2.1 Hamiltonian double quasi-Poisson algebras

Hereafter, we follow [Reference Van den Bergh41, Reference Crawley-Boevey, Etingof and Ginzburg24, Reference Fairon29].

2.1.1 Double derivations

We fix a finitely generated associative unital algebra A over a field $\Bbbk $ of characteristic zero, and we use the unadorned notations $\otimes =\otimes _{\Bbbk }$ , $\operatorname {Hom}=\operatorname {Hom}_{\Bbbk }$ for brevity. The opposite algebra and the enveloping algebra of A will be denoted $A^{\operatorname {op}}$ and $A^{\operatorname {e}}:=A\otimes A^{\operatorname {op}}$ , respectively. We shall identify the category of A-bimodules and the category of (left) $A^{\operatorname {e}}$ -modules. Note that the underlying A-bimodule of A carries two A-bimodule structures with the same underlying vector space $A\otimes A$ , namely the outer and the inner A-bimodule structures, respectively denoted by $(A\otimes A)_{\operatorname {out}}$ and $(A\otimes A)_{\operatorname {inn}}$ ; they are given by

$$ \begin{align*} a_1(a'\otimes a'')a_2&:=(a_1a')\otimes (a''a_2)\quad \text{on }(A\otimes A)_{\operatorname{out}}\,; \\ a_1*(a'\otimes a'')*a_2&:=(a'a_2)\otimes (a_1a'')\quad \text{on }(A\otimes A)_{\operatorname{inn}}\,, \end{align*} $$

for all $a',a'',a_1,a_2\in A$ .

Given another unital associative $\Bbbk $ -algebra B, A is called a B-algebra if there exists a unit preserving $\Bbbk $ -algebra morphism $B\to A$ . Let M be an arbitrary A-bimodule and $\operatorname {Der}_B(A,M)$ be the vector space of B-derivations of A into M, that is, additive maps $\theta \colon A\to M$ satisfying the Leibniz rule and $\theta (b)=0$ , for all $b\in B$ . Following [Reference Cuntz and Quillen26], we define the A-bimodule of noncommutative differential 1-forms as $\Omega ^1_B A:=\operatorname {ker}(m\colon A\otimes _B A\to A)$ , where m is the product of A. As in the commutative case, it carries a canonical B-derivation $\operatorname {d}\colon A\to \Omega ^1_B A$ , $a\mapsto \operatorname {d}{a}:=a\otimes _B 1-1\otimes _B a$ , and the pair $(\Omega ^1_B A,\operatorname {d})$ has a universal property ([Reference Quillen39, §2]), which can be rephrased as

(2.1) $$ \begin{align} \operatorname{Der}_B (A,M)\cong \operatorname{Hom}_{A^{\operatorname{e}}}(\Omega^1_B A,M)\,. \end{align} $$

By taking $M=(A\otimes A)_{\operatorname {out}}$ in equation (2.1), we define the A-bimodule of double derivations

$$\begin{align*}\operatorname{\mathbb{D}er}_B A:=\operatorname{Hom}_{A^{\operatorname{e}}}(\Omega^1_B A,(A\otimes A)_{\operatorname{out}})\cong \operatorname{Der}_B(A,(A\otimes A)_{\operatorname{out}})\,. \end{align*}$$

If $\delta \in \operatorname {\mathbb {D}er}_B A$ , we systematically use Sweedler’s notation: $\delta \colon A\to A\otimes A$ , $a\mapsto \delta (a)^{\prime }\otimes \delta (a)''$ . Note that the A-bimodule structure on $\operatorname {\mathbb {D}er}_B A$ is induced from the inner bimodule structure: $(a_1\delta a_2)(a):=\delta (a)'a_2\otimes a_1\delta (a)^{\prime \prime }$ , for all $a,a_1,a_2\in A$ .

The A-bimodule $\operatorname {\mathbb {D}er}_BA$ carries a distinguished element, called the gauge element E, defined by

(2.2) $$ \begin{align} E\colon A\longrightarrow A\otimes A,\quad a\,\longmapsto\, a\otimes 1-1\otimes a. \end{align} $$

Note that this element was also introduced in [Reference Crawley-Boevey, Etingof and Ginzburg24, §3] and denoted by $\Delta $ .

Following [Reference Van den Bergh41, §3.1], we define the algebra of poly-vector fields as $D_BA=T_A(\operatorname {\mathbb {D}er}_B A)$ , which is the tensor algebra of the A-bimodule $\operatorname {\mathbb {D}er}_B A$ (over A). The n-th component is the n-fold $(D_BA)_n=(\operatorname {\mathbb {D}er}_B A)^{\otimes _ An}$ for $n\geq 1$ .

2.1.2 Double derivations for quivers

A quiver Q is an oriented graph. We denote its set of vertices by I and its set of arrows (oriented edges) by Q. We can define on Q the tail (resp. head) map $t\colon Q\to I$ (resp. $h\colon Q\to I$ ) assigning to any arrow $a\in Q$ its tail/starting vertex $t(a)\in Q$ (resp. its head/ending vertex $h(a)\in Q$ ). We can then form the double $\overline {Q}$ of Q with the same vertex set I by adding an opposite arrow $a^{\ast } : h(a)\to t(a)$ for each $a\in Q$ . We naturally extend $h,t$ to $\overline {Q}$ .

Now, given a quiver Q (not necessarily a double quiver), we form the path algebra $\Bbbk Q$ which is the $\Bbbk $ -algebra generated by the arrows $a\in Q$ and idempotents $(e_s)_{s\in I}$ labelled by the vertices such that

(2.3) $$ \begin{align} a=e_{h(a)}a e_{t(a)}\,, \quad e_s e_t = \delta_{st}\,e_s\,, \end{align} $$

and the product is given by concatenation of paths (if possible; otherwise, the idempotent decomposition makes the product vanish). In this article, we will regard $\Bbbk Q$ as a B-algebra with $B=\oplus _{s\in I}\Bbbk e_s$ . Then, if $A=\Bbbk Q$ , the A-bimodule of double derivations $\operatorname {\mathbb {D}er}_B A$ is generated as an A-bimodule by $\big \{\frac {\partial }{\partial a}\mid a\in Q\big \}$ , where if $b\in Q$ ,

$$\begin{align*}\frac{\partial}{\partial a}(b):=\delta_{ab}e_{h(a)}\otimes e_{t(a)}\,. \end{align*}$$

Remark 2.1. Note that equation (2.3) implies that we read paths from right to left. This is the convention followed by Boalch in [Reference Boalch14, Reference Boalch15] and Crawley-Boevey and Shaw [Reference Crawley-Boevey and Shaw25], but it contrasts with [Reference Van den Bergh41], where Van den Bergh uses the opposite convention (i.e., $a=e_{t(a)}a e_{h(a)}$ ).

2.1.3 Double quasi-Poisson brackets

Let A be a finitely generated associative unital B-algebra. A B-linear n-bracket [Reference Van den Bergh41] is a map $\{\mkern -6mu\{-,\cdots ,-\}\mkern -6mu\}\colon A^{\times n}\to A^{\otimes n}$ which

1. (a) is linear in all its arguments;

2. (b) is cyclically antisymmetric, that is,

   $$ \begin{align*} \tau_{(1\cdots n)}\circ\{\mkern-6mu\{-,\cdots,-\}\mkern-6mu\}\circ \tau^{-1}_{(1\cdots n)}=(-1)^{n+1}\{\mkern-6mu\{-,\cdots,-\}\mkern-6mu\} \end{align*} $$
   for the cyclic permutation $\tau _{(1\cdots n)}(a_1\otimes \cdots \otimes a_{n-1}\otimes a_n)=a_n\otimes a_1\otimes \cdots \otimes a_{n-1}$ ;

3. (c) vanishes when its last argument (hence any argument) is in the image of B;

4. (d) is a derivation $A\to A^{\otimes n}$ in its last argument for the outer bimodule structure on $A^{\otimes n}$ given by $b(a_1\otimes a_2 \otimes \cdots \otimes a_{n-1}\otimes a_n)c= (ba_1)\otimes a_2 \otimes \cdots \otimes a_{n-1}\otimes (a_nc)$ .

We call a 2- and 3-bracket a double and triple bracket, respectively. For the reader’s convenience, we explicitly write the definition of the former.

Definition 2.2 [Reference Van den Bergh41]

A B-linear double bracket on the B-algebra A is a $\Bbbk $ -bilinear map $\{\mkern -6mu\{-,-\}\mkern -6mu\}:A\times A \to A \otimes A$ , which satisfies for any $a,b,c \in A$ ,

(2.4) $$ \begin{align} \{\mkern-6mu\{a,b\}\mkern-6mu\}&=-\tau_{(12)}\{\mkern-6mu\{b,a\}\mkern-6mu\}\,, &&(cyclic \ antisymmetry), \end{align} $$

(2.5) $$ \begin{align} \{\mkern-6mu\{a,bc\}\mkern-6mu\}&=\{\mkern-6mu\{a,b\}\mkern-6mu\}c+b\{\mkern-6mu\{a,c\}\mkern-6mu\}\,, &&(right \ Leibniz \ rule), \end{align} $$

and which is such that $\{\mkern -6mu\{-,e\}\mkern -6mu\}=0$ for all $e\in B$ .

Assuming that equation (2.4) holds, it is easy to check that equation (2.5) is equivalent to

(2.6) $$ \begin{align} \{\mkern-6mu\{bc,a\}\mkern-6mu\}=\{\mkern-6mu\{b,a\}\mkern-6mu\}\ast c+b\ast\{\mkern-6mu\{c,a\}\mkern-6mu\} \,,\qquad \quad {(left \ Leibniz \ rule)}. \end{align} $$

From the derivation rules (2.5) and (2.6), observe that it suffices to define double brackets on generators of A. From now on, if the context is clear, by a double bracket we will mean a B-linear double bracket. The following result will be used in §3.3.

Proposition 2.3 ([Reference Van den Bergh41], Proposition 4.1.1)

There exists a well-defined linear map

$$ \begin{align*} \mu\colon (D_B A)_n\longrightarrow&\, \{B\text{-linear }n\text{-brackets on }A\} \\ Q\,\,\longmapsto&\, \{\mkern-6mu\{-,\cdots,-\}\mkern-6mu\}_Q=\sum^{n-1}_{i=0}(-1)^{(n-1)i}\,\tau^{i}_{(1\cdots n)}\circ\{\mkern-6mu\{-,\cdots,-\}\mkern-6mu\}^{\sim}_Q\circ\tau^{-i}_{(1\cdots n)}\, , \end{align*} $$

where we set for $Q=\delta _1\cdots \delta _n$ and $a_1,\dots ,a_n\in A$ ,

$$\begin{align*}\{\mkern-6mu\{a_1,\cdots,a_n\}\mkern-6mu\}^{\sim}_Q=\delta_n(a_n)'\delta_1(a_1)''\otimes \delta_1(a_1)'\delta_2(a_2)''\otimes\cdots\otimes\delta_{n-1}(a_{n-1})'\delta_n(a_n)'\,.\\[-15pt] \end{align*}$$

Note that, if $\Omega ^1_B A$ is a projective $A^{\operatorname {e}}$ -module—for instance, if A is the path algebra of a quiver— $\mu $ is an isomorphism (see [Reference Van den Bergh41, Proposition 4.1.2]). Also, since it will be used in § 3.3, we specialise the previous formula to noncommutative bivectors $Q=\delta _1\delta _2\in (D_B A)_2=(\operatorname {\mathbb {D}er}_B A) \otimes _A (\operatorname {\mathbb {D}er}_B A)$ :

(2.7) $$ \begin{align} \{\mkern-6mu\{a_1,a_2\}\mkern-6mu\}_{\delta_1 \delta_2}=\delta_2(a_2)' \delta_1(a_1)'' \otimes \delta_1(a_1)' \delta_2(a_2)'' - \delta_1(a_2)' \delta_2(a_1)'' \otimes \delta_2(a_1)' \delta_1(a_2)'' \,,\\[-15pt]\nonumber \end{align} $$

for any $a_1,a_2\in A$ .

To develop a noncommutative version of quasi-Poisson geometry, as introduced in [Reference Alekseev, Kosmann-Schwarzbach and Meinrenken2], we need a nonvanishing noncommutative Jacobi identity. Firstly, given a double bracket $\{\mkern -6mu\{-,-\}\mkern -6mu\}$ on A and $a,b,c\in A$ , Van den Bergh [Reference Van den Bergh41] introduced a suitable extension of the double bracket, given by $\{\mkern -6mu\{a,b\otimes c\}\mkern -6mu\}_L:=\{\mkern -6mu\{a,b\}\mkern -6mu\}\otimes c\in A^{\otimes 3}$ . Next, he defined a natural triple bracket associated with $\{\mkern -6mu\{-,-\}\mkern -6mu\}$ by setting

(2.8) $$ \begin{align} \{\mkern-6mu\{a,b,c\}\mkern-6mu\}:= \{\mkern-6mu\{a,\{\mkern-6mu\{b,c\}\mkern-6mu\}\}\mkern-6mu\}_L+\tau_{(123)} \{\mkern-6mu\{b,\{\mkern-6mu\{c,a\}\mkern-6mu\}\}\mkern-6mu\}_L+\tau_{(132)}\{\mkern-6mu\{c,\{\mkern-6mu\{a,b\}\mkern-6mu\}\}\mkern-6mu\}_L\,.\\[-15pt]\nonumber \end{align} $$

Note that $\tau _{(123)}(a_1\otimes a_2\otimes a_3)=a_{3}\otimes a_1 \otimes a_{2}$ and $\tau _{(132)}(a_1\otimes a_2\otimes a_3)=a_2\otimes a_3\otimes a_1$ , for all $a_1,a_2,a_3\in A$ . Secondly, we assume that the unit in A admits a decomposition $1=\sum _{s\in I} e_s$ in terms of a finite set of orthogonal idempotents, that is, $|I|\in \mathbb {N}^{\times }$ and $e_s e_t = \delta _{st} e_s$ . In that case, we view A as a B-algebra for $B=\oplus _{s\in I} \Bbbk e_s$ . Now, using the distinguished double derivation E defined in equation (2.2), we have $E_s=e_s E e_s\in D_B A$ and we can apply Proposition 2.3 to $\sum _{s\in I}E_s^3\in (D_B A)_3$ and define the following triple bracket:

(2.9) $$ \begin{align} \begin{aligned} &\quad \{\mkern-6mu\{a,b,c\}\mkern-6mu\}_{\mathrm{qP}}:=\frac{1}{12}\sum_{s\in I} \{\mkern-6mu\{a,b,c\}\mkern-6mu\}_{E^3_s} \\ &=\frac14 \sum_{s\in I} \Big( c e_s a \otimes e_s b \otimes e_s - c e_s a \otimes e_s \otimes b e_s - c e_s \otimes a e_s b \otimes e_s + c e_s \otimes a e_s \otimes b e_s \\ &\qquad \quad - e_s a \otimes e_s b \otimes e_s c + e_s a \otimes e_s \otimes b e_s c + e_s \otimes a e_s b \otimes e_s c - e_s \otimes a e_s \otimes b e_s c \Big)\,, \end{aligned}\\[-15pt]\nonumber \end{align} $$

for any $a,b,c\in A$ .

Definition 2.4 [Reference Van den Bergh41]

Let A be an associative B-algebra endowed with a double bracket $\{\mkern -6mu\{-,-\}\mkern -6mu\}$ . We say that the double bracket is quasi-Poisson if the triple brackets (2.8) and (2.9) coincide:

(2.10) $$ \begin{align} \{\mkern-6mu\{a,b,c\}\mkern-6mu\}=\{\mkern-6mu\{a,b,c\}\mkern-6mu\}_{\mathrm{qP}}\,,\\[-15pt]\nonumber \end{align} $$

for all $a,b,c\in A$ . The pair $(A,\{\mkern -6mu\{-,-\}\mkern -6mu\})$ is called a double quasi-Poisson algebra.

Note that $\{\mkern -6mu\{-,-,-\}\mkern -6mu\}$ and $\{\mkern -6mu\{-,-,-\}\mkern -6mu\}_{\mathrm {qP}}$ are triple brackets according to the terminology given at the beginning of this subsection. Thus, since triple brackets are cyclic and enjoy a derivation property, it suffices to check equation (2.10) on generators of A.

2.1.4 Multiplicative moment maps

In addition to the introduction of a noncommutative version of quasi-Poisson brackets, Van den Bergh adapted the notion of Lie group valued moment maps from [Reference Alekseev, Malkin and Meinrenken1, Reference Alekseev, Kosmann-Schwarzbach and Meinrenken2] as follows.

Definition 2.5 [Reference Van den Bergh41]

Let $(A,\{\mkern -6mu\{-,-\}\mkern -6mu\})$ be a double quasi-Poisson algebra over $B=\oplus _{s\in I} \Bbbk e_s$ . An invertible element $\Phi \in A$ is a multiplicative moment map if it can be decomposed as $\Phi =\sum _{s\in I}\Phi _s$ with $\Phi _s\in e_sAe_s$ , and it satisfies that for all $a\in A$ and $s\in I$

(2.11) $$ \begin{align} \{\mkern-6mu\{\Phi_s,a\}\mkern-6mu\}=\frac12 \Big(ae_s\otimes \Phi_s-e_s \otimes \Phi_s a + a \Phi_s \otimes e_s-\Phi_s \otimes e_s a\Big)\,. \end{align} $$

Then we call the triple $(A,\{\mkern -6mu\{-,-\}\mkern -6mu\},\Phi )$ a Hamiltonian double quasi-Poisson algebra.Footnote ³

The invertibility of $\Phi $ implies that $\Phi _s^{-1}:=e_s\Phi ^{-1}e_s$ is an inverse for $\Phi _s$ in $e_s A e_s$ . Then the identity (2.11) yields for all $a\in A$ that

(2.12) $$ \begin{align} \{\mkern-6mu\{\Phi_s^{-1},a\}\mkern-6mu\}=-\frac12 \Big(a \Phi_s^{-1}\otimes e_s-\Phi_s^{-1} \otimes e_s a + a e_s \otimes \Phi_s^{-1}-e_s \otimes \Phi_s^{-1} a\Big)\,. \end{align} $$

Finally, we can define morphisms between such algebras, which we use in Section 5.

Definition 2.6 [Reference Fairon30]

Let $(A,\{\mkern -6mu\{-,-\}\mkern -6mu\},\Phi )$ and $(A',\{\mkern -6mu\{-,-\}\mkern -6mu\}',\Phi ')$ be two Hamiltonian double quasi-Poisson algebras over B. An isomorphism $\psi :A\to A'$ of B-algebras is said to be an isomorphism of Hamiltonian double quasi-Poisson algebras if it preserves the double quasi-Poisson brackets and the multiplicative moment maps, that is,

(2.13) $$ \begin{align} \{\mkern-6mu\{\psi(a_1),\psi(a_2)\}\mkern-6mu\}'=(\psi\otimes \psi)(\{\mkern-6mu\{a_1,a_2\}\mkern-6mu\})\,, \quad \text{ and }\quad \psi(\Phi)=\Phi'\,, \end{align} $$

for all $a_1,a_2\in A$ .

2.2 Fusion of Hamiltonian double quasi-Poisson algebras

Given an algebra A, the method of fusion allows to construct a new algebra $A^f$ by identifying two orthogonal idempotents in A. More importantly, if A is a Hamiltonian double quasi-Poisson algebra, it is possible to obtain such a structure on $A^f$ after performing fusion. Below, we recall the main results associated with this method, which are due to Van den Bergh [Reference Van den Bergh41, §2.5, 5.3] (see also [Reference Brüstle18, §2.1] without the perspective of double brackets); alternative presentations can be found in [Reference Fairon29, 30]. These results are noncommutative analogues of [Reference Alekseev, Kosmann-Schwarzbach and Meinrenken2, §5].

As in §2.1, we assumeFootnote ⁴ that there exist orthogonal idempotents $e_i,e_j\in B$ . First, we extend the algebra A along the (ordered) idempotents $(e_i,e_j)$ as

(2.14) $$ \begin{align} \bar{A}:=A \ast_{\Bbbk e_i \oplus \Bbbk e_j \oplus \Bbbk \hat{e}} \Big(\operatorname{Mat}_2(\Bbbk)\oplus \Bbbk \hat{e}\Big ) = A \ast_B \bar{B}\,. \end{align} $$

Here, we have set $\hat {e}=1-e_i-e_j$ , and $\operatorname {Mat}_2(\Bbbk )$ is the $\Bbbk $ -algebra generated by the elements $e_i=e_{ii},e_{ij},e_{ji},e_j=e_{jj}$ subject to the matrix relations $e_{st}e_{uv}=\delta _{tu}e_{sv}$ . Second, we can introduce the fusion algebra $A^f_{e_j \to e_i}$ obtained by fusing $e_j$ onto $e_i$ , which we abbreviate as $A^f$ . It is simply defined as

(2.15) $$ \begin{align} A^f:=\, \epsilon \bar{A} \epsilon\,, \quad \text{for } \epsilon=1-e_j\,. \end{align} $$

The fusion algebra $A^f$ can be seen as dismissing the elements of $e_j \bar {A} + \bar {A} e_j$ inside $\bar {A}$ . Note that $A^f$ is a $B^f$ -algebra, where $B^f:=\epsilon \bar {B} \epsilon $ . Furthermore, there exists a projection onto the fusion algebra

(2.16) $$ \begin{align} A \longrightarrow A^f\,, \quad a \longmapsto a^f:=\epsilon a \epsilon + e_{ij}ae_{ji} + e_{ij}a \epsilon + \epsilon a e_{ji}\,, \end{align} $$

from which we can get a set of generators in $A^f$ , which we split into four types. Namely,

(2.17) $$ \begin{align} (\text{first type})&\qquad\qquad t^f=t\,, \quad & &\text{for }t \in \epsilon A \epsilon\,, \end{align} $$

(2.18) $$ \begin{align} (\text{second type})&\qquad\qquad u^f=e_{ij}u\,, \quad & &\text{for }u \in e_j A \epsilon\,, \end{align} $$

(2.19) $$ \begin{align} (\text{third type})&\qquad\qquad v^f=v e_{ji}\,, \quad & &\text{for }v \in \epsilon A e_j\,, \end{align} $$

(2.20) $$ \begin{align} (\text{fourth type})&\qquad\qquad w^f=e_{ij} w e_{ji}\,, \quad & &\text{for }w \in e_j A e_j\,. \end{align} $$

We assume that $(A,\{\mkern -6mu\{-,-\}\mkern -6mu\},\Phi )$ is a Hamiltonian double quasi-Poisson algebra over $B=\oplus _{s\in I} \Bbbk e_s$ from now on. Then, the double bracket uniquely extends from A to $\bar A$ by requiring it to be $\bar {B}$ -linear, and it can then be restricted to $A^f$ . We also denote the double bracket hence induced on $A^f$ by $\{\mkern -6mu\{-,-\}\mkern -6mu\}$ . The image in $A^f$ of the component $\Phi _s$ of the multiplicative moment map is either $\Phi _s^f=\epsilon \Phi _s\epsilon =\Phi _s$ if $s\neq j$ , or $\Phi _s^f=e_{ij}\Phi _j e_{ji}$ if $s=j$ . However, the data $(\{\mkern -6mu\{-,-\}\mkern -6mu\},\Phi ^f)$ do not turn $A^f$ into a Hamiltonian double quasi-Poisson algebra.

Lemma 2.7. There exists a unique double bracket $\{\mkern -6mu\{-,-\}\mkern -6mu\}_{\mathrm {fus}}$ on $A^f$ such that for any

(2.21) $$ \begin{align} t,\tilde{t} \in \epsilon A \epsilon,\quad u,\tilde{u} \in e_j A \epsilon,\quad v,\tilde{v} \in \epsilon A e_j, \quad w,\tilde{w} \in e_j A e_j, \end{align} $$

we have that

(2.22a) $$ \begin{align} & \{\mkern-6mu\{\epsilon t \epsilon , \epsilon \tilde{t} \epsilon\}\mkern-6mu\}_{\mathrm{fus}}=0\,, \end{align} $$

(2.22b) $$ \begin{align} &\{\mkern-6mu\{\epsilon t \epsilon , e_{ij} u \epsilon\}\mkern-6mu\}_{\mathrm{fus}}=\frac12 \big( e_i \otimes t e_{ij}u - e_i t \otimes e_{ij}u\big)\,, \end{align} $$

(2.22c) $$ \begin{align} &\{\mkern-6mu\{\epsilon t \epsilon ,\epsilon v e_{ji} \}\mkern-6mu\}_{\mathrm{fus}}=\frac12 \left( v e_{ji} t \otimes e_i - v e_{ji} \otimes t e_i\right)\,, \end{align} $$

(2.22d) $$ \begin{align} &\{\mkern-6mu\{\epsilon t \epsilon , e_{ij} w e_{ji}\}\mkern-6mu\}_{\mathrm{fus}}=\frac12 \left( e_{ij}we_{ji}t \otimes e_i + e_i \otimes t e_{ij} w e_{ji} - e_{ij} w e_{ji} \otimes t e_i - e_i t \otimes e_{ij} w e_{ji}\right), \end{align} $$

(2.22e) $$ \begin{align} &\{\mkern-6mu\{e_{ij} u \epsilon , e_{ij} \tilde{u} \epsilon\}\mkern-6mu\}_{\mathrm{fus}}= \frac12 (e_i \otimes e_{ij}u e_{ij}\tilde{u}-e_{ij}\tilde{u} e_{ij}u \otimes e_i) \,, \end{align} $$

(2.22f) $$ \begin{align} &\{\mkern-6mu\{e_{ij} u \epsilon ,\epsilon v e_{ji} \}\mkern-6mu\}_{\mathrm{fus}}=\frac12(e_{ij}u \otimes e_i ve_{ji}-v e_{ji} \otimes e_{ij}u e_i)\,, \end{align} $$

(2.22g) $$ \begin{align} &\{\mkern-6mu\{e_{ij} u\epsilon , e_{ij} w e_{ji}\}\mkern-6mu\}_{\mathrm{fus}}=\frac12(e_i \otimes e_{ij}u e_{ij}we_{ji} - e_{ij}w e_{ji} \otimes e_{ij}u e_i) \,, \end{align} $$

(2.22h) $$ \begin{align} &\{\mkern-6mu\{\epsilon v e_{ji},\epsilon \tilde{v} e_{ji} \}\mkern-6mu\}_{\mathrm{fus}}=\frac12( \tilde{v} e_{ji}v e_{ji}\otimes e_i - e_i \otimes v e_{ji}\tilde{v} e_{ji})\,, \end{align} $$

(2.22i) $$ \begin{align} &\{\mkern-6mu\{\epsilon v e_{ji} , e_{ij} w e_{ji}\}\mkern-6mu\}_{\mathrm{fus}}=\frac12(e_{ij}we_{ji}ve_{ji}\otimes e_i - e_i v e_{ji}\otimes e_{ij}we_{ji} ) \,, \end{align} $$

(2.22j) $$ \begin{align} &\{\mkern-6mu\{e_{ij} w e_{ji} , e_{ij} \tilde{w} e_{ji}\}\mkern-6mu\}_{\mathrm{fus}}=0 \,. \end{align} $$

The double bracket from Lemma 2.7 is due to Van den Bergh [Reference Van den Bergh41, Theorem 5.3.1]; it was introduced to get a Hamiltonian double quasi-Poisson structure on $A^f$ .

Theorem 2.8 [Reference Van den Bergh41, 29]

Let $A^f=A^f_{e_j \to e_i}$ be the fusion algebra. Then $A^f$ is a Hamiltonian double quasi-Poisson algebra, whose double quasi-Poisson bracket is given by

(2.23) $$ \begin{align} \{\mkern-6mu\{-,-\}\mkern-6mu\}^f:= \{\mkern-6mu\{-,-\}\mkern-6mu\} + \{\mkern-6mu\{-,-\}\mkern-6mu\}_{\mathrm{fus}}\,, \end{align} $$

where $\{\mkern -6mu\{-,-\}\mkern -6mu\}$ is induced in $A^f$ by the double quasi-Poisson bracket of A, while $\{\mkern -6mu\{-,-\}\mkern -6mu\}_{\mathrm {fus}}$ is defined in Lemma 2.7. Its multiplicative moment map is given by

(2.24) $$ \begin{align} \Phi^{f\!f}=\Phi^f_i \Phi_j^f+\sum_{s\neq i,j}\Phi_s^f = \Phi_i e_{ij}\Phi_j e_{ji} + \sum_{s\neq i,j} \Phi_s\,.\\[-15pt]\nonumber \end{align} $$

This result was derived in [Reference Van den Bergh41, Theorems 5.3.1 and 5.3.2] under mild assumptions, which were removed in [Reference Fairon29, Theorems 2.14 and 2.15].

Let us note that fusing the two idempotents $e_i,e_j$ in the opposite order yields an isomorphic algebra. Indeed, each $a\in A$ has a unique image in both $A^f_{e_j \to e_i}$ and $A^f_{e_i \to e_j}$ through the projection map (2.16). By mapping one projection onto the other, we get an algebra isomorphism $A^f_{e_j \to e_i}\longrightarrow A^f_{e_i \to e_j}$ which is induced by the identity map on A. However, this is not an isomorphism of Hamiltonian double quasi-Poisson algebras as in Definition 2.6. To preserve the double quasi-Poisson structure, one needs a different isomorphism $A^f_{e_j \to e_i}\longrightarrow A^f_{e_i \to e_j}$ that involves (images of) $\Phi _j\in A$ ; see [Reference Fairon30, §4.1.1].

2.3 $H_0$ -Poisson algebras

The zeroth Hochschild homology of A, denoted $H_0(A)$ , is the vector space obtained by identifying all the commutators to zero. This means that $H_0(A)=A/[A,A]$ , where $[A,A]\subset A$ is spanned by commutators. We write $\overline {a}$ for the image of $a\in A$ under the natural projection map $A\to A/[A,A]$ .

Interestingly, double quasi-Poisson brackets induce $H_0$ -Poisson structures as follows. By [Reference Van den Bergh41, §2.4 and §5.1], the associated bracket $\{-,-\}:=m\circ \{\mkern -6mu\{-,-\}\mkern -6mu\}:A\to A$ obtained by multiplication descends to a Lie bracket on $H_0(A)$ . Moreover, since the associated bracket on A is a derivation in its second argument, we get an $H_0$ -Poisson structure on A. Furthermore, $H_0$ -Poisson structures naturally arise after performing quotients of Hamiltonian double quasi-Poisson algebras using the corresponding multiplicative moment map as in Definition 2.5.

2.4 Example from a one-arrow quiver

Let $\Gamma _1:=A_2$ be the quiver with one arrow $a:1\to 2$ and $\overline {\Gamma }_1$ be its double with the additional arrow $b=a^{\ast }:2\to 1$ . We form the path algebra $\Bbbk \overline {\Gamma }_1$ as explained in § 2.1.2. Now, let $\mathcal {A}(\Gamma _1)$ be the algebra obtained by universal localisation from the set $S=\{1+ab,1+ba \}$ . This is equivalent to adding the local inverse $(e_2+ab)^{-1}$ to $e_2+ab$ in $e_{2}\mathcal {A}(\Gamma _1) e_{2}$ , as well as to adding $(e_1+ba)^{-1}$ to $e_1+ba$ in $e_{1}\mathcal {A}(\Gamma _1) e_{1}$ . Finally, note that $\mathcal {A}(\Gamma _1)$ is a B-algebra for $B=\Bbbk e_1\oplus \Bbbk e_2$ .

Following [Reference Crawley-Boevey and Shaw25], we define the multiplicative preprojective algebra Footnote ⁵ of $\Gamma _1$ with parameter $q=(q_1,q_2)\in (\Bbbk ^{\times })^2$ as the algebra

(2.25) $$ \begin{align} \Lambda^q(\Gamma_1):=\mathcal{A}(\Gamma_1)/R^q\,, \end{align} $$

where $R^q$ is the ideal generated by

(2.26) $$ \begin{align} e_2+ab=q_2 e_2\,, \quad (e_1+ba)^{-1}=q_1 e_1\,. \end{align} $$

3.1 Euler continuants with idempotents

Given an integer $n\geq 1$ , let $\Gamma _n$ be the quiver whose set of vertices is $\{1,2\}$ and whose set of arrows is $\{a_i\mid 1\leq i \leq n\}$ , where $a_i$ is an arrow from the vertex $1$ to $2$ . We form the double $\overline {\Gamma }_n$ of $\Gamma _n$ by adding the arrows $\{b_i \mid 1\leq i \leq n\}$ , where $b_i$ is an arrow from the vertex $2$ to $1$ ; see Figure 1.

We form the path algebra $\Bbbk \overline {\Gamma }_n$ of $\overline {\Gamma }_n$ as in §2.1.2. This allows us to see this algebra as being generated by the orthogonal idempotents $e_1,e_2$ and the symbols $\{a_i,b_i\mid 1\!\leq \! i\!\leq \! n\}$ subject to the relations

(3.1) $$ \begin{align} e_1+e_2=1\,, \quad a_i=e_2a_ie_1\,, \quad b_i=e_1b_ie_2,\,\, \text{ for }\, 1\leq i\leq n\,. \end{align} $$

Since $a_ib_j$ and $b_ja_i$ are not trivially vanishing in $\Bbbk \overline {\Gamma }_n$ , we can slightly adapt the definition of Euler continuants from Section 1 as follows. We introduce the first and second Euler continuants by setting

(3.2) $$ \begin{align} \begin{aligned} (a_i)&:=a_i\in e_2(\Bbbk \overline{\Gamma}_n)e_1\,,&\quad (b_i)&:=b_i \in e_1(\Bbbk \overline{\Gamma}_n)e_2\,, \\ (a_i,b_j)&:= e_2+a_ib_j\in e_2(\Bbbk \overline{\Gamma}_n)e_2\,,&\quad (b_i,a_j)&:= e_1+b_ia_j\in e_1(\Bbbk \overline{\Gamma}_n)e_1\,, \end{aligned} \end{align} $$

for $i,j=1,\ldots ,n$ . We also put $(a_i,a_j)=0=(b_i,b_j)$ as $a_ia_j=0=b_ib_j$ in $\Bbbk \overline {\Gamma }_n$ . This ensures that the first and second Euler continuants respect the idempotent decomposition, that is, if $y_1,y_2\in \{a_i,b_i\mid 1\leq i\leq n\}$ , we have $(y_1,y_2)=e_{h(y_1)}(y_1,y_2)e_{t(y_2)}$ . Then, given elements $y_1,\ldots ,y_k\in \{a_i,b_i\mid 1\leq i\leq n\}$ for $k\geq 3$ , we have that the k-th Euler continuant is obtained recursively using the rule

(3.3) $$ \begin{align} (y_1,\dots,y_k)=(y_1,\dots,y_{k-1})y_k+(y_1,\dots,y_{k-2})\,. \end{align} $$

We will only consider the Euler continuant $(y_1,\dots ,y_k)$ when the elements $\{y_{\ell }\}$ are alternating with respect to the generators $\{a_i,b_i\mid 1\leq i\leq n\}$ ; that is, if $y_{\ell }=a_i$ (resp. $y_{\ell }=b_i$ ), then $y_{\ell +1}=b_{j}$ (resp. $y_{\ell +1}=a_{j}$ ) for some $1\leq i,j\leq n$ .

Remark 3.1. Alternatively to equation (1.1), observe that the recursive rule used to define Euler continuants can be expressed as

(3.4) $$ \begin{align} (x_1,\dots,x_k)=x_1(x_2,\dots,x_k)+(x_3,\dots,x_k)\,. \end{align} $$

Then, when dealing with idempotents, we can also use equation (3.4) instead of equation (3.3) to define the k-th Euler continuants for $k\geq 3$ .

We close this subsection by proving several identities involving Euler continuants that will be important for later purposes.

Lemma 3.2. For $n\geq 1$ , we have that

(3.5) $$ \begin{align} (a_1,b_1,\dots, b_{n-1},a_n)=\sum^{n+1}_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}\Big]+a_1\, , \end{align} $$

and for $n\geq 2$ , we have that

(3.6) $$ \begin{align} (a_1,b_1,\dots,a_n ,b_n)=(a_1,\dots ,b_{n-1})(a_n,b_n)+\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big]+a_1b_n\,. \end{align} $$

Proof. We prove equation (3.5) by induction. The base case occurs when $n=1$ , which follows by definition: $(a_1)=a_1$ . Now, by equation (3.3) and the inductive hypothesis, we have that

$$ \begin{align*} (a_1,b_1,&\dots ,a_{n-1},b_{n-1},a_n)=(a_1,b_1,\dots ,a_{n-1},b_{n-1})a_n+(a_1,b_1,\dots,a_{n-1}) \\ &=(a_1,b_1,\dots ,a_{n-1},b_{n-1})a_n+\sum^{n}_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}\Big]+a_1 \\ &=\sum^{n+1}_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}\Big]+a_1\,. \end{align*} $$

To prove (3.6), it suffices to use the recursive relation (3.3) and (3.5):

$$ \begin{align*} (a_1,b_1,&\dots,a_n ,b_n)=(a_1,\dots ,a_n)b_n+(a_1,\dots ,b_{n-1}) \\ &=\Big[(a_1,\dots ,b_{n-1})a_n+ (a_1,\dots ,a_{n-1})\Big]b_n+(a_1,\dots ,b_{n-1}) \\ &=(a_1,\dots ,b_{n-1})(a_n,b_n)+ (a_1,\dots ,a_{n-1})b_n \\ &=(a_1,\dots ,b_{n-1})(a_n,b_n)+\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big]+a_1b_n\, .\\[-48pt] \end{align*} $$

Lemma 3.3. For $n\geq 1$ , we have

$$ \begin{align*} (a_1,b_1,\ldots, b_{n-1},a_{n})(b_{n},a_n,\ldots, b_1,a_1)=(a_1,b_1, \ldots,a_n,b_{n})(a_{n},b_{n-1}, \ldots, b_1, a_1)\,. \end{align*} $$

Proof. For $n=1$ , we can write that

$$ \begin{align*} (a_1)(b_1,a_1)=a_1(e_1+b_1a_1)=(e_2+a_1b_1)a_1=(a_1,b_1)(a_1)\,. \end{align*} $$

Now, using the induction hypothesis and the recursive relations (3.3) and (3.4) of the Euler continuants, we have

$$ \begin{align*} (a_1,\ldots,&a_{n})(b_{n},\ldots,a_1) \\ &=(a_1,\ldots,b_{n-1})a_{n}(b_{n},\ldots,a_1)+(a_1,\ldots,a_{n-1})(b_{n},\ldots,a_1) \\ &=(a_1,\ldots,b_{n-1})a_{n}(b_{n},\ldots,a_1)+(a_1,\ldots,a_{n-1})\Big[b_{n}(a_{n},\ldots,a_1)+(b_{n-1},\ldots,a_1)\Big] \\ &=(a_1,\ldots,b_{n-1})\Big[a_{n}(b_{n},\ldots,a_1)+(a_{n-1},\ldots,a_1) \Big]+(a_1,\ldots,a_{n-1})b_{n}(a_{n},\ldots,a_1) \\ &=(a_1,\ldots,b_{n-1})(a_{n},b_{n},\ldots,a_1) + (a_1,\ldots,a_{n-1})b_{n}(a_{n},\ldots,a_1) \\ &=(a_1,\ldots,b_{n})(a_{n},\ldots,a_1)\,. \\[-38pt] \end{align*} $$

3.2 The result

We consider the $(2n)$ -th Euler continuants

(3.7) $$ \begin{align} \begin{aligned} (a_1,b_1,a_2,b_2,\dots,a_n,b_n)&\in e_2(\Bbbk\overline{\Gamma}_n)e_2\, , \\ (b_n,a_n,b_{n-1},a_{n-1},\dots,b_1,a_1)&\in e_1(\Bbbk\overline{\Gamma}_n)e_1\, , \end{aligned} \end{align} $$

and we form the Boalch algebra $\mathcal {B}(\Gamma _n)$ as the path algebra $\Bbbk \overline {\Gamma }_n$ with $(a_1,\dots ,b_n)$ and $ (b_n,\dots ,a_1)$ inverted. As in [Reference Van den Bergh41, §6.5], this means that we introduce elements $A,B$ such that $A=Ae_2=e_2A$ , $B=Be_1=e_1B$ and $A(a_1,\dots ,b_n)=(a_1,\dots ,b_n)A=e_2$ and $B(b_n,\dots ,a_1)=(b_n,\dots ,a_1)B=e_1$ . From now on, we use the notation $(a_1,\dots ,b_n)^{-1}$ and $(b_n,\dots ,a_1)^{-1}$ for A and B, respectively. This entails that $e_1+(a_1,\dots ,b_n)$ and $e_2+(b_n,\dots ,a_1)$ are invertible in $\mathcal {B}(\Gamma _n)$ in the usual sense.

Next, we follow Boalch [Reference Boalch15, Remark 17] and define the fission algebra

(3.8) $$ \begin{align} \mathcal{F}^q(\Gamma_n):=\mathcal{B}(\Gamma_n)/R^q\,, \quad q=(q_1,q_2)\in (\Bbbk^{\times})^2\,, \end{align} $$

where $R^q$ is the ideal generated by the two relations

(3.9) $$ \begin{align} (b_n,\dots,a_1)^{-1}=q_1 e_1\,, \quad (a_1,\dots,b_n)=q_2e_2\,. \end{align} $$

Note that $\mathcal {B}(\Gamma _n)$ is a generalisation of $\mathcal {B}(\Gamma _1)=\mathcal {A}(\Gamma _1)$ introduced in § 2.4, while $\mathcal {F}^q(\Gamma _n)$ is a generalisation of the multiplicative preprojective algebra $\mathcal {F}^q(\Gamma _1)=\Lambda ^q(\Gamma _1)$ of [Reference Crawley-Boevey and Shaw25].

Finally, let $B=\Bbbk e_1\oplus \Bbbk e_2$ , and we introduce a B-linear double bracket $\{\mkern -6mu\{-,-\}\mkern -6mu\}$ on the arrows of $\overline {\Gamma }_n$ (seen as generators of $\mathcal {B}(\Gamma _n)$ ) as

(3.10a) $$ \begin{align} \{\mkern-6mu\{a_i,a_j\}\mkern-6mu\}&= \begin{cases} -\frac{1}{2}\big(a_i\otimes a_j+a_j\otimes a_i\big)\,, &\mbox{if } i<j \\ \;\;\; 0\, , &\mbox{if } i=j \\ \;\;\; \frac{1}{2}\big(a_i\otimes a_j+a_j\otimes a_i\big), & \mbox{if } i>j \end{cases}\,; \end{align} $$

(3.10b) $$ \begin{align} \{\mkern-6mu\{b_i,b_j\}\mkern-6mu\}&= \begin{cases} -\frac{1}{2}\big(b_i\otimes b_j+b_j\otimes b_i\big)\,, &\mbox{if } i<j \\ \;\;\; 0\,, &\mbox{if } i=j \\ \;\;\; \frac{1}{2}\big(b_i\otimes b_j+b_j\otimes b_i\big)\,, & \mbox{if } i>j \end{cases}\,; \end{align} $$

(3.10c) $$ \begin{align} \{\mkern-6mu\{a_i,b_j\}\mkern-6mu\}&= \begin{cases} \;\;\; \frac{1}{2}\big(e_1\otimes a_ib_j+b_ja_i\otimes e_2\big)\,, &\mbox{if } i<j \\ \;\;\; \frac{1}{2}\big(b_ia_i\otimes e_2+e_1\otimes a_ib_i\big)+e_1\otimes e_2\,, &\mbox{if } i=j \\ - \frac{1}{2}\big(b_ja_{i}\otimes e_2+e_1\otimes a_{i}b_j\big)-\delta_{i-j,1}(e_1\otimes e_2)\,, &\mbox{if } i>j \end{cases}\,; \end{align} $$

(3.10d) $$ \begin{align} \{\mkern-6mu\{b_i,a_j\}\mkern-6mu\}&= \begin{cases} \;\;\; \frac{1}{2}\big(e_2\otimes b_ia_j+ a_jb_i\otimes e_1\big)+\delta_{j-i,1}(e_2\otimes e_1)\, , &\mbox{if } i<j \\ - \frac{1}{2}\big(e_2\otimes b_ia_i+ a_ib_i\otimes e_1\big)-e_2\otimes e_1\,, &\mbox{if } i=j \\ -\frac{1}{2}\big(a_jb_i\otimes e_1+ e_2\otimes b_ia_j)\,, &\mbox{if } i>j \end{cases}\,; \end{align} $$

where $i,j\in \{1,\dots ,n\}$ , and we extend $\{\mkern -6mu\{-,-\}\mkern -6mu\}$ to $\mathcal {B}(\Gamma _n)$ by the Leibniz rules (2.5) and (2.6).

It is worth noting how the double bracket can be determined on the inverses of the elements (3.7) so that it is is uniquely defined on $\mathcal {B}(\Gamma _n)$ and not only on $\Bbbk \overline {\Gamma }_n$ . By B-linearity $\{\mkern -6mu\{c,(b_n,\ldots ,a_1)(b_n,\ldots ,a_1)^{-1}\}\mkern -6mu\}=\{\mkern -6mu\{c,e_1\}\mkern -6mu\}=0$ for any $c\in \mathcal {B}(\Gamma _n)$ , hence by equation (2.5)

$$ \begin{align*} \{\mkern-6mu\{c,(b_n,\ldots,a_1)^{-1}\}\mkern-6mu\}=-(b_n,\ldots,a_1)^{-1} \{\mkern-6mu\{c,(b_n,\ldots,a_1)\}\mkern-6mu\} (b_n,\ldots,a_1)^{-1}\,. \end{align*} $$

A similar relation holds for $\{\mkern -6mu\{c,(a_1,\ldots ,b_n)^{-1}\}\mkern -6mu\}$ .

Theorem 3.4. Given an integer $n\geq 1$ , let $\overline {\Gamma }_n$ be the quiver with vertices $\{1,2\}$ and arrows $\{a_i,b_i\mid 1\leq i\leq n\}$ such that $a_i=e_2a_ie_1$ and $b_i=e_1b_ie_2$ . Let $B:=\Bbbk e_1\oplus \Bbbk e_2$ and $\Bbbk \overline {\Gamma }_n$ be the path algebra of $\overline {\Gamma }_n$ over B. Finally, if $(a_1,\dots ,b_n)$ and $ (b_n,\dots ,a_1)$ are $(2n)$ -th Euler continuants, let $\mathcal {B}(\Gamma _n)$ denote $\Bbbk \overline {\Gamma }_n$ with $(a_1,\dots ,b_n)$ and $ (b_n,\dots ,a_1)$ inverted.

Let $\{\mkern -6mu\{-,-\}\mkern -6mu\}$ be the B-linear double bracket defined on $\mathcal {B}(\Gamma _n)$ by equation (3.10). In addition, we introduce the following element of $\mathcal {B}(\Gamma _n)$

(3.11) $$ \begin{align} \Phi:= (a_1,\dots,b_n) + (b_n,\dots,a_1)^{-1}. \end{align} $$

Then the triple $\big (\mathcal {B}(\Gamma _n),\{\mkern -6mu\{-,-\}\mkern -6mu\},\Phi \big )$ is a Hamiltonian double quasi-Poisson algebra.

We postpone the proof of Theorem 3.4 to Section 4. We remark that, in the case $n=1$ where $\mathcal {B}(\Gamma _1)=\mathcal {A}(\Gamma _1)$ , Theorem 3.4 recovers Van den Bergh’s double quasi-Poisson bracket described in Theorem 2.12(i). Furthermore, using Proposition 2.10, we can obtain the following generalisation of Theorem 2.12(ii).

Using the Kontsevich–Rosenberg principle, Theorem 3.4 turns the representation space $\operatorname {Rep}\big (\mathcal {B}(\Gamma _n),(d_1,d_2)\big )$ into a Hamiltonian quasi-Poisson space for the canonical action of $H:=\operatorname {GL}(d_1)\times \operatorname {GL}(d_2)$ . Similarly, Corollary 3.5 induces a Poisson structure on the reduced spaces $\operatorname {Rep}\big (\mathcal {F}^q(\Gamma _n),(d_1,d_2)\big )/\!/H$ , which coincides with the structure obtained by quasi-Hamiltonian reduction; we refer to [Reference Fairon and Fernández31, §2.3] for a review of this general construction based on [Reference Crawley-Boevey23, Reference Van den Bergh41].

As pointed out in the introduction, if $\Bbbk =\mathbb {C}$ and $(d_1,d_2)=(1,1)$ , we are in the situation of the Sibuya spaces [Reference Sibuya40] whose symplectic geometry was studied by Boalch in [Reference Boalch15]. In § 3.4, we give the precise form of the bracket obtained from equation (3.10) in this specific case, and we explain its connection to the Flaschka–Newell Poisson bracket [Reference Flaschka and Newell33]. More generally, if $\Bbbk =\mathbb {C}$ and the dimension vector $(d_1,d_2)$ is arbitrary, we get that the Poisson structure induced on any reduced space $\operatorname {Rep}\big (\mathcal {F}^q(\Gamma _n),(d_1,d_2)\big )/\!/H$ corresponds (on its smooth locus) to the symplectic structure defined by Boalch. This follows from Proposition 5.16 presented in §5.6, where we prove that the Hamiltonian quasi-Poisson structure that we obtain on the representation space $\operatorname {Rep}\big (\mathcal {B}(\Gamma _n),(d_1,d_2)\big )$ corresponds to the quasi-Hamiltonian structure introduced by Boalch [Reference Boalch9, 13] and conveniently studied by Paluba [Reference Paluba38, Ch. 5].

3.3 The noncommutative bivector

In the case of Theorem 3.4, it is possible to write explicitly the bivector defining the double bracket on the Boalch algebra $\mathcal {B}(\Gamma _n)$ through Proposition 2.3. As in §2.1.2, we consider the double derivations

(3.12) $$ \begin{align} \frac{\partial}{\partial a_i}\,,\,\quad \frac{\partial}{\partial b_i} \,, \quad i=1,\ldots,n\,, \end{align} $$

which are defined for $i,k=1,\ldots ,n$ by

(3.13) $$ \begin{align} \frac{\partial a_k}{\partial a_i}=\delta_{ik} \, e_2\otimes e_1\,,\quad \frac{\partial b_k}{\partial a_i}=0\,, \quad \frac{\partial b_k}{\partial b_i}=\delta_{ik}\, e_1\otimes e_2 \,,\,\,\,\frac{\partial a_k}{\partial b_i}=0 \,. \end{align} $$

Then the following result is a direct application of the equivalence given by Proposition 2.3 since path algebras of quivers are smooth.

Proposition 3.6. The double quasi-Poisson bracket from Theorem 3.4 is associated with the following bivector:

(3.14) $$ \begin{align} \begin{aligned} {\mathrm{P}_n}&=\frac12 \sum_{i>j} \left[ \frac{\partial}{\partial a_i}a_i \frac{\partial}{\partial a_j} a_j + \frac{\partial}{\partial a_i} a_j \frac{\partial}{\partial a_j} a_i \right] +\frac12 \sum_{i>j} \left[ \frac{\partial}{\partial b_i}b_i \frac{\partial}{\partial b_j} b_j + \frac{\partial}{\partial b_i} b_j \frac{\partial}{\partial b_j} b_i \right] \\ &\quad+\frac12 \sum_{i\neq j} \operatorname{sgn}(j-i) \left[ \frac{\partial}{\partial a_i}a_i b_j \frac{\partial}{\partial b_j} + a_i \frac{\partial}{\partial a_i} \frac{\partial}{\partial b_j} b_j \right] -\sum_{i=2}^n \frac{\partial}{\partial a_i} \frac{\partial}{\partial b_{i-1}} \\ &\quad +\frac12 \sum_{i=1}^n\left[ \frac{\partial}{\partial a_i}a_i b_i \frac{\partial}{\partial b_i} + a_i\frac{\partial}{\partial a_i} \frac{\partial}{\partial b_i} b_i \right] +\sum_{i=1}^n \frac{\partial}{\partial a_i} \frac{\partial}{\partial b_{i}}\,. \end{aligned} \end{align} $$

Proof. It suffices to show that the double bracket $\{\mkern -6mu\{-,-\}\mkern -6mu\}_{\mathrm {P}_n}$ defined through equation (2.7) using ${\mathrm {P}_n}$ gives equations (3.10a)–(3.10c) when evaluated on generators. If $i>j$ , we note that the only terms contributing to $\{\mkern -6mu\{a_i,a_j\}\mkern -6mu\}_{\mathrm {P}_n}$ come from the first term of equation (3.14) as

$$ \begin{align*} \begin{aligned} \frac12 \left(\frac{\partial}{\partial a_i}a_i\right) \left( \frac{\partial}{\partial a_j} a_j\right) + \frac12 \left(\frac{\partial}{\partial a_i} a_j\right) \left(\frac{\partial}{\partial a_j} a_i \right)\,, \end{aligned} \end{align*} $$

so that equation (2.7) yields

(3.15) $$ \begin{align} \begin{aligned} \{\mkern-6mu\{a_i,a_j\}\mkern-6mu\}_{\mathrm{P}_n}= \frac12\big(a_j e_1 \otimes a_i e_1 - 0\big) + \frac12\big(a_i e_1 \otimes a_j e_1 - 0\big) = \frac12\Big(a_j \otimes a_i + a_i \otimes a_j\Big) \,. \end{aligned} \end{align} $$

Clearly, $\{\mkern -6mu\{a_i,a_i\}\mkern -6mu\}_{\mathrm {P}_n}=0$ , so by cyclic antisymmetry (2.4) we get that $\{\mkern -6mu\{a_i,a_j\}\mkern -6mu\}_{\mathrm {P}_n}$ coincides with equation (3.10a) for all $1\leq i,j \leq n$ . We can show in the same way that $\{\mkern -6mu\{b_i,b_j\}\mkern -6mu\}_{\mathrm {P}_n}$ coincides with equation (3.10b) for all indices.

We now compute $\{\mkern -6mu\{a_i,b_j\}\mkern -6mu\}_{\mathrm {P}_n}$ . If $i<j$ , the only terms that contribute come from the third term of equation (3.14) as

$$ \begin{align*} \begin{aligned} \frac12 \left(\frac{\partial}{\partial a_i}a_i\right) \left(b_j \frac{\partial}{\partial b_j}\right) + \frac12 \left(a_i \frac{\partial}{\partial a_i} \right) \left( \frac{\partial}{\partial b_j} b_j \right)\,, \end{aligned} \end{align*} $$

and we get from equation (2.7) that

(3.16) $$ \begin{align} \begin{aligned} \{\mkern-6mu\{a_i,b_j\}\mkern-6mu\}_{\mathrm{P}_n}= \frac12 e_1 e_1 \otimes a_i b_j + \frac12 b_j a_i \otimes e_2 e_2 = \frac12(e_1 \otimes a_i b_j + b_j a_i \otimes e_2) \,. \end{aligned} \end{align} $$

If $i=j$ , we only need the following terms from the last line in equation (3.14)

$$ \begin{align*} \begin{aligned} \frac12 \left(\frac{\partial}{\partial a_i}a_i\right) \left(b_i \frac{\partial}{\partial b_i}\right) + \frac12 \left(a_i \frac{\partial}{\partial a_i} \right) \left( \frac{\partial}{\partial b_i} b_i \right) + \left( \frac{\partial}{\partial a_i} \right) \left( \frac{\partial}{\partial b_i} \right)\,, \end{aligned} \end{align*} $$

and equation (2.7) yields

(3.17) $$ \begin{align} \begin{aligned} \{\mkern-6mu\{a_i,b_i\}\mkern-6mu\}_{\mathrm{P}_n}= \frac12\Big(e_1 \otimes a_i b_i + b_i a_i \otimes e_2\Big) + e_1 \otimes e_2 \,. \end{aligned} \end{align} $$

In the case $i>j$ , the contributing terms come from the third and fourth sums in equation (3.14) as

$$ \begin{align*} \begin{aligned} -\frac12 \left(\frac{\partial}{\partial a_i}a_i\right) \left(b_j \frac{\partial}{\partial b_j}\right) - \frac12 \left(a_i \frac{\partial}{\partial a_i} \right) \left( \frac{\partial}{\partial b_j} b_j \right) -\delta_{i-j,1} \left( \frac{\partial}{\partial a_i} \right) \left( \frac{\partial}{\partial b_j} \right)\,, \end{aligned}\\[-15pt] \end{align*} $$

and equation (2.7) yields

(3.18) $$ \begin{align} \begin{aligned} \{\mkern-6mu\{a_i,b_j\}\mkern-6mu\}_{\mathrm{P}_n}= -\frac12\Big(e_1 \otimes a_i b_j + b_j a_i \otimes e_2\Big) -\delta_{i-j,1} e_1 \otimes e_2 \,. \end{aligned}\\[-15pt]\nonumber \end{align} $$

Gathering the three cases, we get that $\{\mkern -6mu\{a_i,b_j\}\mkern -6mu\}_{\mathrm {P}_n}$ coincides with equation (3.10c) for all indices.

Remark 3.7. In the case $n=1$ , the noncommutative bivector ${\mathrm {P}}_1$ can directly be compared with the one of Van den Bergh [Reference Van den Bergh41, Theorem 6.5.1]. Setting $a:=a_1$ , $b:=b_1$ , we have that modulo graded-commutator ${\mathrm {P}}_1$ equals

(3.19) $$ \begin{align} {\mathrm{P}}_{\operatorname{VdB}}:=\frac12(1+ba)\frac{\partial}{\partial a}\frac{\partial}{\partial b} -\frac12 (1+ab)\frac{\partial}{\partial b} \frac{\partial}{\partial a}\,,\\[-15pt]\nonumber \end{align} $$

which was constructed by Van den Bergh. It is an easy exercise (see [Reference Van den Bergh41, §4.1] in full generalities) to check that the double bracket induced by an element $Q\in (D_BA)_2$ by Proposition 2.3 only depends on Q modulo graded commutators. Thus, ${\mathrm {P}}_1$ and ${\mathrm {P}}_{\operatorname {VdB}}$ induce the same double quasi-Poisson bracket on $\mathcal {B}(\Gamma _1)= \mathcal {A}(\Gamma _1)$ .

3.4 The Flaschka–Newell Poisson bracket

The representation space of dimension $(1,1)$ , denoted $\operatorname {Rep}\big (\mathcal {B}(\Gamma _n),(1,1)\big )$ , is parametrised by assigning $A_i,B_i\in \mathbb {C}$ to the generators $a_i,b_i$ for each $ 1\leq i\leq n$ , where we require the invertibility of the (scalar) matrices representing the two elements (3.7). Denoting by $A_i,B_i$ the corresponding evaluation functions on $\operatorname {Rep}\big (\mathcal {B}(\Gamma _n),(1,1)\big )$ , this gives precisely the parametrisation (1.9) of the space $\widehat {\mathcal {B}}^{n+1}$ from the introduction. It is endowed with the action of $\mathbb {C}^{\times } \times \mathbb {C}^{\times }$ through

$$ \begin{align*} t\cdot \{A_i,B_i\mid 1\leq i\leq n\} = \{t_2A_i t_1^{-1},t_1 B_i t_2^{-1}\mid 1\leq i\leq n\}\,, \quad t=(t_1,t_2)\in \mathbb{C}^{\times} \times \mathbb{C}^{\times} \,.\\[-15pt] \end{align*} $$

This is a Hamiltonian quasi-Poisson space using Theorem 3.4 and [Reference Van den Bergh41, Proposition 7.13.2], in application of the Kontsevich–Rosenberg principle. Using [Reference Van den Bergh41, Proposition 7.5.1], the quasi-Poisson bracket is given byFootnote ⁶

(3.20a) $$ \begin{align} \{ A_i,A_j \}=&\operatorname{sgn}(i-j) A_iA_j\,, \quad \{ B_i,B_j \}=\operatorname{sgn}(i-j)B_iB_j\,, \end{align} $$

(3.20b) $$ \begin{align} \{ A_i,B_j \}=&\left\{ \begin{array}{cc} A_iB_j \,, &\mbox{if }i<j\\ A_iB_i+1\,, &\mbox{if }i=j\\ -A_iB_j-\delta_{i-j,1}\,, &\mbox{if }i>j \end{array} \right.\,.\\[-15pt]\nonumber \end{align} $$

The corresponding moment map $\widehat {\mu }$ is given by equation (1.10) and, noting that the variables $A_i,B_i$ commute, it can be simply written as

(3.21) $$ \begin{align} \widehat{\mu}=(\lambda^{-1},\lambda)\,, \quad \lambda:=(A_1,B_1,\ldots,A_n,B_n)=(B_n,A_n,\ldots,B_1,A_1)\,.\\[-15pt]\nonumber \end{align} $$

Since $\lambda $ represents the component $\Phi _2$ of the (noncommutative) moment map from Theorem 3.4, we have from [Reference Van den Bergh41, Proposition 7.5.1] and equations (4.12) and (4.13) that

(3.22) $$ \begin{align} \{ \lambda,A_i \}=-\lambda A_i\,, \quad \{ \lambda,B_i \}=\lambda B_i\,. \end{align} $$

Let us note that, since the group $\mathbb {C}^{\times } \times \mathbb {C}^{\times }$ acting on $\widehat {\mathcal {B}}^{n+1}$ is abelian, the quasi-Poisson bracket is in fact a Poisson bracket. This will be important in view of the following change of variables, motivated by a standard parametrisation connected to Stokes matrices [Reference Bertola and Tarricone3, Reference Boalch15]. We set

(3.23) $$ \begin{align} A_i=s_{2n+3-2i}\,, \quad B_i=s_{2n+2-2i}\,, \end{align} $$

for $s_j\in \mathbb {C}$ with $2\leq j\leq 2n+1$ . (The parameters $s_k$ with odd/even indices are the $A_i$ / $B_i$ respectively, and they appear with decreasing indices.) We also denote by $s_j$ the corresponding evaluation function on $\widehat {\mathcal {B}}^{n+1}$ . Then, using this change of coordinates, the moment map $\lambda $ in equation (3.21) becomes $\lambda =(s_2,\ldots ,s_{2n+1})$ . It is an easy exercise using equation (3.20a) to see that we can write

$$ \begin{align*} \{ s_k,s_l \}=s_k s_l \quad \text{ for }k<l \text{ both odd or both even}\,. \end{align*} $$

We can also note that equation (3.20b) yields

$$ \begin{align*} \{ s_k,s_l \}=\left\{ \begin{array}{ll} s_ks_l \,, &\mbox{if } k>l\\ s_ks_l+1\,, &\mbox{if }k=l+1\\ -s_ks_l-1\,, &\mbox{if } k=l-1\\ -s_ks_l\,, &\mbox{if }k<l-1 \end{array} \right. \quad \text{ for } k \text{ odd and } l \text{ even}\,. \end{align*} $$

This is equivalent to

$$ \begin{align*} \{ s_k,s_l \}=\left\{ \begin{array}{ll} -s_ks_l\,, &\mbox{if }k<l\\ -s_ks_l-1\,, &\mbox{if }k=l-1\\ s_ks_l+1\,, &\mbox{if }k=l+1\\ s_ks_l\,, &\mbox{if }k>l+1 \end{array} \right. \quad \text{ for } k \text{ even and } l \text{ odd}\,. \end{align*} $$

Gathering these identities with the moment map identity (3.22), we get that the Poisson bracket is completely determined by

(3.24a) $$ \begin{align} \{ s_k,s_l \}&=-\delta_{k,l-1}+(-1)^{k-l}s_ks_l\,, \quad \text{ for }1<k<l<2n+2\,, \end{align} $$

(3.24b) $$ \begin{align} \{ s_k,\lambda \}&=(-1)^{k+1}s_k\lambda\,. \end{align} $$

Using equation (1.7) and the subsequent Gauss decomposition, we note that we can equivalently parametrise $\widehat {\mathcal {B}}^{n+1}$ in terms of the elements $\{s_j \mid 2\leq j\leq 2n+1\}$ such that

(3.25) $$ \begin{align} \begin{aligned} &\left(\begin{array}{cc} 1&0\\s_2&1 \end{array} \right) \left(\begin{array}{cc} 1&s_3\\0&1 \end{array} \right) \ldots \left(\begin{array}{cc} 1&0\\s_{2n}&1 \end{array} \right) \left(\begin{array}{cc} 1&s_{2n+1}\\0&1 \end{array} \right) \\ =& \left(\begin{array}{cc} 1&(s_3,\ldots,s_{2n+1})\lambda^{-1}\\0&1 \end{array} \right) \left(\begin{array}{cc} \lambda^{-1}&0\\0&\lambda \end{array} \right) \left(\begin{array}{cc} 1&0\\ \lambda^{-1} (s_2,\ldots,s_{2n})&1 \end{array} \right)\,. \end{aligned} \end{align} $$

Consequently, $\widehat {\mathcal {B}}^{n+1}$ can be written as the following subspace of $\mathbb {C}^{2n+2}\times \mathbb {C}^{\times }$ :

(3.26) $$ \begin{align} \left\{ \left(\begin{array}{cc} 1&s_1\\0&1 \end{array} \right)\left(\begin{array}{cc} 1&0\\s_2&1 \end{array} \right) \ldots \left(\begin{array}{cc} 1&s_{2n+1}\\0&1 \end{array} \right) \left(\begin{array}{cc} 1&0\\s_{2n+2}&1 \end{array} \right) \left(\begin{array}{cc} \lambda&0\\0&\lambda^{-1} \end{array} \right) = \operatorname{Id}_2 \right\}\,. \end{align} $$

Indeed, we recover equation (3.25) from equation (3.26) because the Gauss decomposition yields

(3.27) $$ \begin{align} s_1=- (s_3,\ldots,s_{2n+1})\lambda^{-1}\,, \quad s_{2n+2}=-\lambda^{-1} (s_2,\ldots,s_{2n})\,. \end{align} $$

Using the description of $\widehat {\mathcal {B}}^{n+1}$ given in equation (3.26), Bertola and Tarricone [Reference Bertola and Tarricone3] recently wrote the following Poisson bracket

(3.28a) $$ \begin{align} \{ s_k,s_l \}_{FN}=&\delta_{k,l-1}-\frac{\delta_{k,1}\delta_{l,2n+2}}{\lambda^2}+(-1)^{k-l+1}s_ks_l\,, \quad \text{ for }1\leq k<l\leq 2n+2\,, \end{align} $$

(3.28b) $$ \begin{align} \{ s_k,\lambda \}_{FN}=&(-1)^{k}s_k\lambda\,, \end{align} $$

which was originally introduced by Flaschka and Newell [Reference Flaschka and Newell33]. By comparing the expressions (3.24a) and (3.28a) on the generators $\{s_j \mid 2\leq j\leq 2n+1\}$ , we directly see that the double quasi-Poisson bracket from Theorem 3.4 induces the Flaschka–Newell Poisson bracket on the representation space $\operatorname {Rep}\big (\mathcal {B}(\Gamma _n),(1,1)\big )$ (up to an irrelevant factor). In particular, equation (3.28b) is nothing else than the moment map condition.

4.1 Preparation for the proof

In this subsection we shall prove two results that we will extensively use in the proof of Theorem 3.4 below. We recall $(a_n,b_n)=e_2+a_nb_n$ , and $\ast $ denotes the inner bimodule multiplication from §2.1.1.

Proof. To prove (i), if $1\leq i\leq n-1$ , by equations (2.6), (3.10a) and (3.10d), we obtain

$$ \begin{align*} \{\mkern-6mu\{(a_n,b_n),a_i\}\mkern-6mu\}&=a_n*\{\mkern-6mu\{b_n,a_i\}\mkern-6mu\}+\{\mkern-6mu\{a_n,a_i\}\mkern-6mu\}*b_n \\ &=-a_n*\Big(\frac{1}{2}\big(a_ib_n\otimes e_1+e_2\otimes b_na_i\big)\Big)+\Big(\frac{1}{2}\big(a_n\otimes a_i+a_i\otimes a_n\big)\Big)*b_n \\ &=\frac{1}{2}a_nb_n\otimes a_i-\frac{1}{2}e_2\otimes a_nb_na_i =\frac{1}{2}\Big((a_n,b_n)\otimes a_i-e_2\otimes (a_n,b_n)a_i\Big)\,, \end{align*} $$

where we used $(a_n,b_n)=a_nb_n+e_2$ . Similarly, in (ii), the reader can prove the identity $2\{\mkern -6mu\{(a_n,b_n),b_i\}\mkern -6mu\}=b_i\otimes (a_n,b_n)-b_i(a_n,b_n)\otimes e_2 $ , for $1\leq i\leq n-2$ .

Now, to prove the second formula in (i), since $\{\mkern -6mu\{a_n,a_n\}\mkern -6mu\}=0$ , by equation (3.10d) we have

$$ \begin{align*} \{\mkern-6mu\{(a_n,b_n),a_n\}\mkern-6mu\}&=a_n*\{\mkern-6mu\{b_n,a_n\}\mkern-6mu\} \\ &=-a_n*\Big(\frac{1}{2}\big(e_2\otimes b_na_n+a_nb_n\otimes e_2\big)+e_2\otimes e_1\Big) \\ &=-\frac{1}{2}\Big( (a_n,b_n)\otimes a_n+e_2\otimes (a_n,b_n)a_n\Big)\, , \end{align*} $$

where we used again that $(a_n,b_n)=a_nb_n+e_2$ . Similarly, the reader can check that $2\{\mkern -6mu\{(a_n,b_n),b_n\}\mkern -6mu\}=b_n\otimes (a_n,b_n)+b_n(a_n,b_n)\otimes e_2$ holds in (ii).

Finally, we prove the remaining identity in (ii). By equations (3.10b) and (3.10c),

$$ \begin{align*} \{\mkern-6mu\{(&a_n,b_n),b_{n-1}\}\mkern-6mu\}=a_n*\{\mkern-6mu\{b_n,b_{n-1}\}\mkern-6mu\}+\{\mkern-6mu\{a_n,b_{n-1}\}\mkern-6mu\}*b_n \\ &=a_n*\Big(\frac{1}{2}\big(b_n\otimes b_{n-1}+b_{n-1}\otimes b_n\big)\Big) -\Big(\frac{1}{2}\big(b_{n-1}a_n\otimes e_2+e_1\otimes a_nb_{n-1}\big)+e_1\otimes e_2\Big)*b_n \\ &=\frac{1}{2}b_{n-1}\otimes a_nb_n-\frac{1}{2}b_{n-1}a_nb_n\otimes e_2-b_n\otimes e_2 \\ &=\frac{1}{2}\Big(b_{n-1}\otimes (a_n,b_n)-b_{n-1}(a_n,b_n)\otimes e_2\Big)-b_n\otimes e_2\,.\\[-45pt] \end{align*} $$

Proof. We will prove these formulas by strong induction. To prove (i), we start noting that, by equations (3.10b) and (3.10c),

(4.1) $$ \begin{align} \begin{aligned} \{\mkern-6mu\{(a_i,b_i),b_j\}\mkern-6mu\}&=a_i*\{\mkern-6mu\{b_i,b_j\}\mkern-6mu\}+\{\mkern-6mu\{a_i,b_j\}\mkern-6mu\}*b_i \\ &=\frac{1}{2}b_ja_ib_i\otimes e_2-\frac{1}{2}b_j\otimes a_ib_i =\frac{1}{2}\Big(b_j(a_i,b_i)\otimes e_2-b_j\otimes (a_i,b_i)\Big)\,. \end{aligned} \end{align} $$

Then the base case arises when $i=1$ . Now, if $i\geq 2$ , by equation (3.6), we have

$$ \begin{align*} \{\mkern-6mu\{(a_1,\dots, b_i),\,b_j\}\mkern-6mu\}&= \left\{\!\!\!\left\{\Big((a_1,\dots ,b_{i-1})(a_i,b_i)+\sum^i_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_i\Big]+a_1b_i\Big),b_j\right\}\!\!\!\right\} \\ &=(a_1,\dots,b_{i-1})*\{\mkern-6mu\{(a_i,b_i),b_j\}\mkern-6mu\}+\{\mkern-6mu\{(a_1,\dots,b_{i-1}),b_j\}\mkern-6mu\}*(a_i,b_i) \\ &\quad +\sum^i_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}*\{\mkern-6mu\{b_i,b_j\}\mkern-6mu\}+(a_1,\dots,b_{\ell-2})*\{\mkern-6mu\{a_{\ell-1},b_j\}\mkern-6mu\}*b_i\Big] \\ &\quad +\sum^i_{\ell=3}\Big[ \{\mkern-6mu\{(a_1,\dots,b_{\ell-2}),b_j\}\mkern-6mu\}*a_{\ell-1}b_i\Big]+a_1*\{\mkern-6mu\{b_i,b_j\}\mkern-6mu\}+\{\mkern-6mu\{a_1,b_j\}\mkern-6mu\}*b_i\,. \end{align*} $$

If we apply equation (4.1) in the first term and the inductive hypothesis in the second and fifth terms, we obtain

$$ \begin{align*} &=\frac{1}{2}b_j(a_i,b_i)\otimes (a_1,\dots,b_{i-1})-\frac{1}{2}b_j\otimes (a_1,\dots,b_{i-1})(a_i,b_i) \\ &\quad +\frac{1}{2}\Big(b_j(a_1,\dots,b_{i-1})(a_i,b_i)\otimes e_2-b_j(a_i,b_i)\otimes (a_1,\dots,b_{i-1})\Big) \\ &\quad -\frac{1}{2}\sum^i_{\ell=3}\Big[b_i\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_j+b_j\otimes (a_1,\dots ,b_{\ell-2})a_{\ell-1}b_i\Big] \\ &\quad+ \frac{1}{2}\sum^i_{\ell=3}\Big[b_i\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_j+b_ja_{\ell-1}b_i\otimes (a_1,\dots,b_{\ell-2})\Big] \\ &\quad+ \frac{1}{2}\sum^i_{\ell=3}\Big[b_j (a_1,\dots,b_{\ell-2})a_{\ell-1}b_i\otimes e_2-b_ja_{\ell-1}b_i\otimes (a_1,\dots,b_{\ell-2})\Big] \\ &\quad -\frac{1}{2}\Big(b_i\otimes a_1b_j+b_j\otimes a_1b_i\Big)+\frac{1}{2}\Big(b_i\otimes a_1b_j+b_ja_1b_i\otimes e_2\Big) \\ &=\frac{1}{2}\Big(b_j (a_1,\dots,b_{i-1})(a_i,b_i)\otimes e_2+\sum^i_{\ell=3}\Big[b_j(a_1,\dots ,b_{\ell-2})a_{\ell-1}b_i\otimes e_2\Big]+b_ja_1b_i\otimes e_2\Big) \\ &\quad -\frac{1}{2}\Big(b_j\otimes (a_1,\dots,b_{i-1})(a_i,b_i)+\sum^i_{\ell=3}\Big[b_j\otimes(a_1,\dots ,b_{\ell-2})a_{\ell-1}b_i\Big]+b_j\otimes a_1b_i\Big) \\ &=\frac{1}{2}\Big(b_j(a_1,\dots,b_i)\otimes e_2-b_j\otimes (a_1,\dots, b_i)\Big)\,, \end{align*} $$

where we used equation (3.6).

The proof of the formula in (ii) is pretty similar to (i), so it is left to the reader. Nevertheless, the formula in (iii) is slightly more delicate, so we prove it now. If $1\leq i\leq n-1$ , note that by equations (3.10a) and (3.10d), we get

(4.2) $$ \begin{align} \{\mkern-6mu\{(a_i,b_i),a_{i+1}\}\mkern-6mu\}&=a_i*\{\mkern-6mu\{b_i,a_{i+1}\}\mkern-6mu\}+\{\mkern-6mu\{a_i,a_{i+1}\}\mkern-6mu\}*b_{i} \nonumber\\ &=a_i*\Big(\frac{1}{2}\big(e_2\otimes b_ia_{i+1}+a_{i+1}b_i\otimes e_1\big)+e_2\otimes e_1\Big)-\frac{1}{2}\Big(a_i\otimes a_{i+1}+a_{i+1}\otimes a_i\Big)*b_i \nonumber\\ &=\frac{1}{2}e_2\otimes a_ib_ia_{i+1}+e_2\otimes a_{i}-\frac{1}{2}a_ib_i\otimes a_{i+1} \\ &=\frac{1}{2}\Big(e_2\otimes (a_i,b_i)a_{i+1}-(a_i,b_i)\otimes a_{i+1}\Big)+e_2\otimes (a_i)\,.\nonumber \end{align} $$

In particular, if $i=1$ , we have the base case of (iii).

Next, we address the general case by assuming that $i\geq 2$ . By equation (3.6),

$$ \begin{align*} \{\mkern-6mu\{(a_1,\dots, b_i),a_{i+1}\}\mkern-6mu\}&= \left\{\!\!\!\left\{\Big((a_1,\dots ,b_{i-1})(a_i,b_i)+\sum^i_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_i\Big]+a_1b_i\Big),a_{i+1}\right\}\!\!\!\right\} \\ &=(a_1,\dots, b_{i-1})*\{\mkern-6mu\{(a_i,b_i),a_{i+1}\}\mkern-6mu\}+\{\mkern-6mu\{(a_1,\dots,b_{i-1}),a_{i+1}\}\mkern-6mu\}*(a_i,b_i) \\ &\quad +\sum^i_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}*\{\mkern-6mu\{b_i,a_{i+1}\}\mkern-6mu\}+(a_1,\dots,b_{\ell-2})*\{\mkern-6mu\{a_{\ell-1},a_{i+1}\}\mkern-6mu\}*b_i\Big] \\ &\quad +\sum^i_{\ell=3}\Big[\{\mkern-6mu\{(a_1,\dots,b_{\ell-2}),a_{i+1}\}\mkern-6mu\}*a_{\ell-1}b_i\Big] +a_1*\{\mkern-6mu\{b_i,a_{i+1}\}\mkern-6mu\}+\{\mkern-6mu\{a_1,a_{i+1}\}\mkern-6mu\}*b_i\,. \end{align*} $$

Now, we apply equation (4.2) in the first summand and Lemma 4.2(ii) in the second and fifth summands to obtain

$$ \begin{align*} &=\frac{1}{2}\Big(e_2\otimes (a_1,\dots,b_{i-1})(a_i,b_i)a_{i+1}-(a_i,b_i)\otimes (a_1,\dots,b_{i-1})a_{i+1}\Big)+e_2\otimes (a_1,\dots,b_{i-1})a_i \\ &\; +\frac{1}{2}\Big((a_i,b_i)\otimes (a_1,\dots,b_{i-1})a_{i+1}-(a_1,\dots,b_{i-1})(a_i,b_i)\otimes a_{i+1}\Big) \\ &\; +\sum^i_{\ell=3}\Big[\frac{1}{2}\Big(e_2\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_ia_{i+1}+a_{i+1}b_i\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}\Big)+e_2\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}\Big] \\ &\; -\frac{1}{2}\sum^i_{\ell=3}\Big[a_{\ell-1}b_i\otimes (a_1,\dots,b_{\ell-2})a_{i+1}+a_{i+1}b_i\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}\Big] \\ &\; +\frac{1}{2}\sum^i_{\ell=3}\Big[a_{\ell-1}b_i\otimes (a_1,\dots,b_{\ell-2})a_{i+1}-(a_1,\dots,b_{\ell-2})a_{\ell-1}b_i\otimes a_{i+1}\Big] \\ &\; +\frac{1}{2}\Big(e_2\otimes a_1b_ia_{i+1}+a_{i+1}b_i\otimes a_1\Big)+e_2\otimes a_1 -\frac{1}{2}\Big(a_1b_i\otimes a_{i+1}+a_{i+1}b_i\otimes a_1\Big) \\ &=\frac{1}{2}\Big(e_2\otimes (a_1,\dots,b_{i-1})(a_i,b_i)a_{i+1}+\sum^i_{\ell=3}\Big[e_2\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_ia_{i+1}\Big]+e_2\otimes a_1b_ia_{i+1}\Big) \\ &\;-\frac{1}{2}\Big((a_1,\dots,b_{i-1})(a_i,b_i)\otimes a_{i+1}+\sum^i_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_i\otimes a_{i+1}+a_1b_i\otimes a_{i+1}\Big) \\ &\;+e_2\otimes (a_1,\dots,b_{i-1})a_i+\sum^i_{\ell=3}\Big[e_2\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}\Big]+e_2\otimes a_1 \\ &=\frac{1}{2}\Big(e_2\otimes (a_1,\dots,b_i)a_{i+1}-(a_1,\dots,b_i)\otimes a_{i+1}\Big)+e_2\otimes (a_1,\dots,a_i)\,, \end{align*} $$

where in the last identity we used equations (3.6) and (3.5).

4.2 The double bracket (3.10) is quasi-Poisson

By its definition, it is clear that $\{\mkern -6mu\{-,-\}\mkern -6mu\}$ , as given in equation (3.10), is a B-linear double bracket on $\mathcal {B}(\Gamma _n)$ . In order to prove that the pair $(\mathcal {B}(\Gamma _n),\{\mkern -6mu\{-,-\}\mkern -6mu\})$ is a double quasi-Poisson algebra, we need to show that $\{\mkern -6mu\{-,-\}\mkern -6mu\}$ satisfies equation (2.10) for every triple of generators $\{x_i,y_j,z_k\}$ , with $x_i,y_j,z_k\in \{a_{\ell },b_{\ell }\mid 1\leq \ell \leq n\}$ , by distinguishing every possible case. To do that, we need to attend to the number of a’s and b’s occurring in such a triple. We essentially have four cases:Footnote ⁷ $(a_i,a_j,a_k)$ , $(a_i,a_j,b_k)$ , $(b_i,b_j,a_k)$ and $(b_i,b_j,b_k)$ . Then, we shall keep track of the relations of the indices $i,j,k$ with respect to the natural total order $\leq $ on $\mathbb {N}$ . By cyclicity of the triple bracket without loss of generality, we will assume that the distinct element, if any, is in the third position.

Firstly, we note the following general result.

Lemma 4.3. Let $A_0$ be an algebra over $B_0$ , where $B_0$ contains orthogonal idempotents $e_1,e_2$ . Assume that $A_0$ is endowed with a $B_0$ -linear double bracket $\{\mkern -6mu\{-,-\}\mkern -6mu\}$ and that there exist elements $c_i\in e_2 A_0 e_1$ , $i\in \{1,\ldots ,n\}$ , such that

(4.3) $$ \begin{align} \{\mkern-6mu\{c_i,c_j\}\mkern-6mu\}=\epsilon_{ij} (c_i \otimes c_j + c_j \otimes c_i)\,, \quad i,j\in \{1,\ldots,n\}\,, \end{align} $$

with $\epsilon _{ij}\in \Bbbk $ . Then $\epsilon _{ij}=-\epsilon _{ji}$ , and we have that the quasi-Poisson property (2.10) is satisfied when evaluated on any triples from $\{c_i\}$ if and only if

(4.4) $$ \begin{align} C_{ijk}:= \epsilon_{ij}\epsilon_{jk}+\epsilon_{jk}\epsilon_{ki}+\epsilon_{ki}\epsilon_{ij}=-\frac14\,, \end{align} $$

for all $i,j,k\in \{1,\ldots ,n\}$ with $(i,j,k)\neq (i,i,i)$ , that is, the three indices are not all equal.

Proof. Clearly, the cyclic antisymmetry implies that $\epsilon _{ij}=-\epsilon _{ji}$ . The quasi-Poisson property (2.10) on $c_i,c_j,c_k$ reads:

(4.5) $$ \begin{align} \{\mkern-6mu\{c_i,c_j,c_k\}\mkern-6mu\} =\frac{1}{4}\big(c_k\otimes c_i\otimes c_j-c_i\otimes c_j\otimes c_k\big)\,. \end{align} $$

Now, we can compute that

$$ \begin{align*} \{\mkern-6mu\{c_i,\{\mkern-6mu\{c_j,c_k\}\mkern-6mu\}\}\mkern-6mu\}_L &= \epsilon_{jk} [\epsilon_{ij} (c_i \otimes c_j \otimes c_k + c_j \otimes c_i \otimes c_k) + \epsilon_{ik} (c_i \otimes c_k \otimes c_j + c_k \otimes c_i \otimes c_j) ] \\ \tau_{(123)}\{\mkern-6mu\{c_j,\{\mkern-6mu\{c_k,c_i\}\mkern-6mu\}\}\mkern-6mu\}_L &= \epsilon_{ki} [\epsilon_{jk} (c_i \otimes c_j \otimes c_k + c_i \otimes c_k \otimes c_j) + \epsilon_{ji} (c_k \otimes c_j \otimes c_i + c_k \otimes c_i \otimes c_j) ] \\ \tau_{(132)}\{\mkern-6mu\{c_k,\{\mkern-6mu\{c_i,c_j\}\mkern-6mu\}\}\mkern-6mu\}_L &= \epsilon_{ij} [\epsilon_{ki} (c_i \otimes c_j \otimes c_k + c_k \otimes c_j \otimes c_i) + \epsilon_{kj} (c_j \otimes c_i \otimes c_k + c_k \otimes c_i \otimes c_j) ]. \end{align*} $$

Using the identity $\epsilon _{ij}=-\epsilon _{ji}$ , we obtain

(4.6) $$ \begin{align} \{\mkern-6mu\{c_i,c_j,c_k\}\mkern-6mu\}= C_{ijk} \, (c_i \otimes c_j \otimes c_k - c_k \otimes c_i \otimes c_j)\,, \end{align} $$

which satisfies equation (4.5) if and only if $C_{ijk}=-\frac 14$ when $i,j,k$ are not all equal.

In our case, we note that, if we consider $A_0=\Bbbk \overline {\Gamma }_n$ (or $\mathcal {B}(\Gamma _n)$ directly), $B_0=B$ and the elements $c_i=a_i$ , then equation (3.10a) takes the form (4.3) for

(4.7) $$ \begin{align} \epsilon_{ij}=\frac12 \, \operatorname{sgn}(j-i)\,. \end{align} $$

We can then directly verify that the condition (4.4) holds in the cases

(4.8) $$ \begin{align} i=j=k,\quad i=j \neq k,\quad i<j<k,\quad i<k<j,\quad i>j>k,\quad i>k>j\,. \end{align} $$

Due to the cyclic antisymmetry (2.4), these are the only cases to check, and we obtain from Lemma 4.3 that equation (2.10) holds on any triple $(a_i,a_j, a_k)$ with $i,j,k\in \{1,\dots , n\}$ .

Secondly, we shall show that equation (2.10) holds for every triple $(b_i,b_j,b_k)$ , where $1\leq i,j,k\leq n$ . In fact, this can be obtained by an application of Lemma 4.3, but we shall use a different method that we will employ below. As in the proof of [Reference Van den Bergh41, Theorem 6.5.1], we note the existence of an automorphism of order two on $\Bbbk \overline {\Gamma }_n$ , namely S, given by

(4.9) $$ \begin{align} e_1\longleftrightarrow e_2\,,\quad a_{\ell} \longleftrightarrow b_{n+1-\ell}\,,\quad b_{\ell}\longleftrightarrow a_{n+1-\ell}\,, \end{align} $$

for all $1\leq \ell \leq n$ . By inspection on equation (3.10), we can see that this automorphism has the effect $\{\mkern -6mu\{-,-\}\mkern -6mu\}\mapsto -\{\mkern -6mu\{-,-\}\mkern -6mu\}$ . Moreover, this automorphism has no effect on the triple bracket $\{\mkern -6mu\{-,-,-\}\mkern -6mu\}$ defined by equation (2.8); in other words, $\{\mkern -6mu\{-,-,-\}\mkern -6mu\}\mapsto \{\mkern -6mu\{-,-,-\}\mkern -6mu\}$ . Once we have just proved that the identity $4\{\mkern -6mu\{a_i,a_j,a_k\}\mkern -6mu\}=a_k\otimes a_i\otimes a_j-a_i\otimes a_j\otimes a_k$ is satisfied for all $1\leq i,j,k\leq n$ , we can apply the automorphism S to this identity to obtain

$$\begin{align*}\{\mkern-6mu\{b_{n+1-i},b_{n+1-j},b_{n+1-k}\}\mkern-6mu\}=\frac{1}{4}\Big(b_{n+1-k}\otimes b_{n+1-i}\otimes b_{n+1-j}-b_{n+1-i}\otimes b_{n+1-j}\otimes b_{n+1-k}\Big)\,, \end{align*}$$

for all $1\leq i,j,k\leq n$ . In this way, we can conclude that equation (2.10) also holds for every triple $(b_i,b_j,b_k)$ , where $1\leq i,j,k\leq n$ .

Thirdly, we will prove that the bracket $\{\mkern -6mu\{-,-\}\mkern -6mu\}$ defined in equation (3.10) satisfies equation (2.10) for triples $(a_i,a_j,b_k)$ , where $1\leq i,j,k\leq n$ . To achieve this purpose, we shall distinguish the different cases attending to the different values of the indices $i,j,k$ . To begin with, note that in this case the right-hand side of equation (2.10) reduces to

(4.10) $$ \begin{align} \frac{1}{4}\Big(b_ka_i\otimes a_j\otimes e_2-e_1\otimes a_i\otimes a_jb_k\Big)\,. \end{align} $$

Now, the simplest case occurs when the three indices are equal, namely $i=j=k$ . Then, since $\{\mkern -6mu\{a_i,a_i\}\mkern -6mu\}=0$ , and by equations (3.10a), (3.10c) and (3.10d), at the left-hand side of equation (2.10) we obtain

$$ \begin{align*} \{\mkern-6mu\{a_i,a_i,b_i\}\mkern-6mu\}&=\{\mkern-6mu\{a_i,\{\mkern-6mu\{a_i,b_i\}\mkern-6mu\}\}\mkern-6mu\}_L+\tau_{(123)}\{\mkern-6mu\{a_i,\{\mkern-6mu\{b_i,a_i\}\mkern-6mu\}\}\mkern-6mu\}_L \\ &=\frac{1}{2}\Big(\{\mkern-6mu\{a_i,b_i\}\mkern-6mu\}a_i\otimes e_2\Big)-\frac{\tau_{(123)}}{2}\Big(a_i\{\mkern-6mu\{a_i,b_i\}\mkern-6mu\}\otimes e_1\Big) \\ &=\frac{1}{4}\Big(b_ia_i\otimes a_i\otimes e_2+e_1\otimes a_ib_ia_i\otimes e_2+2e_1\otimes a_i\otimes e_2\Big) \\ &\quad-\frac{\tau_{(123)}}{4}\Big(a_ib_ia_i\otimes e_2\otimes e_1+a_i\otimes a_ib_i\otimes e_1+2a_i\otimes e_2\otimes e_1\Big) \\ &=\frac{1}{4}\Big(b_ia_i\otimes a_i\otimes e_2-e_1\otimes a_i\otimes a_ib_i\Big)\,, \end{align*} $$

which coincides with equation (4.10).

Now, we address the cases when two indices coincide. If $i=j$ and $i<k$ , by equations (3.10c) and (3.10d),

$$ \begin{align*} \{\mkern-6mu\{a_i,a_i,b_k\}\mkern-6mu\}&=\{\mkern-6mu\{a_i,\{\mkern-6mu\{a_i,b_k\}\mkern-6mu\}\}\mkern-6mu\}_L+\tau_{(123)}\{\mkern-6mu\{a_i,\{\mkern-6mu\{b_k,a_i\}\mkern-6mu\}\}\mkern-6mu\}_L \\ &=\frac{1}{2}\Big(\{\mkern-6mu\{a_i,b_k\}\mkern-6mu\}a_i\otimes e_2\Big)-\frac{\tau_{(123)}}{2}\Big(a_i\{\mkern-6mu\{a_i,b_k\}\mkern-6mu\}\otimes e_1\Big) \\ &=\frac{1}{4}\Big(e_1\otimes a_ib_ka_i\otimes e_2+b_ka_i\otimes a_i\otimes e_2\Big)-\frac{\tau_{(123)}}{4}\Big(a_i\otimes a_ib_k\otimes e_1+a_ib_ka_i\otimes e_2\otimes e_1\Big) \\ &=\frac{1}{4}\Big(b_ka_i\otimes a_i\otimes e_2-e_1\otimes a_i\otimes a_ib_k\Big)\,, \end{align*} $$

as we wished. Similarly, the reader can check that the four cases occurring when $i=k$ and $j\neq k$ , and $j=k$ and $i\neq k$ are completely analogous to the previous ones.

Finally, we address the six possible cases of triples $(a_i,a_j,b_k)$ when $i,j,k$ are all distinct. Let $(a_i,a_j,b_k)$ with $i<j<k$ . Then, by equations (3.10a), (3.10c) and (3.10d), we obtain

$$ \begin{align*} \{\mkern-6mu\{a_i,a_j,b_k\}\mkern-6mu\}&=\{\mkern-6mu\{a_i,\{\mkern-6mu\{a_j,b_k\}\mkern-6mu\}\}\mkern-6mu\}_L+\tau_{(123)}\{\mkern-6mu\{a_j,\{\mkern-6mu\{b_k,a_i\}\mkern-6mu\}\}\mkern-6mu\}_L+\tau_{(132)}\{\mkern-6mu\{b_k,\{\mkern-6mu\{a_i,a_j\}\mkern-6mu\}\}\mkern-6mu\}_L \\ &=\frac{1}{2}\Big(\big(b_k\{\mkern-6mu\{a_i,a_j\}\mkern-6mu\}+\{\mkern-6mu\{a_i,b_k\}\mkern-6mu\}a_j\big)\otimes e_2\Big) -\frac{\tau_{(123)}}{2}\Big(\big(a_i\{\mkern-6mu\{a_j,b_k\}\mkern-6mu\}+\{\mkern-6mu\{a_j,a_i\}\mkern-6mu\}b_k\big)\otimes e_1\Big) \\ &\quad -\frac{\tau_{(132)}}{2}\Big(\{\mkern-6mu\{b_k,a_i\}\mkern-6mu\}\otimes a_j+\{\mkern-6mu\{b_k,a_j\}\mkern-6mu\}\otimes a_i\Big) \\ &=-\frac{1}{4}\Big(b_ka_i\otimes a_j\otimes e_2+b_ka_j\otimes a_i\otimes e_2-e_1\otimes a_ib_ka_j\otimes e_2-b_ka_i\otimes a_j\otimes e_2\Big) \\ &\quad -\frac{\tau_{(123)}}{4}\Big(a_i\otimes a_jb_k\otimes e_1+a_ib_ka_j\otimes e_2\otimes e_1+a_j\otimes a_ib_k\otimes e_1+a_i\otimes a_jb_k\otimes e_1\Big) \\ &\quad +\frac{\tau_{(132)}}{4}\Big(a_ib_k\otimes e_1\otimes a_j+e_2\otimes b_ka_i\otimes a_j+a_jb_k\otimes e_1\otimes a_i+e_2\otimes b_ka_j\otimes a_i\Big) \\ &=\frac{1}{4}\Big(b_ka_i\otimes a_j\otimes e_2-e_1\otimes a_i\otimes a_jb_k\Big)\,; \end{align*} $$

consequently, equation (2.10) holds. The following case occurs when $i<j$ , $i<k$ and $j>k$ :

$$ \begin{align*} &\{\mkern-6mu\{a_i,a_j,b_k\}\mkern-6mu\} \\ &=-\frac{1}{2}\Big(\big(b_k\{\mkern-6mu\{a_i,a_j\}\mkern-6mu\}+\{\mkern-6mu\{a_i,b_k\}\mkern-6mu\}a_j\big)\otimes e_2\Big)-\frac{\tau_{(123)}}{2}\Big(\big(a_i\{\mkern-6mu\{a_j,b_k\}\mkern-6mu\}+\{\mkern-6mu\{a_j,a_i\}\mkern-6mu\}b_k \big)\otimes e_1\Big) \\ &\quad -\frac{\tau_{(132)}}{2}\Big(\{\mkern-6mu\{b_k,a_i\}\mkern-6mu\}\otimes a_j+\{\mkern-6mu\{b_k,a_j\}\mkern-6mu\}\otimes a_i\Big) \\ &=\frac{1}{4}\Big(b_ka_i\otimes a_j\otimes e_2+b_ka_j\otimes a_i\otimes e_2-e_1\otimes a_ib_ka_j\otimes e_2-b_ka_i\otimes a_j\otimes e_2\Big) \\ &\quad +\frac{\tau_{(123)}}{4}\Big(a_ib_ka_j\otimes e_2\otimes e_1+a_i\otimes a_jb_k\otimes e_1+2\delta_{j-k,1}(a_i\otimes e_2\otimes e_1)-a_j\otimes a_ib_k\otimes e_1 \\ &\qquad \qquad -a_i\otimes a_jb_k\otimes e_1\Big) \\ &\quad +\frac{\tau_{(132)}}{4}\Big(a_ib_k\otimes e_1\otimes a_j+e_2\otimes b_ka_i\otimes a_j-e_2\otimes b_ka_j\otimes a_i-a_jb_k\otimes e_1\otimes a_i \\ &\qquad \qquad -2\delta_{j-k,1}(e_2\otimes e_1\otimes a_i)\Big) \\ &=\frac{1}{4}\Big(b_ka_i\otimes a_j\otimes e_2-e_1\otimes a_i\otimes a_jb_k\Big)\,, \end{align*} $$

which is tantamount to equation (4.10). The next case occurs when $i<j$ , $i>k$ and $j>k$ . It is important to note that with these constraints $j-k\neq 1$ because the condition $j-k=1$ would imply that $0<i-k<1$ , which is a contradiction since $i,k\in \mathbb {N}$ . Then, by equations (3.10a), (3.10c) and (3.10d), we have

$$ \begin{align*} &\{\mkern-6mu\{a_i,a_j,b_k\}\mkern-6mu\} \\ &=-\frac{1}{2}\Big(\big(b_k\{\mkern-6mu\{a_i,a_j\}\mkern-6mu\}+\{\mkern-6mu\{a_i,b_k\}\mkern-6mu\}a_j\big)\otimes e_2\Big) +\frac{\tau_{(123)}}{2}\Big(\big(a_i\{\mkern-6mu\{a_j,b_k\}\mkern-6mu\}+\{\mkern-6mu\{a_j,a_i\}\mkern-6mu\}b_k\big)\otimes e_1\Big) \\ &\quad -\frac{\tau_{(132)}}{2}\Big(\{\mkern-6mu\{b_k,a_i\}\mkern-6mu\}\otimes a_j+\{\mkern-6mu\{b_k,a_j\}\mkern-6mu\}\otimes a_i\Big) \\ &=\frac{1}{4}\Big(b_ka_i\otimes a_j\otimes e_2+b_ka_j\otimes a_i\otimes e_2+b_ka_i\otimes a_j\otimes e_2+e_1\otimes a_ib_ka_j\otimes e_2+2\delta_{i-k,1}(e_1\otimes a_j\otimes e_2)\Big) \\ &\quad -\frac{\tau_{(123)}}{4}\Big(a_ib_ka_j\otimes e_2\otimes e_1+a_i\otimes a_jb_k\otimes e_1-a_j\otimes a_ib_k\otimes e_1-a_i\otimes a_jb_k\otimes e_1\Big) \\ &\quad -\frac{\tau_{(132)}}{4}\Big(e_2\otimes b_ka_i\otimes a_j+a_ib_k\otimes e_1\otimes a_j+2\delta_{i-k,1}(e_2\otimes e_1\otimes a_j)+e_2\otimes b_ka_j\otimes a_i \\ &\qquad \qquad +a_jb_k\otimes e_1\otimes a_i\Big) \\ &=\frac{1}{4}\Big(b_ka_i\otimes a_j\otimes e_2-e_1\otimes a_i\otimes a_jb_k\Big)\,, \end{align*} $$

which is equation (4.10). Next, to prove that the bracket $\{\mkern -6mu\{-,-\}\mkern -6mu\}$ as defined in equation (3.10) is a double quasi-Poisson bracket, we need to check that equation (2.10) holds for the triple $(a_i,a_j,b_k)$ , whose indices are subject to the constraints $i>j$ , $j>k$ and $i>k$ . Note that $i-k\neq 1$ . If not, $0<j-k<1$ that contradicts the fact that $j,k\in \mathbb {N}$ . Then,

$$ \begin{align*} &\{\mkern-6mu\{a_i,a_j,b_k\}\mkern-6mu\} \\ &=-\frac{1}{2}\Big(\big(b_k\{\mkern-6mu\{a_i,a_j\}\mkern-6mu\}+\{\mkern-6mu\{a_i,b_k\}\mkern-6mu\}a_j\big)\otimes e_2\Big)+\frac{\tau_{(123)}}{2}\Big(\big(a_i\{\mkern-6mu\{a_j,b_k\}\mkern-6mu\}+\{\mkern-6mu\{a_j,a_i\}\mkern-6mu\}b_k\big)\otimes e_1\Big) \\ &\quad +\frac{\tau_{(132)}}{2}\Big(\{\mkern-6mu\{b_k,a_i\}\mkern-6mu\}\otimes a_j+\{\mkern-6mu\{b_k,a_j\}\mkern-6mu\}\otimes a_i\Big) \\ &=-\frac{1}{4}\Big(b_ka_i\otimes a_j\otimes e_2+b_ka_j\otimes a_i\otimes e_2-b_ka_i\otimes a_j\otimes e_2-e_1\otimes a_ib_ka_j\otimes e_2\Big) \\ &\quad - \frac{\tau_{(123)}}{4}\Big(a_ib_ka_j\otimes e_2\otimes e_1+a_i\otimes a_jb_k\otimes e_1+2\delta_{j-k,1}(a_i\otimes e_2\otimes e_1)+a_j\otimes a_ib_k\otimes e_1 \\ &\qquad \qquad +a_i\otimes a_jb_k\otimes e_1\Big) \\ &\quad +\frac{\tau_{(132)}}{4}\Big(e_2\otimes b_ka_i\otimes a_j+a_ib_k\otimes e_1\otimes a_j+e_2\otimes b_ka_j\otimes a_i+a_jb_k\otimes e_1\otimes a_i \\ &\qquad \qquad +2\delta_{j-k,1}(e_2\otimes e_1\otimes a_i)\Big) \\ &=\frac{1}{4}\Big(b_ka_i\otimes a_j\otimes e_2-e_1\otimes a_i\otimes a_jb_k\Big) \end{align*} $$

that is just the right-hand side of equation (2.10). Next, the fifth case that we need to study comes from the triple $(a_i,a_j,b_k)$ such that $i>j$ , $j<k$ and $i>k$ . Then

$$ \begin{align*} &\{\mkern-6mu\{a_i,a_j,b_k\}\mkern-6mu\} \\ &=\frac{1}{2}\Big(\big(b_k\{\mkern-6mu\{a_i,a_j\}\mkern-6mu\}+\{\mkern-6mu\{a_i,b_k\}\mkern-6mu\}a_j\big)\otimes e_2\Big) +\frac{\tau_{(123)}}{2}\Big(\big(a_i\{\mkern-6mu\{a_j,b_k\}\mkern-6mu\}+\{\mkern-6mu\{a_j,a_i\}\mkern-6mu\}b_k\big)\otimes e_1\Big) \\ &\quad +\frac{\tau_{(132)}}{2}\Big(\{\mkern-6mu\{b_k,a_i\}\mkern-6mu\}\otimes a_j+\{\mkern-6mu\{b_k,a_j\}\mkern-6mu\}\otimes a_i\Big) \\ &=\frac{1}{4}\Big(b_ka_i\otimes a_j\otimes e_2+b_ka_j\otimes a_i\otimes e_2-b_ka_i\otimes a_j\otimes e_2-e_1\otimes a_ib_ka_j\otimes e_2 \\ &\qquad \qquad -2\delta_{i-k,1}(e_1\otimes a_j\otimes e_2)\Big) \\ &\quad +\frac{\tau_{(123)}}{4}\Big(a_i\otimes a_jb_k\otimes e_1+a_ib_ka_j\otimes e_2\otimes e_1-a_j\otimes a_ib_k\otimes e_1-a_i\otimes a_jb_k\otimes e_1\Big) \\ &\quad +\frac{\tau_{(132)}}{4}\Big(e_2\otimes b_ka_i\otimes a_j+a_ib_k\otimes e_1\otimes a_j+2\delta_{i-k,1}(e_2\otimes e_1\otimes a_j)-a_jb_k\otimes e_1\otimes a_i \\ &\qquad\qquad -e_2\otimes b_ka_j\otimes a_i\Big) \\ &=\frac{1}{4}\Big(b_ka_i\otimes a_j\otimes e_2-e_1\otimes a_i\otimes a_jb_k\Big)\,, \end{align*} $$

which proves equation (2.10). The final case for a triple $(a_i,a_j,b_k)$ , with $i,j,k$ distinct occurs when $i>j$ , $j<k$ and $i<k$ . So, by equations (3.10a), (3.10c) and (3.10d), we can write

$$ \begin{align*} &\{\mkern-6mu\{a_i,a_j,b_k\}\mkern-6mu\} \\ &=\frac{1}{2}\Big(\big(b_k\{\mkern-6mu\{a_i,a_j\}\mkern-6mu\}+\{\mkern-6mu\{a_i,b_k\}\mkern-6mu\}a_j\big)\otimes e_2\Big) -\frac{\tau_{(123)}}{2}\Big(\big(a_i\{\mkern-6mu\{a_j,b_k\}\mkern-6mu\}+\{\mkern-6mu\{a_j,a_i\}\mkern-6mu\}b_k\big)\otimes e_1\Big) \\ &\quad +\frac{\tau_{(132)}}{2}\Big(\{\mkern-6mu\{b_k,a_i\}\mkern-6mu\}\otimes a_j+\{\mkern-6mu\{b_k,a_j\}\mkern-6mu\}\otimes a_i\Big) \\ &=\frac{1}{4}\Big(b_ka_i\otimes a_j\otimes e_2+b_ka_j\otimes a_i\otimes e_2+e_1\otimes a_ib_ka_j\otimes e_2+b_ka_i\otimes a_j\otimes e_2\Big) \\ &\quad -\frac{\tau_{(123)}}{4}\Big(a_i\otimes a_jb_k\otimes e_1+a_ib_ka_j\otimes e_2\otimes e_1-a_j\otimes a_ib_k\otimes e_1-a_i\otimes a_jb_k\otimes e_1\Big) \\ &\quad -\frac{\tau_{(132)}}{4}\Big(a_ib_k\otimes e_1\otimes a_j+e_2\otimes b_ka_i\otimes a_j+a_jb_k\otimes e_1\otimes a_i+e_2\otimes b_ka_j\otimes a_i\Big) \\ &=\frac{1}{4}\Big(b_ka_i\otimes a_j\otimes e_2-e_1\otimes a_i\otimes a_jb_k\Big)\,, \end{align*} $$

as we wanted. So we can conclude that equation (2.10) holds for every triple $(a_i,a_j,b_k)$ , where $1\leq i,j,k\leq n$ .

Finally, we deal with the case $(b_i,b_j,a_k)$ , with $1\leq i,j,k\leq n$ . Since we proved that equation (2.10) is fulfilled for every triple $(a_i,a_j,b_k)$ , that is, $4\{\mkern -6mu\{a_i,a_j,b_k\}\mkern -6mu\}=b_ka_i\otimes a_j\otimes e_2-e_1\otimes a_i\otimes a_jb_k$ , we can apply the automorphism S on $\Bbbk \overline {\Gamma }_n$ defined in equation (4.9) to obtain the identity

$$\begin{align*}\{\mkern-6mu\{b_{n+1-i},b_{n+1-j},a_{n+1-k}\}\mkern-6mu\}=\frac{1}{4}\Big(a_{n+1-k}b_{n+1-i}\otimes b_{n+1-j}\otimes e_1-e_2\otimes b_{n+1-i}\otimes b_{n+1-j} a_{n+1-k}\Big)\,,\\[-15pt] \end{align*}$$

which holds for all indices $i,j,k\in \{1,\dots ,n\}$ . Hence, equation (2.10) is also satisfied for a given triple $(b_i,b_j,a_k)$ with $1\leq i,j,k\leq n$ .

To sum up, we proved that the pair $(\mathcal {B}(\Gamma _n),\{\mkern -6mu\{-,-\}\mkern -6mu\})$ (in fact, the pair $(\Bbbk \overline {\Gamma }_n,\{\mkern -6mu\{-,-\}\mkern -6mu\})$ ), where $\{\mkern -6mu\{-,-\}\mkern -6mu\}$ was defined in equation (3.10), is a double quasi-Poisson algebra (over B).

4.3 The element (3.11) is a multiplicative moment map

In this subsection, we proceed to prove that equation (3.11) is a multiplicative moment map; in other words, we show that equation (3.11) satisfies equation (2.11) for $s\in \{1,2\}$ and $a_1,\dots ,a_n,b_1,\dots , b_n$ , which are the generators of $\mathcal {B}(\Gamma _n)$ . We write $\Phi =\Phi _1+\Phi _2$ , where

$$\begin{align*}\Phi_1=(b_n,\dots,a_1)^{-1}\in e_1 \mathcal{B}(\Gamma_n) e_1\,,\quad \Phi_2=(a_1,\dots,b_n)\in e_2 \mathcal{B}(\Gamma_n) e_2\,.\\[-15pt] \end{align*}$$

Since Euler continuants are defined by the recursive relation (3.3), we shall use strong induction. Note that the base case is given by Theorem 2.12(i). Next, to fix ideas, the inductive hypothesis states that lower Euler continuants satisfy equation (2.11). In particular, using the orthogonality of the idempotents, the inductive hypothesis gives rise to the formulas

(4.11a) $$ \begin{align} \{\mkern-6mu\{(a_1,\dots b_{\ell}),a_{i}\}\mkern-6mu\}&=-\frac{1}{2}\Big(e_2\otimes (a_1,\dots b_{\ell})a_i+(a_1,\dots,b_{\ell})\otimes a_i\Big)\, , \\[-15pt]\nonumber\end{align} $$

(4.11b) $$ \begin{align} \{\mkern-6mu\{(a_1,\dots b_{\ell}),b_{i}\}\mkern-6mu\}&=\;\frac{1}{2}\Big(b_i\otimes (a_1,\dots,b_{\ell})+b_i(a_1,\dots,b_{\ell})\otimes e_2\Big)\, , \end{align} $$

where $\ell \in \{1,\dots ,n-1\}$ and $1\leq i\leq \ell $ .

By the very definition of the double quasi-Poisson bracket (3.10), we need to distinguish different cases, depending on the use of the inductive hypothesis and Lemma 4.2.

Let $b_i\in \{b_1,\dots ,b_n\}$ . Then, since $e_ie_j=\delta _{i,j}e_i$ for $i,j\in \{1,2\}$ , the identity (2.11) reduces to

(4.12) $$ \begin{align} \begin{aligned} \{\mkern-6mu\{\Phi_2,b_i\}\mkern-6mu\}&=\frac{1}{2}\Big(b_i\otimes \Phi_2+b_i\Phi_2\otimes e_2\Big) \\ &=\frac{1}{2}\Big(b_i\otimes(a_1,\dots ,b_{n-1})(a_n,b_n)+\sum^n_{\ell=3}\Big[b_i\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big]+b_i\otimes a_1b_n \\ &\quad + b_i(a_1,\dots,b_{n-1})(a_n,b_n)\otimes e_2+\sum^n_{\ell=3}\Big[b_i(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes e_2\Big]+b_ia_1b_n\otimes e_2\Big)\,, \end{aligned}\\[-15pt]\nonumber \end{align} $$

where in the last identity we applied equation (3.6).

To begin with, we shall check equation (2.11) for $b_1$ . Using the left Leibniz rule (2.6) in the first argument (that makes the inner action appear) and separating the case $\ell =3$ in the sum to keep track of $\{\mkern -6mu\{a_2,b_1\}\mkern -6mu\}$ , we can write

$$ \begin{align*} \{\mkern-6mu\{&\Phi_2,b_1\}\mkern-6mu\}=\{\mkern-6mu\{(a_1,\dots,b_n),b_1\}\mkern-6mu\} \\ &= \left\{\!\!\!\left\{\Big((a_1,\dots ,b_{n-1})(a_n,b_n)+\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big]+a_1b_n\Big),b_1\right\}\!\!\!\right\} \\ &=(a_1,\dots,b_{n-1})*\{\mkern-6mu\{(a_n,b_n),b_1\}\mkern-6mu\}+\{\mkern-6mu\{(a_1,\dots ,b_{n-1}),b_1\}\mkern-6mu\}*(a_n,b_n) \\ &\quad +\sum^n_{\ell=3}\Big[(a_1,\dots, b_{\ell-2})a_{\ell-1}*\{\mkern-6mu\{b_n,b_1\}\mkern-6mu\}\Big] +(a_1,b_1)*\{\mkern-6mu\{a_2,b_1\}\mkern-6mu\}*b_n \\ &\quad +\sum^{n}_{\ell=4}\Big[(a_1,\dots,b_{\ell-2})*\{\mkern-6mu\{a_{\ell-1},b_1\}\mkern-6mu\}*b_n \Big] +\sum^n_{\ell=3}\Big[\{\mkern-6mu\{(a_1,\dots,b_{\ell-2}),b_1\}\mkern-6mu\}*a_{\ell-1}b_n\Big] \\ &\quad + a_1*\{\mkern-6mu\{b_n,b_1\}\mkern-6mu\}+\{\mkern-6mu\{a_1,b_1\}\mkern-6mu\}*b_n\,. \end{align*} $$

By Lemma 4.1(ii), the inductive hypothesis (4.11b) and equation (3.10), we have

$$ \begin{align*} &=\frac{1}{2}\Big(b_1\otimes (a_1,\dots,b_{n-1})(a_n,b_n)-b_1(a_n,b_n)\otimes (a_1,\dots,b_{n-1})\Big) \\ &\quad +\frac{1}{2}\Big(b_1(a_n,b_n)\otimes (a_1,\dots,b_{n-1})+b_1(a_1,\dots,b_{n-1})(a_n,b_n)\otimes e_2\Big) \\ &\quad +\frac{1}{2}\sum^n_{\ell=3}\Big[b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_1+b_1\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big] \\ &\quad -\frac{1}{2}\Big(b_1a_2b_n\otimes (a_1,b_1)+b_n\otimes (a_1,b_1)a_2b_1\Big)-b_n\otimes (a_1,b_1) \\ &\quad -\frac{1}{2}\sum^n_{\ell=4}\Big[b_1a_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})+b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_1\Big] \\ &\quad +\frac{1}{2}\sum^n_{\ell=3}\Big[b_1a_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})+b_1(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes e_2\Big] \\ &\quad +\frac{1}{2}\Big(b_n\otimes a_1b_1+b_1\otimes a_1b_n\Big)+\frac{1}{2}\Big(b_1a_1b_n\otimes e_2+b_n\otimes a_1b_1\Big)+b_n\otimes e_2 \\ &=\frac{1}{2}\Big(b_1\otimes (a_1,\dots,b_{n-1})(a_n,b_n)+\sum^n_{\ell=3}\Big[b_1\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big]+b_1\otimes a_1b_n\Big) \\ &\quad +\frac{1}{2}\Big(b_1(a_1,\dots,b_{n-1})(a_n,b_n)\otimes e_2+\sum^n_{\ell=3}\Big[b_1(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes e_2\Big]+b_1a_1b_n\otimes e_2\Big] \\ &\quad -b_n\otimes (a_1,b_1)+b_n\otimes a_1b_1+b_n\otimes e_2 \\ &=\frac{1}{2}\big(b_1\otimes \Phi_2+b_1\Phi_2\otimes e_2\big)\,, \end{align*} $$

where the last identity is due to equation (3.6).

Now, we need to check equation (2.11) for an arrow $b_k$ , where $k\in \{2,\dots ,n-2\}$ is fixed. Then,

$$ \begin{align*} \{\mkern-6mu\{&\Phi_2,b_k\}\mkern-6mu\}= \left\{\!\!\!\left\{\Big((a_1,\dots ,b_{n-1})(a_n,b_n)+\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big]+a_1b_n\Big),b_k\right\}\!\!\!\right\} \\ &=(a_1,\dots,b_{n-1})*\{\mkern-6mu\{(a_n,b_n),b_k\}\mkern-6mu\}+\{\mkern-6mu\{(a_1,\dots,b_{n-1}),b_k\}\mkern-6mu\}*(a_n,b_n) \\ &\quad +\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}*\{\mkern-6mu\{b_n,b_k\}\mkern-6mu\}\Big]+\sum^k_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})*\{\mkern-6mu\{a_{\ell-1},b_k\}\mkern-6mu\}*b_n\Big] \\ &\quad +(a_1,\dots,b_{k-1})*\{\mkern-6mu\{a_k,b_k\}\mkern-6mu\}*b_n+(a_1,\dots,b_k)*\{\mkern-6mu\{a_{k+1},b_k\}\mkern-6mu\}*b_n \\ &\quad + \sum^n_{\ell=k+3}\Big[(a_1,\dots,b_{\ell-2})*\{\mkern-6mu\{a_{\ell-1},b_k\}\mkern-6mu\}*b_n\Big] +\sum^{k+1}_{\ell=3}\Big[\{\mkern-6mu\{(a_1,\dots,b_{\ell-2}),b_k\}\mkern-6mu\}*a_{\ell-1}b_n\Big] \\ &\quad +\sum^n_{\ell=k+2}\Big[\{\mkern-6mu\{(a_1,\dots,b_{\ell-2}),b_k\}\mkern-6mu\}*a_{\ell-1}b_n\Big] +a_1*\{\mkern-6mu\{b_n,b_k\}\mkern-6mu\}+\{\mkern-6mu\{a_1,b_k\}\mkern-6mu\}*b_n\,. \end{align*} $$

Applying Lemma 4.1(ii) in the first summand, the inductive hypothesis (4.11b) in the second and ninth summands and Lemma 4.2(i) in the eighth one, we obtain:

$$ \begin{align*} &=\frac{1}{2}\Big(b_k\otimes (a_1,\dots,b_{n-1})(a_n,b_n)-b_k(a_n,b_n)\otimes (a_1,\dots,b_{n-1})\Big) \\ &\quad +\frac{1}{2}\Big(b_k(a_n,b_n)\otimes (a_1,\dots,b_{n-1})+b_k(a_1,\dots,b_{n-1})(a_n,b_n)\otimes e_2\Big) \\ &\quad +\frac{1}{2}\sum^n_{\ell=3}\Big[b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_k+b_k\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big] \\ &\quad +\frac{1}{2}\sum^k_{\ell=3}\Big[b_ka_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})+b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_k\Big] \\ &\quad +\frac{1}{2}\Big(b_ka_kb_n\otimes (a_1,\dots ,b_{k-1}) +b_n\otimes (a_1,\dots,b_{k-1})a_kb_k\Big)+b_n\otimes (a_1,\dots,b_{k-1}) \\ &\quad-\frac{1}{2}\Big(b_ka_{k+1}b_n\otimes (a_1,\dots ,b_k)+b_n\otimes (a_1,\dots,b_k)a_{k+1}b_k\Big)-b_n\otimes (a_1,\dots ,b_k) \\ &\quad -\frac{1}{2}\sum^n_{\ell=k+3}\Big[b_ka_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})+b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_k\Big] \\ &\quad +\frac{1}{2}\sum^{k+1}_{\ell=3}\Big[b_k(a_1,\dots, b_{\ell-2})a_{\ell-1}b_n\otimes e_2-b_ka_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})\Big] \\ &\quad +\frac{1}{2}\sum^{n}_{\ell=k+2}\Big[b_k(a_1,\dots, b_{\ell-2})a_{\ell-1}b_n\otimes e_2+b_ka_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})\Big] \\ &\quad +\frac{1}{2}\Big(b_n\otimes a_1b_k+b_k\otimes a_1b_n\Big)+\frac{1}{2}\Big(b_ka_1b_n\otimes e_2+b_n\otimes a_1b_k\Big). \end{align*} $$

At this point, if in the third line we rewrite

$$ \begin{align*} \begin{split} \sum^n_{\ell=3}\Big[b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_k\Big]=\sum^k_{\ell=3}\Big[b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_k\Big]+b_n\otimes (a_1,\dots,b_{k-1})a_kb_k \\ +\ b_n\otimes (a_1,\dots,b_{k})a_{k+1}b_k+\sum^n_{\ell=k+3}\Big[b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_k\Big]\,, \end{split} \end{align*} $$

we can cancel some terms, obtaining

$$ \begin{align*} &=\frac{1}{2}\Big(b_k\otimes (a_1,\dots,b_{n-1})(a_n,b_n)+\sum^n_{\ell=3}\Big[b_k\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big]+b_k\otimes a_1b_n\Big) \\ &\quad +\frac{1}{2}\Big(b_k(a_1,\dots,b_{n-1})(a_n,b_n)\otimes e_2+\sum^n_{\ell=3}\Big[b_k(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes e_2\Big]+b_ka_1b_n\otimes e_2\Big) \\ &\quad -b_n\otimes (a_1,,\dots,b_k)+b_n\otimes (a_1,\dots,b_{k-1})(a_k,b_k)\\&\quad +\sum^k_{\ell=3}\Big[b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_k\Big]+b_n\otimes a_1b_k \\ &=\frac{1}{2}\big(b_k\otimes \Phi_2+b_k\Phi_2\otimes e_2\big)\,, \end{align*} $$

where in the last identity we used equation (3.6).

Now, we shall show the validity of equation (2.11) for $b_{n-1}$ . By equation (3.5) and the left Leibniz rule (2.6) of the double bracket, we can write

$$ \begin{align*} &\{\mkern-6mu\{\Phi_2,b_{n-1}\}\mkern-6mu\}= \left\{\!\!\!\left\{\Big((a_1,\dots,b_{n-1})(a_n,b_n)+\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big]+a_1b_n\Big),b_{n-1}\right\}\!\!\!\right\} \\ &=(a_1,\dots,b_{n-1})*\{\mkern-6mu\{(a_n,b_n),b_{n-1}\}\mkern-6mu\}+\{\mkern-6mu\{(a_1,\dots,b_{n-1}),b_{n-1}\}\mkern-6mu\}*(a_n,b_n) \\ &\quad +\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}*\{\mkern-6mu\{b_n,b_{n-1}\}\mkern-6mu\}\Big] \\ &\quad +\sum^{n-1}_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})*\{\mkern-6mu\{a_{\ell-1},b_{n-1}\}\mkern-6mu\}*b_n\Big]+(a_1,\dots,b_{n-2})*\{\mkern-6mu\{a_{n-1},b_{n-1}\}\mkern-6mu\}*b_n \\ &\quad +\sum^n_{\ell=3}\Big[\{\mkern-6mu\{(a_1,\dots,b_{\ell-2}),b_{n-1}\}\mkern-6mu\}*a_{\ell-1}b_n\Big]+a_1*\{\mkern-6mu\{b_n,b_{n-1}\}\mkern-6mu\}+\{\mkern-6mu\{a_1,b_{n-1}\}\mkern-6mu\}*b_n\,. \end{align*} $$

Applying Lemma 4.1(ii) in the first term, the inductive hypothesis in the second one and Lemma 4.2(i), we obtain

$$ \begin{align*} &=\frac{1}{2}\Big(b_{n-1}\otimes (a_1,\dots,b_{n-1})(a_n,b_n)-b_{n-1}(a_n,b_n)\otimes (a_1,\dots,b_{n-1})\Big)-b_n\otimes (a_1,\dots,b_{n-1}) \\ &\quad +\frac{1}{2}\Big(b_{n-1}(a_n,b_n)\otimes (a_1,\dots,b_{n-1})+b_{n-1}(a_1,\dots,b_{n-1})(a_n,b_n)\otimes e_2\Big) \\ &\quad +\frac{1}{2}\sum^n_{\ell=3}\Big[b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_{n-1}+b_{n-1}\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big] \\ &\quad +\frac{1}{2}\sum^{n-1}_{\ell=3}\Big[b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_{n-1}+b_{n-1}a_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})\Big] \\ &\quad + \frac{1}{2}\Big(b_{n-1}a_{n-1}b_n\otimes (a_1,\dots,b_{n-2})+b_n\otimes (a_1,\dots,b_{n-2})a_{n-1}b_{n-1}\Big)+b_n\otimes (a_1,\dots,b_{n-2}) \\ &\quad +\frac{1}{2}\sum^n_{\ell=3}\Big[b_{n-1}(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes e_2-b_{n-1}a_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})\Big] \\ &\quad +\frac{1}{2}\Big(b_n\otimes a_1b_{n-1}+b_{n-1}\otimes a_1b_n\Big)+\frac{1}{2}\Big(b_n\otimes a_1b_{n-1}+b_{n-1}a_1b_n\otimes e_2\Big)\,. \end{align*} $$

We can cancel some terms to obtain

$$ \begin{align*} &=\frac{1}{2}\Big(b_{n-1}\otimes(a_1,\dots,b_{n-1})(a_n,b_n)+\sum^n_{\ell=3}\Big[b_{n-1}\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big]+b_{n-1}\otimes a_1b_n\Big) \\ &\quad +\frac{1}{2}\Big(b_{n-1}(a_1,\dots,b_{n-1})(a_n,b_n)\otimes e_2 +\sum^n_{\ell=3}\Big[b_{n-1}(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes e_2 +b_{n-1}a_1b_n\otimes e_2\Big) \\ &\quad -b_n\otimes (a_1,\dots,b_{n-1})+b_n\otimes (a_1,\dots,b_{n-2})(a_{n-1},b_{n-1})+\sum^{n-1}_{\ell=3}\Big[b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_{n-1}\Big] \\ &\quad +b_n\otimes a_1b_{n-1} \\ &=\frac{1}{2}\big(b_{n-1}\otimes \Phi_2+b_{n-1}\Phi_2\otimes e_2\big)\,, \end{align*} $$

which is tantamount to equation (4.12).

Finally, we shall prove that equation (2.11) holds for $b_n$ via equation (4.12). Then,

$$ \begin{align*} &\{\mkern-6mu\{\Phi_2,b_{n}\}\mkern-6mu\}= \left\{\!\!\!\left\{\Big((a_1,\dots,b_{n-1})(a_n,b_n)+\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big]+a_1b_n\Big),b_{n}\right\}\!\!\!\right\} \\ &=(a_1,\dots,b_{n-1})*\{\mkern-6mu\{(a_n,b_n),b_{n}\}\mkern-6mu\}+\{\mkern-6mu\{(a_1,\dots,b_{n-1}),b_{n}\}\mkern-6mu\}*(a_n,b_n) \\ &\quad +\sum^{n}_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})*\{\mkern-6mu\{a_{\ell-1},b_{n}\}\mkern-6mu\}*b_n+\{\mkern-6mu\{(a_1,\dots,b_{\ell-2}),b_{n}\}\mkern-6mu\}*a_{\ell-1}b_n\Big]+\{\mkern-6mu\{a_1,b_{n}\}\mkern-6mu\}*b_n\,, \end{align*} $$

where we used that $\{\mkern -6mu\{b_n,b_n\}\mkern -6mu\}=0$ . Now, by Lemma 4.1(ii) and Lemma 4.2(i), we have

$$ \begin{align*} &=\frac{1}{2}\Big(b_n\otimes (a_1,\dots,b_{n-1})(a_n,b_n)+b_n(a_n,b_n)\otimes (a_1,\dots, b_{n-1})\Big) \\ &\quad +\frac{1}{2}\Big(b_n(a_1,\dots,b_{n-1})(a_n,b_n)\otimes e_2-b_n(a_n,b_n)\otimes (a_1,\dots,b_{n-1})\Big) \\ &\quad +\frac{1}{2}\sum^n_{\ell=3}\Big[b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1} b_n+b_na_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})\Big] \\ &\quad +\frac{1}{2}\sum^n_{\ell=3}\Big[b_n(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes e_2-b_na_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})\Big] \\ &\quad +\frac{1}{2}\Big(b_n\otimes a_1b_n+b_na_1b_n\otimes e_2\Big) \\ &=\frac{1}{2}\big(b_{n}\otimes \Phi_2+b_{n}\Phi_2\otimes e_2\big)\,, \end{align*} $$

where, once again, in the last identity we applied equation (3.6). To sum up, so far, we proved that equation (2.11) holds for $\Phi _2$ and $b_i\in \{b_1,\dots ,b_n\}$ .

Next, let $a_i\in \{a_1,\dots ,a_n\}$ . Then, since $e_ie_j=\delta _{i,j}e_i$ for $i,j\in \{1,2\}$ , equation (2.11) reduces to

(4.13) $$ \begin{align} \begin{aligned} \{\mkern-6mu\{&\Phi_2,a_i\}\mkern-6mu\}=-\frac{1}{2}\Big(e_2\otimes \Phi_2a_i+\Phi_2\otimes a_i\Big) \\ &=-\frac{1}{2}\Big(e_2\otimes(a_1,\dots ,b_{n-1})(a_n,b_n)a_i+ \sum^n_{\ell=3} \Big[e_2\otimes(a_1,\dots,b_{\ell-2})a_{\ell-1}b_na_i\Big]+e_2\otimes a_1b_na_i \\ &\quad + (a_1,\dots,b_{n-1})(a_n,b_n)\otimes a_i+\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes a_i\Big]+a_1b_n\otimes a_i\Big)\,, \end{aligned} \end{align} $$

where we used equation (3.6).

Now, applying the left Leibniz rule (2.6) and using that $\{\mkern -6mu\{a_1,a_1\}\mkern -6mu\}=0$ , we can write

$$ \begin{align*} &\{\mkern-6mu\{\Phi_2,a_1\}\mkern-6mu\}= \left\{\!\!\!\left\{\Big((a_1,\dots,b_{n-1})(a_n,b_n)+\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big]+a_1b_n\Big),a_1\right\}\!\!\!\right\} \\ &=(a_1,\dots,b_{n-1})*\{\mkern-6mu\{(a_n,b_n),a_1\}\mkern-6mu\}+\{\mkern-6mu\{(a_1,\dots, b_{n-1}),a_1\}\mkern-6mu\}*(a_n,b_n) \\ &\quad+ \sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}*\{\mkern-6mu\{b_n,a_1\}\mkern-6mu\} +(a_1,\dots,b_{\ell-2})*\{\mkern-6mu\{a_{\ell-1},a_1\}\mkern-6mu\}*b_n\Big] \\ &\quad + \sum^n_{\ell=3}\Big[ \{\mkern-6mu\{(a_1,\dots,b_{\ell-2}),a_1\}\mkern-6mu\}*a_{\ell-1}b_n\Big] +a_1*\{\mkern-6mu\{b_n,a_1\}\mkern-6mu\}\,. \end{align*} $$

Using Lemma 4.1(i) in the first term and the inductive hypothesis (4.11a) in the second and fifth terms,

$$ \begin{align*} &=\frac{1}{2}\Big((a_n,b_n)\otimes (a_1,\dots,b_{n-1})a_1-e_2\otimes (a_1,\dots,b_{n-1})(a_n,b_n)a_1\Big) \\ &\quad -\frac{1}{2}\Big((a_n,b_n)\otimes (a_1,\dots,b_{n-1})a_1+(a_1,\dots,b_{n-1})(a_n,b_n)\otimes a_1\Big) \\ &\quad -\frac{1}{2}\sum^n_{\ell=3}\Big[a_1b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}+e_2\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_na_1\Big] \\ &\quad +\frac{1}{2}\sum^n_{\ell=3}\Big[a_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})a_1+a_1b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}\Big] \\ &\quad -\frac{1}{2}\sum^n_{\ell=3}\Big[a_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})a_1+(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes a_1 \Big] \\ &\quad -\frac{1}{2}\Big(a_1b_n\otimes a_1+e_2\otimes a_1b_na_1\Big) \\ &=-\frac{1}{2}\Big(e_2\otimes \Phi_2a_1+\Phi_2\otimes a_1\Big)\,, \end{align*} $$

where we used equation (3.6) in the last identity. Thus, equation (2.11) holds for $a_1$ .

Next, we need to prove first that equation (2.11) holds for $a_2,\dots , a_{n-2}$ . Since the proof is similar, we are going to fix $k\in \{2,\dots , n-2\}$ , and we shall show that equation (2.11) is satisfied for the arrow $a_k$ . By equation (3.6),

$$ \begin{align*} \{\mkern-6mu\{\Phi_2,a_k\}\mkern-6mu\}&= \left\{\!\!\!\left\{\Big((a_1,\dots, b_{n-1})(a_n,b_n)+\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big]+a_1b_n\Big),a_k\right\}\!\!\!\right\} \\ &=(a_1,\dots,b_{n-1})*\{\mkern-6mu\{(a_n,b_n),a_k\}\mkern-6mu\}+\{\mkern-6mu\{(a_1,\dots,b_{n-1}),a_k\}\mkern-6mu\}*(a_n,b_n) \\ &\quad +\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}*\{\mkern-6mu\{b_n,a_k\}\mkern-6mu\}\Big] +\sum^k_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})*\{\mkern-6mu\{a_{\ell-1},a_k\}\mkern-6mu\}*b_n\Big] \\ &\quad +\sum^n_{\ell=k+2}\Big[(a_1,\dots,b_{\ell-2})*\{\mkern-6mu\{a_{\ell-1},a_k\}\mkern-6mu\}*b_n\Big] +\sum^k_{\ell=3}\Big[\{\mkern-6mu\{(a_1,\dots,b_{\ell-2}),a_k\}\mkern-6mu\}*a_{\ell-1}b_n\Big] \\ &\quad +\{\mkern-6mu\{(a_1,\dots,b_{k-1}),a_k\}\mkern-6mu\}*a_kb_n +\sum^n_{\ell=k+2}\Big[\{\mkern-6mu\{(a_1,\dots,b_{\ell-2}),a_k\}\mkern-6mu\}*a_{\ell-1}b_n\Big] \\ &\quad +a_1*\{\mkern-6mu\{b_n,a_k\}\mkern-6mu\}+\{\mkern-6mu\{a_1,a_k\}\mkern-6mu\}*b_n\,, \end{align*} $$

where we used that $\{\mkern -6mu\{a_k,a_k\}\mkern -6mu\}=0$ . Now, we use Lemma 4.1(i) in the first summand, the inductive hypothesis in the second and eighth terms, Lemma 4.2(ii) in the sixth term and Lemma 4.2(iii) in the seventh summand.

$$ \begin{align*} &=\frac{1}{2}\Big((a_n,b_n)\otimes (a_1,\dots,b_{n-1})a_k-e_2\otimes (a_1,\dots,b_{n-1})(a_n,b_n)a_k\Big) \\ &\quad -\frac{1}{2}\Big((a_1,\dots,b_{n-1})(a_n,b_n)\otimes a_k+(a_n,b_n)\otimes (a_1,\dots,b_{n-1})a_k\Big) \\ &\quad -\frac{1}{2}\sum^n_{\ell=3}\Big[a_kb_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}+e_2\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_na_k\Big] \\ &\quad -\frac{1}{2}\sum^k_{\ell=3}\Big[a_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})a_k+a_kb_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}\Big] \\ &\quad +\frac{1}{2}\sum^n_{\ell=k+2}\Big[a_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})a_k+a_kb_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}\Big] \\ &\quad +\frac{1}{2}\sum^k_{\ell=3}\Big[a_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})a_k-(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes a_k\Big] \\ &\quad +\frac{1}{2}\Big(a_kb_n\otimes (a_1,\dots,b_{k-1})a_k-(a_1,\dots,b_{k-1})a_kb_n\otimes a_k\Big)+a_kb_n\otimes (a_1,\dots,a_{k-1}) \\ &\quad-\frac{1}{2}\sum^n_{\ell=k+2}\Big[a_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})a_k+(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes a_k\Big] \\ &\quad -\frac{1}{2}\Big(a_kb_n\otimes a_1+e_2\otimes a_1b_na_k\Big)-\frac{1}{2}\Big(a_1b_n\otimes a_k+a_kb_n\otimes a_1\Big)\,. \end{align*} $$

Some terms cancel each other, giving

$$ \begin{align*} &=-\frac{1}{2}\Big(e_2\otimes (a_1,\dots,b_{n-1})(a_n,b_n)a_k+\sum^n_{\ell=3}\Big[e_2\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_na_k\Big]+e_2\otimes a_1b_na_k\Big) \\ &\quad -\frac{1}{2}\Big((a_1,\dots,b_{n-1})(a_n,b_n)\otimes a_k+\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes a_k\Big]+a_1b_n\otimes a_k\Big) \\ &\quad +a_kb_n\otimes (a_1,\dots,a_{k-1})-\sum^k_{\ell=3}\Big[a_kb_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}\Big]-a_kb_n\otimes a_1 \\ &=-\frac{1}{2}\Big(e_2\otimes \Phi_2a_k+\Phi_2\otimes a_k\Big)\,. \end{align*} $$

Consequently, we can conclude that equation (2.11) holds for $a_2,\dots ,a_{n-2}$ .

Now, we proceed to prove that $a_{n-1}$ satisfies equation (2.11), that is, we need to show that $2\{\mkern -6mu\{\Phi _2,a_{n-1}\}\mkern -6mu\}=-\big (e_2\otimes \Phi _2a_{n-1}+\Phi _2\otimes a_{n-1}\big )$ . By equation (3.6),

$$ \begin{align*} \{\mkern-6mu\{\Phi_2,a_{n-1}\}\mkern-6mu\}&= \left\{\!\!\!\left\{\Big((a_1,\dots,b_{n-1})(a_n,b_n)+\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big]+a_1b_n\Big),a_{n-1}\right\}\!\!\!\right\} \\ &=(a_1,\dots,b_{n-1})*\{\mkern-6mu\{(a_n,b_n),a_{n-1}\}\mkern-6mu\}+\{\mkern-6mu\{(a_1,\dots,b_{n-1}),a_{n-1}\}\mkern-6mu\}*(a_n,b_n) \\ &\quad +\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}*\{\mkern-6mu\{b_n,a_{n-1}\}\mkern-6mu\}\Big]+\sum^{n-1}_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})*\{\mkern-6mu\{a_{\ell-1},a_{n-1}\}\mkern-6mu\}*b_n\Big] \\ &\quad+ \sum^{n-1}_{\ell=3}\Big[\{\mkern-6mu\{(a_1,\dots,b_{\ell-2}),a_{n-1}\}\mkern-6mu\}*a_{\ell-1}b_n\Big]+\{\mkern-6mu\{(a_1,\dots,b_{n-2}),a_{n-1}\}\mkern-6mu\}*a_{n-1}b_n \\ &\quad + a_1*\{\mkern-6mu\{b_n,a_{n-1}\}\mkern-6mu\}+\{\mkern-6mu\{a_1,a_{n-1}\}\mkern-6mu\}*b_n\,, \end{align*} $$

where we used that $\{\mkern -6mu\{a_{n-1},a_{n-1}\}\mkern -6mu\}=0$ . Next, we apply Lemma 4.1(i) to the first summand, the inductive hypothesis to the second summand, Lemma 4.2(ii) to the fifth term and Lemma 4.2(iii) in the sixth one.

$$ \begin{align*} &=\frac{1}{2}\Big((a_n,b_n)\otimes (a_1,\dots,b_{n-1})a_{n-1}-e_2\otimes (a_1,\dots,b_{n-1})(a_n,b_n)a_{n-1}\Big) \\ &\quad -\frac{1}{2}\Big((a_n,b_n)\otimes (a_1,\dots,b_{n-1})a_{n-1}+(a_1,\dots,b_{n-1})(a_n,b_n)\otimes a_{n-1}\Big) \\ &\quad -\frac{1}{2}\sum^{n}_{\ell=3}\Big[a_{n-1}b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}+e_2\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_na_{n-1}\Big] \\ &\quad -\frac{1}{2}\sum^{n-1}_{\ell=3}\Big[a_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})a_{n-1}+a_{n-1}b_n\otimes (a_1,\dots, b_{\ell-2})a_{\ell-1}\Big] \\ &\quad +\frac{1}{2}\sum^{n-1}_{\ell=3}\Big[a_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})a_{n-1}-(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes a_{n-1}\Big] \\ &\quad +\frac{1}{2}\Big(a_{n-1}b_n\otimes (a_1,\dots,b_{n-2})a_{n-1}-(a_1,\dots,b_{n-2})a_{n-1}b_n\otimes a_{n-1}\Big)+a_{n-1}b_n\otimes (a_1,\dots, a_{n-2}) \\ &\quad -\frac{1}{2}\Big(a_{n-1}b_n\otimes a_1+e_2\otimes a_1b_na_{n-1}\Big)-\frac{1}{2}\Big(a_1b_n\otimes a_{n-1}+a_{n-1}b_n\otimes a_1\Big)\,. \end{align*} $$

If we cancel some terms and use both equations (3.6) and (3.5), we obtain

$$ \begin{align*} &=-\frac{1}{2}\Big(e_2\otimes (a_1,\dots,b_{n-1})(a_n,b_n)a_{n-1} +\sum^n_{\ell=3}\Big[e_2\otimes(a_1,\dots,b_{\ell-2})a_{\ell-1}b_na_{n-1}\Big] +e_2\otimes a_1b_na_{n-1} \Big) \\ &\quad -\frac{1}{2} \Big( (a_1,\dots,b_{n-1})(a_n,b_n)\otimes a_{n-1} +\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes a_{n-1}\Big] + a_1b_n\otimes a_{n-1} \Big) \\ &\quad +a_{n-1}b_n\otimes (a_1,\dots,a_{n-2})-\sum^{n-1}_{\ell=3}\Big[a_{n-1}b_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}\Big]- a_{n-1}b_n\otimes a_1 \\ &=-\frac{1}{2}\Big(e_2\otimes \Phi_2a_{n-1}+\Phi_2\otimes a_{n-1}\Big)\,. \end{align*} $$

Finally, we check equation (2.11) for $a_{n}$ via equation (4.13). Once again, by equation (3.6),

$$ \begin{align*} &\{\mkern-6mu\{\Phi_2,a_{n}\}\mkern-6mu\}= \left\{\!\!\!\left\{\Big((a_1,\dots,b_{n-1})(a_n,b_n)+\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\Big]+a_1b_n\Big),a_{n}\right\}\!\!\!\right\} \\ &=(a_1,\dots,b_{n-1})*\{\mkern-6mu\{(a_n,b_n),a_n\}\mkern-6mu\}+\{\mkern-6mu\{(a_1,\dots,b_{n-1}),a_n\}\mkern-6mu\}*(a_n,b_n) \\ &\quad +\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}*\{\mkern-6mu\{b_n,a_n\}\mkern-6mu\}+(a_1,\dots,b_{\ell-2})*\{\mkern-6mu\{a_{\ell-1},a_n\}\mkern-6mu\}*b_n\Big] \\ &\quad + \sum^n_{\ell=3}\Big[\{\mkern-6mu\{(a_1,\dots,b_{\ell-2}),a_n\}\mkern-6mu\}*a_{\ell-1}b_n\Big]+ a_1*\{\mkern-6mu\{b_n,a_n\}\mkern-6mu\}+\{\mkern-6mu\{a_1,a_n\}\mkern-6mu\}*b_n\,. \end{align*} $$

Now, applying Lemma 4.1(i) to the first summand, Lemma 4.2(iii) in the second term, and Lemma 4.2(ii) in the fifth term, we get

$$ \begin{align*} &=-\frac{1}{2}\Big(e_2\otimes (a_1,\dots ,b_{n-1})(a_n,b_n)a_n+(a_n,b_n)\otimes (a_1,\dots,b_{n-1})a_n\Big) \\ &\quad +\frac{1}{2}\Big((a_n,b_n)\otimes (a_1,\dots,b_{n-1})a_n-(a_1,\dots,b_{n-1})(a_n,b_n)\otimes a_n\Big)+(a_n,b_n)\otimes (a_1,\dots,a_{n-1}) \\ &\quad -\sum^n_{\ell=3}\Big[\frac{1}{2}\big(e_2\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}b_na_n+a_nb_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}\big)+e_2\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}\Big] \\ &\quad -\frac{1}{2}\sum^n_{\ell=3}\Big[a_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})a_n+a_nb_n\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}\Big] \\ &\quad +\frac{1}{2}\sum^n_{\ell=3}\Big[a_{\ell-1}b_n\otimes (a_1,\dots,b_{\ell-2})a_n-(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes a_n\Big] \\ &\quad -\frac{1}{2}\Big(e_2\otimes a_1b_na_n+a_nb_n\otimes a_1\Big)-e_2\otimes a_1 -\frac{1}{2}\Big(a_1b_n\otimes a_n+a_nb_n\otimes a_1\Big)\,. \end{align*} $$

Using that $(a_n,b_n)=a_nb_n+e_2$ and equations (3.6) and (3.5), we finally obtain

$$ \begin{align*} &=-\frac{1}{2}\Big(e_2\otimes (a_1,\dots,b_{n-1})(a_n,b_n)a_n+\sum^n_{\ell=3}\Big[e_2\otimes(a_1,\dots,b_{\ell-2})a_{\ell-1}b_na_n\Big]+e_2\otimes a_1b_na_n\Big) \\ &\quad -\frac{1}{2} \Big( (a_1,\dots,b_{n-1})(a_n,b_n)\otimes a_n+\sum^n_{\ell=3}\Big[(a_1,\dots,b_{\ell-2})a_{\ell-1}b_n\otimes a_n\Big]+a_1b_n\otimes a_n\Big) \\ &\quad+ (a_n,b_n)\otimes (a_1,\dots,a_{n-1})-\sum^n_{\ell=3}\Big[(a_n,b_n)\otimes (a_1,\dots,b_{\ell-2})a_{\ell-1}\Big]-(a_n,b_n)\otimes a_1 \\ &=-\frac{1}{2}\Big(e_2\otimes \Phi_2a_{n}+\Phi_2\otimes a_{n}\Big)\,. \end{align*} $$

To sum up, we proved that $\Phi _2=(a_1,\dots , b_n)$ satisfies equation (2.11) in $\mathcal {B}(\Gamma _n)$ . Now, we need to prove that $\Phi _1=(b_n,\dots , a_1)^{-1}\in e_1 \mathcal {B}(\Gamma _n) e_1$ satisfies equation (2.11) as well. The automorphism S of order two (see equation (4.9)) has the effect $\{\mkern -6mu\{-,-\}\mkern -6mu\}\mapsto -\{\mkern -6mu\{-,-\}\mkern -6mu\}$ and $(a_1,\dots ,b_n)\mapsto (b_n,\dots , a_1)$ . By applying S to the just proved identity $2\{\mkern -6mu\{(a_1,\dots , b_n),a_i\}\mkern -6mu\}=-\big (e_2\otimes \Phi _2a_{i}+\Phi _2\otimes a_{i}\big )$ —see equation (4.13)—for $1\leq i\leq n$ , we obtain

$$\begin{align*}\{\mkern-6mu\{(b_n,\dots, a_1),b_{n+1-i}\}\mkern-6mu\}=\frac{1}{2}\Big(e_1\otimes (b_n,\dots,a_1)b_{n+1-i}+(b_n,\dots, a_1)\otimes b_{n+1-i}\Big)\,, \end{align*}$$

for $1\leq i\leq n$ . Then, by equation (2.6) and noting that $\Phi _1(b_n,\dots ,a_1)=(b_n,\dots ,a_1)\Phi _1=e_1$ , we finally obtain

$$ \begin{align*} \{\mkern-6mu\{\Phi_1,b_{n+1-i}\}\mkern-6mu\}&=-\Phi_1*\{\mkern-6mu\{(b_n,\dots, a_1),b_{n+1-i}\}\mkern-6mu\}*\Phi_1 \\ &=-\Phi_1*\Big(\frac{1}{2}\big(e_1\otimes (b_n,\dots,a_1)b_{n+1-i}+(b_n,\dots, a_1)\otimes b_{n+1-i}\big)\Big)*\Phi_1 \\ &=-\frac{1}{2}\Big(\Phi_1\otimes b_{n+1-i}+e_1\otimes \Phi_1b_{n+1-i}\Big)\,, \end{align*} $$

for all $1\leq i\leq n$ , which is tantamount to equation (2.11) since $b_{n+1-i}e_1=0$ . Similarly, applying the automorphism S to equation (4.12), we get the identity

$$\begin{align*}-\{\mkern-6mu\{(b_n,\dots, a_1),a_{n+1-i}\}\mkern-6mu\}=\frac{1}{2}\Big(a_{n+1-i}\otimes (b_n,\dots, a_1)+a_{n+1-i}(b_n,\dots, a_1)\otimes e_1\Big)\,. \end{align*}$$

By equation (2.6) and the fact $e_1a_{n+1-i}=0$ , we have

$$ \begin{align*} \{\mkern-6mu\{\Phi_1,a_{n+1-i}\}\mkern-6mu\}&=-\Phi_1*\{\mkern-6mu\{(b_n,\dots, a_1),a_{n+1-i}\}\mkern-6mu\}*\Phi_1 \\ &=\Phi_1*\Big(\frac{1}{2}\big(a_{n+1-i}\otimes (b_n,\dots, a_1)+a_{n+1-i}(b_n,\dots, a_1)\otimes e_1\big)\Big)*\Phi_1 \\ &=\frac{1}{2}\Big(a_{n+1-i}\Phi_1\otimes e_1+a_{n+1-i}\otimes \Phi_1\Big)\,, \end{align*} $$

as we wished. So, we conclude that equation (2.11) holds for $\Phi _1$ ; therefore, $\Phi $ , as defined in equation (3.11), is a multiplicative moment map for the double quasi-Poisson bracket (3.10). This finishes the proof of Theorem 3.4.

5.1 The result

Recall from § 3.2 that $\mathcal {B}(\Gamma _n)$ denotes the algebra obtained from $\Bbbk \overline {\Gamma }_n$ after localisation at the Euler continuants (3.7). We now consider the algebra $\mathcal {B}_{n}^{\operatorname {loc}}$ obtained by further localising $\mathcal {B}(\Gamma _n)$ at the elements

(5.1) $$ \begin{align} \begin{aligned} &(a_1,b_1)\,,\ldots\,,(a_1,b_1,\dots,a_{n-1},b_{n-1})\in e_2(\Bbbk\overline{\Gamma}_n)e_2\,, \\ &(b_1,a_1)\,,\ldots,(b_{n-1},a_{n-1}\,,\dots,b_1,a_1)\in e_1(\Bbbk\overline{\Gamma}_n)e_1\,. \end{aligned} \end{align} $$

In other words, $\mathcal {B}_{n}^{\operatorname {loc}}$ is obtained from $\Bbbk \overline {\Gamma }_n$ by requiring that the following elements are inverted (in the sense of §3.2):

(5.2) $$ \begin{align} (a_1,b_1,\dots,a_k,b_k)\in e_2(\Bbbk\overline{\Gamma}_n)e_2\,,\quad (b_k,a_k,\dots,b_1,a_1)\in e_1(\Bbbk\overline{\Gamma}_n)e_1\,,\quad k=1,\ldots,n\,. \end{align} $$

The structure on $\mathcal {B}(\Gamma _n)$ given in Theorem 3.4 descends to $\mathcal {B}_{n}^{\operatorname {loc}}$ by localisation.

The algebra $\mathcal {A}_{n}^{\operatorname {fus}}$ is obtained by fusion which, as we recalled in § 2.2, is a method to get a double quasi-Poisson bracket and a multiplicative moment map from a Hamiltonian double quasi-Poisson algebra by identifying several idempotents. We describe the structure of $\mathcal {A}_{n}^{\operatorname {fus}}$ in § 5.2. In particular, we will notice in Remark 5.5 that the algebra $\mathcal {A}_{n}^{\operatorname {fus}}$ is, by construction, an example of Hamiltonian double quasi-Poisson algebra associated with the quiver $\Gamma _n$ (where we see $\overline {\Gamma }_n$ as its double quiver for $b_k=a_k^{\ast }$ ) by Van den Bergh; see Remark 2.13. Thus, the next result is a direct consequence of Theorem 5.1.

5.2 The algebra $\mathcal {A}_{n}^{\operatorname {fus}}$

We consider n copies of the algebra $\mathcal {B}(\Gamma _1)$ , that we denote by $\mathcal {A}(\Gamma _1)$ for the rest of this subsection, since the Hamiltonian double quasi-Poisson structures of Theorems 2.12 and 3.4 (with $n=1$ ) coincide. We respectively denote the two generators $a_1,b_1$ in the $\ell $ -th copy as $c_{\ell },d_{\ell }$ and the two idempotents $e_1,e_2$ in the $\ell $ -th copy as $e_{1,\ell },e_{2,\ell }$ . In particular, the continuants $(c_{\ell },d_{\ell })$ and $(d_{\ell },c_{\ell })$ are inverted in each copy. We can naturally see $\oplus _{\ell =1}^n \mathcal {A}(\Gamma _1)$ as a Hamiltonian double quasi-Poisson algebra over $\tilde {B}$ , where $\tilde B = (\oplus _{\ell =1}^n \Bbbk e_{1,\ell }) \bigoplus (\oplus _{\ell =1}^n \Bbbk e_{2,\ell })$ ; see [Reference Fairon29, Remark 2.13].

Next, we identify idempotents in $\oplus _{\ell =1}^n \mathcal {A}(\Gamma _1)$ . We let $e_1:=e_{1,1}$ and $e_2:=e_{2,1}$ , and we inductively perform fusion (2.2) of the idempotents $e_{1,\ell }$ onto $e_1$ and $e_{2,\ell }$ onto $e_2$ for $\ell =2,\ldots ,n$ . The algebra $\mathcal {A}_{n}^{\operatorname {fus}}$ obtained in this way can be described as being generated by the orthogonal idempotents $e_1,e_2$ satisfying $e_1+e_2=1$ , together with the elements

(5.3) $$ \begin{align} c_1=e_2c_1 e_1\,,\quad \ldots\,,\quad c_n=e_2c_n e_1\,;\quad d_1=e_1 d_1 e_2\,,\quad \ldots\,,\quad d_n=e_1d_n e_2\,, \end{align} $$

and the following elements are inverted:

(5.4) $$ \begin{align} (c_{\ell},d_{\ell})=e_2+c_{\ell} d_{\ell} \in e_2\mathcal{A}_{n}^{\operatorname{fus}} e_2\,,\quad (d_{\ell} ,c_{\ell})=e_1+d_{\ell} c_{\ell}\in e_1\mathcal{A}_{n}^{\operatorname{fus}} e_1\,,\quad \ell=1,\ldots,n\,. \end{align} $$

The choice of order in which we perform fusion gives a Hamiltonian double quasi-Poisson algebra structure on $\mathcal {A}_{n}^{\operatorname {fus}}$ by [Reference Van den Bergh41, §5.3].

Proposition 5.4. The Hamiltonian double quasi-Poisson structure on $\mathcal {A}_{n}^{\operatorname {fus}}$ is given by the B-linear double bracket $\{\mkern -6mu\{-,-\}\mkern -6mu\}^{(n)}$ which is defined on generators by

(5.5a) $$ \begin{align} \{\mkern-6mu\{c_i,c_j\}\mkern-6mu\}^{(n)}&= \begin{cases} -\frac{1}{2}\big(c_i\otimes c_j+c_j\otimes c_i\big)\,, &\mbox{if } i<j \\ \;\;\; 0\,, &\mbox{if } i=j \\ \;\;\; \frac{1}{2}\big(c_i\otimes c_j+c_j\otimes c_i\big)\,, & \mbox{if } i>j \end{cases}; \end{align} $$

(5.5b) $$ \begin{align} \{\mkern-6mu\{d_i,d_j\}\mkern-6mu\}^{(n)}&= \begin{cases} -\frac{1}{2}\big(d_i\otimes d_j+d_j\otimes d_i\big)\,, &\mbox{if } i<j \\ \;\;\; 0\,, &\mbox{if } i=j \\ \;\;\; \frac{1}{2}\big(d_i\otimes d_j+d_j\otimes d_i\big)\,, & \mbox{if } i>j \end{cases}; \end{align} $$

(5.5c) $$ \begin{align} \{\mkern-6mu\{c_i,d_j\}\mkern-6mu\}^{(n)}&= \begin{cases} \;\;\; \frac{1}{2}\big(e_1\otimes c_id_j+d_jc_i\otimes e_2\big)\,, &\mbox{if } i<j \\ \;\;\; \frac{1}{2}\big(e_1\otimes c_id_i+ d_ic_i\otimes e_2\big)+e_1\otimes e_2\,, &\mbox{if } i=j \\ - \frac{1}{2}\big(e_1\otimes c_id_j+d_jc_i\otimes e_2\big)\,, &\mbox{if } i>j \end{cases}; \end{align} $$

and which we extend to $\mathcal {A}_{n}^{\operatorname {fus}}$ by the Leibniz rule and cyclic antisymmetry. The multiplicative moment map $\Phi ^{(n)}=\Phi _1^{(n)}+\Phi _2^{(n)}$ is given by

(5.6) $$ \begin{align} \Phi_1^{(n)}=(e_1+d_1 c_1)^{-1}\cdots (e_1+d_n c_n)^{-1},\,\quad \Phi_2^{(n)}=(e_2+c_1 d_1)\cdots(e_2+c_n d_n)\,. \end{align} $$

Proof. The Hamiltonian double quasi-Poisson structure on $\oplus _{\ell =1}^n \mathcal {A}(\Gamma _1)$ is given by the double quasi-Poisson bracket $\{\mkern -6mu\{-,-\}\mkern -6mu\}^{\operatorname {sep}}$ satisfying for $1\leq i,j\leq n$

(5.7) $$ \begin{align} \begin{aligned} &\{\mkern-6mu\{c_i,c_j\}\mkern-6mu\}^{\operatorname{sep}}=0\,, \quad\{\mkern-6mu\{d_i,d_j\}\mkern-6mu\}^{\operatorname{sep}}=0\,, \\ &\{\mkern-6mu\{c_i,d_j\}\mkern-6mu\}^{\operatorname{sep}}=\delta_{ij} \,e_{1,i}\otimes e_{2,i}+\frac12 \delta_{ij} (e_{1,i}\otimes c_i d_i + d_i c_i \otimes e_{2,i})\,, \end{aligned} \end{align} $$

and the multiplicative moment map $\Phi ^{\operatorname {sep}}$ for

(5.8) $$ \begin{align} \Phi^{\operatorname{sep}}=\sum_{\ell=1}^n (\Phi_{1,\ell}+\Phi_{2,\ell})\,, \quad \Phi_{1,\ell}:=(e_{1,\ell}+d_{\ell} c_{\ell})^{-1},\quad \Phi_{2,\ell}:=e_{2,\ell}+c_{\ell} d_{\ell}\,. \end{align} $$

We inductively perform fusion of $e_{1,\ell }$ onto $e_1=e_{1,1}$ and of $e_{2,\ell }$ onto $e_2=e_{2,1}$ for $\ell =2,\ldots ,n$ . When such fusions have been performed for all $\ell =1,\ldots ,k$ , $2\leq k\leq n$ , we have by Theorem 2.8 that the multiplicative moment map $\Phi ^{(k)}$ becomes

(5.9) $$ \begin{align} \begin{aligned} & \Phi^{(k)}=\Phi_1^{(k)}+ \Phi_2^{(k)} +\sum_{\ell=k+1}^n (\Phi_{1,\ell}+\Phi_{2,\ell})\,, \\ &\Phi_{1}^{(k)}=(e_{1}+d_1 c_1)^{-1}\cdots (e_{1}+d_k c_k)^{-1},\quad \Phi_{2}^{(k)}=(e_{2}+c_1 d_1)\cdots (e_{2}+c_k d_k)\,. \end{aligned} \end{align} $$

In particular, $\Phi ^{(n)}_1=e_1 \Phi ^{(n)}e_1$ and $\Phi ^{(n)}_2=e_2\Phi ^{(n)}e_2$ are the components of the multiplicative moment map on $\mathcal {A}_{n}^{\operatorname {fus}}$ by construction, and this is just equation (5.6).

Using again Theorem 2.8, we get that the double quasi-Poisson bracket $\{\mkern -6mu\{-,-\}\mkern -6mu\}^{(k)}$ obtained after fusing together the first k copies of $\mathcal {A}(\Gamma _1)$ in $\oplus _{\ell =1}^n \mathcal {A}(\Gamma _1)$ is defined by adding an extra ‘fusion’ double bracket $\{\mkern -6mu\{-,-\}\mkern -6mu\}^{(k)}_{\operatorname {fus}}$ to the previously obtained one $\{\mkern -6mu\{-,-\}\mkern -6mu\}^{(k-1)}$ (where $\{\mkern -6mu\{-,-\}\mkern -6mu\}^{(0)}=\{\mkern -6mu\{-,-\}\mkern -6mu\}^{\operatorname {sep}}$ ). It can be read from Lemma 2.7 that the fusions of $e_{1,k}$ onto $e_1$ and of $e_{2,k}$ onto $e_2$ result in the additional fusion double bracket $\{\mkern -6mu\{-,-\}\mkern -6mu\}^{(k)}_{\operatorname {fus}}$ which satisfies the following relations

(5.10a) $$ \begin{align} \{\mkern-6mu\{c_i,c_k\}\mkern-6mu\}^{(k)}_{\operatorname{fus}}=&-\frac12 (c_i \otimes c_k + c_k \otimes c_i)\,, \quad i<k\,, \end{align} $$

(5.10b) $$ \begin{align} \{\mkern-6mu\{d_i,d_k\}\mkern-6mu\}^{(k)}_{\operatorname{fus}}=&-\frac12 (d_i \otimes d_k + d_k \otimes d_i)\,, \quad i<k\,, \end{align} $$

(5.10c) $$ \begin{align} \{\mkern-6mu\{c_i,d_k\}\mkern-6mu\}^{(k)}_{\operatorname{fus}}=&\;\;\;\;\,\frac12 (e_1 \otimes c_i d_k + d_k c_i \otimes e_2)\,, \quad i<k\,, \end{align} $$

(5.10d) $$ \begin{align} \{\mkern-6mu\{d_i,c_k\}\mkern-6mu\}^{(k)}_{\operatorname{fus}}=&\;\;\;\;\,\frac12 (e_1 \otimes c_i d_k + d_k c_i \otimes e_2)\,, \quad i<k\,, \end{align} $$

(5.10e) $$ \begin{align} \{\mkern-6mu\{f,f'\}\mkern-6mu\}^{(k)}_{\operatorname{fus}}=&\;\;\;\;\, 0\,, \quad \text{if either }f,f'\in \{c_i,d_i \mid i \neq k\} \text{ or if }f,f'\in \{c_k,d_k\}\,. \end{align} $$

Let us derive equation (5.10a). When we fuse $e_{1,k}$ onto $e_1$ , according to the terminology of § 2.2 the element $c_i$ (for $i<k$ ) is a generator of first type, and $c_k$ is a generator of third type. Hence, this fusion results in the following added term in $\{\mkern -6mu\{c_i,c_k\}\mkern -6mu\}^{(k)}_{\operatorname {fus}}$ by equation (2.22c):

(5.11) $$ \begin{align} \frac12 (c_k c_i \otimes e_1 - c_k \otimes c_i e_1 ) = -\frac12 c_k\otimes c_i \,. \end{align} $$

When we fuse $e_{2,k}$ onto $e_2$ , the element $c_i$ (for $i<k$ ) is again of first type, but $c_k$ is of second type, so we add the following term in $\{\mkern -6mu\{c_i,c_k\}\mkern -6mu\}^{(k)}_{\operatorname {fus}}$ by equation (2.22b):

(5.12) $$ \begin{align} \frac12 (e_1 \otimes c_i c_k - e_1 c_i \otimes c_k) = -\frac12 c_i \otimes c_k \,. \end{align} $$

Putting together equations (5.11) and (5.12), we get the two terms in equation (5.10a).

To conclude, note that the double bracket $\{\mkern -6mu\{-,-\}\mkern -6mu\}^{(n)}$ on $\mathcal {A}_{n}^{\operatorname {fus}}$ is obtained by adding all the double brackets $\{\mkern -6mu\{-,-\}\mkern -6mu\}_{\operatorname {fus}}^{(\ell )}$ for $2\leq \ell \leq n$ to equation (5.7). It is easy to see that it coincides with equation (5.5).

Remark 5.5. We can relate Proposition 5.4 to Remark 2.13 as follows. Recall that the Hamiltonian double quasi-Poisson structure on $\mathcal {A}(\Gamma _1)$ is the structureFootnote ⁸ constructed by Van den Bergh on $\Bbbk \overline {\Gamma }_1$ localised at $e_2+ab$ and $e_1+ba$ .

Denote by $\Gamma _n$ the quiver consisting of n arrows $c_{\ell }:1\to 2$ , and denote by $d_{\ell }=c_{\ell }^{\ast }:2\to 1$ the corresponding arrows in the double $\overline {\Gamma }_n$ . For a fixed ordering of the arrows starting at each vertex, Van den Bergh obtained a Hamiltonian double quasi-Poisson structure on $\mathcal {A}(\Gamma _n)$ , which is defined as the algebra $\Bbbk \overline {\Gamma }_n$ localised at all the $e_2+c_{\ell } d_{\ell }$ and $e_1+d_{\ell } c_{\ell }$ . This structure is obtained by fusion, starting with n copies of $\Bbbk \overline {\Gamma }_1$ . By construction, it was noticed by Van den Bergh that the order in which we perform fusion can be encoded in an ordering taken at each vertex on the arrows which start at that vertex. In our case the Hamiltonian double quasi-Poisson structure for the orderings

(5.13) $$ \begin{align} c_1<c_2<\ldots <c_n \quad \text{at the vertex }1\,, \qquad d_1<d_2< \ldots <d_n \quad \text{at the vertex }2\, \end{align} $$

is precisely the result obtained in Proposition 5.4. The comparison is easily made using the explicit form of Van den Bergh’s double bracket written in [Reference Fairon29, Theorem 3.3] (keeping in mind footnote 8). Note that, up to isomorphism, Van den Bergh’s Hamiltonian double quasi-Poisson algebra structure associated with $\Gamma _n$ only depends on the quiver seen as an undirected graph [Reference Fairon30, Theorem 4.12].

5.3 Alternative description of the algebra $\mathcal {B}_{n}^{\operatorname {loc}}$

By construction, $\mathcal {B}_{n}^{\operatorname {loc}}$ is generated by the orthogonal idempotents $e_1,e_2$ satisfying $e_1+e_2=1$ , together with the elements

(5.14) $$ \begin{align} a_1=e_2a_1 e_1\,,\quad \ldots \,,\quad\, a_n=e_2a_n e_1\,;\quad b_1=e_1 b_1 e_2\,,\quad \ldots\,,\quad\, b_n=e_1b_n e_2\,, \end{align} $$

while the elements in equation (5.2) are inverted. By localisation, the double quasi-Poisson bracket and the multiplicative moment map can be written as in Theorem 3.4 in terms of the generators (5.14).

Let us introduce the following elements of $\mathcal {B}_{n}^{\operatorname {loc}}$ :

(5.15) $$ \begin{align} \begin{aligned} b_{\ell}'&=b_{\ell}\,, & &\text{for }1\leq \ell\leq n\,, \\ a_1'&=a_1\,, & &a_{\ell}'=a_{\ell}+(a_1,\ldots,b_{\ell-1})^{-1}(a_1,\ldots,a_{\ell-1})\,,\quad \text{for }2\leq \ell\leq n\,. \end{aligned} \end{align} $$

Though the formulae are cumbersome, we can see by induction that the elements (5.14) can be expressed in terms of the elements $\{a_{\ell }',b_{\ell }'\}$ in equation (5.15). Hence, the elements $\{a_{\ell }',b_{\ell }'\}$ can be used as a set of generators for $\mathcal {B}_{n}^{\operatorname {loc}}$ (where the elements in equation (5.2) written in terms of $\{a_{\ell }',b_{\ell }'\}$ are inverted). We can also write

(5.16) $$ \begin{align} a_{\ell}'=a_{\ell}+(a_{\ell-1},\ldots,a_{1}) (b_{\ell-1},\ldots,a_{1})^{-1}\,,\quad \text{for }2\leq \ell\leq n\, \end{align} $$

due to Lemma 3.3.

Proposition 5.7. In terms of the generators $\{a_k',b_k'\mid 1\leq k \leq n\}$ , the Hamiltonian double quasi-Poisson algebra structure on $\mathcal {B}_{n}^{\operatorname {loc}}$ is given by the B-linear double bracket

(5.17a) $$ \begin{align} \{\mkern-6mu\{a_i',a_j'\}\mkern-6mu\}&= \begin{cases} -\frac{1}{2}\big(a_i'\otimes a_j'+a_j'\otimes a_i'\big)\,, &\mbox{if } i<j \\ \;\;\; 0\,, &\mbox{if } i=j \\ \;\;\; \frac{1}{2}\big(a_i'\otimes a_j'+a_j'\otimes a_i'\big)\,, & \mbox{if } i>j \end{cases}; \end{align} $$

(5.17b) $$ \begin{align} \{\mkern-6mu\{b_i',b_j'\}\mkern-6mu\}&= \begin{cases} -\frac{1}{2}\big(b_i'\otimes b_j'+b_j'\otimes b_i'\big)\,, &\mbox{if } i<j \\ \;\;\; 0\,, &\mbox{if } i=j \\ \;\;\; \frac{1}{2}\big(b_i'\otimes b_j'+b_j'\otimes b_i'\big)\,, & \mbox{if } i>j \end{cases}; \end{align} $$

(5.17c) $$ \begin{align} \{\mkern-6mu\{a_i',b_j'\}\mkern-6mu\}&= \begin{cases} \;\;\; \frac{1}{2}\big(e_1\otimes a_i'b_j'+b_j'a_i'\otimes e_2\big)\,, &\mbox{if } i<j \\ \;\;\; \frac{1}{2}\big(e_1\otimes a_i'b_i'+ b_i'a_i'\otimes e_2\big)+e_1\otimes e_2\,, &\mbox{if } i=j \\ - \frac{1}{2}\big(e_1\otimes a_i'b_j'+b_j'a_i'\otimes e_2\big)\,, &\mbox{if } i>j \end{cases}; \end{align} $$

and the multiplicative moment map $\Phi =\Phi _1+\Phi _2$ for

(5.18) $$ \begin{align} \Phi_1=(e_1+b_1' a_1')^{-1}\cdots (e_1+b_n'a_n')^{-1},\qquad \Phi_2=(e_2+a_1'b_1')\cdots(e_2+a_n' b_n')\,. \end{align} $$

The proof of Proposition 5.7 is presented in § 5.5.

5.4 Proof of Theorem 5.1

We consider the generators introduced in §5.2 for $\mathcal {A}_{n}^{\operatorname {fus}}$ and §5.3 for $\mathcal {B}_{n}^{\operatorname {loc}}$ . We first note that we have a B-linear isomorphism given by

(5.19) $$ \begin{align} \psi:\mathcal{B}_{n}^{\operatorname{loc}}\to \mathcal{A}_{n}^{\operatorname{fus}}\,,\qquad \psi(a_{\ell}')=c_{\ell}\,,\,\quad \psi(b_{\ell}')=d_{\ell}\,, \end{align} $$

for $\ell =1,\ldots ,n$ . To see that this map is well-defined, note that the invertibility of the elements (5.2) in $\mathcal {B}_{n}^{\operatorname {loc}}$ is equivalent to the invertibility of the elements $(a_{\ell }',b_{\ell }')=e_2+a_{\ell }'b_{\ell }'$ and $(b_{\ell }',a_{\ell }')=e_1+b_{\ell }' a_{\ell }'$ in $\mathcal {B}_{n}^{\operatorname {loc}}$ for $1\leq \ell \leq n$ due to equation (5.24), which we obtain as part of the proof of Proposition 5.7. But these elements are precisely mapped onto the invertible elements $(c_{\ell },d_{\ell })$ and $(d_{\ell },c_{\ell })$ in $\mathcal {A}_{n}^{\operatorname {fus}}$ .

Next, by comparing Propositions 5.4 and 5.7, we can see that

(5.20) $$ \begin{align} \begin{aligned} & \{\mkern-6mu\{\psi(a_i'),\psi(a_j')\}\mkern-6mu\}^{(n)}=\psi^{\otimes 2}(\{\mkern-6mu\{a_i',a_j'\}\mkern-6mu\})\,, \\ &\{\mkern-6mu\{\psi(b_i'),\psi(b_j')\}\mkern-6mu\}^{(n)}=\psi^{\otimes 2}(\{\mkern-6mu\{b_i',b_j'\}\mkern-6mu\})\,, \\ & \{\mkern-6mu\{\psi(a_i'),\psi(b_j')\}\mkern-6mu\}^{(n)}=\psi^{\otimes 2}(\{\mkern-6mu\{a_i',b_j'\}\mkern-6mu\})\,, \quad \end{aligned} \end{align} $$

for any $i,j=1,\ldots ,n$ . We thus get that, for all $a,b\in \mathcal {B}_{n}^{\operatorname {loc}}$ ,

(5.21) $$ \begin{align} \{\mkern-6mu\{\psi(a),\psi(b)\}\mkern-6mu\}^{(n)}=\psi^{\otimes 2}(\{\mkern-6mu\{a,b\}\mkern-6mu\})\,. \end{align} $$

It is also clear that $\psi (\Phi _1)=\Phi _1^{(n)}$ and $\psi (\Phi _2)=\Phi _2^{(n)}$ , so $\psi $ is an isomorphism of Hamiltonian double quasi-Poisson algebras.

5.5 Proof of Proposition 5.7

5.5.1 Preparation

Recall that the double quasi-Poisson bracket on $\mathcal {B}_{n}^{\operatorname {loc}}$ takes the form (3.10) on the elements $\{a_i,b_i\}$ since it is obtained by localisation of $\mathcal {B}(\Gamma _n)$ .

Lemma 5.8. The double bracket on $\mathcal {B}(\Gamma _n)$ is such that

(5.22a) $$ \begin{align} \{\mkern-6mu\{(a_1,\dots, b_i),b_j\}\mkern-6mu\}&= \begin{cases} \;\;\; \frac12\Big(b_j(a_1,\dots, b_i)\otimes e_2-b_j\otimes (a_1,\dots, b_i)\Big) \,, &\mbox{if } i<j, \\ \;\;\;\frac12\Big(b_j(a_1,\dots, b_i)\otimes e_2+b_j\otimes (a_1,\dots, b_i)\Big) \,, &\mbox{if } i\geq j , \end{cases} \end{align} $$

(5.22b) $$ \begin{align} \{\mkern-6mu\{(a_1,\dots, b_i),a_j\}\mkern-6mu\}&= \begin{cases} \;\;\; \frac12 \Big(e_2\otimes (a_1,\dots, b_i)a_j- (a_1,\dots, b_i)\otimes a_j\Big)\,, &\mbox{if } i<j\!-\!1 , \\ \;\;\; \frac{1}{2}\Big(e_2\otimes (a_1,\dots, b_i)a_{j}- (a_1,\dots, b_i)\otimes a_{j}\Big)& \\ \qquad\; +\,e_2\otimes (a_1,\dots, a_i)\,, &\mbox{if } i=j\!-\!1 ,\\ -\frac12 \Big(e_2\otimes (a_1,\dots, b_i)a_j+ (a_1,\dots, b_i)\otimes a_j\Big)\,, & \mbox{if } i\geq j. \end{cases}\\[-15pt]\nonumber \end{align} $$

In particular, these identities hold in $\mathcal {B}_{n}^{\operatorname {loc}}$ .

Proof. If $i<j$ , these relations have been derived in Lemma 4.2.

If $i\geq j$ , note by inspecting Theorem 3.4 that the double brackets involving the elements $\{a_{\ell },b_{\ell }\mid 1\leq \ell \leq i\}$ of $\mathcal {B}(\Gamma _n)$ are exactly the same as the double brackets in $\mathcal {B}(\Gamma _i)$ . But since $(a_1,\dots , b_i)$ is the component in $e_2 \mathcal {B}(\Gamma _i) e_2$ of the multiplicative moment map of $\mathcal {B}(\Gamma _i)$ , the claim followsFootnote ⁹ from equation (2.11).

We compute double brackets with the $(2i-1)$ -th Euler continuant $(a_1,b_1,\ldots ,a_{i-1},b_{i-1},a_i)$ . By equation (3.5), we can use the decomposition

(5.23) $$ \begin{align} (a_1,\ldots,a_i)=\sum_{\ell=2}^i (a_1,\ldots,b_{\ell-1})a_{\ell}\,+a_1\,.\\[-20pt]\nonumber \end{align} $$

(The formula holds for $i=1$ since the sum is over an empty set, hence we omit it).

Lemma 5.9. The double bracket on $\mathcal {B}(\Gamma _n)$ is such that

$$ \begin{align*} \begin{aligned} \{\mkern-6mu\{(a_1,\dots, a_i),b_j\}\mkern-6mu\}&= \begin{cases} \frac12\Big(b_j(a_1,\dots, a_i)\otimes e_2+ e_1 \otimes (a_1,\dots, a_i)b_j\Big)\, , &\mbox{if } i<j, \\ \frac12\Big(b_i(a_1,\dots, a_i)\otimes e_2+ e_1 \otimes (a_1,\dots, a_i)b_i\Big) & \\ \qquad +\, e_1 \otimes (a_1,\ldots,b_{i-1}) \,, &\mbox{if } i=j, \\ \frac12\Big(b_j(a_1,\dots, a_i)\otimes e_2 - e_1 \otimes (a_1,\dots, a_i)b_j\Big) \,, &\mbox{if } i>j. \end{cases} \end{aligned} \end{align*} $$

In particular, these identities hold in $\mathcal {B}_{n}^{\operatorname {loc}}$ .

Proof. If $i< j$ , by equation (5.22a), we have

$$ \begin{align*} \{\mkern-6mu\{(a_1,\dots, a_i),b_j\}\mkern-6mu\}=& \sum_{\ell=2}^i \Big[\{\mkern-6mu\{(a_1,\ldots,b_{\ell-1}),b_j\}\mkern-6mu\}\ast a_{\ell} + (a_1,\ldots,b_{\ell-1})\ast \{\mkern-6mu\{a_{\ell},b_j\}\mkern-6mu\}\Big]\,+\{\mkern-6mu\{a_1,b_j\}\mkern-6mu\} \\ =& \frac12\sum_{\ell=2}^i \Big(b_j(a_1,\dots, b_{\ell-1})\otimes e_2-b_j\otimes (a_1,\dots, b_{\ell-1})\Big) \ast a_{\ell}\\ &+\frac12\sum_{\ell=2}^i (a_1,\ldots,b_{\ell-1})\ast \Big(e_1 \otimes a_{\ell} b_j + b_j a_{\ell} \otimes e_2\Big) +\frac12 (e_1 \otimes a_1 b_j + b_j a_1 \otimes e_2) \\ =&\frac12b_j\left[\sum_{\ell=2}^i (a_1,\ldots,b_{\ell-1})a_{\ell}\,+a_1\right]\otimes e_2 +\frac12 e_1 \otimes \left[\sum_{\ell=2}^i (a_1,\ldots,b_{\ell-1})a_{\ell}\,+a_1\right] b_j\,, \end{align*} $$

which is the desired result by equation (5.23).

If $i=j=1$ , this is just $\{\mkern -6mu\{a_1,b_1\}\mkern -6mu\}$ equation (3.10c). If $i=j$ is distinct from $1$ , we use again equation (5.22a) to get

$$ \begin{align*} &\{\mkern-6mu\{(a_1,\dots, a_i),b_i\}\mkern-6mu\} \\ =& \sum_{\ell=2}^i \Big[\{\mkern-6mu\{(a_1,\ldots,b_{\ell-1}),b_i\}\mkern-6mu\}\ast a_{\ell} + (a_1,\ldots,b_{\ell-1})\ast \{\mkern-6mu\{a_{\ell},b_i\}\mkern-6mu\}\Big]\,+\{\mkern-6mu\{a_1,b_i\}\mkern-6mu\} \\ =& \frac12\sum_{\ell=2}^i \Big(b_i(a_1,\dots, b_{\ell-1})\otimes e_2-b_i\otimes (a_1,\dots, b_{\ell-1})\Big) \ast a_{\ell} \\ &+\frac12\sum_{\ell=2}^{i-1} (a_1,\ldots,b_{\ell-1})\ast \Big(e_1 \otimes a_{\ell} b_i + b_i a_{\ell} \otimes e_2\Big) \\ &+\frac12 (a_1,\ldots,b_{i-1})\ast \Big(e_1 \otimes a_i b_i + b_i a_i \otimes e_2 + 2 e_1 \otimes e_2\Big) +\frac12 (e_1 \otimes a_1 b_j + b_j a_1 \otimes e_2) \\ =&\frac12b_i\left[\sum_{\ell=2}^i (a_1,\ldots,b_{\ell-1})a_{\ell}\,+a_1\right]\otimes e_2 +\frac12 e_1 \otimes \left[\sum_{\ell=2}^i (a_1,\ldots,b_{\ell-1})a_{\ell}\,+a_1\right] b_i + e_1 \otimes (a_1,\ldots,b_{i-1}), \end{align*} $$

which is the claimed result by equation (5.23).

If $i>j$ , we only prove the result for $j\neq 1$ and leave the easier case $j=1$ to the reader. We can use equation (5.22a) one last time to obtain

$$ \begin{align*} \{\mkern-6mu\{(a_1,\dots, a_i),b_j\}\mkern-6mu\}= & \sum_{\ell=2}^i \Big[\{\mkern-6mu\{(a_1,\ldots,b_{\ell-1}),b_j\}\mkern-6mu\}\ast a_{\ell} + (a_1,\ldots,b_{\ell-1})\ast \{\mkern-6mu\{a_{\ell},b_j\}\mkern-6mu\}\Big]\,+\{\mkern-6mu\{a_1,b_j\}\mkern-6mu\} \\ =& \frac12\sum_{\ell=2}^j \Big(b_j(a_1,\dots, b_{\ell-1})\otimes e_2-b_j\otimes (a_1,\dots, b_{\ell-1})\Big) \ast a_{\ell} \\ &+\frac12\sum_{\ell=j+1}^i \Big(b_j(a_1,\dots, b_{\ell-1})\otimes e_2 + b_j\otimes (a_1,\dots, b_{\ell-1})\Big) \ast a_{\ell} \\ &+\frac12\sum_{\ell=2}^{j-1} (a_1,\ldots,b_{\ell-1})\ast \Big(e_1 \otimes a_{\ell} b_j + b_j a_{\ell} \otimes e_2\Big) \\ &+\frac12 (a_1,\ldots,b_{j-1})\ast \Big(e_1 \otimes a_j b_j + b_j a_j \otimes e_2 + 2 e_1 \otimes e_2\Big) \\ &-\frac12 (a_1,\ldots,b_{j})\ast \Big(e_1 \otimes a_{j+1} b_j + b_j a_{j+1} \otimes e_2 + 2 e_1 \otimes e_2\Big) \\ &-\frac12\sum_{\ell=j+2}^{i} (a_1,\ldots,b_{\ell-1})\ast \Big(e_1 \otimes a_{\ell} b_j + b_j a_{\ell} \otimes e_2\Big) +\frac12 (e_1 \otimes a_1 b_j + b_j a_1 \otimes e_2) \\ =&\frac12b_j\left[\sum_{\ell=2}^i (a_1,\ldots,b_{\ell-1})a_{\ell}\,+a_1\right]\otimes e_2 -\frac12 e_1 \otimes \left[\sum_{\ell=2}^i (a_1,\ldots,b_{\ell-1})a_{\ell}\,+a_1\right] b_j \\ &+ e_1 \otimes \Big[ (a_1,\ldots,b_{j-1}) + (a_1,\ldots,a_{j})b_j - (a_1,\ldots, b_j) \Big], \end{align*} $$

which is the claimed result if we use equation (5.23) as well as the recursive relation for $(a_1,\ldots , b_j)$ , so that the last term disappears.

Lemma 5.10. The double bracket on $\mathcal {B}(\Gamma _n)$ is such that

$$ \begin{align*} \begin{aligned} \{\mkern-6mu\{(a_1,\dots, a_i),a_j\}\mkern-6mu\}&= \begin{cases} \;\;\, \frac12\Big(a_j \otimes (a_1,\dots, a_i) - (a_1,\dots, a_i) \otimes a_j\Big)\, , &\mbox{if } i\geq j, \\ -\frac12\Big(a_j \otimes (a_1,\dots, a_i) + (a_1,\dots, a_i) \otimes a_j\Big)\, , &\mbox{if } i=j-1,\, j\neq 1. \\ \end{cases} \end{aligned} \end{align*} $$

In particular, these identities hold in $\mathcal {B}_{n}^{\operatorname {loc}}$ .

Proof. We first assume that $i>j>1$ , and we leave the case $i>j$ with $j=1$ to the reader. Using equations (5.23) and (5.22b), we can write $\{\mkern -6mu\{(a_1,\dots , a_i),a_j\}\mkern -6mu\}$ as

$$ \begin{align*} \sum_{\ell=2}^i \Big[\{\mkern-6mu\{(a_1,&\ldots,b_{\ell-1}),a_j\}\mkern-6mu\}\ast a_{\ell} + (a_1,\ldots,b_{\ell-1})\ast \{\mkern-6mu\{a_{\ell},a_j\}\mkern-6mu\}\Big]\,+\{\mkern-6mu\{a_1,a_j\}\mkern-6mu\} \\ =& \frac12\sum_{\ell=2}^{j-1} \Big(e_2 \otimes (a_1,\dots, b_{\ell-1}) a_j - (a_1,\dots, b_{\ell-1})\otimes a_j \Big) \ast a_{\ell} \\ &+\frac12 \Big(e_2 \otimes (a_1,\dots, b_{j-1}) a_j - (a_1,\dots, b_{j-1})\otimes a_j + 2 e_2 \otimes (a_1,\ldots,a_{j-1}) \Big) \ast a_j \\ &-\frac12\sum_{\ell=j+1}^i \Big(e_2 \otimes (a_1,\dots, b_{\ell-1}) a_j + (a_1,\dots, b_{\ell-1})\otimes a_j \Big) \ast a_{\ell} \\ &-\frac12\sum_{\ell=2}^{j-1} (a_1,\ldots,b_{\ell-1})\ast \Big( a_{\ell} \otimes a_j + a_j \otimes a_{\ell}\Big) \\ &+\frac12\sum_{\ell=j+1}^{i} (a_1,\ldots,b_{\ell-1})\ast \Big( a_{\ell} \otimes a_j + a_j \otimes a_{\ell}\Big) -\frac12 (a_1 \otimes a_j + a_j \otimes a_1) \\ =&\frac12a_j\otimes \left[\sum_{\ell=2}^i (a_1,\ldots,b_{\ell-1})a_{\ell}\,+a_1\right] -\frac12 \left[\sum_{\ell=2}^i (a_1,\ldots,b_{\ell-1})a_{\ell}\,+a_1\right] \otimes a_j \\ &+ a_j \otimes \left[ (a_1,\ldots,a_{j-1}) -\sum_{\ell=2}^{j-1} (a_1,\ldots,b_{\ell-1})a_{\ell} -a_1 \right] \\ =& \frac12\Big(a_j \otimes (a_1,\dots, a_i) - (a_1,\dots, a_i) \otimes a_j\Big)\,. \end{align*} $$

Next, let $i=j$ . If $j=1$ , $\{\mkern -6mu\{a_1,a_1\}\mkern -6mu\}=0$ , and we are done. If $j\neq 1$ , we compute $\{\mkern -6mu\{(a_1,\dots , a_j),a_j\}\mkern -6mu\}$ as in the previous case and find

$$ \begin{align*} \sum_{\ell=2}^j \Big[\{\mkern-6mu\{(a_1,&\ldots,b_{\ell-1}),a_j\}\mkern-6mu\}\ast a_{\ell} + (a_1,\ldots,b_{\ell-1})\ast \{\mkern-6mu\{a_{\ell},a_j\}\mkern-6mu\}\Big]\,+\{\mkern-6mu\{a_1,a_j\}\mkern-6mu\} \\ =& \frac12\sum_{\ell=2}^{j-1} \Big(e_2 \otimes (a_1,\dots, b_{\ell-1}) a_j - (a_1,\dots, b_{\ell-1})\otimes a_j \Big) \ast a_{\ell} \\ &+\frac12 \Big(e_2 \otimes (a_1,\dots, b_{j-1}) a_j - (a_1,\dots, b_{j-1})\otimes a_j + 2 e_2 \otimes (a_1,\ldots,a_{j-1}) \Big) \ast a_j \\ &-\frac12\sum_{\ell=2}^{j-1} (a_1,\ldots,b_{\ell-1})\ast \Big( a_{\ell} \otimes a_j + a_j \otimes a_{\ell}\Big) -\frac12 (a_1 \otimes a_j + a_j \otimes a_1) \\ =&\frac12a_j\otimes \left[(a_1,\dots, b_{j-1}) a_j + 2 (a_1,\ldots,a_{j-1})-\sum_{\ell=2}^{j-1} (a_1,\ldots,b_{\ell-1})a_{\ell}\,-a_1\right] \\ & -\frac12 \left[\sum_{\ell=2}^j (a_1,\ldots,b_{\ell-1})a_{\ell}\,+a_1\right] \otimes a_j \\ =&\frac12a_j\otimes \Big[(a_1,\dots, b_{j-1}) a_j + (a_1,\ldots,a_{j-1}) \Big] -\frac12 (a_1,\dots, a_j) \otimes a_j \\ =& \frac12\Big(a_j \otimes (a_1,\dots, a_j) - (a_1,\dots, a_j) \otimes a_j\Big)\,, \end{align*} $$

where we needed to use the recursive relation for $(a_1,\ldots ,a_{j})$ in the last line.

Finally, for $i=j-1$ with $j\neq 1$ , we get for $\{\mkern -6mu\{(a_1,\dots , a_{j-1}),a_j\}\mkern -6mu\}$ that

$$ \begin{align*} \sum_{\ell=2}^{j-1} \Big[\{\mkern-6mu\{(a_1,&\ldots,b_{\ell-1}),a_j\}\mkern-6mu\}\ast a_{\ell} + (a_1,\ldots,b_{\ell-1})\ast \{\mkern-6mu\{a_{\ell},a_j\}\mkern-6mu\}\Big]\,+\{\mkern-6mu\{a_1,a_j\}\mkern-6mu\} \\ =& \frac12\sum_{\ell=2}^{j-1} \Big(e_2 \otimes (a_1,\dots, b_{\ell-1}) a_j - (a_1,\dots, b_{\ell-1})\otimes a_j \Big) \ast a_{\ell} \\ &-\frac12\sum_{\ell=2}^{j-1} (a_1,\ldots,b_{\ell-1})\ast \Big( a_{\ell} \otimes a_j + a_j \otimes a_{\ell}\Big) -\frac12 (a_1 \otimes a_j + a_j \otimes a_1) \\ =&-\frac12a_j\otimes \Big[ \sum_{\ell=2}^{j-1} (a_1,\ldots,b_{\ell-1})a_{\ell}\,+ a_1\Big] -\frac12 \left[\sum_{\ell=2}^{j-1} (a_1,\ldots,b_{\ell-1})a_{\ell}\,+a_1\right] \otimes a_j \\ =& - \frac12\Big(a_j \otimes (a_1,\dots, a_{j-1}) + (a_1,\dots, a_{j-1}) \otimes a_j\Big)\,. \\[-38pt] \end{align*} $$

We now perform computations in $\mathcal {B}_{n}^{\operatorname {loc}}$ involving the elements $\{a_j'\mid 1\leq j\leq n\}$ for which we use the definition in equation (5.15).

Lemma 5.11. The double bracket on $\mathcal {B}_{n}^{\operatorname {loc}}$ is such that

$$ \begin{align*} \begin{aligned} \{\mkern-6mu\{a_i',a_j\}\mkern-6mu\}&= \begin{cases} \frac12\Big(a_i' \otimes a_j + a_j \otimes a_i' \Big)\, , &\mbox{if } i> j, \\ \frac12\Big(a_j' \otimes a_j + a_j \otimes a_j' \Big) - a_j' \otimes a_j' \,, &\mbox{if } i=j. \\ \end{cases} \end{aligned} \end{align*} $$

Proof. We will repeatedly use equations (3.10a), (5.22b) and Lemma 5.10. If $i>j>1$ , we have that $\{\mkern -6mu\{a_i',a_j\}\mkern -6mu\}$ equals

$$ \begin{align*} & \left\{\!\!\left\{\Big(a_i+(a_1,\ldots,b_{i-1})^{-1}(a_1,\ldots,a_{i-1})\Big),a_j\right\}\!\!\right\} \\ =& \frac12 (a_i \otimes a_j + a_j \otimes a_i) +\frac12 (a_1,\ldots,b_{i-1})^{-1}\ast \Big(a_j \otimes (a_1,\ldots,a_{i-1}) - (a_1,\ldots,a_{i-1}) \otimes a_j \Big) \\ &-\!\frac12 (a_1,\ldots,b_{i-1})^{-1}\!\!\ast \!\big(\!- e_2\! \otimes \!(a_1,\ldots,b_{i-1})a_j \!- (a_1,\ldots,b_{i-1}) \!\otimes\! a_j \big) \!\ast\! (a_1,\ldots,b_{i-1})^{-1}(a_1,\ldots,a_{i-1}) \\ =&\frac12 \Big[a_i+(a_1,\ldots,b_{i-1})^{-1}(a_1,\ldots,a_{i-1})\Big]\otimes a_j + \frac12 a_j \otimes \Big[a_i+(a_1,\ldots,b_{i-1})^{-1}(a_1,\ldots,a_{i-1})\Big]\\ =& \frac12\Big(a_i' \otimes a_j + a_j \otimes a_i' \Big) \,. \end{align*} $$

If $i=j=1$ , we have $\{\mkern -6mu\{a_1',a_1\}\mkern -6mu\}=\{\mkern -6mu\{a_1,a_1\}\mkern -6mu\}=0$ . If $i=j$ is distinct from $1$ , we have that $\{\mkern -6mu\{a_j',a_j\}\mkern -6mu\}$ can be computed as follows:

$$ \begin{align*} & \left\{\!\!\left\{\Big(a_j+(a_1,\ldots,b_{j-1})^{-1}(a_1,\ldots,a_{j-1})\Big),a_j\right\}\!\!\right\} \\ =& -\frac12 (a_1,\ldots,b_{j-1})^{-1}\ast \Big(a_j \otimes (a_1,\ldots,a_{j-1}) + (a_1,\ldots,a_{j-1}) \otimes a_j \Big) \\ &-\frac12 (a_1,\ldots,b_{j-1})^{-1}\ast \Big( e_2 \otimes (a_1,\ldots,b_{j-1})a_j - (a_1,\ldots,b_{j-1}) \otimes a_j \\ &\qquad \qquad \qquad \qquad \qquad \quad + 2 e_2 \otimes (a_1,\ldots,a_{j-1})\Big) \ast (a_1,\ldots,b_{j-1})^{-1}(a_1,\ldots,a_{j-1}) \\ =&-\frac12 (a_1,\ldots,b_{j-1})^{-1}(a_1,\ldots,a_{j-1})\otimes a_j - \frac12 a_j \otimes (a_1,\ldots,b_{j-1})^{-1}(a_1,\ldots,a_{j-1})\\ &-(a_1,\ldots,b_{j-1})^{-1}(a_1,\ldots,a_{j-1}) \otimes (a_1,\ldots,b_{j-1})^{-1}(a_1,\ldots,a_{j-1}) \\ =& -\frac12\Big((a_j' - a_j) \otimes a_j + a_j \otimes (a_j'-a_j) \Big) - (a_j'-a_j)\otimes (a_j'-a_j) \\ =&\frac12\Big(a_j' \otimes a_j + a_j \otimes a_j' \Big) - a_j' \otimes a_j' \,.\\[-38pt] \end{align*} $$

5.5.2 Proof of Proposition 5.7

We show that we can write $\Phi $ as in equation (5.18), then we show that the double bracket takes the form (5.17) on the generators $\{ a_{\ell }',b_{\ell }'\}$ .

Step 1: Rewriting $\Phi $ . For $1\leq \ell \leq n$ , we prove that,

(5.24) $$ \begin{align} \begin{aligned} (a_1,\ldots,b_{\ell})&=(e_2+a_1'b_1')\cdots(e_2+a_{\ell}' b_{\ell}')\,, \\ (b_{\ell},\ldots,a_1)&= (e_1+b_{\ell}'a_{\ell}')\cdots (e_1+b_1' a_1')\,. \end{aligned} \end{align} $$

This gives in particular that $\Phi $ takes the form (5.18) when $\ell =n$ . We prove equation (5.24) by induction. The case $\ell =1$ is obvious. If we assume that equation (5.24) holds for $\ell -1$ , we have

$$ \begin{align*} \begin{aligned} (a_1,\ldots,b_{\ell})&=(a_1,\ldots,b_{\ell-1})(e_2+a_{\ell} b_{\ell}) + (a_1,\ldots,a_{\ell-1})b_{\ell}\\ &=(a_1,\ldots,b_{\ell-1}) (e_2+\underbrace{[a_{\ell}+(a_1,\ldots,b_{\ell-1})^{-1}(a_1,\ldots,a_{\ell-1})]}_{=\,a_{\ell}'\text{ by equation } (5.15) }b_{\ell})\\ &=(e_2+a_1'b_1')\cdots(e_2+a_{\ell-1}' b_{\ell-1}')(e_2+a_{\ell}' b_{\ell}')\,, \\ (b_{\ell},\ldots,a_1)&=(e_1+b_{\ell} a_{\ell}) (b_{\ell-1},\ldots,a_1) + (b_{\ell-1},\ldots,a_1)\\ &=(e_1+b_{\ell} \underbrace{[a_{\ell}+(a_{\ell-1},\ldots,a_1)(b_{\ell-1},\ldots,a_1)^{-1}]}_{=\,a_{\ell}'\text{ by equation } (5.16)} (b_{\ell-1},\ldots,a_1) \\ &=(e_1+b_{\ell}'a_{\ell}')(e_1+b_{\ell-1}'a_{\ell-1}')\cdots (e_1+b_1'a_1')\,. \end{aligned} \end{align*} $$

Step 2: The double bracket $\{\mkern -6mu\{b_i',b_j'\}\mkern -6mu\}$ . It is clear that equation (5.17b) holds since it is just equation (3.10b) because by definition $b^{\prime }_{\ell }=b_{\ell }$ .

Step 3: The double bracket $\{\mkern -6mu\{a_i',b_j'\}\mkern -6mu\}$ . We have to check equation (5.17c). It directly holds for $i=1$ by equation (3.10c) since $\{\mkern -6mu\{a_1',b_j'\}\mkern -6mu\}=\{\mkern -6mu\{a_1,b_j\}\mkern -6mu\}$ . So, hereafter, we assume $i\neq 1$ . Then, we need to compute

$$ \begin{align*} \begin{aligned} \{\mkern-6mu\{a_i',b_j'\}\mkern-6mu\}=& \left\{\!\!\left\{\Big(a_i+(a_1,\ldots,b_{i-1})^{-1}(a_1,\ldots,a_{i-1})\Big), b_j\right\}\!\!\right\} \\ =& \{\mkern-6mu\{a_i,b_j\}\mkern-6mu\} + (a_1,\ldots,b_{i-1})^{-1}\ast \{\mkern-6mu\{(a_1,\ldots,a_{i-1}) ,b_j\}\mkern-6mu\} \\ &- (a_1,\ldots,b_{i-1})^{-1}\ast \{\mkern-6mu\{(a_1,\ldots,b_{i-1}),b_j\}\mkern-6mu\} \ast (a_1,\ldots,b_{i-1})^{-1} (a_1,\ldots,a_{i-1}) \,, \end{aligned} \end{align*} $$

and this can be obtained from equation (3.10c) and Lemmas 5.8 and 5.9. If $i<j$ , we find

$$ \begin{align*} \{\mkern-6mu\{a_i',b_j'\}\mkern-6mu\}=&\frac12 (e_1 \otimes a_ib_j+b_j a_i \otimes e_2) \\ &+ \frac12 \Big(b_j (a_1,\ldots,a_{i-1}) \otimes (a_1,\ldots,b_{i-1})^{-1} + e_1 \otimes (a_1,\ldots,b_{i-1})^{-1}(a_1,\ldots,a_{i-1}) b_j \Big) \\ &-\frac12 \Big(b_j(a_1,\ldots,a_{i-1}) \otimes (a_1,\ldots,b_{i-1})^{-1} - b_j (a_1,\ldots,b_{i-1})^{-1}(a_1,\ldots,a_{i-1}) \otimes e_2 \Big) \\ =&\frac12 e_1 \otimes \Big[a_i+(a_1,\ldots,b_{i-1})^{-1}(a_1,\ldots,a_{i-1})\Big] b_j \\ &+\frac12\, b_j \Big[a_i+(a_1,\ldots,b_{i-1})^{-1}(a_1,\ldots,a_{i-1}) \Big] \otimes e_2\\ =& \frac12 (e_1 \otimes a_i' b_j' + b_j' a_i'\otimes e_2)\,, \end{align*} $$

as expected. The proof is basically the same for $i=j$ as we add a term $+e_1 \otimes e_2$ coming from $\{\mkern -6mu\{a_i,b_i\}\mkern -6mu\}$ .

If $i=j+1$ , we get

$$ \begin{align*} \{\mkern-6mu\{a_{j+1}',b_j'\}\mkern-6mu\} =&-\frac12 (e_1 \otimes a_{j+1}b_j+b_j a_{j+1} \otimes e_2) - e_1 \otimes e_2 \\ &+ \frac12 \Big(b_j (a_1,\ldots,a_{j}) \otimes (a_1,\ldots,b_{j})^{-1} + e_1 \otimes (a_1,\ldots,b_{j})^{-1}(a_1,\ldots,a_{j}) b_j\Big) \\ &+ e_1 \otimes (a_1,\ldots,b_j)^{-1} (a_1,\ldots,b_{j-1})\\ &-\frac12 \Big(b_j(a_1,\ldots,a_{j}) \otimes (a_1,\ldots,b_{j})^{-1} + b_j (a_1,\ldots,b_{j})^{-1}(a_1,\ldots,a_{j}) \otimes e_2 \Big) \\ =&-\frac12 e_1 \!\otimes \!\Big[a_{j+1}+(a_1,\ldots,b_{j})^{-1}(a_1,\ldots,a_{j})\Big] b_j \\ &-\frac12\, b_j \Big[a_{j+1}+(a_1,\ldots,b_{j})^{-1}(a_1,\ldots,a_{j}) \Big]\! \otimes e_2\\ &+ e_1 \otimes (a_1,\ldots,b_{j})^{-1}\Big[ (a_1,\ldots,a_{j}) b_j + (a_1,\ldots,b_{j-1}) - (a_1,\ldots,b_{j})\Big] \\ =&- \frac12 (e_1 \otimes a_i' b_j' + b_j' a_i'\otimes e_2)\,, \end{align*} $$

where we used the recursive relation for $(a_1,\ldots , b_j)$ .

If $i>j+1$ , we finally obtain

$$ \begin{align*} \{\mkern-6mu\{a_i',b_j'\}\mkern-6mu\} =&-\frac12 (e_1 \otimes a_ib_j+b_j a_i \otimes e_2) \\ &+ \frac12 \Big(b_j (a_1,\ldots,a_{i-1}) \otimes (a_1,\ldots,b_{i-1})^{-1} - e_1 \otimes (a_1,\ldots,b_{i-1})^{-1}(a_1,\ldots,a_{i-1}) b_j \Big) \\ &-\frac12 \Big(b_j(a_1,\ldots,a_{i-1}) \otimes (a_1,\ldots,b_{i-1})^{-1} + b_j (a_1,\ldots,b_{i-1})^{-1}(a_1,\ldots,a_{i-1}) \otimes e_2 \Big) \\ =&-\frac12 e_1 \!\otimes \Big[a_i+(a_1,\ldots,b_{i-1})^{-1}(a_1,\ldots,a_{i-1})\Big] b_j\\ &-\frac12\, b_j \Big[a_i+(a_1,\ldots,b_{i-1})^{-1}(a_1,\ldots,a_{i-1}) \Big]\! \otimes e_2\\ =&- \frac12 (e_1 \otimes a_i' b_j' + b_j' a_i'\otimes e_2)\,, \end{align*} $$

as we wanted.

Step 4: Interlude (some double brackets involving $a_i'$ ). For the sake of clarity, let us rewrite the following two identities which were obtained in Lemma 5.11 and in equation (5.17b) with $b_j=b_j'$ (the latter has just been checked):

(5.25a) $$ \begin{align} \{\mkern-6mu\{a_i',b_j\}\mkern-6mu\}=&-\frac12 \big(e_1 \otimes a_i' b_j + b_j a_i' \otimes e_2\big)\,, & &\text{ for } i>j\,, \end{align} $$

(5.25b) $$ \begin{align} \{\mkern-6mu\{a_i',a_j\}\mkern-6mu\}=&\frac12 \big(a_i' \otimes a_j + a_j \otimes a_i'\big)\,, & &\text{ for } i>j\,. \end{align} $$

Lemma 5.12. The double bracket on $\mathcal {B}_{n}^{\operatorname {loc}}$ is such that

(5.26) $$ \begin{align} \{\mkern-6mu\{a_i',(a_j,b_j)\}\mkern-6mu\}= \frac12 \Big( a_i' \otimes (a_j,b_j) - (a_j,b_j) a_i' \otimes e_2 \Big)\,, \quad \text{ for } i>j\,. \end{align} $$

Proof. Using equation (5.25), we get for $i>j$ that

$$ \begin{align*} \{\mkern-6mu\{a_i',(a_j,b_j)\}\mkern-6mu\}=& \{\mkern-6mu\{a_i',a_jb_j\}\mkern-6mu\} \\ =& \frac12 (a_i' \otimes a_j + a_j \otimes a_i')b_j -\frac12 a_j(e_1 \otimes a_i' b_j + b_j a_i' \otimes e_2) \\ =& \frac12 a_i' \otimes (e_2+a_j b_j) -\frac12 (e_2+a_j b_j) a_i' \otimes e_2\,.\\[-38pt] \end{align*} $$

Lemma 5.13. The double bracket on $\mathcal {B}_{n}^{\operatorname {loc}}$ is such that

(5.27) $$ \begin{align} \{\mkern-6mu\{a_i',(a_1,\ldots,b_j)\}\mkern-6mu\}= \frac12 \Big( a_i' \otimes (a_1,\ldots,b_j) - (a_1,\ldots,b_j) a_i' \otimes e_2 \Big)\,, \quad \text{ for } i>j\,. \end{align} $$

Proof. We prove the result by induction on j. Note that the case $j=1$ is directly obtained from Lemma 5.12. If $i>j>1$ , by equation (3.6),

$$ \begin{align*} \{\mkern-6mu\{a_i',(a_1,\ldots,b_j)\}\mkern-6mu\}=\{\mkern-6mu\{a_i',(a_1,\ldots,b_{j-1})(a_j,b_j)\}\mkern-6mu\} + \sum_{\ell=2}^{j-1}\{\mkern-6mu\{a_i',(a_1,\ldots,b_{\ell-1})a_{\ell} b_j \}\mkern-6mu\} + \{\mkern-6mu\{a_i',a_1 b_j\}\mkern-6mu\}\\[-20pt] \end{align*} $$

which can be computed using the induction hypothesis together with Lemma 5.12 and equation (5.25) as

$$ \begin{align*} =&\frac12 \Big( a_i' \otimes (a_1,\ldots,b_{j-1}) - (a_1,\ldots,b_{j-1}) a_i' \otimes e_2 \Big) (a_j,b_j) \\ &+\frac12 (a_1,\ldots,b_{j-1})\Big( a_i' \otimes (a_j,b_j) - (a_j,b_j) a_i' \otimes e_2 \Big) \\ &+\frac12 \sum_{\ell=2}^{j-1} \Big( a_i' \otimes (a_1,\ldots,b_{\ell-1}) - (a_1,\ldots,b_{\ell-1}) a_i' \otimes e_2 \Big) a_{\ell} b_j \\ &+\frac12 \sum_{\ell=2}^{j-1} (a_1,\ldots,b_{\ell-1}) \Big(a_i' \otimes a_{\ell} + a_{\ell} \otimes a_i'\Big) b_j - \frac12 \sum_{\ell=2}^{j-1} (a_1,\ldots,b_{\ell-1}) a_{\ell} \Big(e_1 \otimes a_i' b_j + b_j a_i' \otimes e_2\Big) \\ &+ \frac12 (a_i' \otimes a_1 + a_1 \otimes a_i') b_j - \frac12 a_1 (e_1 \otimes a_i' b_j + b_j a_i' \otimes e_2) \\ =&\frac12 a_i' \otimes \Big[(a_1,\ldots,b_{j-1})(a_j,b_j) + \sum_{\ell=2}^{j-1} (a_1,\ldots,b_{\ell-1})a_{\ell} b_j + a_1 b_j \Big] \\ &-\frac12 \Big[ (a_1,\ldots,b_{j-1})(a_j,b_j) + \sum_{\ell=2}^{j-1} (a_1,\ldots,b_{\ell-1})a_{\ell} b_j + a_1 b_j \Big] a_i' \otimes e_2\,, \end{align*} $$

and this is the claimed result due to equation (3.6).

Lemma 5.14. The double bracket on $\mathcal {B}_{n}^{\operatorname {loc}}$ is such that

(5.28) $$ \begin{align} \{\mkern-6mu\{a_i',(a_1,\ldots,a_j)\}\mkern-6mu\}= \frac12 \Big( a_i' \otimes (a_1,\ldots,a_j) + (a_1,\ldots,a_j) \otimes a_i' \Big)\,, \quad \text{ for } i>j\,. \end{align} $$

Proof. We prove the result by induction on j, and the case $j=1$ is a special case of equation (5.25b). We can use equation (5.23) to write

$$ \begin{align*} \{\mkern-6mu\{a_i',(a_1,\ldots,a_j)\}\mkern-6mu\}=\sum_{\ell=2}^{j}\{\mkern-6mu\{a_i',(a_1,\ldots,b_{\ell-1})a_{\ell}\}\mkern-6mu\} + \{\mkern-6mu\{a_i',a_1\}\mkern-6mu\}\,, \end{align*} $$

and due to equation (5.25b) and Lemma 5.13 we find

$$ \begin{align*} =&\frac12 \sum_{\ell=2}^{j} \Big( a_i' \otimes (a_1,\ldots,b_{\ell-1}) - (a_1,\ldots,b_{\ell-1}) a_i' \otimes e_2 \Big) a_{\ell} \\ &+\frac12\sum_{\ell=2}^{j} (a_1,\ldots,b_{\ell-1}) \Big( a_i' \otimes a_{\ell} + a_{\ell} \otimes a_i' \Big) + \Big( a_i' \otimes a_1 + a_1 \otimes a_i' \Big) \\ =&\frac12 a_i' \otimes \Big[\sum_{\ell=2}^{j}(a_1,\ldots,b_{\ell-1}) a_{\ell} + a_1 \Big] + \frac12 \Big[\sum_{\ell=2}^{j}(a_1,\ldots,b_{\ell-1}) a_{\ell} + a_1 \Big] \otimes a_i'\,, \end{align*} $$

which is the desired result by equation (5.23).

Step 5: The double bracket $\{\mkern -6mu\{a_i',a_j'\}\mkern -6mu\}$ . It remains to prove that $\{\mkern -6mu\{a_i',a_j'\}\mkern -6mu\}$ takes the form (5.17a), and it suffices to do so for $i\geq j$ due to the cyclic antisymmetry.

Firstly, we consider the case when $i>j$ . We write $a_j'$ using equation (5.15). The case $j=1$ is just equation (5.25b), while for $i>j>1$ we have that

$$ \begin{align*} \{\mkern-6mu\{a_i',a_j'\}\mkern-6mu\}= \left\{\!\!\left\{a_i',\Big(a_j+(a_1,\ldots,b_{j-1})^{-1}(a_1,\ldots,a_{j-1})\Big)\right\}\!\!\right\} \,. \end{align*} $$

This can be computed using equation (5.25b) and Lemmas 5.13 and 5.14, as

$$ \begin{align*} =&\frac12 (a_i'\otimes a_j + a_j \otimes a_i') +\frac12(a_1,\ldots,b_{j-1})^{-1} \Big( a_i' \otimes (a_1,\ldots,a_{j-1}) + (a_1,\ldots,a_{j-1}) \otimes a_i' \Big) \\ &-\frac12(a_1,\ldots,b_{j-1})^{-1} \Big( a_i' \otimes (a_1,\ldots,b_{j-1}) - (a_1,\ldots,b_{j-1}) a_i' \otimes e_2 \Big) (a_1,\ldots,b_{j-1})^{-1}(a_1,\ldots,a_{j-1}) \\ =&\frac12 a_i' \otimes \Big[ a_j+(a_1,\ldots,b_{j-1})^{-1}(a_1,\ldots,a_{j-1}) \Big] +\frac12 \Big[a_j+(a_1,\ldots,b_{j-1})^{-1}(a_1,\ldots,a_{j-1}) \Big]\otimes a_i' \\ =&\frac12( a_i' \otimes a_j' + a_j' \otimes a_i')\,. \end{align*} $$

When $i=j$ , this is obviously zero if $j=1$ by equation (3.10a). If $j> 1$ we use Lemma 5.11 to get $\{\mkern -6mu\{a_j',a_j\}\mkern -6mu\}$ so that as in the previous case we can compute

$$ \begin{align*} \{\mkern-6mu\{a&_j',a_j'\}\mkern-6mu\} \\ =&\frac12 (a_j'\otimes a_j + a_j \otimes a_j') - a_j' \otimes a_j'\\ &+\frac12(a_1,\ldots,b_{j-1})^{-1} \Big( a_j' \otimes (a_1,\ldots,a_{j-1}) + (a_1,\ldots,a_{j-1}) \otimes a_j' \Big) \\ &-\frac12(a_1,\ldots,b_{j-1})^{-1} \Big( a_j' \otimes (a_1,\ldots,b_{j-1}) - (a_1,\ldots,b_{j-1}) a_j' \otimes e_2 \Big) (a_1,\ldots,b_{j-1})^{-1}(a_1,\ldots,a_{j-1}) \\ =&\frac12 a_j' \otimes \Big[a_j+(a_1,\ldots,b_{j-1})^{-1}(a_1,\ldots,a_{j-1}) \Big] \\ &+\frac12 \Big[a_j+(a_1,\ldots,b_{j-1})^{-1}(a_1,\ldots,a_{j-1}) \Big] \otimes a_j' - a_j' \otimes a_j' \,, \end{align*} $$

which is zero, as expected. So $\{\mkern -6mu\{a^{\prime }_j,a^{\prime }_j\}\mkern -6mu\}$ takes the form (5.17a), and Proposition 5.17 follows.

5.6 Towards the quasi-bisymplectic form

Working in full generalities as in §2.1, we consider a finitely generated unital algebra A over $B=\oplus _{s\in I} \Bbbk e_s$ and the A-bimodule $\Omega ^1_B A$ of (B-relative) noncommutative differential $1$ -forms. Following [Reference Cuntz and Quillen26], we put $\Omega _B A:=T_A \Omega ^1_B A$ and note that $\operatorname {d}\colon A\to \Omega ^1_B A$ can be extended to a differential on $\Omega _B A$ . If A is freely generated by elements $\{a_j\mid j\in J\}$ , with $|J|<\infty $ and $a_j=e_{h_j} a_j e_{t_j}$ , we can introduce the map

(5.29) $$ \begin{align} \tilde{\operatorname{d}}:=\sum_{j\in J}\operatorname{d}\! a_j\frac{\partial}{\partial a_j}:A\longrightarrow A\otimes \Omega_B^1A\,, \quad \tilde{\operatorname{d}}(b)= \sum_{j\in J}\left(\frac{\partial b}{\partial a_j}\right)'\otimes \operatorname{d}\! a_j\left(\frac{\partial b}{\partial a_j}\right)''\, \end{align} $$

for the double derivations $\partial /\partial a_j\in \operatorname {\mathbb {D}er}_BA$ given by $\partial a_i/\partial a_j=\delta _{ij} \, e_{h_j}\otimes e_{t_j}$ . Since we have that $A\otimes \Omega _B^1A\subset \Omega _BA\otimes \Omega _BA$ , we can decompose the differential as

$$ \begin{align*} \operatorname{d}:A\stackrel{\tilde{\operatorname{d}}}{\longrightarrow} \Omega_BA\otimes \Omega_BA \stackrel{\tilde{\operatorname{m}}}{\longrightarrow} \Omega_BA\, \end{align*} $$

through the multiplication map $\tilde {\operatorname {m}}$ on $\Omega _BA$ . By iterating $\tilde {\operatorname {d}}$ , we can introduce

(5.30) $$ \begin{align} \begin{aligned} \operatorname{D}^2:=&\tilde{\operatorname{m}}\circ (\operatorname{d}\otimes 1)\circ \tilde{\operatorname{d}}:A\longrightarrow \Omega^2_BA \subset \Omega_BA\,,\\ \operatorname{D}^2(b)=&\sum_{i,j\in J} \left(\frac{\partial}{\partial a_i}\left(\frac{\partial b}{\partial a_j}\right)'\right)' \operatorname{d}\! a_i \left(\frac{\partial}{\partial a_i}\left(\frac{\partial b}{\partial a_j}\right)'\right)'' \operatorname{d}\! a_j\left(\frac{\partial b}{\partial a_j}\right)''\,. \end{aligned} \end{align} $$

The above construction can be applied to the algebra $\Bbbk \overline {\Gamma }_n$ and after localisation to $\mathcal {B}(\Gamma _n)$ . Using the double derivations given in equation (3.12), we have

$$ \begin{align*} \tilde{\operatorname{d}}=\sum_{j=1}^n \left( \operatorname{d}\! a_j\frac{\partial}{\partial a_j}+ \operatorname{d}\! b_j\frac{\partial}{\partial b_j}\right)\,. \end{align*} $$

In particular, we can define the map $\operatorname {D}^2:\mathcal {B}(\Gamma _n)\to \Omega _B^2(\mathcal {B}(\Gamma _n))$ by equation (5.30). Motivated by the construction of Paluba [Reference Paluba38, §5.6] (see below), we introduce the element

(5.31) $$ \begin{align} \omega_n=\frac12 (b_n,\ldots,a_1)^{-1} \operatorname{D}^2(b_n,\ldots,a_1) - \frac12 (a_1,\ldots,b_n)^{-1} \operatorname{D}^2 (a_1,\ldots,b_n)\,. \end{align} $$

Note that we use the two Euler continuants that are inverted in $\mathcal {B}(\Gamma _n)$ and which occur in the moment map $\Phi $ , as defined in equation (3.11).

Since we are not going to prove this result here,Footnote ¹⁰ the precise definitions (of quasi-bisymplectic algebra, nondegeneracy or compatibility) are omitted, and the reader can find them in [Reference Van den Bergh42]. Let us nevertheless remark that by [Reference Van den Bergh42, Proposition 6.1], if the conjecture holds, we will have that $\omega _n$ turns any representation space $\operatorname {Rep}(\overline {\Gamma }_n,(d_1,d_2))$ (with its natural $\operatorname {GL}_{d_1}(\Bbbk )\times \operatorname {GL}_{d_2}(\Bbbk )$ -action) into a quasi-Hamiltonian space in the sense of [Reference Alekseev, Malkin and Meinrenken1]. Moreover, the induced $2$ -form will be compatible with the quasi-Poisson bracket on $\operatorname {Rep}(\overline {\Gamma }_n,(d_1,d_2))$ induced by the double quasi-Poisson bracket from Theorem 3.4, which will be nondegenerate by [Reference Van den Bergh42, §6–7]. We should emphasise that, in the previous sentence, compatibility and nondegeneracy are meant in the geometric sense of [Reference Alekseev, Kosmann-Schwarzbach and Meinrenken2], not at the noncommutative/associative algebra level stated in the conjecture. To strengthen Conjecture 5.15, we explain how the induced results on $\operatorname {Rep}(\overline {\Gamma }_n,(d_1,d_2))$ hold in the particular case of $\Bbbk =\mathbb {C}$ thanks to the factorisation presented in this section.

Fix $\Bbbk =\mathbb {C}$ and $d_1,d_2\geq 1$ . Let ${\mathrm {P}}_{\operatorname {Rep}}$ and $\omega _{\operatorname {Rep}}$ be the (matrix) bivectors and $2$ -forms induced on $\operatorname {Rep}(\overline {\Gamma }_n,(d_1,d_2))$ by ${\mathrm {P}_n}$ and $\omega _n$ , respectively. Their explicit expressions can be obtained using the formalism of [Reference Van den Bergh41, §7].