&m_{M \, \widetilde{\otimes} \, N} : (M \otimes N) \otimes (M \otimes N) \xrightarrow{\mathrm{id}_M \,\otimes \,c_{N,M} \,\otimes\, \mathrm{id}_N} M \otimes M \otimes N \otimes N \xrightarrow{m_M \otimes m_N} M \otimes N, \\
&1_{M \, \widetilde{\otimes} \, N} = 1_M \otimes 1_N.
\end{align*}

This endows $M \otimes N$ with a structure of (right) $H$-module-algebra, denoted by $M \, \widetilde{\otimes} \, N$ and called the braided tensor product of $M$ and $N$ <cit.>. It is easy to see that $\widetilde{\otimes}$ is associative. We often write a pure tensor $m \otimes n \in M \, \widetilde{\otimes} \, N$ as $m \, \widetilde{\otimes} \, n$ to stress that we use the braided product in $M \otimes N$. Explicitly:

\begin{equation}\label{multiplicationInBraidedTensorProduct}
(m \,\widetilde{\otimes}\, n)(m' \,\widetilde{\otimes}\, n') = \sum_{(R)}m (m' \cdot R_{(1)}) \,\widetilde{\otimes}\, (n \cdot R_{(2)})n'.
\end{equation}

The embeddings of $k$-vector spaces

\begin{equation}\label{embeddingsBraided}
\fonc{j_M}{M}{M \, \widetilde{\otimes} \, N}{m}{m \,\widetilde{\otimes}\, 1} \quad \fonc{j_N}{N}{M \, \widetilde{\otimes} \, N}{n}{1 \,\widetilde{\otimes}\, n}
\end{equation}

are morphisms of $H$-module-algebras and we have

\begin{align}
j_M(m) j_N(n) &= m \,\widetilde{\otimes}\, n,\label{factorisationBraidedTensors}\\
j_N(n)j_M(m) &= \sum_{(R)} j_M(m \cdot R_{(1)}) j_N(n \cdot R_{(2)}).\label{productBraided}
\end{align}

$\mathcal{L}_{g,n}(H) = \mathcal{L}_{1,0}(H)^{\widetilde{\otimes} g} \: \widetilde{\otimes} \: \mathcal{L}_{0,1}(H)^{\widetilde{\otimes} n}$ as right $H$-module-algebras.

Note that $K_g = \bigl( \Delta^{(3)} \otimes \Delta^{(4g-1)} \bigr)(R')$; then by definition of the multiplication $m_{\mathcal{L}_{g+1,0}(H)}$ in $\mathcal{L}_{g+1,0}(H)$ (recall (<ref>) and the definition of the twisted tensor product) we have for $u,v \in \mathcal{L}_{g,0}(H)$ and $x,y \in \mathcal{L}_{1,0}(H)$:

\begin{align*}
&m_{\mathcal{L}_{g+1,0}(H)}\bigl( (u \otimes x) \otimes (v \otimes y) \bigr)\\
=\:&\sum_{(K_g)} m_{\mathcal{L}_{g,0}(H)}\bigl( u \otimes \bigl(v\cdot (K_g)_{(2)}\bigr) \bigr) \otimes m_{\mathcal{L}_{1,0}(H)}\bigl( \bigl(x \cdot (K_g)_{(1)}\bigr) \otimes y \bigr)\\
=\:&\sum_{(R)} m_{\mathcal{L}_{g,0}(H)}\bigl( u \otimes \bigl(v\cdot \Delta^{(4g-1)}(R_{(1)})\bigr) \bigr) \otimes m_{\mathcal{L}_{1,0}(H)}\bigl( \bigl(x \cdot \Delta^{(3)}(R_{(2)})\bigr) \otimes y \bigr)\\
=\:&\sum_{(R)} m_{\mathcal{L}_{g,0}(H)}\bigl( u \otimes \mathrm{coad}^r(R_{(1)})(v) \bigr) \otimes m_{\mathcal{L}_{1,0}(H)}\bigl( \mathrm{coad}^r(R_{(2)})(x) \otimes y \bigr) =(u \, \widetilde{\otimes} \, x)(v \, \widetilde{\otimes} \, y).
\end{align*}

For the third equality we used (<ref>). Hence $\mathcal{L}_{g+1,0}(H) = \mathcal{L}_{g,0}(H) \, \widetilde{\otimes} \, \mathcal{L}_{1,0}(H)$ and it follows that $\mathcal{L}_{g,0}(H) = \mathcal{L}_{1,0}(H)^{\widetilde{\otimes} g}$. One shows similarly that $\mathcal{L}_{0,n}(H) = \mathcal{L}_{0,1}(H)^{\widetilde{\otimes} n}$ and that $\mathcal{L}_{g,0}(H) \otimes^{F_{g,n}} \mathcal{L}_{0,n}(H) = \mathcal{L}_{g,0}(H) \, \widetilde{\otimes} \, \mathcal{L}_{0,n}(H)$.

Thanks to Proposition <ref> and (<ref>) we see that there are embeddings of $H$-module-algebras $j_i : \mathcal{L}_{1,0}(H) \to \mathcal{L}_{g,n}(H)$ for $1 \leq i \leq g$ and $j_i : \mathcal{L}_{0,1}(H) \to \mathcal{L}_{g,n}(H)$ for $g+1 \leq i \leq g+n$, given by

\begin{equation}\label{embeddingsL01L10InLgn}
j_i(x) = 1^{\widetilde{\otimes} (i-1)} \, \widetilde{\otimes} \, x \, \widetilde{\otimes} \, 1^{\widetilde{\otimes}(g+n-i)}
\end{equation}

where $1$ is the unit of $\mathcal{L}_{1,0}(H)$ (i.e. $\varepsilon \otimes \varepsilon$) or of $\mathcal{L}_{0,1}(H)$ (i.e. $\varepsilon$) depending on the position.
Recall the embeddings $\mathfrak{i}_A, \mathfrak{i}_B$ from (<ref>); we denote

\begin{equation}\label{embeddingsFrakI}
\begin{array}{l}
\mathfrak{i}_{A(i)} = j_i \circ \mathfrak{i}_A, \:\: \mathfrak{i}_{B(i)} = j_i \circ \mathfrak{i}_B \quad \text{ for } 1 \leq i \leq g,\\[.2em]
\mathfrak{i}_{M(i)} = j_i \quad \text{ for } g+1 \leq i \leq g+n
\end{array}
\end{equation}

which are all embeddings $\mathcal{L}_{0,1}(H) \to \mathcal{L}_{g,n}(H)$. Then from (<ref>) and (<ref>) we find

\begin{equation}\label{monomialsLgn}
\begin{array}{rl}
&\mathfrak{i}_{B(1)}(\varphi_1) \, \mathfrak{i}_{A(1)}(\varphi_2) \ldots \mathfrak{i}_{B(g)}(\varphi_{2g-1}) \, \mathfrak{i}_{A(g)}(\varphi_{2g}) \, \mathfrak{i}_{M(g+1)}(\varphi_{2g+1}) \ldots \mathfrak{i}_{M(g+n)}(\varphi_{2g+n})\\[.5em]
= \!\!\!\!& j_1(\varphi_1 \otimes \varphi_2) \ldots j_g(\varphi_{2g-1} \otimes \varphi_{2g}) \, j_{g+1}(\varphi_{2g+1}) \ldots j_{g+n}(\varphi_{2g+n})\\[.5em]
=\!\!\!\!& (\varphi_1 \otimes \varphi_2) \, \widetilde{\otimes} \, \ldots \, \widetilde{\otimes} \, (\varphi_{2g-1} \otimes \varphi_{2g}) \, \widetilde{\otimes} \, \varphi_{2g+1} \, \widetilde{\otimes} \, \ldots \, \widetilde{\otimes} \, \varphi_{2g+n}
\end{array}
\end{equation}

and any element in $\mathcal{L}_{g,n}(H)$ is a linear combination of such elements.

The product in $\mathcal{L}_{g,n}(H)$ is entirely determined by the following equalities:

\begin{align*}
\mathfrak{i}_{X(i)}(\varphi) \, \mathfrak{i}_{X(i)}(\psi) &= \mathfrak{i}_{X(i)}(\varphi \psi)\\
\mathfrak{i}_{A(i)}(\varphi) \, \mathfrak{i}_{B(i)}(\psi) &= \sum_{(R^1),\ldots,(R^4)} \!\!\!\mathfrak{i}_{B(i)}\bigl( R_{(2)}^4 R_{(1)}^3 \rhd \psi \lhd R_{(1)}^1 R_{(1)}^2 \bigr) \, \mathfrak{i}_{A(i)}\bigl( R_{(2)}^3 S(R_{(2)}^1) \rhd \varphi \lhd R_{(2)}^2 R_{(1)}^4  \bigr)\\
\mathfrak{i}_{Y(j)}(\varphi) \, \mathfrak{i}_{X(i)}(\psi) &= \sum_{(R^1),\ldots,(R^4)} \!\!\! \mathfrak{i}_{X(i)}\bigl( S(R^3_{(1)}R^4_{(1)}) \rhd \psi \lhd R^1_{(1)}R^2_{(1)} \bigr) \, \mathfrak{i}_{Y(j)}\bigl( S(R^1_{(2)}R^3_{(2)}) \rhd \varphi \lhd R^2_{(2)}R^4_{(2)} \bigr)
\end{align*}

for all $\varphi, \psi \in \mathcal{L}_{0,1}(H)$, where $1 \leq i < j \leq g+n$ and $X(i)$ (resp. $Y(j)$) is indifferently $A(i)$, $B(i)$ or $M(i)$ (resp. $A(j)$, $B(j)$ or $M(j)$) and $R^1, \ldots, R^4$ are four copies of $R \in H^{\otimes 2}$.

It is clear that these formulas determine the product since they allow us to compute the product of any two elements from (<ref>). The first formula simply expresses the fact that $\mathfrak{i}_{X(i)}$ is a morphism of algebras. The second formula is obtained from (<ref>) and (<ref>) thanks to the algebra embedding $j_i : \mathcal{L}_{1,0}(H) \to \mathcal{L}_{g,n}(H)$. Let us prove the third formula. Assume for instance that $i < j \leq g$. Using Proposition <ref> and the associativity of $\widetilde{\otimes}$ we can write $\mathcal{L}_{g,n}(H) = \mathcal{L}_{i,0}(H) \, \widetilde{\otimes} \, \mathcal{L}_{g-i,n}(H)$; let $j_{\mathcal{L}_{i,0}(H)}$ and $j_{\mathcal{L}_{g-i,n}(H)}$ be the corresponding $H$-module-algebra embeddings (see (<ref>)). It is easy to see that

\[ \mathfrak{i}^{(g,n)}_{X(i)} = j_{\mathcal{L}_{i,0}(H)} \circ \mathfrak{i}^{(i,0)}_{X(i)}, \qquad \mathfrak{i}^{(g,n)}_{Y(j)} = j_{\mathcal{L}_{g-i,n}(H)} \circ \mathfrak{i}^{(g-i,n)}_{Y(j-i)} \]

where $\mathfrak{i}^{(g,n)}_{X(l)}$, $\mathfrak{i}^{(i,0)}_{X(l)}$ and $\mathfrak{i}^{(g-i,n)}_{X(l)}$ are the $H$-module-algebra embeddings from (<ref>) for $\mathcal{L}_{g,n}(H)$, $\mathcal{L}_{i,0}(H)$ and $\mathcal{L}_{g-i,n}(H)$ respectively. Hence we have

\begin{align*}
&\mathfrak{i}^{(g,n)}_{Y(j)}(\varphi) \, \mathfrak{i}^{(g,n)}_{X(i)}(\psi) =  j_{\mathcal{L}_{g-i,n}(H)}\bigl(\mathfrak{i}^{(g-i,n)}_{Y(j-i)}(\varphi)\bigr) j_{\mathcal{L}_{i,0}(H)}\bigl(\mathfrak{i}^{(i,0)}_{X(i)}(\psi)\bigr)\\
=\:& \sum_{(R)} j_{\mathcal{L}_{i,0}(H)}\!\left(\mathrm{coad}^r(R_{(1)})\bigl(\mathfrak{i}^{(i,0)}_{X(i)}(\psi)\bigr)\right) j_{\mathcal{L}_{g-i,n}(H)}\!\left(\mathrm{coad}^r(R_{(2)})\bigr(\mathfrak{i}^{(g-i,n)}_{Y(j-i)}(\varphi)\bigl)\right)\\
=\:& \sum_{(R), (R_{(1)}), (R_{(2)})} j_{\mathcal{L}_{i,0}(H)}\!\left(\mathfrak{i}^{(i,0)}_{X(i)}\!\left( S((R_{(1)})_{(2)}) \rhd \psi \lhd (R_{(1)})_{(1)}\right)\right)\\[-1.3em]
&\hspace{13em} \times j_{\mathcal{L}_{g-i,n}(H)}\!\left(\mathfrak{i}^{(g-i,n)}_{Y(j-i)}\!\left( S((R_{(2)})_{(2)}) \rhd \psi \lhd (R_{(2)})_{(1)}\right)\right)\\
=\:&\sum_{(R^1),\ldots,(R^4)} \!\!\! \mathfrak{i}_{X(i)}^{(g,n)}\bigl( S(R^3_{(1)}R^4_{(1)}) \rhd \psi \lhd R^1_{(1)}R^2_{(1)} \bigr) \, \mathfrak{i}^{(g,n)}_{Y(j)}\bigl( S(R^1_{(2)}R^3_{(2)}) \rhd \varphi \lhd R^2_{(2)}R^4_{(2)} \bigr).
\end{align*}

For the second equality we used (<ref>), for the third equality we used the $H$-linearity of $\mathfrak{i}^{(i,0)}_{X(i)}$, $\mathfrak{i}^{(g-i,n)}_{Y(j-i)}$ and the definition (<ref>) of the $H$-action on $\mathcal{L}_{0,1}(H)$, and for the last equality we used that $(\Delta \otimes \Delta)(R) = R_{14}R_{13}R_{24}R_{23}$, which follows from the axioms of a $R$-matrix. The other cases for $i$ and $j$ are treated similarly.

Recall the notations introduced before Proposition <ref>, in particular the matrix $\overset{V}{M} \in \mathcal{L}_{0,1}(H) \otimes \mathrm{End}(V)$ for any finite-dimensional $H$-module $V$. Let

\begin{equation}\label{matricesABMgn}
\begin{array}{l}
\overset{V}{A}(i) = (\mathfrak{i}_{A(i)} \otimes \mathrm{id})\bigl( \overset{V}{M} \bigr), \:\: \overset{V}{B}(i) = (\mathfrak{i}_{B(i)} \otimes \mathrm{id})\bigl( \overset{V}{M} \bigr) \quad \text{for } 1 \leq i \leq g\\[.4em]
\overset{V}{M}(i) = (\mathfrak{i}_{M(i)} \otimes \mathrm{id})\bigl( \overset{V}{M} \bigr) \quad \text{for } g+1 \leq i \leq g+n
\end{array}
\end{equation}

which are elements in $\mathcal{L}_{g,n}(H) \otimes \mathrm{End}(V)$, or in other words matrices of size $\dim(V)$ with coefficients in $\mathcal{L}_{g,n}(H)$. In mathematical physics $\overset{V}{A}(i), \overset{V}{B}(i), \overset{V}{M}(i)$ are seen as quantum holonomy matrices associated to the usual generators $a_i, b_i, m_i$ of the fundamental group of the surface $\Sigma_{g,n}$ with an open disk removed.

The following relations hold true for any finite-dimensional $H$-modules $V,W$:

\begin{align*}
& \overset{V \otimes W}{X}\!\!\!\!(i) = \overset{V}{X}\hspace{-1pt}(i)_1\,(R')_{VW}\,\overset{W}{X}\hspace{-1pt}(i)_2\, (R')_{VW}^{-1} \quad \text{ for } 1 \leq i \leq g+n,\\
& R_{VW}\,\overset{V}{B}(i)_1\, (R')_{VW}\,\overset{W}{A}(i)_2 = \overset{W}{A}(i)_2\, R_{VW}\,\overset{V}{B}(i)_1\, R_{VW}^{-1} \quad \text{ for } 1 \leq i \leq g,\\
& R_{VW}\,\overset{V}{X}\hspace{-1pt}(i)_1\, R_{VW}^{-1} \,\overset{W}{Y}\hspace{-1pt}(j)_2 = \overset{W}{Y}\hspace{-1pt}(j)_2\, R_{VW}\,\overset{V}{X}\hspace{-1pt}(i)_1\, R_{VW}^{-1} \quad  \text{ for } 1 \leq i < j \leq g+n,
\end{align*}

where $X(i)$ (resp. $Y(j)$) is indifferently $A(i)$, $B(i)$ or $M(i)$ (resp. $A(j)$, $B(j)$ or $M(j)$).

These matrix equalities entirely describe the product in $\mathcal{L}_{g,n}(H)$.

This is completely parallel to the proof of Proposition <ref>. By the same arguments we already get that the first and second matrix relations are equivalent to the first and second formulas in Proposition <ref>. It remains to show that the third matrix relation is equivalent to the third formula in Proposition <ref>; this is similar to the computation in the proof of Proposition <ref> and is left as an exercise for the reader.

In many papers the matrix relations in Proposition <ref> are taken as a definition for the product in $\mathcal{L}_{g,n}(H)$.

§.§ The case $H = U_q^{\mathrm{ad}}(\mathfrak{g})$

Since $U_q^{\mathrm{ad}}(\mathfrak{g})$ is not quasitriangular in the usual sense (<ref>), the above definition of $\mathcal{L}_{g,n}(H)$ must be slightly adapted. We use the notations from sections <ref> and <ref>. Let us put by convention $\mathbb{A}_{0,0} = \mathcal{L}_{0,0}(q^{1/D}) = \mathbb{C}(q^{1/D})$. Recall from <ref> that $\mathcal{L}_{0,1}$ and $\mathcal{L}_{1,0}$ are module-algebras over the Hopf algebras $\mathbb{A}_{0,1}$ and $\mathbb{A}_{1,0}$ respectively. For $n \geq 0$ we define

\begin{align*}
&\mathbb{A}_{0,n+1} = \mathbb{A}_{0,n} \otimes^{J_n} \mathbb{A}_{0,1} \quad \text{with } J_n = \bigl( \Delta_{\mathbb{U}_q} \otimes \Delta_{\mathbb{U}_q}^{\odot n} \bigr)(R')\\
&\mathcal{L}_{0,n+1}(q^{1/D}) = \mathcal{L}_{0,n}(q^{1/D}) \otimes^{J_n} \mathcal{O}_q(q^{1/D})_F
\end{align*}

where $\textstyle R' = \sum_{(R)} R_{(2)} \otimes R_{(1)}$ is the flip of the $R$-matrix $R \in \mathbb{U}_q^{\otimes 2}$. For $g \geq 0$ let

\begin{align*}
&\mathbb{A}_{g+1,0} = \mathbb{A}_{g,0} \otimes^{K_g} \mathbb{A}_{1,0} \quad \text{with } K_g = \bigl( \Delta_{\mathbb{U}_q}^{(3)} \otimes (\Delta_{\mathbb{U}_q}^{(3)})^{\odot g} \bigr)(R')\\
&\mathcal{L}_{g+1,0} = \mathcal{L}_{g,0} \otimes^{K_g} \mathcal{L}_{1,0}.
\end{align*}

In <cit.> it is shown that restricting the product of $\mathcal{L}_{0,n}(q^{1/D})$ on the $\mathbb{C}(q)$-subspace $\mathcal{L}_{0,1}^{\otimes n}$ gives a $\mathbb{C}(q)$-subalgebra, which is denoted by $\mathcal{L}_{0,n}$. However, as we have seen in section <ref>, the algebra $\mathcal{L}_{1,0}$ (and hence $\mathcal{L}_{g,0}$) can only be defined over $\mathbb{C}(q^{1/D})$. This forces us to define the general case over $\mathbb{C}(q^{1/D})$, as follows:

\begin{align*}
\mathbb{A}_{g,n} &= \mathbb{A}_{g,0} \otimes^{F_{g,n}} \mathbb{A}_{0,n} \quad \text{with } F_{g,n} = \bigl(\Delta_{\mathbb{U}_q}^{\odot n} \otimes (\Delta_{\mathbb{U}_q}^{(3)})^{\odot g}\bigr)(R')\\
\mathcal{L}_{g,n} &= \mathcal{L}_{g,0} \otimes^{F_{g,n}} \mathcal{L}_{0,n}(q^{1/D}).
\end{align*}

Explicitly, $\mathcal{L}_{g,n}$ is the $\mathbb{C}(q^{1/D})$-vector space $\mathcal{O}_q(q^{1/D})^{\otimes (2g+n)}$ endowed with the product described in Proposition <ref>.

By construction $\mathcal{L}_{g,n}$ is a right $\mathbb{A}_{g,n}$-module-algebra. As in <ref>, we have a morphism of Hopf algebras $\Delta_{\mathbb{U}_q}^{(4g+2n-1)} : \mathbb{U}_q \to \mathbb{A}_{g,n}$. Combining this with the morphism of Hopf algebras $\iota : U_q \to \mathbb{U}_q$ introduced below (<ref>), we obtain that $\mathcal{L}_{g,n}$ is a right $U_q$-module-algebra for the action $\mathrm{coad}^r$ in (<ref>).

Note that the factorization of $\mathcal{L}_{g,n}$ as the braided tensor product $\mathcal{L}_{1,0}^{\widetilde{\otimes} g} \:\widetilde{\otimes} \: \mathcal{L}_{0,1}^{\widetilde{\otimes} n}$ (Prop. <ref>) holds true for $H = U_q^{\mathrm{ad}}$ as well, though the whole category of $U_q^{\mathrm{ad}}$-modules (not necessarily finite-dimensional) is not braided. This is because all the elements of $\mathcal{L}_{0,1}$ and $\mathcal{L}_{1,0}$ are locally-finite, and so the pairing (<ref>) and (<ref>) ensure that the formula (<ref>) for the product in $M \,\widetilde{\otimes} \, N$ is well-defined when $M = \mathcal{L}_{g,n}$, $N = \mathcal{L}_{g',n'}$ and the right action $\cdot$ is $\mathrm{coad}^r$.

The algebra $\mathcal{L}_{g,n}$ is finitely generated.

Recall the algebra $\mathrm{gr}_{\mathcal{F}_l}(\mathcal{L}_{0,1})$ from <ref>. It is proved in <cit.> that $\mathrm{gr}_{\mathcal{F}_l}(\mathcal{L}_{0,1})$ is finitely generated, thanks to the first item in Lemma <ref>. As a result $\mathcal{L}_{0,1}$ is finitely generated, see Lemma <ref>. Let $x_1, \ldots, x_m$ be the generators of $\mathcal{L}_{0,1}$. We deduce from (<ref>) that the elements $\mathfrak{i}_{B(k)}(x_s)$, $\mathfrak{i}_{A(k)}(x_s)$, $\mathfrak{i}_{M(l)}(x_s)$ with $1 \leq k \leq g$, $g+1 \leq l \leq g+n$, $1 \leq s \leq m$ generate $\mathcal{L}_{g,n}$.

§.§ Noetherianity of $\mathcal{L}_{g,n}$

Here are the steps of the proof:

* we introduce a filtration $\mathcal{G}_1$ of $\mathcal{L}_{g,n}$ whose graded algebra $\mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n})$ simplifies the third relation in Proposition <ref>,

* we construct a filtration $\mathcal{G}_2$ of $\mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n})$ thanks to the filtrations on $\mathcal{L}_{0,1}$ and on $\mathcal{L}_{1,0}$ from <ref>,

* using Lemma <ref>, we show that $\mathrm{gr}_{\mathcal{G}_2}\bigl( \mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n}) \bigr)$ is Noetherian,

* it follows from Lemma <ref> that $\mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n})$ is Noetherian and that $\mathcal{L}_{g,n}$ is Noetherian.

We write $\mathcal{L}_{0,1}$ instead of $\mathcal{L}_{0,1}(q^{1/D})$ for simplicity of notation. For $\mu, \omega \in P_+$ and $\lambda \in P$ consider the subspaces

\begin{align*}
&(\mathcal{L}_{0,1})_{\mu, \lambda} = \bigl\{ \varphi \in C(\mu) \, \big| \, \forall \, \nu \in P, \: \mathrm{coad}^r(K_{\nu}^{-1})(\varphi) = q^{(\lambda,\nu)}\varphi \bigr\},\\
&(\mathcal{L}_{1,0})_{\mu, \omega, \lambda} = \bigl\{ x \in C(\mu) \otimes C(\omega) \, \big| \, \forall \, \nu \in P, \: \mathrm{coad}^r(K_{\nu}^{-1})(x) = q^{(\lambda,\nu)}x \bigr\}
\end{align*}

where $\mathrm{coad}^r$ is defined in (<ref>) for $\mathcal{L}_{0,1}$ and in (<ref>) for $\mathcal{L}_{1,0}$, and $C(\mu)$ is the subspace of matrix coefficients of the irreducible $U_q^{\mathrm{ad}}$-module with highest weight $\mu$. For a finite sequence $\bigl( [\mu], [\lambda] \bigr)$, with $[\mu] = (\mu_1, \ldots, \mu_{2g+n}) \in (P_+)^{2g+n}$ and $[\lambda] = (\lambda_1, \ldots, \lambda_{g+n})\in P^{g+n}$, we put

\[ (\mathcal{L}_{g,n})_{[\mu], [\lambda]} = (\mathcal{L}_{1,0})_{\mu_1,\mu_2, \lambda_1} \otimes \ldots \otimes (\mathcal{L}_{1,0})_{\mu_{2g-1}, \mu_{2g}, \lambda_g} \otimes (\mathcal{L}_{0,1})_{\mu_{2g+1}, \lambda_{g+1}} \otimes \ldots \otimes (\mathcal{L}_{0,1})_{\mu_{2g+n}, \lambda_{g+n}}. \]

Recall the partial order $\preceq$ on $P$ defined in <ref>. For $\mu, \mu', \omega, \omega' \in P_+$ and $\lambda, \lambda' \in P$ we write $(\mu',\lambda') \preceq (\mu,\lambda)$ (resp. $(\mu',\omega',\lambda') \preceq (\mu,\omega,\lambda)$) to mean that $\mu' \preceq \mu$ and $\lambda' \preceq \lambda$ (resp. $\mu' \preceq \mu$ and $\omega' \preceq \omega$ and $\lambda' \preceq \lambda$). We say that $\bigl( [\mu'], [\lambda'] \bigr) \prec_1 \bigl( [\mu], [\lambda] \bigr)$ if and only if

\begin{align*}
&\:(\mu'_{2g+n}, \lambda'_{g+n}) \prec (\mu_{2g+n}, \lambda_{g+n})\\
\text{ or } &\: \Big( (\mu'_{2g+n}, \lambda'_{g+n}) = (\mu_{2g+n}, \lambda_{g+n}) \text{ and } (\mu'_{2g+n-1}, \lambda'_{g+n-1}) \prec (\mu_{2g+n-1}, \lambda_{g+n-1}) \Big) \text{ or } \ldots
\end{align*}

This defines a partial order $\preceq_1$ on $(P_+)^{2g+n} \times P^{g+n}$, which is a block lexicographic order starting from the right. For instance if $g=1$ and $n=1$ we have $\bigl( [\mu'_1,\mu'_2,\mu'_3], [\lambda'_1,\lambda'_2] \bigr) \prec_1 \bigl( [\mu_1,\mu_2,\mu_3], [\lambda_1,\lambda_2] \bigr)$ if and only if

\[ (\mu'_3, \lambda'_2) \prec (\mu_3, \lambda_2) \text{ or } \Big( (\mu'_3, \lambda'_2) = (\mu_3, \lambda_2) \text{ and } (\mu'_1, \mu'_2, \lambda'_1) \prec (\mu_1, \mu_2, \lambda_1) \Big). \]

Now let

\[ \Xi = \bigl\{ \bigl( [\mu], [\lambda] \bigr) \in (P_+)^{2g+n} \times P^{g+n} \, \big| \, (\mathcal{L}_{g,n})_{[\mu], [\lambda]} \neq 0 \bigr\}. \]

Then $(\Xi, \preceq_1)$ is an ordered abelian monoid and the order $\preceq_1$ is well-founded on $\Xi$. Moreover $\preceq_1$ satisfies the condition (<ref>) and it follows that

\[ \mathcal{G}_1^{[\mu], [\lambda]} = \bigoplus_{( [\mu'], [\lambda']) \preceq_1 ([\mu], [\lambda])} (\mathcal{L}_{g,n})_{[\mu'], [\lambda']} \qquad (\text{with } \bigl( [\mu'], [\lambda']\bigr) \in \Xi) \]

defines a filtration $\mathcal{G}_1$ of the vector space $\mathcal{L}_{g,n}$ indexed by $\Xi$. For any $1 \leq k \leq g$ and $g+1 \leq l \leq g+n$ it is convenient to use the obvious embeddings

\begin{equation}\label{specialSequenceForLgn}
S_k : P_+^2 \times P \to P_+^{2g+n} \times P^{g+n}, \qquad S_l : P_+ \times P \to P_+^{2g+n} \times P^{g+n}
\end{equation}

which are such that

\begin{equation}\label{propertySpecialSequences}
(\mathcal{L}_{g,n})_{S_k(\mu,\omega,\lambda)} = j_k\bigl( (\mathcal{L}_{1,0})_{\mu,\omega,\lambda} \bigr), \qquad (\mathcal{L}_{g,n})_{S_l(\mu, \lambda)} = j_{l}\bigl( (\mathcal{L}_{0,1})_{\mu, \lambda} \bigr)
\end{equation}

for $\mu,\omega \in P_+$, $\lambda \in P$, and where $j_k$, $j_l$ are the embeddings from (<ref>).

$\mathcal{G}_1$ is a filtration of the algebra $\mathcal{L}_{g,n}$.

\[ y_1 \otimes \ldots \otimes y_{g+n} = j_1(y_1) \ldots j_{g+n}(y_{g+n}) \in (\mathcal{L}_{g,n})_{[\mu'], [\lambda']} \]

where by definition we have $y_k \in (\mathcal{L}_{1,0})_{\mu'_{2k-1}, \mu'_{2k}, \lambda'_k}$ for $1 \leq k \leq g$ and $y_l \in (\mathcal{L}_{0,1})_{\mu'_{g+l}, \lambda'_l}$ for $g+1 \leq l \leq g+n$. Take an element $x \in (\mathcal{L}_{0,1})_{\mu, \lambda}$; so $j_m(x) \in (\mathcal{L}_{g,n})_{S_m(\mu, \lambda)}$ for any $m \in \{g+1, \ldots, g+n\}$ (see (<ref>) and (<ref>)). By Proposition <ref> and (<ref>) we have

\begin{align*}
&j_m(x) \, j_1(y_1) \ldots j_{g+n}(y_{g+n})\\
=\:\,&\sum_{(R^1), \ldots, (R^{m-1})} j_1\!\left( \mathrm{coad}^r(R^1_{(1)})(y_1) \right) \ldots j_{m-1}\!\left( \mathrm{coad}^r(R^{m-1}_{(1)})(y_{m-1}) \right)\\[-1em]
&\hphantom{\sum_{(R^1), \ldots, (R^{m-1})}} \:\:\times j_m\!\left( \mathrm{coad}^r\bigl( R^1_{(2)} \ldots R^{m-1}_{(2)} \bigr)(x)y_m \right) j_{m+1}(y_{m+1}) \ldots j_{g+n}(y_{g+n}).
\end{align*}

For any term in this sum and any $\nu \in P$, the fact that $\mathcal{L}_{0,1}$ is a $U_q$-module-algebras for $\mathrm{coad}^r$ together with (<ref>) gives

\begin{multline*}
\mathrm{coad}^r(K_{\nu}^{-1})\!\left( \mathrm{coad}^r\bigl(R^1_{(2)} \ldots R^{m-1}_{(2)}\bigr)(x) \, y_m \right)\\
= q^{(\nu,\, \lambda + \lambda'_m - \gamma_1 - \ldots - \gamma_{m-1})} \mathrm{coad}^r\bigl(R^1_{(2)} \ldots R^{m-1}_{(2)}\bigr)(x) \, y_m
\end{multline*}

for some $\gamma_1, \ldots, \gamma_{m-1} \in Q_+$. Moreover due to (<ref>) and (<ref>) we have

\[ \textstyle \mathrm{coad}^r\bigl(R^1_{(2)} \ldots R^{m-1}_{(2)}\bigr)(x) \, y_m \in \bigoplus_{\kappa \preceq \mu + \mu_{g+m}'} C(\kappa). \]

It follows from these two facts that $j_m(x) \, j_1(y_1) \ldots j_{g+n}(y_{g+n}) \in \mathcal{G}_1^{S_m(\mu, \lambda) + ([\mu'],[\lambda'])}$ by definition of $\mathcal{G}_1$. If now $x$ is an element in $(\mathcal{L}_{1,0})_{\mu, \omega, \lambda}$, $j_m(x) \in (\mathcal{L}_{g,n})_{S_m(\mu, \omega, \lambda)}$ for any $m \in \{1, \ldots, g\}$, and we have an analogous proof. These two particular cases imply the result.

Let us identify $\mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n})$ and $\mathcal{L}_{g,n}$ as vector spaces; namely $x \in (\mathcal{L}_{g,n})_{[\mu],[\lambda]}$ is identified with the coset $x + \mathcal{G}_1^{\prec_1 ([\mu],[\lambda])} \in \mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n})$. Denote by $\bullet$ the resulting product on $\mathcal{L}_{g,n}$. Consider the canonical projections

\[ \pi_{\eta} : \bigoplus_{\kappa \preceq \eta} C(\kappa) \to C(\eta), \quad \pi_{(\eta,\eta')} : \bigoplus_{\substack{\kappa \preceq \eta\\\kappa' \preceq \eta'}} C(\kappa) \otimes C(\kappa') \to C(\eta) \otimes C(\eta') \]

which allow us to define the truncated product $\overline{\,\cdot\,}$ on $\mathcal{L}_{0,1}$ and $\mathcal{L}_{1,0}$:

\[ x \, \overline{\,\cdot\,} \, y = \pi_{\mu+\mu'}(xy) \qquad (\text{resp. } x \, \overline{\,\cdot\,} \, y = \pi_{(\mu+\mu', \omega + \omega')}(xy)) \]

for $x \in C(\mu)$ and $y \in C(\mu')$ (resp. $x \in C(\mu) \otimes C(\omega)$ and $y \in C(\mu') \otimes C(\omega')$). We denote by $(\mathcal{L}_{0,1}, \overline{\,\cdot\,})$ and $(\mathcal{L}_{1,0}, \overline{\,\cdot\,})$ the vector spaces $\mathcal{L}_{0,1}$ and $\mathcal{L}_{1,0}$ endowed with the products $\overline{\,\cdot\,}$ above. The next lemma describes $\bullet$ thanks to the embeddings of vector spaces $j_k$ from (<ref>).

Let $k,l \in \{1, \ldots, g+n\}$, $\mu,\omega,\mu',\omega' \in P_+$ and $\lambda, \lambda' \in P$. Take $x$ in $(\mathcal{L}_{1,0})_{\mu,\omega,\lambda}$ or $(\mathcal{L}_{0,1})_{\mu, \lambda}$, depending if $k$ is $\leq g$ or $>g$. Similarly, take $y$ in $(\mathcal{L}_{1,0})_{\mu', \omega', \lambda'}$ or $(\mathcal{L}_{0,1})_{\mu',\lambda'}$, depending if $l$ is $\leq g$ or $>g$. Then

\begin{align*}
&j_k(x) \bullet j_k(y) = j_k(x \,\overline{\,\cdot\,}\, y) \quad \text{ for all } k,\\
&j_k(x) \bullet j_l(y) = j_k(x) j_l(y) \quad \text{if } k<l,\\
&j_k(x) \bullet j_l(y) = q^{(\lambda,\lambda')} j_l(y)j_k(x) \quad \text{if } k>l.
\end{align*}

Let $\simeq$ be the identification $\mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n}) \simeq \mathcal{L}_{g,n}$. Let us prove the first formula. Take for instance $k > g$. By assumption $j_k(x) \in (\mathcal{L}_{g,n})_{S_k(\mu, \lambda)}$ and $j_k(y) \in (\mathcal{L}_{g,n})_{S_k(\mu',\lambda')}$. Then $\textstyle xy \in \bigoplus_{\kappa \preceq \mu + \mu'} (\mathcal{L}_{0,1})_{\kappa,\lambda+\lambda'}$ and we have

\[ j_k(x) \bullet j_k(y) \simeq \left( j_k(x) + \mathcal{G}_1^{\prec_1 S_k(\mu,\lambda)} \right) \left( j_k(y) + \mathcal{G}_1^{\prec_1 S_k(\mu',\lambda')} \right) = j_k(xy) + \mathcal{G}_1^{\prec_1 S_k(\mu+\mu',\lambda+\lambda')} \simeq j_k(x\, \overline{\,\cdot\,} \,y). \]

The second formula is obtained similarly. For the third formula, recall the expression of $R$ in (<ref>) and its commutations rules with $K_{\nu}$ in (<ref>). Then from (<ref>) and the definition of $\mathcal{G}_1$ we get

\begin{align*}
&j_k(x) \bullet j_l(y) \simeq \left( j_k(x) + \mathcal{G}_1^{\prec_1 S_k(\mu,\lambda)} \right) \left( j_l(y) + \mathcal{G}_1^{\prec_1 S_l(\mu',\lambda')} \right) = j_k(x)j_l(y) + \mathcal{G}_1^{\prec_1 S_k(\mu,\lambda) + S_l(\mu',\lambda')}\\
=\:\,& \sum_{(R)} j_l\bigl( \mathrm{coad}^r(R_{(1)})(y) \bigr) j_k\bigl( \mathrm{coad}^r(R_{(2)})(x) \bigr) + \mathcal{G}_1^{\prec_1 S_k(\mu,\lambda) + S_l(\mu',\lambda')}\\
=\:\,& j_l\bigl( \mathrm{coad}^r(\Theta_{(1)})(y) \bigr) j_k\bigl( \mathrm{coad}^r(\Theta_{(2)})(x) \bigr) + \mathcal{G}_1^{\prec_1 S_k(\mu,\lambda) + S_l(\mu',\lambda')}\\
=\:\,& q^{(\lambda,\lambda')}j_l(y)j_k(x) + \mathcal{G}_1^{\prec_1 S_k(\mu,\lambda) + S_l(\mu',\lambda')} \simeq q^{(\lambda,\lambda')}j_l(y)j_k(x). \qedhere
\end{align*}

Recall from <ref> the filtration $\mathcal{F}_\ell$ on $\mathcal{L}_{0,1}$ indexed by $\Lambda$ (where $\Lambda$ is defined in (<ref>)) and the filtration $\mathcal{F}$ on $\mathcal{L}_{1,0}$ indexed by $\Lambda \times \Lambda$. Since we identified $\mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n})$ with $\mathcal{L}_{1,0}^{\otimes g} \otimes \mathcal{L}_{0,1}^{\otimes n}$ as a vector space, we can consider the family of subspaces $\mathcal{G}_2$ indexed by $\Lambda^{2g+n}$ and defined by

\begin{multline*}
\mathcal{G}_2^{\mu_1,\lambda_1,\eta_1,\kappa_1, \ldots, \mu_g,\lambda_g,\eta_g,\kappa_g, \mu_{g+1},\lambda_{g+1}, \ldots, \mu_{g+n},\lambda_{g+n}}\\
= \mathcal{F}^{\mu_1,\lambda_1,\eta_1,\kappa_1} \otimes \ldots \otimes \mathcal{F}^{\mu_g,\lambda_g,\eta_g,\kappa_g} \otimes \mathcal{F}_\ell^{\mu_{g+1},\lambda_{g+1}} \otimes \ldots \otimes \mathcal{F}_\ell^{\mu_{g+n},\lambda_{g+n}}.
\end{multline*}

We endow $P^{4g+2n}$ with the partial order $\preceq_2$ given by

\[ (p_1, \ldots, p_{4g+2n}) \preceq_2 (p'_1, \ldots, p'_{4g+2n}) \quad \iff \quad \forall \, i, \: p_i \preceq p'_i  \]

which gives by restriction a partial order on $\Lambda^{2g+n}$ which satisfies (<ref>) and is well-founded. Then $\mathcal{G}_2$ becomes a filtration of the vector space $\mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n})$.

$\mathcal{G}_2$ is a filtration of the algebra $\mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n})$.

Take $j_1(x_1) \ldots j_{g+n}(x_{g+n}) \in \mathcal{G}_2^S$ and $j_1(y_1) \ldots j_{g+n}(y_{g+n}) \in \mathcal{G}_2^{S'}$ for some $S,S' \in \Lambda^{2g+n}$, where $j_k$ is the embedding of vector spaces (<ref>). We can assume that these elements are homogeneous in $\mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n})$. Then according to Lemma <ref>:

\[ \bigl( j_1(x_1) \ldots j_{g+n}(x_{g+n}) \bigr) \bullet \bigl( j_1(y_1) \ldots j_{g+n}(y_{g+n}) \bigr) = q^Nj_1(x_1\,\overline{\,\cdot\,}\,y_1) \ldots j_{g+n}(x_{g+n}\,\overline{\,\cdot\,}\,y_{g+n}) \]

for some $N \in (1/D)\mathbb{Z}$. It is easy to see that $\mathcal{F}$ and $\mathcal{F}_\ell$ are also filtrations of $(\mathcal{L}_{1,0}, \overline{\,\cdot\,})$ and $(\mathcal{L}_{0,1}, \overline{\,\cdot\,})$ respectively, which gives the result.

For $1 \leq k \leq g$ and $g+1 \leq l \leq g+n$ we will use the obvious embeddings

\[ \mathbf{S}_k : \Lambda^2 \to (\Lambda^2)^g \times \Lambda^n, \qquad \mathbf{S}_l : \Lambda \to (\Lambda^2)^g \times \Lambda^n. \]

The algebra $\mathcal{L}_{g,n}$ is Noetherian.

We are going to show that $\mathrm{gr}_{\mathcal{G}_2}\bigl( \mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n}) \bigr)$ is Noetherian. Recall that $\mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n})$ is identified with $\mathcal{L}_{g,n}$ endowed with the product $\bullet$. To avoid confusion we do not identify again $\mathrm{gr}_{\mathcal{G}_2}\bigl( \mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n}) \bigr)$ with $\mathcal{L}_{g,n}$ as a vector space and instead we use cosets $x + \mathcal{G}_2^{\prec_2 S}$.

Note first that thanks to the first formula in Lemma <ref> the linear maps $\mathfrak{i}_{B(k)}, \mathfrak{i}_{A(k)}, \mathfrak{i}_{M(s)} : \mathcal{L}_{0,1} \to \mathcal{L}_{g,n}$ from (<ref>) are morphisms of algebras

\[ \overline{\mathfrak{i}}_{B(k)}, \overline{\mathfrak{i}}_{A(k)}, \overline{\mathfrak{i}}_{M(s)} : \bigl( \mathcal{L}_{0,1}, \overline{\,\cdot\,} \, \bigr) \to \mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n}). \]

It is easy to see that the filtration $\mathcal{F}_x$ ($x \in \{\ell,r\}$) of $\mathcal{L}_{0,1}$ is also a filtration of $\bigl( \mathcal{L}_{0,1}, \overline{\,\cdot\,} \, \bigr)$ and that $\mathrm{gr}_{\mathcal{F}_x}\bigl( \mathcal{L}_{0,1}, \overline{\,\cdot\,} \, \bigr) = \mathrm{gr}_{\mathcal{F}_x}(\mathcal{L}_{0,1})$. If $\varphi, \psi \in \mathcal{L}_{0,1}$ are such that $\mathfrak{i}_B(\varphi) \in \mathcal{F}^{\mu,\lambda,0,0}$ and $\mathfrak{i}_A(\psi) \in \mathcal{F}^{0,0,\eta,\kappa}$ we let

\[ \widehat{\mathfrak{i}}_{B(k)}(\varphi) = \overline{\mathfrak{i}}_{B(k)}(\varphi) + \mathcal{G}_2^{\prec_2 \mathbf{S}_k(\mu,\lambda,0,0)}, \qquad
\widehat{\mathfrak{i}}_{A(k)}(\psi) = \overline{\mathfrak{i}}_{A(k)}(\psi) + \mathcal{G}_2^{\prec_2 \mathbf{S}_k(0,0,\eta,\kappa)} \]

for $1 \leq k \leq g$. Similarly, for $\varphi \in \mathcal{F}^{\mu,\lambda}_\ell \subset \mathcal{L}_{0,1}$ and $g+1 \leq s \leq g+n$ we let

\[ \widehat{\mathfrak{i}}_{M(s)}(\varphi) = \overline{\mathfrak{i}}_{M(s)}(\varphi) + \mathcal{G}_2^{\prec_2 \mathbf{S}_s(\mu,\lambda)}. \]

Since $\overline{\mathfrak{i}}_{B(k)}, \overline{\mathfrak{i}}_{A(k)}, \overline{\mathfrak{i}}_{M(s)}$ are morphisms of algebras and

\[ \overline{\mathfrak{i}}_{B(k)}\bigl( \mathcal{F}_r^{\prec \mu,\lambda} \bigr) = \mathcal{G}_2^{\prec_2 \mathbf{S}_k(\mu,\lambda,0,0)}, \quad \overline{\mathfrak{i}}_{A(k)}\bigl( \mathcal{F}_\ell^{\prec \eta,\kappa} \bigr) = \mathcal{G}_2^{\prec_2 \mathbf{S}_k(0,0,\eta,\kappa)}, \quad \overline{\mathfrak{i}}_{M(s)}\bigl( \mathcal{F}_\ell^{\prec \mu,\lambda} \bigr) = \mathcal{G}_2^{\prec_2 \mathbf{S}_s(\mu,\lambda)} \]

we obtain morphisms of algebras

\begin{align*}
\widehat{\mathfrak{i}}_{B(k)} :&\:\, \mathrm{gr}_{\mathcal{F}_r}\bigl( \mathcal{L}_{0,1}, \overline{\,\cdot\,} \, \bigr) = \mathrm{gr}_{\mathcal{F}_r}(\mathcal{L}_{0,1}) \to \mathrm{gr}_{\mathcal{G}_2}\bigl( \mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n}) \bigr),\\
\widehat{\mathfrak{i}}_{A(k)}, \: \widehat{\mathfrak{i}}_{M(s)} :&\:\, \mathrm{gr}_{\mathcal{F}_\ell}\bigl( \mathcal{L}_{0,1}, \overline{\,\cdot\,} \, \bigr) = \mathrm{gr}_{\mathcal{F}_\ell}(\mathcal{L}_{0,1}) \to \mathrm{gr}_{\mathcal{G}_2}\bigl( \mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n}) \bigr).
\end{align*}

Recall from <ref> that $\mathrm{gr}_{\mathcal{F}_x}(\mathcal{L}_{0,1})$ with $x \in \{\ell,r\}$ is generated by a finite number of matrix coefficients $u_1, \ldots, u_p$. Thanks to the second formula in Lemma <ref>, we deduce that $\mathrm{gr}_{\mathcal{G}_2}\bigl( \mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n}) \bigr)$ is generated by the finite set of elements $\widehat{\mathfrak{i}}_{B(k)}(u_r)$, $\widehat{\mathfrak{i}}_{A(k)}(u_r)$, $\widehat{\mathfrak{i}}_{M(s)}(u_r)$ for $1 \leq k \leq g$, $g+1 \leq s \leq g+n$ and $1 \leq r \leq p$. By applying the morphism $\widehat{\mathfrak{i}}_{X(r)}$ where $X(r)$ is indifferently $A(r)$, $B(r)$ or $M(r)$ to (<ref>), we obtain

\[ \forall \, 1 \leq b < a \leq p, \quad \widehat{\mathfrak{i}}_{X(r)}(u_a)\,\widehat{\mathfrak{i}}_{X(r)}(u_b) = q_{ab} \, \widehat{\mathfrak{i}}_{X(r)}(u_b)\, \widehat{\mathfrak{i}}_{X(r)}(u_a) + \sum_{s=1}^{b-1} \sum_{t=1}^p \alpha^{ab}_{st} \, \widehat{\mathfrak{i}}_{X(r)}(u_s) \, \widehat{\mathfrak{i}}_{X(r)}(u_t) \]

with certain scalars $q_{ab} \in \mathbb{C}(q^{1/D})^{\times}$ and $\alpha^{ab}_{st} \in \mathbb{C}(q^{1/D})$. Similarly, it follows from (<ref>) that

\[ \forall \, 1 \leq b < a \leq p, \quad \widehat{\mathfrak{i}}_{A(r)}(u_a) \, \widehat{\mathfrak{i}}_{B(r)}(u_b) = q'_{ab} \, \widehat{\mathfrak{i}}_{B(r)}(u_b) \, \widehat{\mathfrak{i}}_{A(r)}(u_a) + \sum_{s=1}^{b-1} \sum_{t=1}^p \lambda^{ab}_{st} \, \widehat{\mathfrak{i}}_{B(r)}(u_s) \, \widehat{\mathfrak{i}}_{A(r)}(u_t) \]

for any $1 \leq r \leq g$ and with certain scalars $q'_{ab} \in \mathbb{C}(q^{1/D})^{\times}$ and $\lambda^{ab}_{st} \in \mathbb{C}(q^{1/D})$.
Finally, we see from the third formula in Lemma <ref> that

\[ \forall \, 1 \leq a, b \leq p, \quad \widehat{\mathfrak{i}}_{Y(s)}(u_a) \, \widehat{\mathfrak{i}}_{X(r)}(u_b) = q''_{ab} \, \widehat{\mathfrak{i}}_{X(r)}(u_b) \, \widehat{\mathfrak{i}}_{Y(s)}(u_a) \]

where $r<s$ and $X(r)$ (resp. $Y(s)$) is indifferently $A(r)$, $B(r)$ or $M(r)$ (resp. $A(s)$, $B(s)$ or $M(s)$) and $q''_{ab} \in \mathbb{C}(q^{1/D})^{\times}$.

We define a sequence $(\widehat{x}_1, \ldots, \widehat{x}_{(2g+n)p})$ by

\begin{align*}
&\widehat{x}_1 = \widehat{\mathfrak{i}}_{B(1)}(u_1), \ldots, \widehat{x}_p = \widehat{\mathfrak{i}}_{B(1)}(u_p), \: \widehat{x}_{p+1} = \widehat{\mathfrak{i}}_{A(1)}(u_1), \ldots, \widehat{x}_{2p} = \widehat{\mathfrak{i}}_{A(1)}(u_p),\\
&\widehat{x}_{2(g-1)p+1} = \widehat{\mathfrak{i}}_{B(g)}(u_1), \ldots, \widehat{x}_{2(g-1)p+p} = \widehat{\mathfrak{i}}_{B(g)}(u_p), \: \widehat{x}_{2(g-1)p+p+1} = \widehat{\mathfrak{i}}_{A(g)}(u_1), \ldots, \widehat{x}_{2gp} = \widehat{\mathfrak{i}}_{A(g)}(u_p),\\
&\widehat{x}_{2gp+1} = \widehat{\mathfrak{i}}_{M(g+1)}(u_1), \ldots, \widehat{x}_{(2g+1)p} = \widehat{\mathfrak{i}}_{M(g+1)}(u_p),\\
&\widehat{x}_{(2g+n-1)p+1} = \widehat{\mathfrak{i}}_{M(g+n)}(u_1), \ldots, \widehat{x}_{(2g+n)p} = \widehat{\mathfrak{i}}_{M(g+n)}(u_p).
\end{align*}

The relations above show that the generators $\widehat{x}_s$ satisfy the relations required by Lemma <ref>. It follows that $\mathrm{gr}_{\mathcal{G}_2}\bigl( \mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n}) \bigr)$ is Noetherian. Using Lemma <ref> two times, we obtain that $\mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n})$ is Noetherian and that $\mathcal{L}_{g,n}$ is Noetherian.

One can wonder why we did not use the simpler filtration by the subspaces $\textstyle \widetilde{\mathcal{G}}_1^{[\lambda]} = \cup_{[\mu] \in P^{2g+n}} \mathcal{G}_1^{[\mu],[\lambda]}$ indexed by $P^{g+n}$ with lexicographic order from the right, for which all the above proofs seem simpler since the product $\overline{\,\cdot\,}$ does not appear in $\mathrm{gr}_{\widetilde{\mathcal{G}}_1}(\mathcal{L}_{g,n})$. The problem is that the lexicographic order on $P^{g+n}$ is not well-founded, hence Lemma <ref> does not apply to $\mathrm{gr}_{\widetilde{\mathcal{G}}_1}(\mathcal{L}_{g,n})$ and the last argument in the proof of Theorem <ref> becomes wrong.

§.§ Noetherianity and finiteness of $\mathcal{L}_{g,n}^{U_q}$

The subalgebra of $U_q$-invariant elements is

\[ \mathcal{L}_{g,n}^{U_q} = \bigl\{ x \in \mathcal{L}_{g,n} \, \big| \, \forall \, h \in U_q, \: \mathrm{coad}^r(h)(x) = \varepsilon(h)x \bigr\} \]

where the right action $\mathrm{coad}^r$ was defined in (<ref>). More generally if $V$ is a module over a Hopf algebra $H$, the subspace of invariant elements is

\[ V^H =  \bigl\{ v \in V \: | \: \forall \, h \in H, \: h\cdot v = \varepsilon(h)v \bigr\} \]

where $\varepsilon$ is the counit of $H$.

We start with a general result about the structure of invariant elements in module-algebras; it is just an abstract formulation of the discussion below Theorem 3.2 in [9], which is itself inspired by <cit.>. Denote by $\mathcal{I}$ an additive abelian monoid endowed with a partial order $\leq$ such that

\begin{equation}\label{conditionsOrdreHilbertNagata}
\begin{array}{l}
\forall \, i \in \mathcal{I}, \:\: 0 \leq i \\
\forall \, i,j,k \in \mathcal{I}, \:\: i \leq j \implies i+k \leq j+k.
\end{array}
\end{equation}

We assume moreover that $\leq$ is well-founded. For instance, $\mathcal{I} = P_+$ endowed with the order $\preceq$ (see <ref>) satisfies these assumptions.
[Hilbert–Nagata theorem for module-algebras]
Let $H$ be a Hopf algebra and let $\textstyle A = \bigoplus_{i \in \mathcal{I}} A_i$ be a graded $k$-algebra which is a $H$-module-algebra. Assume that

* $A_0 = k\, 1$,

* $A$ is Noetherian and finitely generated,

* for each $i \in \mathcal{I}$, $A_i$ is stable under the action of $H$ and is a semisimple $H$-module.

Then the subalgebra $A^H$ is Noetherian and finitely generated.

The proof resorts on three basic lemmas. Note that since $A_i$ is semisimple we can write

\[ \textstyle A_i = (A_i)^H \oplus \bigoplus_{n} V_{i,n}\]

where the $V_{i,n}$ are some simple $H$-modules not isomorphic to the trivial module $k$. Projecting along this decomposition gives a $H$-morphism $\mathfrak{R}_i : A_i \to (A_i)^H$. Since each $A_i$ is $H$-stable we have $\textstyle A^H = \bigoplus_{i \in \mathcal{I}} (A_i)^H$ and we can define

\[ \mathfrak{R} = \bigoplus_{i \in \mathcal{I}} \mathfrak{R}_i : A \to A^H. \]

The morphism $\mathfrak{R}$ is called the Reynolds operator.

For $x \in A^H$ and $y \in A$, it holds $\mathfrak{R}(xy) = x\mathfrak{R}(y)$ and $\mathfrak{R}(yx) = \mathfrak{R}(y)x$.

We can assume without loss of generality that $x \in (A_i)^H$ and $y \in A_j$. Consider the map

\[ \fonc{m_x}{A_j}{A_{i+j}}{y}{xy} \]

It is $H$-linear:

\[ h \cdot m_x(y) = h\cdot(xy) = \sum_{(h)} (h_{(1)} \cdot x)(h_{(2)} \cdot y) = \sum_{(h)} (\varepsilon(h_{(1)})x)(h_{(2)} \cdot y) = x(h \cdot y) = m_x(h \cdot y). \]

Let $A_j = (A_j)^H \oplus V_j$ and $A_{i+j} = (A_{i+j})^H \oplus V_{i+j}$, where the semisimple modules $V_j$ and $V_{i+j}$ do not have any direct summand isomorphic to the trivial module $k$. Then by Schur's lemma

\[ \mathrm{Hom}_H(A_j, A_{i+j}) = \mathrm{Hom}_k\bigl( (A_j)^H, (A_{i+j})^H \bigr) \oplus \mathrm{Hom}_H(V_j, V_{i+j}). \]

It follows that if we write $y = \mathfrak{R}(y) \oplus v \in (A_j)^H \oplus V_j$, we have

\[ xy = m_x(y) = m_x(\mathfrak{R}(y) \oplus v) = m_x\bigl(\mathfrak{R}(y)\bigr) \oplus m_x(v) = x\mathfrak{R}(y) \oplus xv \in (A_{i+j})^H \oplus V_{i+j}. \]

But by definition of $\mathfrak{R}$ we also have $xy = \mathfrak{R}(xy) \oplus w$, which gives the desired equality. The second one is shown similarly.

Let $A$ be a Noetherian algebra and $X \subset A$. There exists a finite subset $X_{\mathrm{fin}} \subset X$ such that $AX = AX_{\mathrm{fin}}$, where $AY$ denotes the left ideal generated by $Y \subset A$. The same is true for right ideals.

Since $A$ is Noetherian, $AX$ is finitely generated. So there exists elements

\[ g_1 = \sum_{i \in I_1} a_i^{(1)} x_i^{(1)}, \: \ldots, \: g_n = \sum_{i \in I_n} a_i^{(n)} x_i^{(n)} \]

(where $a_i^{(j)} \in A$, $x_i^{(j)} \in X$ and $I_j$ is finite for each $j$) such that $AX = Ag_1 + \ldots + Ag_n$. Take $X_{\mathrm{fin}} = \{ x_i^{(1)} \}_{i \in I_1} \cup \ldots \cup \{ x_i^{(n)} \}_{i \in I_n}$. Since $X_{\mathrm{fin}}$ is a (finite) subset of $X$, we have $AX_{\mathrm{fin}} \subset AX$. On the other hand, since $g_1, \ldots, g_n \in AX_{\mathrm{fin}}$, we have $AX \subset A(AX_{\mathrm{fin}}) = AX_{\mathrm{fin}}$.

The last lemma is a well-known fact:

Let $\textstyle A = \bigoplus_{i \in \mathcal{I}} A_i$ be a graded $k$-algebra such that $A_0 = k \, 1$. If $A$ is Noetherian, then it is finitely generated.

Let $\textstyle A_+ = \bigoplus_{i \in \mathcal{I}\setminus \{0\}} A_i$. Since $A_+$ is an ideal, it is finitely generated: there exists $g_1, \ldots, g_n$ such that $A_+ = Ag_1 + \ldots + Ag_n$. We can assume that the elements $g_s$ are homogeneous (i.e. $g_s \in A_{j_s}$ with $j_s > 0$). We will show that $A_i \subset k\langle g_1, \ldots, g_n \rangle$ by well-founded induction on $i$. Firstly, $A_0 = k\,1 \subset k\langle g_1, \ldots, g_n \rangle$. Now assume that $\textstyle \bigoplus_{j < i} A_j \subset k\langle g_1, \ldots, g_n \rangle$ for some $i \in \mathcal{I}\setminus \{0\}$ and take $x \in A_i$. Since $A_i \subset A_+$ we have $\textstyle x = \sum_{s=1}^n a_s g_s$. The elements $x$ and $g_s$ being homogeneous, the $a_s$ can be assumed to be homogeneous as well: $a_s \in A_{l_s}$. Then $j_s + l_s = i$ for all $s$, which implies $l_s < i$ due to the assumptions (<ref>) on $\leq$. Hence $a_s \in \textstyle \bigoplus_{j < i} A_j$ so that $a_s \in k\langle g_1, \ldots, g_n \rangle$ by the induction hypothesis. It follows that $x \in k\langle g_1, \ldots, g_n \rangle$, as desired.

Let $I$ be a left ideal in $A^H$; we want to show that it is finitely generated. Lemma <ref> gives us elements $x_1, \ldots, x_n \in I$ such that $AI = Ax_1 + \ldots + Ax_n$. We have $\mathfrak{R}(AI) = I$. Indeed, the inclusion $I \subset \mathfrak{R}(AI)$ is obvious since $\mathfrak{R}(I) = I$. Conversely, take $\textstyle x = \sum_{s=1}^n a_s x_s \in AI$. Applying Lemma <ref>, we get $\textstyle \mathfrak{R}(x) = \sum_{s=1}^n \mathfrak{R}(a_s)x_s \in I$, so that $\mathfrak{R}(AI) \subset I$. Now, using Lemma <ref> again, we find

\[ I = \mathfrak{R}(AI) = \mathfrak{R}(Ax_1 + \ldots + Ax_n) = \mathfrak{R}(A)x_1 + \ldots + \mathfrak{R}(A)x_n = A^H x_1 + \ldots + A^H x_n \]

which means that $I$ is generated by $x_1, \ldots, x_n$. The same arguments show that right ideals of $A^H$ are finitely generated as well. Thus $A^H$ is Noetherian. To show that the algebra $A^H$ is finitely generated, note that it is graded by $\mathcal{I}$ if we define $(A^H)_i = (A_i)^H$; hence Lemma <ref> applies.

We will now use Theorem <ref> to prove that the subalgebra $\mathcal{L}^{U_q}_{g,n}$ is Noetherian and finitely generated, which is our main result in this section. Recall that $\mathcal{L}_{g,n}$ is $\mathcal{O}_q(q^{1/D})^{\otimes (2g+n)}$ as a vector space. For $[\mu] = (\mu_1, \ldots, \mu_{2g+n}) \in P_+^{2g+n}$ let

\[ C([\mu]) = C(\mu_1) \otimes \ldots \otimes C(\mu_{2g+n}) \subset \mathcal{L}_{g,n} \]

where $C(\mu_i)$ is the subspace of matrix coefficients of the irreducible $U_q^{\mathrm{ad}}$-module with highest weight $\mu_i$. We say that $[\mu'] \preceq [\mu]$ if and only if $\mu'_i \preceq \mu_i$ for all $1 \leq i \leq 2g+n$, where we recall that $\mu'_i \preceq \mu_i$ means that $\mu_i - \mu'_i \in D^{-1}Q_+$. This gives an order on $P_+^{2g+n}$ which satisfies the conditions (<ref>) and (<ref>) and is well-founded. We see from the formulas in Proposition <ref> and (<ref>) that the subspaces

\[ \mathcal{W}^{[\mu]} = \bigoplus_{[\mu'] \preceq [\mu]} C([\mu']) \]

form a filtration $\mathcal{W}$ of the algebra $\mathcal{L}_{g,n}$ indexed by $P_+^{2g+n}$. Let

\begin{equation}\label{canoProjCmu}
\textstyle \pi_{[\mu]} : \bigoplus_{[\mu'] \preceq [\mu]} C([\mu']) \to C([\mu])
\end{equation}

be the canonical projection and put $x \,\overline{\,\cdot\,}\,y = \pi_{[\mu]+[\eta]}(xy)$ for $x \in C([\mu])$, $y \in C([\eta])$. Then $\mathrm{gr}_{\mathcal{W}}(\mathcal{L}_{g,n})$ is identified with the vector space $\mathcal{L}_{g,n}$ endowed with the product $\overline{\,\cdot\,}$.

The algebra $\mathcal{L}_{g,n}^{U_q}$ is Noetherian and finitely generated.

Recall the filtration $\mathcal{G}_1$ of $\mathcal{L}_{g,n}$ that we used in <ref>. We see that $\mathcal{G}_1$ is a filtration of $\mathrm{gr}_{\mathcal{W}}(\mathcal{L}_{g,n})$ as well and that

\[ \mathrm{gr}_{\mathcal{G}_1}\bigl(\mathrm{gr}_{\mathcal{W}}(\mathcal{L}_{g,n})\bigr) = \mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n}). \]

In the proof of Theorem <ref> we showed that $\mathrm{gr}_{\mathcal{G}_2}\bigl( \mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n}) \bigr)$ is Noetherian and finitely generated. It follows from Lemmas <ref> and <ref> that $\mathrm{gr}_{\mathcal{G}_1}(\mathcal{L}_{g,n})$ also has these properties. Using again these lemmas, we obtain that $\mathrm{gr}_{\mathcal{W}}(\mathcal{L}_{g,n})$ is Noetherian and finitely generated.

Since $\mathcal{L}_{g,n}$ is a $U_q$-module-algebra and the canonical projections $\pi_{[\mu]}$ in (<ref>) are $U_q$-linear for $\mathrm{coad}^r$, we see that $\mathrm{gr}_{\mathcal{W}}(\mathcal{L}_{g,n})$ is a $U_q$-module-algebra for $\mathrm{coad}^r$. It satisfies all the assumptions of Theorem <ref> (recall that $U_q\text{-}\mathrm{mod}$ is semisimple). Hence $\mathrm{gr}_{\mathcal{W}}(\mathcal{L}_{g,n})^{U_q}$ is Noetherian and finitely generated. To conclude, let

\[ \mathcal{W}^{[\mu]}_{\mathrm{inv}} = \bigl( \mathcal{W}^{[\mu]} \bigr)^{U_q} = \mathcal{W}^{[\mu]} \cap \mathcal{L}_{g,n}^{U_q} \subset \mathcal{L}_{g,n}^{U_q}. \]

Then $\mathcal{W}_{\mathrm{inv}}$ is a filtration of $\mathcal{L}_{g,n}^{U_q}$ and we have

\[ \mathrm{gr}_{\mathcal{W}}(\mathcal{L}_{g,n})^{U_q} = \mathrm{gr}_{\mathcal{W}_{\mathrm{inv}}}\bigl(\mathcal{L}_{g,n}^{U_q}\bigr). \]

The result then follows from Lemma <ref> for Noetherianity and from Lemma <ref> for finiteness.

§ THE ALEKSEEV MORPHISM

Let $H$ be a quasitriangular Hopf algebra with an invertible antipode $S$. Usually the Alekseev morphism [1] is expected to be a morphism of algebras

\[ \mathcal{L}_{g,n}(H) \to \mathcal{H}(H^{\circ})^{\otimes g} \otimes H^{\otimes n} \]

where $\mathcal{H}(H^{\circ})$ is the Heisenberg double (<ref>). However, without further assumptions on $H$ (e.g. finite-dimensionality, as in [28]) we have to introduce a bigger algebra than $\mathcal{H}(H^{\circ})$ in order to make sense of the formulas given in [1]; we call this algebra the two-sided Heisenberg double. We construct it in <ref>, and define the Alekseev morphism in <ref>. The case of finite-dimensional $H$ is discussed in <ref>. Then in <ref> we focus on the quantum group $H = U_q^{\mathrm{ad}}(\mathfrak{g})$ and we use the Alekseev morphism to prove that $\mathcal{L}_{g,n}\bigl( U_q^{\mathrm{ad}}(\mathfrak{g}) \bigr)$ does not have non-trivial zero divisors.

§.§ The two-sided Heisenberg double
This section does not use the $R$-matrix of $H$. Recall that $H^{\mathrm{cop}}$ is $H$ with the opposite coproduct. The restricted dual $H^{\circ}$ endowed with its usual product $\star$ is a left $(H \otimes H^{\mathrm{cop}})$-module-algebra for the action

\[ (x \otimes y) \to \varphi = x\rhd \varphi \lhd S^{-1}(y). \]

As a result we can make the following definition:

The two-sided Heisenberg double is the smash product $H^{\circ} \# (H \otimes H^{\mathrm{cop}})$. It is an associative algebra which we denote by $\mathcal{HH}(H^{\circ})$.

By the general definition of a smash product (see e.g. <cit.>), $\mathcal{HH}(H^{\circ})$ is the vector space $H^{\circ} \otimes (H \otimes H^{\mathrm{cop}})$ endowed with the multiplication

\begin{equation}\label{produitSurHH}
\begin{array}{rl}
\bigl( \varphi \,\#\, (x \otimes y) \bigr) \bigl( \psi \,\#\, (z \otimes t) \bigr) \!\!&= \sum_{(x \,\otimes\, y)} \varphi \star \bigl( (x \otimes y)_{(1)} \to \psi \bigr) \,\#\, (x \otimes y)_{(2)}(z \otimes t)\\[1.5em]
&= \sum_{(x),(y)} \varphi \star \bigl(x_{(1)} \rhd \psi \lhd S^{-1}(y_{(2)})\bigr) \,\#\, (x_{(2)}z \otimes y_{(1)}t).
\end{array}
\end{equation}

where we write $\varphi \,\#\, (x \otimes y)$ for the element $\varphi \otimes (x \otimes y) \in H^{\circ} \otimes (H \otimes H)$. From this formula we see that $H^{\circ} \#\, (1 \otimes 1)$, $\varepsilon \,\#\, (H \otimes 1)$ and $\varepsilon \,\#\, (1 \otimes H)$ are subalgebras of $\mathcal{HH}(H^{\circ})$ which are canonically isomorphic to $H^{\circ}$, $H$ and $H$ respectively. Moreover $H^{\circ} \# (H \otimes 1)$ is a subalgebra of $\mathcal{HH}(H^{\circ})$ which is canonically isomorphic to the Heisenberg double $\mathcal{H}(H^{\circ})$ from <ref>.

We note that $\mathcal{HH}(H^{\circ})$ has a natural representation:

There is a representation $\blacktriangleright$ of $\mathcal{HH}(H^{\circ})$ on $H^{\circ}$ given by

\[ \varphi \,\#\, (x \otimes y) \blacktriangleright \psi = \varphi \star \bigl( x \rhd \psi \lhd S^{-1}(y) \bigr). \]

This is actually a general fact: for any left $H$-module-algebra $A$ there is a representation $\blacktriangleright$ of $A \,\#\, H$ on $A$ defined by $(a \,\#\, h) \blacktriangleright x = a(h \cdot x)$. The proof of this claim is straightforward.

The representation of $\mathcal{H}(H^{\circ})$ on $H^{\circ}$ (obtained by restricting $\blacktriangleright$ to this subalgebra) is known to be faithful <cit.>. In general this is not true for the representation of $\mathcal{HH}(H^{\circ})$ on $H^{\circ}$. For instance if $H$ is finite-dimensional we have $H^{\circ} = H^*$ and $\dim\bigl( \mathcal{HH}(H^*) \bigr) = \dim(H)^3$ while $\dim\bigl( \mathrm{End}_k(H^*) \bigr) = \dim(H)^2$, which forces the representation morphism $\rho : \mathcal{HH}(H^*) \to \mathrm{End}_k(H^*)$ to have a non-zero kernel.

Finally, let us mention another construction of $\mathcal{HH}(H^{\circ})$. We know from <ref> that $\mathcal{H}(H^{\circ})$ is a right $H$-module-algebra; it follows that $\mathcal{H}(H^{\circ})$ is a left $H^{\mathrm{cop}}$-module-algebra for the action

\[ h \cdot (\varphi \,\#\, x) = (\varphi\,\#\, x) \cdot S^{-1}(h) \]

where the $\cdot$ on the right-hand side is defined in (<ref>). As a result we can consider the smash product $\mathcal{H}(H^{\circ}) \# H^{\mathrm{cop}}$ and this algebra is isomorphic to $\mathcal{HH}(H^{\circ})$:

\[ \flecheIso{\mathcal{H}(H^{\circ}) \# H^{\mathrm{cop}}}{\mathcal{HH}(H^{\circ})}{(\varphi \,\#\, x) \,\#\, y}{\sum_{(y)} \varphi \,\#\, \bigl(xy_{(2)} \otimes S(y_{(1)})\bigr)} \]

§.§ Definition of the Alekseev morphism

Recall that $H$ is a quasitriangular Hopf algebra with an invertible antipode $S$. We use the following notations in $\mathcal{HH}(H^{\circ})$:

* we write $\varphi$ instead of $\varphi \,\#\, (1 \otimes 1)$,

* we write $h$ instead of $\varepsilon \,\#\, (h \otimes 1)$,

* we write $\widetilde{h}$ instead of $\varepsilon \,\#\, (1 \otimes h)$,

where $\varphi \in H^{\circ}$ and $h \in H$. The multiplication in $\mathcal{HH}(H^{\circ})$ is then described by

\begin{equation}\label{commutationsHH}
h \varphi = \sum_{(h)} (h_{(1)} \rhd \varphi) \, h_{(2)}, \qquad \widetilde{h} \varphi = \sum_{(h)} \bigl(\varphi \lhd S^{-1}(h_{(2)})\bigr) \, \widetilde{h_{(1)}}, \qquad \widetilde{h}g = g \widetilde{h}
\end{equation}

for all $\varphi \in H^{\circ}$ and $h,g \in H$. In particular the Heisenberg double $\mathcal{H}(H^{\circ})$ is identified with the subalgebra of elements of the form $\textstyle \sum_i \varphi_i \, h_i$. We note also that by the very definition $\widetilde{gh} = \widetilde{g}\,\widetilde{h}$. Finally, we endow $\mathcal{HH}(H^{\circ})^{\otimes g} \otimes H^{\otimes n}$ with the usual multiplication on a tensor product of algebras.

We will use the embeddings $j_k : \mathcal{L}_{1,0}(H) \to \mathcal{L}_{g,n}(H)$ for $1 \leq k \leq g$ and $j_{g+l} : \mathcal{L}_{0,1}(H) \to \mathcal{L}_{g,n}(H)$ for $1 \leq l \leq n$ from (<ref>) and the morphisms $\Phi_{0,1} : \mathcal{L}_{0,1}(H) \to H$ and $\Phi_{1,0} : \mathcal{L}_{1,0}(H) \to \mathcal{H}(H^{\circ}) \subset \mathcal{HH}(H^{\circ})$ from (<ref>) and Proposition <ref>. Let

\[ \fonc{\mathsf{D}_{g,n}}{H}{\mathcal{HH}(H^{\circ})^{\otimes g} \otimes H^{\otimes n}}{h}{\sum_ {(h)} \widetilde{h_{(1)}}\,h_{(2)} \otimes \ldots \otimes \widetilde{h_{(2g-1)}}\,h_{(2g)} \otimes h_ {(2g+1)} \otimes \ldots \otimes h_{(2g+n)}} \]

which is a morphism of algebras. We use the convention $\mathsf{D}_{0,0}(h) = \varepsilon(h)$. Note for further use that

\begin{equation}\label{inductionGamma}
\mathsf{D}_{0,n+1}(h) = \sum_{(h)} h_{(1)} \otimes \mathsf{D}_{0,n}(h_{(2)}), \qquad \mathsf{D}_{g+1,n}(h) = \sum_{(h)} \widetilde{h_{(1)}} \, h_{(2)} \otimes \mathsf{D}_{g,n}(h_{(3)}).
\end{equation}

There is a morphism of algebras

\[ \Phi_{g,n} : \mathcal{L}_{g,n}(H) \to \mathcal{HH}(H^{\circ})^{\otimes g} \otimes H^{\otimes n} \]

defined by

\begin{align*}
&\Phi_{g,n}\bigl(j_{g+l}(\varphi)\bigr) = \sum_{(R)} 1^{\otimes g} \, \otimes 1^{\otimes (l-1)} \otimes \Phi_{0,1}\!\left( \mathrm{coad}^r\bigl(R_{(1)}\bigr)(\varphi)\right) \otimes \mathsf{D}_{0,n-l}\bigl(R_{(2)}\bigr)\\
&\Phi_{g,n}\bigl( j_k(x) \bigr) = \sum_{(R)} 1^{\otimes (k-1)} \otimes \Phi_{1,0} \!\left( \mathrm{coad}^r\bigl( R_{(1)} \bigr)(x) \right) \otimes \mathsf{D}_{g-k,n}\bigl(R_{(2)}\bigr)
\end{align*}

for any $\varphi \in \mathcal{L}_{0,1}(H)$, $1 \leq l \leq n$, $x \in \mathcal{L}_{1,0}(H)$, $1 \leq k \leq g$, and where $\mathrm{coad}^r$ is defined in (<ref>) and (<ref>).

By (<ref>) the value of $\Phi_{g,n}$ on general elements is

\begin{align*}
&\Phi_{g,n}\bigl( x_1 \, \widetilde{\otimes} \, \ldots \, \widetilde{\otimes} \, x_g \, \widetilde{\otimes} \, \varphi_1 \, \widetilde{\otimes} \, \ldots \, \widetilde{\otimes} \, \varphi_n \bigr)\\
=\:& \Phi_{g,n}\bigl( j_1(x_1) \bigr) \ldots \Phi_{g,n}\bigl( j_g(x_g) \bigr) \Phi_{g,n}\bigl( j_{g+1}(\varphi_1) \bigr) \ldots \Phi_{g,n}\bigl( j_{g+n}(\varphi_n) \bigr).
\end{align*}

Note that for $g>0$ the morphism $\Phi_{g,n}$ actually takes values in $\mathcal{H}(H^{\circ}) \otimes \mathcal{HH}(H^{\circ})^{\otimes (g-1)} \otimes H^{\otimes n}$ where $\mathcal{H}(H^{\circ})$ is identified with the subalgebra $H^{\circ} \,\#\, (H \otimes 1)$ in $\mathcal{HH}(H^{\circ})$. This remark will be important in the proof of Theorem <ref> in <ref> below.

The following lemma is the key point for the proof of Theorem <ref>:

For all $h \in H$ and $x \in \mathcal{L}_{g,n}(H)$ we have

\[ \mathsf{D}_{g,n}(h) \, \Phi_{g,n}(x) = \sum_ {(h)}\Phi_{g,n}\!\left( \mathrm{coad}^r\bigl( S^{-1}(h_ {(2)}) \bigr)(x) \right) \mathsf{D}_{g,n}(h_ {(1)}) .\]

The proof is by induction. First we prove the result for $\Phi_{0,n}$ by induction on $n$. Then we fix $n$ and we prove the result for $\Phi_{g,n}$ by induction on $g$. So let us consider the case $g=0$. Thanks to the relation $R\Delta = \Delta^{\mathrm{op}}R$ we get for all $h \in H$ and $\varphi \in \mathcal{L}_{0,1}(H)$

\begin{equation}\label{commutationHPhi01}
\begin{array}{rl}
h\,\Phi_{0,1}(\varphi) \!\!\!&= \sum_{(R^1),(R^2)} \varphi\bigl( R^1_{(1)}R^2_{(2)} \bigr) \, hR^1_{(2)} R^2_{(1)}\\[1.4em]
& = \sum_{(R^1),(R^2),(h)} \varphi\!\left( S^{-1}(h_{(3)})h_{(2)}R^1_{(1)}R^2_{(2)} \right) \, h_{(1)}R^1_{(2)} R^2_{(1)}\\[1.4em]
& = \sum_{(R^1),(R^2),(h)} \varphi\!\left( S^{-1}(h_{(3)})R^1_{(1)}R^2_{(2)} h_{(2)}\right) \, R^1_{(2)} R^2_{(1)} h_{(1)}\\[1.4em]
& = \sum_{(h)}\Phi_{0,1}\!\left( \mathrm{coad}^r\bigl( S^{-1}(h_{(2)})\bigr)(\varphi) \right) h_{(1)}
\end{array}
\end{equation}

which is the desired formula for $(g,n) = (0,1)$. Now assume that the formula is true for some $n \geq 1$ (still with $g=0$) and let $\varphi \,\widetilde{\otimes}\, x \in \mathcal{L}_{0,n+1}(H) = \mathcal{L}_{0,1}(H) \,\widetilde{\otimes}\, \mathcal{L}_{0,n}(H)$. We have in $H^{\otimes (n+1)}$

\begin{align*}
&\mathsf{D}_{0,n+1}(h) \Phi_{0,n+1}(\varphi \,\widetilde{\otimes}\, x) = \sum_{(R),(h)} h_{(1)}\Phi_{0,1}\!\left( \mathrm{coad}^r\bigl(R_{(1)}\bigr)(\varphi)\right) \otimes \mathsf{D}_{0,n}\bigl(h_{(2)}R_{(2)}\bigr)\Phi_{0,n}(x)\\
=\:& \sum_{(R),(h)} \Phi_{0,1}\!\left( \mathrm{coad}^r\bigl(R_{(1)}S^{-1}(h_ {(2)})\bigr)(\varphi)\right)h_ {(1)} \otimes \mathsf{D}_{0,n}\bigl(h_{(3)}R_{(2)}\bigr)\Phi_{0,n}(x)\\
=\:& \sum_{(R),(h)} \Phi_{0,1}\!\left( \mathrm{coad}^r\bigl(S^{-1}(h_ {(3)})R_{(1)} \bigr)(\varphi)\right)h_ {(1)} \otimes \mathsf{D}_{0,n}\bigl(R_{(2)}h_{(2)} \bigr)\Phi_{0,n}(x)\\
=\:& \sum_{(R),(h)} \Phi_{0,1}\!\left( \mathrm{coad}^r\bigl(S^{-1}(h_ {(4)})R_{(1)} \bigr)(\varphi)\right)h_ {(1)} \otimes \mathsf{D}_{0,n}\bigl(R_{(2)}\bigr)\Phi_{0,n}\bigl(\mathrm{coad}^r\bigl( S^{-1}(h_ {(3)}) \bigr)(x) \bigr)\mathsf{D}_{0,n}\bigl(h_{(2)} \bigr)\\
=\:& \sum_{(h)} \Phi_{0,n+1}\bigl( \mathrm{coad}^r\bigl( S^{-1}(h_{(2)}) \bigr)(\varphi \,\widetilde{\otimes}\, x) \bigr) \mathsf{D}_{0,n+1}(h_{(1)}).
\end{align*}

For the first equality we used (<ref>), for the second we used (<ref>), for the third we used that $(S \otimes \mathrm{id})(R) = R^{-1}$ together with $R^{-1}\Delta^{\mathrm{op}} = \Delta R^{-1}$, for the fourth we used the induction hypothesis and for the fifth we used (<ref>), the definition of $\Phi_{0,n+1}$ and the definition of $\mathrm{coad}^r$ in (<ref>). Now we treat the case $g > 0$. Observe first that for all $h \in H$ and $\beta \otimes \alpha \in \mathcal{L}_{1,0}(H)$, we have in $\mathcal{HH}(H^{\circ})$:

\begin{align*}
&\mathsf{D}_{1,0}(h) \Phi_{1,0}(\beta \otimes \alpha) = \sum_{(R^1),(R^2),(R^3),(h)} \widetilde{h_{(1)}}h_{(2)} \left( R^1_{(2)}R^2_{(2)} \rhd \beta \lhd R^3_{(1)}R^1_{(1)} \right)R^3_{(2)}R^2_{(1)}\Phi_{0,1}(\alpha)\\
=\:&\sum_{(R^1),(R^2),(R^3),(h)} \left( h_ {(3)}R^1_{(2)}R^2_{(2)} \rhd \beta \lhd R^3_{(1)}R^1_{(1)}S^{-1}(h_{(2)}) \right) h_{(4)}R^3_{(2)}R^2_{(1)} \Phi_{0,1}(\alpha) \widetilde{h_{(1)}}\\
=\:&\sum_{(R^1),(R^2),(R^3),(h)} \left( R^1_{(2)}R^2_{(2)}h_{(3)} \rhd \beta \lhd S^{-1}(h_{(4)})R^3_{(1)}R^1_{(1)} \right) R^3_{(2)}R^2_{(1)} h_{(2)} \Phi_{0,1}(\alpha)\widetilde{h_{(1)}}\\
=\:&\sum_{(h)} \Phi_{1,0}\bigl( \mathrm{coad}^r\bigl( S^{-1}(h_ {(3)}) \bigr)(\beta) \otimes \varepsilon \bigr) \Phi_{0,1}\bigl( \mathrm{coad}^r\bigl( S^{-1}(h_ {(2)})\bigr)(\alpha) \bigr) \mathsf{D}_{1,0}(h_{(1)})\\
=\:&\sum_{(h)} \Phi_{1,0}\left( \mathrm{coad}^r\bigl(S^{-1}(h_{(2)})\bigr)(\beta \otimes \alpha)\right) \mathsf{D}_{1,0}(h_{(1)})
\end{align*}

which is the formula for the case $(g,n)=(1,0)$. The first equality is the definition of $\Phi_{1,0}$ (Proposition <ref>), the second equality uses (<ref>), the third equality uses $R\Delta = \Delta^{\mathrm{op}}R$ three times, the fourth equality uses (<ref>), the definition of $\Phi_{1,0}$ and the definition of $\mathsf{D}_{1,0}$ and the fifth equality uses the definition of $\Phi_{1,0}$ and (<ref>). For the general case, fix $n$ and make an induction on $g$; this is completely similar to the computation for $\Phi_{0,n+1}$ above, using (<ref>) and the result for $\Phi_{1,0}$ just proved.

We will obtain the result by the same type of induction as in the proof of Lemma <ref>, i.e. first one proves the result for $\Phi_{0,n}$ by induction on $n$ and then one fixes $n$ and proves the result for $\Phi_{g,n}$ by induction on $g$. Let us do for instance the second part, the first part being completely similar and left to the reader. So assume that $\Phi_{g,n}$ is a morphism of algebras and let $x \,\widetilde{\otimes}\, u, y \,\widetilde{\otimes}\, v \in \mathcal{L}_{g+1,n}(H) = \mathcal{L}_{1,0}(H) \,\widetilde{\otimes}\, \mathcal{L}_{g,n}(H)$. We have

\begin{align*}
&\Phi_{g+1,n}\bigl( (x \,\widetilde{\otimes}\, u)(y \,\widetilde{\otimes}\, v) \bigr) = \sum_{(R)} \Phi_{g+1,n}\bigl( x\,\mathrm{coad}^r(R_ {(1)})(y) \,\widetilde{\otimes}\, \mathrm{coad}^r(R_ {(2)})(u)\,v \bigr)\\
=\:& \sum_{(R^1), (R^2)} \Phi_{1,0}\bigl( \mathrm{coad}^r(R^2_{(1)})\bigl( x\,\mathrm{coad}^r(R^1_{(1)})(y)\bigr) \bigr) \otimes \mathsf{D}_{g,n}(R^2_{(2)})\,\Phi_{g,n}\bigl( \mathrm{coad}^r(R^1_{(2)})(u)\,v \bigr)\\
=\:& \sum_{(R^1), (R^2), (R^3)} \Phi_{1,0}\bigl( \mathrm{coad}^r(R^2_{(1)})(x) \bigr) \,\Phi_{1,0}\bigl( \mathrm{coad}^r(R^1_{(1)}R^3_{(1)})(y) \bigr)\\[-1.2em]
&\hspace{17.7em} \otimes \mathsf{D}_{g,n}(R^2_{(2)}R^3_{(2)})\,\Phi_{g,n}\bigl( \mathrm{coad}^r(R^1_{(2)})(u)\bigr)\,\Phi_{g,n}(v)\\[.5em]
=\:& \sum_{\substack{(R^1), (R^2),\\(R^3), (R^4)}} \Phi_{1,0}\bigl( \mathrm{coad}^r(R^2_{(1)})(x) \bigr)\, \Phi_{1,0}\bigl( \mathrm{coad}^r(R^1_{(1)}R^3_{(1)}R^4_{(1)})(y) \bigr)\\[-1.9em]
&\hspace{11em} \otimes \mathsf{D}_{g,n}(R^2_{(2)}) \,\Phi_{g,n}\bigl( \mathrm{coad}^r\bigl(R^1_{(2)}S^{-1}(R^3_{(2)})\bigr)(u) \bigr) \, \mathsf{D}_{g,n}(R^4_{(2)})\,\Phi_{g,n}(v)\\[.5em]
=\:& \sum_{(R^2), (R^4)} \Phi_{1,0}\bigl( \mathrm{coad}^r(R^2_{(1)})(x) \bigr)\, \Phi_{1,0}\bigl( \mathrm{coad}^r(R^4_{(1)})(y) \bigr) \otimes \mathsf{D}_{g,n}(R^2_{(2)}) \,\Phi_{g,n}(u) \, \mathsf{D}_{g,n}(R^4_{(2)})\,\Phi_{g,n}(v)\\
=\:& \Phi_{g+1,n}(x \,\widetilde{\otimes}\, u)\,\Phi_{g+1,n}(y \,\widetilde{\otimes}\, v).
\end{align*}

The first equality uses (<ref>), the second uses the definition of $\Phi_{g+1,n}$, the third uses that $\mathcal{L}_{1,0}(H)$ is a module-algebra for $\mathrm{coad}^r$ together with the formula $(\Delta \otimes \mathrm{id})(R) = R_{13}R_ {23}$ and then uses that $\Phi_{1,0}$ and $\Phi_{g,n}$ are morphisms of algebras (by Proposition <ref> and the induction hypothesis), the fourth uses Lemma <ref> and the formula $(\mathrm{id} \otimes \Delta)(R)= R_{13}R_{12}$, the fifth uses $(\mathrm{id} \otimes S^{-1})(R) = R^{-1}$ and the sixth uses the definition of $\Phi_{g+1,n}$.

For completeness let us provide an alternative definition of $\Phi_{g,n}$ based on the matrices $\overset{V}{B}(i)$, $\overset{V}{A}(i)$, $\overset{V}{M}(j)$ from (<ref>). Let $R^{(+)} = R$ and $\textstyle R^{(-)} = (R')^{-1} = \sum_{(R)} R_{(2)} \otimes S(R_{(1)})$ and write as usual $\textstyle R^{(\pm)} = \sum_{(R^{(\pm)})} R^{(\pm)}_{(1)} \otimes R^{(\pm)}_{(2)}$. Note that $R^{(-)}$ is also a $R$-matrix. For a finite-dimensional $H$-module $V$ we define

\begin{align*}
&\overset{V}{L}{^{(\pm)}} = \sum_{(R^{(\pm)})} R^{(\pm)}_{(2)} \otimes \rho_V\!\left(R^{(\pm)}_{(1)}\right) \in H \otimes \mathrm{End}_k(V)\\
&\overset{V}{\widetilde{L}}{^{(\pm)}} = \sum_{(R^{(\pm)})} \widetilde{R^{(\pm)}_{(2)}} \otimes \rho_V\!\left(R^{(\pm)}_{(1)}\right) \in \mathcal{HH}(H^{\circ}) \otimes \mathrm{End}_k(V)\\
&\overset{V}{T} = \sum_{i,j} {_V\phi^{e^i}_{e_j}} \otimes E_{ij} \in H^{\circ} \otimes \mathrm{End}(V)
\end{align*}

where $\rho_V : H \to \mathrm{End}_k(V)$ is the representation morphism, ${_V\phi^{e^i}_{e_j}} : h \mapsto e^i(h \cdot e_j)$ are the matrix coefficients of $V$ in some basis $(e_i)$ and $E_{ij}(e_k) = \delta_{jk}e_i$. These elements can be seen as matrices of size $\dim(V)$ with coefficients in $\mathcal{HH}(H^{\circ})$. They satisfy various exchange relations; for instance it is an exercise to show that (<ref>) implies

\[ \overset{V}{L}{^{(\pm)}_1} \overset{W}{T}_2 = \overset{W}{T}_2 \, \overset{V}{L}{^{(\pm)}_1} \, R^{(\pm)}_{VW}, \quad R^{(\pm)}_{VW} \, \overset{V}{\widetilde{L}}{^{(\pm)}_1} \, \overset{W}{T}_2 = \overset{W}{T}_2 \, \overset{V}{\widetilde{L}}{^{(\pm)}_1}, \quad \overset{V}{L}{^{(\pm)}_1} \, \overset{W}{\widetilde{L}}{^{(\pm)}_2} = \overset{W}{\widetilde{L}}{^{(\pm)}_2}\, \overset{V}{L}{^{(\pm)}_1} \]

where we use the notations introduced before Proposition <ref>. Let $\overset{V}{M}$ (resp. $\overset{V}{A}$, $\overset{V}{B}$) be the matrix with coefficients in $\mathcal{L}_{0,1}(H)$ (resp. in $\mathcal{L}_{1,0}(H)$) introduced before Proposition <ref>. We have

\begin{equation}\label{matrixFormPhi01Phi10}
\Phi_{0,1}\bigl( \overset{V}{M} \bigr) = \overset{V}{L}{^{(+)}} \overset{V}{L}{^{(-)-1}}, \quad \Phi_{1,0}\bigl( \overset{V}{A} \bigr) = \overset{V}{L}{^{(+)}} \overset{V}{L}{^{(-)-1}}, \quad \Phi_{1,0}\bigl( \overset{V}{B} \bigr) = \overset{V}{L}{^{(+)}} \, \overset{V}{T} \, \overset{V}{L}{^{(-)-1}}
\end{equation}

where $\Phi_{0,1}\bigl( \overset{V}{M} \bigr)$ is viewed as a matrix with coefficients in $H$ while $\Phi_{1,0}\bigl( \overset{V}{A} \bigr)$ and $\Phi_{1,0}\bigl( \overset{V}{B} \bigr)$ are viewed as matrices with coefficients in $\mathcal{H}(H^{\circ})$. Finally for $1 \leq k \leq g$ let $J_k :\mathcal{HH}(H^{\circ}) \to \mathcal{HH}(H^{\circ})^{\otimes g} \otimes H^{\otimes n}$ and for $g+1 \leq l \leq g+n$ let $J_{g+l} : H \to \mathcal{HH}(H^{\circ})^{\otimes g} \otimes H^{\otimes n}$ be the obvious embeddings. We define

\begin{align*}
&\overset{V}{L}{^{(\pm)}}(i) = (J_i \otimes \mathrm{id})\bigl( \overset{V}{L}{^{(\pm)}}  \bigr) \quad \text{for } 1 \leq i \leq g+n,\\
&\overset{V}{\widetilde{L}}{^{(\pm)}}(k) = (J_k \otimes \mathrm{id})\bigl( \overset{V}{\widetilde{L}}{^{(\pm)}}  \bigr), \quad \overset{V}{T}(k) = (J_k \otimes \mathrm{id})\bigl( \overset{V}{T}  \bigr) \quad \text{for } 1 \leq k \leq g.
\end{align*}

A straightforward computation based on the axiom $(\mathrm{id} \otimes \Delta)(R) = R_{13}R_{12}$ reveals that

\begin{equation}\label{matrixFormD}
\sum_{(R)} \left(1^{\otimes g} \, \otimes 1^{\otimes l} \otimes \mathsf{D}_{0,n-l}\bigl(R_{(2)}\bigr) \right) \otimes \rho_V\bigl(R_ {(1)}\bigr) = \overset{V}{L}{^{(+)}}(g+n) \ldots \overset{V}{L}{^{(+)}}(g+l+1)
\end{equation}

and similarly

\begin{equation}\label{matrixFormDBis}
\begin{array}{l}
\sum_{(R)} \left(1^{\otimes k} \otimes \mathsf{D}_{g-k,n}\bigl(R_{(2)}\bigr)\right) \otimes \rho_V\bigl(R_ {(1)}\bigr)\\
= \overset{V}{L}{^{(+)}}(g+n) \ldots \overset{V}{L}{^{(+)}}(g+1) \, \overset{V}{L}{^{(+)}}(g) \overset{V}{\widetilde{L}}{^{(+)}}(g) \ldots \overset{V}{L}{^{(+)}}(k+1) \overset{V}{\widetilde{L}}{^{(+)}}(k+1).
\end{array}
\end{equation}

Let us denote the matrices in (<ref>) and (<ref>) by $\Lambda_{0,n-l}$ and $\Lambda_{g-k,n}$ respectively, with the convention $\Lambda_{0,0} = 1$. A computation which combines (<ref>) with  (<ref>) and (<ref>) yields

\begin{equation}\label{matrixFormAlekseev}
\begin{array}{l}
\Phi_{g,n}\bigl( \overset{V}{M}(g+l) \bigr) = \Lambda_{0, n-l} \, \overset{V}{L}{^{(+)}}(g+l) \, \overset{V}{L}{^{(-)}}(g+l)^{-1} \Lambda_{0, n-l}^{-1},\\
\Phi_{g,n}\bigl( \overset{V}{A}(k) \bigr) = \Lambda_{g-k,n} \, \overset{V}{L}{^{(+)}}(k) \overset{V}{L}{^{(-)}}(k)^{-1} \, \Lambda_{g-k, n}^{-1},\\
\Phi_{g,n}\bigl( \overset{V}{B}(k) \bigr) = \Lambda_{g-k,n} \, \overset{V}{L}{^{(+)}}(k) \, \overset{V}{T}(k)  \,\overset{V}{L}{^{(-)}}(k)^{-1} \, \Lambda_{g-k, n}^{-1}
\end{array}
\end{equation}

for $1 \leq l \leq n$ and $1 \leq k \leq g$. This is the matrix definition of $\Phi_{g,n}$.

Another possible definition of $\Phi_{g,n}$ uses $\overset{V}{L}{^{(-)}}$, $\overset{V}{\widetilde{L}}{^{(-)}}$ instead of $\overset{V}{L}{^{(+)}}$, $\overset{V}{\widetilde{L}}{^{(+)}}$ to define different conjugation matrices $\Lambda$, where the conjugation starts from the first tensorand (here it starts from the last tensorand). See e.g. <cit.>. For $g=0$ our definition here agrees with  the one in <cit.>, as can be seen from the first equality in (<ref>).

§.§ Finite-dimensional case

Here we quickly discuss the case where $H$ is finite-dimensional, which is treated in detail in [28]. In this case one can use $\mathcal{H}(H^{\circ})$ instead of $\mathcal{HH}(H^{\circ})$.

Recall from <ref> the representation of $\mathcal{H}(H^*)$ on $H^*$, which is faithful. Since $\dim\bigl( \mathcal{H}(H^*) \bigr) = \dim\bigl( \mathrm{End}_k(H^*) \bigr)$ the representation morphism $\rho : \mathcal{H}(H^*) \to \mathrm{End}_k(H^*)$ is an isomorphism. For $h \in H$ let $Q_h \in \mathcal{H}(H^*)$ be defined by $\rho(Q_h)(\psi) = \psi \lhd S^{-1}(h)$ for all $\psi \in H^*$ (be aware that $Q_h$ was denoted by $\widetilde{h}$ in [28]). We have $Q_{hg} = Q_h\,Q_g$ so that $\{ Q_h \,|\, h \in H\}$ is a subalgebra of $\mathcal{H}(H^*)$ isomorphic to $H$. These elements obey the following commutation rules in $\mathcal{H}(H^*)$:

\begin{equation}\label{commutationQHeisenberg}
Q_h \, \varphi = \sum_{(h)} \bigl(\varphi \lhd S^{-1}(h_{(2)})\bigr) \, Q_{h_{(1)}}, \qquad Q_h \, g = g \, Q_h
\end{equation}

for all $\varphi \in H^*$ and $h,g \in H$.

\[ \fonc{\mathsf{D}^{\mathrm{fin}}_{g,n}}{H}{\mathcal{H}(H^*)^{\otimes g} \otimes H^{\otimes n}}{h}{\sum_ {(h)} Q_ {h_{(1)}}\,h_{(2)} \otimes \ldots \otimes Q_{h_{(2g-1)}}\,h_{(2g)} \otimes h_ {(2g+1)} \otimes \ldots \otimes h_{(2g+n)}}. \]

which is a morphism of algebras and define

\[ \Phi_{g,n}^{\mathrm{fin}} : \mathcal{L}_{g,n}(H) \to \mathcal{H}(H^*)^{\otimes g} \otimes H^{\otimes n} \]

by the formulas of Theorem <ref> but with $\mathsf{D}^{\mathrm{fin}}$ instead of $\mathsf{D}$. Thanks to (<ref>), the proof of Theorem <ref> remains valid with the elements $Q_h$ instead of $\widetilde{h}$ and $\Phi_{g,n}^{\mathrm{fin}}$ is a morphism of algebras.

Hence we have two possible different morphisms in the finite-dimensional case: $\Phi_{g,n}$ and $\Phi_{g,n}^{\mathrm{fin}}$. They are related as follows. Let $r_{V_i} : H \to \mathrm{End}_k(V_i)$ be any representations of $H$ for $1 \leq i \leq n$ and let $\rho : \mathcal{HH}(H^*) \to \mathrm{End}_k(H^*)$ be the representation from Proposition <ref>, which by restriction gives the isomorphism $\mathcal{H}(H^*) \cong \mathrm{End}_k(H^*)$. Then the two representations of $\mathcal{L}_{g,n}(H)$

\begin{align*}
&\bigl( \rho^{\otimes g} \otimes r_{V_1} \otimes \ldots \otimes r_{V_n} \bigr) \circ \Phi_{g,n}^{\mathrm{fin}} : \mathcal{L}_{g,n}(H) \to \mathrm{End}_k\bigl( (H^*)^{\otimes g} \otimes V_1 \otimes \ldots \otimes V_n \bigr),\\
&\bigl( \rho^{\otimes g} \otimes r_{V_1} \otimes \ldots \otimes r_{V_n} \bigr) \circ \Phi_{g,n} : \mathcal{L}_{g,n}(H) \to \mathrm{End}_k\bigl( (H^*)^{\otimes g} \otimes V_1 \otimes \ldots \otimes V_n \bigr)
\end{align*}

are equal, simply because $\rho(Q_h) = \rho(\widetilde{h})$. This property is satisfying, as one of the main applications of $\Phi_{g,n}$ is to construct representations of $\mathcal{L}_{g,n}(H)$.

When $H$ is infinite dimensional we cannot define the elements $Q_h \in \mathcal{H}(H^{\circ})$, because we do not know if the representation morphism $\rho : \mathcal{H}(H^*) \to \mathrm{End}_k(H^{\circ})$ is surjective. This is why we introduced the bigger algebra $\mathcal{HH}(H^{\circ})$ in which the elements $\widetilde{h}$ exist by construction for all $h$ and are a substitute for $Q_h$.

§.§ The case $H = U_q^{\mathrm{ad}}(\mathfrak{g})$

The left and right coregular actions of $U_q$ on $\mathcal{O}_q$ allow us to define the two-sided Heisenberg double $\mathcal{H}_q = \mathcal{H}_q(\mathfrak{g})$ as the smash product

\[ \mathcal{H}_q = \mathcal{O}_q \,\#\, \bigl( U_q \otimes U_q^{\mathrm{cop}}\bigr). \]

where the scalars are implicitly extended to $\mathbb{C}(q^{1/D})$. It is the $\mathbb{C}(q^{1/D})$-vector space $\mathcal{O}_q \otimes \bigl( U_q \otimes U_q\bigr)$ endowed with the product (<ref>).

Let $G$ be the connected, complex, semisimple algebraic group $G$ with Lie algebra $\mathfrak{g}$. Recall from <ref> that $\mathcal{O}_q = \mathcal{O}_q(G)$ is the subalgebra of $U_q^{\circ}$ generated over $\mathbb{C}(q)$ by the family of all matrix coefficients of the irreducible $U_q^{\mathrm{ad}}(\mathfrak{g})$-modules of type $1$.

For all $k \geq 1$ the algebra $\mathcal{O}_q(G)^{\otimes k}$ does not have non-trivial zero divisors.

The direct product $G^{\times k}$ is a connected, complex, semisimple algebraic group with Lie algebra $\mathfrak{g}^{\oplus k}$. By definition, $\mathcal{O}_q(G^{\times k})$ is generated over $\mathbb{C}(q)$ by the family of all matrix coefficients of the irreducible $U_q^{\mathrm{ad}}(\mathfrak{g}^{\oplus k})$-modules of type $1$. Let us note that $U_q^{\mathrm{ad}}(\mathfrak{g}^{\oplus k}) \cong U_q^{\mathrm{ad}}(\mathfrak{g})^{\otimes k}$ as Hopf algebras. It is a general fact that given two algebras $A$ and $B$ the irreducible $(A \otimes B)$-modules are all of the form $V \otimes W$ where $V$ is an irreducible $A$-module and $W$ is an irreducible $B$-module, see e.g. <cit.>. Hence we have an isomorphism of algebras

\[ \flecheIso{\mathcal{O}_q(G^{\times k})}{\mathcal{O}_q(G)^{\otimes k}}{_{V_1 \otimes \ldots \otimes V_k}\phi^{i_1, \ldots, i_k}_{j_1, \ldots, j_k}}{_{V_1}\phi^{i_1}_{j_1} \otimes \ldots \otimes {_{V_k}\phi^{i_k}_{j_k}}} \]

where each $V_l$ is an irreducible $U_q^{\mathrm{ad}}(\mathfrak{g})$-module of type $1$, $(v_{l,i})$ is a basis of $V_l$ with dual basis $(v^{l,j})$ and the $\phi$'s are the matrix coefficients in these bases. By <cit.> the algebra $\mathcal{O}_q(G^{\times k})$ has non-trivial zero divisors and we are done.

The algebra $\mathcal{HH}_q^{\otimes g} \otimes U_q^{\otimes n}$ does not have non-trivial zero divisors.

We first introduce a filtration $\mathcal{G}$ on $\mathcal{HH}_q$ which straightforwardly generalizes the filtration of $\mathcal{H}_q$ used in the proof of Proposition <ref>. It is defined by

\[ \mathcal{G}^{\mathbf{m}, \mathbf{n}} = \mathcal{O}_q \,\#\, (\mathcal{F}_{\mathrm{DCK}}^{\mathbf{m}} \otimes \mathcal{F}_{\mathrm{DCK}}^{\mathbf{n}}) \]

for $(\mathbf{m}, \mathbf{n}) \in (\mathbb{N}^{2N+1})^2$. We endow the monoid $(\mathbb{N}^{2N+1})^2$ with the partial order defined by $(\mathbf{m}, \mathbf{n}) \leq (\mathbf{m}', \mathbf{n}')$ if and only if $\mathbf{m} \leq \mathbf{m}'$ and $\mathbf{n} \leq \mathbf{n}'$, where $\leq$ on $\mathbb{N}^{2N+1}$ is the lexicographic order from the right (<ref>). Thanks to formula (<ref>) for the product in $\mathcal{HH}_q$ and to Corollary <ref> we see that $\mathcal{G}$ is a filtration of $\mathcal{HH}_q$. We have

\[ \mathrm{gr}_{\mathcal{G}}(\mathcal{HH}_q) = \mathcal{O}_q \,\#\, \bigl( \mathrm{gr}_{\mathcal{F}_{\mathrm{DCK}}}(U_q) \otimes \mathrm{gr}_{\mathcal{F}_{\mathrm{DCK}}}(U_q) \bigr). \]

Let us explain this equality more precisely. Let $_V\phi^i_j : x \mapsto v^i(x \cdot v_j)$ be the matrix coefficients (see <ref>) of a finite-dimensional $U_q^{\mathrm{ad}}$-module $V$ in a basis of weight vectors $(v_i)$ with weights $(\epsilon_i)$. Write

\[ _V\phi^i_j, \quad \overline{E_{\beta_k}}, \quad \overline{F_{\beta_k}}, \quad \overline{K_{\mu}}, \quad \overline{\widetilde{E_{\beta_k}}}, \quad \overline{\widetilde{F_{\beta_k}}}, \quad \overline{\widetilde{K_{\mu}}} \]

instead of

\begin{align*}
_V\phi^i_j + \mathcal{G}^{<(\mathbf{0},\mathbf{0})}, \quad &E_{\beta_k} + \mathcal{G}^{<(d(E_{\beta_k}),\mathbf{0})}, \quad F_{\beta_k} + \mathcal{G}^{<(d(F_{\beta_k}),\mathbf{0})}, \quad K_{\mu} + \mathcal{G}^{<(\mathbf{0}, \mathbf{0})}\\
&\widetilde{E_{\beta_k}} + \mathcal{G}^{<(\mathbf{0},d(E_{\beta_k}))}, \quad \widetilde{F_{\beta_k}} + \mathcal{G}^{<(\mathbf{0},d(F_{\beta_k}))}, \quad \widetilde{K_{\mu}} + \mathcal{G}^{<(\mathbf{0}, \mathbf{0})}
\end{align*}

in $\mathrm{gr}_{\mathcal{G}}(\mathcal{HH}_q)$. Note that $\mathcal{G}^{<(\mathbf{0},\mathbf{0})}=0$. The subalgebra generated by $\overline{E_{\beta_k}}$, $\overline{F_{\beta_k}}$, $\overline{K_{\mu}}$, $\overline{\widetilde{E_{\beta_k}}}$, $\overline{\widetilde{F_{\beta_k}}}$, $\overline{\widetilde{K_{\mu}}}$ for $1 \leq k \leq N$ and $\mu \in P$ is isomorphic to $\mathrm{gr}_{\mathcal{F}_{\mathrm{DCK}}}(U_q)^{\otimes 2}$ and is thus a quasi-polynomial ring (over $\mathbb{C}(q)$). Moreover by Proposition <ref> and (<ref>) we get

\begin{equation}\label{commutationGHHq}
\begin{array}{lll}
\overline{E_{\beta_k}} \, {_V\phi^i_j} = {_V\phi^i_j} \, \overline{E_{\beta_k}}, & \overline{F_{\beta_k}} \, {_V\phi^i_j} = q^{-(\beta_k, \epsilon_j)} {_V\phi^i_j} \, \overline{F_{\beta_k}}, & \overline{K_{\mu}} \, {_V\phi^i_j} = q^{(\mu, \epsilon_j)} {_V\phi^i_j} \, \overline{K_{\nu}},\\[.5em]
\overline{\widetilde{E_{\beta_k}}} \, {_V\phi^i_j} = q^{-(\beta_k, \epsilon_i)} {_V\phi^i_j} \, \overline{\widetilde{E_{\beta_k}}}, & \overline{\widetilde{F_{\beta_k}}} \, {_V\phi^i_j} = {_V\phi^i_j} \, \overline{\widetilde{F_{\beta_k}}}, & \overline{K_{\mu}} \, {_V\phi^i_j} = q^{-(\mu, \epsilon_i)} {_V\phi^i_j} \, \overline{K_{\nu}}
\end{array}
\end{equation}

for all $\mu \in P$. Hence $\mathrm{gr}_{\mathcal{G}}(\mathcal{HH}_q)$ is a quasi-polynomial ring over $\mathcal{O}_q(q^{1/D})$, generated over $\mathcal{O}_q(q^{1/D})$ by $\overline{E_{\beta_k}}$, $\overline{F_{\beta_k}}$, $\overline{K_{\mu}}$, $\overline{\widetilde{E_{\beta_k}}}$, $\overline{\widetilde{F_{\beta_k}}}$, $\overline{\widetilde{K_{\mu}}}$ for $1 \leq k \leq N$ and $\mu \in P$.

Now for $(\mathbf{m}_1, \mathbf{n}_1, \ldots, \mathbf{m}_g, \mathbf{n}_g, \mathbf{p}_{1}, \ldots, \mathbf{p}_{n}) \in (\mathbb{N}^{2N+1})^{2g+n}$ let

\[ \mathcal{T}^{\mathbf{m}_1, \mathbf{n}_1, \ldots, \mathbf{m}_g, \mathbf{n}_g, \mathbf{p}_{1}, \ldots, \mathbf{p}_{n}} = \mathcal{G}^{\mathbf{m}_1, \mathbf{n}_1} \otimes \ldots \otimes \mathcal{G}^{\mathbf{m}_g, \mathbf{n}_g} \otimes \mathcal{F}_{\mathrm{DCK}}^{\mathbf{p}_1} \otimes \ldots \otimes \mathcal{F}_{\mathrm{DCK}}^{\mathbf{p}_n} \subset \mathcal{HH}_q^{\otimes g} \otimes U_q^{\otimes n}. \]

We endow the monoid $(\mathbb{N}^{2N+1})^{2g+n}$ with the direct product order:

\begin{align*}
&(\mathbf{m}_1, \mathbf{n}_1, \ldots, \mathbf{m}_g, \mathbf{n}_g, \mathbf{p}_{1}, \ldots, \mathbf{p}_{n}) \leq (\mathbf{m}'_1, \mathbf{n}'_1, \ldots, \mathbf{m}'_g, \mathbf{n}'_g, \mathbf{p}'_{1}, \ldots, \mathbf{p}'_{n})\\
&\iff \forall\, i,j,\quad \mathbf{m}_i \leq \mathbf{m}'_i, \quad \mathbf{n}_i \leq \mathbf{n}'_i, \quad \mathbf{p}_j \leq \mathbf{p}'_j.
\end{align*}

It is clear that $\mathcal{T}$ is a filtration of $\mathcal{HH}_q^{\otimes g} \otimes U_q^{\otimes n}$ and that

\[ \mathrm{gr}_{\mathcal{T}}\bigl( \mathcal{HH}_q^{\otimes g} \otimes U_q^{\otimes n} \bigr) = \mathrm{gr}_{\mathcal{G}}(\mathcal{HH}_q)^{\otimes g} \otimes \mathrm{gr}_{\mathcal{F}_{\mathrm{DCK}}}(U_q)^{\otimes n}. \]

We have seen that $\mathrm{gr}_{\mathcal{G}}(\mathcal{HH}_q)$ is a quasipolynomial ring over $\mathcal{O}_q(q^{1/D})$ and we know that $\mathrm{gr}_{\mathcal{F}_{\mathrm{DCK}}}(U_q)$ is a quasipolynomial ring over $\mathbb{C}(q)$ (<ref>). Hence $\mathrm{gr}_{\mathcal{T}}\bigl( \mathcal{HH}_q^{\otimes g} \otimes U_q^{\otimes n} \bigr)$ is a quasipolynomial ring over $\mathcal{O}_q(q^{1/D})^{\otimes g}$. Since $\mathcal{O}_q(q^{1/D})^{\otimes g}$ does not have non-trivial zero divisors by Lemma <ref>, it follows from a general result (see e.g. <cit.>) that $\mathrm{gr}_{\mathcal{T}}(\mathcal{HH}_q^{\otimes g} \otimes U_q^{\otimes n})$ does not have non-trivial zero divisors. The conclusion then follows from Lemma <ref>.

1. The morphism $\Phi_{g,n} : \mathcal{L}_{g,n} \to \mathcal{HH}_q^{\otimes g} \otimes U_q^{\otimes n}$ is injective.

2. The algebra $\mathcal{L}_{g,n}$ does not have non-trivial zero divisors.

1. The proof is by induction. First we prove the result for $\Phi_{0,n}$ by induction on $n$. Then we fix $n$ and we prove the result for $\Phi_{g,n}$ by induction on $g$.

It has been shown in <cit.> that the morphism $\Phi_{0,1}$ is injective (see <cit.> for this statement in the present framework). Now assume that $\Phi_{0,n}$ is injective for some $n$. Let $\mathrm{ad}^r$ be the right adjoint action of $U_q$ on itself: $\textstyle \mathrm{ad}^r(h)(x) = \sum_{(h)} S(h_{(1)})xh_{(2)}$. For $\lambda \in P$ consider the subspaces

\begin{align*}
(\mathcal{L}_{0,1})_{\lambda} &= \bigl\{ \varphi \in \mathcal{L}_{0,1} \, \big| \, \forall \, \nu \in P, \:\: \mathrm{coad}^r(K_{\nu})(\varphi) = q^{(\lambda,\nu)}\varphi \bigr\}, \\
(U_q)_{\lambda} &= \bigl\{ x \in U_q \, \big| \, \forall \, \nu \in P, \:\: \mathrm{ad}^r(K_{\nu})(x) = q^{(\lambda,\nu)}x \bigr\}.
\end{align*}

We have $\textstyle \mathcal{L}_{0,1} = \bigoplus_{\lambda \in P} (\mathcal{L}_{0,1})_{\lambda}$ and $\textstyle U_q = \bigoplus_{\lambda \in P} (U_q)_{\lambda}$, thus

\[ \mathcal{L}_{0,n+1} = \bigoplus_{\lambda \in P} \, (\mathcal{L}_{0,1})_{\lambda} \,\widetilde{\otimes}\, \mathcal{L}_{0,n}, \qquad U_q^{\otimes (n+1)} =  \bigoplus_{\lambda \in P} \, (U_q)_{\lambda} \otimes (U_q)^{\otimes n} \]

where for the first equality we used that $\mathcal{L}_{0,n+1} = \mathcal{L}_{0,1} \,\widetilde{\otimes}\, \mathcal{L}_{0,n}$ (Prop. <ref>). Since

\[ \forall\, h \in U_q, \: \forall \, \psi \in \mathcal{L}_{0,1}, \quad\Phi_{0,1}\bigl( \mathrm{coad}^r(h)(\psi) \bigr) = \mathrm{ad}^r(h)\bigl( \Phi_{0,1}(\psi) \bigr) \]

we have $\Phi_{0,1}\bigl( (\mathcal{L}_{0,1})_{\lambda} \bigr) \subset (U_q)_{\lambda}$. Take $\varphi \in (\mathcal{L}_{0,1})_{\lambda}$ and $x \in \mathcal{L}_{0,n}$. By (<ref>) we have

\[ \mathrm{coad}^r(\Theta_{(1)})(\varphi) \otimes \Theta_{(2)} = \varphi \otimes K_{\lambda}.  \]

It follows from (<ref>) and the expression of $R$ in (<ref>) that

\begin{align*}
\Phi_{0,n+1}(\varphi \,\widetilde{\otimes}\, x)&= \sum_{(R)}\Phi_{0,1}\!\left( \mathrm{coad}^r\bigl(R_{(1)}\bigr)(\varphi)\right) \otimes \mathsf{D}_{0,n}\bigl(R_{(2)}\bigr)\Phi_{0,n}(x)\\
&\in \Phi_{0,1}\!\left( \mathrm{coad}^r\bigl(\Theta_{(1)}\bigr)(\varphi)\right) \otimes \mathsf{D}_{0,n}\bigl(\Theta_{(2)}\bigr)\Phi_{0,n}(x) + \bigoplus_{\lambda' < \lambda} \, (U_q)_{\lambda'} \otimes (U_q)^{\otimes n}\\
&=\Phi_{0,1}(\varphi) \otimes (K_{\lambda})^{\otimes n}\Phi_{0,n}(x) + \bigoplus_{\lambda' < \lambda} \, (U_q)_{\lambda'} \otimes (U_q)^{\otimes n}
\end{align*}

where $\lambda' < \lambda$ means that $\lambda - \lambda' \in Q_+\!\setminus\!\{0\}$ and we used that $K_ {\lambda}$ is grouplike for the last equality. We are now in position to show our result. Let $y \in \mathcal{L}_{0,n+1}$ be a non-zero element and write it as $\textstyle y = \sum_{\lambda \in P} \sum_{i \in I_{\lambda}} \varphi_{\lambda,i} \,\widetilde{\otimes}\, x_{\lambda,i}$ with $\varphi_{\lambda,i} \in (\mathcal{L}_{0,1})_{\lambda}$ for all $i \in I_{\lambda}$ and such that for each $\lambda$ the elements $\bigl(\varphi_{\lambda,i}\bigr)_{i \in I_{\lambda}}$ are linearly independent over $\mathbb{C}(q^{1/D})$. Take a $\lambda$ maximal for the order $\leq$ on $P$ such that there exists at least one $i \in I_{\lambda}$ with $\varphi_{\lambda,i} \neq 0$ and $x_{\lambda,i} \neq 0$. Then

\[ \Phi_{0,n+1}(y) \in \sum_{i \in I_{\lambda}} \Phi_{0,1}(\varphi_{\lambda,i}) \otimes (K_{\lambda})^{\otimes n}\Phi_{0,n}(x_{\lambda,i}) + \bigoplus_{\lambda' \neq \lambda} \, (U_q)_{\lambda} \otimes (U_q)^{\otimes n} \]

By the induction hypothesis the morphism $\Phi_{0,n}$ is injective so $(K_{\lambda})^{\otimes n}\Phi_{0,n}(x_{\lambda,i}) \neq 0$ ($K_{\lambda}$ being invertible) for at least one $i \in I_{\lambda}$. Moreover the elements $\varphi_{\lambda,i}$ are linearly independent so the elements $\Phi_{0,1}(\varphi_{\lambda,i})$ are linearly independent as well and it follows that $\Phi_{0,n+1}(y) \neq 0$.

Let us now examine the case of $\Phi_{g,n}$. Recall first from Theorem <ref> that $\Phi_{1,0}$ is injective. We see easily from the formulas in Theorem <ref> that $\Phi_{g,n}$ actually takes values in $\mathcal{H}_q \otimes \mathcal{HH}_q^{\otimes (g-1)} \otimes U_q^{\otimes n}$. Recall the right action (<ref>) of $U_q$ on $\mathcal{H}_q$ and consider the subspaces

\begin{align*}
(\mathcal{L}_{1,0})_{\lambda} &= \bigl\{ x \in \mathcal{L}_{1,0} \, \big| \, \forall \, \nu \in P, \:\: \mathrm{coad}^r(K_{\nu})(x) = q^{(\lambda,\nu)}x \bigr\}, \\
(\mathcal{H}_q)_{\lambda} &= \bigl\{ y \in \mathcal{H}_q \,\big|\, \forall \, \nu \in P, \: y \cdot K_{\nu} = q^{(\lambda,\nu)}y \bigr\}.
\end{align*}

One sees easily that $\textstyle \mathcal{H}_q = \bigoplus_{\lambda \in P} (\mathcal{H}_q)_{\lambda}$ and we obtain the decompositions

\[ \mathcal{L}_{g+1,n} = \bigoplus_{\lambda \in P} \, (\mathcal{L}_{1,0})_{\lambda} \,\widetilde{\otimes}\, \mathcal{L}_{g,n}, \quad \mathcal{H}_q \otimes \mathcal{HH}_q^{\otimes g} \otimes U_q^{\otimes n} =  \bigoplus_{\lambda \in P} \, (\mathcal{H}_q)_{\lambda} \otimes \mathcal{HH}_q^{\otimes g} \otimes U_q^{\otimes n}. \]

Moreover by Proposition <ref> we have $\Phi_{1,0}\bigl( (\mathcal{L}_{1,0})_{\lambda} \bigr) \subset (\mathcal{H}_q)_{\lambda}$. After these preliminary remarks the proof (by induction) of the injectivity of $\Phi_{g+1,n}$ is completely similar to the proof for $\Phi_{0,n+1}$ above and is thus left to the reader.

2. Follows from Proposition <ref> and item 1.

§ TOPOLOGICAL INTERPRETATION OF $\MATHCAL{L}_{G,N}(H)$

Let $H$ be a ribbon Hopf algebra with an invertible antipode. In this last section we show that the algebra $\mathcal{L}_{g,n}(H)$ is isomorphic to the stated skein algebra of a surface as defined in [21]. If moreover $H\text{-}\mathrm{mod}$ is semisimple we show that the subalgebra of invariant elements $\mathcal{L}_{g,n}^H(H)$ is isomorphic to the skein algebra of the surface. Finally we explain why this last property is false when $H$-mod is not semisimple.

§.§ Stated skein algebras of surfaces

In this subsection we recall the definitions of the topological objects that we will use. For details, see [40, 22, 21].

Denote by $\Sigma_g$ the compact closed oriented surface of genus $g \geq 0$, by $\Sigma_{g,n}$ the surface obtained from $\Sigma_g$ by removing $n$ points, by $\Sigma_{g,n}^{\circ}$ the punctured bordered surface obtained from $\Sigma_{g,n}$ by removing an open disk $D$, and then by $\Sigma_{g,n}^{\circ,\bullet}$ the surface obtained from $\Sigma_{g,n}^{\circ}$ by removing a point from the boundary component $\partial(\Sigma_{g,n}^{\circ,\bullet})$. Here is a picture of $\Sigma_{g,n}^{\circ,\bullet}$:

The puncture on the boundary is represented by a big black dot (•) while the $n$ inner punctures are represented by the cross marks ($\mathsf{x}$). For our purposes it is much more convenient to use a less familiar view of $\Sigma_{g,n}^{\circ,\bullet}$:

\begin{equation}\label{surfaceRuban}
\input{surface2.pdf_tex}
\end{equation}

Let $\boldsymbol{\Sigma} = \Sigma_{g,n}^{\circ,\bullet} \times [0,1]$ be the thickening of $\Sigma_{g,n}^{\circ,\bullet}$. A point $P \in \boldsymbol{\Sigma}$ is a couple $(p,h)$ and the number $h$ is called the height of $P$. Recall from <cit.> that a $H$-colored $\partial\boldsymbol{\Sigma}$-tangle is an oriented, framed, compact, properly embedded
1-dimensional submanifold $\mathbf{T} \subset \boldsymbol{\Sigma}$ such that

* the framing is vertical at any point of $\partial\mathbf{T} = \mathbf{T} \cap \partial\boldsymbol{\Sigma}$,

* the points of $\partial\mathbf{T}$ have distincts heights,

* each connected component of $\mathbf{T}$ is colored (i.e. labelled) by a finite-dimensional $H$-module.

Here we moreover allow the $\partial\boldsymbol{\Sigma}$-tangles to contain coupons, which are squares colored by $H$-morphisms in the usual coherent way. The coupons are not allowed to intersect $\partial\boldsymbol{\Sigma}$. This gives $\partial\boldsymbol{\Sigma}$-ribbon graphs, which are the generalization to surfaces of the homogeneous colored directed ribbon graphs from [51].

An isotopy of $\partial\boldsymbol{\Sigma}$-ribbon graphs is an isotopy which preserves all the properties listed above. We represent a $\partial\boldsymbol{\Sigma}$-ribbon graph $\mathbf{T}$ by its diagram, which is the projection of $\mathbf{T}$ on $\Sigma_{g,n}^{\circ,\bullet}$. We use isotopy to ensure that the projection has at most double points, and at double points we record the over/under-passing information as usual. Moreover, the height information of the boundary points $\partial\mathbf{T}$ must be recorded with the diagram; thanks to isotopy it is enough to record the order of the boundary points with respect to the relation $<$ defined by $(p,h)<(p',h')$ if and only $h < h'$. It is customary to identify a $\partial\boldsymbol{\Sigma}$-ribbon graph with its isotopy class.

Let $\mathfrak{o}$ be an orientation of $\partial\Sigma_{g,n}^{\circ,\bullet}$. We say that a diagram of the $\partial\boldsymbol{\Sigma}$-ribbon graph $\mathbf{T}$ is $\mathfrak{o}$-ordered if the boundary points of $\mathbf{T}$ have increasing heights when one goes along $\partial\Sigma_{g,n}^{\circ,\bullet}$ according to $\mathfrak{o}$ and starting from the point $\bullet$. In a $\mathfrak{o}$-ordered diagram the height relation $<$ of the boundary points of $\mathbf{T}$ is given by $\mathfrak{o}$. Using isotopy it is clear that any $\partial\boldsymbol{\Sigma}$-ribbon graph can be represented by a $\mathfrak{o}$-ordered diagram. Here is an example of such a diagram for $(g,n)=(1,1)$:

\begin{equation}\label{exempleTangle}
\input{exempleTangle.pdf_tex}
\end{equation}

The colors (i.e. labels) $U,V,W,X,Y$ are finite-dimensional $H$-modules and $f : W \otimes V^* \to V$, $g : U \to Y$ are $H$-morphisms. With this choice of orientation of the boundary, the diagram implicitly means that the heights of the boundary points labeled by $U$, $W$ and $Y$ respectively form a strictly increasing sequence. In the sequel we always use ordered diagrams with respect to this orientation.

Let $\mathbf{T}$ be a $\partial\boldsymbol{\Sigma}$-ribbon graph, let $p_1, \ldots, p_r$ be the boundary points of $\mathbf{T}$ and let $V_i$ be the color of the strand to which belongs $p_i$ for each $i$. A state of $\mathbf{T}$ is a map which to each $p_i$ associates an element of $V_i$ or $V_i^*$, depending if the orientation of the strand is incoming or outcoming at $p_i$ respectively. In practice we label $p_i$ by an element of $V_i$ or $V_i^*$ on the ribbon graph diagram. Here is an example of a state for a ribbon graph like in (<ref>):

with $u \in U$, $w \in W$, $\gamma \in Y^*$; in this picture we just show $\mathbf{T}$ in a disk which intersects the boundary. A $\partial\boldsymbol{\Sigma}$-ribbon graph together with a state is called a stated ribbon graph. If $s = (s_1, \ldots, s_l)$ is a state of the ribbon graph $\mathbf{T}$ (i.e. a labelling of the boundary points ordered by increasing height), we denote by $\mathbf{T}^s$ the ribbon graph $\mathbf{T}$ with the state $s$.

Denote by $M^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet})$ the vector space freely generated by the (isotopy classes of) stated ribbon graphs, over the base field $k$ of $H$. In $M^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet})$ we “multilinearize” the states, i.e we put the relations $\mathbf{T}^{(s_1, \ldots, \lambda s_i + \mu s'_i, \ldots, s_n)} = \lambda \mathbf{T}^{(s_1, \ldots, s_i, \ldots, s_n)} + \mu \mathbf{T}^{(s_1, \ldots, s'_i, \ldots, s_n)}$ for all $i$, all $\lambda,\mu \in k$ and any $\partial\boldsymbol{\Sigma}$-ribbon graph $\mathbf{T}$ which can be labelled with these states.

Let $\mathbf{T}^s$ be a stated ribbon graph which looks as follows in a disk $D$ which intersects the boundary:

\begin{equation}\label{tangleInDisk}
\input{planarTangle.pdf_tex}
\end{equation}

where $m,l \geq 0$ and $T$ is any oriented and colored $(m,l)$-ribbon graph in $\mathbb{R}^3$. We emphasize that $m$ or $l$ may be equal to $0$. Each strand carries an orientation. For $1 \leq i \leq l$ let $\epsilon(i) \in \{\pm\}$ be $+$ (resp. $-$) if the orientation of the strand colored by $V_i$ points inward (resp. outward) the surface at the boundary. Set $V_i^+ = V_i$ and $V_i^- = V_i^*$ so that $s_i \in V_i^{-\epsilon(i)}$. Similarly for $1 \leq j \leq m$ let $\eta(j) \in \{\pm\}$ be $+$ (resp. $-$) if the strand colored by $W_j$ is outcoming (resp. incoming) at the bottom of $T$ and set $W_j^+ = W_j$ and $W_j^- = W_j^*$. The Reshetikhin–Turaev functor $F_{\mathrm{RT}}$ [51] applied to $T$ gives a linear map

\[ F_{\mathrm{RT}}(T) : W_1^{\eta(1)} \otimes \ldots \otimes W_m^{\eta(m)} \to V_1^{\epsilon(1)} \otimes \ldots \otimes V_l^{\epsilon(l)}. \]

Denote as usual $\langle -, - \rangle$ for the evaluation pairing: $\langle \varphi, v \rangle = \langle v, \varphi \rangle = \varphi(v)$ for any $v \in V_i$, $\varphi \in V_i^*$. We extend this pairing straightforwardly:

\begin{equation}\label{dualityBracket}
\begin{array}{rrcl}
\langle -, - \rangle : &\left( V_1^{-\epsilon(1)} \otimes \ldots \otimes V_l^{-\epsilon(l)} \right) \otimes \left( V_1^{\epsilon(1)} \otimes \ldots \otimes V_l^{\epsilon(l)} \right) & \to & k\\[.5em]
& (s_1 \otimes \ldots \otimes s_l) \otimes (x_1 \otimes \ldots \otimes x_l) & \mapsto & \langle s_1, x_1 \rangle \ldots \langle s_l, x_l \rangle
\end{array}
\end{equation}

Let $\bigl(w^{j,+}_r\bigr)_r$ be a basis of $W_j$ and $\bigl(w^{j,-}_r\bigr)_r$ be its dual basis. Then we identify $\mathbf{T}^s$ with the following element of $M^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet})$:

\begin{equation}\label{skeinRelation}
\input{skeinRelation.pdf_tex}
\end{equation}

We stress that the stated ribbon graphs appearing in the linear combination (<ref>) are equal to $\mathbf{T}$ outside the disk $D$ represented in (<ref>). Such an identification is called a stated skein relation.

The stated skein algebra of $\Sigma_{g,n}^{\circ,\bullet}$ associated to $H$, which we denote by $\mathcal{S}^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet})$, is the quotient of $M^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet})$ by all the stated skein relations (<ref>).

Given two stated ribbon graphs $\mathbf{T}_1^s$, $\mathbf{T}_2^t$ we can use isotopy to transform $\mathbf{T}_1^s$ into $\textstyle (\mathbf{T}_1^s)^- \subset \Sigma_{g,n}^{\circ,\bullet} \times [0,\frac{1}{2} [$ and $\mathbf{T}_2$ into $\textstyle (\mathbf{T}_2^t)^+ \subset \Sigma_{g,n}^{\circ,\bullet} \times ]\frac{1}{2}, 1]$. The disjoint union $(\mathbf{T}_1^s)^- \cup (\mathbf{T}_2^t)^+$ is denoted by $\mathbf{T}_1^s \ast \mathbf{T}_2^t$. Each boundary point keeps its state in $(\mathbf{T}_1^s)^- \cup (\mathbf{T}_2^t)^+$ so that the result is naturally a stated ribbon graph. This operation $\ast$ defines an associative algebra structure on $M^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet})$ which descends to $\mathcal{S}^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet})$. The unit is the empty $\partial\boldsymbol{\Sigma}$-ribbon graph.

Here is an example of stated skein relation which we will use later:

\begin{equation}\label{evalBraiding}
\input{evalBraiding.pdf_tex}
\end{equation}

where $v \in X$, $\varphi \in Y^*$, $c$ is the braiding in $H\text{-}\mathrm{mod}$, $(x_i)$ is a basis of $X$, $(y_j)$ is a basis of $Y$, $(x^i)$, $(y^j)$ are their dual bases and $\varphi(R_{(1)}?)$ denotes the linear form on $Y$ defined by $y \mapsto \varphi(R_{(1)}y)$. The detail of the second equality is

\begin{align*}
&\sum_{i,j} \left\langle v \otimes \varphi, c_{Y,X^*}(y_i \otimes x^j) \right\rangle y^i \otimes x_j = \sum_{(R),i,j} \left\langle v \otimes \varphi, R_{(2)}x^j \otimes R_{(1)}y_i \right\rangle y^i \otimes x_j\\
=\:&\sum_{(R),i,j} \left\langle x^j, S(R_{(2)})v\right\rangle \left\langle \varphi, R_{(1)}y_i \right\rangle y^i \otimes x_j =\sum_{(R)} \varphi(R_{(1)}?) \otimes S(R_{(2)})v.
\end{align*}

One proves similarly that

\begin{equation}\label{evalBraiding2}
\input{evalBraiding2.pdf_tex}
\end{equation}

§.§ Isomorphism $\mathcal{S}^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet}) \cong \mathcal{L}_{g,n}(H)$

This subsection is a generalization of <cit.>, which dealt with the case $H = U_q^{\mathrm{ad}}(\mathfrak{sl}_2)$. The main ingredient in the definition of the isomorphism will be the holonomy map

\[ \mathrm{hol} : \bigl\{\partial\boldsymbol{\Sigma}\text{-ribbon graphs} \bigr\} \to \bigl\{\text{tensors with coefficients in } \mathcal{L}_{g,n}(H) \bigr\} \]

defined in <cit.>.

Let $\mathbf{T}^s \subset \Sigma_{g,n}^{\circ, \bullet} \times [0,1]$ be a stated ribbon graph with state $s = (s_1, \ldots, s_l)$ which looks as follows near the boundary:

As before, let $\epsilon(i) \in \{\pm\}$ be $+$ (resp. $-$) if the orientation of the strand colored by $V_i$ points inward (resp. outward) the surface at the boundary, and set $V_i^+ = V_i$, $V_i^- = V_i^*$ so that $s_i \in V^{-\epsilon(i)}$. By definition $\mathbf{T}$ is the $\partial\boldsymbol{\Sigma}$-ribbon graph obtained from $\mathbf{T}^s$ by forgetting the states. Then the holonomy map gives an element

\[ \mathrm{hol}(\mathbf{T}) \in \mathcal{L}_{g,n}(H) \otimes V_1^{\epsilon(1)} \otimes \ldots \otimes V_l^{\epsilon(l)}. \]

Let $\langle -, - \rangle$ be the evaluation pairing as in (<ref>). Then $\langle s_1 \otimes \ldots \otimes s_l, - \rangle$ is a linear map $V_1^{\epsilon(1)} \otimes \ldots \otimes V_l^{\epsilon(l)} \to k$ and we define

\[ \mathrm{hol}^{\mathrm{st}}(\mathbf{T}^s) = \left(\mathrm{id}_ {\mathcal{L}_{g,n}(H)} \otimes \langle s_1 \otimes \ldots \otimes s_l, - \rangle\right)\!(\mathrm{hol}(\mathbf{T})) \in \mathcal{L}_{g,n}(H). \]

Extending linearly, this gives a map $\mathrm{hol}^{\mathrm{st}} : M^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet}) \to \mathcal{L}_{g,n}(H)$. It is immediate from the definition of $\mathrm{hol}$ in [29] that $\mathrm{hol}^{\mathrm{st}}$ is compatible with the stated skein relations, so it descends to $\mathcal{S}^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet})$.

We call $\mathrm{hol}^{\mathrm{st}} : \mathcal{S}^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet}) \to \mathcal{L}_{g,n}(H)$ the stated holonomy map.

The stated holonomy map is a morphism of algebras.

This follows from one of the main properties of $\mathrm{hol}$ proven in [29]. Let $\mathbf{T}^s_1$, $\mathbf{T}^t_2$ be stated ribbon graphs and let

be their boundary points and states. By definition of the product of stated ribbon graphs and of an ordered $\partial\boldsymbol{\Sigma}$-ribbon graph diagram, the boundary points of $\mathbf{T}^s_1 \ast \mathbf{T}_2^t$ are arranged as follows:

Let $\mathbf{T}_1$, $\mathbf{T}_2$ be the $\partial\boldsymbol{\Sigma}$-ribbon graphs without states and write $\textstyle \mathrm{hol}(\mathbf{T}_1) = \sum_i x_i \otimes \mathbf{v}_i$, $\textstyle \mathrm{hol}(\mathbf{T}_2) = \sum_j y_j \otimes \mathbf{w}_j$ with $x_i, y_j \in \mathcal{L}_{g,n}(H)$, $\mathbf{v}_i \in V_1^{\epsilon(1)} \otimes \ldots \otimes V_l^{\epsilon(l)}$ and $\mathbf{w}_j \in W_1^{\eta(1)} \otimes \ldots \otimes W_m^{\eta(m)}$, where the $\epsilon$'s and $\eta$'s are signs as above.
We know from <cit.> that $\textstyle \mathrm{hol}(\mathbf{T}_1 \ast \mathbf{T}_2) = \sum_{i,j} x_iy_j \otimes \mathbf{v}_i \otimes \mathbf{w}_j$. Hence:

\begin{align*}
\mathrm{hol}^{\mathrm{st}}(\mathbf{T}^s_1 \ast \mathbf{T}^t_2) &= \left(\mathrm{id}_ {\mathcal{L}_{g,n}(H)} \otimes \langle s_1 \otimes \ldots \otimes s_l \otimes t_1 \otimes \ldots \otimes t_m, - \rangle\right)\!(\mathrm{hol}(\mathbf{T}_1 \ast \mathbf{T}_2))\\
&= \sum_{i,j} x_iy_j \, \langle s_1 \otimes \ldots \otimes s_l, \mathbf{v}_i \rangle \, \langle t_1 \otimes \ldots \otimes t_m, \mathbf{w}_j\rangle = \mathrm{hol}^{\mathrm{st}}(\mathbf{T}^s_1) \, \mathrm{hol}^{\mathrm{st}}(\mathbf{T}^t _2). \qedhere
\end{align*}

We now define what will be a right inverse to $\mathrm{hol}^{\mathrm{st}}$. Recall that $H^{\circ}$ is spanned by the matrix coefficients $_V\phi^f_v$ of finite-dimensional $H$-modules and that $\mathcal{L}_{g,n}(H)$ is $(H^{\circ})^{\otimes (2g+n)}$ as a vector space. We define a linear map $\xi_{g,n} : \mathcal{L}_{g,n}(H) \to \mathcal{S}^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet})$ by

\begin{equation}\label{defXign}
\input{isoXign.pdf_tex}
\end{equation}

The linear map $\xi_{g,n} : \mathcal{L}_{g,n}(H) \to \mathcal{S}^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet})$ is surjective.

Due to the stated skein relations (<ref>) in $\mathcal{S}^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet})$ we see that every element is equal to a linear combination of “standard” stated tangles of the form

In this picture each strand carries some orientation and color and $s = (s_1, \ldots, s_N)$ is some state. We put the crossings in view of the definition (<ref>); it is clear that we can always add such crossings since they correspond to the braiding isomorphism through the Reshetikhin–Turaev functor. Now we observe that the stated skein relations imply an orientation reversal relation:

where $\langle -, v \rangle : \varphi \mapsto \varphi(v)$ and $f(g^{-1} \cdot ?) : x \mapsto f(g^{-1}\cdot x)$ with $g \in H$ the pivotal element. We similarly have an all-in-one relation:

In these pictures the handle represents either one of the handles or one of the punctured half-disks in (<ref>). As a result the “standard” stated tangles above can be rewritten as the ones in (<ref>), which by definition span the image of $\xi_{g,n}$.

The stated holonomy map $\mathrm{hol}^{\mathrm{st}} : \mathcal{S}^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet}) \to \mathcal{L}_{g,n}(H)$ is an isomorphism of algebras.

We already know from Proposition <ref> that $\mathrm{hol}^{\mathrm{st}}$ is a morphism of algebras. It is immediate from the definition of $\mathrm{hol}$ in [29] that $\mathrm{hol}^{\mathrm{st}} \circ \xi_{g,n} = \mathrm{id}_{\mathcal{L}_{g,n}(H)}$ (the role of the crossings in the definition (<ref>) of $\xi_{g,n}$ is precisely to obtain this equality). This implies that the linear map $\xi_{g,n}$ is injective. It follows from Lemma <ref> that $\xi_{g,n}$ is an isomorphism of vector spaces. Hence $\mathrm{hol}^{\mathrm{st}} = \xi^{-1}_{g,n}$ is an isomorphism as well.

A corollary of the previous proof is that $\xi_{g,n}$ is a morphism of algebras. It is nevertheless worthwile to mention that the direct proof (i.e. without resorting on $\mathrm{hol}^{\mathrm{st}}$) of this fact allows one to recover the product in $\mathcal{L}_{g,n}(H)$ by means of stated skein relations, thus giving a topological flavour to the formulas in Proposition <ref>. We explain this for $\mathcal{S}^{\mathrm{st}}_H(\Sigma_{0,1}^{\circ,\bullet})$ and $\mathcal{S}^{\mathrm{st}}_H(\Sigma_{1,0}^{\circ,\bullet})$.

So let us first discuss the case of $\xi_{0,1} : \mathcal{L}_{0,1}(H) \to \mathcal{S}^{\mathrm{st}}_H(\Sigma_{0,1}^{\circ,\bullet})$. It is useful to note that

\begin{equation}\label{coregularOnMatrixCoeffs}
h \rhd {_V\phi^f_v} = {_V\phi^f_{hv}}, \qquad {_V\phi^f_v} \lhd h = {_V\phi^{f(h?)}_v}
\end{equation}

where we recall that $\rhd$ and $\lhd$ are the left and right coregular actions of $H$ on $H^{\circ}$, see  (<ref>). We then have the following stated skein computation, which recovers the formula (<ref>) for the product in $\mathcal{L}_{0,1}(H)$:

The first equality is by definition of the product in $\mathcal{S}^{\mathrm{st}}_H(\Sigma_{0,1}^{\circ,\bullet})$, the second equality uses the stated skein relations (<ref>) and (<ref>), the third equality uses the all-in-one relation (see the proof of Lemma <ref>), the fourth equality is the definition of $\xi_{0,1}$ and the last equality uses (<ref>) and (<ref>).

Let us now consider $\xi_{1,0} : \mathcal{L}_{1,0}(H) \to \mathcal{S}^{\mathrm{st}}_H(\Sigma_{1,0}^{\circ,\bullet})$. Recall that the product in $\mathcal{L}_{1,0}(H)$ is fully described by the formulas (<ref>). Let us write

\[ \xi_b = \xi_{1,0} \circ \mathfrak{i}_B, \qquad \xi_a = \xi_{1,0} \circ \mathfrak{i}_A \]

which are morphisms of algebras $\mathcal{L}_{0,1}(H) \to \mathcal{S}^{\mathrm{st}}_H(\Sigma_{1,0}^{\circ,\bullet})$, and we recall that $\mathfrak{i}_B$ and $\mathfrak{i}_A$ are defined in (<ref>). Then the definition of $\xi_{1,0}$ in (<ref>) is such that

\begin{equation}\label{Xi10XiaXib}
\xi_{1,0}(\beta \otimes \alpha) = \xi_b(\beta) \, \xi_a(\alpha).
\end{equation}

We have two natural morphisms of algebras $\mathfrak{e}_b, \mathfrak{e}_a : \mathcal{S}^{\mathrm{st}}_H(\Sigma_{0,1}^{\circ,\bullet}) \to \mathcal{S}^{\mathrm{st}}_H(\Sigma_{1,0}^{\circ,\bullet})$ defined by

(we use that every element of $\mathcal{S}^{\mathrm{st}}_H(\Sigma_{0,1}^{\circ,\bullet})$ can be written as a linear combination of such stated tangles, see the proof of Lemma <ref>). They satisfy $\xi_b = \mathfrak{e}_b \circ \xi_{0,1}$, $\xi_a = \mathfrak{e}_a \circ \xi_{0,1}$. Hence by the previous computation for $\xi_{0,1}$ we find

\[ \xi_b(\varphi)\,\xi_b(\psi) = \mathfrak{e}_b\bigl( \xi_{0,1}(\varphi)\,\xi_{0,1}(\psi) \bigr) = \mathfrak{e}_b\bigl( \xi_{0,1}(\varphi\psi) \bigr) = \xi_b(\varphi\psi) \]

and similarly for $\xi_a$; these two equalities correspond to the first and second formulas of (<ref>). Finally:

The first equality is by definition of the product in $\mathcal{S}^{\mathrm{st}}_H(\Sigma_{1,0}^{\circ,\bullet})$, the second equality is a trick, the third equality uses stated skein relations like (<ref>) and (<ref>), the fourth equality uses the definition of $\xi_{1,0}$, the last equality uses (<ref>) and (<ref>). This equality corresponds to the third formula of (<ref>).

For general $g$ and $n$ write

\[ \xi_{b_i} = \xi_{g,n} \circ \mathfrak{i}_{B(i)}, \qquad \xi_{a_i} = \xi_{g,n} \circ \mathfrak{i}_{A(i)}, \qquad \xi_{m_j} = \xi_{g,n} \circ \mathfrak{i}_{M(j)}. \]

for all $1 \leq i \leq g$, $g+1 \leq j \leq g+n$ and observe that by definition of $\xi_{g,n}$ (see (<ref>)) we have

\begin{align*}
\xi_{g,n}\bigl( \varphi_1 \otimes \ldots \otimes \varphi_{2g+n} \bigr) = \xi_{b_1}(\varphi_1) \, \xi_{a_1}(\varphi_2) \ldots \xi_{b_g}(\varphi_{2g-1}) \, \xi_{a_g}(\varphi_g) \, \xi_{m_{g+1}}(\varphi_{2g+1}) \ldots \xi_{m_{g+n}}(\varphi_{2g+n}).
\end{align*}

As above, one can show using the stated skein relations that the elements $\xi_{b_i}(\varphi)$, $\xi_{a_i}(\varphi)$, $\xi_{m_j}(\varphi) \in \mathcal{S}^{\mathrm{st}}_H(\Sigma_{g,n}^{\circ,\bullet})$ with $\varphi \in H^{\circ}$ obey the same product than the elements $\mathfrak{i}_{B(i)}(\varphi)$, $\mathfrak{i}_{A(i)}(\varphi)$, $\mathfrak{i}_{M(j)}(\varphi) \in \mathcal{L}_{g,n}(H)$. The reader can treat the case $(g,n)=(0,2)$ as an exercise.

The computations above reveal the topological relevance of the elements $\mathfrak{i}_{B(i)}(\varphi)$, $\mathfrak{i}_{A(i)}(\varphi)$, $\mathfrak{i}_{M(j)}(\varphi)$ and of the formulas in Proposition <ref> for their products.

§.§ Isomorphism $\mathcal{S}_H(\Sigma_{g,n}^{\circ}) \cong \mathcal{L}_{g,n}^H(H)$ for semisimple $H\text{-}\mathrm{mod}$

This part is a generalization of <cit.>, which dealt with the case $H = U_q^{\mathrm{ad}}(\mathfrak{sl}_2)$.

Recall that $\Sigma_{g,n}^{\circ}$ is the compact oriented surface of genus $g$ with one boundary component and $n$ punctures not belonging to the boundary. Recall that $F_{\mathrm{RT}}$ denotes the Reshetikhin–Turaev functor [51], and $k$ is the base field of $H$.

The skein algebra $\mathcal{S}_H(\Sigma_{g,n}^{\circ})$ is the $k$-vector space generated by the isotopy classes of ribbon links with coupons (i.e. ribbon graphs without boundary points) in $\Sigma_{g,n}^{\circ} \times [0,1]$, modulo the skein relations:

\begin{equation}\label{usualSkeinRelation}
\input{usualSkeinRelations.pdf_tex}
\end{equation}

where $\lambda_i \in k$ and the $T_i$'s are any ribbon graphs in $[0,1]^3$ having common colors on the incoming and outcoming strands. The sum on the left represents a linear combination of ribbon links which are equal outside of some cube in $\Sigma_{g,n}^{\circ} \times [0,1]$ which is depicted in grey.

Note that up to introducing identity coupons, we can always assume that the incoming and outcoming strands of $T_i$ have this orientation. Then $F_{\mathrm{RT}}(T_i) \in \mathrm{Hom}_H\bigl( X_1 \otimes \ldots \otimes X_k, Y_1 \otimes \ldots \otimes Y_l \bigr)$. The product in $\mathcal{S}_H(\Sigma_{g,n}^{\circ})$ is given by stacking (see after Definition <ref>).

Recall that $\Sigma_{g,n}^{\circ, \bullet}$ is $\Sigma_{g,n}^{\circ}$ with a puncture on the boundary. This puncture is irrelevant when we restrict to ribbon links, so if $L$ is the isotopy class of a ribbon link in $\Sigma_{g,n}^{\circ} \times [0,1]$ we have the corresponding isotopy class $I(L)$ in $\Sigma_{g,n}^{\circ, \bullet} \times [0,1]$. Note that $I(L)$ can be seen as a stated ribbon graph without boundary points (and thus without states): $I(L) \in \mathcal{S}_H^{\mathrm{st}}(\Sigma_{g,n}^{\circ, \bullet})$.

The map $L \mapsto I(L)$ induces a well-defined morphism of algebras $I : \mathcal{S}_H(\Sigma_{g,n}^{\circ}) \to \mathcal{S}_H^{\mathrm{st}}(\Sigma_{g,n}^{\circ, \bullet})$.

Let $L_1, \ldots, L_k$ be links in $\Sigma_{g,n}^{\circ} \times [0,1]$ and $\textstyle L = \sum_i \lambda_i L_i \in \mathcal{S}_H(\Sigma_{g,n}^{\circ})$. We have to show that if $L = 0$ is a skein relation (<ref>) in $\mathcal{S}_H(\Sigma_{g,n}^{\circ})$ then $\textstyle I(L) = \sum_i \lambda_i I(L_i) = 0$ follows from the stated skein relations in $\mathcal{S}_H^{\mathrm{st}}(\Sigma_{g,n}^{\circ, \bullet})$. For notational simplicity we take ribbon graphs $T_i$ which have $2$ incoming strands and $1$ outcoming strand in (<ref>). Using isotopy we can assume that the cube in (<ref>) is very close to $\partial(\Sigma_{g,n}^{\circ})$. Therefore in $\mathcal{S}_H^{\mathrm{st}}(\Sigma_{g,n}^{\circ, \bullet})$ we have

where all ribbon graphs are equal outside of the grey cube and $(y_j)$, $(x^{(1)}_k)$, $(x^{(2)}_l)$ are bases of $Y$, $X_1$, $X_2$ with dual bases $(y^j)$, $(x^{(1), k})$, $(x^{(2), l})$. For the second equality we used (<ref>) and for the third equality we used that $\textstyle \sum_i \lambda_i \, F_{\mathrm{RT}}(T_i) = 0$ by assumption. It is clear that $I$ is a morphism of algebras since both products are by stacking.

Note that since $I(L)$ is a stated ribbon graph without boundary points (and thus without states), $\mathrm{hol}^{\mathrm{st}}\bigl( I(L) \bigr)$ is just an element of $\mathcal{L}_{g,n}(H)$. Thus we can make the following definition:

The morphism $W = \mathrm{hol}^{\mathrm{st}} \circ I :\mathcal{S}_H(\Sigma_{g,n}^{\circ}) \to \mathcal{L}_{g,n}(H)$ is called the Wilson loop map.

Recall that the subalgebra of $H$-invariant elements is

\[ \mathcal{L}_{g,n}^H(H) = \bigl\{ x \in \mathcal{L}_{g,n}(H) \, \big| \, \forall \, h \in H, \: \mathrm{coad}^r(h)(x) = \varepsilon(h)x \bigr\}. \]

If the ribbon Hopf algebra $H$ has semisimple category $H\text{-}\mathrm{mod}$, the Wilson loop map takes values in $\mathcal{L}^H_{g,n}(H)$ and provides an isomorphism of algebras $\mathcal{S}_H(\Sigma_{g,n}^{\circ}) \overset{\sim}{\to} \mathcal{L}^H_{g,n}(H)$.

It is well-known that the ribbon category $\mathcal{C}$ of finite dimensional $U_q^{\mathrm{ad}}(\mathfrak{g})$-modules of type $1$ is semisimple. Therefore we can apply this theorem to $H=U_q^{\mathrm{ad}}(\mathfrak{g})$, meaning that the ribbon links with coupons defining $\mathcal{S}_H(\Sigma_{g,n}^{\circ})$ are colored by objects and morphisms in $\mathcal{C}$. Since $\mathcal{C}$ is equivalent to the category of finite dimensional modules of type $1$ over the simply connected quantum group $U_q(\mathfrak{g})$, we can indifferently take $H=U_q(\mathfrak{g})$.

It is true that $W$ takes values in $\mathcal{L}^H_{g,n}(H)$ even if $H$-mod is not semisimple, but in general it is not an isomorphism. See <ref> for more details on the non-semisimple case.

In all the remaining of this section we assume that $H$-mod is semisimple. In order to prove the theorem we first define two auxiliary maps $\mathcal{E}$ and $\mathcal{F}$.

Let $\mathrm{Irr}(H)$ be the set of irreducible finite-dimensional $H$-modules up to isomorphisms. Since $H\text{-}\mathrm{mod}$ is semisimple the matrix coefficients of the modules in $\mathrm{Irr}(H)$ form a basis of $H^{\circ}$, i.e. $\textstyle H^{\circ} = \bigoplus_{X \in \mathrm{Irr}(H)} C(X)$. Hence

\begin{equation}\label{decompositionLgnMatCoeffs}
\mathcal{L}_{g,n}(H) = \bigoplus_{\mathbf{X} \in \mathrm{Irr}(H)^{2g+n}} C(X_1) \otimes C(X_2) \otimes \ldots \otimes C(X_{2g+n})
\end{equation}

where $\mathbf{X} = (X_1, \ldots, X_{2g+n})$. For any such $\mathbf{X} \in \mathrm{Irr}(H)^{2g+n}$, let

\[ \textstyle \mathcal{E}_{\mathbf{X}} : \mathrm{Hom}_k\!\left( \bigotimes_{i=1}^{2g+n} X_i \otimes X_i^*, k \right) \overset{\sim}{\longrightarrow} C(X_1) \otimes \ldots \otimes C(X_{2g+n}) \]

($\mathrm{Hom}_k$ denotes the space of all linear maps) be the isomorphism of vector spaces defined by

\[ \mathcal{E}_{\mathbf{X}}(f) = \sum_{\substack{i_1, \ldots, i_N\\j_1, \ldots, j_N}} f\!\left( x_{i_1}^{(1)} \otimes x^{(1),j_1} \otimes \ldots \otimes x_{i_N}^{(N)} \otimes x^{(N),j_N} \right) \, {_{X_1}\phi^{i_1}_{j_1}} \otimes \ldots \otimes {_{X_N}\phi^{i_N}_{j_N}} \]

where $N = 2g+n$, $\bigl( x_{i_l}^{(l)} \bigr)$ is a basis of $X_l$ with dual basis $\bigl( x^{(l), j_l} \bigr)$ and $_{X_l}\phi^{i_l}_{j_l}$ denotes the matrix coefficients of $X_l$ in this basis. When $f$ is $H$-linear, it is useful to represent the element $\mathcal{E}_{\mathbf{X}}(f)$ by using the diagrammatic calculus of <cit.> as follows:

\begin{equation}\label{diagramInvariant}
\input{diagramInvariant.pdf_tex}
\end{equation}

By restriction to $H$-linear maps, the $k$-linear isomorphism $\mathcal{E}_{\mathbf{X}}$ gives an isomorphism of vector spaces $\textstyle \mathrm{Hom}_H\!\left( \bigotimes_{i=1}^{2g+n} X_i \otimes X_i^*, k \right) \overset{\sim}{\longrightarrow} \bigl( C(X_1) \otimes \ldots \otimes C(X_{2g+n}) \bigr)^H$. It follows that we have an isomorphism of vector spaces

\[ \mathcal{E} : \bigoplus_{\mathbf{X} \in \mathrm{Irr}(H)^{2g+n}} \mathrm{Hom}_H\!\left( \bigotimes_{i=1}^{2g+n} X_i \otimes X_i^*, k \right) \overset{\sim}{\longrightarrow} \mathcal{L}_{g,n}^H(H). \]

Let $\mathrm{act}^r$ be the right action of $H$ on $\textstyle \mathrm{Hom}_k\!\left( \bigotimes_{i=1}^{2g+n} X_i \otimes X_i^*, k \right)$ defined by

\[ \mathrm{act}^r(h)(f)\bigl( x_1 \otimes \varphi_1 \otimes \ldots \otimes x_N \otimes \varphi_N \bigr) = \sum_{(h)} f\bigl( h_{(1)} \cdot x_1 \otimes h_{(2)} \cdot \varphi_1 \otimes \ldots \otimes h_{(2N-1)} \cdot x_N \otimes h_{(2N)} \cdot \varphi_N \bigr) \]

for all $x_l \in X_l$, $\varphi_l \in X_l^*$, where $N = 2g+n$ and $h \in H$. Then $\mathcal{E}_{\mathbf{X}}$ intertwines the right actions $\mathrm{act}^r$ and $\mathrm{coad}^r$. This is due to the definition of $\mathrm{coad}^r$ (see (<ref>) and (<ref>)) and to the following formulas, which come from (<ref>):

\begin{equation}\label{balancingActions}
\begin{array}{l}
\sum_{i_l,j_l} \bigl( h \rhd {_{X_l}\phi^{i_l}_{j_l}} \bigr) \otimes x_{i_l} \otimes x^{j_l} = \sum_{i_l,j_l} {_{X_l}\phi^{i_l}_{j_l}} \otimes x_{i_l} \otimes S^{-1}(h) \cdot x^{j_l},\\[1.5em]
\sum_{i_l,j_l} \bigl( {_{X_l}\phi^{i_l}_{j_l}} \lhd h \bigr) \otimes x_{i_l} \otimes x^{j_l} = \sum_{i_l,j_l} {_{X_l}\phi^{i_l}_{j_l}} \otimes h \cdot x_{i_l} \otimes x^{j_l}.
\end{array}
\end{equation}

For the first formula we used that if $V$ is a $H$-module, then the action of $h \in H$ on $\psi \in V^*$ is defined by $\langle h \cdot \psi, v \rangle = \langle \psi, S(h) \cdot v \rangle$ for all $v \in V$. The lemma follows from this intertwining property because

\[ \textstyle f \in \mathrm{Hom}_H\!\left( \bigotimes_{i=1}^{2g+n} X_i \otimes X_i^*, k \right) \quad \iff \quad \forall \, h \in H, \:\: \mathrm{act}^r(h)(f) = \varepsilon(h)f. \qedhere \]

Now, for $\mathbf{X} = (X_1, \ldots, X_{2g+n}) \in \mathrm{Irr}(H)^{2g+n}$, let

\[ \textstyle \mathcal{F}_{\mathbf{X}} : \mathrm{Hom}_H\!\left( \bigotimes_{i=1}^{2g+n} X_i \otimes X_i^*, k \right) \to \mathcal{S}_H(\Sigma_{g,n}^{\circ}) \]

defined by

The map $\mathcal{F}_{\mathbf{X}}$ is $k$-linear thanks to the skein relations (<ref>). Let

\[ \mathcal{F} : \bigoplus_{\mathbf{X} \in \mathrm{Irr}(H)^{2g+n}} \mathrm{Hom}_H\!\left( \bigotimes_{i=1}^{2g+n} X_i \otimes X_i^*, k \right) \to \mathcal{S}_H(\Sigma_{g,n}^{\circ}) \]

be the sum of the linear maps $\mathcal{F}_{\mathbf{X}}$.

The linear map $\mathcal{F}$ is surjective.

As in the proof of Lemma <ref> we observe that every element of $\mathcal{S}_H(\Sigma_{g,n}^{\circ})$ is a linear combination of links of the form

where $T_{\mathbf{X}}$ is some oriented and colored $(4g+2n, 0)$-ribbon graph in $[0,1]^3$. We can assume that the colors $\mathbf{X} = (X_1, \ldots, X_{2g+n})$ are irreducible. Indeed, if a strand is colored by a non-irreducible module $V$, then write $\textstyle V = \bigoplus_l S_l$ where the $S_l$ are irreducible. We can introduce coupons containing the injections $I_l : S_l \to V$ and projections $\pi_l : V \to S_l$ thanks to the relation $\textstyle \mathrm{id}_V = \sum_l I_l \circ \pi_l$. By sliding these coupons along the strand we get a sum of links where the strands under consideration are colored by the modules $S_l$.

Thanks to the skein relations (<ref>) we can replace the ribbon graph $T_{\mathbf{X}}$ by a coupon colored by $\textstyle F_{\mathrm{RT}}(T_{\mathbf{X}}) \in \mathrm{Hom}_H\!\left( \bigotimes_{i=1}^{2g+n} X_i \otimes X_i^*, k \right)$ and it follows that $L_{\mathbf{X}} = \mathcal{F}_{\mathbf{X}}\bigl( F_{\mathrm{RT}}(T_{\mathbf{X}}) \bigr)$.

By definition of $\mathrm{hol}$ <cit.> and (<ref>), we have a commutative diagram

\[ \xymatrix{
\bigoplus_{\mathbf{X} \in \mathrm{Irr}(H)^{2g+n}} \mathrm{Hom}_H\!\left( \bigotimes_{i=1}^{2g+n} X_i \otimes X_i^*, k \right) \ar[r]^-{\mathcal{E}} \ar[d]_{\mathcal{F}} & \mathcal{L}_{g,n}^H(H) \ar@{^{(}->}[d]\\
\mathcal{S}_H(\Sigma_{g,n}^{\circ}) \ar[r]_W & \mathcal{L}_{g,n}(H)
} \]

The surjectivity of $\mathcal{F}$ (Lemma <ref>) implies that $W$ takes values in $\mathcal{L}_{g,n}^H(H)$. Since $\mathcal{E}$ is an isomorphism (Lemma <ref>), $\mathcal{F}$ is necessarily injective, so it is an isomorphism of vector spaces. It follows that $W$ is an isomorphism between $\mathcal{S}_H(\Sigma_{g,n}^{\circ})$ and $\mathcal{L}^H_{g,n}(H)$.

Recall the map $I$ introduced before Lemma <ref>. Our results imply the following important fact, which is not obvious despite seeming completely natural:

Assume that $H\text{-}\mathrm{mod}$ is semisimple. Then the morphism of algebras $I : \mathcal{S}_H(\Sigma_{g,n}^{\circ}) \to \mathcal{S}_H^{\mathrm{st}}(\Sigma_{g,n}^{\circ, \bullet})$ is injective.

By the definition of $W$ (Definition <ref>), Theorem <ref> and Theorem <ref> we have a commutative diagram where the rows are isomorphisms:

\[<EMAIL_ADDRESS>\mathcal{S}_H^{\mathrm{st}}(\Sigma_{g,n}^{\circ,\bullet}) \ar[rr]_{\mathrm{hol}^{\mathrm{st}}}^{\sim} & & \mathcal{L}_{g,n}(H)\\
& \circlearrowleft & \\
\mathcal{S}_H(\Sigma_{g,n}^{\circ}) \ar[rr]_W^{\sim} \ar[uu]^I & & \mathcal{L}_{g,n}^H(H) \ar@{^{(}->}[uu]_{\text{(subalgebra)}}
} \]

It immediately follows that $I$ is injective.

This corollary means that the skein algebra $\mathcal{S}_H(\Sigma_{g,n}^{\circ})$ is isomorphic to the subalgebra of ribbon links (with coupons) in the stated skein algebra $\mathcal{S}_H^{\mathrm{st}}(\Sigma_{g,n}^{\circ, \bullet})$.

§.§ Remarks on semisimplicity

If the category $H\text{-}\mathrm{mod}$ is not semisimple:

* the matrix coefficients of irreducible modules do not form a generating family of $H^{\circ}$,

* the matrix coefficients of indecomposable modules form a generating family of $H^{\circ}$ but not a free family (even if one restricts to projective modules).
Hence $\textstyle H^{\circ} = \sum_{X \in \mathrm{Ind}(H)} C(X)$, where $\mathrm{Ind}(H)$ denotes the set of indecomposable $H$-modules up to isomorphisms.

Consequently the decomposition (<ref>) is no longer true, and instead one has

\[ \mathcal{L}_{g,n}(H) = \sum_{\mathbf{X} \in \mathrm{Ind}(H)^{2g+n}} C(X_1) \otimes C(X_2) \otimes \ldots \otimes C(X_{2g+n}). \]

So there is a surjection $\textstyle \bigoplus_{\mathbf{X} \in \mathrm{Ind}(H)^{2g+n}} \mathrm{Hom}_k\!\left( \bigotimes_{i=1}^{2g+n} X_i \otimes X_i^*, k \right) \twoheadrightarrow \mathcal{L}_{g,n}(H)$ and a commutative diagram

\[ \xymatrix{
\bigoplus_{\mathbf{X} \in \mathrm{Ind}(H)^{2g+n}} \mathrm{Hom}_H\!\left( \bigotimes_{i=1}^{2g+n} X_i \otimes X_i^*, k \right) \ar[r]^-{\mathcal{E}} \ar[d]_{\mathcal{F}} & \mathcal{L}_{g,n}^H(H) \ar@{^{(}->}[d]\\
\mathcal{S}_H(\Sigma_{g,n}^{\circ}) \ar[r]_W & \mathcal{L}_{g,n}(H)
} \]

Moreover $\mathcal{F}$ is still surjective, thus $W$ takes values in $\mathcal{L}_{g,n}^H(H)$. But in general nothing can be said about the surjectivity of $\mathcal{E}$. So when $H$ is not semisimple, Theorem <ref> will in general not be true:

\[ \mathrm{im}(W) \subsetneq \mathcal{L}_{g,n}^H(H). \]

Non-semisimplicity arises notably at roots of unity. For instance take $(g,n)=(0,1)$ and $H = \overline{U}_{\epsilon} = \overline{U}_{\epsilon}(\mathfrak{sl}_2)$, the restricted quantum group of $\mathfrak{sl}(2)$ at $\epsilon = e^{i\pi/p}$ with $p \geq 2$ (see e.g. <cit.>). We have $\mathcal{L}_{0,1}^{\overline{U}_{\epsilon}} \cong \mathcal{Z}(\overline{U}_{\epsilon})$ (see e.g. <cit.>) and it is known that $\dim\bigl( \mathcal{Z}(\overline{U}_{\epsilon}) \bigr) = 3p-1$ <cit.>. On the other hand one can check that $\dim\bigl( \mathrm{im}(W) \bigr) = 2p$, showing that the inclusion is indeed strict.

In <cit.> a generalization of Wilson loops was defined in order to produce more $H$-invariant elements. The idea is roughly to use other symmetric linear forms than the traces; for $\Sigma^{\circ}_{0,1}$ these generalized Wilson loops recover the whole of $\mathcal{L}^H_{0,1}(H)$.

Another interesting situation where non-semisimplicity arises is when working over the ring $A=\mc[q^{1/D},q^{-1/D}]$ and specializing $q$ to a root of unity. To define an integral version $\mathcal{L}_{g,n}^A$ of $\mathcal{L}_{g,n}$ defined over $A$, such that $\mathcal{L}_{g,n}^A\otimes_A \mc(q^{1/D}) = \mathcal{L}_{g,n}$, one considers Lusztig's restricted quantum group $U_{\mathcal{A}}^{\mathrm{res}}(\mathfrak{g})\subset U_q^{\mathrm{ad}}(\mathfrak{g})$, which is defined over $\mathcal{A}:=\mc[q,q^{-1}]$, and its restricted dual $\mathcal{O}_{\mathcal{A}}(G)$, associated to the category $\mathcal{C}_{\mathcal{A}}$ of finite rank $U_{\mathcal{A}}^{\mathrm{res}}(\mathfrak{g})$-modules of type $1$. The algebra $\mathcal{L}_{g,n}^A$ in the case $g=0$ is described in [8].

§ THE CASE OF SURFACES WITHOUT BOUNDARY
By Theorem <ref>, we know that if the ribbon Hopf algebra $H$ has semisimple category $H\text{-}\mathrm{mod}$, the Wilson loop map provides an isomorphism of algebras $\mathcal{S}_H(\Sigma_{g,n}^{\circ}) \overset{\sim}{\to} \mathcal{L}^H_{g,n}(H)$. Here we address the question of defining algebras which are related to the surface $\Sigma_{g,n}$ obtained by closing $\Sigma_{g,n}^{\circ}$, and derived from $\mathcal{S}_H(\Sigma_{g,n}^{\circ}) $ and $\mathcal{L}^H_{g,n}(H)$ respectively. We stress that the construction and results  presented below works for any $n\geq0$, including in particular the closed surface of genus $g$.

§.§ Quantum moment maps and quantum reduction

Let $H$ be a Hopf algebra over a field $k$, and $H'\subset H$ a subalgebra and right $H$-coideal (so $\Delta(H')\subset H'\otimes H$). On $H$ we have the right adjoint action,

$$\mathrm{ad}^r(h)(g) = \sum_{(h)} S(h_{(1)})gh_{(2)}, \quad (h,g\in H),$$

and we assume that $H'$ is a stable subspace under this action. Let $A$ be a left $H$-module algebra. We denote by $\smallblacktriangle$ the action of $H$ on $A$.
A morphism of algebras $\mu\colon H'\ra A$ is a quantum moment map (QMM) if for every $a\in A$ and $h'\in H'$ it satisfies

$$\sum_{(h')}(h_{(2)}'\smallblacktriangle a) \mu(h_{(1)}')= \mu(h') a.$$

This definition is implicit in [41], and explicit in <cit.> (we use an opposite coproduct, which is suited to our conventions). The following facts follow readily from the definitions and are properties we can expect of a quantum analog of moment map (see [41], Theorem 3.10).
Assume we are given a QMM $\mu\colon H'\ra A$. Then:

$(i)$ (Equivariance) For every $h\in H$ and $h'\in H'$ we have
$$\mu\bigl(\mathrm{ad}^r(h)(h')\bigr) = S(h)\smallblacktriangle \mu(h').$$

$(ii)$ (Recovering the action)  For every $a\in A$ and $h\in H$ such that $(S\otimes id)\Delta(h)\in H'\otimes H'$ (eg. when $H'=H$), we have
$$\textstyle \sum_{(h)} \mu(S(h_{(1)})) a \mu(h_{(2)}) = S(h)\smallblacktriangle a.$$

Given a morphism of algebras $\chi\colon H'\ra k$, put

\begin{equation}\label{defiIchi}
I_\chi := A\mu({\rm Ker}(\chi)).
\end{equation}

This is a left ideal of $A$ (but in general not a two-sided ideal). Since ${\rm Ker}(\chi) = \{h'-\chi(h')1, h'\in H'\}$, and $A$ is a $H$-module algebra and $H'$ is stable under $\mathrm{ad}^r$, $(i)$ in the lemma implies that $I_\chi$ is $H'$-stable. Therefore we can consider the left $(A/I_\chi)$-submodule $(A/I_\chi)^{H'}\subset A/I_\chi$ formed by the $H'$-invariant elements. Define the map $\varphi_\chi\colon H'\ra H$, $\textstyle h'\mapsto \sum_{(h')} \chi(h_{(1)}')h_{(2)}'$, and denote by $\varepsilon_{\vert H'}\colon H'\ra k$ the restriction of the counit of $H$.
The product of $A$ descends to the space $(A/I_\chi)^{H'}$ and gives it a structure of algebra. Moreover, if ${\rm Im}(\varphi_\chi)\subset H'$ and $\varphi_\chi({\rm Ker}(\chi))= {\rm Ker}(\varepsilon_{\vert H'})$, then the algebra $(A/I_\chi)^{H'}$ is isomorphic to $End_A(A/I_\chi)^{op}$.

Both claims follow from arguments in Proposition 1.5.2 of [53], to which we refer. The first claim is also detailed in Proposition 3.12 of [32]. For completeness let us explain it. The point is to observe that a coset $a+I_\chi \in A/I_\chi$ is $H'$-invariant if and only if ${\rm Ker}(\varepsilon_{\vert H'}) \smallblacktriangle a \subset I_\chi$. But ${\rm Ker}(\varepsilon_{\vert H'})$ obviously contains the elements $\textstyle \sum_{(h')} \chi(h_{(1)}')h_{(2)}'\in H$, where $h'\in {\rm Ker}(\chi)$, and

\begin{align*}\left(\sum_{(h')} \chi(h_{(1)}')h_{(2)}'\right)\smallblacktriangle a
& = \sum_{(h')} (h_{(2)}'\smallblacktriangle a)(\mu(h'_{(1)})) + \sum_{(h')} (h_{(2)}'\smallblacktriangle a)(\chi(h'_{(1)}) - \mu(h_{(1)}'))
\\ & = \sum_{(h')} (h_{(2)}'\smallblacktriangle a)\mu(h_{(1)}') \quad {\rm modulo}\ I_\chi \\
& = \mu(h') a \quad {\rm modulo}\ I_\chi.
\end{align*}

The third equality follows from the QMM equation in Definition <ref>. Therefore, for every $h'\in {\rm Ker}(\chi)$, $\mu(h') a \in {\rm Ker}(\varepsilon_{\vert H'}) \smallblacktriangle a + I_\chi \subset I_\chi$, whence $I_\chi a \subset I_\chi$, which shows $I_\chi$ is stable under right multiplication by elements which are $H'$-invariant modulo $I_\chi$. It follows that the product of $A$ descends to $(A/I_\chi)^{H'}$.

From [53] we take:

The algebra $A/\!\!/_{\!\chi} H' := (A/I_\chi)^{H'}$ is the quantum reduction of $A$ by the character $\chi$.

By the defining property of $\mu$, it is immediate that given an $A$-module $V$ the subspace

$$V^{H',\chi} :=\{v\in V, \mu(h') \cdot v = \chi(h')v, \forall h'\in H'\}$$

is a $(A/\!\!/_{\!\chi} H')$-module. Therefore we have a functor $A$-Mod$\ra (A/\!\!/_{\!\chi} H')$-Mod.

Let us observe the following fact in the case of the quantum reduction by the counit $\varepsilon$, which will be important in Section <ref>. Consider the canonical quotient map of $H$-modules, $p \colon A \ra  A/I_\varepsilon$. The restriction $\pi$ of $p$ to $A^{H'}$ has image contained in $A/\!\!/_{\!\varepsilon} H'$, and as explained after Proposition <ref> it is a morphism of algebras

\begin{equation}\label{defPi}
\pi\colon A^{H'} \ra  A/\!\!/_{\!\varepsilon} H'.
\end{equation}

Denote by $Z(H')$ the center of $H'$.
We will say that $Z(H')$ separates the simple types if the following property holds: for any finite collection of non isomorphic and non trivial simple $H$-modules $X_1,\ldots,X_k$, there exists an element $z\in Z(H')$ such that $\varepsilon(z)=0$ and $z\smallblacktriangle v  = z_i v$ for every $v\in X_i$, where $z_i \in k$ satisfy $z_i\ne z_j$ for $i\ne j$.

Assume that the $H$-module $A$ is completely reducible and $Z(H')$ separates the simple types. Then the map  $\pi$ is surjective.

Let $a+I_\varepsilon \in A/\!\!/_{\!\varepsilon} H'$. By definition we have $h'\smallblacktriangle a - \varepsilon(h')a \in I_\varepsilon$, for every $h'\in H'$. Since $A$ is completely reducible, we can decompose $a$ in the form $a=\textstyle \sum_{i=0}^k a_i$ with $a_i$ the isotypical component of $a$ of type $X_i$, $i\in \{1,\ldots,k\}$, and $a_0$ the isotypical component of $a$ of trivial type, ie. the $H$-invariant component of $a$. Then

$$h'\smallblacktriangle a - \varepsilon(h')a = \sum_{i=1}^k  (h'\smallblacktriangle a_i - \varepsilon(h')a_i).$$

Let us apply this to an element $z\in Z(H')$, separating the components of type $X_i$ as in Definition <ref>. For every integer $s\geq 1$ we have

$$z^s\smallblacktriangle a - \varepsilon(z^s)a = \sum_{i=1}^k  z_i^s a_i.$$

Therefore $\textstyle \sum_{i=1}^k  z_i^s a_i \in I_\varepsilon$ for every $s\geq 1$, with $z_i\ne z_j$ for $i\ne j$. Because the Vandermonde matrix $(z_i^s)$ is invertible for $1\leq s\leq k$, we find $a_i \in I_\varepsilon$ for every $i\in \{1,\ldots,k\}$. Therefore $a+I_\varepsilon = a_0 +I_\varepsilon$, showing that $\pi$ is surjective.

§.§ A quantum moment map for $\mathcal{L}_{g,n}(H)$
Let $H$ be a ribbon Hopf algebra with an invertible antipode; we denote the ribbon element of $H$ by $v$.

Assumption: the morphism $\Phi_{0,1} : \mathcal{L}_{0,1}(H) \to H$ from (<ref>) is injective.