Appendix A
1.1 A.1. Computations with double brackets
We gather some results that we need when performing computations with double brackets in the other appendices.
Firstly, we have noted that if \((\mathcal {A}, \left\{ \left\{ -,-\right\} \right\} )\) is a double quasi-Poisson algebra, its double quasi-Poisson bracket satisfies the cyclic antisymmetry rule \( \left\{ \left\{ b,a\right\} \right\} =- \left\{ \left\{ a,b\right\} \right\} ^\circ \) and the derivation property \( \left\{ \left\{ a,bc\right\} \right\} =b \left\{ \left\{ a,c\right\} \right\} + \left\{ \left\{ a,b\right\} \right\} c\), i.e. \( \left\{ \left\{ a,bc\right\} \right\} =b \left\{ \left\{ a,c\right\} \right\} ' \otimes \left\{ \left\{ a,c\right\} \right\} ''+ \left\{ \left\{ a,b\right\} \right\} ' \otimes \left\{ \left\{ a,b\right\} \right\} '' c\) using Sweedler’s notation. (This is true for the less restrictive assumption that \(\mathcal {A}\) has a double bracket.) There is a similar derivation property in the first argument for the inner bimodule structure \(*\), see [44, (2.4)], which gives \( \left\{ \left\{ bc,a\right\} \right\} = \left\{ \left\{ b,a\right\} \right\} *c+b * \left\{ \left\{ c,a\right\} \right\} \) or more explicitly, \( \left\{ \left\{ bc,a\right\} \right\} = \left\{ \left\{ b,a\right\} \right\} ' c \otimes \left\{ \left\{ b,a\right\} \right\} '' + \left\{ \left\{ c,a\right\} \right\} ' \otimes b \left\{ \left\{ c,a\right\} \right\} ''\).
Secondly, note that from the above properties we get that for any \(a,b \in \mathcal {A}\) where a has an inverse \(a^{-1}\)
$$\begin{aligned} \left\{ \left\{ b,a^{-1}\right\} \right\} =-a^{-1} \left\{ \left\{ b,a\right\} \right\} a^{-1},\quad \left\{ \left\{ a^{-1},b\right\} \right\} =- a^{-1} * \left\{ \left\{ a,b\right\} \right\} *a^{-1}. \end{aligned}$$
(A.1)
Thirdly, if \(a=a_1 \ldots a_k\) and \(a'=a_1' \ldots a_l'\) are two elements of \(\mathcal {A}\) written in terms of generators, then the double bracket between them is given by
$$\begin{aligned} \begin{aligned} \left\{ \left\{ a, a'\right\} \right\}&=\sum _{s=1}^k\sum _{t=1}^l (a_1 \ldots a_{s-1}) *(a_1' \ldots a_{t-1}') \left\{ \left\{ a_s,a_t'\right\} \right\} (a_{t+1} \ldots a_{l}) *(a_{s+1} \ldots a_{k}) \\&=\sum _{s=1}^k\sum _{t=1}^l (a_1' \ldots a_{t-1}') \left\{ \left\{ a_s,a_t'\right\} \right\} ' (a_{s+1} \ldots a_{k}) \otimes (a_1 \ldots a_{s-1}) \left\{ \left\{ a_s,a_t'\right\} \right\} '' (a_{t+1} \ldots a_{l}) , \end{aligned} \end{aligned}$$
using the derivation properties as above.
1.2 A.2. Proof of Lemma 3.1
We show the claim by descending induction, starting from \(\alpha =d\). So, the first step is to show that
$$\begin{aligned} \left\{ \left\{ s_d,z\right\} \right\}&=\frac{1}{2} (s_d \otimes z - zs_d \otimes e_0 + e_0 \otimes s_dz - z \otimes s_d) \end{aligned}$$
(A.2a)
$$\begin{aligned} \left\{ \left\{ s_d,x\right\} \right\}&=\frac{1}{2} (s_d\otimes x - xs_d \otimes e_0 - e_0 \otimes s_d x - x \otimes s_d) \end{aligned}$$
(A.2b)
$$\begin{aligned} \left\{ \left\{ s_d,v_\beta \right\} \right\}&= -\frac{1}{2} (v_\beta s_d \otimes e_0 + v_\beta \otimes s_d) \quad \left\{ \left\{ s_d,w_\beta \right\} \right\} =\frac{1}{2} (s_d \otimes w_\beta + e_0 \otimes s_d w_\beta ). \end{aligned}$$
(A.2c)
To compute such double brackets, we use the relation \(s_d=(\Phi _0)^{-1}\phi z\) and obtain
$$\begin{aligned} \left\{ \left\{ s_d,a\right\} \right\} = \left\{ \left\{ \Phi _0^{-1},a\right\} \right\} *\phi z+ \Phi _0^{-1}* \left\{ \left\{ \phi ,a\right\} \right\} *z + \Phi _0^{-1}\phi * \left\{ \left\{ z,a\right\} \right\} . \end{aligned}$$
(A.3)
The first term can be calculated with the help of (3.6):
$$\begin{aligned} \left\{ \left\{ \Phi _0^{-1},a\right\} \right\}= & {} -\Phi _0^{-1}* \left\{ \left\{ \Phi _0,a\right\} \right\} *\Phi _0^{-1}\\= & {} -\frac{1}{2} (a \Phi _0^{-1} \otimes e_0- \Phi _0^{-1} \otimes e_0 a + a e_0 \otimes \Phi _0^{-1} - e_0 \otimes \Phi _0^{-1} a). \end{aligned}$$
(Note that in the case when \(a=v_\beta \) or \(a=w_\beta \), some of the terms in this expression vanish due to \(e_0 v_\beta =w_\beta e_0=0\).) The second term in (A.3) is calculated using (3.7)–(3.8), while for the third term we use (3.3a)–(3.3b). Doing this for each of the cases \(a=x, z, v_\beta , w_\beta \) verifies (A.2a)–(A.2c). We leave the details to the reader.
For the induction step, recall that \(s_\alpha =u_{\alpha +1} s_{\alpha +1}\) for \(u_{\alpha }=(1+w_{\alpha }v_{\alpha })^{-1}\). Therefore,
$$\begin{aligned} \left\{ \left\{ s_\alpha ,a\right\} \right\} = \left\{ \left\{ u_{\alpha +1},a\right\} \right\} *s_{\alpha +1}+ u_{\alpha +1}* \left\{ \left\{ s_{\alpha +1},a\right\} \right\} . \end{aligned}$$
The second term is given by the induction hypothesis, while we can find the first term using the easily verified formulas
$$\begin{aligned} \left\{ \left\{ u_{\alpha +1},z\right\} \right\}&= \frac{1}{2} (u_{\alpha +1} \otimes z - z u_{\alpha +1} \otimes e_0 - e_0 \otimes u_{\alpha +1} z + z \otimes u_{\alpha +1}),\\ \left\{ \left\{ u_{\alpha +1},x\right\} \right\}&= \frac{1}{2} (u_{\alpha +1} \otimes x - x u_{\alpha +1} \otimes e_0 - e_0 \otimes u_{\alpha +1} x + x \otimes u_{\alpha +1}), \\ \left\{ \left\{ u_{\alpha +1},v_\beta \right\} \right\}&=\frac{1}{2} \delta _{(\alpha +1, \beta )}(v_\beta u_{\alpha +1} \otimes e_0 + v_\beta \otimes u_{\alpha +1})\\&\quad + \frac{1}{2} o(\alpha +1,\beta ) (v_\beta u_{\alpha +1} \otimes e_0 - v_\beta \otimes u_{\alpha +1}), \\ \left\{ \left\{ u_{\alpha +1},w_\beta \right\} \right\}&=-\frac{1}{2} \delta _{(\alpha +1, \beta )}(e_0 \otimes _{\alpha +1} w_\beta + u_{\alpha +1} \otimes w_\beta )\\&\quad +\frac{1}{2} o(\alpha +1,\beta ) (e_0 \otimes u_{\alpha +1} w_\beta - u_{\alpha +1} \otimes w_\beta ). \end{aligned}$$
In order to prove (3.10a)–(3.10d), we need to consider the cases \(a=x, z, v_\beta , w_\beta \). We will do the case \(a=w_\beta \), leaving the other cases to the reader.
First, if \(\beta \le \alpha \), we can use (3.10c) and since \(o(\alpha +1,\beta )=-1\) we get
$$\begin{aligned} \begin{aligned} \left\{ \left\{ s_\alpha ,w_\beta \right\} \right\}&=-\frac{1}{2} (s_{\alpha +1} \otimes u_{\alpha +1} w_\beta - u_{\alpha +1}s_{\alpha +1} \otimes w_\beta )\\&\quad +\frac{1}{2} (s_{\alpha +1} \otimes u_{\alpha +1} w_\beta + e_0 \otimes u_{\alpha +1}s_{\alpha +1} w_\beta ) \\&=\frac{1}{2} (e_0 \otimes s_{\alpha } w_\beta +s_\alpha \otimes w_\beta ). \end{aligned} \end{aligned}$$
Next, if \(\beta =\alpha +1\), we still use (3.10c) and find
$$\begin{aligned} \begin{aligned} \left\{ \left\{ s_\alpha ,w_{\alpha +1}\right\} \right\}&=-\frac{1}{2} (s_{\alpha +1} \otimes u_{\alpha +1} w_\beta + u_{\alpha +1}s_{\alpha +1} \otimes w_\beta )\\&\quad +\frac{1}{2} (s_{\alpha +1} \otimes u_{\alpha +1} w_{\alpha +1} + e_0 \otimes u_{\alpha +1}s_{\alpha +1} w_{\alpha +1}) \\&=\frac{1}{2} (e_0 \otimes s_{\alpha } w_{\alpha +1} - s_\alpha \otimes w_{\alpha +1}). \end{aligned} \end{aligned}$$
Finally, if \(\beta > \alpha +1\), we need (3.10d) and since \(o(\alpha +1,\beta )=+1\) we get
$$\begin{aligned} \begin{aligned} \left\{ \left\{ s_\alpha ,w_{\alpha +1}\right\} \right\}&=\frac{1}{2} (s_{\alpha +1} \otimes u_{\alpha +1} w_\beta - u_{\alpha +1}s_{\alpha +1} \otimes w_\beta )\\&\quad +\quad \frac{1}{2} (e_0 \otimes u_{\alpha +1} s_{\alpha +1} w_\beta - s_{\alpha +1} \otimes u_{\alpha +1} w_\beta ) \\&= \frac{1}{2} (e_0 \otimes s_{\alpha } w_\beta - s_{\alpha } \otimes w_\beta ).&\quad \end{aligned} \end{aligned}$$
\(\square \)
1.3 A.3. Proof of Lemma 3.2
Using the cyclic antisymmetry of the double bracket, we only need to show that for any \(\alpha \ge \beta \),
$$\begin{aligned} \left\{ \left\{ s_\alpha ,s_\beta \right\} \right\} = \frac{1}{2} (e_0 \otimes s_\alpha s_\beta - s_\beta s_\alpha \otimes e_0 + s_\alpha \otimes s_\beta - s_\beta \otimes s_\alpha ) . \end{aligned}$$
(A.4)
Using the elements \(r_\gamma =1+w_\gamma v_\gamma \) and the definition of \(s_\beta \) as (3.9), we can write \(s_\beta =r_\beta \ldots r_1 z\). Therefore, we find that
$$\begin{aligned} \left\{ \left\{ s_\alpha ,s_\beta \right\} \right\} =r_\beta \dots r_1 \left\{ \left\{ s_\alpha ,z\right\} \right\} + \sum _{\gamma =1}^\beta r_\beta \dots r_{\gamma +1} \left\{ \left\{ s_\alpha ,r_\gamma \right\} \right\} r_{\gamma -1} \dots r_1 z. \end{aligned}$$
(A.5)
To find the double brackets \( \left\{ \left\{ s_\alpha ,r_\gamma \right\} \right\} \), we use (3.10c) and we obtain
$$\begin{aligned} \left\{ \left\{ s_\alpha ,r_\gamma \right\} \right\} =\frac{1}{2} (e_0 \otimes s_\alpha r_\gamma - r_\gamma \otimes s_\alpha s_\alpha \otimes r_\gamma - r_\gamma s_\alpha \otimes e_0 ), \quad 1\le \gamma \le \alpha . \end{aligned}$$
For any \(\beta \le \alpha \), it remains to substitute these double brackets together with (3.10a) back in (A.5), and we obtain (A.4) after simplification. \(\square \)
1.4 A.4. Proof of Proposition 4.1
Recall that \(f_k:= {\text {tr}}(X^k)\) and \(g^k_{\gamma \epsilon }={\text {tr}}(A^\gamma B^\epsilon X^k)\). First, we need to compute the Poisson brackets between those functions. We have remarked in Sect. 3.2 that the Poisson bracket \(\{ -,- \}\) on \(\mathcal {C}_{n, d, q}\) is (globally) defined from the corresponding Lie bracket \(\{ -,- \}\) on \(\Lambda ^q/[\Lambda ^q,\Lambda ^q]\) by (3.23b). In fact, it is sufficient to compute that bracket in \(\mathcal {A}/[\mathcal {A},\mathcal {A}]\), then projects into \(\Lambda ^q/[\Lambda ^q,\Lambda ^q]\). Assuming that x is invertible, the same holds in \(\mathcal {A}^{\times }\). Therefore, we need the following lemma.
Lemma A.1
For any \(k,l\ge 1\), the following identities hold in \(\mathcal {A}^{\times }/[\mathcal {A}^{\times },\mathcal {A}^{\times }]\)
$$\begin{aligned} \{ x^k,x^l \}&=0, \quad \{ x^k,a_\alpha b_\beta x^l \}=k\, a_\alpha b_\beta x^{k+l}, \end{aligned}$$
(A.6a)
$$\begin{aligned} \{ a_\gamma b_\epsilon x^k,a_\alpha b_\beta x^l \}&=\frac{1}{2} \left( \sum _{r=1}^k-\sum _{r=1}^l \right) \left( a_\alpha b_\beta x^r a_\gamma b_\epsilon x^{k+l-r} +a_\alpha b_\beta x^{k+l-r} a_\gamma b_\epsilon x^{r} \right) \nonumber \\&\quad +\frac{1}{2} o(\alpha ,\gamma ) (a_\gamma b_\epsilon x^k a_\alpha b_\beta x^l +a_\alpha b_\epsilon x^k a_\gamma b_\beta x^l) \nonumber \\&\quad +\frac{1}{2} o(\epsilon ,\beta ) (a_\alpha b_\beta x^k a_\gamma b_\epsilon x^l -a_\alpha b_\epsilon x^k a_\gamma b_\beta x^l) \nonumber \\&\quad +\frac{1}{2} [o(\epsilon ,\alpha )+\delta _{\alpha \epsilon }]\,a_\alpha b_\epsilon x^k a_\gamma b_\beta x^l -\frac{1}{2} [o(\beta ,\gamma )+\delta _{\beta \gamma }]\,a_\alpha b_\epsilon x^k a_\gamma b_\beta x^l \nonumber \\&\quad +\,\delta _{\alpha \epsilon } \left( zx^k+\sum _{\lambda =1}^{\epsilon -1} a_\lambda b_\lambda x^k \right) a_\gamma b_\beta x^l - \delta _{\beta \gamma }\,\, a_\alpha b_\epsilon x^k \left( zx^l + \sum _{\mu =1}^{\beta -1} a_\mu b_\mu x^l \right) . \end{aligned}$$
(A.6b)
The proof can be seen as a special case of [16, Lemma 3.2]. By taking the traces and using the identity (3.23b), we obtain the Poisson brackets between the functions \((f_k, g^{k}_{\alpha \beta })\). To write them in terms of \(f^k, g^k_{\alpha \beta }\) and \(h_{\gamma \epsilon }^{k,l}={\text {tr}}(A_\gamma B_\epsilon X^k Z X^l)\), we use
$$\begin{aligned} {\text {tr}}(A_\alpha B_\beta X^{k} A_\gamma B_\epsilon X^{l})= (B_\beta X^{k} A_\gamma ) (B_\epsilon X^{l}A_\alpha )=g_{\gamma \beta }^k g_{\alpha \epsilon }^l, \end{aligned}$$
and similar variants.
Lemma A.2
For any \(\alpha ,\beta =1,\ldots ,d\) and \(k,l\ge 1\),
$$\begin{aligned} \{ f_k,f_l \}&=0, \end{aligned}$$
(A.7a)
$$\begin{aligned} \{ f_k,g_{\alpha \beta }^l \}&=k\, g_{\alpha \beta }^{k+l}, \end{aligned}$$
(A.7b)
$$\begin{aligned} \{ g_{\gamma \epsilon }^k,g_{\alpha \beta }^l \}&= \frac{1}{2} \left( \sum _{r=1}^k-\sum _{r=1}^l \right) \left( g_{\gamma \beta }^r g_{\alpha \epsilon }^{k+l-r} + g_{\gamma \beta }^{k+l-r} g_{\alpha \epsilon }^{r}\right) \nonumber \\&\quad +\frac{1}{2} o(\alpha ,\gamma ) \left( g_{\gamma \beta }^l g_{\alpha \epsilon }^{k} + g_{\gamma \epsilon }^{k} g_{\alpha \beta }^{l}\right) +\frac{1}{2} o(\epsilon ,\beta ) \left( g_{\gamma \beta }^k g_{\alpha \epsilon }^l - g_{\gamma \epsilon }^{k} g_{\alpha \beta }^{l}\right) \nonumber \\&\quad +\frac{1}{2} [o(\epsilon ,\alpha )+\delta _{\alpha \epsilon }-o(\beta ,\gamma )-\delta _{\beta \gamma }]\, g_{\gamma \epsilon }^k g_{\alpha \beta }^{l} \nonumber \\&\quad +\,\delta _{\alpha \epsilon } h_{\gamma \beta }^{l,k} +\delta _{\alpha \epsilon } \sum _{\lambda =1}^{\epsilon -1} g_{\gamma \lambda }^k g_{\lambda \beta }^l -\delta _{\beta \gamma } h_{\alpha \epsilon }^{k,l} - \delta _{\beta \gamma }\sum _{\mu =1}^{\beta -1} g_{\alpha \mu }^l g_{\mu \epsilon }^k. \end{aligned}$$
(A.7c)
Our goal is to show that for the functions \((f_k,g^k_{\gamma \epsilon })\) generating the ring of functions at a generic point, the following equalities hold
$$\begin{aligned}&\xi ^*\{ f_k,f_l \}=\{ \xi ^* f_k , \xi ^* f_l \}, \quad \xi ^*\{ f_k,g^l_{\alpha \beta } \}=\{ \xi ^* f_k , \xi ^* g^l_{\alpha \beta } \},\\&\quad \xi ^*\{ g^k_{\gamma \epsilon },g^l_{\alpha \beta } \}=\{ \xi ^* g^k_{\gamma \epsilon }, \xi ^* g^l_{\alpha \beta } \}, \end{aligned}$$
where we compose with \(\xi \) the identities from Lemma A.2 in the left-hand sides, and use the expressions (4.9) in the right-hand sides. In fact, we will be quite pedantic and prove these identities also after summing over \(\alpha \) and/or \(\gamma \) ranging from 1 to d. This allows us to show that the Poisson brackets given in Proposition 4.1 are correct one at a time. Note that in local coordinates, we use \(\xi ^*X_{ij}=\delta _{ij} x_i\), \(\xi ^*(A_\alpha B_\beta )_{ij}= a_i^\alpha b_j^\beta \) while we simply write \(\xi ^*Z_{ij}=Z_{ij}\).
To show that the brackets in (2.10a) are correct, first notice that \(\{ x_i,x_j \}=0\) implies \(\xi ^*\{ f_k,f_l \}=\{ \xi ^* f_k , \xi ^* f_l \}\) as both expressions vanish. Second, recall that by assumption \(\sum _\alpha a_i^\alpha =1\) for all i. Thus, from \(\{ x_i, b_j^\beta \}=\delta _{ij} x_i b_j^\beta \),
$$\begin{aligned} \begin{aligned} \sum _{\alpha } \{ \xi ^* f_k , \xi ^* g^l_{\alpha \beta } \}&=\sum _{i,j=1}^n\{ x_i^k, b_j^\beta x_j^l \} =\sum _{i=1}^n k\, x_i^{k+l} b_i^\beta , \\ \sum _{\alpha } \xi ^*\{ f_k,g^l_{\alpha \beta } \}&= k\, \sum _{\alpha } \xi ^*{\text {tr}}(A_\alpha B_\beta X^{k+l}) =k\, \sum _{i=1}^n b_i^\beta x_i^{k+l}, \end{aligned} \end{aligned}$$
and we get \(\xi ^*\{ f_k, \sum _{\alpha } g^l_{\alpha \beta } \}=\{ \xi ^* f_k , \xi ^* \sum _{\alpha } g^l_{\alpha \beta } \}\). Third, without summing, we get again that \(\xi ^*\{ f_k,g^l_{\alpha \beta } \}=\{ \xi ^* f_k , \xi ^* g^l_{\alpha \beta } \}\) using \(\{ a_i^\alpha ,x_j \}=0\).
To see that we need (2.10d), (2.10c) and (2.10b), we will, respectively, sum over all values of \(\alpha \) and \(\gamma \), all values of \(\gamma \) and finally not sum at all the functions \(\xi ^*\{ g^k_{\gamma \epsilon },g^l_{\alpha \beta } \}\) and \(\{ \xi ^* g^k_{\gamma \epsilon }, \xi ^* g^l_{\alpha \beta } \}\), to show that they agree. We get from Lemma A.2 that we can write
$$\begin{aligned}&\xi ^*\{ g^k_{\gamma \epsilon },g^l_{\alpha \beta } \}= \frac{1}{2} \left( \sum _{r=1}^k-\sum _{r=1}^l \right) \sum _{i,j=1}^n \left( b_i^\beta x_i^r a_i^\gamma b_j^\epsilon x_j^{k+l-r} a_j^\alpha +b_i^\beta x_i^{k+l-r} a_i^\gamma b_j^\epsilon x_j^r a_j^\alpha \right) \nonumber \\&\quad +\,\frac{1}{2} o(\alpha ,\gamma ) \sum _{i,j=1}^n \left( b_j^\epsilon x_j^k a_j^\alpha b_i^\beta x_i^l a_i^\gamma + b_j^\epsilon x_j^k a_j^\gamma b_i^\beta x_i^l a_i^\alpha \right) \nonumber \\&\quad +\,\frac{1}{2} o(\epsilon ,\beta ) \sum _{i,j=1}^n \left( b_i^\beta x_i^k a_i^\gamma b_j^\epsilon x_j^l a_j^\alpha - b_j^\epsilon x_j^k a_j^\gamma b_i^\beta x_i^l a_i^\alpha \right) \nonumber \\&\quad +\,\frac{1}{2} [o(\epsilon ,\alpha )+\delta _{\alpha \epsilon }]\, \sum _{i,j=1}^n b_j^\epsilon x_j^k a_j^\gamma b_i^\beta x_i^la_i^\alpha -\frac{1}{2} [o(\beta ,\gamma )+\delta _{\beta \gamma }]\, \sum _{i,j=1}^n b_j^\epsilon x_j^k a_j^\gamma b_i^\beta x_i^l a_i^\alpha \nonumber \\&\quad +\,\delta _{\alpha \epsilon } \sum _{i,j=1}^n \left( Z_{ij} + \sum _{\lambda =1}^{\epsilon -1} a_i^\lambda b_j^\lambda \right) x_j^k a^\gamma _j b_i^\beta x_i^l -\delta _{\beta \gamma } \sum _{i,j=1}^n \left( Z_{ji}+\sum _{\mu =1}^{\beta -1} a_j^\mu b_i^\mu \right) x_i^l a_i^\alpha b_j^\epsilon x_j^k.\nonumber \\ \end{aligned}$$
(A.8)
In the first case, we have to prove
$$\begin{aligned} \sum _{\gamma ,\alpha =1}^d \xi ^*\{ g^k_{\gamma \epsilon },g^l_{\alpha \beta } \}=\sum _{i,j=1}^n\{ b_j^\epsilon x_j^k, b_i^\beta x_i^l \}. \end{aligned}$$
(A.9)
The right-hand side of (A.9) can be read as
$$\begin{aligned} \begin{aligned} (\mathrm{A.9})_\mathrm{{RHS}}&= \sum _{i,j=1}^n\left( \{ b_j^\epsilon , x_i^l \} x_j^k b_i^\beta +\{ x_j^k, b_i^\beta \} b_j^\epsilon x_i^l +\{ b_j^\epsilon , b_i^\beta \} x_j^k x_i^l \right) \\&=(k-l) \sum _{i=1}^n b_i^\epsilon , b_i^\beta x_i^{k+l} +\frac{1}{2} \sum _{\begin{array}{c} i,j=1\\ i\ne j \end{array}}^n x_j^k x_i^l \frac{x_j+x_i}{x_j-x_i} (b_j^\epsilon b_i^\beta + b_i^\epsilon b_j^\beta ) \\&\quad +\,\sum _{i,j=1}^n x_j^k x_i^l(b_i^\beta Z_{ij} - b_j^\epsilon Z_{ji}) +\frac{1}{2} o(\epsilon ,\beta ) \sum _{i,j=1}^n x_j^k x_i^l(b_i^\epsilon b_j^\beta -b_j^\epsilon b_i^\beta ) \\&\quad +\,\sum _{i,j=1}^n x_j^k x_i^l b_i^\beta \sum _{\lambda =1}^{\epsilon -1}a_i^\lambda (b_j^\lambda -b_j^\epsilon ) -\sum _{i,j=1}^n x_j^k x_i^l b_j^\epsilon \sum _{\mu =1}^{\beta -1}a_j^\mu (b_i^\mu -b_i^\beta ). \end{aligned} \end{aligned}$$
Now, the left-hand side of (A.9) can be written from (A.8), after summing over \(\alpha ,\gamma \) and using the condition \(\sum _\gamma a_i^\gamma =1\) when possible. We get
$$\begin{aligned}&(\mathrm{A.9})_\mathrm{{LHS}}= \frac{1}{2} \sum _{i,j=1}^n b_i^\beta b_j^\epsilon \left( \sum _{r=1}^k-\sum _{r=1}^l \right) \left( x_i^r x_j^{k+l-r}+ x_i^{k+l-r} x_j^r\right) \end{aligned}$$
(A.10a)
$$\begin{aligned}&\quad +\,\frac{1}{2} \sum _{i,j=1}^n b_j^\epsilon b_i^\beta x_j^k x_i^l \sum _{\alpha ,\gamma =1}^d o(\alpha ,\gamma ) \left( a_i^\gamma a_j^\alpha + a_i^\alpha a_j^\gamma \right) +\frac{1}{2} o(\epsilon ,\beta ) \sum _{i,j=1}^n x_j^k x_i^l \left( b_j^\beta b_i^\epsilon - b_j^\epsilon b_i^\beta \right) \end{aligned}$$
(A.10b)
$$\begin{aligned}&\quad +\,\frac{1}{2} \sum _{\alpha =1}^d[o(\epsilon ,\alpha )+\delta _{\alpha \epsilon }]\, \sum _{i,j=1}^n a_i^\alpha b_j^\epsilon x_j^k b_i^\beta x_i^l -\frac{1}{2} \sum _{\gamma =1}^d [o(\beta ,\gamma )+\delta _{\beta \gamma }]\, \sum _{i,j=1}^n b_j^\epsilon x_j^k a_j^\gamma b_i^\beta x_i^l \end{aligned}$$
(A.10c)
$$\begin{aligned}&\quad +\,\sum _{i,j=1}^n x_j^k x_i^l (Z_{ij} b_i^\beta - Z_{ji} b_j^\epsilon ) +\sum _{i,j=1}^n \sum _{\lambda =1}^{\epsilon -1} a_i^\lambda b_j^\lambda x_j^k b_i^\beta x_i^l - \sum _{i,j=1}^n \sum _{\mu =1}^{\beta -1} a_j^\mu b_i^\mu x_i^l b_j^\epsilon x_j^k , \end{aligned}$$
(A.10d)
To reduce this expression further, remark that by definition of the ordering function \(o(-,-)\)
$$\begin{aligned} \sum _{\alpha ,\gamma =1}^d o(\alpha ,\gamma ) \left( a_i^\gamma a_j^\alpha + a_i^\alpha a_j^\gamma \right) = \sum _{\alpha <\gamma } \left( a_i^\gamma a_j^\alpha + a_i^\alpha a_j^\gamma \right) - \sum _{\alpha >\gamma } \left( a_i^\gamma a_j^\alpha + a_i^\alpha a_j^\gamma \right) =0, \end{aligned}$$
after relabelling the indices in the second sum, so that the first term of (A.10b) disappears. Then, write (A.10a) as
$$\begin{aligned} \begin{aligned} (\mathrm{A.10a})&= (k-l) \sum _{i=1}^n b_i^\beta b_i^\epsilon x_i^{k+l} + \frac{1}{2} \sum _{\begin{array}{c} i,j=1\\ i\ne j \end{array}}^n b_i^\beta b_j^\epsilon \left( \sum _{r=1}^k-\sum _{r=1}^l \right) \left( x_i^r x_j^{k+l-r}+ x_i^{k+l-r} x_j^r\right) , \end{aligned} \end{aligned}$$
so that the sum for \(i\ne j\) can be written as (here, we assume \(k>l\), the case \(k<l\) is exactly the same)
$$\begin{aligned} \begin{aligned}&\frac{1}{2} \sum _{\begin{array}{c} i,j=1\\ i\ne j \end{array}}^n b_i^\beta b_j^\epsilon \sum _{r=l+1}^k \frac{x_i-x_j}{x_i-x_j}\left( x_i^r x_j^{k+l-r}+ x_i^{k+l-r} x_j^r\right) \\&\quad =\frac{1}{2} \sum _{\begin{array}{c} i,j=1\\ i\ne j \end{array}}^n b_i^\beta b_j^\epsilon \frac{x_i+x_j}{x_i-x_j} \left( x_i^{k} x_j^{l}-x_i^{l} x_j^{k}\right) \, =\,-\frac{1}{2} \sum _{\begin{array}{c} i,j=1\\ i\ne j \end{array}}^n x_i^{l} x_j^{k} \frac{x_i+x_j}{x_i-x_j} \left( b_j^\beta b_i^\epsilon +b_i^\beta b_j^\epsilon \right) , \end{aligned} \end{aligned}$$
(A.11)
after relabelling indices to obtain last equality. Finally, let’s look at the terms in (A.10c) and (A.10d) with no factor \(Z_{ij}\). They can be written as
$$\begin{aligned} \frac{1}{2} \sum _{i,j=1}^n x_j^k x_i^l \left( \left[ \sum _{\alpha \ge \epsilon }-\sum _{\alpha =1}^{\epsilon -1}\right] a_i^\alpha b_j^\epsilon b_i^\beta -\left[ \sum _{\gamma \ge \beta }-\sum _{\gamma =1}^{\beta -1}\right] b_j^\epsilon a_j^\gamma b_i^\beta + 2\sum _{\lambda =1}^{\epsilon -1} a_i^\lambda b_j^\lambda b_i^\beta - 2 \sum _{\mu =1}^{\beta -1} a_j^\mu b_i^\mu b_j^\epsilon \right) , \end{aligned}$$
and if we split the sum \(\sum _{\alpha \ge \epsilon }\) as \(\sum _{\alpha =1}^d-\sum _{\alpha =1}^{\epsilon -1}\) and do the same with the sum over \(\gamma \ge \beta \), we get after using the conditions \(\sum _\alpha a_i^\alpha =1\) (and the same for \(\gamma \))
$$\begin{aligned} \begin{aligned}&\sum _{i,j=1}^n x_j^k x_i^l \left( -\sum _{\alpha =1}^{\epsilon -1}a_i^\alpha b_j^\epsilon b_i^\beta +\sum _{\gamma =1}^{\beta -1} b_j^\epsilon a_j^\gamma b_i^\beta + \sum _{\lambda =1}^{\epsilon -1} a_i^\lambda b_j^\lambda b_i^\beta - \sum _{\mu =1}^{\beta -1} a_j^\mu b_i^\mu b_j^\epsilon \right) \\&\quad =\sum _{i,j=1}^n x_j^k x_i^l \left( \sum _{\lambda =1}^{\epsilon -1} a_i^\lambda (b_j^\lambda -b_j^\epsilon ) b_i^\beta - \sum _{\mu =1}^{\beta -1} a_j^\mu (b_i^\mu - b_i^\beta ) b_j^\epsilon \right) . \end{aligned} \end{aligned}$$
Summing together all the terms, we have reduced the left-hand side of (A.9) to the form
$$\begin{aligned} \begin{aligned} (\mathrm{A.9})_\mathrm{{LHS}}&=(k-l) \sum _{i=1}^n b_i^\beta b_i^\epsilon x_i^{k+l} -\frac{1}{2} \sum _{\begin{array}{c} i,j=1\\ i\ne j \end{array}}^n x_i^{l} x_j^{k} \frac{x_i+x_j}{x_i-x_j} \left( b_j^\beta b_i^\epsilon +b_i^\beta b_j^\epsilon \right) \\&\quad +\frac{1}{2} o(\epsilon ,\beta ) \sum _{i,j=1}^n x_j^k x_i^l \left( b_j^\beta b_i^\epsilon - b_j^\epsilon b_i^\beta \right) \\&\quad +\sum _{i,j=1}^n x_j^k x_i^l (Z_{ij} b_i^\beta - Z_{ji} b_j^\epsilon )\\&\quad +\sum _{i,j=1}^n x_j^k x_i^l \left( \sum _{\lambda =1}^{\epsilon -1} a_i^\lambda (b_j^\lambda -b_j^\epsilon ) b_i^\beta - \sum _{\mu =1}^{\beta -1} a_j^\mu (b_i^\mu - b_i^\beta ) b_j^\epsilon \right) . \end{aligned} \end{aligned}$$
This is precisely the right-hand side of (A.9). In the second case, we show
$$\begin{aligned} \sum _{\gamma =1}^d \xi ^*\{ g^k_{\gamma \epsilon },g^l_{\alpha \beta } \}=\sum _{i,j=1}^n\{ b_j^\epsilon x_j^k, a_i^\alpha b_i^\beta x_i^l \}. \end{aligned}$$
(A.12)
The right-hand side of (A.12) can be read as
$$\begin{aligned} (\mathrm{A.12})_\mathrm{{RHS}}= & {} \sum _{i,j=1}^n\left( \{ b_j^\epsilon , x_i^l \} x_j^k a_i^\alpha b_i^\beta +\{ x_j^k, b_i^\beta \} b_j^\epsilon a_i^\alpha x_i^l +\{ b_j^\epsilon , b_i^\beta \} x_j^k x_i^l a_i^\alpha + \{ b_j^\epsilon ,a_i^\alpha \} x_j^k x_i^l b_i^\beta \right) \\= & {} \,(k-l) \sum _{i=1}^n b_i^\epsilon a_i^\alpha b_i^\beta x_i^{k+l} +\frac{1}{2} \sum _{\begin{array}{c} i,j=1\\ i\ne j \end{array}}^n x_j^k x_i^l \frac{x_j+x_i}{x_j-x_i} (b_i^\epsilon b_j^\beta a_i^\alpha + b_j^\epsilon b_i^\beta a_j^\alpha ) \\&\quad +\,\frac{1}{2} \sum _{\kappa =1}^d o(\alpha ,\kappa ) \sum _{i,j=1}^n x_j^k x_i^l b_i^\beta b_j^\epsilon (a_j^\kappa a_i^\alpha +a_i^\kappa a_j^\alpha )\\&\quad +\,\frac{1}{2} o(\epsilon ,\beta ) \sum _{i,j=1}^n x_j^k x_i^l(b_i^\epsilon b_j^\beta -b_j^\epsilon b_i^\beta ) a_i^\alpha \\&\quad -\,\sum _{i,j=1}^n x_j^k x_i^l a_i^\alpha b_j^\epsilon \sum _{\mu =1}^{\beta -1}a_j^\mu (b_i^\mu -b_i^\beta )-\delta _{(\alpha <\epsilon )} \sum _{i,j=1}^n x_j^k x_i^l b_i^\beta a_i^\alpha b_j^\epsilon \\&\quad +\,\delta _{\epsilon \alpha }\, \sum _{i,j=1}^n x_j^k x_i^l b_i^\beta \left( Z_{ij} + \sum _{\lambda =1}^{\epsilon -1} a_i^\lambda b_j^\lambda \right) -\sum _{i,j=1}^n x_j^k x_i^l b_j^\epsilon Z_{ji} a_i^\alpha , \end{aligned}$$
after some easy simplifications. To get the left-hand side, we sum (A.8) over \(\gamma \) and we write
$$\begin{aligned}&(\mathrm{A.12})_\mathrm{{LHS}}\nonumber \\&\quad = (k-l)\sum _{i=1}^n a_i^\alpha b_i^\beta b_j^\epsilon x_i^{k+l} - \frac{1}{2} \sum _{\begin{array}{c} i,j=1\\ i\ne j \end{array}}^n x_j^k x_i^l \frac{x_i+x_j}{x_i-x_j}\left( a_i^\alpha b_j^\beta b_i^\epsilon + a_j^\alpha b_i^\beta b_j^\epsilon \right) \end{aligned}$$
(A.13a)
$$\begin{aligned}&\quad \quad +\frac{1}{2} \sum _{\gamma =1}^d o(\alpha ,\gamma ) \sum _{i,j=1}^n x_j^k x_i^l b_j^\epsilon b_i^\beta \left( a_i^\gamma a_j^\alpha + a_i^\alpha a_j^\gamma \right) +\frac{1}{2} o(\epsilon ,\beta ) \sum _{i,j=1}^n x_j^k x_i^l a_i^\alpha \left( b_j^\beta b_i^\epsilon - b_j^\epsilon b_i^\beta \right) \end{aligned}$$
(A.13b)
$$\begin{aligned}&\qquad +\frac{1}{2} [o(\epsilon ,\alpha )+\delta _{\alpha \epsilon }]\, \sum _{i,j=1}^n x_j^k x_i^l b_j^\epsilon a_i^\alpha b_i^\beta -\frac{1}{2} \sum _{\gamma =1}^d [o(\beta ,\gamma )+\delta _{\beta \gamma }]\, \sum _{i,j=1}^n x_j^k x_i^l a_i^\alpha b_j^\epsilon a_j^\gamma b_i^\beta \end{aligned}$$
(A.13c)
$$\begin{aligned}&\qquad +\,\delta _{\alpha \epsilon } \sum _{i,j=1}^n x_j^k x_i^l b_i^\beta \left( Z_{ij} + \sum _{\lambda =1}^{\epsilon -1} a_i^\lambda b_j^\lambda \right) - \sum _{i,j=1}^n x_j^k x_i^l \left( Z_{ji}+\sum _{\mu =1}^{\beta -1} a_j^\mu b_i^\mu \right) a_i^\alpha b_j^\epsilon , \end{aligned}$$
(A.13d)
where we used an argument similar to (A.11) to rewrite the first line of (A.8) in order to obtain (A.13a). Next, we can write after rearranging terms
$$\begin{aligned} \begin{aligned}&(\mathrm{A.13c})+(\mathrm{A.13d}) \\&\quad = +\frac{1}{2} [1-2\delta _{(\alpha<\epsilon )}]\, \sum _{i,j=1}^n x_j^k x_i^l b_j^\epsilon a_i^\alpha b_i^\beta -\frac{1}{2} \sum _{\gamma =1}^d [1-2\delta _{(\beta >\gamma )}]\, \sum _{i,j=1}^n x_j^k x_i^l a_i^\alpha b_j^\epsilon a_j^\gamma b_i^\beta \\&\qquad +\,\delta _{\alpha \epsilon } \sum _{i,j=1}^n x_j^k x_i^l b_i^\beta \left( Z_{ij} + \sum _{\lambda =1}^{\epsilon -1} a_i^\lambda b_j^\lambda \right) - \sum _{i,j=1}^n x_j^k x_i^l Z_{ji} a_i^\alpha b_j^\epsilon - \sum _{i,j=1}^n x_j^k x_i^l a_i^\alpha b_j^\epsilon \sum _{\mu =1}^{\beta -1} a_j^\mu b_i^\mu \\&\quad = -\delta _{(\alpha <\epsilon )}\, \sum _{i,j=1}^n x_j^k x_i^l b_j^\epsilon a_i^\alpha b_i^\beta + \sum _{i,j=1}^n x_j^k x_i^l a_i^\alpha b_j^\epsilon \sum _{\mu =1}^{\beta -1} a_j^\mu (b_i^\beta -b_i^\mu ) \\&\qquad +\,\delta _{\alpha \epsilon } \sum _{i,j=1}^n x_j^k x_i^l b_i^\beta \left( Z_{ij} + \sum _{\lambda =1}^{\epsilon -1} a_i^\lambda b_j^\lambda \right) - \sum _{i,j=1}^n x_j^k x_i^l Z_{ji} a_i^\alpha b_j^\epsilon , \end{aligned} \end{aligned}$$
where we used again the condition \(\sum _{\gamma =1}^d a_j^\gamma =1\). It is not hard to see that replacing the terms in (A.13c) and (A.13d) by this last expression gives that (A.12)\(_\mathrm{{LHS}}\) and (A.12)\(_\mathrm{{RHS}}\) coincide. In the third case, we need to prove that
$$\begin{aligned} \xi ^*\{ g^k_{\gamma \epsilon },g^l_{\alpha \beta } \}=\sum _{i,j=1}^n\{ a_j^\gamma b_j^\epsilon x_j^k, a_i^\alpha b_i^\beta x_i^l \}. \end{aligned}$$
(A.14)
By antisymmetry in (2.10c), we can write
$$\begin{aligned} \begin{aligned} \{ a_j^\gamma ,b_i^\beta \}&=-\delta _{\beta \gamma }Z_{ji}+a_j^\gamma Z_{ji}- \frac{1}{2} \delta _{(j \ne i)}\frac{x_i+x_j}{x_i-x_j}b_i^\beta (a_i^\gamma -a_j^\gamma ) +\delta _{(\gamma <\beta )}a_j^\gamma b_i^\beta \\&+\,a_j^\gamma \sum _{\mu =1}^{\beta -1}a_j^\mu (b_i^\mu -b_i^\beta ) -\delta _{\beta \gamma } \sum _{\mu =1}^{\beta -1} a_j^\mu b_i^\mu -\frac{1}{2} \sum _{\sigma =1}^d o(\gamma ,\sigma )b_i^\beta (a_i^\sigma a_j^\gamma +a_j^\sigma a_i^\gamma ) , \end{aligned} \end{aligned}$$
so that the right-hand side yields
$$\begin{aligned} \begin{aligned}&(\mathrm{A.14})_\mathrm{{RHS}}\\&\quad =\,(k-l) \sum _{i=1}^n a_i^\gamma b_i^\epsilon a_i^\alpha b_i^\beta x_i^{k+l}\\&\qquad +\, \sum _{i,j=1}^nx_j^k x_i^l \left( \{ a_j^\gamma , a_i^\alpha \}b_j^\epsilon b_i^\beta + \{ a_j^\gamma , b_i^\beta \}b_j^\epsilon a_i^\alpha + \{ b_j^\epsilon ,a_i^\alpha \} a_j^\gamma b_i^\beta +\{ b_j^\epsilon , b_i^\beta \} a_j^\gamma a_i^\alpha \right) \\&\quad = \,(k-l) \sum _{i=1}^n a_i^\gamma b_i^\epsilon a_i^\alpha b_i^\beta x_i^{k+l} +\frac{1}{2} \sum _{\begin{array}{c} i,j=1\\ i\ne j \end{array}}^n x_j^k x_i^l \frac{x_j+x_i}{x_j-x_i} (b_i^\epsilon b_j^\beta a_j^\gamma a_i^\alpha +b_j^\epsilon b_i^\beta a_i^\gamma a_j^\alpha ) \\&\qquad +\frac{1}{2}\,o(\alpha ,\gamma ) \sum _{i,j=1}^n x_j^k x_i^l b_j^\epsilon b_i^\beta (a_j^\gamma a_i^\alpha +a_i^\gamma a_j^\alpha ) +\,\frac{1}{2} o(\epsilon ,\beta ) \sum _{i,j=1}^n x_j^k x_i^l a_i^\alpha a_j^\gamma (b_i^\epsilon b_j^\beta -b_j^\epsilon b_i^\beta )\\&\qquad +\,[\delta _{(\gamma<\beta )} -\delta _{(\alpha <\epsilon )}] \sum _{i,j=1}^n x_j^k x_i^l a_j^\gamma b_j^\epsilon b_i^\beta a_i^\alpha \\&\qquad +\,\delta _{\epsilon \alpha }\sum _{i,j=1}^n x_j^k x_i^l b_i^\beta a_j^\gamma \left( Z_{ij}+\sum _{\lambda =1}^{\epsilon -1} a_i^\lambda b_j^\lambda \right) \\&\qquad -\,\delta _{\beta \gamma }\sum _{i,j=1}^n x_j^k x_i^l b_j^\epsilon a_i^\alpha \left( Z_{ji}+\sum _{\mu =1}^{\beta -1} a_j^\mu b_i^\mu \right) . \end{aligned} \end{aligned}$$
This is obtained by simplifying terms without any non-obvious manipulation. Now, remark that we can write \(o(\epsilon ,\alpha )=\delta _{(\epsilon <\alpha )}-\delta _{(\epsilon>\alpha )}=1-\delta _{\epsilon \alpha }-2\delta _{(\epsilon >\alpha )}\) so that
$$\begin{aligned} \frac{1}{2} [o(\epsilon ,\alpha )+\delta _{\epsilon \alpha }- o(\beta ,\gamma )- \delta _{\beta \gamma }] =[\delta _{(\gamma<\beta )} -\delta _{(\alpha <\epsilon )}]. \end{aligned}$$
We can also repeat the argument in (A.11), to get
$$\begin{aligned} \begin{aligned}&\frac{1}{2} \left( \sum _{r=1}^k-\sum _{r=1}^l \right) \sum _{i,j=1}^n a_j^\alpha b_i^\beta a_i^\gamma b_j^\epsilon \left( x_i^r x_j^{k+l-r}+ x_i^{k+l-r} x_j^r \right) \\&\quad =(k-l) \sum _{i=1}^n a_i^\gamma b_i^\epsilon a_i^\alpha b_i^\beta x_i^{k+l} -\frac{1}{2} \sum _{\begin{array}{c} i,j=1\\ i\ne j \end{array}}^n x_i^{l} x_j^{k} \frac{x_i+x_j}{x_i-x_j} \left( a_j^\gamma b_j^\beta a_i^\alpha b_i^\epsilon +a_i^\gamma b_i^\beta a_j^\alpha b_j^\epsilon \right) . \end{aligned} \end{aligned}$$
Incorporating these two facts in (A.14)\(_\mathrm{{RHS}}\) gives us
$$\begin{aligned} \begin{aligned} (\mathrm{A.14})_\mathrm{{RHS}}&= \frac{1}{2} \left( \sum _{r=1}^k-\sum _{r=1}^l \right) \sum _{i,j=1}^n a_j^\alpha b_i^\beta a_i^\gamma b_j^\epsilon \left( x_i^r x_j^{k+l-r}+ x_i^{k+l-r} x_j^r \right) \\&\quad +\frac{1}{2} o(\alpha ,\gamma ) \sum _{i,j=1}^n x_j^k x_i^l b_j^\epsilon b_i^\beta (a_j^\gamma a_i^\alpha +a_i^\gamma a_j^\alpha )\\&\quad +\frac{1}{2} o(\epsilon ,\beta ) \sum _{i,j=1}^n x_j^k x_i^l a_i^\alpha a_j^\gamma (b_i^\epsilon b_j^\beta -b_j^\epsilon b_i^\beta )\\&\quad +\, \frac{1}{2} [o(\epsilon ,\alpha )+\delta _{\epsilon \alpha }- o(\beta ,\gamma )- \delta _{\beta \gamma }] \sum _{i,j=1}^n x_j^k x_i^l a_j^\gamma b_j^\epsilon b_i^\beta a_i^\alpha \\&\quad +\,\delta _{\epsilon \alpha }\sum _{i,j=1}^n x_j^k x_i^l b_i^\beta a_j^\gamma \left( Z_{ij}+\sum _{\lambda =1}^{\epsilon -1} a_i^\lambda b_j^\lambda \right) \\&\quad -\,\delta _{\beta \gamma }\sum _{i,j=1}^n x_j^k x_i^l b_j^\epsilon a_i^\alpha \left( Z_{ji}+\sum _{\mu =1}^{\beta -1} a_j^\mu b_i^\mu \right) . \end{aligned} \end{aligned}$$
This is nothing else than (A.8), which is (A.14)\(_\mathrm{{LHS}}\) as desired. \(\square \)
1.5 A.5. Proof of Proposition 4.3
We need the following lemma.
Lemma A.3
For any \(\epsilon ,\gamma =1,\ldots ,d\) and \(j,k,l=1,\ldots ,n\),
$$\begin{aligned} \{ b_j^\epsilon , f_{kl} \}&=(Z_{kj}b^\epsilon _l-Z_{jl}b_j^\epsilon ) +(Z_{lj}-Z_{kj}) f_{kl}+\frac{1}{2} \delta _{(j\ne k)}\frac{x_j+x_k}{x_j-x_k}b_j^\epsilon ( f_{jl}- f_{kl}) \nonumber \\&\quad +\frac{1}{2} \delta _{(j\ne l)}\frac{x_j+x_l}{x_j-x_l}(b_j^\epsilon f_{kl}+b_l^\epsilon f_{kj}) +\frac{1}{2} b_l^\epsilon f_{kj}-\frac{1}{2} b_j^\epsilon f_{jl} \nonumber \\&\quad +\, f_{kl} \sum _{\lambda =1}^{\epsilon -1}(b_j^\lambda -b_j^\epsilon ) (a_l^\lambda -a_k^\lambda ) \end{aligned}$$
(A.15a)
$$\begin{aligned} \{ a_i^\gamma , f_{kl} \}&= a_i^\gamma Z_{il}-a_k^\gamma Z_{il} +\frac{1}{2} \delta _{(i\ne k)}\frac{x_i+x_k}{x_i-x_k}(a_k^\gamma -a_i^\gamma )( f_{il}- f_{kl}) \nonumber \\&\quad +\frac{1}{2} \delta _{(i\ne l)}\frac{x_i+x_l}{x_i-x_l} f_{kl}(a_{l}^\gamma -a_i^\gamma ) +\frac{1}{2} a_i^\gamma f_{il}-\frac{1}{2} a_k^\gamma f_{il} \nonumber \\&\quad +\frac{1}{2} \sum _{\sigma =1}^d o(\gamma ,\sigma ) f_{kl}[a_i^\gamma (a_k^\sigma -a_l^\sigma ) +a_i^\sigma (a_k^\gamma -a_l^\gamma )]. \end{aligned}$$
(A.15b)
Proof
As usual, we use the normalisation \(\sum _\alpha a_k^\alpha =1\), and we compute from (2.10c) and (2.10d) that
$$\begin{aligned} \begin{aligned} \{ b_j^\epsilon , f_{kl} \}&=\sum _{\alpha =1}^d \left( \{ b_j^\epsilon ,a^\alpha _k \}b^\alpha _l + a^\alpha _k \{ b_j^\epsilon ,b^\alpha _l \} \right) \\&=b^\epsilon _l Z_{kj} - f_{kl} Z_{kj} +\frac{1}{2} \delta _{(j\ne k)}\frac{x_j+x_k}{x_j-x_k} b_j^\epsilon ( f_{jl}- f_{kl}) - \sum _{\alpha =1}^{\epsilon -1} b^\alpha _l a_k^\alpha b_j^\epsilon \\&\quad -\, f_{kl} \sum _{\lambda =1}^{\epsilon -1}a_k^\lambda (b_j^\lambda -b_j^\epsilon ) + b^\epsilon _l \sum _{\lambda =1}^{\epsilon -1} a_k^\lambda b_j^\lambda +\frac{1}{2} \sum _{\alpha =1}^d \sum _{\kappa =1}^d o(\alpha ,\kappa ) b^\alpha _l b_j^\epsilon (a_j^\kappa a_k^\alpha +a_k^\kappa a_j^\alpha ) \\&\quad +\frac{1}{2} \delta _{(j\ne l)}\frac{x_j+x_l}{x_j-x_l} (b_j^\epsilon f_{kl} + b_l^\epsilon f_{kj}) + f_{kl} Z_{lj} - b_j^\epsilon Z_{jl} \\&\quad +\frac{1}{2} \sum _{\alpha =1}^d o(\epsilon ,\alpha ) a^\alpha _k(b_l^\epsilon b_j^\alpha -b_j^\epsilon b_l^\alpha ) + f_{kl} \sum _{\lambda =1}^{\epsilon -1} a_l^\lambda (b_j^\lambda -b_j^\epsilon ) \\&\quad -\sum _{\alpha =1}^d \sum _{\mu =1}^{\alpha -1} a^\alpha _k b_j^\epsilon a_j^\mu (b_l^\mu -b_l^\alpha ). \end{aligned} \end{aligned}$$
(A.16)
Our aim is to reduce some of these thirteen terms, mostly using properties of the ordering function \(o(-,-)\). Summing the fourth, sixth and eleventh terms of (A.16) together yields
$$\begin{aligned} \begin{aligned}&- \sum _{\lambda =1}^{\epsilon -1} a_k^\lambda b^\lambda _l b_j^\epsilon + \sum _{\lambda =1}^{\epsilon -1} a_k^\lambda b_j^\lambda b^\epsilon _l +\frac{1}{2} \left[ \sum _{\lambda =\epsilon +1}^d - \sum _{\lambda =1}^{\epsilon -1} \right] (a^\lambda _kb_j^\lambda b_l^\epsilon - a^\lambda _k b_l^\lambda b_j^\epsilon ) \\&\quad =\frac{1}{2} \sum _{\begin{array}{c} \lambda =1\\ \lambda \ne \epsilon \end{array}}^d (a^\lambda _kb_j^\lambda b_l^\epsilon - a^\lambda _k b_l^\lambda b_j^\epsilon ) =\frac{1}{2} ( f_{kj}b_l^\epsilon - f_{kl} b_j^\epsilon ). \end{aligned} \end{aligned}$$
The fifth and twelfth terms of (A.16) give
$$\begin{aligned} - f_{kl} \sum _{\lambda =1}^{\epsilon -1}a_k^\lambda (b_j^\lambda -b_j^\epsilon ) + f_{kl} \sum _{\lambda =1}^{\epsilon -1} a_l^\lambda (b_j^\lambda -b_j^\epsilon ) = f_{kl} \sum _{\lambda =1}^{\epsilon -1} (a_l^\lambda -a_k^\lambda ) (b_j^\lambda -b_j^\epsilon ). \end{aligned}$$
Relabelling indices, we transform the seventh terms from (A.16) as
$$\begin{aligned} \begin{aligned}&\frac{1}{2} \left[ \sum _{\alpha =1}^d \sum _{\kappa =\alpha +1}^d - \sum _{\alpha =1}^d \sum _{\kappa =1}^{\alpha -1} \right] b^\alpha _l b_j^\epsilon (a_j^\kappa a_k^\alpha +a_k^\kappa a_j^\alpha )\\&\quad =\frac{1}{2} \sum _{\alpha =1}^d \sum _{\mu =1}^{\alpha -1} b_j^\epsilon (b^\mu _l a_j^\alpha a_k^\mu +b^\mu _l a_k^\alpha a_j^\mu -b^\alpha _l a_j^\mu a_k^\alpha -b^\alpha _l a_k^\mu a_j^\alpha ) \\&\quad =\frac{1}{2} \sum _{\alpha =1}^d \sum _{\mu =1}^{\alpha -1} b_j^\epsilon (b^\mu _l-b^\alpha _l) ( a_j^\alpha a_k^\mu + a_k^\alpha a_j^\mu ), \end{aligned} \end{aligned}$$
which can be summed with the thirteen terms in (A.16) to yield
$$\begin{aligned} \begin{aligned}&\frac{1}{2} \sum _{\alpha =1}^d \sum _{\mu =1}^{\alpha -1} b_j^\epsilon (b_l^\alpha -b_l^\mu ) \left( 2 a^\alpha _k a_j^\mu -a_j^\alpha a_k^\mu - a_k^\alpha a_j^\mu \right) =\frac{1}{2} \sum _{\alpha =1}^d \sum _{\mu =1}^{\alpha -1} b_j^\epsilon (b_l^\alpha -b_l^\mu ) \left( a^\alpha _k a_j^\mu -a_j^\alpha a_k^\mu \right) \\&\quad =\frac{1}{2} \sum _{\alpha =1}^d \sum _{\mu =1}^{\alpha -1} b_j^\epsilon (b_l^\alpha -b_l^\mu ) a^\alpha _k a_j^\mu -\frac{1}{2} \sum _{\alpha =1}^d \sum _{\mu =\alpha +1}^{d} b_j^\epsilon (b_l^\mu -b_l^\alpha ) a_j^\mu a_k^\alpha \\&\quad =\frac{1}{2} \sum _{\alpha =1}^d \sum _{\begin{array}{c} \mu =1\\ \mu \ne \alpha \end{array}}^{d} b_j^\epsilon (b_l^\alpha -b_l^\mu ) a^\alpha _k a_j^\mu =\frac{1}{2} \sum _{\alpha =1}^d \sum _{\mu =1}^{d} b_j^\epsilon (a^\alpha _k b_l^\alpha a_j^\mu -a_j^\mu b_l^\mu a^\alpha _k) = \frac{1}{2} b_j^\epsilon \left( f_{kl} - f_{jl} \right) . \end{aligned} \end{aligned}$$
Introducing the different terms back in (A.16), we find
$$\begin{aligned} \begin{aligned} \{ b_j^\epsilon , f_{kl} \}&= (b^\epsilon _l Z_{kj} - b_j^\epsilon Z_{jl}) + (Z_{lj} - Z_{kj}) f_{kl} +\frac{1}{2} \delta _{(j\ne k)}\frac{x_j+x_k}{x_j-x_k} b_j^\epsilon ( f_{jl}- f_{kl}) \\&\quad +\frac{1}{2} \delta _{(j\ne l)}\frac{x_j+x_l}{x_j-x_l} (b_j^\epsilon f_{kl} + b_l^\epsilon f_{kj}) +\frac{1}{2} (b_l^\epsilon f_{kj} - b_j^\epsilon f_{jl} ) \\&\quad +\, f_{kl} \sum _{\lambda =1}^{\epsilon -1}(b_j^\lambda -b_j^\epsilon ) (a_l^\lambda -a_k^\lambda ) , \end{aligned} \end{aligned}$$
as desired. For the second identity, we need (2.10b) and (A.15a), then the same kind of manipulations allow to find \(\{ a_i^\gamma , f_{kl} \}\). \(\square \)
To establish Proposition 4.3, we have from Lemma A.3 and the identity \(\sum _{\gamma =1}^d a_i^\gamma =1\) that
$$\begin{aligned} \{ f_{ij},f_{kl} \}&= \sum _{\gamma =1}^d\left( \{ a_i^\gamma , f_{kl} \}b_j^\gamma +a_i^\gamma \{ b_j^\gamma , f_{kl} \} \right) \nonumber \\&=(f_{ij}- f_{kj}) Z_{il} + \frac{1}{2} \delta _{(i\ne k)}\frac{x_i+x_k}{x_i-x_k} ( f_{kj}- f_{ij}) ( f_{il}- f_{kl}) \nonumber \\&\quad +\frac{1}{2} \delta _{(i \ne l)}\frac{x_i+x_l}{x_i-x_l} ( f_{lj}- f_{ij}) f_{kl} +\frac{1}{2} f_{ij} f_{il} - \frac{1}{2} f_{kj} f_{il} \nonumber \\&\quad +\frac{1}{2} \sum _{\gamma =1}^d \sum _{\sigma =1}^d o(\gamma ,\sigma ) f_{kl} \big (a_i^\gamma b_j^\gamma (a_k^\sigma -a_l^\sigma )+a_i^\sigma (a_k^\gamma - a_l^\gamma ) b_j^\gamma \big ) \nonumber \\&\quad +\, Z_{kj} f_{il}-Z_{jl} f_{ij} +(Z_{lj}-Z_{kj}) f_{kl} \nonumber \\&\quad +\frac{1}{2} \delta _{(j\ne k)}\frac{x_j+x_k}{x_j-x_k} f_{ij} ( f_{jl}- f_{kl}) +\frac{1}{2} \delta _{(j\ne l)}\frac{x_j+x_l}{x_j-x_l}( f_{ij} f_{kl}+ f_{il} f_{kj})\nonumber \\&\quad +\frac{1}{2} f_{il} f_{kj} -\frac{1}{2} f_{ij} f_{jl} + f_{kl} \sum _{\gamma =1}^d \sum _{\lambda =1}^{\gamma -1}a_i^\gamma (b_j^\lambda -b_j^\gamma )(a_l^\lambda -a_k^\lambda ). \end{aligned}$$
(A.17)
The sums in the third line of (A.17) can be re-expressed as follows :
$$\begin{aligned} \begin{aligned}&\frac{1}{2} f_{kl}\, \sum _{\gamma =1}^d \left[ \sum _{\sigma =\gamma +1}^d- \sum _{\sigma =1}^{\gamma -1}\right] \big (a_i^\gamma b_j^\gamma (a_k^\sigma -a_l^\sigma )+a_i^\sigma (a_k^\gamma - a_l^\gamma ) b_j^\gamma \big ) \\&\quad = \frac{1}{2} f_{kl}\, \sum _{\gamma =1}^d \left[ \sum _{\sigma =\gamma +1}^d- \sum _{\sigma =1}^{\gamma -1}\right] \big (a_i^\gamma b_j^\gamma (a_k^\sigma -a_l^\sigma ) - a_i^\gamma (a_k^\sigma - a_l^\sigma ) b_j^\sigma \big ), \end{aligned} \end{aligned}$$
after swapping the labels \(\sigma \leftrightarrow \gamma \) for the second term in the sums. This is nothing else that
$$\begin{aligned} \frac{1}{2} f_{kl}\, \sum _{\gamma =1}^d \left[ \sum _{\sigma =\gamma +1}^d- \sum _{\sigma =1}^{\gamma -1}\right] a_i^\gamma ( b_j^\gamma - b_j^\sigma ) (a_k^\sigma -a_l^\sigma ). \end{aligned}$$
Summing with the last term of (A.17), we get
$$\begin{aligned} \begin{aligned}&\frac{1}{2} f_{kl}\, \sum _{\gamma =1}^d \left[ \sum _{\lambda =\gamma +1}^d- \sum _{\lambda =1}^{\gamma -1}\right] a_i^\gamma (b_j^\gamma - b_j^\lambda ) (a_k^\lambda -a_l^\lambda ) + f_{kl} \sum _{\gamma =1}^d \sum _{\lambda =1}^{\gamma -1}a_i^\gamma (b_j^\lambda -b_j^\gamma )(a_l^\lambda -a_k^\lambda )\\&\quad =\frac{1}{2} f_{kl}\, \sum _{\gamma =1}^d \sum _{\begin{array}{c} \lambda =1\\ \lambda \ne \gamma \end{array}}^d a_i^\gamma (b_j^\gamma - b_j^\lambda ) (a_k^\lambda -a_l^\lambda ) =\frac{1}{2} f_{kl}\, \sum _{\gamma =1}^d \sum _{\lambda =1}^d a_i^\gamma \left( b_j^\gamma a_k^\lambda - b_j^\gamma a_l^\lambda - a_k^\lambda b_j^\lambda +a_l^\lambda b_j^\lambda \right) \\&\quad =\frac{1}{2} f_{kl} \big ( f_{ij} - f_{ij} - f_{kj} + f_{lj} \big ) = \frac{1}{2} f_{kl} f_{lj} - \frac{1}{2} f_{kl} f_{kj}. \end{aligned} \end{aligned}$$
We can thus rewrite (A.17) as
$$\begin{aligned} \begin{aligned} \{ f_{ij},f_{kl} \}&=f_{ij} Z_{il} - f_{kj} Z_{il} + \frac{1}{2} \delta _{(i\ne k)}\frac{x_i+x_k}{x_i-x_k} (f_{kj}f_{il}- f_{kj}f_{kl} -f_{ij} f_{il} +f_{ij} f_{kl}) \\&\quad +\frac{1}{2} \delta _{(i \ne l)}\frac{x_i+x_l}{x_i-x_l} (f_{lj}f_{kl}-f_{ij}f_{kl}) +\frac{1}{2} f_{ij} f_{il} - \frac{1}{2} f_{kj} f_{il} +\frac{1}{2} f_{kl}f_{lj} - \frac{1}{2} f_{kl} f_{kj} \\&\quad +\, Z_{kj} f_{il}-Z_{jl} f_{ij}+ Z_{lj}f_{kl} - Z_{kj}f_{kl} +\frac{1}{2} \delta _{(j\ne k)}\frac{x_j+x_k}{x_j-x_k} (f_{ij} f_{jl} - f_{ij} f_{kl}) \\&\quad +\frac{1}{2} \delta _{(j\ne l)}\frac{x_j+x_l}{x_j-x_l}(f_{ij} f_{kl}+f_{il}f_{kj}) +\frac{1}{2} f_{il} f_{kj}-\frac{1}{2} f_{ij} f_{jl} . \end{aligned} \end{aligned}$$
Now, remark that we can rearrange terms to obtain
$$\begin{aligned} \begin{aligned} \{ f_{ij},f_{kl} \}&= f_{ij} \left( Z_{il}+\frac{1}{2} f_{il}\right) - f_{kj} \left( Z_{il} +\frac{1}{2} f_{il} \right) + f_{il} \left( Z_{kj} + \frac{1}{2} f_{kj} \right) - f_{ij} \left( Z_{jl} + \frac{1}{2} f_{jl} \right) \\&\quad +\, f_{kl} \left( Z_{lj} + \frac{1}{2} f_{lj}\right) - f_{kl}\left( Z_{kj} + \frac{1}{2} f_{kj} \right) +\frac{1}{2} \delta _{(i \ne l)}\frac{x_i+x_l}{x_i-x_l}(f_{lj}f_{kl}-f_{ij}f_{kl}) \\&\quad + \frac{1}{2} \delta _{(i\ne k)}\frac{x_i+x_k}{x_i-x_k} (f_{kj}f_{il}- f_{kj}f_{kl} -f_{ij} f_{il} +f_{ij} f_{kl}) \\&\quad +\frac{1}{2} \delta _{(j\ne k)}\frac{x_j+x_k}{x_j-x_k} (f_{ij} f_{jl} - f_{ij} f_{kl}) +\frac{1}{2} \delta _{(j\ne l)}\frac{x_j+x_l}{x_j-x_l}(f_{ij} f_{kl}+f_{il}f_{kj}). \end{aligned} \end{aligned}$$
After that, a simple rearrangement using
$$\begin{aligned} Z_{ij}+\frac{1}{2} f_{ij}=\frac{1}{2} \frac{x_i+qx_j}{x_i-q x_j} f_{ij} \end{aligned}$$
(A.18)
leads to (2.4b). This proves the first claim of Proposition 4.3. To prove the second claim, we use Lemma A.3 to get
$$\begin{aligned} \begin{aligned} \{ a_i^\gamma ,f_{kk} \}&= \frac{1}{2} \delta _{(i\ne k)}\frac{x_i+x_k}{x_i-x_k}(a_k^\gamma -a_i^\gamma )f_{ik} + (a_i^\gamma -a_k^\gamma ) \left( Z_{ik}+\frac{1}{2}f_{ik}\right) , \\ \{ b_j^\epsilon ,f_{kk} \}&=\frac{1}{2}\delta _{(j\ne k)}\frac{x_j+x_k}{x_j-x_k}(b_j^\epsilon f_{jk}+b_k^\epsilon f_{kj}) + b_k^\epsilon \left( Z_{kj} + \frac{1}{2} f_{kj}\right) - b_j^\epsilon \left( Z_{jk} + \frac{1}{2} f_{jk}\right) . \end{aligned} \end{aligned}$$
Using (A.18) we get, after summation over k, the relations (4.10b) and (4.10c). The remaining relation (4.10a) is obvious. \(\square \)
1.6 A.6. Proof of Lemma 5.1
Let \(u\in \{y,z\}\), and remark that we can write \( \left\{ \left\{ u,u\right\} \right\} =-\frac{1}{2} [u^2 \otimes e_0 - e_0 \otimes u^2]\), together with
$$\begin{aligned} \left\{ \left\{ u, w_\alpha \right\} \right\} = \frac{1}{2} e_{0}\otimes uw_\alpha -\frac{1}{2} u\otimes w_\alpha ,\quad \left\{ \left\{ u, v_\alpha \right\} \right\} = \frac{1}{2} v_\alpha u\otimes e_0-\frac{1}{2} v_\alpha \otimes u. \end{aligned}$$
(A.19)
Now, consider the elements \(w_\alpha v_\beta u^l\in \mathcal {A}\) for any \(l\in \mathbb {N}\) and \(\alpha ,\beta =1,\ldots ,d\). The following statement holds in \(\mathcal {A}/[\mathcal {A},\mathcal {A}]\) if \(u=y\), and \(\mathcal {A}^{\times }/[\mathcal {A}^{\times },\mathcal {A}^{\times }]\) if \(u=z\).
Lemma A.4
For any \(k,l\ge 1\) and \(\alpha ,\beta ,\gamma ,\epsilon =1,\ldots ,d\), we have that
$$\begin{aligned} \{ u^k,w_\alpha v_\beta u^l \}&=0, \nonumber \\ \{ w_\gamma v_\epsilon u^k,w_\alpha v_\beta u^l \}&=\frac{1}{2} \left[ o(\gamma ,\beta )+o(\epsilon ,\alpha )-o(\epsilon ,\beta ) -o(\gamma ,\alpha )\right] w_\alpha v_\epsilon u^k\,w_\gamma v_\beta u^l \nonumber \\&\quad +\frac{1}{2}o(\gamma ,\beta )\,\, w_\alpha v_\beta \, w_\gamma v_\epsilon u^{k+l} +\frac{1}{2}o(\epsilon ,\alpha )\,\, w_\alpha v_\beta u^{k+l}\,w_\gamma v_\epsilon \nonumber \\&\quad -\frac{1}{2}o(\epsilon ,\beta )\,\, w_\alpha v_\beta u^{k}\,w_\gamma v_\epsilon u^l -\frac{1}{2}o(\gamma ,\alpha )\,\, w_\alpha v_\beta u^{l}\,w_\gamma v_\epsilon u^{k} \nonumber \\&\quad -\,\delta _{\gamma \beta }\left[ w_\alpha v_\epsilon u^{k+l} +\frac{1}{2} w_\alpha v_\beta \,w_\gamma v_\epsilon u^{k+l} +\frac{1}{2}w_\alpha v_\epsilon u^k\,w_\gamma v_\beta u^l \right] \nonumber \\&\quad +\,\delta _{\alpha \epsilon }\left[ w_\gamma v_\beta u^{k+l} +\frac{1}{2} w_\alpha v_\beta u^{k+l}\,w_\gamma v_\epsilon +\frac{1}{2}w_\alpha v_\epsilon u^k\,w_\gamma v_\beta u^l \right] \nonumber \\&\quad -\frac{1}{2} \,\left[ \sum _{\tau =1}^{k-1} w_\alpha v_\beta u^{k+l-\tau } w_\gamma v_\epsilon u^{\tau } + \sum _{\sigma =1}^l w_\alpha v_\beta u^{k+\sigma } w_\gamma v_\epsilon u^{l-\sigma } \right] \nonumber \\&\quad +\frac{1}{2} \,\left[ \sum _{\sigma =1}^{l-1} w_\alpha v_\beta u^{\sigma } w_\gamma v_\epsilon u^{k+l-\sigma } + \sum _{\tau =1}^k w_\alpha v_\beta u^{k-\tau } w_\gamma v_\epsilon u^{l+\tau }\right] . \end{aligned}$$
(A.20)
The same holds without the sums if \(k=0\), \(l=0\) or both \(k=l=0\).
We delay the proof of this lemma until “Appendix A.6.1” to explain how we can conclude from this result. Denote by U the matrix representing \(u\in \{z,y\}\). Then, \(t_{\alpha \epsilon }^l={\text {tr}}( W_\alpha V_\epsilon U^l)\), and Lemma 5.1 is deduced from Lemma A.4 and (3.23b) by using that \({\text {tr}}(W_\alpha V_\beta U^k W_\gamma V_\epsilon U^l)=(V_\epsilon U^l W_\alpha ) ( V_\beta U^k W_\gamma )=t_{\alpha \epsilon }^l t_{\gamma \beta }^k\). \(\square \)
Remark 5.8
A similar result also holds for \(u\in \{x,e_0+xy\}\), and with \(u=x+y^{-1}\) if we decide to localise at y. We have in those cases \( \left\{ \left\{ u,u\right\} \right\} =+\frac{1}{2} [u^2 \otimes e_0 - e_0 \otimes u^2]\), and (A.19) also holds. Then, Lemma A.4 can also be proved, except that we need to change the signs in front of the last two lines in (A.20). We do not discuss these cases any further.
1.6.1 Proof of Lemma A.4
First, we note from the discussion at the beginning of “Appendix A.1” that
$$\begin{aligned} \left\{ \left\{ u^k, u^l\right\} \right\} =-\frac{1}{2}\sum _{\tau =1}^k\sum _{\sigma =1}^l \left( u^{k-\tau +\sigma +1} \otimes u^{l-\sigma +\tau -1} - u^{k-\tau +\sigma -1} \otimes u^{l-\sigma +\tau +1}\right) . \end{aligned}$$
Next, using (A.19), we remark that
$$\begin{aligned} \begin{aligned} \left\{ \left\{ u, w_\alpha v_\beta \right\} \right\}&= w_\alpha \left\{ \left\{ u, v_\beta \right\} \right\} + \left\{ \left\{ u, w_\alpha \right\} \right\} v_\beta \\&= \frac{1}{2} \left( w_\alpha v_\beta u\otimes e_0 - w_\alpha v_\beta \otimes u +e_{0}\otimes uw_\alpha v_\beta - u\otimes w_\alpha v_\beta \right) . \end{aligned} \end{aligned}$$
(A.21)
Therefore
$$\begin{aligned} \begin{aligned} \left\{ \left\{ u^k,w_\alpha v_\beta u^l\right\} \right\}&=\frac{1}{2} \sum _{\tau =1}^k \Big (w_\alpha v_\beta u^{k-\tau +1} \otimes u^{l+\tau -1} - w_\alpha v_\beta u^{k-\tau }\otimes u^{l+\tau } \\&\quad +\,u^{k-\tau }\otimes u^{\tau }w_\alpha v_\beta u^{l} - u^{k-\tau +1}\otimes u^{\tau -1}w_\alpha v_\beta u^{l} \Big )\\&-\frac{1}{2} \sum _{\tau =1}^k \sum _{\sigma =1}^l \left( w_\alpha v_\beta u^{k-\tau +\sigma +1} \otimes u^{l-\sigma +\tau -1} -w_\alpha v_\beta u^{k-\tau +\sigma -1} \otimes u^{l-\sigma +\tau +1} \right) . \end{aligned} \end{aligned}$$
(A.22)
After application of the multiplication map, we get 0 and the first equality follows. To prove that (A.20) holds, write
$$\begin{aligned}&\left\{ \left\{ w_\gamma v_\epsilon u^k, w_\alpha v_\beta u^l\right\} \right\} = w_\gamma v_\epsilon * \left\{ \left\{ u^k,w_\alpha v_\beta u^l\right\} \right\} \nonumber \\&+ w_\alpha v_\beta \left\{ \left\{ w_\gamma v_\epsilon , u^l\right\} \right\} *u^k + \left\{ \left\{ w_\gamma v_\epsilon ,w_\alpha v_\beta \right\} \right\} u^l *u^k . \end{aligned}$$
(A.23)
Again, we begin by reducing the two first terms. Using (A.21) and (A.22) gives
$$\begin{aligned} \begin{aligned} T&:= w_\gamma v_\epsilon * \left\{ \left\{ u^k,w_\alpha v_\beta u^l\right\} \right\} - \sum _{\sigma =1}^l w_\alpha v_\beta u^{\sigma -1} \left\{ \left\{ u,w_\gamma v_\epsilon \right\} \right\} ^\circ u^{l-\sigma } *u^k \\&=- \frac{1}{2} \sum _{\tau =1}^k \sum _{\sigma =1}^l \left( w_\alpha v_\beta u^{k-\tau +\sigma +1} \otimes w_\gamma v_\epsilon u^{l+\tau -\sigma -1} -w_\alpha v_\beta u^{k-\tau +\sigma -1} \otimes w_\gamma v_\epsilon u^{l+\tau -\sigma +1} \right) \\&\quad + \frac{1}{2} \sum _{\tau =1}^k \left( w_\alpha v_\beta u^{k-\tau +1} \otimes w_\gamma v_\epsilon u^{l+\tau -1} - w_\alpha v_\beta u^{k-\tau }\otimes w_\gamma v_\epsilon u^{l+\tau }\right) \\&\quad + \frac{1}{2} \sum _{\tau =1}^k \left( u^{k-\tau }\otimes w_\gamma v_\epsilon u^{\tau }w_\alpha v_\beta u^{l} - u^{k-\tau +1}\otimes w_\gamma v_\epsilon u^{\tau -1}w_\alpha v_\beta u^{l} \right) \\&\quad -\frac{1}{2} \sum _{\sigma =1}^l \left( w_\alpha v_\beta u^\sigma w_\gamma v_\epsilon u^k \otimes u^{l-\sigma } - w_\alpha v_\beta u^{\sigma -1} w_\gamma v_\epsilon u^k \otimes u^{l-\sigma +1} \right) \\&\quad -\frac{1}{2} \sum _{\sigma =1}^l \left( w_\alpha v_\beta u^{k+\sigma -1} \otimes w_\gamma v_\epsilon u^{l-\sigma +1} - w_\alpha v_\beta u^{k+\sigma } \otimes w_\gamma v_\epsilon u^{l-\sigma } \right) . \end{aligned} \end{aligned}$$
This gives, after multiplication and modulo commutators
$$\begin{aligned} \begin{aligned}&m \circ T= -\frac{1}{2} \sum _{\tau =1}^k \sum _{\sigma =1}^l \left( w_\alpha v_\beta u^{k-\tau +\sigma +1} w_\gamma v_\epsilon u^{l+\tau -\sigma -1} -w_\alpha v_\beta u^{k-\tau +\sigma -1} w_\gamma v_\epsilon u^{l+\tau -\sigma +1} \right) , \end{aligned} \end{aligned}$$
because the last four sums cancel out. Relabelling indices, we write
$$\begin{aligned} \begin{aligned} m \circ T&=- \frac{1}{2} \left[ \sum _{\tau =1}^{k-1} \sum _{\sigma =l}+\sum _{\tau =0} \sum _{\sigma =1}^l - \sum _{\tau =k} \sum _{\sigma =1}^{l-1} - \sum _{\tau =1}^{k} \sum _{\sigma =0} \right] w_\alpha v_\beta u^{k-\tau +\sigma } w_\gamma v_\epsilon u^{l+\tau -\sigma } \\&=-\frac{1}{2} \,\left[ \sum _{\tau =1}^{k-1} w_\alpha v_\beta u^{k+l-\tau } w_\gamma v_\epsilon u^{\tau } + \sum _{\sigma =1}^l w_\alpha v_\beta u^{k+\sigma } w_\gamma v_\epsilon u^{l-\sigma } \right] \\&\quad +\frac{1}{2} \,\left[ \sum _{\sigma =1}^{l-1} w_\alpha v_\beta u^{\sigma } w_\gamma v_\epsilon u^{k+l-\sigma } + \sum _{\tau =1}^k w_\alpha v_\beta u^{k-\tau } w_\gamma v_\epsilon u^{l+\tau }\right] . \end{aligned} \end{aligned}$$
It remains to compute \( \left\{ \left\{ w_\gamma v_\epsilon ,w_\alpha v_\beta \right\} \right\} \). We can find from (3.2e)–(3.2g)
$$\begin{aligned} \left\{ \left\{ w_\gamma v_\epsilon ,w_\alpha v_\beta \right\} \right\}= & {} -\frac{1}{2} o(\gamma ,\alpha ) \left( w_\alpha v_\epsilon \otimes w_\gamma v_\beta + w_\gamma v_\epsilon \otimes w_\alpha v_\beta \right) \\&-\frac{1}{2} o(\beta ,\gamma ) \left( w_\alpha v_\beta w_\gamma v_\epsilon \otimes e_0 + w_\alpha v_\epsilon \otimes w_\gamma v_\beta \right) \\&-\,\delta _{\beta \gamma } \left( w_\alpha v_\epsilon \otimes e_0 + \frac{1}{2} w_\alpha v_\epsilon \otimes w_\gamma v_\beta + \frac{1}{2} w_\alpha v_\beta w_\gamma v_\epsilon \otimes e_0 \right) \\&+\,\delta _{\alpha \epsilon } \left( e_0 \otimes w_\gamma v_\beta + \frac{1}{2} w_\alpha v_\epsilon \otimes w_\gamma v_\beta + \frac{1}{2} e_0 \otimes w_\gamma v_\epsilon w_\alpha v_\beta \right) \\&+\frac{1}{2} o(\epsilon ,\alpha ) \left( e_0 \otimes w_\gamma v_\epsilon w_\alpha v_\beta + w_\alpha v_\epsilon \otimes w_\gamma v_\beta \right) \\&- \frac{1}{2} o(\epsilon ,\beta ) \left( w_\alpha v_\beta \otimes w_\gamma v_\epsilon + w_\alpha v_\epsilon \otimes w_\gamma v_\beta \right) . \end{aligned}$$
By applying the multiplication map m on \( \left\{ \left\{ w_\gamma v_\epsilon ,w_\alpha v_\beta \right\} \right\} u^l *u^k \), we get
$$\begin{aligned} \begin{aligned} m\circ ( \left\{ \left\{ w_\gamma v_\epsilon ,w_\alpha v_\beta \right\} \right\} u^l *u^k )=&-\frac{1}{2} o(\gamma ,\alpha ) \left( w_\alpha v_\epsilon u^k w_\gamma v_\beta u^l + w_\gamma v_\epsilon u^k w_\alpha v_\beta u^l \right) \\&\quad -\frac{1}{2} o(\beta ,\gamma ) \left( w_\alpha v_\beta w_\gamma v_\epsilon u^k u^l + w_\alpha v_\epsilon u^k w_\gamma v_\beta u^l \right) \\&\quad -\,\delta _{\beta \gamma } \left( w_\alpha v_\epsilon u^k u^l + \frac{1}{2} w_\alpha v_\epsilon u^k w_\gamma v_\beta u^l + \frac{1}{2} w_\alpha v_\beta w_\gamma v_\epsilon u^k u^l \right) \\&\quad +\,\delta _{\alpha \epsilon } \left( u^k w_\gamma v_\beta u^l + \frac{1}{2} w_\alpha v_\epsilon u^k w_\gamma v_\beta u^l + \frac{1}{2} u^k w_\gamma v_\epsilon w_\alpha v_\beta u^l \right) \\&\quad +\frac{1}{2} o(\epsilon ,\alpha ) \left( u^k w_\gamma v_\epsilon w_\alpha v_\beta u^l + w_\alpha v_\epsilon u^k w_\gamma v_\beta u^l \right) \\&\quad - \frac{1}{2} o(\epsilon ,\beta ) \left( w_\alpha v_\beta u^k w_\gamma v_\epsilon u^l + w_\alpha v_\epsilon u^k w_\gamma v_\beta u^l \right) . \end{aligned} \end{aligned}$$
Adding \(m \circ T\) to this last expression finishes the proof. \(\square \)
1.7 A.7. Computations with S
To simplify notations in this appendix, we denote by s the element \(s_d\) given by (3.9) with \(\alpha =d\).
1.7.1 A.7.1. Some brackets
Recall that (5.7) holds if we can show that the relation \(\{{\text {tr}}S^k, t_{\alpha \beta }^i\}=0\) on \({\text {Rep}}(\mathcal {A}^{\times }, \bar{\alpha })\). Moreover, we have that \(t_{\alpha \beta }^i\) and \({\text {tr}}S^k\) are traces of the matrices representing the elements \(w_\alpha v_\beta z^i\) and \(s^k\). Using (3.23b), the desired relation follows from the following lemma.
Lemma A.5
For any \(k,l\in \mathbb {N}\) and \(1\le \alpha ,\beta \le d\), \(\{ s^k,w_\alpha v_\beta z^i \}=0\) in \(\mathcal {A}^{\times }/[\mathcal {A}^{\times },\mathcal {A}^{\times }]\).
Proof
We use (3.10a) and (3.10c) to get
$$\begin{aligned} \begin{aligned} \left\{ \left\{ s,w_\alpha v_\beta z^i\right\} \right\}&= \left\{ \left\{ s,w_\alpha \right\} \right\} v_\beta z^i + w_\alpha \left\{ \left\{ s,v_\beta \right\} \right\} z^i +\sum _{\tau =1}^i w_\alpha v_\beta z^{\tau -1} \left\{ \left\{ s,z\right\} \right\} z^{i-\tau } \\&=\frac{1}{2} (s \otimes w_\alpha v_\beta z^i + e_0 \otimes s w_\alpha v_\beta z^i - w_\alpha v_\beta z^i s \otimes e_0 - w_\alpha v_\beta z^i \otimes s). \end{aligned} \end{aligned}$$
Hence,
$$\begin{aligned} \{ s^k,w_\alpha v_\beta z^i \} = m \circ \left( \sum _{\tau =1}^k s^{\tau -1} * \left\{ \left\{ s,w_\alpha v_\beta z^i\right\} \right\} *s^{k-\tau } \right) = l\, (s^k w_\alpha v_\beta z^i - w_\alpha v_\beta z^i s^k), \end{aligned}$$
which vanishes modulo commutators. \(\square \)
1.7.2 A.7.2. Proof of Theorem 5.10
Consider the following result.
Lemma A.6
Let \(z_{\eta }=z+\eta s\) for arbitrary \(\eta \in \mathbb {C}\) playing the role of a spectral parameter. Then,
$$\begin{aligned} \{ z_{\mu }^k,z_{\eta }^l \}=0 \quad \text { mod }[\mathcal {A}^{\times },\mathcal {A}^{\times }], \qquad \text { for any }\mu ,\eta \in \mathbb {C}, k,l \in \mathbb {N}^\times . \end{aligned}$$
Using this lemma together with (3.23b), we get the property \(\{ {\text {tr}}Z_{\mu }^k,\, {\text {tr}}Z_{\eta }^l \}=0\) since the matrix \(Z_\eta \) represents the element \(z_\eta \in \mathcal {A}^{\times }\). Thus, Theorem 5.10 follows directly from this intermediate result.
To prove Lemma A.6, we use the derivation properties of the double bracket to see that
$$\begin{aligned} \frac{1}{kl}\{ z_{\mu }^k,z_{\eta }^l \}= \left\{ \left\{ z_{\mu },z_{\eta }\right\} \right\} ' z_\mu ^{k-1} \left\{ \left\{ z_{\mu },z_{\eta }\right\} \right\} '' z_\eta ^{l-1} \quad \text { mod }[\mathcal {A}^{\times },\mathcal {A}^{\times }]. \end{aligned}$$
(A.24)
Hence, the first step is to compute the double bracket \( \left\{ \left\{ z_{\mu },z_{\eta }\right\} \right\} \). Using that \(z_{\eta }=z+\eta s\) and the same with \(\mu \), we need \( \left\{ \left\{ z,z\right\} \right\} , \left\{ \left\{ s,z\right\} \right\} \) given in (3.3a), (3.10a) together with
$$\begin{aligned} \left\{ \left\{ s,s\right\} \right\} =\frac{1}{2} (e_0 \otimes s^2 - s^2 \otimes e_0), \end{aligned}$$
(A.25)
which is a special case of Lemma 3.2. This yields
$$\begin{aligned} \begin{aligned} \left\{ \left\{ z_{\mu },z_{\eta }\right\} \right\}&= \left\{ \left\{ z,z\right\} \right\} +\mu \left\{ \left\{ s,z\right\} \right\} - \eta \left\{ \left\{ s,z\right\} \right\} ^\circ + \mu \eta \left\{ \left\{ s,s\right\} \right\} \\&=\frac{1}{2} (e_0 \otimes z^2- z^2 \otimes e_0) + \frac{1}{2} \mu (s \otimes z - z s \otimes e_0 + e_0 \otimes sz - z \otimes s) \\&-\frac{1}{2} \eta (z \otimes s - e_0 \otimes zs + sz \otimes e_0 - s \otimes z) +\frac{1}{2} \mu \eta (e_0 \otimes s^2 - s^2 \otimes e_0). \end{aligned} \end{aligned}$$
By grouping terms together, we can write
$$\begin{aligned} \begin{aligned} \left\{ \left\{ z_{\mu },z_{\eta }\right\} \right\}&=\frac{1}{2} (e_0 \otimes z z_\eta - z z_\mu \otimes e_0) +\frac{1}{2} \mu (e_0 \otimes s z_\eta + s \otimes z - z \otimes s) \\&\quad - \frac{1}{2} \eta (s z_\mu \otimes e_0 + z \otimes s - s \otimes z). \end{aligned} \end{aligned}$$
We can use that \(\mu s = z_\mu - z\) for the terms with a factor \(\mu \), and do the same with \(\eta \). We can write in this way
$$\begin{aligned} \begin{aligned} \left\{ \left\{ z_{\mu },z_{\eta }\right\} \right\}&=\frac{1}{2} (e_0 \otimes z_\mu z_\eta - z_\eta z_\mu \otimes e_0) +\frac{1}{2} (z_\mu \otimes z - z \otimes z_\mu ) + \frac{1}{2} (z_\eta \otimes z - z \otimes z_\eta ). \end{aligned} \end{aligned}$$
Substituting back in (A.24), we get modulo commutators
$$\begin{aligned} \frac{1}{kl}\{ z_{\mu }^k,z_{\eta }^l \}= \frac{1}{2}\left( z_\mu ^{k}z z_\eta ^{l-1} - z z_\mu ^{k} z_\eta ^{l-1}\right) + \frac{1}{2} \left( z_\mu ^{k-1} z z_\eta ^{l} - z z_\mu ^{k-1} z_\eta ^{l}\right) . \end{aligned}$$
This is clearly zero when \(\mu =\eta \). If \(\mu \ne \eta \), we can substitute \(z=\frac{1}{\mu -\eta }(\mu z_\eta - \eta z_\mu )\) in the two groups of terms, which then vanish modulo commutators. \(\square \)
