definition of $E_{1^{m}}$ , we see that

	$\displaystyle E(X)$	$\displaystyle=\sum_{m=0}^{\infty}\frac{1}{(m+1)!}\sum_{s=0}^{m}g_{1}^{\mathcyr{sh}_{\hbar}(m-s)}\,\mathcyr{sh}_{\hbar}\,(g_{1}+\hbar b)^{\mathcyr{sh}_{\hbar}s}X^{m}$
		$\displaystyle=\sum_{m\geq s\geq j\geq 0}\frac{1}{(m+1)!}\binom{s}{j}g_{1}^{\mathcyr{sh}_{\hbar}(m-j)}(\hbar b)^{j}X^{m}.$

Using

\displaystyle\sum_{s=j}^{m}\binom{s}{j}=\sum_{s=j}^{m}\left(\binom{s+1}{j+1}-\binom{s}{j+1}\right)=\binom{m+1}{j+1}

and (B.2), we see that

\displaystyle E(X)=\sum_{m\geq j\geq 0}\frac{g_{1}^{\mathcyr{sh}_{\hbar}(m-j)}}{(m-j)!}X^{m-j}\frac{(\hbar b)^{j}}{(j+1)!}X^{j}=\exp_{\mathcyr{sh}_{\hbar}}{(g_{1}X)}\frac{e^{\hbar bX}-1}{\hbar bX}=\frac{1}{1-R(X)g_{1}}\frac{R(X)}{X}.

∎

Corollary B.6.

For any index $\mathbf{k}$ , it holds that $\psi^{\mathcyr{sh}}(E_{\mathbf{k}})\equiv(-1)^{\mathrm{wt}(\mathbf{k})}E_{\overline{\mathbf{k}}}$ modulo $\hbar\mathfrak{e}$ . Moreover, if $\mathbf{k}$ (resp. $\overline{\mathbf{k}}$ ) is admissible, then $\psi^{\mathcyr{sh}}(E_{\mathbf{k}})-(-1)^{\mathrm{wt}(\mathbf{k})}E_{\overline{\mathbf{k}}}$ belongs to $\hbar\sum_{j\geq 2}e_{j}\mathfrak{e}$ (resp. $\hbar\sum_{j\geq 2}\mathfrak{e}e_{j}$ ).

Proof.

For $k\geq 2$ , using (3.1), we see that

(B.3)

\displaystyle\psi^{\mathcyr{sh}}(e_{k})=\psi^{\mathcyr{sh}}(g_{k}+\hbar g_{k-1})=(-1)^{k}ba(a+\hbar)^{k-1}=(-1)^{k}\sum_{j=2}^{k}\binom{k-2}{j-2}\hbar^{k-j}e_{j}.

Next we calculate $\psi^{\mathcyr{sh}}(E_{1^{m}})$ using Proposition B.5. Since $\psi^{\mathcyr{sh}}(g_{1})=-e_{1}$ and $\psi^{\mathcyr{sh}}(\hbar b)=\hbar b$ , it holds that

\displaystyle\psi^{\mathcyr{sh}}(E(X))=\frac{R(X)}{X}\frac{1}{1+e_{1}R(X)}=\frac{1}{1+R(X)e_{1}}\frac{R(X)}{X}.

We have

(B.4)

\displaystyle 1+R(X)e_{1}=1+R(X)(e_{1}-g_{1})+R(X)g_{1}=e^{\hbar bX}+R(X)g_{1}=e^{\hbar bX}\left(1-R(-X)g_{1}\right).

Thus we see that $\psi^{\mathcyr{sh}}(E(X))=E(-X)$ . Hence

(B.5)

\displaystyle\psi^{\mathcyr{sh}}(E_{1^{m}})=(-1)^{m}E_{1^{m}}

for $m\geq 0$ . Therefore, for an index $\mathbf{k}$ of the form (5.1), we have

\displaystyle\psi^{\mathcyr{sh}}(E_{\mathbf{k}})=(-1)^{\mathrm{wt}(\mathbf{k})}\sum_{j_{1}=0}^{t_{1}}\cdots\sum_{j_{r}=0}^{t_{r}}\left\{\prod_{l=1}^{r}\hbar^{t_{l}-j_{l}}\binom{t_{l}}{j_{l}}\right\}E_{1^{s_{r}}}e_{j_{r}+2}\cdots E_{1^{s_{1}}}e_{j_{1}+2}E_{1^{s_{0}}},

and it implies the statement. ∎

B.3. Proof of Proposition 5.4

Note that

(B.6)

\displaystyle(\mathcal{C}\langle A\rangle\setminus\mathcal{C})a\subset\mathfrak{H}^{0}

because $(e_{1}-g_{1})a=\hbar g_{1}$ and $g_{k}a=g_{k+1}$ for $k\geq 1$ .

We define the $\mathcal{C}$ -trilinear map $K_{\hbar}^{\mathcyr{sh}}:\mathcal{C}\langle A\rangle\times\mathcal{C}\langle A\rangle\times\mathcal{C}\langle A\rangle\rightarrow\widehat{\mathfrak{H}^{1}}$ by

\displaystyle K_{\hbar}^{\mathcyr{sh}}(w,u_{1}\cdots u_{r},w^{\prime})=\sum_{i=0}^{r}wu_{1}\cdots u_{i}\mathcyr{sh}_{\hbar}\psi^{\mathcyr{sh}}(u_{i+1}\cdots u_{r}w^{\prime})

for $w,w^{\prime}\in\mathcal{C}\langle A\rangle$ and $u_{1},\ldots,u_{r}\in A$ . It suffices to show that $K_{\hbar}^{\mathcyr{sh}}(E_{\mathbf{m}},E_{1^{r}},E_{\overline{\mathbf{m}^{\prime}}})$ belongs to $\mathfrak{H}^{0}$ for any admissible index $\mathbf{m},\mathbf{m}^{\prime}$ and $r\geq 0$ .

Lemma B.7.

For any $w,w^{\prime}\in\mathcal{C}\langle A\rangle$ , it holds that

(B.7)

\displaystyle K_{\hbar}^{\mathcyr{sh}}(w,E(X),w^{\prime})=w\frac{1}{1-R(X)g_{1}}\,\mathcyr{sh}_{\hbar}\,\psi^{\mathcyr{sh}}(w^{\prime})\frac{1}{1-R(-X)g_{1}}.

Proof.

Decompose $(1-R(X)g_{1})^{-1}=1+(1-R(X)g_{1})^{-1}R(X)g_{1}$ . Then we obtain

\displaystyle K_{\hbar}^{\mathcyr{sh}}(w,E(X),w^{\prime})=K_{\hbar}^{\mathcyr{sh}}(w,\frac{R(X)}{X},w^{\prime})+K_{\hbar}^{\mathcyr{sh}}(w,\frac{1}{1-R(X)g_{1}}R(X)g_{1}\frac{R(X)}{X},w^{\prime}).

From the definition of $K_{\hbar}^{\mathcyr{sh}}$ , we see that the second term in the right hand side is equal to

	$\displaystyle K_{\hbar}^{\mathcyr{sh}}(w,\frac{1}{1-R(X)g_{1}}R(X),g_{1}\frac{R(X)}{X}w^{\prime})+K_{\hbar}^{\mathcyr{sh}}(w\frac{1}{1-R(X)g_{1}}R(X)g_{1},\frac{R(X)}{X},w^{\prime})$
	$\displaystyle=K_{\hbar}^{\mathcyr{sh}}(w,E(X),g_{1}R(X)w^{\prime})+K_{\hbar}^{\mathcyr{sh}}(w(\frac{1}{1-R(X)g_{1}}-1),\frac{R(X)}{X},w^{\prime}).$

Hence, it holds that

\displaystyle K_{\hbar}^{\mathcyr{sh}}(w,E(X),(1-g_{1}R(X))w^{\prime})=K_{\hbar}^{\mathcyr{sh}}(w\frac{1}{1-R(X)g_{1}},\frac{R(X)}{X},w^{\prime}).

Since $\psi^{\mathcyr{sh}}(\hbar b)=\hbar b$ , we see that the right hand side above is equal to

\displaystyle(w\frac{1}{1-R(X)g_{1}}\,\mathcyr{sh}_{\hbar}\,\psi^{\mathcyr{sh}}(w^{\prime}))e^{\hbar bX}=w\frac{1}{1-R(X)g_{1}}\,\mathcyr{sh}_{\hbar}\,(\psi^{\mathcyr{sh}}(w^{\prime})e^{\hbar bX}).

Therefore, by replacing $w^{\prime}$ with $(1-g_{1}R(X))^{-1}w^{\prime}$ , we obtain

\displaystyle K_{\hbar}^{\mathcyr{sh}}(w,E(X),w^{\prime})=w\frac{1}{1-R(X)g_{1}}\,\mathcyr{sh}_{\hbar}\,(\psi^{\mathcyr{sh}}(\frac{1}{1-g_{1}R(X)}w^{\prime})e^{\hbar bX}).

Since $\psi^{\mathcyr{sh}}(R(X))=R(X)$ and $\psi^{\mathcyr{sh}}(g_{1})=-e_{1}$ , it holds that

\displaystyle\psi^{\mathcyr{sh}}(\frac{1}{1-g_{1}R(X)}w^{\prime})=\psi^{\mathcyr{sh}}(w^{\prime})\frac{1}{1+R(X)e_{1}}.

Now the desired equality (B.7) follows from (B.4). ∎

Since $R(X)+R(-X)+\hbar b\,R(X)R(-X)=0$ , from Proposition B.1, we see that the right hand side of (B.7) is equal to

(B.8)		$\displaystyle{}-w\,\mathcyr{sh}_{\hbar}\,\psi^{\mathcyr{sh}}(w^{\prime})+\left(w\frac{1}{1-R(X)g_{1}}\,\mathcyr{sh}_{\hbar}\,\psi^{\mathcyr{sh}}(w^{\prime})\right)(1-R(X)g_{1})$
	$\displaystyle+\left(w\,\mathcyr{sh}_{\hbar}\,\psi^{\mathcyr{sh}}(w^{\prime})\frac{1}{1-R(-X)g_{1}}\right)(1-R(-X)g_{1}).$

Now set $w=E_{\mathbf{m}}$ and $w^{\prime}=E_{\overline{\mathbf{m}^{\prime}}}$ with admissible indices $\mathbf{m}$ and $\mathbf{m}^{\prime}$ . We have $E_{\mathbf{m}}\mathcyr{sh}_{\hbar}\psi^{\mathcyr{sh}}(E_{\overline{\mathbf{m}^{\prime}}})\equiv(-1)^{\mathrm{wt}(\mathbf{m}^{\prime})}E_{\mathbf{m}}\mathcyr{sh}_{\hbar}E_{\mathbf{m}^{\prime}}$ modulo $\hbar\widehat{\mathfrak{H}^{0}}$ because of Proposition 3.3, Corollary B.6 and $E_{\mathbf{m}}\in\widehat{\mathfrak{H}^{0}}$ . From the definition of the shuffle product and (B.6), we see that $E_{\mathbf{m}}\mathcyr{sh}_{\hbar}E_{\mathbf{m}^{\prime}}$ belongs to $\mathfrak{H}^{0}$ , hence so does the first term of (B.8) with $w=E_{\mathbf{m}}$ and $w^{\prime}=E_{\overline{\mathbf{m}^{\prime}}}$ . Since $\mathbf{m}$ and $\mathbf{m}^{\prime}$ are admissible, we can write $E_{\mathbf{m}}=ua$ and $\psi^{\mathcyr{sh}}(E_{\overline{\mathbf{m}^{\prime}}})=va$ with some $u,v\in\widehat{\mathfrak{H}^{0}}$ because of (B.3). Then, from Proposition B.2, the second term of (B.8) is equal to

\displaystyle\left\{u\,\mathcyr{sh}_{\hbar}\,va+\hbar\,u\,\mathcyr{sh}_{\hbar}\,v+\left(ua\frac{1}{1-R(X)g_{1}}\,\mathcyr{sh}_{\hbar}\,v\right)e^{\hbar bX}\right\}a,

which belongs to $\mathfrak{H}^{0}[[X]]$ because of (B.6). Similarly, the third term of (B.8) also belongs to $\mathfrak{H}^{0}[[X]]$ . Thus we find that $K_{\hbar}^{\mathcyr{sh}}(E_{\mathbf{m}},E(X),E_{\overline{\mathbf{m}^{\prime}}})$ belongs to $\mathfrak{H}^{0}[[X]]$ , and this completes the proof of Proposition 5.4.

B.4. Proof of Lemma 5.9

We use the map $\rho_{R(X)}$ and $\rho_{X}$ defined in Corollary B.3 with $U(X)=R(X)$ and $U(X)=X$ , respectively. Let $w\in\mathfrak{H}$ . Since $\rho_{R(X)}$ is a $\mathcal{C}$ -algebra homomorphism and $\rho_{R(X)}(g_{1})=(1-R(X)g_{1})^{-1}e^{\hbar bX}g_{1}$ , we have

	$\displaystyle\frac{1}{1-R(X)g_{1}}\,\mathcyr{sh}_{\hbar}\,w\frac{1}{1-g_{1}X}$	$\displaystyle=\rho_{R(X)}\left(w\frac{1}{1-g_{1}X}\right)\frac{1}{1-R(X)g_{1}}$
		$\displaystyle=\rho_{R(X)}(w)\left(1-\frac{1}{1-R(X)g_{1}}e^{\hbar bX}g_{1}X\right)^{-1}\left(1-R(X)g_{1}\right)^{-1}$
		$\displaystyle=\rho_{R(X)}(w)\left(1-(R(X)+e^{\hbar bX}X)g_{1}\right)^{-1}.$

Similarly, we have

	$\displaystyle\frac{1}{1-g_{1}X}\,\mathcyr{sh}_{\hbar}\,w\frac{1}{1-R(X)g_{1}}$	$\displaystyle=\rho_{X}\left(w\frac{1}{1-R(X)g_{1}}\right)\frac{1}{1-g_{1}X}$
		$\displaystyle=\rho_{X}(w)\left(1-\frac{1}{1-g_{1}X}R(X)(1+\hbar bX)g_{1}\right)^{-1}(1-g_{1}X)^{-1}$
		$\displaystyle=\rho_{X}(w)\left(1-(R(X)+e^{\hbar bX}X)g_{1}\right)^{-1}.$

Note that $\mathfrak{n}$ is a two-sided ideal of $\widehat{\mathfrak{H}^{0}}$ with respect to the concatenation product. Hence, to prove Lemma 5.9, it suffices to show that

\displaystyle\rho_{R(X)}(g_{\mathbf{k}})\equiv\rho_{X}(E_{\mathbf{k}})

modulo $\mathfrak{n}[[X]]$ for any non-empty admissible index $\mathbf{k}$ .

First we calculate $\rho_{R(X)}(g_{\mathbf{k}})$ . For $k\geq 1$ , we have

	$\displaystyle\rho_{R(X)}(g_{k})$	$\displaystyle=\rho_{R(X)}(ba)\left(\rho_{R(X)}(a)\right)^{k-1}$
		$\displaystyle=\frac{1}{1-R(X)g_{1}}e^{\hbar bX}g_{1}\left(\frac{1}{1-R(X)g_{1}}(a+\hbar R(X)g_{1})\right)^{k-1}$
		$\displaystyle\equiv\frac{1}{1-R(X)g_{1}}e^{\hbar bX}g_{1}\left(\frac{1}{1-R(X)g_{1}}a\right)^{k-1}$

modulo $\hbar\,\widehat{\mathfrak{H}^{0}}[[X]]$ . Moreover,

\displaystyle\frac{1}{1-R(X)g_{1}}a=a+\sum_{n\geq 1}(R(X)g_{1})^{n-1}R(X)g_{2}\equiv a+\sum_{n\geq 1}(Xg_{1})^{n-1}Xg_{2}=\frac{1}{1-g_{1}X}a

modulo $\mathfrak{n}_{0}[[X]]$ . Therefore, it holds that

\displaystyle\rho_{R(X)}(g_{k})\equiv\frac{1}{1-R(X)g_{1}}e^{\hbar bX}g_{1}\left(\frac{1}{1-g_{1}X}a\right)^{k-1}

modulo $\mathfrak{n}[[X]]$ for $k\geq 1$ . If $k\geq 2$ , each coefficient of the formal power series $g_{1}((1-g_{1}X)^{-1}a)^{k-1}$ belongs to $\sum_{j\geq 2}\mathcal{C}\langle A\rangle g_{j}$ . For any $w\in\mathcal{C}\langle A\rangle$ and $j\geq 2$ , we see that $(\hbar b)wg_{j}=(e_{1}-g_{1})wg_{j}\in\mathfrak{n}$ . Therefore, if $\mathbf{k}=(k_{1},\ldots,k_{r})$ is a non-empty admissible index, it holds that

\displaystyle\rho_{R(X)}(g_{\mathbf{k}})\equiv\prod_{1\leq i\leq r}^{\curvearrowright}\left(\frac{1}{1-g_{1}X}\,g_{1}\left(\frac{1}{1-g_{1}X}a\right)^{k_{i}-1}\right)

modulo $\mathfrak{n}[[X]]$ , where $\prod_{1\leq i\leq r}^{\curvearrowright}A_{i}=A_{1}\cdots A_{r}$ denotes the ordered product.

Next we calculate $\rho_{X}(E_{\mathbf{k}})$ . For $k\geq 1$ , we have

	$\displaystyle\rho_{X}(e_{k})$	$\displaystyle=\rho_{X}(b(a+\hbar))(\rho_{X}(a))^{k-1}=\frac{1}{1-g_{1}X}\,e_{1}\left(\frac{1}{1-g_{1}X}(a+\hbar g_{1}X)\right)^{k-1}$
		$\displaystyle\equiv\frac{1}{1-g_{1}X}\,e_{1}\left(\frac{1}{1-g_{1}X}a\right)^{k-1}$

modulo $\hbar\,\widehat{\mathfrak{H}^{0}}[[X]]$ . Note that

\displaystyle(e_{1}-g_{1})\frac{1}{1-g_{1}X}a=(e_{1}-g_{1})\left(1+\frac{1}{1-g_{1}X}g_{1}X\right)a=\hbar g_{1}+(e_{1}-g_{1})\frac{1}{1-g_{1}X}g_{2}X,

which belongs to $\mathfrak{n}[[X]]$ . Since $\mathfrak{n}a\subset\mathfrak{n}$ , it holds that

(B.9)

\displaystyle\rho_{X}(e_{k})\equiv\frac{1}{1-g_{1}X}g_{1}\left(\frac{1}{1-g_{1}X}a\right)^{k-1}

modulo $\mathfrak{n}[[X]]$ for $k\geq 2$ . Now we set $\mathbf{k}=(k_{1},\ldots,k_{r})$ and $\mathbf{k}^{\prime}=(k_{1},\ldots,k_{r-1})$ . Since $\mathbf{k}$ is admissible, it holds that $\rho_{X}(E_{\mathbf{k}})=\rho_{X}(E_{\mathbf{k}^{\prime}})\rho_{X}(e_{k_{r}})$ , and each coefficient of $\rho_{X}(e_{k_{r}})$ belongs to $\sum_{j\geq 2}\mathcal{C}\langle A\rangle g_{j}$ because of (B.9). Therefore, we may calculate $\rho_{X}(E_{\mathbf{k}^{\prime}})$ modulo $(\mathcal{C}\langle A\rangle(e_{1}-g_{1})\mathcal{C}\langle A\rangle)[[X]]$ . Then we see that

\displaystyle\rho_{X}(E(Y))\equiv\left(1-\frac{1}{1-g_{1}X}Y\right)^{-1}

modulo $(\mathcal{C}\langle A\rangle(e_{1}-g_{1})\mathcal{C}\langle A\rangle)[[X]]$ , and hence

\displaystyle\rho_{X}(E_{1^{m}})\equiv\left(\frac{1}{1-g_{1}X}g_{1}\right)^{m}

for $m\geq 0$ . As a result we find that

\displaystyle\rho_{X}(E_{\mathbf{k}})\equiv\prod_{1\leq i\leq r}^{\curvearrowright}\left(\frac{1}{1-g_{1}X}\,g_{1}\left(\frac{1}{1-g_{1}X}a\right)^{k_{i}-1}\right)

modulo $\mathfrak{n}[[X]]$ . This completes the proof of Lemma 5.9.

B.5. Proof of Proposition 6.4

We set

\displaystyle e(X)=\sum_{k\geq 1}e_{k}X^{k-1}=e_{1}\frac{1}{1-aX},\quad e^{0}(X)=\sum_{k\geq 2}e_{k}X^{k-2}=e_{2}\frac{1}{1-aX}=e(X)a.

For $w,w^{\prime}\in\mathfrak{H}$ , we set $\Xi(w,w^{\prime})$

\displaystyle\Xi(w,w^{\prime})=w\,\mathcyr{sh}_{\hbar}\,w^{\prime}E_{1}+wE_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime},

where $E_{1}=(e_{1}+g_{1})/2$ . Note that $\Xi(w,w^{\prime})$ is symmetric with respect to $w$ and $w^{\prime}$ , and

(B.10)

\displaystyle we_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}e_{1}=\Xi(w,w^{\prime})\,e_{1}.

We also set

	$\displaystyle\partial(w,w^{\prime})=wg_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}-(w\,\mathcyr{sh}_{\hbar}\,w^{\prime})g_{1}=we_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}-(w\,\mathcyr{sh}_{\hbar}\,w^{\prime})e_{1},$
	$\displaystyle\Lambda_{Y}(w,w^{\prime})=\left(w\frac{1}{1-R(Y)g_{1}}\,\mathcyr{sh}_{\hbar}\,w^{\prime}\right)\left(1-R(Y)g_{1}\right)=\left(wE(Y)\mathcyr{sh}_{\hbar}\,w^{\prime}\right)E(Y)^{-1}.$

Proposition B.8.

Set

\displaystyle E(Y_{1},Y_{2})=\frac{1}{Y_{1}Y_{2}}\sum_{j=1}^{2}Y_{j}\left(E(Y_{1}+Y_{2})-E(Y_{j})\right).

It holds that

	$\displaystyle\Xi(wE(Y_{1}),w^{\prime}E(Y_{2}))$	$\displaystyle=\partial(w^{\prime}E(Y_{2}),w)E(Y_{1})+\partial(wE(Y_{1}),w^{\prime})E(Y_{2})$
		$\displaystyle+\left(-w\,\mathcyr{sh}_{\hbar}w^{\prime}+\Lambda_{Y_{1}}(w,w^{\prime})+\Lambda_{Y_{2}}(w^{\prime},w)\right)E(Y_{1},Y_{2}).$

for any $w,w^{\prime}\in\mathfrak{H}$ .

Proof.

Set

\displaystyle P(Y_{1},Y_{2})=w\frac{1}{1-R(Y_{1})g_{1}}\,\mathcyr{sh}_{\hbar}\,w^{\prime}\frac{1}{1-R(Y_{2})g_{1}}.

Since $E_{1}=\hbar b/2+ba$ , we have

	$\displaystyle wE(Y_{1})\,\mathcyr{sh}_{\hbar}\,w^{\prime}E(Y_{2})E_{1}$	$\displaystyle=P(Y_{1},Y_{2})\frac{1}{2Y_{1}Y_{2}}\hbar bR(Y_{1})R(Y_{2})$
		$\displaystyle+\left(w\frac{1}{1-R(Y_{1})g_{1}}\,\mathcyr{sh}_{\hbar}\,w^{\prime}E(Y_{2})ba\right)(1-R(Y_{1})g_{1})E(Y_{1}).$

We calculate the second term of the right hand side by using Proposition B.2 and

(B.11)

\displaystyle R(Y_{1})+R(Y_{2})+\hbar bR(Y_{1})R(Y_{2})=R(Y_{1}+Y_{2}).

Then we obtain

\displaystyle\left\{\partial(w^{\prime}E(Y_{2}),w)+P(Y_{1},Y_{2})\frac{1}{Y_{2}}(R(Y_{1}+Y_{2})-R(Y_{1}))g_{1}\right\}E(Y_{1}).

By changing $w\leftrightarrow w^{\prime}$ and $Y_{1}\leftrightarrow Y_{2}$ , we obtain a similar formula for $wE(Y_{1})E_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}E(Y_{2})$ . Thus we get

\displaystyle\Xi(wE(Y_{1}),w^{\prime}E(Y_{2}))=\partial(w^{\prime}E(Y_{2}),w)E(Y_{1})+\partial(wE(Y_{1}),w^{\prime})E(Y_{2})+\frac{1}{Y_{1}Y_{2}}P(Y_{1},Y_{2})Q(Y_{1},Y_{2}),

where

\displaystyle Q(Y_{1},Y_{2})=\hbar bR(Y_{1})R(Y_{2})+\sum_{j=1}^{2}Y_{j}(R(Y_{1}+Y_{2})-R(Y_{j}))g_{1}E(Y_{j}).

From Proposition B.1 and (B.11), we see that

\displaystyle P(Y_{1},Y_{2})=\left(-w\,\mathcyr{sh}_{\hbar}w^{\prime}+\Lambda_{Y_{1}}(w,w^{\prime})+\Lambda_{Y_{2}}(w^{\prime},w)\right)\frac{1}{1-R(Y_{1}+Y_{2})g_{1}}.

Using (B.11) we see that

\displaystyle Q(Y_{1},Y_{2})=R(Y_{1}+Y_{2})\left(1+g_{1}\sum_{j=1}^{2}Y_{j}E(Y_{j})\right)-\sum_{j=1}^{2}R(Y_{j})(1+Y_{j}g_{1}E(Y_{j})).

Because $R(Y_{j})(1+Y_{j}g_{1}E(Y_{j}))=Y_{j}E(Y_{j})$ , we have

\displaystyle Q(Y_{1},Y_{2})=R(Y_{1}+Y_{2})+(R(Y_{1}+Y_{2})g_{1}-1)\sum_{j=1}^{2}Y_{j}E(Y_{j})=(1-R(Y_{1}+Y_{2})g_{1})Y_{1}Y_{2}E(Y_{1},Y_{2}).

Thus we get the desired formula. ∎

Lemma B.9.

For any $w,w^{\prime}\in\mathfrak{H}$ , the following formulas hold.

(B.12)		$\displaystyle\partial(w,w^{\prime}e^{0}(X))=\left(\Xi(w,w^{\prime})+Xw\,\mathcyr{sh}_{\hbar}\,w^{\prime}e^{0}(X)\right)e^{0}(X),$
(B.13)		$\displaystyle\Lambda_{Y}(w,w^{\prime}e^{0}(X))=w\,\mathcyr{sh}_{\hbar}\,w^{\prime}e^{0}(X)+Y\partial(wE(Y),w^{\prime}e^{0}(X)).$

Proof.

Note that $e^{0}(X)=e(X)a$ and $\partial(w,w^{\prime}a)=(we_{1}\,\mathcyr{sh}_{\hbar}w^{\prime})a$ for any $w,w^{\prime}\in\mathfrak{H}$ . Hence

(B.14)

\displaystyle\partial(w,w^{\prime}e^{0}(X))=\partial(w,w^{\prime}e(X)a)=(we_{1}\,\mathcyr{sh}_{\hbar}w^{\prime}e(X))a.

Since $e(X)=e_{1}+e(X)aX$ , we see that

	$\displaystyle we_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}e(X)$	$\displaystyle=\Xi(w,w^{\prime})e_{1}+Xwe_{1}\,\mathcyr{sh}_{\hbar}w^{\prime}e(X)a$
		$\displaystyle=\Xi(w,w^{\prime})e_{1}+X\left\{(w\,\mathcyr{sh}_{\hbar}\,w^{\prime}e(X)a)e_{1}+(we_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}e(X))a\right\}$
		$\displaystyle=\left(\Xi(w,w^{\prime})e_{1}+Xw\,\mathcyr{sh}_{\hbar}\,w^{\prime}e^{0}(X)\right)e_{1}+(we_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}e(X))aX.$

Therefore we have

\displaystyle we_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}e(X)=\left(\Xi(w,w^{\prime})e_{1}+Xw\,\mathcyr{sh}_{\hbar}\,w^{\prime}e^{0}(X)\right)e(X)

since $e_{1}(1-aX)^{-1}=e(X)$ . Thus we get (B.12).

Next we prove (B.13). Using $e^{0}(X)=e(X)a$ and Proposition B.2, we see that

	$\displaystyle\Lambda_{Y}(w,w^{\prime}e^{0}(X))$	$\displaystyle=w\,\mathcyr{sh}_{\hbar}w^{\prime}e^{0}(X)-(w\,\mathcyr{sh}_{\hbar}^{,}w^{\prime}e(X))a+\left(w\frac{1}{1-R(Y)g_{1}}(1+\hbar bR(Y))\,\mathcyr{sh}_{\hbar}\,w^{\prime}e(X)\right)a$
		$\displaystyle=w\,\mathcyr{sh}_{\hbar}w^{\prime}e^{0}(X)+\left(w\frac{1}{1-R(Y)g_{1}}R(Y)e_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}e(X)\right)a.$

The second term of the right hand side is equal to

\displaystyle Y\left(wE(Y)e_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}e(X)\right)a=Y\partial(wE(Y),w^{\prime}e^{0}(X))

because of (B.14). Thus we get (B.13). ∎

Proposition B.10.

For any $w,w^{\prime}\in\mathfrak{H}$ , it holds that

	$\displaystyle wE(Y)\,\mathcyr{sh}_{\hbar}\,w^{\prime}e^{0}(X)$
	$\displaystyle=\left\{w\,\mathcyr{sh}_{\hbar}\,w^{\prime}e^{0}(X)+Y\Xi(wE(Y),w^{\prime})e^{0}(X)\right\}E(Y)\frac{1}{1-XYe^{0}(X)E(Y)}.$

Proof.

We denote the left hand side by $J(X,Y)$ . We see that

	$\displaystyle J(X,Y)$	$\displaystyle=\Lambda_{Y}(w,w^{\prime}e^{0}(X))E(Y)$
		$\displaystyle=\left\{w\,\mathcyr{sh}_{\hbar}\,w^{\prime}e^{0}(X)+Y\left(\Xi(wE(Y),w^{\prime})+XJ(X,Y)\right)e^{0}(X)\right\}E(Y)$

using (B.12) and (B.13). It implies the desired equality. ∎

Proposition B.11.

For any $w,w^{\prime}\in\mathfrak{H}$ , it holds that

	$\displaystyle we^{0}(X_{1})\,\mathcyr{sh}_{\hbar}\,w^{\prime}e^{0}(X_{2})$
	$\displaystyle=\frac{1}{X_{1}X_{2}}\biggl{\{}-(X_{1}+X_{2}+\hbar X_{1}X_{2})\Xi(w,w^{\prime})e_{2}$
	$\displaystyle\qquad\qquad\qquad{}+X_{1}(1+\hbar X_{2})\,\partial(w,w^{\prime}e^{0}(X_{2}))+X_{2}(1+\hbar X_{1})\,\partial(w^{\prime},we^{0}(X_{1}))\biggr{\}}$
	$\displaystyle{}\times\frac{1}{1-(X_{1}+X_{2}+\hbar X_{1}X_{2})a}.$

Proof.

Since $e^{0}(X)=(e(X)-e_{1})/X$ , we have

\displaystyle X_{1}X_{2}\,we^{0}(X_{1})\,\mathcyr{sh}_{\hbar}\,w^{\prime}e^{0}(X_{2})=w(e(X_{1})-e_{1})\,\mathcyr{sh}_{\hbar}\,w^{\prime}(e(X_{2})-e_{1}).

Calculate $we(X_{1})\,\mathcyr{sh}_{\hbar}\,w^{\prime}e(X_{2})$ using Proposition B.1, and we see that

	$\displaystyle X_{1}X_{2}\,(we^{0}(X_{1})\,\mathcyr{sh}_{\hbar}\,w^{\prime}e^{0}(X_{2}))(1-(X_{1}+X_{2}+\hbar X_{1}X_{2})a)$
	$\displaystyle=-(X_{1}+X_{2}+\hbar X_{1}X_{2})(we_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}e_{1})a$
	$\displaystyle+X_{1}(1+\hbar X_{2})(we_{1}\,\mathcyr{sh}_{\hbar}\,w^{\prime}e(X_{2}))a+X_{2}(1+\hbar X_{1})(we(X_{1})\,\mathcyr{sh}_{\hbar}\,w^{\prime}e_{1})a.$

Now the desired equality follows from (B.10) and (B.14). ∎

Now we prove Proposition 6.4. For a subset $\mathfrak{a}$ of $\mathfrak{H}$ and a formal power series $f(X_{1},\ldots,X_{n})\in\mathfrak{H}[[X_{1},\ldots,X_{n}]]$ , we say that $f(X_{1},\ldots,X_{n})$ belongs to $\mathfrak{a}$ if all the coefficients of $f(X_{1},\ldots,X_{n})$ belong to $\mathfrak{a}$ . We show that

(i)

$\Xi(w,w^{\prime})$ belongs to $\mathfrak{e}$ ,
(ii)

$w\,\mathcyr{sh}_{\hbar}\,w^{\prime}e^{0}(X)$ and $we^{0}(X)\,\mathcyr{sh}_{\hbar}\,w^{\prime}$ belong to $\mathfrak{e}$ ,
(iii)

$we^{0}(X_{1})\,\mathcyr{sh}_{\hbar}\,w^{\prime}e^{0}(X_{2})$ belongs to $\mathfrak{e}^{0}$

for any $w,w^{\prime}\in\mathfrak{e}$ . Proposition 6.4 follows from (ii) and (iii). The third property (iii) follows from (i) and (ii) because of Proposition B.11 and (B.12). Hence it suffices to prove (i) and (ii). Note that the $\mathcal{C}$ -module $\mathfrak{e}$ is generated by the coefficients of the formal power series

\displaystyle C_{r}=C_{r}(X_{1},\ldots,X_{r};Y_{0},\ldots,Y_{r})=E(Y_{0})\prod_{1\leq j\leq r}^{\curvearrowright}\left(e^{0}(X_{j})E(Y_{j})\right)

with $r\geq 0$ . Hence we may assume that $w=C_{r}$ and $w^{\prime}=C_{s}^{\prime}$ for some $r,s\geq 0$ , where the prime symbol indicates that $w^{\prime}$ contains a different family of variables from $w$ . We proceed the proof of (i) and (ii) by induction on $r+s$ .

First we consider the case of $r=0$ . From Proposition B.8, we see that $\Xi(E(Y_{1}),E(Y_{2}))=E(Y_{1},Y_{2})$ belongs to $\mathfrak{e}$ , and it implies that $E(Y_{1})\,\mathcyr{sh}_{\hbar}\,E(Y_{2})e^{0}(X)$ also belongs to $\mathfrak{e}$ because of Proposition B.10. Hence (i) and (ii) are true in the case of $r=s=0$ . Now we consider the case where $r=0$ and $s\geq 1$ . Set $w=C_{0}=E(Y_{1})$ and $w^{\prime}=C_{s}^{\prime}=C_{s-1}^{\prime}e^{0}(X_{2})E(Y_{2})$ . Proposition B.8 and Proposition B.10 imply that

	$\displaystyle\Xi(C_{0},C_{s}^{\prime})$	$\displaystyle=\partial(C_{0},C_{s-1}^{\prime}e^{0}(X_{2}))E(Y_{2})+\Lambda_{Y_{1}}(1,C_{s-1}^{\prime}e^{0}(X_{2}))E(Y_{1},Y_{2}),$
	$\displaystyle C_{0}\,\mathcyr{sh}_{\hbar}\,C_{s}^{\prime}e^{0}(X)$	$\displaystyle=\left\{C_{s}^{\prime}+Y_{1}\,\Xi(C_{0},C_{s}^{\prime})\right\}e^{0}(X)E(Y_{1})\frac{1}{1-XY_{1}e^{0}(X)E(Y_{1})},$
	$\displaystyle C_{0}e^{0}(X)\,\mathcyr{sh}_{\hbar}\,C_{s}^{\prime}$	$\displaystyle=\left\{C_{0}e^{0}(X)\,\mathcyr{sh}_{\hbar}\,C_{s-1}^{\prime}e^{0}(X_{2})+Y_{2}\,\Xi(C_{0},C_{s}^{\prime})e^{0}(X)\right\}E(Y_{2})$
		$\displaystyle{}\times\frac{1}{1-XY_{2}e^{0}(X)E(Y_{2})}.$

From Lemma B.9 and the induction hypothesis, we see that $\partial(C_{0},C_{s-1}^{\prime}e^{0}(X_{2}))$ and $\Lambda_{Y_{1}}(1,C_{s-1}^{\prime}e^{0}(X_{2}))$ belong to $\mathfrak{e}^{0}$ . Hence $\Xi(C_{0},C_{s}^{\prime})$ belongs to $\mathfrak{e}$ , and it implies that $C_{0}\,\mathcyr{sh}_{\hbar}\,C_{s}^{\prime}e^{0}(X)$ belongs to $\mathfrak{e}$ . Moreover, since $C_{0}e^{0}(X)\,\mathcyr{sh}_{\hbar}\,C_{s-1}^{\prime}e^{0}(X_{2})$ belongs to $\mathfrak{e}^{0}$ because of the induction hypothesis (iii), we see that $C_{0}e^{0}(X)\,\mathcyr{sh}_{\hbar}\,C_{s}^{\prime}$ belongs to $\mathfrak{e}$ . Thus we obtain (i), (ii) with $w=C_{0}$ and $w^{\prime}=C_{s}^{\prime}$ for any $s\geq 0$ . Since $\Xi(w,w^{\prime})$ is symmetric with respect to $w$ and $w^{\prime}$ , they are also true in the case where $w=C_{r}$ with $r\geq 0$ and $w^{\prime}=C_{0}^{\prime}$ .

Next we consider the case where $r,s\geq 1$ . Set $w=C_{r}=C_{r-1}e^{0}(X_{1})E(Y_{1})$ and $w^{\prime}=C_{s}^{\prime}=C_{s-1}^{\prime}e^{0}(X_{2})E(Y_{2})$ . From Proposition B.8, Proposition B.10 and (B.13), we see that

	$\displaystyle\Xi(C_{r},C_{s}^{\prime})$	$\displaystyle=\partial(C_{r},C_{s-1}^{\prime}e^{0}(X_{2}))\left\{E(Y_{1})+Y_{2}E(Y_{1},Y_{2})\right\}$
		$\displaystyle+\partial(C_{s}^{\prime},C_{r-1}e^{0}(X_{1}))\left\{E(Y_{2})+Y_{1}E(Y_{1},Y_{2})\right\}$
		$\displaystyle+\left(C_{r-1}e^{0}(X_{1})\,\mathcyr{sh}_{\hbar}\,C_{s-1}^{\prime}e^{0}(X_{2})\right)E(Y_{1},Y_{2})$

and

\displaystyle C_{r}\,\mathcyr{sh}_{\hbar}\,C_{s}^{\prime}e^{0}(X)=\left\{C_{r-1}e^{0}(X_{1})\,\mathcyr{sh}_{\hbar}\,C_{s}^{\prime}e^{0}(X)+Y_{1}\,\Xi(C_{r},C_{s}^{\prime})e^{0}(X)\right\}E(Y_{1})\frac{1}{1-XY_{1}e^{0}(X)E(Y_{1})}.

From (B.12) and the induction hypothesis, we see that $\partial(C_{r},C_{s-1}^{\prime}e^{0}(X_{2})),\partial(C_{s}^{\prime},C_{r-1}e^{0}(X_{1}))$ and $C_{r-1}e^{0}(X_{1})\,\mathcyr{sh}_{\hbar}\,C_{s-1}^{\prime}e^{0}(X_{2})$ belong to $\mathfrak{e}^{0}$ . Hence $\Xi(C_{r},C_{s}^{\prime})$ belongs to $\mathfrak{e}$ . The induction hypothesis says that $C_{r-1}e^{0}(X_{1})\,\mathcyr{sh}_{\hbar}\,C_{s}^{\prime}e^{0}(X)$ belongs to $\mathfrak{e}^{0}$ , and hence $C_{r}\,\mathcyr{sh}_{\hbar}\,C_{s}^{\prime}e^{0}(X)$ belongs to $\mathfrak{e}$ . Similarly we find that $C_{r}e^{0}(X)\,\mathcyr{sh}_{\hbar}\,C_{s}^{\prime}$ belongs to $\mathfrak{e}$ . Therefore (i) and (ii) hold for $w=C_{r}$ and $w^{\prime}=C_{s}$ , and this completes the proof of Proposition 6.4.

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A qq-analogue of symmetric multiple zeta value

Abstract.

1. Introduction

2. Symmetric multiple zeta value

2.1. Multiple zeta value

Proposition 2.1.

2.2. Symmetric multiple zeta value

2.3. Truncated symmetric multiple zeta value

Theorem 2.2 ([4, 6]).

2.4. Relations of SMZVs

Theorem 2.3.

Theorem 2.4.

Theorem 2.5 (Ohno-type relation [7]).

3. A qq-analogue of multiple zeta value

3.1. Algebraic formulation

Remark 3.1.

3.2. Double shuffle relation of qqMZVs

Proposition 3.2.

Proposition 3.3.

Proof.

3.3. Limit of qqMZVs as q→1−0q\to 1-0

Proposition 3.4.

Corollary 3.5.

Remark 3.6.

3.4. Restoration of finite double shuffle relation of MZVs

Proposition 3.7.

Proof.

Corollary 3.8.

Proof.

Proposition 3.9.

Proof.

Lemma 3.10.

4. A qq-analogue of truncated SMZV

Proposition 4.1.

Proof.

Proposition 4.2.

Proof.

5. A qq-analogue of SMZV

Proposition 5.1.

Proof.

Definition 5.2.

Example 5.3.

Proposition 5.4.

Definition 5.5.

Example 5.6.

Example 5.7.

Theorem 5.8.

Proof.

Lemma 5.9.

Corollary 5.10.

Proof.

Definition 5.11.

Example 5.12.

Example 5.13.

Theorem 5.14.

Proof.

6. Relations of the qq-analogue of symmetric multiple zeta value

6.1. Reversal relation

Theorem 6.1.

Proof.

6.2. Double shuffle relation

Proposition 6.2.

Example 6.3.

Proposition 6.4.

Proposition 6.5.

Proof.

Theorem 6.6.

Proof.

6.3. Ohno-type relation

Theorem 6.7.

Proof.

Appendix A Proof of Proposition 3.4

Lemma A.1.

Proof.

Corollary A.2.

Proof.

Proposition A.3.

Proof.

Proposition A.4.

Proof.

A $q$ -analogue of symmetric multiple zeta value

3. A $q$ -analogue of multiple zeta value

3.2. Double shuffle relation of $q$ MZVs

3.3. Limit of $q$ MZVs as $q\to 1-0$

4. A $q$ -analogue of truncated SMZV

5. A $q$ -analogue of SMZV

6. Relations of the $q$ -analogue of symmetric multiple zeta value

B.2. Generating function of $E_{1^{m}}$