A q-analogue of symmetric multiple zeta value

Abstract

We construct a q-analogue of the 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 reversal relation and a part of the double shuffle relation and the Ohno-type relation.

Appendices

Appendix A. Proof of Proposition 3.4

Lemma A.1

Suppose that $\alpha \ge 0$ and $\max {(1, \alpha )}\le \beta \le 2\alpha +1$. 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 $\alpha \ge (\beta -1)/2>0$. Set

$$\begin{aligned} g(x)=x^{1-\alpha }(1-x^{\beta })^{2}f'(x)= \alpha -(\alpha +1)x+(\beta -\alpha )x^{\beta }+(\alpha -\beta +1)x^{\beta +1}. \end{aligned}$$

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

$$\begin{aligned} h(x)=x^{-\beta }g'(x)=-(\alpha +1)x^{-\beta }+\beta (\beta -\alpha )x^{-1}+ (\alpha -\beta +1)(\beta +1). \end{aligned}$$

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

$$\begin{aligned} h'(x)=\beta x^{-\beta -1}(\alpha +1-(\beta -\alpha )x^{\beta -1})> \beta x^{-\beta -1}(2\alpha -\beta +1)\ge 0, \end{aligned}$$

for $0<x<1$. Therefore we see that $h(x)<0$, and hence $g'(x)<0$, on the interval (0, 1). $\square$

Corollary A.2

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

$$\begin{aligned} x^{\alpha }\frac{1-x}{1-x^{\beta }} \le \frac{1}{\beta }, \end{aligned}$$

for $0<x<1$.

Proof

It is because $f(x) \rightarrow 1/\beta$ as $x \rightarrow 1$. $\square$

Proposition A.3

Suppose that $0<q<1$. Set

$$\begin{aligned} B_{q}(\alpha _{1}, \ldots , \alpha _{r}; \beta _{1}, \ldots , \beta _{r})= \sum _{l_{1}, \ldots , l_{r}\ge 1} \prod _{j=1}^{r}\left( {\begin{array}{c}l_{j}\\ \alpha _{j}\end{array}}\right) \frac{(1-q)^{\alpha _{j}} q^{l_{j}/2}}{(l_{1}+\cdots +l_{j})^{\beta _{j}+1}}. \end{aligned}$$

Then, for any non-negative integer $\alpha _{1}, \ldots , \alpha _{r}, \beta _{1}, \ldots , \beta _{r}$, it holds that

$$\begin{aligned} 0\le Z_{q}((e_{1}-g_{1})^{\alpha _{1}}g_{\beta _{1}+1} \cdots (e_{1}-g_{1})^{\alpha _{r}}g_{\beta _{r}+1}) \le B_{q}(\alpha _{1}, \ldots , \alpha _{r}; \beta _{1}, \ldots , \beta _{r}). \end{aligned}$$

Proof

From Corollary A.2, we see that

$$\begin{aligned} 0\le \frac{q^{n}}{[n]}=q^{(n-1)/2}\frac{1-q}{1-q^{n}}q^{(n+1)/2} \le \frac{q^{(n+1)/2}}{n}, \end{aligned}$$

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

$$\begin{aligned} 0&\le Z_{q}((e_{1}-g_{1})^{\alpha _{1}}g_{\beta _{1}+1} \cdots (e_{1}-g_{1})^{\alpha _{r}}g_{\beta _{r}+1}) \\&=\sum _{0=n_{0}<n_{1}<\cdots<n_{r}}\prod _{j=1}^{r} (1-q)^{\alpha _{j}}\left( {\begin{array}{c}n_{j}-n_{j-1}-1\\ \alpha _{j}\end{array}}\right) \left( \frac{q^{n_{j}}}{[n_{j}]}\right) ^{\beta _{j}+1} \\&\le \sum _{0=n_{0}<n_{1}<\cdots <n_{r}}\prod _{j=1}^{r} (1-q)^{\alpha _{j}}\left( {\begin{array}{c}n_{j}-n_{j-1}\\ \alpha _{j}\end{array}}\right) \left( \frac{q^{(n_{j}+1)/2}}{n_{j}}\right) ^{\beta _{j}+1}. \end{aligned}$$

Set $l_{j}=n_{j}-n_{j-1}$ for $1\le j \le r$. Then the right-hand side is dominated by $B_{q}(\alpha _{1}, \ldots , \alpha _{r}; \beta _{1}, \ldots , \beta _{r})$ since $0<q<1$. $\square$

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

$$\begin{aligned} B_{q}(\alpha _{1}, \ldots , \alpha _{r}; \beta _{1}, \ldots , \beta _{r})= O((-\log {(1-q)})^{r}) \qquad (q \rightarrow 1). \end{aligned}$$

(A.1)

Moreover, if $\alpha _{s} \ge 1$ and $\beta _{t}\ge 1$ for some $1\le s\le t \le r$, then we have

$$\begin{aligned} B_{q}(\alpha _{1}, \ldots , \alpha _{r}; \beta _{1}, \ldots , \beta _{r})= (1-q)\, O((-\log {(1-q)})^{r}) \qquad (q \rightarrow 1). \end{aligned}$$

(A.2)

Proof

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

$$\begin{aligned} \varphi _{\alpha }(x)=(1-x)^{\alpha }\sum _{l\ge 1}\left( {\begin{array}{c}l\\ \alpha \end{array}}\right) \frac{x^{l/2}}{l}, \qquad \tilde{\varphi }_{\alpha }(x)=(1-x)^{\alpha } \sum _{l\ge 1}\left( {\begin{array}{c}l\\ \alpha \end{array}}\right) \frac{x^{l/2}}{l^2}. \end{aligned}$$

We have

$$\begin{aligned} \varphi _{0}(x)=-\log {(1-x^{1/2})}, \qquad \varphi _{\alpha }(x)=\frac{x^{\alpha /2}}{\alpha }(1+x^{1/2})^{\alpha } \quad (\alpha \ge 1). \end{aligned}$$

Hence $\varphi _{\alpha }(x)=O(-\log {(1-x)})$ as $x \rightarrow 1-0$ for any $\alpha \ge 0$. If $\alpha \ge 1$, we see that

$$\begin{aligned} 0\le \tilde{\varphi }_{\alpha }(x)=(1-x)^{\alpha } \sum _{l\ge 1}\frac{1}{\alpha }\left( {\begin{array}{c}l-1\\ \alpha -1\end{array}}\right) \frac{x^{l/2}}{l} \le \frac{1-x}{\alpha }\varphi _{\alpha -1}(x), \end{aligned}$$

for $0\le x \le 1$. Hence $\tilde{\varphi }_{\alpha }(x)=(1-x)\,O(-\log {(1-x)})$ as $x \rightarrow 1-0$ for any $\alpha \ge 1$.

Since $\beta _{j}\ge 0$ for $1\le j \le r$, we see that

$$\begin{aligned} 0\le B_{q}(\alpha _{1}, \ldots , \alpha _{r}; \beta _{1}, \ldots , \beta _{r}) \le \prod _{j=1}^{r}\varphi _{\alpha _{j}}(q). \end{aligned}$$

Hence we have (A.1). Now assume further that $\alpha _{s} \ge 1$ and $\beta _{t}\ge 1$ for some $1\le s\le t \le r$. If $s=t$, it holds that

$$\begin{aligned} 0 \le B_{q}(\alpha _{1}, \ldots , \alpha _{r}; \beta _{1}, \ldots , \beta _{r}) \le \tilde{\varphi }_{\alpha _{s}}(q) \prod _{\begin{array}{c} 1\le j \le r \\ j\not =s \end{array}}\varphi _{\alpha _{j}}(q). \end{aligned}$$

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

$$\begin{aligned} 0&\le B_{q}(\alpha _{1}, \ldots , \alpha _{r}; \beta _{1}, \ldots , \beta _{r}) \\&\le \sum _{l_{1}, \ldots , l_{r}\ge 1} \prod _{\begin{array}{c} 1\le j \le r \\ j\not =s, t \end{array}} \left\{ (1-q)^{\alpha _{j}} \left( {\begin{array}{c}l_{j}\\ \alpha _{j}\end{array}}\right) \frac{q^{l_{j}/2}}{l_{j}}\right\} \\&\quad \times (1-q)^{\alpha _{s}+\alpha _{t}} \left( {\begin{array}{c}l_{s}\\ \alpha _{s}\end{array}}\right) \left( {\begin{array}{c}l_{t}\\ \alpha _{t}\end{array}}\right) \frac{q^{l_{s}/2}}{(l_{1}+\cdots +l_{s})^{\beta _{s}+\beta _{t}+1}} \frac{q^{l_{t}/2}}{l_{1}+\cdots +l_{t}} \\&\le \sum _{l_{1}, \ldots , l_{r}\ge 1} \prod _{\begin{array}{c} 1\le j \le r \\ j\not =s \end{array}} \left\{ (1-q)^{\alpha _{j}} \left( {\begin{array}{c}l_{j}\\ \alpha _{j}\end{array}}\right) \frac{q^{l_{j}/2}}{l_{j}}\right\} (1-q)^{\alpha _{s}}\left( {\begin{array}{c}l_{s}\\ \alpha _{s}\end{array}}\right) \frac{q^{l_{s}/2}}{l_{s}^{2}} \\&=\tilde{\varphi }_{\alpha _{s}}(q) \prod _{\begin{array}{c} 1\le j \le r \\ j\not =s \end{array}} \varphi _{\alpha _{j}}(q), \end{aligned}$$

since $\beta _{t} \ge 1$. Thus we get (A.2) in the case where $s<t$ either. $\square$

Proof of Proposition 3.4

From Lemma A.1 and the monotone convergence theorem, we see that $Z_{q}(g_{\textbf{k}}) \rightarrow \zeta (\textbf{k})$ as $q \rightarrow 1$ for any admissible index $\textbf{k}$. 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 $q \rightarrow 1$. Hence $Z_{q}(\hbar w)\rightarrow 0$ in the limit as $q \rightarrow 1$. If $w \in {\mathfrak {n}}_{0}$, then there exist s and t such that $\alpha _{s}\ge 1, \beta _{t}\ge 1$ and $1\le s\le t \le r$. Then Proposition A.3 and (A.2) imply that $Z_{q}(w)\rightarrow 0$ in the limit as $q \rightarrow 1$. Thus we see that $\lim _{q\rightarrow 1}Z_{q}(w)=0$ for any $w \in {\mathfrak {n}}$. $\square$

Remark A.5

For a polynomial P(x) and a positive integer n, we set $(\partial P)(n)=P(n)-P(n-1)$. We also set $(\partial P)(0)=P(0)$. Then it holds that

$$\begin{aligned}&Z_{q}((e_{1}-g_{1})^{\alpha _{1}}g_{\beta _{1}+1} \cdots (e_{1}-g_{1})^{\alpha _{r}}g_{\beta _{r}+1}) \\&\quad =(1-q)^{\sum _{j=1}^{r}(\alpha _{j}+\beta _{j}+1)} \sum _{0<m_{1}<\cdots <m_{r}}\prod _{j=1}^{r}\left( \left( {\begin{array}{c}m_{j}-m_{j-1}-1\\ \alpha _{j}\end{array}}\right) \sum _{0\le n}(\partial P_{\beta _{j}+1}(n)) q^{nm_{j}} \right) , \end{aligned}$$

where we set $m_{0}=0$ and $P_{k}(x)=\left( {\begin{array}{c}x\\ k\end{array}}\right)$ for $k \ge 1$. In [1], Bachmann and van-Ittersum discuss the asymptotics as $q \rightarrow 1$ of infinite sums of the following form:

$$\begin{aligned} \sum _{0<m_{1}<\cdots <m_{r}}\prod _{j=1}^{r}\left( m_{j}^{d_{j}}\sum _{0\le n}(\partial f_{k_{j}}(n)) q^{nm_{j}} \right) , \end{aligned}$$

where $d_{j}$ and $k_{j} \, (1\le j \le k)$ are non-negative integers and $\{f_{k}(x)\}_{k=0}^{\infty }$ is a family of polynomials such that $\textrm{deg}f_{k}(x)=k$ for $k \ge 0$ and $f_{k}(0)=0$ for $k\ge 1$.

Appendix B. Proofs

1.1 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

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

Proof

Set

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

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

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

Thus we get the desired equality. $\square$

Proposition B.2

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

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

Proof

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

Proof of Lemma 3.10

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

Using (3.9), we get the desired equality. $\square$

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

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

$$\begin{aligned} \rho _{U(X)}(a)&=\frac{1}{1-U(X)g_{1}}\left( 1+\hbar b\, U(X)\right) a, \nonumber \\ \rho _{U(X)}(b)&=\frac{1}{1-U(X)g_{1}}b \left( 1-U(X) g_{1}\right) . \end{aligned}$$

(B.1)

Proof

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

$$\begin{aligned} \rho _{U(X)}(w'a)=\rho _{U(X)}(w')\frac{1}{1-U(X)g_{1}}(1+\hbar bU(X))a. \end{aligned}$$

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

$$\begin{aligned} \rho _{U(X)}(w'b)=\rho _{U(X)}(w')\frac{1}{1-U(X)g_{1}}b(1-U(X)g_{1}). \end{aligned}$$

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

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

Here we calculate the generating function

$$\begin{aligned} E(X)=\sum _{m=0}^{\infty }E_{1^{m}}X^{m}. \end{aligned}$$

Recall that

$$\begin{aligned} R(X)=\frac{e^{\hbar b X}-1}{\hbar b}=\sum _{n=1}^{\infty }\frac{X^{n}}{n!}(e_{1}-g_{1})^{n-1}. \end{aligned}$$

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

Proposition B.4

It holds that

(B.2)

Proof

The power series is the unique solution of the differential equation satisfying $\Psi (0)=1$. Since the right-hand side with $X=0$ is equal to one, it suffices to show that

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

since . $\square$

Proposition B.5

It holds that

$$\begin{aligned} E(X)=\frac{1}{1-R(X)g_{1}}\frac{R(X)}{X}. \end{aligned}$$

Proof

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

Using

$$\begin{aligned} \sum _{s=j}^{m}\left( {\begin{array}{c}s\\ j\end{array}}\right) =\sum _{s=j}^{m}\left( \left( {\begin{array}{c}s+1\\ j+1\end{array}}\right) -\left( {\begin{array}{c}s\\ j+1\end{array}}\right) \right) = \left( {\begin{array}{c}m+1\\ j+1\end{array}}\right) \end{aligned}$$

and (B.2), we see that

$\square$

Corollary B.6

For any index $\textbf{k}$, it holds that modulo $\hbar {\mathfrak {e}}$. Moreover, if $\textbf{k}$ (resp. $\overline{\textbf{k}}$) is admissible, then belongs to $\hbar \sum _{j\ge 2}e_{j}{\mathfrak {e}}$ (resp. $\hbar \sum _{j\ge 2}{\mathfrak {e}}e_{j}$).

Proof

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

(B.3)

Next we calculate using Proposition B.5. Since and , it holds that

We have

$$\begin{aligned} 1+R(X)e_{1}=&1+R(X)(e_{1}-g_{1})+R(X)g_{1}=e^{\hbar b X}+R(X)g_{1}\nonumber \\&\quad = e^{\hbar b X}\left( 1-R(-X)g_{1}\right) . \end{aligned}$$

(B.4)

Thus we see that . Hence

(B.5)

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

and it implies the statement. $\square$

1.3 B.3. Proof of Proposition 5.4

Note that

$$\begin{aligned} ({\mathcal {C}}\langle A \rangle \setminus {\mathcal {C}})a \subset {\mathfrak {H}}^{0} \end{aligned}$$

(B.6)

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

We define the ${\mathcal {C}}$-trilinear map by

for $w, w' \in {\mathcal {C}}\langle A \rangle$ and $u_{1}, \ldots , u_{r} \in A$. It suffices to show that belongs to ${\mathfrak {H}}^{0}$ for any admissible index $\textbf{m}, \textbf{m}'$ and $r \ge 0$.

Lemma B.7

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

(B.7)

Proof

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

From the definition of , we see that the second term in the right-hand side is equal to

Hence, it holds that

Since , we see that the right-hand side above is equal to

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

Since and , it holds that

Now the desired equality (B.7) follows from (B.4). $\square$

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)

Now set $w=E_{\textbf{m}}$ and $w'=E_{\overline{\textbf{m}'}}$ with admissible indices $\textbf{m}$ and $\textbf{m}'$. We have modulo $\hbar \widehat{{\mathfrak {H}}^{0}}$ because of Proposition 3.3, Corollary B.6 and $E_{\textbf{m}} \in \widehat{{\mathfrak {H}}^{0}}$. From the definition of the shuffle product and (B.6), we see that belongs to ${\mathfrak {H}}^{0}$, hence so does the first term of (B.8) with $w=E_{\textbf{m}}$ and $w'=E_{\overline{\textbf{m}'}}$. Since $\textbf{m}$ and $\textbf{m}'$ are admissible, we can write $E_{\textbf{m}}=ua$ and 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

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 belongs to ${\mathfrak {H}}^{0}[[X]]$, and this completes the proof of Proposition 5.4.

1.4 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 $\rho _{R(X)}(g_{1})=(1-R(X)g_{1})^{-1}e^{\hbar b X}g_{1}$, we have

Similarly, we have

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

$$\begin{aligned} \rho _{R(X)}(g_{\textbf{k}}) \equiv \rho _{X}(E_{\textbf{k}}), \end{aligned}$$

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

First we calculate $\rho _{R(X)}(g_{\textbf{k}})$. For $k \ge 1$, we have

$$\begin{aligned} \rho _{R(X)}(g_{k})&=\rho _{R(X)}(ba) \left( \rho _{R(X)}(a)\right) ^{k-1} \\&\quad =\frac{1}{1-R(X)g_{1}}e^{\hbar b X}g_{1} \left( \frac{1}{1-R(X)g_{1}} (a+\hbar R(X)g_{1})\right) ^{k-1} \\&\quad \equiv \frac{1}{1-R(X)g_{1}}e^{\hbar b X}g_{1} \left( \frac{1}{1-R(X)g_{1}} a \right) ^{k-1}, \end{aligned}$$

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

$$\begin{aligned} \frac{1}{1-R(X)g_{1}}a=a+\sum _{n \ge 1}(R(X)g_{1})^{n-1}R(X)g_{2} \equiv a+\sum _{n \ge 1}(Xg_{1})^{n-1}Xg_{2}=\frac{1}{1-g_{1}X}a, \end{aligned}$$

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

$$\begin{aligned} \rho _{R(X)}(g_{k})\equiv \frac{1}{1-R(X)g_{1}}e^{\hbar b X}g_{1} \left( \frac{1}{1-g_{1} X} a \right) ^{k-1}, \end{aligned}$$

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

$$\begin{aligned} \rho _{R(X)}(g_{\textbf{k}})\equiv \prod _{1\le i\le r}^{\curvearrowright } \left( \frac{1}{1-g_{1}X}\, g_{1}\left( \frac{1}{1-g_{1}X}a \right) ^{k_{i}-1} \right) , \end{aligned}$$

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

Next we calculate $\rho _{X}(E_{\textbf{k}})$. For $k \ge 1$, we have

$$\begin{aligned} \rho _{X}(e_{k})&=\rho _{X}(b(a+\hbar ))(\rho _{X}(a))^{k-1}= \frac{1}{1-g_{1}X}\,e_{1} \left( \frac{1}{1-g_{1}X}(a+\hbar g_{1}X) \right) ^{k-1} \\&\equiv \frac{1}{1-g_{1}X}\,e_{1} \left( \frac{1}{1-g_{1}X}a \right) ^{k-1}, \end{aligned}$$

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

$$\begin{aligned} (e_{1}-g_{1})\frac{1}{1-g_{1}X}a&=(e_{1}-g_{1})\left( 1+\frac{1}{1-g_{1}X}g_{1}X\right) a\\&= \hbar g_{1}+(e_{1}-g_{1})\frac{1}{1-g_{1}X}g_{2}X, \end{aligned}$$

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

$$\begin{aligned} \rho _{X}(e_{k})\equiv \frac{1}{1-g_{1}X}g_{1}\left( \frac{1}{1-g_{1}X}a\right) ^{k-1}, \end{aligned}$$

(B.9)

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

$$\begin{aligned} \rho _{X}(E(Y))\equiv \left( 1-\frac{1}{1-g_{1}X}Y\right) ^{-1}, \end{aligned}$$

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

$$\begin{aligned} \rho _{X}(E_{1^{m}})\equiv \left( \frac{1}{1-g_{1}X}g_{1}\right) ^{m}, \end{aligned}$$

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

$$\begin{aligned} \rho _{X}(E_{\textbf{k}})\equiv \prod _{1\le i\le r}^{\curvearrowright } \left( \frac{1}{1-g_{1}X}\, g_{1}\left( \frac{1}{1-g_{1}X}a \right) ^{k_{i}-1} \right) , \end{aligned}$$

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

1.5 B.5. Proof of Proposition 6.4

We set

$$\begin{aligned} e(X)=\sum _{k \ge 1}e_{k}X^{k-1}=e_{1}\frac{1}{1-aX}, \quad e^{0}(X)=\sum _{k \ge 2}e_{k}X^{k-2}=e_{2}\frac{1}{1-aX}=e(X)a. \end{aligned}$$

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

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

(B.10)

We also set

Proposition B.8

Set

$$\begin{aligned} E(Y_{1}, Y_{2})=\frac{1}{Y_{1}Y_{2}} \sum _{j=1}^{2}Y_{j}\left( E(Y_{1}+Y_{2})-E(Y_{j})\right) . \end{aligned}$$

It holds that

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

Proof

Set

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

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

$$\begin{aligned} R(Y_{1})+R(Y_{2})+\hbar b R(Y_{1})R(Y_{2})=R(Y_{1}+Y_{2}). \end{aligned}$$

(B.11)

Then we obtain

$$\begin{aligned} \left\{ \partial (w'E(Y_{2}), w)+ P(Y_{1}, Y_{2})\frac{1}{Y_{2}}(R(Y_{1}+Y_{2})-R(Y_{1}))g_{1} \right\} E(Y_{1}). \end{aligned}$$

By changing $w \leftrightarrow w'$ and $Y_{1} \leftrightarrow Y_{2}$, we obtain a similar formula for . Thus we get

$$\begin{aligned} \Xi (wE(Y_{1}), w'E(Y_{2}))&= \partial (w'E(Y_{2}), w)E(Y_{1})+\partial (wE(Y_{1}), w')E(Y_{2})\\&\quad + \frac{1}{Y_{1}Y_{2}}P(Y_{1}, Y_{2})Q(Y_{1}, Y_{2}), \end{aligned}$$

where

$$\begin{aligned} Q(Y_{1}, Y_{2})= \hbar b R(Y_{1})R(Y_{2})+\sum _{j=1}^{2}Y_{j}(R(Y_{1}+Y_{2})-R(Y_{j}))g_{1}E(Y_{j}). \end{aligned}$$

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

Using (B.11) we see that

$$\begin{aligned} Q(Y_{1}, Y_{2})=R(Y_{1}+Y_{2})\left( 1+g_{1}\sum _{j=1}^{2}Y_{j}E(Y_{j})\right) - \sum _{j=1}^{2}R(Y_{j})(1+Y_{j}g_{1}E(Y_{j})). \end{aligned}$$

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

$$\begin{aligned} Q(Y_{1}, Y_{2})&=R(Y_{1}+Y_{2})+(R(Y_{1}+Y_{2})g_{1}-1)\sum _{j=1}^{2}Y_{j}E(Y_{j})\\&= (1-R(Y_{1}+Y_{2})g_{1})Y_{1}Y_{2}E(Y_{1}, Y_{2}). \end{aligned}$$

Thus we get the desired formula. $\square$

Lemma B.9

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

(B.12)

(B.13)

Proof

Note that $e^{0}(X)=e(X)a$ and for any $w, w' \in {\mathfrak {H}}$. Hence

(B.14)

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

Therefore we have

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

The second term of the right-hand side is equal to

because of (B.14). Thus we get (B.13). $\square$

Proposition B.10

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

Proof

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

using (B.12) and (B.13). It implies the desired equality. $\square$

Proposition B.11

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

Proof

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

Calculate using Proposition B.1, and we see that

Now the desired equality follows from (B.10) and (B.14). $\square$

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) $\Xi (w, w')$ belongs to ${\mathfrak {e}}$,

• (ii) and belong to ${\mathfrak {e}}$,

• (iii) belongs to ${\mathfrak {e}}^{0}$

for any $w, w' \in {\mathfrak {e}}$. 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

$$\begin{aligned} C_{r}=C_{r}(X_{1}, \ldots , X_{r}; Y_{0}, \ldots , Y_{r})= E(Y_{0})\prod _{1\le j \le r}^{\curvearrowright }\left( e^{0}(X_{j})E(Y_{j})\right) , \end{aligned}$$

with $r \ge 0$. Hence we may assume that $w=C_{r}$ and $w'=C_{s}'$ for some $r, s\ge 0$, where the prime symbol indicates that $w'$ 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 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 $s \ge 1$. Set $w=C_{0}=E(Y_{1})$ and $w'=C_{s}'=C_{s-1}'e^{0}(X_{2})E(Y_{2})$. Proposition B.8 and Proposition B.10 imply that

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

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

and

From (B.13) and the induction hypothesis, we see that $\partial (C_{r}, C_{s-1}' e^{0}(X_{2})), \partial (C_{s}', C_{r-1} e^{0}(X_{1}))$ and belong to ${\mathfrak {e}}^{0}$. Hence $\Xi (C_{r}, C_{s}')$ belongs to ${\mathfrak {e}}$. The induction hypothesis says that , and hence belongs to ${\mathfrak {e}}$. Similarly we find that belongs to ${\mathfrak {e}}$. Therefore (i) and (ii) hold for $w=C_{r}$ and $w'=C_{s}$, and this completes the proof of Proposition 6.4.