$\displaystyle\displaystyle=q^{-(\nu,\gamma)}\sum_{(\beta),(\beta)}\beta_{(1)}(R^{2}_{(2)}R^{4}_{(2)})\left(S(R^{1}_{(2)}R^{3}_{(2)})K_{\nu}\rhd\beta_{(2)}\right)$
$\displaystyle\displaystyle=q^{(\nu,\lambda_{2}-\gamma)}\sum_{(\beta),(\beta)}\beta_{(1)}(R^{2}_{(2)}R^{4}_{(2)})\left(S(R^{1}_{(2)}R^{3}_{(2)})\rhd\beta_{(2)}\right)$
$\displaystyle\displaystyle=q^{(\nu,\lambda_{2}-\gamma)}\left(S(R^{1}_{(2)}R^{3}_{(2)})\rhd\beta\lhd
R^{2}_{(2)}R^{4}_{(2)}\right).$

for some positive root $\displaystyle\gamma\in Q_{+}$. Therefore
$\displaystyle S(R^{1}_{(2)}R^{3}_{(2)})\rhd\beta\lhd
R^{2}_{(2)}R^{4}_{(2)}\in C(\mu_{2})_{\lambda_{2}-\gamma}$; by a similar
computation we find that $\displaystyle
S(R^{3}_{(1)}R^{4}_{(1)})\rhd\alpha^{\prime}\lhd R^{1}_{(1)}R^{2}_{(1)}\in
C(\mu_{1}^{\prime})_{\lambda_{1}^{\prime}+\gamma}$ for the complementary
components $\displaystyle R^{1}_{(1)},\ldots,R^{4}_{(1)}$. Then we always have
$\displaystyle S(R^{1}_{(2)}R^{3}_{(2)})\rhd\beta\lhd
R^{2}_{(2)}R^{4}_{(2)}\in{\mathcal{F}}_{2}^{\mu_{2},\lambda_{2}}$, and if
$\displaystyle S(R^{1}_{(2)}R^{3}_{(2)})\rhd\beta\lhd
R^{2}_{(2)}R^{4}_{(2)}\in C(\mu_{2})_{\lambda_{2}}$ then $\displaystyle
S(R^{3}_{(1)}R^{4}_{(1)})\rhd\alpha^{\prime}\lhd R^{1}_{(1)}R^{2}_{(1)}\in
C(\mu_{1}^{\prime})_{\lambda_{1}^{\prime}}$. Since the product of
$\displaystyle{\mathcal{L}}_{0,n}$ is componentwise that of
$\displaystyle{\mathcal{L}}_{0,1}$, by (35) we have

(42) $\left(S(R^{1}_{(2)}R^{3}_{(2)})\rhd\beta\lhd
R^{2}_{(2)}R^{4}_{(2)}\right)\beta^{\prime}\in{\mathcal{F}}_{2}^{\mu_{2}+\mu_{2}^{\prime},\lambda_{2}+\lambda_{2}^{\prime}}$

and if $\displaystyle(S(R^{1}_{(2)}R^{3}_{(2)})\rhd\beta\lhd
R^{2}_{(2)}R^{4}_{(2)})\beta^{\prime}\in
C(\mu_{2}+\mu_{2}^{\prime})_{\lambda_{2}+\lambda_{2}^{\prime}}$ then

(43) $\alpha\left(S(R^{3}_{(1)}R^{4}_{(1)})\rhd\alpha^{\prime}\lhd
R^{1}_{(1)}R^{2}_{(1)}\right)\in
C(\mu_{1})_{\lambda_{1}}C(\mu_{1}^{\prime})_{\lambda_{1}^{\prime}}\subset{\mathcal{F}}_{2}^{\mu_{1}+\mu_{1}^{\prime},\lambda_{1}+\lambda_{1}^{\prime}}.$

In conclusion

$\displaystyle(\alpha\otimes\beta)(\alpha^{\prime}\otimes\beta^{\prime})\in{\mathcal{F}}_{2}^{[\mu],[\lambda]}.$

Similar arguments work for any $\displaystyle n\geq 2$. This proves that
$\displaystyle Gr_{{\mathcal{F}}_{2}}({\mathcal{L}}_{0,n})$ is a graded
algebra with the product inherited from $\displaystyle{\mathcal{L}}_{0,n}$,
which we denote by $\displaystyle\circ_{n}$. Recall that it is defined on
homogeneous elements $\displaystyle\overline{\alpha\otimes\beta}\in
Gr_{{\mathcal{F}}_{2}}({\mathcal{L}}_{0,n})_{[\mu],[\lambda]}$,
$\displaystyle\overline{\alpha^{\prime}\otimes\beta^{\prime}}\in
Gr_{{\mathcal{F}}_{2}}({\mathcal{L}}_{0,n})_{[\mu^{\prime}],[\lambda^{\prime}]}$
by

$\displaystyle\overline{\alpha\otimes\beta}\circ_{n}\overline{\alpha^{\prime}\otimes\beta^{\prime}}=(\alpha\otimes\beta)(\alpha^{\prime}\otimes\beta^{\prime})+{\mathcal{F}}_{2}^{<[\mu+\mu^{\prime}],[\lambda+\lambda^{\prime}]}.$

Next we show that (36) implies the same property for the product
$\displaystyle\circ$ of $\displaystyle
Gr_{{\mathcal{F}}_{2}}({\mathcal{L}}_{0,n})$. First it gives
$\displaystyle\left(C(\mu_{1})\circ
C(\mu_{1}^{\prime})\right)\otimes\left(C(\mu_{2})\circ
C(\mu_{2}^{\prime})\right)=C([\mu+\mu^{\prime}])$. Now, by the previous remark
and (42), (43) we have

$\displaystyle C([\mu])\circ_{n}C([\mu^{\prime}])\subset\left(C(\mu_{1})\circ
C(\mu_{1}^{\prime})\right)\otimes\left(C(\mu_{2})\circ
C(\mu_{2}^{\prime})\right).$

The converse inclusion holds true as well, as one can see by reasoning as
above, starting with the standard (componentwise) product of
$\displaystyle{\mathcal{L}}_{0,1}^{\otimes n}$ expressed in terms of the
product of $\displaystyle{\mathcal{L}}_{0,n}$ by the formula (16). In
conclusion

(44) $C([\mu])\circ_{n}C([\mu^{\prime}])=C([\mu+\mu^{\prime}]).$

We are left to show that (37) generalizes to
$\displaystyle{\mathcal{L}}_{0,n}$. Again we note that this cannot be deduced
from the case $\displaystyle n=1$, because for the vector space isomorphism
$\displaystyle Gr_{{\mathcal{F}}_{2}}({\mathcal{L}}_{0,n})\rightarrow
Gr_{{\mathcal{F}}_{2}}({\mathcal{L}}_{0,1})^{\otimes n}$ induced from the
equality $\displaystyle{\mathcal{L}}_{0,n}={\mathcal{L}}_{0,1}^{\otimes n}$ is
not a morphism of algebras with respect to the products
$\displaystyle\circ_{n}$ and $\displaystyle\circ^{\otimes n}$. Therefore one
cannot take the filtration on $\displaystyle
Gr_{{\mathcal{F}}_{2}}({\mathcal{L}}_{0,n})$ which is componentwise
$\displaystyle\mathcal{F}_{3}$, and that we will denote again by
$\displaystyle\mathcal{F}_{3}$, to deduce that $\displaystyle
Gr_{{\mathcal{F}}_{3}}(Gr_{{\mathcal{F}}_{2}}({\mathcal{L}}_{0,n}))$ is a
quotient of a quasi-polynomial algebra, whence Noetherian. However, we can
proceed in essentially the same way. We give the details when $\displaystyle
n=2$, the general case being similar. Let us write the twist $\displaystyle F$
in (15) as

$\displaystyle F=\sum_{(F)}F_{(1)}\otimes F_{(2)}=\sum_{(F)}F_{(1)1}\otimes
F_{(1)2}\otimes F_{(2)1}\otimes F_{(2)2}$

that is, setting $\displaystyle F_{(1)1}:=R^{2}_{(2)}R^{4}_{(2)}$,
$\displaystyle F_{(1)2}:=R^{1}_{(2)}R^{3}_{(2)}$, $\displaystyle
F_{(2)1}:=R^{1}_{(1)}R^{2}_{(1)}$, $\displaystyle
F_{(2)2}:=R^{3}_{(1)}R^{4}_{(1)}$. Keep the notations of (37), and put
$\displaystyle d(\mu):=dim(V_{\mu})$, $\displaystyle\mu\in P_{+}$, and

$\displaystyle\Delta^{(2)}({}_{{\mu_{2}}}\phi_{k_{2}}^{l_{2}})=\sum_{p,s=1}^{\scriptscriptstyle{d(\mu_{2})}}{}_{{\mu_{2}}}\phi_{k_{2}}^{p}\otimes{}_{{\mu_{2}}}\phi_{p}^{s}\otimes{}_{{\mu_{2}}}\phi_{s}^{l_{2}}\
,\
\Delta^{(2)}({}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{l_{1}^{\prime}})=\sum_{p^{\prime},s^{\prime}=1}^{\scriptscriptstyle{d(\mu_{1}^{\prime})}}{}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{p^{\prime}}\otimes{}_{{\mu_{1}^{\prime}}}\phi_{p^{\prime}}^{s^{\prime}}\otimes{}_{{\mu_{1}^{\prime}}}\phi_{s^{\prime}}^{l_{1}^{\prime}}.$

Assume that
$\displaystyle{}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}<{}_{{\mu_{2}}}\phi_{k_{2}}^{l_{2}}$
and
$\displaystyle{}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{l_{1}^{\prime}}<{}_{{\mu_{1}}}\phi_{k_{1}}^{l_{1}}$.
From (14) and then (37)-(38) in the second equality, one obtains

$\displaystyle\displaystyle\left({}_{{\mu_{1}}}\phi_{k_{1}}^{l_{1}}\otimes{}_{{\mu_{2}}}\phi_{k_{2}}^{l_{2}}\right)\circ_{2}\left({}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{l_{1}^{\prime}}\otimes{}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}\right)$
$\displaystyle\displaystyle\hskip
42.67912pt=\sum_{(F)}\sum_{p,s=1}^{\scriptscriptstyle{d(\mu_{2})}}\sum_{p^{\prime},s^{\prime}=1}^{\scriptscriptstyle{d(\mu_{1}^{\prime})}}\left({}_{{\mu_{1}}}\phi_{k_{1}}^{l_{1}}\circ{}_{{\mu_{1}^{\prime}}}\phi_{p^{\prime}}^{s^{\prime}}\left({}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{p^{\prime}}(F_{(2)1}){}_{{\mu_{1}^{\prime}}}\phi_{s^{\prime}}^{l_{1}^{\prime}}(S(F_{(2)2}))\right)\right)$
$\displaystyle\displaystyle\hskip
199.16928pt\otimes\left({}_{{\mu_{2}}}\phi_{p}^{s}\circ{}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}\left({}_{{\mu_{2}}}\phi_{k_{2}}^{p}(F_{(1)1}){}_{{\mu_{2}}}\phi_{s}^{l_{2}}(S(F_{(1)2}))\right)\right)$
$\displaystyle\displaystyle\hskip
42.67912pt=\sum_{(F)}\sum_{p,s=1}^{\scriptscriptstyle{d(\mu_{2})}}\sum_{p^{\prime},s^{\prime}=1}^{\scriptscriptstyle{d(\mu_{1}^{\prime})}}q_{p^{\prime}s^{\prime}k_{1}l_{1}}\
q_{k_{2}^{\prime}l_{2}^{\prime}ps}\
\left({}_{{\mu_{1}^{\prime}}}\phi_{p^{\prime}}^{s^{\prime}}\circ{}_{{\mu_{1}}}\phi_{k_{1}}^{l_{1}}\left({}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{p^{\prime}}(F_{(2)1}){}_{{\mu_{1}^{\prime}}}\phi_{s^{\prime}}^{l_{1}^{\prime}}(S(F_{(2)2}))\right)\right)$
$\displaystyle\displaystyle\hskip
170.71652pt\otimes\left({}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}\circ{}_{{\mu_{2}}}\phi_{p}^{s}\left({}_{{\mu_{2}}}\phi_{k_{2}}^{p}(F_{(1)1}){}_{{\mu_{2}}}\phi_{s}^{l_{2}}(S(F_{(1)2}))\right)\right)+\ldots$

Here the dots are sums of tensors of the form $\displaystyle(x_{1}\circ
x_{2})\otimes(y_{1}\circ y_{2})$ where $\displaystyle
x_{1}<{}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{l_{1}^{\prime}}$ and
$\displaystyle
y_{1}<{}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}$. In fact,
by the expression of $\displaystyle R=\Theta\hat{R}$ we have
$\displaystyle{}_{{\mu_{2}}}\phi_{k_{2}}^{p}(F_{(1)1}){}_{{\mu_{2}}}\phi_{s}^{l_{2}}(S(F_{(1)2}))=0$
unless $\displaystyle k_{2}\geq p$ and $\displaystyle s\geq l_{2}$, and
$\displaystyle{}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{p^{\prime}}(F_{(2)1}){}_{{\mu_{1}^{\prime}}}\phi_{s^{\prime}}^{l_{1}^{\prime}}(S(F_{(2)2})=0$
unless $\displaystyle k_{1}^{\prime}\leq p^{\prime}$ and $\displaystyle
s^{\prime}\leq l_{1}^{\prime}$. It is immediate that
$\displaystyle{}_{{\mu_{2}}}\phi_{p}^{s}\in C(\mu_{2})_{w(e_{s}^{\mu_{2}})}$.
By definition $\displaystyle s>l_{2}$ implies $\displaystyle
w(e_{s}^{\mu_{2}})\leq w(e_{l_{2}}^{\mu_{2}})$, and by (35) if $\displaystyle
w(e_{s}^{\mu_{2}})<w(e_{l_{2}}^{\mu_{2}})$ then
$\displaystyle{}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}\circ{}_{{\mu_{2}}}\phi_{p}^{s}\in\mathcal{F}_{2}^{<\mu^{\prime\prime}_{2},\lambda_{2}^{\prime\prime}}$,
where $\displaystyle\mu_{2}^{\prime\prime}:=\mu_{2}+\mu_{2}^{\prime}$,
$\displaystyle\lambda_{2}^{\prime\prime}:=w(e_{l_{2}}^{\mu_{2}})+w(e_{l_{2}^{\prime}}^{\mu_{2}^{\prime}})$.
In that case the term
$\displaystyle({}_{{\mu_{1}^{\prime}}}\phi_{p^{\prime}}^{s^{\prime}}\circ{}_{{\mu_{1}}}\phi_{k_{1}}^{l_{1}})\otimes({}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}\circ{}_{{\mu_{2}}}\phi_{p}^{s})$
in the sum above vanishes in $\displaystyle
Gr_{\mathcal{F}_{2}}({\mathcal{L}}_{0,2})$. Moreover, the computations before
(43) show that such a summand achieves the maximal weight $\displaystyle
w(e_{l_{1}}^{\mu_{1}})+w(e_{l_{1}^{\prime}}^{\mu_{1}^{\prime}})+w(e_{l_{2}}^{\mu_{2}})+w(e_{l_{2}^{\prime}}^{\mu_{2}^{\prime}})$
only if $\displaystyle
w(e_{s^{\prime}}^{\mu_{1}^{\prime}})=w(e_{l_{1}^{\prime}}^{\mu_{1}^{\prime}})$
and $\displaystyle w(e_{s}^{\mu_{2}})=w(e_{l_{2}}^{\mu_{2}})$, which occurs
when $\displaystyle F_{(1)2}$, $\displaystyle F_{(2)2}$ have no component of
$\displaystyle\hat{R}$ but only of $\displaystyle\Theta$. Also, if
$\displaystyle p=k_{2}$, $\displaystyle s=l_{2}$, $\displaystyle
p^{\prime}=k_{1}^{\prime}$, $\displaystyle s^{\prime}=l_{1}^{\prime}$ then

$\displaystyle\displaystyle\sum_{(F)}\ $
$\displaystyle\displaystyle{}_{{\mu_{2}}}\phi_{k_{2}}^{k_{2}}(F_{(1)1}){}_{{\mu_{2}}}\phi_{l_{2}}^{l_{2}}(S(F_{(1)2})){}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{k_{1}^{\prime}}(F_{(2)1}){}_{{\mu_{1}^{\prime}}}\phi_{l_{1}^{\prime}}^{l_{1}^{\prime}}(S(F_{(2)2})$
$\displaystyle\displaystyle\hskip
113.81102pt=\left\langle{}_{{\mu_{2}}}\phi_{k_{2}}^{k_{2}}\otimes{}_{{\mu_{2}}}\phi_{l_{2}}^{l_{2}}\otimes{}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{k_{1}^{\prime}}\otimes{}_{{\mu_{1}^{\prime}}}\phi_{l_{1}^{\prime}}^{l_{1}^{\prime}},\Theta_{13}\Theta_{14}^{-1}\Theta_{24}\Theta_{23}^{-1}\right\rangle$
$\displaystyle\displaystyle\hskip
113.81102pt=q^{\left(w(e_{k_{2}}^{\mu_{2}})-w(e_{l_{2}}^{\mu_{2}}),w(e_{k_{1}^{\prime}}^{\mu_{1}^{\prime}})-w(e_{l_{1}^{\prime}}^{\mu_{1}^{\prime}})\right)}.$

Denoting by $\displaystyle
q^{\prime}_{k_{2}l_{2}k_{1}^{\prime}l_{1}^{\prime}}$ this scalar, it follows

$\displaystyle\displaystyle\left({}_{{\mu_{1}}}\phi_{k_{1}}^{l_{1}}\otimes{}_{{\mu_{2}}}\phi_{k_{2}}^{l_{2}}\right)\circ_{2}\left({}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{l_{1}^{\prime}}\otimes{}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}\right)$
$\displaystyle\displaystyle\hskip
42.67912pt=q_{k_{1}^{\prime}l_{1}^{\prime}k_{1}l_{1}}\
q_{k_{2}^{\prime}l_{2}^{\prime}k_{2}l_{2}}\
q^{\prime}_{k_{2}l_{2}k_{1}^{\prime}l_{1}^{\prime}}\left(\left({}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{l_{1}^{\prime}}\circ{}_{{\mu_{1}}}\phi_{k_{1}}^{l_{1}}\right)\otimes\left({}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}\circ{}_{{\mu_{2}}}\phi_{k_{2}}^{l_{2}}\right)\right)$
$\displaystyle\displaystyle\hskip
113.81102pt+\sum_{p=1}^{k_{2}-1}\sum_{p^{\prime}=k_{1}^{\prime}+1}^{\scriptscriptstyle{d(\mu_{1}^{\prime})}}\alpha_{p^{\prime}k_{1}k_{2}^{\prime}p}^{l_{1}^{\prime}l_{1}l_{2}^{\prime}l_{2}}\left(\left({}_{{\mu_{1}^{\prime}}}\phi_{p^{\prime}}^{l_{1}^{\prime}}\circ{}_{{\mu_{1}}}\phi_{k_{1}}^{l_{1}}\right)\otimes\left({}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}\circ{}_{{\mu_{2}}}\phi_{p}^{l_{2}}\right)\right)+\ldots$

for some scalars
$\displaystyle\alpha_{p^{\prime}k_{1}k_{2}^{\prime}p}^{l_{1}^{\prime}l_{1}l_{2}^{\prime}l_{2}}\in\mathbb{C}(q^{1/D})$,
with the dots as above. Moreover
$\displaystyle\alpha_{p^{\prime}k_{1}k_{2}^{\prime}p}^{l_{1}^{\prime}l_{1}l_{2}^{\prime}l_{2}}=0$
unless $\displaystyle w(e_{p}^{\mu_{2}})>w(e_{k_{2}}^{\mu_{2}})$ and
$\displaystyle
w(e_{p^{\prime}}^{\mu_{1}^{\prime}})<w(e_{k_{1}^{\prime}}^{\mu_{1}^{\prime}})$.
Now, recall (16). In a similar way we find for all $\displaystyle
p\in\\{1,\ldots,k_{2}\\}$, $\displaystyle
p^{\prime}\in\\{k_{1}^{\prime},\ldots,d(\mu_{1}^{\prime})\\}$ that

$\displaystyle\displaystyle\left({}_{{\mu_{1}^{\prime}}}\phi_{p^{\prime}}^{l_{1}^{\prime}}\circ{}_{{\mu_{1}}}\phi_{k_{1}}^{l_{1}}\right)\otimes\left({}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}\circ{}_{{\mu_{2}}}\phi_{p}^{l_{2}}\right)$
$\displaystyle\displaystyle\hskip
42.67912pt=\sum_{(\bar{F})}\sum_{r,t=1}^{\scriptscriptstyle{d(\mu_{1})}}\sum_{r^{\prime},t^{\prime}=1}^{\scriptscriptstyle{d(\mu_{2}^{\prime})}}\left({}_{{\mu_{1}^{\prime}}}\phi_{p^{\prime}}^{l_{1}^{\prime}}\otimes{}_{{\mu_{2}^{\prime}}}\phi_{r^{\prime}}^{t^{\prime}}\left({}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{r^{\prime}}(\bar{F}_{(2)1}){}_{{\mu_{2}^{\prime}}}\phi_{t^{\prime}}^{l_{2}^{\prime}}(S(\bar{F}_{(2)2}))\right)\right)$
$\displaystyle\displaystyle\hskip
199.16928pt\circ_{2}\left({}_{{\mu_{1}}}\phi_{r}^{t}\left({}_{{\mu_{1}}}\phi_{k_{1}}^{r}(\bar{F}_{(1)1}){}_{{\mu_{1}}}\phi_{t}^{l_{1}}(S(\bar{F}_{(1)2}))\right)\otimes{}_{{\mu_{2}}}\phi_{p}^{l_{2}}\right)$
$\displaystyle\displaystyle\hskip
42.67912pt=q^{\prime}_{k_{1}l_{1}k_{2}^{\prime}l_{2}^{\prime}}{}^{-1}\
\left(\left({}_{{\mu_{1}^{\prime}}}\phi_{p^{\prime}}^{l_{1}^{\prime}}\otimes{}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}\right)\circ_{2}\left({}_{{\mu_{1}}}\phi_{k_{1}}^{l_{1}}\otimes{}_{{\mu_{2}}}\phi_{p}^{l_{2}}\right)\right)$
$\displaystyle\displaystyle\hskip
128.0374pt+\sum_{r=1}^{k_{1}-1}\sum_{r^{\prime}=k_{2}^{\prime}+1}^{\scriptscriptstyle{d(\mu_{2}^{\prime})}}\beta_{p^{\prime}r^{\prime}rp}^{l_{1}^{\prime}l_{2}^{\prime}l_{1}l_{2}}\left(\left({}_{{\mu_{1}^{\prime}}}\phi_{p^{\prime}}^{l_{1}^{\prime}}\otimes{}_{{\mu_{2}^{\prime}}}\phi_{r^{\prime}}^{l_{2}^{\prime}}\right)\circ_{2}\left({}_{{\mu_{1}}}\phi_{r}^{l_{1}}\otimes{}_{{\mu_{2}}}\phi_{p}^{l_{2}}\right)\right)$

for some scalars
$\displaystyle\beta_{p^{\prime}r^{\prime}rp}^{l_{1}^{\prime}l_{2}^{\prime}l_{1}l_{2}}\in\mathbb{C}(q^{1/D})$
such that
$\displaystyle\beta_{p^{\prime}r^{\prime}rp}^{l_{1}^{\prime}l_{2}^{\prime}l_{1}l_{2}}=0$
unless $\displaystyle w(e_{r}^{\mu_{1}})>w(e_{k_{1}}^{\mu_{1}})$ and
$\displaystyle
w(e_{r^{\prime}}^{\mu_{2}^{\prime}})<w(e_{k_{2}^{\prime}}^{\mu_{2}^{\prime}})$.
Summing up we obtain

$\displaystyle\displaystyle\left({}_{{\mu_{1}}}\phi_{k_{1}}^{l_{1}}\otimes{}_{{\mu_{2}}}\phi_{k_{2}}^{l_{2}}\right)\circ_{2}\left({}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{l_{1}^{\prime}}\otimes{}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}\right)$
$\displaystyle\displaystyle\hskip
14.22636pt=q_{k_{1}^{\prime}l_{1}^{\prime}k_{1}l_{1}}\
q_{k_{2}^{\prime}l_{2}^{\prime}k_{2}l_{2}}\
q^{\prime}_{k_{2}l_{2}k_{1}^{\prime}l_{1}^{\prime}}\
q^{\prime}_{k_{1}l_{1}k_{2}^{\prime}l_{2}^{\prime}}{}^{-1}\left(\left({}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{l_{1}^{\prime}}\otimes{}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}\right)\circ_{2}\left({}_{{\mu_{1}}}\phi_{k_{1}}^{l_{1}}\otimes{}_{{\mu_{2}}}\phi_{k_{2}}^{l_{2}}\right)\right)$
$\displaystyle\displaystyle\hskip
28.45274pt+\sum_{p=1}^{k_{2}}\sum_{p^{\prime}=k_{1}^{\prime}}^{\scriptscriptstyle{d(\mu_{1}^{\prime})}}\sum_{r=1}^{k_{1}-1}\sum_{r^{\prime}=k_{2}^{\prime}+1}^{\scriptscriptstyle{d(\mu_{2}^{\prime})}}\alpha_{p^{\prime}k_{1}k_{2}^{\prime}p}^{l_{1}^{\prime}l_{1}l_{2}^{\prime}l_{2}}\beta_{p^{\prime}r^{\prime}rp}^{l_{1}^{\prime}l_{2}^{\prime}l_{1}l_{2}}\left(\left({}_{{\mu_{1}^{\prime}}}\phi_{p^{\prime}}^{l_{1}^{\prime}}\otimes{}_{{\mu_{2}^{\prime}}}\phi_{r^{\prime}}^{l_{2}^{\prime}}\right)\circ_{2}\left({}_{{\mu_{1}}}\phi_{r}^{l_{1}}\otimes{}_{{\mu_{2}}}\phi_{p}^{l_{2}}\right)\right)+\ldots$

where at $\displaystyle p=k_{2}$, $\displaystyle p^{\prime}=k_{1}^{\prime}$ we
set
$\displaystyle\alpha_{k_{1}^{\prime}k_{1}k_{2}^{\prime}k_{2}}^{l_{1}^{\prime}l_{1}l_{2}^{\prime}l_{2}}:=q_{k_{1}^{\prime}l_{1}^{\prime}k_{1}l_{1}}\
q_{k_{2}^{\prime}l_{2}^{\prime}k_{2}l_{2}}\
q^{\prime}_{k_{2}l_{2}k_{1}^{\prime}l_{1}^{\prime}}$, and the dots are sums of
tensors of the form $\displaystyle(x_{1}\otimes y_{1})\circ_{2}(x_{2}\otimes
y_{2})$ where $\displaystyle
y_{1}<{}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}$. Recall
that in (38) we denoted by $\displaystyle u_{1},\ldots,u_{M}$ the ordered list
of matrix coefficients $\displaystyle{}_{\varpi_{k}}\phi_{i}^{j}$. Let us
order in a lexicographic way the elements $\displaystyle u_{i}\otimes u_{j}$,
ie. as a sequence $\displaystyle u_{1}^{(2)},\ldots,u_{M^{2}}^{(2)}$ such that
the following condition holds: if
$\displaystyle{}_{\varpi_{l^{\prime}}}\phi_{s^{\prime}}^{t^{\prime}}<{}_{\varpi_{k^{\prime}}}\phi_{i^{\prime}}^{j^{\prime}}$,
or
$\displaystyle{}_{\varpi_{l^{\prime}}}\phi_{s^{\prime}}^{t^{\prime}}={}_{\varpi_{k^{\prime}}}\phi_{i^{\prime}}^{j^{\prime}}$
and $\displaystyle{}_{\varpi_{l}}\phi_{s}^{t}<{}_{\varpi_{k}}\phi_{i}^{j}$,
then $\displaystyle
u_{a}^{(2)}:={}_{\varpi_{k}}\phi_{i}^{j}\otimes{}_{\varpi_{k^{\prime}}}\phi_{i^{\prime}}^{j^{\prime}}$
and $\displaystyle
u_{b}^{(2)}:={}_{\varpi_{l}}\phi_{s}^{t}\otimes{}_{\varpi_{l^{\prime}}}\phi_{s^{\prime}}^{t^{\prime}}$
satisfy $\displaystyle u_{b}^{(2)}<u_{a}^{(2)}$. Then, by the conditions
ensuring when
$\displaystyle\alpha_{p^{\prime}k_{1}k_{2}^{\prime}p}^{l_{1}^{\prime}l_{1}l_{2}^{\prime}l_{2}}$
and
$\displaystyle\beta_{p^{\prime}r^{\prime}rp}^{l_{1}^{\prime}l_{2}^{\prime}l_{1}l_{2}}$
are non zero, the last identity takes the form of (38) by replacing
$\displaystyle u_{i}$, $\displaystyle u_{j}$ with $\displaystyle
u_{i}^{(2)}:={}_{{\mu_{1}}}\phi_{k_{1}}^{l_{1}}\otimes{}_{{\mu_{2}}}\phi_{k_{2}}^{l_{2}}$,
$\displaystyle
u_{j}^{(2)}:={}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{l_{1}^{\prime}}\otimes{}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}$.
At the beginning of this computation we assumed
$\displaystyle{}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}<{}_{{\mu_{2}}}\phi_{k_{2}}^{l_{2}}$
and
$\displaystyle{}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{l_{1}^{\prime}}<{}_{{\mu_{1}}}\phi_{k_{1}}^{l_{1}}$,
but the same result occurs (in a simpler fashion) if
$\displaystyle{}_{{\mu_{2}^{\prime}}}\phi_{k_{2}^{\prime}}^{l_{2}^{\prime}}={}_{{\mu_{2}}}\phi_{k_{2}}^{l_{2}}$
and
$\displaystyle{}_{{\mu_{1}^{\prime}}}\phi_{k_{1}^{\prime}}^{l_{1}^{\prime}}<{}_{{\mu_{1}}}\phi_{k_{1}}^{l_{1}}$,
so eventually we find that (38) holds true for all cases $\displaystyle 1\leq
u_{j}^{(2)}<u_{i}^{(2)}\leq M^{2}$.

As in the case of $\displaystyle{\mathcal{L}}_{0,1}$, by using Proposition
I.8.17 of [20] one can therefore conclude that there is a filtration
$\displaystyle\mathcal{F}_{3}$ of $\displaystyle
Gr_{\mathcal{F}_{2}}({\mathcal{L}}_{0,n})$ such that $\displaystyle
Gr_{\mathcal{F}_{3}}(Gr_{\mathcal{F}_{2}}({\mathcal{L}}_{0,n}))$ is a quotient
of a quasi-polynomial algebra, and finally that
$\displaystyle{\mathcal{L}}_{0,n}$ is Noetherian. $\displaystyle\Box$

######  Theorem 3.2.

The algebra $\displaystyle{\mathcal{M}}_{0,n}={\mathcal{L}}_{0,n}^{U_{q}}$
(respectively $\displaystyle{\mathcal{M}}_{0,n}^{A}$, and
$\displaystyle{\mathcal{M}}_{0,n}^{A,{\epsilon^{\prime}}}$,
$\displaystyle{\epsilon^{\prime}}\in\mathbb{C}^{\times}$) is Noetherian, and
generated over $\displaystyle{\mathbb{C}}(q)$ (resp. $\displaystyle A$,
$\displaystyle\mathbb{C}$) by a finite number of elements.

Our method of proof follows closely that of the Hilbert-Nagata theorem (see
[32]). Let us recall one version of this theorem, which is enough for our
purposes. Let $\displaystyle A=K[a_{1},...,a_{n}]$ be a finitely generated
commutative algebra over an arbitrary field $\displaystyle K$, and
$\displaystyle G$ a group of algebra automorphisms of $\displaystyle A$.

######  Theorem 3.3.

If the action of $\displaystyle G$ on $\displaystyle A$ is completely
reducible on finite dimensional representations, then the ring $\displaystyle
A^{G}$ of invariants of $\displaystyle A$ with respect to $\displaystyle G$ is
Noetherian and a finitely generated algebra over $\displaystyle K$.

We recall here the main steps of the proof that we will adapt in order to
prove Theorem 3.2:

(a) From the complete reducibility of the action of $\displaystyle G$ on
$\displaystyle A$ one can define a linear map

$\displaystyle R:A\rightarrow A^{G}$

namely the projection onto the space of invariants along the space of non-
trivial isotypical components of $\displaystyle A$. This linear map is called
the Reynolds operator; it satisfies

(45) $R(hf)=hR(f)$

for every $\displaystyle f\in A$, $\displaystyle h\in A^{G}$.

(b) Let $\displaystyle I$ be an ideal of $\displaystyle A^{G}$. Then
$\displaystyle I=R(AI)=AI\cap A^{G}$. Because $\displaystyle AI$ is an ideal
of $\displaystyle A$, and $\displaystyle A$ is Noetherian, there exist
elements $\displaystyle b_{1},...,b_{s}$, that can be chosen in $\displaystyle
I\subset A^{G}$, such that $\displaystyle AI=Ab_{1}+\ldots+Ab_{s}$. Since
$\displaystyle I=R(AI)=R(Ab_{1}+\ldots+Ab_{s})=A^{G}b_{1}+\ldots+A^{G}b_{s}$,
$\displaystyle I$ is finitely generated over $\displaystyle A^{G}$. Therefore
$\displaystyle A^{G}$ is Noetherian.

(c) Let $\displaystyle B$ be an algebra graded over $\displaystyle\mathbb{N}$
(for simplicity of notations): $\displaystyle\textstyle
B=\bigoplus_{i=0}^{+\infty}B_{n}$, with $\displaystyle B_{m}.B_{n}\subset
B_{m+n}$. The augmentation ideal of $\displaystyle B$ is
$\displaystyle\textstyle B^{+}=\bigoplus_{i=1}^{+\infty}B_{n}$. If
$\displaystyle B^{+}$ is a Noetherian ideal of $\displaystyle B$, then
$\displaystyle B$ is a finitely generated algebra over $\displaystyle B_{0}$.
This is Lemma 2.4.5 of [65] (in that statement $\displaystyle B$ is
commutative, but this hypothesis is not necessary for the proof).

(d) Assume that $\displaystyle A^{G}$ is graded over $\displaystyle\mathbb{N}$
(for simplicity of notations): $\displaystyle\textstyle
A^{G}=\bigoplus_{i=0}^{+\infty}A^{G}_{n}$ with $\displaystyle A^{G}_{0}=K$.
Then $\displaystyle\textstyle A^{G}{}^{+}=\bigoplus_{i=1}^{+\infty}A^{G}_{n}$
is an ideal of $\displaystyle A^{G}$, which is Noetherian by (b) above.
Applying $\displaystyle(c)$ we deduce that $\displaystyle A^{G}$ is a finitely
generated algebra over $\displaystyle K$.

Proof of Theorem 3.2. As for Theorem 3.1 the result for
$\displaystyle{\mathcal{M}}_{0,n}^{A,{\epsilon^{\prime}}}$ follows from that
for $\displaystyle{\mathcal{M}}_{0,n}^{A}$ and
$\displaystyle{\mathcal{M}}_{0,n}$, which are proved in the same way. Let us
consider $\displaystyle{\mathcal{M}}_{0,n}$. Consider the filtration
$\displaystyle{\mathcal{F}}$ of $\displaystyle{\mathcal{L}}_{0,n}$ by the
subspaces

$\displaystyle{\mathcal{F}}^{[\mu]}=\bigoplus_{[\mu^{\prime}]\leq[\mu]}C([\mu^{\prime}]),\mu\in
P_{+}^{n}$

where $\displaystyle P_{+}^{n}$ is given the lexicographic partial order
induced from $\displaystyle[\Lambda]$. It is easily seen that
$\displaystyle\mathcal{F}$ is an algebra filtration: the coregular actions
$\displaystyle\rhd$, $\displaystyle\lhd$ fix globally each component
$\displaystyle C(\mu)$ of $\displaystyle{\mathcal{L}}_{0,1}$, so the claim
follows from (11), (14) and the fact that $\displaystyle C(\mu)\star
C(\nu)\subset C(\mu+\nu)$ for all $\displaystyle\mu,\nu\in P_{+}$. Denote by
$\displaystyle Gr_{\mathcal{F}}({\mathcal{L}}_{0,n})$ the corresponding graded
algebra. Again

(46)
$Gr_{{\mathcal{F}}}({\mathcal{L}}_{0,n})={\mathcal{L}}_{0,n}=\bigoplus_{[\mu]\in
P_{+}^{n}}C([\mu]).$

Because each space $\displaystyle C([\mu])$ is stabilized by the coadjoint
action of $\displaystyle U_{q}$, the decomposition (46) has a key advantage on
(41). Indeed, since $\displaystyle{\mathcal{L}}_{0,n}$ is a $\displaystyle
U_{q}$-module algebra, the action of $\displaystyle U_{q}$ is well-defined on
$\displaystyle Gr_{\mathcal{F}}({\mathcal{L}}_{0,n})$ and it gives it a
structure of $\displaystyle U_{q}$-module algebra. As vector spaces we have

(47) $Gr_{{\mathcal{F}}}({\mathcal{L}}_{0,n})^{U_{q}}=\bigoplus_{[\mu]\in
P_{+}^{n}}C([\mu])^{U_{q}}={\mathcal{L}}_{0,n}^{U_{q}}.$

Now we can adapt the differents steps (a)–(d) recalled above:

(a’) The action of $\displaystyle U_{q}$ on $\displaystyle{\mathcal{L}}_{0,n}$
is completely reducible. This follows from Theorem 2.2 (1) (noting that the
summands, being isomorphic by (22) to spaces $\displaystyle C(\mu)$, are
finite-dimensional and thus completely reducible $\displaystyle
U_{q}$-modules), and the isomorphism of $\displaystyle U_{q}$-modules (see
(19)):

(48)
${\mathcal{L}}_{0,n}\stackrel{{\scriptstyle\Phi_{n}}}{{\longrightarrow}}(U_{q}({\mathfrak{g}})^{\otimes
n})^{lf}\stackrel{{\scriptstyle\psi_{n}^{-1}}}{{\longrightarrow}}U_{q}^{lf}({\mathfrak{g}})^{\otimes
n}$

where $\displaystyle lf$ means respectively locally finite for the action
$\displaystyle ad_{n}^{r}$ of $\displaystyle U_{q}({\mathfrak{g}})$ on
$\displaystyle U_{q}({\mathfrak{g}})^{\otimes n}$, and locally finite for the
action $\displaystyle ad^{r}$ of $\displaystyle U_{q}({\mathfrak{g}})$ on
$\displaystyle U_{q}({\mathfrak{g}})$. By (46) it follows that $\displaystyle
Gr_{\mathcal{F}}({\mathcal{L}}_{0,n})$ is also completely reducible. We can
therefore define the Reynolds operator $\displaystyle R\colon
Gr_{{\mathcal{F}}}({\mathcal{L}}_{0,n})\rightarrow
Gr_{{\mathcal{F}}}({\mathcal{L}}_{0,n})^{U_{q}}$ as in (a).

(b’) In the proof of Theorem 3.1 we showed that $\displaystyle
Gr_{{\mathcal{F}}_{2}}({\mathcal{L}}_{0,n})$ is Noetherian, and then deduced
that $\displaystyle{\mathcal{L}}_{0,n}$ is Noetherian by a classical argument
(see eg. [60], 1.6.9). This same argument implies that $\displaystyle
Gr_{{\mathcal{F}}}({\mathcal{L}}_{0,n})$ is Noetherian, because (46) shows it
is filtered by $\displaystyle{\mathcal{F}}_{2}$, and $\displaystyle
Gr_{{\mathcal{F}}_{2}}(Gr_{{\mathcal{F}}}({\mathcal{L}}_{0,n}))=Gr_{{\mathcal{F}}_{2}}({\mathcal{L}}_{0,n})$
is Noetherian. As in (b) we deduce that $\displaystyle
Gr_{{\mathcal{F}}}({\mathcal{L}}_{0,n})^{U_{q}}$ is Noetherian. But
$\displaystyle
Gr_{{\mathcal{F}}}({\mathcal{L}}_{0,n})^{U_{q}}=Gr_{{\mathcal{F}}}({\mathcal{L}}_{0,n}^{U_{q}})$,
which implies that $\displaystyle{\mathcal{L}}_{0,n}^{U_{q}}$ is Noetherian.

(c’-d’) Then we can apply the steps (c)-(d). As a result $\displaystyle
Gr_{{\mathcal{F}}}({\mathcal{L}}_{0,n})^{U_{q}}$ is finitely generated, say by
$\displaystyle k$ homogeneous elements
$\displaystyle\bar{x}_{i}\in{\mathcal{F}}^{[\mu_{i}]}/(\oplus_{[\mu^{\prime}]<[\mu_{i}]}C([\mu^{\prime}]))$.

(e’) From (47) we deduce that $\displaystyle{\mathcal{L}}_{0,n}^{U_{q}}$ is
generated by the $\displaystyle x_{i}\in C([\mu_{i}])$ with leading terms
$\displaystyle\bar{x}_{1},\ldots,\bar{x}_{k}$. This follows from the following
elementary fact: if $\displaystyle A$ is a filtered
$\displaystyle\mathbb{K}$-algebra ($\displaystyle\mathbb{K}$ a field) which
graded algebra $\displaystyle Gr(A)$ is finitely generated, then
$\displaystyle A$ is finitely generated by elements which leading terms
generate $\displaystyle Gr(A)$. Indeed, let $\displaystyle A$ have the algebra
filtration $\displaystyle(A_{i})_{i\in\mathbb{N}}$ (we take a filtration over
$\displaystyle\mathbb{N}$ to simplify notations). Put $\displaystyle
Gr(A)=\oplus_{i\in\mathbb{N}}A_{(i)}$, $\displaystyle
A_{(i+1)}:=A_{i+1}/A_{i}$. We have
$\displaystyle\bar{a}+\bar{b}=\overline{a+b}$ and
$\displaystyle\bar{a}\bar{b}=ab+A_{n+m-1}\in A_{(m+n)}$, so
$\displaystyle\bar{a}\bar{b}=0$ if $\displaystyle ab\in A_{n+m-1}$, and
$\displaystyle\bar{a}\bar{b}=\overline{ab}$ otherwise. Now assume that
$\displaystyle Gr(A)$ is finitely generated over $\displaystyle\mathbb{K}$.
Denote by $\displaystyle\bar{f}_{1},\ldots,\bar{f}_{n}$ a finite set of
generators. Let $\displaystyle a\in A$, with $\displaystyle a\in
A_{k}\setminus A_{k-1}$. Then $\displaystyle\bar{a}\in A_{(k)}$ can be written
as $\displaystyle\textstyle\bar{a}=\sum_{(i_{1},...,i_{n})\in
I}\lambda_{(i_{1},...,i_{n})}\bar{f}_{1}^{i_{1}}\ldots\bar{f}_{n}^{i_{n}}$ for
some finite set $\displaystyle I\subset\mathbb{N}^{n}$ and scalars
$\displaystyle\lambda_{(i_{1},...,i_{n})}\in\mathbb{K}$. Either
$\displaystyle\bar{f}_{1}^{i_{1}}\ldots\bar{f}_{n}^{i_{n}}$ is zero, or
$\displaystyle\bar{f}_{1}^{i_{1}}\ldots\bar{f}_{n}^{i_{n}}=\overline{f_{1}^{i_{1}}\ldots
f_{n}^{i_{n}}}$. We therefore have
$\displaystyle\textstyle\overline{a-\sum_{(i_{1},...,i_{n})\in
I}\lambda_{(i_{1},...,i_{n})}f_{1}^{i_{1}}\ldots f_{n}^{i_{n}}}=0\in A_{(k)}$,
that is, $\displaystyle\textstyle a-\sum_{(i_{1},...,i_{n})\in
I}\lambda_{(i_{1},...,i_{n})}f_{1}^{i_{1}}\ldots f_{n}^{i_{n}}\in A_{k-1}$.
The conclusion follows by an easy induction. $\displaystyle\Box$

######  Remark 3.4.

(1) Because $\displaystyle{\mathcal{L}}_{0,1}^{U_{q}}$ is the center of
$\displaystyle{\mathcal{L}}_{0,1}$, (e’) proves it is finitely generated. Of
course this follows also from the isomorphism
$\displaystyle{\mathcal{L}}_{0,1}\cong U_{q}^{lf}$ and the fact that the
center of $\displaystyle U_{q}^{lf}$ is the center of $\displaystyle U_{q}$
(by Theorem 2.2), plus the well-known description of the latter. But the
argument here is elementary and it applies to
$\displaystyle{\mathcal{L}}_{0,n}^{U_{q}}$ for any $\displaystyle n\geq 1$.

(2) In spite of the isomorphism
$\displaystyle\Phi_{n}\colon{\mathcal{M}}_{0,n}\rightarrow(U_{q}^{\otimes
n})^{U_{q}}$, in order to prove Theorem 3.2 one cannot bypass the hard study
of $\displaystyle(U_{q}^{\otimes n})^{lf}$, whence of
$\displaystyle{\mathcal{L}}_{0,1}\cong U_{q}^{lf}$, by working directly with
$\displaystyle U_{q}$. Indeed the adjoint action is not completely reducible
thereon. In fact, $\displaystyle U_{q}^{lf}$ is exactly the socle of this
action (see [43], Lemma 7.1.24).

(3) In the $\displaystyle sl(2)$ case the filtration
$\displaystyle\mathcal{F}$ on $\displaystyle{\mathcal{L}}_{0,n}^{U_{q}}$
should correspond via the Wilson loop isomorphism (defined in [23], Section
8.2) to the filtration of skein algebras of spheres with $\displaystyle n+1$
punctures used in [57].

##  4\. Proof of Theorem 1.2

As usual we let $\displaystyle\epsilon$ be a primitive $\displaystyle l$-th
root of unity with $\displaystyle l$ odd and $\displaystyle l>d_{i}$ for all
$\displaystyle i\in\\{1,\ldots,m\\}$.

Recall that $\displaystyle\mathcal{Z}_{0}(U_{\epsilon})\subset U_{\epsilon}$
is the central polynomial subalgebra generated by $\displaystyle
E_{\beta_{k}}^{l}$, $\displaystyle F_{\beta_{k}}^{l}$, $\displaystyle
L_{i}^{\pm l}$, for $\displaystyle k\in\\{1,\ldots,N\\}$ and $\displaystyle
i\in\\{1,\ldots m\\}$. Define

$\displaystyle\mathcal{Z}_{0}(U_{\epsilon}^{lf})=U_{\epsilon}^{lf}\cap\mathcal{Z}_{0}(U_{\epsilon}).$

Examples show that generating sets of
$\displaystyle\mathcal{Z}_{0}(U_{\epsilon}^{lf})$ have complicated expressions
in general. Nevertheless, specializing $\displaystyle q$ at
$\displaystyle\epsilon$ in Theorem 2.2 (2) we get

(49)
$\mathcal{Z}_{0}(U_{\epsilon})=T_{2-}^{(l)-1}\mathcal{Z}_{0}(U_{\epsilon}^{lf})[T^{(l)}/T_{2}^{(l)}]$

where $\displaystyle T^{(l)}$, $\displaystyle T_{2-}^{(l)}$ and $\displaystyle
T_{2}^{(l)}$ are the subsets of $\displaystyle T$, $\displaystyle T_{2-}$ and
$\displaystyle T_{2}$ formed by the elements $\displaystyle K_{\lambda l}$
with $\displaystyle\lambda\in P$, $\displaystyle\lambda\in-2P_{+}$ and
$\displaystyle\lambda\in 2P$ respectively. Define

$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})=\Phi_{1}^{-1}(\mathcal{Z}_{0}(U_{\epsilon}^{lf})).$

Recall the isomorphism
$\displaystyle\eta^{*}\colon{\mathcal{O}}(G)\rightarrow\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})$
(see Theorem 2.15 (1)).

######  Proposition 4.1.

$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})=\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})$,
and therefore $\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$
is isomorphic to $\displaystyle{\mathcal{O}}(G)$.

Proof. The claim follows from the fact that
$\displaystyle\Phi_{1}\colon{\mathcal{L}}_{0,1}^{\epsilon}\rightarrow
U^{lf}_{\epsilon}$ is an isomorphism (see (26)), the identity
$\displaystyle\Phi_{1}=m\circ(id\otimes S^{-1})\circ\Phi$, and Theorem 2.15
(2).

Here is an alternative proof of the isomorphism
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})\cong{\mathcal{O}}(G)$,
not using $\displaystyle\eta^{*}$. Recall the notations introduced before
Theorem 2.14. As varieties $\displaystyle H=U_{+}T_{G}U_{-}=G^{0}$, so the map
$\displaystyle\sigma$ yields identifications
$\displaystyle{\mathcal{O}}(H)={\mathcal{O}}(U_{+}){\mathcal{O}}(T_{G}){\mathcal{O}}(U_{-})$
and
$\displaystyle{\mathcal{O}}(G^{0})={\mathcal{O}}(U_{+}){\mathcal{O}}(T_{G}/(2)){\mathcal{O}}(U_{-})$;
we can identify $\displaystyle{\mathcal{O}}(G^{0})$ with the subalgebra
$\displaystyle\sigma_{|H}^{*}({\mathcal{O}}(G^{0}))$ of
$\displaystyle{\mathcal{O}}(H)$. Consider the space $\displaystyle
V=\wedge^{N}\mathfrak{g}$, endowed with the action of $\displaystyle G$ given
on each factor by the adjoint representation. Put on
$\displaystyle\mathfrak{g}$ a basis consisting of one element $\displaystyle
e_{\alpha}$ per root space $\displaystyle\mathfrak{g}_{\alpha}$, along with a
basis of $\displaystyle\mathfrak{h}$. Let $\displaystyle v\in V$ be the
exterior power of the $\displaystyle e_{\alpha}$’s for $\displaystyle\alpha$
negative, and $\displaystyle v^{*}$ a dual vector such that $\displaystyle
v^{*}(v)=1$ and $\displaystyle v^{*}$ vanishes on a $\displaystyle
T_{G}$-invariant complement of $\displaystyle v$. It is classical that
$\displaystyle G\setminus G^{0}$ has defining equation
$\displaystyle\delta(g)=0$, where $\displaystyle\delta$ is the matrix
coefficient $\displaystyle\delta(g)=v^{*}(\pi_{V}(g)v)$ (see eg. [42], page
174). Hence $\displaystyle{\mathcal{O}}(G^{0})={\mathcal{O}}(G)[\delta^{-1}]$.
On $\displaystyle G^{0}$ we have
$\displaystyle\delta(u_{+}tu_{-})=\chi_{-\rho}(t)$, where
$\displaystyle\chi_{-\rho}$ is the character of $\displaystyle T_{G}$
associated to the root $\displaystyle-\rho$. Now we can make the connection
with $\displaystyle U_{\epsilon}$. The isomorphism
$\displaystyle\mathcal{Z}_{0}(U_{\epsilon})\cong{\mathcal{O}}(H)$ of Theorem
2.14 (2) identifies $\displaystyle\mathcal{Z}_{0}(U_{\epsilon})\cap
U_{\epsilon}(\mathfrak{h})=\mathbb{C}[T^{(l)}]$ with
$\displaystyle{\mathcal{O}}(T_{G})$ by mapping $\displaystyle K_{\lambda l}$
to the character of $\displaystyle T_{G}$ associated to
$\displaystyle\lambda$. Therefore it maps
$\displaystyle\mathbb{C}[T_{2}^{(l)}]$ to
$\displaystyle{\mathcal{O}}(T_{G}/(2))$, $\displaystyle\ell^{l}=K_{l\rho}$ to
$\displaystyle\chi_{\rho}$, and $\displaystyle
T_{2-}^{(l)-1}\mathcal{Z}_{0}(U_{\epsilon}^{lf})=\mathcal{Z}_{0}(U_{\epsilon}^{lf})[\ell^{l}]$
to $\displaystyle{\mathcal{O}}(G^{0})$ by (49). Since
$\displaystyle{\mathcal{O}}(G^{0})={\mathcal{O}}(G)[\delta^{-1}]$ and
$\displaystyle\delta^{-1}=\chi_{\rho}$ on $\displaystyle G^{0}$, it follows
that $\displaystyle\mathcal{Z}_{0}(U_{\epsilon}^{lf})\cong{\mathcal{O}}(G)$.
Then
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})\cong{\mathcal{O}}(G)$
by injectivity of $\displaystyle\Phi_{1}$. $\displaystyle\Box$

Consider the linear subspace of $\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$
defined by

$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})=\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})^{\otimes
n}.$

By Proposition 4.1 we have an isomorphism of algebras
$\displaystyle(\eta^{*}{}^{-1})^{\otimes
n}\colon\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})\rightarrow{\mathcal{O}}(G)^{\otimes
n}$.

######  Proposition 4.2.

(1) $\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$ is a
central subalgebra of $\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$, and
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}={\mathcal{O}}_{\epsilon}^{\otimes
n}$ as modules over $\displaystyle{\mathcal{O}}(G)^{\otimes n}$. Moreover
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$ is a Noetherian
ring.

(2) The $\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$-module
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ is generated by the elements of
the spaces $\displaystyle C([\mu])$ where
$\displaystyle[\mu]=(\mu_{1},\ldots,\mu_{n})\in P_{+}^{n}$ satisfies
$\displaystyle 0\leq(\mu_{i},\alpha\check{}_{j})<l$ for every $\displaystyle
i\in\\{1,\ldots,n\\}$, $\displaystyle j\in\\{1,\ldots,m\\}$, where as usual
$\displaystyle\alpha\check{}_{j}:=2\alpha_{j}/(\alpha_{j},\alpha_{j})$. In
particular it is a finite Noetherian
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$-module, and
therefore a Noetherian ring.

###### Proof.

(1) Recall the formula (14). It is enough to consider the case where
$\displaystyle\alpha=\beta^{\prime}=1$, and show that
$\displaystyle(1\otimes(\beta)^{(b)})((\alpha)^{(a)}\otimes
1)=(\alpha)^{(a)}\otimes(\beta)^{(b)}$ whenever
$\displaystyle\alpha\in\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$, for
this is trivially equal to $\displaystyle((\alpha)^{(a)}\otimes
1)(1\otimes(\beta)^{(b)})$ (similar arguments apply if instead
$\displaystyle\beta\in\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$ and
$\displaystyle\alpha$ is arbitrary). We have (denoting
$\displaystyle\textstyle\sum_{(\alpha),(\alpha),(\alpha),(\alpha)}$ by
$\displaystyle\textstyle\sum_{(\alpha)^{4}}$,
$\displaystyle\textstyle\Delta(\alpha_{(1)})=\sum_{(\alpha)}\alpha_{(1)(1)}\otimes\alpha_{(1)(2)}$
etc.):

$\displaystyle\displaystyle\left(1\otimes(\beta)^{(b)}\right)\\!\\!\left((\alpha)^{(a)}\otimes
1\right)$
$\displaystyle\displaystyle=\sum_{\scriptscriptstyle(R^{i})}\left(S(R^{3}_{(1)}R^{4}_{(1)})\rhd\alpha\lhd
R^{1}_{(1)}R^{2}_{(1)}\right)^{(a)}$ $\displaystyle\displaystyle\hskip
113.81102pt\otimes\left(S(R^{1}_{(2)}R^{3}_{(2)})\rhd\beta\lhd
R^{2}_{(2)}R^{4}_{(2)}\right)^{(b)}$
$\displaystyle\displaystyle=\sum_{\scriptscriptstyle(R^{i}),(\alpha)^{4},(\beta)^{2}}(\alpha_{(2)})^{(a)}\otimes(\beta_{(2)})^{(b)}$
$\displaystyle\displaystyle\hskip
56.9055pt\times\beta_{(1)}\left(\alpha_{(1)(2)}(R^{2}_{(1)})R^{2}_{(2)}\alpha_{(3)(1)}(S(R^{4}_{(1)}))R^{4}_{(2)}\right)$
$\displaystyle\displaystyle\hskip
85.35826pt\times\beta_{(3)}\left(\alpha_{(3)(2)}(R^{3}_{(1)})R^{3}_{(2)}\alpha_{(1)(1)}(R^{1}_{(1)})S(R^{1}_{(2)})\right).$

By Theorem 2.15 (2) it follows that
$\displaystyle\alpha_{(1)(2)}(R^{2}_{(1)})R^{2}_{(2)}=\Phi^{+}(\alpha_{(1)(2)})\in\mathcal{Z}_{0}(U_{\epsilon})$,
and similarly
$\displaystyle\alpha_{(3)(1)}(S(R^{4}_{(1)}))R^{4}_{(2)},\alpha_{(3)(2)}(R^{3}_{(1)})R^{3}_{(2)},\alpha_{(1)(1)}(R^{1}_{(1)})S(R^{1}_{(2)})\in\mathcal{Z}_{0}(U_{\epsilon})$.
Denote by $\displaystyle z$ any such element. Note that
$\displaystyle{\mathcal{Z}}_{0}(U_{\epsilon})$ acts by the trivial character
on $\displaystyle\Gamma$-modules, and that the expression of $\displaystyle z$
in terms of the corresponding $\displaystyle\alpha_{(i)(j)}$ implies
$\displaystyle\epsilon(z)=\epsilon(\alpha_{(i)(j)})$. Then

$\displaystyle\displaystyle\beta_{(1)}\left(\alpha_{(1)(2)}(R^{2}_{(1)})R^{2}_{(2)}\alpha_{(3)(1)}(S(R^{4}_{(1)}))R^{4}_{(2)}\right)$
$\displaystyle\displaystyle=\epsilon(\alpha_{(1)(2)}\alpha_{(3)(1)})\beta_{(1)}(1)$
$\displaystyle\displaystyle=\epsilon(\alpha_{(1)(2)})\epsilon(\alpha_{(3)(1)})\epsilon(\beta_{(1)})$
$\displaystyle\displaystyle\beta_{(3)}\left(\alpha_{(3)(2)}(R^{3}_{(1)})R^{3}_{(2)}\alpha_{(1)(1)}(R^{1}_{(1)})S(R^{1}_{(2)})\right)$
$\displaystyle\displaystyle=\epsilon(\alpha_{(3)(2)})\epsilon(\alpha_{(1)(1)})\epsilon(\beta_{(3)})$

and finally $\displaystyle(1\otimes(\beta)^{(b)})((\alpha)^{(a)}\otimes
1)=(\alpha)^{(a)}\otimes(\beta)^{(b)}$. Therefore
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})^{(a)}$ is
central in $\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ for all
$\displaystyle a=1,\ldots,n$. These algebras generate
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$ in
$\displaystyle({\mathcal{L}}_{0,1}^{\epsilon})^{\otimes n}$, and hence in
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ (this follows from the comment
before (15)). Therefore
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$ is central in
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$. A computation similar to the
above one, based on the formula (11) instead of (14), shows that
$\displaystyle\alpha\beta=\alpha\star\beta$ whenever
$\displaystyle\alpha\in\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$. Hence
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}$ and
$\displaystyle{\mathcal{O}}_{\epsilon}$ coincide as modules over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})=\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})$.
The second claim follows immediately; for instance when $\displaystyle n=2$,
given
$\displaystyle\alpha^{\prime},\beta^{\prime}\in\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$
we have
$\displaystyle(\alpha^{\prime}\otimes\beta^{\prime})(\alpha\otimes\beta)=(\alpha^{\prime}\otimes
1)(1\otimes\beta^{\prime})(\alpha\otimes 1)(1\otimes\beta)$ immediately by
(14), and $\displaystyle(1\otimes\beta^{\prime})(\alpha\otimes
1)=\alpha\otimes\beta^{\prime}=(\alpha\otimes 1)(1\otimes\beta^{\prime})$ as
above. Then
$\displaystyle(\alpha^{\prime}\otimes\beta^{\prime})(\alpha\otimes\beta)=\alpha^{\prime}\alpha\otimes\beta^{\prime}\beta$.
Finally $\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$ is a
Noetherian ring as it is isomorphic to $\displaystyle{\mathcal{O}}(G^{n})$
(Proposition 4.1), the coordinate ring of an affine algebraic variety.

(2) Let $\displaystyle[\lambda]=(\lambda_{1},\ldots,\lambda_{n})\in
P_{+}^{n}$. For all $\displaystyle i\in\\{1,\ldots,n\\}$ there are unique
$\displaystyle\lambda_{i0},\lambda_{i1}\in P_{+}$ such that
$\displaystyle\lambda_{i}=\lambda_{i0}+l\lambda_{i1}$ and $\displaystyle
0\leq(\lambda_{i0},\alpha\check{}_{j})<l$ for every $\displaystyle
j\in\\{1,\ldots,m\\}$. Then set
$\displaystyle[\lambda_{0}]=(\lambda_{10},\ldots,\lambda_{n0})$,
$\displaystyle[\lambda_{1}]=(\lambda_{11},\ldots,\lambda_{n1})$. We have
$\displaystyle C([\lambda])=C([\lambda_{0}])\circ C([l\lambda_{1}])$ by (44),
and $\displaystyle
C(l\lambda_{i1})\subset\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})=\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$
for all $\displaystyle i\in\\{1,\ldots,n\\}$ by Theorem 2.15 (2) and
Proposition 4.1, so $\displaystyle
C([l\lambda_{1}])\subset\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$. It
follows that $\displaystyle\textstyle
Gr_{{\mathcal{F}}_{2}}({\mathcal{L}}_{0,n}^{\epsilon})=\bigoplus_{[\lambda]\in
P_{+}^{n}}C([\lambda])$ is generated over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$ by the elements
of the spaces $\displaystyle C([\lambda_{0}])$, $\displaystyle[\lambda]\in
P_{+}^{n}$. By step (e’) of the proof of Theorem 3.2, formulated for modules
instead of algebras, it follows that
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ is generated over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$ by the same
elements. The last claims follow in a standard way: since
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ is a finitely generated
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$-module and
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$ is Noetherian,
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ it is a Noetherian
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$-module (eg. by
[5] Proposition 6.5). Moreover $\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$
is a Noetherian ring (by eg. [60], 1.3). ∎

Note that we had already obtained independently the Noetherianity of the ring
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ as a consequence of the
Noetherianity of $\displaystyle{\mathcal{L}}_{0,n}^{A}$ (Proposition 3.1).

We need below explicit descriptions of the $\displaystyle\mathcal{Z}$\- and
$\displaystyle\mathcal{Z}_{0}$-centers for $\displaystyle\mathfrak{g}=sl(2)$.
Let us recall a few facts in this case. Denote by $\displaystyle a$,
$\displaystyle b$, $\displaystyle c$, $\displaystyle d$ the standard
generators of $\displaystyle\mathcal{O}_{q}(SL_{2})$, ie. the matrix
coefficients in the basis of weight vectors $\displaystyle v_{0}$,
$\displaystyle v_{1}=F.v_{0}$ of the $\displaystyle 2$-dimensional irreducible
representation $\displaystyle V_{2}$ of $\displaystyle U_{q}(sl(2))$. Denote
by $\displaystyle x^{\star k}$, $\displaystyle k\in\mathbb{N}$, the
$\displaystyle k$-th power of an element $\displaystyle
x\in{\mathcal{O}}_{A}(SL_{2})$. The algebra
$\displaystyle\mathcal{O}_{A}(SL_{2})$ is generated by $\displaystyle a$,
$\displaystyle b$, $\displaystyle c$, $\displaystyle d$; the monomials
$\displaystyle a^{\star i}\star b^{\star j}\star d^{\star r}$ and
$\displaystyle a^{\star i}\star c^{\star k}\star d^{\star r}$, $\displaystyle
i,j,k,r\in\mathbb{N},k>0$, form an $\displaystyle A$-basis of
$\displaystyle\mathcal{O}_{A}(SL_{2})$. The algebra
$\displaystyle\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon}(SL_{2}))$ is generated
by $\displaystyle a^{\star l}$, $\displaystyle b^{\star l}$, $\displaystyle
c^{\star l}$, $\displaystyle d^{\star l}$; the monomials $\displaystyle
a^{\star il}\star b^{\star jl}\star d^{\star rl}$ and $\displaystyle a^{\star
il}\star c^{\star kl}\star d^{\star rl}$ form a basis of
$\displaystyle\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon}(SL_{2}))$, and
$\displaystyle\mathcal{Z}({\mathcal{O}}_{\epsilon}(SL_{2}))$ is generated by
$\displaystyle\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon}(SL_{2}))$ and the
elements $\displaystyle b^{\star(l-k)}\star c^{\star k}$, $\displaystyle
k=0,\ldots,l$ (see [36], Proposition 1.4 and the Appendix). We have the
relation

(50) $a^{\star l}\star d^{\star l}-b^{\star l}\star c^{\star l}=1$

and the Frobenius isomorphism of Parshall-Wang (see [56], Chapter 7) coincides
with the map

$\displaystyle
Fr_{PW}\colon{\mathcal{O}}(SL_{2})\rightarrow\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon}(SL_{2}))$

induced by $\displaystyle\eta^{*}$; it sends the standard generators
$\displaystyle\underline{a},\underline{b},\underline{c},\underline{d}$ of
$\displaystyle{\mathcal{O}}(SL_{2})=\mathcal{O}_{1}(SL_{2})$ respectively to
$\displaystyle a^{\star l}$, $\displaystyle b^{\star l}$, $\displaystyle
c^{\star l}$, $\displaystyle d^{\star l}$. Finally, let us quote from [37]
that a basis of the rank $\displaystyle l^{3}$ free
$\displaystyle\mathcal{Z}_{0}({\mathcal{O}}(SL_{2}))$-module
$\displaystyle{\mathcal{O}}_{\epsilon}(SL_{2})$ (see Theorem 2.15 (3)) is
formed by the monomials $\displaystyle a^{m}b^{n}c^{s^{\prime}}$ and
$\displaystyle b^{n}c^{s^{\prime\prime}}d^{r}$, with the integers
$\displaystyle m,n,r,s^{\prime},s^{\prime\prime}$ in the range

(51) $1\leq m\leq l-1\ ,0\leq n,r\leq l-1\ ,m\leq s^{\prime}\leq l-1\ ,0\leq
s^{\prime\prime}\leq l-r-1.$

Now consider $\displaystyle{\mathcal{L}}_{0,1}^{A}(sl(2))$. Recall that
$\displaystyle{\mathcal{L}}_{0,1}^{A}={\mathcal{O}}_{A}$ as $\displaystyle
U_{A}$-modules. The algebra $\displaystyle{\mathcal{L}}_{0,1}^{A}(sl(2))$ is
also generated by $\displaystyle a$, $\displaystyle b$, $\displaystyle c$,
$\displaystyle d$; a set of defining relations is (see [23], Section 5):

(52) $\begin{array}[]{ll}ad=da&,\ ab-ba=-(1-q^{-2})bd\\\ db=q^{2}bd&,\ cb-
bc=(1-q^{-2})(da-d^{2})\\\ cd=q^{2}dc&,\ ac-ca=(1-q^{-2})dc\\\ &\ \
ad-q^{2}bc=1.\end{array}$

The element $\displaystyle\omega:=qa+q^{-1}d$ is central. Let $\displaystyle
T_{k}$, $\displaystyle k\in\mathbb{N}$, be such that $\displaystyle
T_{k}(x)/2$ is the $\displaystyle k$-th Chebyshev polynomial of the first type
in the variable $\displaystyle x/2$. We have (see [23], Proposition 7.3, for
the generalization to $\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}(sl(2))$):

######  Lemma 4.3.

$\displaystyle\mathcal{Z}({\mathcal{L}}_{0,1}^{\epsilon}(sl(2)))=\mathbb{C}[\omega,b^{l},c^{l},d^{l}]/\mathcal{I}$
and
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}(sl(2)))=\mathbb{C}[(T_{l}(\omega),b^{l},c^{l},d^{l}]/\mathcal{I}$,
where $\displaystyle\mathcal{I}$ is the ideal of
$\displaystyle\mathbb{C}[\omega,b^{l},c^{l},d^{l}]$ generated by
$\displaystyle(T_{l}(\omega)-d^{l})d^{l}-b^{l}c^{l}-1$.

Here $\displaystyle b^{l},c^{l},d^{l}$ are the $\displaystyle l$-th powers of
$\displaystyle b,c,d$ computed using the product of
$\displaystyle{\mathcal{L}}_{0,1}^{A}(sl(2))$, not the product
$\displaystyle\star$ of
$\displaystyle\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon}(SL_{2}))$. The above
generator of $\displaystyle\mathcal{I}$ can be interpreted as a determinant,
and $\displaystyle\omega$ as a quantum trace on $\displaystyle V_{2}$.

######  Lemma 4.4.

Viewed as element of $\displaystyle{\mathcal{O}}_{A}(SL_{2})$, $\displaystyle
T_{l}(\omega)-d^{l}=a^{\star l}$ and $\displaystyle x^{l}=x^{\star l}$,
$\displaystyle x\in\\{b,c,d\\}$.

Proof. Let $\displaystyle\alpha$ and $\displaystyle\varpi$ be the simple root
and fundamental weight of $\displaystyle sl(2)$. In the notations of (30) we
have $\displaystyle b=\psi^{-\alpha}_{-\varpi}$, $\displaystyle
c=\psi^{\alpha}_{-\varpi}$, $\displaystyle d=\psi_{-\varpi}$; the formulas
give $\displaystyle\Phi_{1}(b^{\star l})=(q-q^{-1})^{l}F^{l}$,
$\displaystyle\Phi_{1}(c^{\star l})=(q-q^{-1})^{l}E^{l}K^{-l}$,
$\displaystyle\Phi_{1}(d^{\star l})=K^{-l}$. These coincide respectively with
$\displaystyle\Phi_{1}(b^{l})$, $\displaystyle\Phi_{1}(c^{l})$,
$\displaystyle\Phi_{1}(d^{l})$ (see (32) in [23]). By passing to the
localization $\displaystyle{\mathcal{O}}_{A}(SL_{2})[d^{-1}]$, and using
Parshall-Wang’s relation (50), one deduces easily
$\displaystyle\Phi_{1}(a^{\star
l})=K^{l}+(q-q^{-1})^{2l}F^{l}E^{l}=T_{l}(\Omega)-K^{-l}$, where
$\displaystyle\Omega$ is $\displaystyle(q-q^{-1})^{2l}$ times the Casimir
element of $\displaystyle U_{q}(sl(2))$, and $\displaystyle T_{l}(x)/2$ is the
$\displaystyle l$-th Chebyshev polynomial of the first type in the variable
$\displaystyle x/2$. We have $\displaystyle\Phi_{1}(\omega)=\Omega$, so
$\displaystyle\Phi_{1}(a^{\star l})=T_{l}(\omega)-d^{l}$. The conclusion
follows from the injectivity of $\displaystyle\Phi_{1}$. $\displaystyle\Box$

This lemma proves that we have a commutative diagram

$\displaystyle\textstyle{{\mathcal{O}}(SL_{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\displaystyle\scriptstyle{Fr_{PW}\
\ \
}$$\displaystyle\scriptstyle{Fr}$$\displaystyle\textstyle{\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon}(SL_{2}))\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\displaystyle\textstyle{{\mathcal{O}}_{\epsilon}(SL_{2})\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\displaystyle\textstyle{\quad\quad\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}(sl(2)))\ignorespaces\ignorespaces\ignorespaces\ignorespaces}$$\displaystyle\textstyle{{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))}$

where $\displaystyle Fr_{PW}$ is Parshall-Wang’s Frobenius isomorphism
recalled above, $\displaystyle Fr$ is the Frobenius isomorphism introduced in
[23], Definition 7.2, and the vertical arrows are the identifications as
vector spaces (the middle one proved by Proposition 4.2).

######  Remark 4.5.

By Lemma 4.3 the monomials $\displaystyle T_{l}(\omega)^{i}b^{jl}d^{rl}$ and
$\displaystyle T_{l}(\omega)^{i}c^{kl}d^{rl}$, for $\displaystyle
i,j,k,r\in\mathbb{N}$ and $\displaystyle k>0$, form an $\displaystyle A$-basis
of $\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}(sl(2)))$. It
is straightforward (though cumbersome) to express these basis elements in
terms of the basis elements $\displaystyle a^{\star il}\star b^{\star jl}\star
d^{\star rl}$ and $\displaystyle a^{\star il}\star c^{\star kl}\star d^{\star
rl}$ of $\displaystyle\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon}(SL_{2}))$, and
conversely; this can be done by using Lemma 4.4, the formula (11) or the
inverse one (expressing $\displaystyle\star$ in terms of the product of
$\displaystyle{\mathcal{L}}_{0,1}$, see (18) in [23]), and the formula of the
coproduct
$\displaystyle\Delta\colon{\mathcal{L}}_{0,1}^{\epsilon}(sl(2)))\rightarrow{\mathcal{L}}_{0,2}^{\epsilon}(sl(2)))$
restricted to
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}(sl(2)))$ (given
in Proposition 6.15 and Lemma 7.7 of [23]).

Since $\displaystyle{\mathcal{L}}_{0,1}^{A}={\mathcal{O}}_{A}$ as an
$\displaystyle A$-module, the functionals $\displaystyle t_{i}$ in Proposition
2.16 can be seen as maps $\displaystyle
t_{i}:{\mathcal{L}}_{0,1}^{A}\rightarrow A$. Though the algebra structures of
$\displaystyle{\mathcal{O}}_{\epsilon}$ and
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}$ are very different, we have the
analogous result:

######  Proposition 4.6.

The maps $\displaystyle\lhd t_{i}$ preserve
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$, and they
satisfy $\displaystyle(f\lhd t_{i})(a)=f(n_{i}a)$ and
$\displaystyle(f\alpha)\lhd t_{i}=(f\lhd t_{i})(\alpha\lhd t_{i})$ for every
$\displaystyle f\in\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$,
$\displaystyle a\in G$,
$\displaystyle\alpha\in{\mathcal{L}}_{0,1}^{\epsilon}$.

Proof. The first two claims follow from Proposition 2.16 and the equality
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})=\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})$
in Proposition 4.1.

The last claim follows from the case $\displaystyle\mathfrak{g}=sl(2)$, as in
the proof of Proposition 7.1 of [36]. In fact it is enough to show that
$\displaystyle t(fg)=t(f)t(g)$ for every $\displaystyle
f\in\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}(sl(2))$, $\displaystyle
g\in{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))$; for completeness we explain this
in the Appendix, see (72). A word of caution is needed: the proof of (72) uses
that
$\displaystyle\Delta\colon{\mathcal{O}}_{\epsilon}\rightarrow{\mathcal{O}}_{\epsilon}\otimes{\mathcal{O}}_{\epsilon}$
is a morphism of algebras. The analogous property for
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}$ is that $\displaystyle\Delta$
yields a morphism of algebras
$\displaystyle\Delta\colon{\mathcal{L}}_{0,1}^{\epsilon}\rightarrow{\mathcal{L}}_{0,2}^{\epsilon}$.
Since the algebra structure of $\displaystyle{\mathcal{L}}_{0,2}^{\epsilon}$
is not the product one on
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}\otimes{\mathcal{L}}_{0,1}^{\epsilon}$,
it is not true in general that
$\displaystyle\textstyle\sum_{(f),(g)}(f_{(1)}\otimes f_{(2)})(g_{(1)}\otimes
g_{(2)})=\sum_{(f),(g)}f_{(1)}g_{(1)}\otimes f_{(2)}g_{(2)}$ for every
$\displaystyle f,g\in{\mathcal{L}}_{0,1}^{\epsilon}$. However it holds
whenever $\displaystyle f\in\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$,
since
$\displaystyle\Delta(\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}))\subset\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})\otimes\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$
and therefore $\displaystyle
f_{(2)}\in\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})=\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})$
commutes in $\displaystyle{\mathcal{L}}_{0,2}^{\epsilon}$ with any
$\displaystyle
g_{(1)}\in{\mathcal{L}}_{0,1}^{\epsilon}={\mathcal{O}}_{\epsilon}$.

It is enough to prove the identity $\displaystyle t(fg)=t(f)t(g)$ when
$\displaystyle f$ ranges in a set of generators of the algebra
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}(sl(2)))$. So one
can take $\displaystyle f$ among, say, $\displaystyle
T_{l}(\omega)-d^{l}=a^{\star l}$ and $\displaystyle x^{l}=x^{\star l}$,
$\displaystyle x\in\\{b,c,d\\}$ (using Lemma 4.3). By (11) and Proposition 6.1
in the Appendix we have

$\displaystyle t(fg)=\sum_{(R),(R)}t\left(R_{(2^{\prime})}{S}(R_{(2)})\rhd
f\right)t\left(R_{(1^{\prime})}\rhd g\lhd R_{(1)}\right).$

Expanding coproducts and using that $\displaystyle R^{-1}=(S\otimes id)(R)$ we
deduce

$\displaystyle\displaystyle t(fg)$
$\displaystyle\displaystyle=\sum_{(f),(R),(R)}t\big{(}f_{(1)}\big{)}\left\langle
f_{(2)},R_{(2^{\prime})}{S}(R_{(2)})\right\rangle\ t\left(R_{(1^{\prime})}\rhd
g\lhd R_{(1)}\right)$
$\displaystyle\displaystyle=\sum_{(f),(R),(R)}t\big{(}f_{(1)}\big{)}\
t\bigg{(}\left\langle f_{(2)},R_{(2^{\prime})}\right\rangle
R_{(1^{\prime})}\rhd g\lhd\left\langle f_{(3)},S(R_{(2)})\right\rangle
R_{(1)}\bigg{)}$ $\displaystyle\displaystyle=\sum_{(f)}t(f_{(1)})\
t\bigg{(}S^{-1}(\Phi^{-}(f_{(2)}))\rhd g\lhd
S^{-2}(\Phi^{-}(f_{(3)}))\bigg{)}$
$\displaystyle\displaystyle=\sum_{(f)}t(f_{(1)})\
\bigg{\langle}g,S^{-2}(\Phi^{-}(f_{(3)}))\underline{w}S^{-1}(\Phi^{-}(f_{(2)}))\bigg{\rangle}$
$\displaystyle\displaystyle=\sum_{(f)}t(f_{(1)})\
\varepsilon\bigg{(}S^{-2}(\Phi^{-}(f_{(3)}))\bigg{)}\varepsilon\bigg{(}S^{-1}(\Phi^{-}(f_{(2)}))\bigg{)}t(g)$

where $\displaystyle\underline{w}\in\mathbb{U}_{\Gamma}$ is the quantum Weyl
group element dual to $\displaystyle t$ (see Section 6.1), and in the last
equality we used that $\displaystyle\Phi^{-}$ maps
$\displaystyle\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})$ into
$\displaystyle\mathcal{Z}_{0}(U_{\epsilon})$ (see Theorem 2.15 (2)), which
acts on $\displaystyle\Gamma$-modules by the trivial character (the counit)
$\displaystyle\varepsilon\colon U_{\epsilon}\rightarrow\mathbb{C}$. By
(64)-(65) in the Appendix we have $\displaystyle t(a^{\star l})=t(d^{\star
l})=0$ and $\displaystyle t(b^{\star l})=1$, $\displaystyle t(c^{\star
l})=-1$. Now the computation of $\displaystyle t(fg)$ follows easily. For
instance, taking $\displaystyle f=b^{l}=b^{\star l}$, by using
$\displaystyle\Delta(b^{\star l})=a^{\star l}\otimes b^{\star l}+b^{\star
l}\otimes d^{\star l}$ and $\displaystyle\Delta(d^{\star l})=c^{\star
l}\otimes b^{\star l}+d^{\star l}\otimes d^{\star l}$ we get

$\displaystyle\displaystyle t(b^{l}g)$
$\displaystyle\displaystyle=\varepsilon\bigg{(}S^{-2}(\Phi^{-}(b^{\star
l}))\bigg{)}\varepsilon\bigg{(}S^{-1}(\Phi^{-}(c^{\star
l}))\bigg{)}t(g)+\varepsilon\bigg{(}S^{-2}(\Phi^{-}(d^{\star
l}))\bigg{)}\varepsilon\bigg{(}S^{-1}(\Phi^{-}(d^{\star l}))\bigg{)}t(g)$

Since $\displaystyle b^{\star l}\in{\mathcal{O}}_{\epsilon}(U_{+})$,
$\displaystyle\Phi^{-}(b^{\star l})=0$. Also, it is immediate from the
definition of $\displaystyle\Phi^{-}$ that $\displaystyle\Phi^{-}(d^{\star
l})=\Phi^{-}(d)^{l}=L^{l}$; alternatively, one can bypass this computation by
observing that $\displaystyle\Phi^{-}$ sets an isomorphism from
$\displaystyle{\mathcal{O}}_{\epsilon}(T_{G})={\mathcal{O}}_{\epsilon}(B_{+})\cap{\mathcal{O}}_{\epsilon}(B_{-})$
to $\displaystyle\mathbb{C}[L^{\pm 1}]=U_{\epsilon}(\mathfrak{b}_{+})\cap
U_{\epsilon}(\mathfrak{b}_{-})$, mapping a generator $\displaystyle d$ to
$\displaystyle L$ or $\displaystyle L^{-1}$. We have
$\displaystyle\varepsilon(L^{l})=1$, and therefore

$\displaystyle\displaystyle t(b^{l}g)=t(g)=t(b^{l})t(g).$

The other cases $\displaystyle f=T_{l}(\omega)-d^{l},c^{l},d^{l}$ are similar.
$\displaystyle\Box$

######  Theorem 4.7.

$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ is a free
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$-module of rank
$\displaystyle l^{n.dim\mathfrak{g}}$.

Proof. By Proposition 4.2 (1), $\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$
and $\displaystyle{\mathcal{O}}_{\epsilon}^{\otimes n}$ coincide as modules
over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})=\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon}^{\otimes
n})$, so the claim follows immediately from Theorem 7.2 of [36], which shows
that $\displaystyle{\mathcal{O}}_{\epsilon}$ is a finitely generated
projective module of rank $\displaystyle l^{dim\mathfrak{g}}$ over
$\displaystyle\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})$, and the arguments of
[21], implying that this module is free.

For the sake of clarity let us provide a self-contained proof of the first
claim, not invoking directly [36, 21]. Since
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ and
$\displaystyle{\mathcal{L}}_{0,1}^{\otimes n}$ coincide as modules over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})=\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})^{\otimes
n}$, the result follows from the case $\displaystyle n=1$. Then we argue in
four steps. First, using Theorem 2.2 we show that a certain localization of
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}$ is a free module of rank
$\displaystyle l^{dim\mathfrak{g}}$. Then, assuming that
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}$ is finitely generated and
projective, we explain why it has constant rank $\displaystyle
l^{dim\mathfrak{g}}$ (this is very classical). Thirdly, we prove that
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}$ is finitely generated and
projective as in Theorem 7.2 of [36]. Finally we obtain that it is a free
module as in [21].

We have $\displaystyle
T_{2-}^{(l)-1}U_{\epsilon}^{lf}=T_{2-}^{-1}U_{\epsilon}^{lf}=U_{\epsilon}^{lf}[\ell]=U_{\epsilon}^{lf}[\ell^{l}]$,
where $\displaystyle\ell$ is the pivotal element. Then Theorem 2.14 (1) and
(49) imply that $\displaystyle U_{\epsilon}$ is a free
$\displaystyle\mathcal{Z}_{0}(U_{\epsilon}^{lf})[\ell^{l}]$-module of rank
$\displaystyle 2^{m}l^{dim\mathfrak{g}}$, and Theorem 2.2 (2) says that it is
also the direct sum of $\displaystyle 2^{m}$ copies of the (free)
$\displaystyle\mathcal{Z}_{0}(U_{\epsilon}^{lf})[\ell^{l}]$-module
$\displaystyle U_{\epsilon}^{lf}[\ell^{l}]$. The decomposition being unique,
it follows that $\displaystyle U_{\epsilon}^{lf}[\ell^{l}]$ is free of rank
$\displaystyle l^{dim\mathfrak{g}}$ over
$\displaystyle\mathcal{Z}_{0}(U_{\epsilon}^{lf})[\ell^{l}]$. Pulling this back
via $\displaystyle\Phi_{1}$, this proves
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}[d^{-l}]$ is free of rank
$\displaystyle l^{dim\mathfrak{g}}$ over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})[d^{-l}]$, where
$\displaystyle d=\psi_{-\rho}=\Phi_{1}^{-1}(\ell)$ (see Corollary 2.4, Theorem
2.7).

Here we note that, by the first formula of Theorem 2.11 (3), taking powers
with respect to the product of $\displaystyle{\mathcal{L}}_{0,1}$ we have

$\displaystyle\psi_{-\rho}^{l}=\psi_{-l\rho}.$

Assume that $\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}$ is finitely
generated and projective. Let us show that its rank is $\displaystyle
l^{dim\mathfrak{g}}$. The localization
$\displaystyle({\mathcal{L}}_{0,1}^{\epsilon})_{P}$ of
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}$ at any prime ideal
$\displaystyle P$ of
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$ is a free
module over $\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})_{P}$
([62], Proposition 2.12.15); the ranks of such modules are finite in number
([62], Proposition 2.12.20). If these ranks are all equal, then, by
definition, it is the rank of $\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}$
over $\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$. This
happens if $\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$ has
no non-trivial (ie. $\displaystyle\neq 1$) idempotent ([62], Corollary
2.12.23), which is the case since it has non non-trivial zero divisors. To
compute the rank, suppose $\displaystyle P$ does not contain $\displaystyle
d^{l}$. Such ideals $\displaystyle P$ are in $\displaystyle 1$-$\displaystyle
1$ correspondence with the prime ideals of
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})[d^{-l}]$ by the
natural ring monomorphism
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})\rightarrow\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})[d^{-l}]$.
The set $\displaystyle
S=\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})\setminus P$ is
multiplicatively closed, and we have also a ring morphism
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})[d^{-l}]\rightarrow
S^{-1}\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$, which is also an
injection (there are no zero divisors in
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$, whence in
$\displaystyle S$). Then

(53)
$({\mathcal{L}}_{0,1}^{\epsilon})_{P}=S^{-1}{\mathcal{L}}_{0,1}^{\epsilon}={\mathcal{L}}_{0,1}^{\epsilon}[d^{-l}]\otimes_{\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})[d^{-l}]}S^{-1}\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$

shows that $\displaystyle({\mathcal{L}}_{0,1}^{\epsilon})_{P}$ has over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})_{P}=S^{-1}\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$
the same rank $\displaystyle l^{dim\mathfrak{g}}$ as
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}[d^{-l}]$ over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})[d^{-l}]$. This
proves our claim.

In order to show that $\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}$ is
finitely generated and projective over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$ it is enough to
show it is finite locally free, ie. there are elements $\displaystyle
d_{i}\in\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$ such that the
localizations $\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}[d_{i}^{-1}]$ are
finite free
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})[d_{i}^{-1}]$-modules,
and SpecM$\displaystyle(\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}))$ is
covered by the open sets $\displaystyle U(d_{i})$ made of the ideals not
containing $\displaystyle d_{i}$ (see Lemma 77.2 of [68]).

We have seen above that $\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}[d^{-l}]$
is free of rank $\displaystyle l^{dim\mathfrak{g}}$ over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})[d^{-l}]$. In the
proof of Proposition 4.1 we saw that there are isomorphisms
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})[d^{-l}]\cong\mathcal{Z}_{0}(U_{\epsilon}^{lf})[\ell^{l}]\cong{\mathcal{O}}(G^{0})$,
and $\displaystyle{\mathcal{O}}(G^{0})={\mathcal{O}}(G)[\delta^{-1}]$. Now,
given $\displaystyle w\in W$ with a reduced expression $\displaystyle
s_{i_{1}}\ldots s_{i_{k}}$, put $\displaystyle t_{w}=t_{i_{1}}\ldots
t_{i_{k}}$. Let $\displaystyle w$ be represented by $\displaystyle
n_{w}=n_{i_{1}}\ldots n_{i_{k}}$ in $\displaystyle N(T_{G})$. By Proposition
4.6 we have $\displaystyle(f\lhd t_{w})(x)=f(n_{w}x)$ for every $\displaystyle
f\in\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$, $\displaystyle x\in G$.
Then

$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})[d^{-l}]\lhd
t_{w}\cong{\mathcal{O}}(n_{w}^{-1}G^{0})\cong{\mathcal{O}}(G)[(\delta\lhd
t_{w})^{-1}].$

If $\displaystyle b_{1},\ldots,b_{r}$ ($\displaystyle r:=l^{dim\mathfrak{g}}$)
is a basis of $\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}[d^{-l}]$ over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})[d^{-l}]$, then
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}[d^{-l}]\lhd t_{w}$ is free over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})[(d\lhd
t_{w})^{-l}]\cong{\mathcal{O}}(n_{w}^{-1}G^{0})$ with basis $\displaystyle
b_{1}\lhd t_{w},\ldots,b_{r}\lhd t_{w}$. Consider the Bruhat decomposition of
$\displaystyle G$: any $\displaystyle g\in G$ can be written in the form
$\displaystyle g=b_{1}nb_{2}$, where $\displaystyle b_{1},b_{2}\in B_{-}$,
$\displaystyle n\in W$. Hence we have $\displaystyle g=nn^{-1}b_{1}nb_{2}\in
nB_{+}B_{-}=nG^{0}$, and therefore

$\displaystyle G=\cup_{w\in W}(B_{-}n_{w}B_{-})=\cup_{w\in W}(n_{w}G^{0}).$

For every $\displaystyle w\in W$ put

$\displaystyle d^{l}_{w}:=d^{l}\lhd t_{w}.$

Under the isomorphism of
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$ with
$\displaystyle G$, we thus get that $\displaystyle{\rm
SpecM}(\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}))$ is covered by the
open sets $\displaystyle U(d^{l}_{w})\cong n_{w}G^{0}$, and
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}[d_{w}^{-l}]$ is finite free over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})[d_{w}^{-l}]$.
Therefore $\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}$ is finitely generated
and projective over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$.

Finally, let us explain why $\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}$ is
free over $\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$,
following the arguments of [21]. Let $\displaystyle R$ be a commutative
Noetherian ring, put $\displaystyle X={\rm SpecM}(R)$, and let $\displaystyle
P$ be an $\displaystyle R$-module. Denote by $\displaystyle R_{I}$,
$\displaystyle P_{I}$ the localizations of $\displaystyle R$, $\displaystyle
P$ at a maximal ideal $\displaystyle I\in X$. Define the f-rank of
$\displaystyle P$ as f-rank$\displaystyle\textstyle(P)=\inf_{I\in X}\\{$
f-rank$\displaystyle{}_{R_{I}}(P_{I})\\}$, where
f-rank$\displaystyle{}_{R_{I}}(P_{I})=\sup\\{r\in\mathbb{N},R_{I}^{\otimes
r}\subset P_{I}\\}\in\mathbb{N}\cup\\{+\infty\\}$ (ie. the maximal dimension
of a free summand of $\displaystyle P_{I}$). Bass’ Cancellation theorem
asserts that if $\displaystyle P$ is projective and
f-rank$\displaystyle(P)>{\rm dim}(X)$, and $\displaystyle P\oplus Q\cong
M\oplus Q$ for some $\displaystyle R$-modules $\displaystyle Q$ and
$\displaystyle M$ such that $\displaystyle Q$ is finitely generated and
projective, then $\displaystyle P\cong M$ (see [8], IV.3.5 and pages 167 and
170, taking $\displaystyle A=R$, or [60], section 11.7.13). Let us apply this
to $\displaystyle R={\mathcal{O}}(G)$ and $\displaystyle
P={\mathcal{L}}_{0,1}^{\epsilon}$. We proved above that
f-rank$\displaystyle{}_{R_{I}}(P_{I})=l^{dim\mathfrak{g}}$, a constant, and we
have $\displaystyle l^{dim\mathfrak{g}}>dim\mathfrak{g}={\rm dim}(G)$. By a
result of Marlin [59], the Grothendieck ring $\displaystyle
K_{0}({\mathcal{O}}(G))$ is isomorphic to $\displaystyle\mathbb{Z}$. Therefore
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}\oplus Q\cong{\mathcal{O}}(G)^{r}$
for some free $\displaystyle{\mathcal{O}}(G)$-module $\displaystyle Q$ and
$\displaystyle r\in\mathbb{N}$. Then Bass’ Cancellation implies
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}$ is free over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1})\cong{\mathcal{O}}(G)$.
$\displaystyle\Box$

##  5\. Proof of Theorem 1.3

We begin with two lemmas, interesting in themselves.

######  Lemma 5.1.

$\displaystyle\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon})$ is a finite
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$-module and a
Noetherian ring. Therefore the ring
$\displaystyle\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon})$ is integral over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$.

Proof. We know that $\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ is finite
over $\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$ (Theorem
4.7), and $\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$ is a
Noetherian ring (Proposition 4.2). Therefore
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ is a Noetherian
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$-module. This
implies that the submodule
$\displaystyle\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon})$ is finitely
generated. But being finite over the Noetherian ring
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$, it is a
Noetherian ring (by eg. Proposition 7.2 of [5]).

Let $\displaystyle x\in\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon})$. The
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$-submodule
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})[x]$ of
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ is finitely generated by the
same argument. Using the fact that an element $\displaystyle x$ is integral
over $\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$ if and
only if $\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})[x]$ is a
finitely generated
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$-module (by eg.
Proposition 5.1 of [5]), this proves the last claim. $\displaystyle\Box$

As usual, denote by $\displaystyle Q(Z)$ the quotient field of a commutative
integral domain $\displaystyle Z$. Then, consider the fields $\displaystyle
Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))$ and $\displaystyle
Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon}))$. Since
$\displaystyle\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon})$ is finite over
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$ and has no non-
trivial zero divisors, the ring
$\displaystyle\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon})\otimes_{\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})}Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon}))$
is a field. Therefore it is equal to $\displaystyle
Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))$.

######  Lemma 5.2.

$\displaystyle[Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon})):Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon}))]=l^{mn}$.

Proof. First consider the case $\displaystyle n=1$. Note that $\displaystyle
Q(\mathcal{Z}(U_{\epsilon}))=Q(\mathcal{Z}(U_{\epsilon}^{lf}))(T^{(l)}/T_{2}^{(l)})$
by (49), and similarly by replacing $\displaystyle\mathcal{Z}$ with
$\displaystyle\mathcal{Z}_{0}$. Then, applying $\displaystyle\Phi_{1}$ and
using that $\displaystyle Q(\mathcal{Z}(U_{\epsilon}))$ has degree
$\displaystyle l^{m}$ over $\displaystyle Q(\mathcal{Z}_{0}(U_{\epsilon}))$
(see the comment after Theorem 2.14) we deduce

(54)
$[Q(\mathcal{Z}({\mathcal{L}}_{0,1}^{\epsilon})):Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}))]=[Q(\mathcal{Z}(U_{\epsilon}^{lf})):Q(\mathcal{Z}_{0}(U_{\epsilon}^{lf}))]=l^{m}.$

Next let $\displaystyle n\geq 1$ be arbitrary. Recall from (24) the matrices
$\displaystyle M_{j}^{(i)}$, for $\displaystyle i=1,\ldots,n$ and
$\displaystyle j=1,\ldots,m$. More generally, for every
$\displaystyle\lambda\in P_{+}$ and $\displaystyle i=1,\ldots,n$ consider the
matrix of matrix coefficients $\displaystyle
M_{\lambda}^{(i)}:=({}_{V_{\lambda}}\phi_{e_{k}}^{e_{l}(i)})_{k,l}\in
End(V_{\lambda})\otimes{\mathcal{L}}_{0,n}^{A}$, where
$\displaystyle{}_{V_{\lambda}}\phi_{e_{k}}^{e_{l}(i)}:=1^{\otimes(i-1)}\otimes{}_{V_{\lambda}}\phi_{e_{k}}^{e_{l}}\otimes
1^{\otimes(n-i)}$, and $\displaystyle\\{e_{k}\\}$ is the canonical basis of
weight vectors of $\displaystyle V_{\lambda}$. In [23], Proposition 6.22, we
showed that the elements
$\displaystyle{}_{\lambda}\omega^{(i)}:=Tr(\pi_{V_{\lambda}}(\ell)M_{\lambda}^{(i)})$
are central in $\displaystyle{\mathcal{L}}_{0,n}^{A}$, where $\displaystyle
Tr$ is the standard trace on $\displaystyle End(V_{\lambda})$; moreover, the
family $\displaystyle\\{{}_{\lambda}\omega^{(i)},\lambda\in P_{+}\\}$ is a
basis of the center of the $\displaystyle i$-th factor
$\displaystyle{\mathcal{L}}_{0,1}^{A}$ of
$\displaystyle{\mathcal{L}}_{0,n}^{A}$. Now, recall the graded algebra
$\displaystyle Gr_{{\mathcal{F}}}({\mathcal{L}}_{0,n})$ in (46). Take
$\displaystyle\lambda=\varpi_{j}$. By (44), for every $\displaystyle
i\in\\{1,\ldots,n\\}$, $\displaystyle j\in\\{1,\ldots,m\\}$ and $\displaystyle
r\in\mathbb{N}$ the leading coefficient of
$\displaystyle({}_{\varpi_{j}}\omega^{(i)})^{r}$ in $\displaystyle
Gr_{{\mathcal{F}}}({\mathcal{L}}_{0,n})$ belongs to the space $\displaystyle
C([0,\ldots,0,r\varpi_{j},0,\ldots,0])$ (with $\displaystyle\varpi_{j}$ in the
$\displaystyle i$-th entry). Take the specialization at $\displaystyle
q=\epsilon$ and denote by
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})^{(i)}$ the
$\displaystyle i$-th factor of
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$. Recall that
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})^{(i)}=\textstyle\oplus_{\mu\in
P_{+}}C([0,\ldots,0,l\mu,0,\ldots,0])$. Then we get that the minimal
polynomial of $\displaystyle{}_{\varpi_{j}}\omega^{(i)}$ over $\displaystyle
Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})^{(i)})$ has degree
$\displaystyle l$. On another hand, denoting by
$\displaystyle\mathcal{Z}({\mathcal{L}}_{0,1}^{\epsilon})^{(i)}$ the center of
the $\displaystyle i$-th factor of
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$, we have
$\displaystyle[Q(\mathcal{Z}({\mathcal{L}}_{0,1}^{\epsilon})^{(i)}):Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})^{(i)})]=l^{m}$
as in the $\displaystyle n=1$ case above. It follows that $\displaystyle
Q(\mathcal{Z}({\mathcal{L}}_{0,1}^{\epsilon})^{(i)})$ is generated over
$\displaystyle Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})^{(i)})$ by
the elements $\displaystyle{}_{\lambda}\omega^{(i)}$, $\displaystyle\lambda\in
P+$. The fields $\displaystyle
Q(\mathcal{Z}({\mathcal{L}}_{0,1}^{\epsilon})^{(i)})$ are supported by
distinct tensor factors, so they are linearly disjoint subfields of
$\displaystyle Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))$. The same is
true of the fields $\displaystyle
Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})^{(i)})$. Therefore, the
compositum of the fields $\displaystyle
Q(\mathcal{Z}({\mathcal{L}}_{0,1}^{\epsilon})^{(i)})$, say $\displaystyle
Q^{\prime}$, has degree $\displaystyle(l^{m})^{n}$ over the compositum of the
fields $\displaystyle
Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})^{(i)})$, which is
$\displaystyle Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon}))$. By the
same argument, $\displaystyle Q(\mathcal{Z}(U_{\epsilon}^{\otimes n}))$ has
degree $\displaystyle l^{mn}$ over $\displaystyle
Q(\mathcal{Z}_{0}(U_{\epsilon}^{\otimes n}))$. Now, recall from (26) the
isomorphism of algebras
$\displaystyle\Phi_{n}\colon{}_{loc^{\prime}}{\mathcal{L}}_{0,n}^{\epsilon}\rightarrow
U^{\otimes n}_{\epsilon}$. Consider the induced isomorphism of fields
$\displaystyle\Phi_{n}\colon
Q(\mathcal{Z}({}_{loc^{\prime}}{\mathcal{L}}_{0,n}^{\epsilon}))\rightarrow
Q(\mathcal{Z}(U^{\otimes n}_{\epsilon}))$, and the field extensions in
$\displaystyle
Q(\mathcal{Z}({}_{loc^{\prime}}{\mathcal{L}}_{0,n}^{\epsilon}))$,

$\displaystyle\displaystyle{}_{loc^{\prime}}Q^{\prime}$
$\displaystyle\displaystyle=Q^{\prime}[\\{\nu_{j_{n}}^{(n)}{}^{-l}\\}][\\{\nu_{j_{n-1}}^{(n-1)}{}^{-l}\\}]\cdots[\\{\nu_{j_{1}}^{(1)}{}^{-l}\\}]$
$\displaystyle\displaystyle{}_{loc^{\prime}}Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon}))$
$\displaystyle\displaystyle=Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon}))[\\{\nu_{j_{n}}^{(n)}{}^{-l}\\}][\\{\nu_{j_{n-1}}^{(n-1)}{}^{-l}\\}]\cdots[\\{\nu_{j_{1}}^{(1)}{}^{-l}\\}].$

Compose $\displaystyle\Phi_{n}$ with the linear automorphism of $\displaystyle
Q(\mathcal{Z}(U^{\otimes n}_{\epsilon}))$ induced by
$\displaystyle\psi_{n}^{-1}=\Phi_{1}^{\otimes n}\circ\Phi_{n}^{-1}$ (see (19)
for the latter). The image of
$\displaystyle{}_{loc^{\prime}}Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon}))$
by $\displaystyle\psi_{n}^{-1}\circ\Phi_{n}$ is $\displaystyle
Q(\mathcal{Z}_{0}(U_{\epsilon}^{\otimes n}))$. Since $\displaystyle
l^{mn}=[Q^{\prime}:Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon}))]=[{}_{loc^{\prime}}Q^{\prime}:{}_{loc^{\prime}}Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon}))]=[\psi_{n}^{-1}\circ\Phi_{n}({}_{loc^{\prime}}Q^{\prime}):Q(\mathcal{Z}_{0}(U_{\epsilon}^{\otimes
n}))]$ and $\displaystyle[Q(\mathcal{Z}(U_{\epsilon}^{\otimes
n})):Q(\mathcal{Z}_{0}(U_{\epsilon}^{\otimes n}))]=l^{mn}$, it follows that
$\displaystyle\psi_{n}^{-1}\circ\Phi_{n}({}_{loc^{\prime}}Q^{\prime})=Q(\mathcal{Z}(U_{\epsilon}^{\otimes
n}))$, whence
$\displaystyle{}_{loc^{\prime}}Q^{\prime}=Q(\mathcal{Z}({}_{loc^{\prime}}{\mathcal{L}}_{0,n}^{\epsilon}))={}_{loc^{\prime}}Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))$,
and then $\displaystyle
Q^{\prime}=Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))$. This eventually
proves $\displaystyle
l^{mn}=[Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon})):Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon}))]$.
$\displaystyle\Box$

Recall the ring

$\displaystyle
Q({\mathcal{L}}_{0,n}^{\epsilon})=Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))\otimes_{\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon})}{\mathcal{L}}_{0,n}^{\epsilon}.$

The center of $\displaystyle Q({\mathcal{L}}_{0,n}^{\epsilon})$ is
$\displaystyle Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))$. By the comment
before Lemma 5.2 we have another description of $\displaystyle
Q({\mathcal{L}}_{0,n}^{\epsilon})$, namely

(55)
$Q({\mathcal{L}}_{0,n}^{\epsilon})=Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon}))\otimes_{\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})}{\mathcal{L}}_{0,n}^{\epsilon}.$

Recall that we denote by $\displaystyle N$ the number of positive roots of
$\displaystyle\mathfrak{g}$.

######  Proposition 5.3.

$\displaystyle Q({\mathcal{L}}_{0,n}^{\epsilon})$ is a central simple algebra
of dimension $\displaystyle l^{2Nn}$ over $\displaystyle
Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))$.

Proof. From its definition $\displaystyle Q({\mathcal{L}}_{0,n}^{\epsilon})$
is a vector space over $\displaystyle
Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))$. Because
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ has no non-trivial divisors and
$\displaystyle Q({\mathcal{L}}_{0,n}^{\epsilon})$ is finite-dimensional over
$\displaystyle Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))$, it is a
division algebra over $\displaystyle
Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon})),$ whence a simple algebra. Its
center being $\displaystyle Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))$,
this proves the first part of the statement. By classical theory (see eg.
Section 13.3.5 of [60], or [62], Corollary 2.3.25), it then follows that there
is a finite extension (a splitting field) $\displaystyle\mathbb{F}$ of
$\displaystyle Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))$ such that

$\displaystyle\mathbb{F}\otimes_{Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))}Q({\mathcal{L}}_{0,n}^{\epsilon})=M_{d}(\mathbb{F})$

where $\displaystyle d\in\mathbb{N}$, the PI degree of $\displaystyle
Q({\mathcal{L}}_{0,n}^{\epsilon})$, is given by $\displaystyle
d^{2}=[Q({\mathcal{L}}_{0,n}^{\epsilon}):Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))]$.
Therefore

$\displaystyle
d^{2}=\frac{[Q({\mathcal{L}}_{0,n}^{\epsilon}):Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon}))]}{[Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon})):Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon}))]}=l^{2Nn}$

where we use Theorem 4.7 and Lemma 5.2, and we recall that $\displaystyle
dim\mathfrak{g}=m+2N$. $\displaystyle\Box$

Let us recall for the sake of clarity different notions of ring theory,
bearing in mind that we will apply them to the case where $\displaystyle
A={\mathcal{L}}_{0,n}^{\epsilon}$. Let $\displaystyle A$ be a ring with no
non-trivial zero divisors. The center $\displaystyle Z=Z(A)$ is a commutative
integral domain. Denote by $\displaystyle Q(Z)$ its field of fractions, and
let

$\displaystyle Q(A)=Q(Z)\otimes_{Z}A.$

It is an algebra over its center $\displaystyle Q(Z)$.

An element $\displaystyle a\in A$ is integral over $\displaystyle Z$ if
$\displaystyle Z[a]$ is a finitely generated $\displaystyle Z$-module.
$\displaystyle A$ is integral over $\displaystyle Z$ if every element of
$\displaystyle A$ is integral over $\displaystyle Z$. An element
$\displaystyle a\in A$ is c-integral over $\displaystyle Z$ if $\displaystyle
Z[a]$ is contained in a finitely generated $\displaystyle Z$-module.
$\displaystyle A$ is c-integral over $\displaystyle Z$ if every element of
$\displaystyle A$ is c-integral over $\displaystyle Z$. When $\displaystyle Z$
is a Noetherian ring these two notions are equivalent ([60], Lemma 5.3.2).

If $\displaystyle Q(A)$ has finite dimension over $\displaystyle Q(Z)$, it is
a division algebra and therefore a central simple algebra. Moreover
$\displaystyle Q(A)$ is integral over $\displaystyle Q(Z)$.

There are different notions of orders, that are equivalent in our context. Let
$\displaystyle B\subset Q(A)$ be a subring. $\displaystyle B$ is said to be an
order of $\displaystyle Q(A)$ ([60], 3.1.2) if every element $\displaystyle q$
of $\displaystyle Q(A)$ can be written as $\displaystyle
q=rs^{-1}={s^{\prime}}{}^{-1}r^{\prime}$, where $\displaystyle
r,s,r^{\prime},s^{\prime}\in B$. In particular, $\displaystyle A$ is an order
of $\displaystyle Q(A)$ and $\displaystyle s,s^{\prime}$ can in this case be
chosen in $\displaystyle Z(A)$. $\displaystyle B$ is a classical order of
$\displaystyle Q(A)$ ([60], 5.3.5) if $\displaystyle Z\subset B$,
$\displaystyle Q(Z)B=Q(A)$, and $\displaystyle B$ is finitely generated as a
$\displaystyle Z$-module. In particular, if $\displaystyle A$ is a finitely
generated $\displaystyle Z$-module then $\displaystyle A$ is a classical order
of $\displaystyle Q(A)$. Finally, $\displaystyle B$ is a $\displaystyle
Z$-order of $\displaystyle Q(A)$ ([60], 5.3.6) if $\displaystyle Z\subset B$,
$\displaystyle Q(Z)B=Q(A)$ and $\displaystyle B$ is $\displaystyle c$-integral
over $\displaystyle Z$.

We now assume that $\displaystyle Z$ is Noetherian and $\displaystyle Q(A)$ is
of finite dimension over $\displaystyle Q(Z)$. Let $\displaystyle B$ be a
subring of $\displaystyle Q(A)$ having center $\displaystyle Z(B)=Z(A)=Z$.
Then the following assertions are equivalent:

  1. (1)

$\displaystyle B$ is an order of $\displaystyle Q(A)$;

  2. (2)

$\displaystyle B$ is a classical order of $\displaystyle Q(A)$;

  3. (3)

$\displaystyle B$ is a $\displaystyle Z$-order of $\displaystyle Q(A)$.

Indeed, from the definitions we trivially have
$\displaystyle(2)\Rightarrow(3)$. From the Noetherianity of $\displaystyle Z$,
Proposition 5.3.14 of [60] gives $\displaystyle(3)\Rightarrow(2)$. Because
$\displaystyle Q(A)$ is a central simple algebra, the equivalence
$\displaystyle(1)\Leftrightarrow(3)$ is part of Proposition 5.3.10 of [60].

In particular, because $\displaystyle A$ is always an order of $\displaystyle
Q(A)$ these equivalences imply that $\displaystyle A$ is a classical order,
whence a finitely generated $\displaystyle Z$-module. Moreover $\displaystyle
A$ is Noetherian by Proposition 5.3.14 of [60].

There are two standard notions of maximality for $\displaystyle Z$-orders
$\displaystyle B$ of $\displaystyle Q(A)$. One applies strictly to
$\displaystyle Z$-orders ([60], 5.3.13): namely, a $\displaystyle Z$-order
$\displaystyle B$ of $\displaystyle Q(A)$ is a maximal $\displaystyle Z$-order
if $\displaystyle B\subset C$ with $\displaystyle C$ a $\displaystyle Z$-order
implies $\displaystyle B=C$. The other notion of maximality applies to
arbitrary orders of $\displaystyle Q(A)$ (see [60], 5.1.1); when
$\displaystyle B$ is a $\displaystyle Z$-order with center $\displaystyle Z$,
$\displaystyle B$ is a maximal $\displaystyle Z$-order if and only if it is
maximal in this latter sense.

If $\displaystyle A$ is a maximal order, then $\displaystyle Z$ contains all
the $\displaystyle c$-integral elements over it, ie. $\displaystyle Z$ is
c-integrally closed ([60], 5.3.13). Since $\displaystyle Z$ is Noetherian, it
is then integrally closed (in the usual sense), and therefore coincides with
the trace ring of $\displaystyle A$ (ie. the subring of $\displaystyle Q(Z)$
generated over $\displaystyle Z$ by the coefficients of the characteristic
polynomials of elements of $\displaystyle A$, represented as endomorphisms by
left multiplication on $\displaystyle Q(A)$, see [60], 13.9.2-13.9.8).

Finally, we say that $\displaystyle A$ is DCK-integrally closed if the
following condition holds: for every subring $\displaystyle R$ of
$\displaystyle Q(A)$ such that $\displaystyle A\subset R\subset z^{-1}A$ for
some non zero $\displaystyle z\in Z(A)$, we have $\displaystyle R=A$. We
borrow this notion from [33]; it is closely related to that of fractional
ideal of $\displaystyle Q(A)$ (see [60], 3.1.11-3.1.12 and 5.1.4), but
simpler. Its relevance comes from the following lemma, which shows that in the
commutative and Noetherian case it is equivalent to the usual definition of
integrally closure.

######  Lemma 5.4.

Let $\displaystyle B$ be a commutative Noetherian ring with no non-trivial
zero divisors. Then $\displaystyle B$ is integrally closed in $\displaystyle
Q(B)$ if and only if $\displaystyle B$ is DCK-integrally closed.

Proof. Assume $\displaystyle B$ is DCK-integrally closed. Let $\displaystyle
x=b/c\in Q(B)$ be integral over $\displaystyle B$. Denote by $\displaystyle n$
the degree of its minimal polynomial over $\displaystyle B$. The ring
$\displaystyle R=B[x]$ is contained in $\displaystyle c^{-n}B$, so
$\displaystyle R=B$ and $\displaystyle x\in B$. (Note that this does not need
$\displaystyle B$ Noetherian.)

Conversely, let $\displaystyle R$ be a subring of $\displaystyle Q(B)$ such
that $\displaystyle B\subset R\subset z^{-1}B$, and let $\displaystyle x\in
R$. Then $\displaystyle M=B[x]$ is a $\displaystyle B$-submodule of
$\displaystyle Q(B)$ such that $\displaystyle xM\subset M$. It is also a
$\displaystyle B$-submodule of $\displaystyle z^{-1}B$, which is free with
basis $\displaystyle z^{-1}$ over $\displaystyle B$. Since $\displaystyle B$
is Noetherian, $\displaystyle z^{-1}B$ is a Noetherian $\displaystyle
B$-module, and therefore $\displaystyle M$ is a finite $\displaystyle
B$-module. It follows that $\displaystyle x$ is integral over $\displaystyle
B$ (Proposition 5.1 of [5]), whence $\displaystyle x\in B$.
$\displaystyle\Box$

######  Lemma 5.5.

Assume that $\displaystyle A$ is a ring with no non-trivial zero divisors,
with center $\displaystyle Z$ Noetherian and such that $\displaystyle Q(A)$
has finite dimension over $\displaystyle Q(Z)$. Then $\displaystyle A$ is DCK-
integrally closed if and only if $\displaystyle A$ is a maximal order.

Proof. Assume that $\displaystyle A$ is DCK-integrally closed. Let
$\displaystyle B$ a $\displaystyle Z$-order of $\displaystyle Q(A)$ such that
$\displaystyle A\subset B$, and let $\displaystyle b\in B$. Since
$\displaystyle b$ is c-integral over $\displaystyle Z$ and $\displaystyle Z$
is Noetherian, $\displaystyle b$ is integral and thus $\displaystyle A[b]$ is
a finitely generated $\displaystyle A$-module. Let $\displaystyle
e_{i}=a_{i}/z_{i}\in Q(A)$, $\displaystyle z_{i}\in Z$ and $\displaystyle
i=1,\ldots,n$, be the generators of $\displaystyle A[b]$. We have
$\displaystyle A[b]\subset z^{-1}A$ with $\displaystyle\textstyle
z=\prod_{i=1}^{n}z_{i}$. Therefore $\displaystyle A[b]=A$, whence
$\displaystyle b\in A$. This proves that $\displaystyle A$ is a maximal order.

Conversely, assume that $\displaystyle A$ is maximal order, and let
$\displaystyle B$ be a subring of $\displaystyle Q(A)$ such that
$\displaystyle A\subset B\subset z^{-1}A$. Since $\displaystyle A$ is a
finitely generated $\displaystyle Z$-module, $\displaystyle B$ is contained in
a finitely generated $\displaystyle Z$-module, which is necessarily a
Noetherian $\displaystyle Z$-module because $\displaystyle Z$ is Noetherian.
Therefore $\displaystyle B$ is a finitely generated $\displaystyle Z$-module.
As clearly $\displaystyle Z\subset B$ and $\displaystyle Q(Z)B=Q(A)$, it is in
fact a classical order of $\displaystyle Q(A)$. Because $\displaystyle A$ is
maximal, we have $\displaystyle A=B$, which proves $\displaystyle A$ is DCK-
integrally closed. $\displaystyle\Box$

######  Theorem 5.6.

$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ is a maximal order of
$\displaystyle Q({\mathcal{L}}_{0,n}^{\epsilon})$.

Note that, by the discussion before Lemma 5.4 it follows that
$\displaystyle\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon})$ is integrally
closed and coincides with the trace ring of $\displaystyle
Q({\mathcal{L}}_{0,n}^{\epsilon})$.

Proof of Theorem 5.6. We derive the result by “twisting” the analogous
statement for $\displaystyle{\mathcal{O}}_{\epsilon}$, obtained in Theorem 7.4
of [36]. We have already proved that
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ satisfies the hypothesis on
$\displaystyle A$ in Lemma 5.5. So let $\displaystyle R\subset
Q({\mathcal{L}}_{0,n}^{\epsilon})$ be a subring such that
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}\subset R\subset
x^{-1}{\mathcal{L}}_{0,n}^{\epsilon}$ for some non zero $\displaystyle
x\in\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$. We have to show that
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}=R$. We know
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}={\mathcal{O}}_{\epsilon}^{\otimes
n}$ as $\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,n}^{\epsilon})$-modules
(Proposition 4.2 (1)), so $\displaystyle
x^{-1}{\mathcal{L}}_{0,n}^{\epsilon}=x^{-1}\star{\mathcal{O}}_{\epsilon}^{\otimes
n}$. Also, the product of $\displaystyle R$ is inherited from that of
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ (for, given $\displaystyle
r_{1}$, $\displaystyle r_{2}\in R$ we have $\displaystyle
r_{1}r_{2}=x^{-2}(xr_{1})(xr_{2})$), and the latter is defined from the one of
$\displaystyle{\mathcal{O}}_{\epsilon}^{\otimes n}$ by two consecutive twists
(see the formulas (11) and (14)). Therefore, by applying the inverse twists on
the inclusions $\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}\subset R\subset
x^{-1}{\mathcal{L}}_{0,n}^{\epsilon}$ we get
$\displaystyle{\mathcal{O}}_{\epsilon}^{\otimes n}\subset R^{\prime}\subset
x^{-1}\star{\mathcal{O}}_{\epsilon}^{\otimes n}$ where $\displaystyle
R^{\prime}$ is the vector space $\displaystyle R$ endowed with the ring
structure inherited from $\displaystyle{\mathcal{O}}_{\epsilon}^{\otimes n}$.
Now $\displaystyle{\mathcal{O}}_{\epsilon}^{\otimes
n}={\mathcal{O}}_{\epsilon}(G^{n})$ is a maximal order of $\displaystyle
Q({\mathcal{O}}_{\epsilon}(G^{n}))$ by Theorem 7.4 in [36]. Therefore
$\displaystyle{\mathcal{O}}_{\epsilon}^{\otimes n}=R^{\prime}$, and finally
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}=R$. $\displaystyle\Box$

A proof of Theorem 5.6 independent of Theorem 7.4 of [36] seems to be
difficult for arbitrary $\displaystyle\mathfrak{g}$, even for $\displaystyle
n=1$. In the case of $\displaystyle{\mathcal{L}}_{0,1}(sl(2))$ we can however
apply a similar reasoning. Let us explain the details.

Identify $\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}(sl(2))$
with $\displaystyle{\mathcal{O}}(SL_{2})$ as described in Lemma 4.4 and before
Lemma 4.3. We proceed in two steps : (a) we show that
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))[d^{-l}]$ and
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))[b^{-l}]$ are maximal
orders, (b) from this we deduce the result.

As for (a), recall that
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))[d^{-l}]\cong
U_{\epsilon}^{lf}(sl(2))[\ell^{l}]$. We have $\displaystyle
U_{\epsilon}^{lf}(sl(2))[\ell^{l}]=U_{\epsilon}^{ad}(sl(2))$, which is a
maximal order by Theorem 1.8 of [33]. Using (52) one computes easily that
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))[b^{-l}]$ is generated by
$\displaystyle\omega$, $\displaystyle b^{\pm 1}$ and $\displaystyle d$, with
defining relations: $\displaystyle\omega$ is central, $\displaystyle
db=q^{2}bd$. Therefore
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))[b^{-l}]$ is the tensor
product of $\displaystyle\mathbb{C}[\omega]$ with a quasi-polynomial algebra.
The $\displaystyle l$-th powers of $\displaystyle d$, $\displaystyle b$ are
central, and $\displaystyle\mathbb{C}[\omega]$ is integrally closed, so by a
direct application of Proposition 1.8 of [33] it follows that
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))[b^{-l}]$ is a maximal
order.

Let us now deduce that $\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))$ is
a maximal order. Let $\displaystyle
x\in\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}(sl(2))$ and $\displaystyle
R$ a subring of $\displaystyle Q({\mathcal{L}}_{0,1}^{\epsilon}(sl(2)))$ such
that $\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))\subset R\subset
x^{-1}{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))$. Let $\displaystyle y\in R$.
Because $\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))[d^{-l}]$ is
maximal, we have
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))[d^{-l}]=R[d^{-l}]$. So
$\displaystyle yd^{l.n_{d}}\in{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))$ for some
non negative integer $\displaystyle n_{d}$. Similarly, $\displaystyle
yb^{l.n_{b}}\in{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))$ for some non negative
integer $\displaystyle n_{b}$. Now, note that $\displaystyle d^{l}\lhd
t_{w}=b^{l}$. Consider the open subsets $\displaystyle U(d^{l})$ and
$\displaystyle U(b^{l})$ of $\displaystyle SL_{2}$ consisting of matrices with
non vanishing lower right entry and non vanishing upper right entry,
respectively. As in the proof of Theorem 4.7 we have
$\displaystyle\textstyle{\rm
SpecM}(\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}(sl(2)))=SL_{2}=U(d^{l})\cup
U(b^{l})$. Since $\displaystyle U(d^{l})=U(d^{l.n_{d}})$, $\displaystyle
U(b^{l})=U(b^{l.n_{b}})$, likewise $\displaystyle\textstyle{\rm
SpecM}(\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}(sl(2)))=U(d^{l.n_{d}})\cup
U(b^{l.n_{b}})$. This implies that the ideal of
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}(sl(2)))$
generated by $\displaystyle d^{l.n_{d}}$ and $\displaystyle b^{l.n_{b}}$ is
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}(sl(2)))$ itself;
hence there are elements $\displaystyle
u_{d},u_{b}\in\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}(sl(2)))$ such
that $\displaystyle u_{d}d^{l.n_{d}}+u_{b}b^{l.n_{b}}=1$, which proves
$\displaystyle
y=u_{d}d^{l.n_{d}}y+u_{b}b^{l.n_{b}}y\in{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))$.
This concludes the proof that
$\displaystyle{\mathcal{L}}_{0,1}^{\epsilon}(sl(2))$ is a maximal order.

Proof of Theorem 1.3 (2). Denote by
$\displaystyle\Delta^{(n)}\colon{\mathcal{O}}_{\epsilon}\rightarrow{\mathcal{O}}_{\epsilon}^{\otimes
n}$, $\displaystyle n\geq 2$, the $\displaystyle n$-fold coproduct, that is
$\displaystyle\Delta^{(n)}:=(id\otimes\Delta)\circ\Delta^{(n-1)}$ for
$\displaystyle n\geq 3$, and $\displaystyle\Delta^{(2)}:=\Delta$, the standard
coproduct of $\displaystyle{\mathcal{O}}_{\epsilon}$. Identifying
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$ with
$\displaystyle{\mathcal{O}}_{\epsilon}^{\otimes n}$ as a vector space, we
consider $\displaystyle\Delta^{(n)}$ as a map
$\displaystyle\Delta^{(n)}\colon{\mathcal{L}}_{0,1}^{\epsilon}\rightarrow{\mathcal{L}}_{0,n}^{\epsilon}$.
It is an algebra morphism ([23], Proposition 6.16), injective because
$\displaystyle(\varepsilon^{\otimes(n-1)}\otimes id)\Delta^{(n)}=id$. Then it
extends uniquely to the fraction algebra $\displaystyle
Q({\mathcal{L}}_{0,1}^{\epsilon})$. As noted above $\displaystyle
Q({\mathcal{L}}_{0,1}^{\epsilon})=Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}))\otimes_{\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})}{\mathcal{L}}_{0,1}^{\epsilon}$.
Since
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})=\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})$
is a Hopf subalgebra of $\displaystyle{\mathcal{O}}_{\epsilon}$ ([36],
Proposition 6.4), $\displaystyle\Delta^{(n)}$ maps
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})$ to
$\displaystyle\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})^{\otimes n}$.
Extending the scalars of $\displaystyle Q({\mathcal{L}}_{0,1}^{\epsilon})$ by
the field $\displaystyle Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))$,
consider

$\displaystyle\displaystyle\tilde{Q}(\Delta^{(n)}({\mathcal{L}}_{0,1}^{\epsilon}))$
$\displaystyle\displaystyle:=Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))\otimes_{\Delta^{(n)}(Q(\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon})))}\Delta^{(n)}(Q({\mathcal{L}}_{0,1}^{\epsilon}))$
$\displaystyle\displaystyle=Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))\otimes_{\Delta^{(n)}(\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}))}\Delta^{(n)}({\mathcal{L}}_{0,1}^{\epsilon}).$

It is a subalgebra of $\displaystyle Q({\mathcal{L}}_{0,n}^{\epsilon})$, and a
simple algebra. The center of
$\displaystyle\tilde{Q}(\Delta^{(n)}({\mathcal{L}}_{0,1}^{\epsilon}))$ is the
field
$\displaystyle\tilde{Q}(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon})):=Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))\otimes_{\Delta^{(n)}(\mathcal{Z}_{0}({\mathcal{L}}_{0,1}^{\epsilon}))}\Delta^{(n)}(\mathcal{Z}({\mathcal{L}}_{0,1}^{\epsilon}))$.
We proved in Proposition 6.17 of [23] that the ring
$\displaystyle({\mathcal{L}}_{0,n}^{A})^{U_{A}}$ is the centralizer of
$\displaystyle\Delta^{(n)}({\mathcal{L}}_{0,1}^{A})$ in
$\displaystyle{\mathcal{L}}_{0,n}^{A}$; the arguments of the proof apply as
well to $\displaystyle\Delta^{(n)}({\mathcal{L}}_{0,1}^{\epsilon})$ in
$\displaystyle{\mathcal{L}}_{0,n}^{\epsilon}$, so
$\displaystyle\tilde{Q}(({\mathcal{L}}_{0,n}^{\epsilon})^{U_{\epsilon}}):=Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))\otimes_{\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon})}({\mathcal{L}}_{0,n}^{\epsilon})^{U_{\epsilon}}$
is the centralizer of
$\displaystyle\tilde{Q}(\Delta^{(n)}({\mathcal{L}}_{0,1}^{\epsilon}))$ in
$\displaystyle Q({\mathcal{L}}_{0,n}^{\epsilon})$. Then the centralizer
theorem (see eg. [69], Theorem 7.1) implies that
$\displaystyle\tilde{Q}(({\mathcal{L}}_{0,n}^{\epsilon})^{U_{\epsilon}})$ is a
simple algebra, and

$\displaystyle\displaystyle[\tilde{Q}(({\mathcal{L}}_{0,n}^{\epsilon})^{U_{\epsilon}}):Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))]$
$\displaystyle\displaystyle=\frac{[Q({\mathcal{L}}_{0,n}^{\epsilon}):Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))]}{[\tilde{Q}(\Delta^{(n)}({\mathcal{L}}_{0,1}^{\epsilon})):Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))]}=l^{2nN-(2N+m)}.$

Moreover the centralizer of
$\displaystyle\tilde{Q}(({\mathcal{L}}_{0,n}^{\epsilon})^{U_{\epsilon}})$ is
$\displaystyle\tilde{Q}(\Delta^{(n)}({\mathcal{L}}_{0,1}^{\epsilon}))$, so
$\displaystyle\tilde{Q}(({\mathcal{L}}_{0,n}^{\epsilon})^{U_{\epsilon}})\cap\tilde{Q}(\Delta^{(n)}({\mathcal{L}}_{0,1}^{\epsilon}))=\tilde{Q}(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))$
is its center. It then follows

$\displaystyle\displaystyle[\tilde{Q}(({\mathcal{L}}_{0,n}^{\epsilon})^{U_{\epsilon}}):\tilde{Q}(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))]$
$\displaystyle\displaystyle=\frac{[\tilde{Q}(({\mathcal{L}}_{0,n}^{\epsilon})^{U_{\epsilon}}):Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))]}{[\tilde{Q}(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon})):Q(\mathcal{Z}({\mathcal{L}}_{0,n}^{\epsilon}))]}$
$\displaystyle\displaystyle=l^{2nN-(2N+m)}.l^{-m}=l^{2(N(n-1)-m)}.$

This concludes the proof. $\displaystyle\Box$

##  6\. Appendix

###  6.1. Quantum Weyl group

We recall some of the formulas of [27]. Let $\displaystyle e_{q}(z)$ be the
formal power series in $\displaystyle z$ with coefficients in
$\displaystyle\mathbb{C}(q)$ defined by:

(56) $e_{q}(z)=\sum_{n=0}^{+\infty}\frac{z^{n}}{(n)_{q}\\!}.$

We first consider the case of $\displaystyle\mathfrak{g}=sl(2)$. As explained
in [23], Section 3, the Cartan element $\displaystyle H\in\mathfrak{g}$
defines an element of $\displaystyle{\mathbb{U}}_{q}(sl(2))$. Viewed as
elements of $\displaystyle{\mathbb{U}}_{q}$ we have $\displaystyle L=q^{H/2}$.
The series $\displaystyle\Theta=q^{H\otimes H/2}$ defines an element of
$\displaystyle{\mathbb{U}}_{q}(sl(2))^{\hat{\otimes}2}$, its image under
multiplication being $\displaystyle q^{H^{2}/2}$. The $\displaystyle R$-matrix
can be expressed as $\displaystyle R=\Theta\hat{R}$ where
$\displaystyle\hat{R}=e_{q^{-1}}((q-q^{-1})E\otimes F)$ is a well defined
element of $\displaystyle{\mathbb{U}}_{q}^{\hat{\otimes}2}$. Consider the
Lusztig [52] braid group automorphism of $\displaystyle U_{q}(sl(2))$, defined
by

(57) $T(L)=L^{-1},T(E)=-FK^{-1},T(F)=-KE.$

For every $\displaystyle x\in U_{q}(sl(2))$ it satisfies:

(58) $\Delta(T(x))=\hat{R}^{-1}(T\otimes T)(\Delta(x))\hat{R}$

Define the quantum Weyl group element
$\displaystyle\hat{w}\in{\mathbb{U}}_{q}(sl(2))$ by Saito’s formula [64]:

(59)
$\hat{w}=e_{q^{-1}}(F)q^{-H^{2}/4}e_{q^{-1}}(-E)q^{-H^{2}/4}e_{q^{-1}}(F)q^{-H/2}.$

For every $\displaystyle x\in U_{q}(sl(2))$ it satisfies:

(60) $\displaystyle\displaystyle T(x)$
$\displaystyle\displaystyle=\hat{w}x\hat{w}^{-1},$ (61)
$\displaystyle\displaystyle\Delta(\hat{w})$
$\displaystyle\displaystyle=\hat{R}^{-1}(\hat{w}\otimes\hat{w}),$ (62)
$\displaystyle\displaystyle\hat{w}^{2}$
$\displaystyle\displaystyle=q^{H^{2}/2}\xi\theta,$

where $\displaystyle\theta\in{\mathbb{U}}_{q}(sl(2))$ is the ribbon element,
and $\displaystyle\xi\in{\mathbb{U}}_{q}(sl(2))$ is the central group element
whose value on the type $\displaystyle 1$ simple module $\displaystyle X$ of
$\displaystyle U_{q}(sl(2))$ of dimension $\displaystyle k+1$ is the scalar
endomorphism $\displaystyle(-1)^{k}id_{X}$.

In order to compare our setting to the one of [36] we need an explicit formula
of $\displaystyle\hat{w}$. Consider the type $\displaystyle 1$ simple module
$\displaystyle V_{r+1}$ of dimension $\displaystyle r+1$, and its basis
vectors $\displaystyle v_{0},\ldots,v_{r}$ such that:

$\displaystyle\begin{array}[]{l}K.v_{j}=\epsilon^{r-2j}v_{j},\\\
F.v_{j}=v_{j+1}\ \mathrm{if}\ j<r,\ F.v_{r}=0,\\\
E.v_{j}=[j]_{\epsilon}[r-j+1]_{\epsilon}v_{j-1}\ \mathrm{if}\ j>0,\
E.v_{0}=0.\end{array}$

Setting $\displaystyle{v}_{j}^{\prime}=v_{j}/[j]!$ and using (57), (60) and
(62), we obtain:

(63) $\hat{w}{v}_{j}^{\prime}=(-1)^{j}q^{-j(k-j-1)-k}{v}_{k-j}^{\prime}.$

In [36] another quantum Weyl group element $\displaystyle\underline{w}$ is
defined. It is dual to the Vaksman-Soibelman functional $\displaystyle
t\colon{\mathcal{O}}_{q}(SL_{2})\rightarrow\mathbb{C}(q)$ of [67, 66], so
$\displaystyle t(\alpha)=\langle\alpha,\underline{w}\rangle$,
$\displaystyle\alpha\in{\mathcal{O}}_{q}(SL_{2})$. By comparing (63) with the
formulas defining the action of $\displaystyle t$ in Section 1.7 of [36], we
find

$\displaystyle\underline{w}=\xi\hat{w}K$

and the basis vectors $\displaystyle w_{r}^{p}$ of [36] are related to the
vectors $\displaystyle v_{j}^{\prime}$ above as follows:

$\displaystyle{v}_{j}^{\prime}=\lambda_{j}w_{r}^{p}$

where $\displaystyle k=2p$, $\displaystyle j=p-r$,
$\displaystyle\lambda_{0}=1$, $\displaystyle\lambda_{1}=[k]q^{-k}$, and

$\displaystyle\lambda_{j}=\frac{[k]!}{[j]![k-(j-2)]!}q^{j(j+1)-j(k+2)},\quad
j\geq 2.$

Explicit formulas of the evaluation of $\displaystyle t$ on basis vectors of
$\displaystyle{\mathcal{O}}_{q}(SL_{2})$ can be computed. We get:

(64) $\displaystyle\displaystyle t(\tilde{a}^{\star m}\star\tilde{b}^{\star
n}\star\tilde{d}^{\star p})=\delta_{m,p}q^{-np}\prod_{i=1}^{p}(1-q^{-2i}),$
(65) $\displaystyle\displaystyle t(\tilde{a}^{\star m}\star\tilde{c}^{\star
n}\star\tilde{d}^{\star
p})=(-1)^{n}\delta_{m,p}q^{-n(p+1)}\prod_{i=1}^{p}(1-q^{-2i})$

where

(66) $\tilde{a}=a\ ,\ \tilde{b}=qb\ ,\ \tilde{c}=q^{-1}c\ ,\ \tilde{d}=d$

and as usual $\displaystyle a$, $\displaystyle b$, $\displaystyle c$,
$\displaystyle d$ are the standard generators of
$\displaystyle\mathcal{O}_{q}(SL_{2})$, ie. the matrix coefficients in the
basis of weight vectors $\displaystyle v_{0}$, $\displaystyle v_{1}=F.v_{0}$
of the $\displaystyle 2$-dimensional irreducible representation $\displaystyle
V_{2}$ of $\displaystyle U_{q}(sl(2))$. Here we have introduced the generators
$\displaystyle\tilde{a},\ldots,\tilde{d}$ to facilitate the comparison with
the formulas in [36]; these generators come naturally in their setup because
they use different generators $\displaystyle E_{i}$ and $\displaystyle F_{i}$
of $\displaystyle U_{q}(\mathfrak{g})$, which in our notations can be written
respectively as $\displaystyle K_{i}^{-1}E_{i}$ and $\displaystyle
F_{i}K_{i}$.

The formulas (64)-(65) can be shown by two independent methods. The first uses
a definition of $\displaystyle t$ as a $\displaystyle GNS$ state associated to
an infinite dimensional representation of
$\displaystyle{\mathcal{O}}_{q}(SL_{2})$, as recalled in Section 1.6 of [36].
The second is to write eg.

(67) $t(\tilde{a}^{\star m}\star\tilde{b}^{\star n}\star\tilde{d}^{\star
p})=\left\langle\tilde{a}^{\otimes m}\otimes\tilde{b}^{\otimes
n}\otimes\tilde{d}^{\otimes p},\Delta^{(m+n+p-1)}(\underline{w})\right\rangle$

and to use explicit expressions of
$\displaystyle\Delta^{(m+n+p-1)}(\underline{w})$ when represented on
$\displaystyle V_{2}^{\otimes(m+n+p)}$. In general one can check that

(68) $\Delta^{(n)}(\hat{\omega})=\left(\Delta^{(n-1)}\otimes
id\right)\left(\widehat{R}^{-1}\right)\left(\left(\Delta^{(n-2)}\otimes
id\right)\left(\widehat{R}^{-1}\right)\otimes id\right)\\\
\ldots\left(\left(\Delta\otimes id\right)\left(\widehat{R}^{-1}\right)\otimes
id^{\otimes(n-3)}\right)\left(\widehat{R}^{-1}\otimes
id^{\otimes(n-2)}\right)\hat{\omega}^{\otimes n}.$

By (63) or (64)-(65) we see that $\displaystyle\hat{w}$ (or
$\displaystyle\underline{w}$) and $\displaystyle t$ are well-defined on the
integral forms,

$\displaystyle\hat{w}\in\mathbb{U}_{\Gamma}\ ,\
t\colon{\mathcal{O}}_{A}(SL_{2})\rightarrow A.$

We now consider the case where $\displaystyle\mathfrak{g}$ is of rank
$\displaystyle m\geq 2$. To each simple root $\displaystyle\alpha_{i}$,
$\displaystyle 1\leq i\leq m$, is associated the subalgebra of $\displaystyle
U_{q}$ generated by $\displaystyle E_{i},F_{i},L_{i},L_{i}^{-1}$. It is a copy
of $\displaystyle U_{q_{i}}(sl(2))$, where $\displaystyle q_{i}=q^{d_{i}}$.
Let $\displaystyle\hat{w}_{i}$ be the corresponding quantum Weyl group element
in $\displaystyle\mathbb{U}_{q}=\mathbb{U}_{q}(\mathfrak{g})$, defined by
Saito’s formula (59), replacing $\displaystyle H$, $\displaystyle E$,
$\displaystyle F$ by $\displaystyle H_{i}$, $\displaystyle E_{i}$ and
$\displaystyle F_{i}$. Also, denote by
$\displaystyle\nu_{i}\colon\mathcal{O}_{q}\rightarrow\mathcal{O}_{q_{i}}(SL_{2})$
the projection map dual to the inclusion $\displaystyle
U_{q_{i}}(sl(2))\otimes_{\mathbb{C}(q_{i})}\mathbb{C}(q)\hookrightarrow U_{q}$
associated to $\displaystyle\alpha_{i}$, and put $\displaystyle
t_{i}=t\circ\nu_{i}$. Let $\displaystyle\underline{w}_{i}$ be the
corresponding quantum Weyl group element in $\displaystyle\mathbb{U}_{q}$, ie.
$\displaystyle t_{i}(\alpha)=\langle\alpha,\underline{w}_{i}\rangle$ for all
$\displaystyle\alpha\in{\mathcal{O}}_{q}$. On integral forms they yield well-
defined elements
$\displaystyle\hat{w}_{i},\underline{w}_{i}\in\mathbb{U}_{\Gamma}$ and
$\displaystyle t_{i}\colon{\mathcal{O}}_{A}\rightarrow A$ (see [36],
Proposition 5.1). They satisfy the defining relations of the braid group
$\displaystyle\mathcal{B}(\mathfrak{g})$ of $\displaystyle\mathfrak{g}$ [47]:

$\displaystyle\displaystyle\hat{w}_{i}\hat{w}_{j}\hat{w}_{i}$
$\displaystyle\displaystyle=\hat{w}_{j}\hat{w}_{i}\hat{w}_{j}\quad{\rm if\
}a_{ij}a_{ji}=1$ $\displaystyle\displaystyle(\hat{w}_{i}\hat{w}_{j})^{k}$
$\displaystyle\displaystyle=(\hat{w}_{j}\hat{w}_{i})^{k}\quad{\rm for\
}k=1,2,3\ {\rm if\ }a_{ij}a_{ji}=0,2,3$

and similarly by replacing $\displaystyle\hat{w}_{i}$ with
$\displaystyle\underline{w}_{i}$, or with $\displaystyle t_{i}$ (see [66] for
the latter). The Weyl group $\displaystyle W=W(\mathfrak{g})=N(T_{G})/T_{G}$
is generated by the reflexions $\displaystyle s_{i}$ associated to the simple
roots $\displaystyle\alpha_{i}$. Denote by $\displaystyle n_{i}\in N(T_{G})$ a
representative of $\displaystyle s_{i}$. Let $\displaystyle w\in W$ and denote
by $\displaystyle w=s_{i_{1}}\ldots s_{i_{k}}$ a reduced expression. Because
of the braid group relations the elements
$\displaystyle\hat{w}=\hat{w}_{i_{1}}\ldots\hat{w}_{i_{k}}$,
$\displaystyle\underline{w}=\underline{w}_{i_{1}}\ldots\underline{w}_{i_{k}}$
and the functional $\displaystyle t_{w}=t_{i_{1}}\ldots t_{i_{k}}$ do not
depend on the choice of reduced expression. The Lusztig [52] braid group
automorphism $\displaystyle T_{w}\colon\Gamma\rightarrow\Gamma$ associated to
$\displaystyle w$ satisfies (see [36]):

$\displaystyle T_{w}(x)=\hat{w}x\hat{w}^{-1},\quad x\in\Gamma.$

Let $\displaystyle w_{0}$ be the longest element in $\displaystyle W$. We have

(69) $\Delta(\hat{w}_{0})=\hat{R}^{-1}(\hat{w}_{0}\otimes\hat{w}_{0})$

where as usual $\displaystyle R={\Theta}{\hat{R}}$.

###  6.2. Regular action on $\displaystyle{\mathcal{O}}_{\epsilon}$

The following result is proved in Section 1.10 of [36]. For completeness let
us give a (different) proof. Recall from (33) that we may identify
$\displaystyle\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})$ with
$\displaystyle{\mathcal{O}}(G)$.

######  Proposition 6.1.

For every $\displaystyle f\in\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})$,
$\displaystyle g\in{\mathcal{O}}_{\epsilon}$ we have

(70) $\displaystyle\displaystyle t_{i}(f)=f(n_{i})$ (71)
$\displaystyle\displaystyle t_{i}(f\star g)=t_{i}(f)t_{i}(g).$

Proof. It is sufficient to prove the results for $\displaystyle SL_{2}$
because
$\displaystyle\nu_{i}:{\mathcal{O}}_{\epsilon}\rightarrow{\mathcal{O}}_{\epsilon}(SL_{2})$
is a morphism of Hopf algebras and
$\displaystyle\nu_{i}(\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon}))\subset\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon}(SL_{2}))$.
In this case (70) can be proved by using (64)-(65), evaluating $\displaystyle
t$ on basis elements of
$\displaystyle\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon}(SL_{2}))$ as is done in
Lemma 1.5 (a) of [36]. Such a basis is formed by monomials like in (64)-(65),
with all exponents divisible by $\displaystyle l$; then for instance

$\displaystyle t(\tilde{a}^{\star ml}\star\tilde{b}^{\star
nl}\star\tilde{d}^{\star
pl})=\delta_{p,0}\delta_{m,0}=\underline{a}^{m}\underline{b}^{n}\underline{d}^{p}(n)$

where $\displaystyle\underline{a},\ldots,\underline{d}$ are the generators of
$\displaystyle{\mathcal{O}}(G)={\mathcal{O}}_{1}(G)$ corresponding to
$\displaystyle a,\ldots,d$, and we take

$\displaystyle n=\left(\begin{array}[]{rr}0&1\\\ -1&0\end{array}\right)$

as representative of the reflexion $\displaystyle s$ generating the Weyl group
$\displaystyle W(sl(2))$. Here is an alternative proof of (70): (71) shows
that $\displaystyle t$ is a homomorphism on
$\displaystyle\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon}(SL_{2}))$, so by
proving (71) at first one is reduced to check (70) on the generators
$\displaystyle a^{\star l},\ldots,d^{\star l}$, which is easy by means of (67)
and (69).

We provide a proof of (71) that we find more conceptual than the one in Lemma
1.5 (b) of [36] (which uses again (64)-(65)). As above let us denote
$\displaystyle\underline{w}=\xi\hat{w}K$. For any $\displaystyle
f,g\in{\mathcal{O}}_{\epsilon}$ we have

$\displaystyle\displaystyle t(f\star g)$ $\displaystyle\displaystyle=(f\otimes
g)(\Delta(\underline{w}))$ $\displaystyle\displaystyle=(f\otimes
g)\left(\widehat{R}^{-1}(\underline{w}\otimes\underline{w})\right)$
$\displaystyle\displaystyle=\sum_{(\widehat{R}^{-1})}f\left((\widehat{R}^{-1})_{(1)}\underline{w}\right)g\left((\widehat{R}^{-1})_{(2)}\underline{w}\right)$
$\displaystyle\displaystyle=\sum_{(\widehat{R}^{-1}),(f)}f_{(1)}\left((\widehat{R}^{-1})_{(1)}\right)f_{(2)}(\underline{w})\
g\left((\widehat{R}^{-1})_{(2)}\underline{w}\right)$
$\displaystyle\displaystyle=\sum_{(f)}f_{(2)}(\underline{w})\
g\left((f_{(1)}\otimes id)(\widehat{R}^{-1})\underline{w}\right).$

Assume now $\displaystyle
f\in\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon}(SL_{2}))$. Since
$\displaystyle\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon}(SL_{2}))$ is a Hopf
subalgebra of $\displaystyle{\mathcal{O}}_{\epsilon}(SL_{2})$ we have
$\displaystyle f_{(1)}\in\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon}(SL_{2}))$.
From Theorem 2.15 (2) we deduce

$\displaystyle(f_{(1)}\otimes id)(\widehat{R}^{-1})\in
U_{\epsilon}(\mathfrak{n}_{-})\cap\mathcal{Z}_{0}(U_{\epsilon}).$

Denote by $\displaystyle z$ this element. Note that from its expression we
have $\displaystyle\epsilon(z)=\epsilon(f_{(1)})$. Now
$\displaystyle\textstyle
g(z\underline{w})=\sum_{(g)}g_{(1)}(z)g_{(2)}(\underline{w})$, but
$\displaystyle g_{(1)}$ is a linear combination of matrix elements of
$\displaystyle\Gamma$-modules, on which
$\displaystyle{\mathcal{Z}}_{0}(U_{\epsilon})$ acts by the trivial character.
Therefore

$\displaystyle\textstyle
g(z\underline{w})=\sum_{(g)}\epsilon(z)g_{(1)}(1)g_{(2)}(\underline{w})=\epsilon(z)g(\underline{w})=\epsilon(f_{(1)})g(\underline{w})$

and eventually

$\displaystyle t(f\star
g)=\sum_{(f)}f_{(2)}(\underline{w})\epsilon(f_{(1)})g(\underline{w})=t(f)t(g).$

This concludes the proof. $\displaystyle\Box$

For the sake of completeness, let us show how this result implies:

Proof of Proposition 2.16 (ie. Proposition 7.1 of [36]). We have
$\displaystyle\textstyle f\lhd t_{i}=\sum_{(f)}t_{i}(f_{(1)})f_{(2)}$,
$\displaystyle f\in\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})$. Since
$\displaystyle\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})$ is a Hopf subalgebra
of $\displaystyle{\mathcal{O}}_{\epsilon}$, $\displaystyle
f_{(2)}\in\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})$ and therefore the maps
$\displaystyle\lhd
t_{i}\colon{\mathcal{O}}_{\epsilon}\rightarrow{\mathcal{O}}_{\epsilon}$
preserve $\displaystyle\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})$. Moreover,
$\displaystyle\textstyle(f\lhd
t_{i})(a)=\sum_{(f)}f_{(1)}(n_{i})f_{(2)}(a)=f(n_{i}a)$, $\displaystyle a\in
G$, by (70).

It remains to show that $\displaystyle(f\star\alpha)\lhd t_{i}=(f\lhd
t_{i})(\alpha\lhd t_{i})$ for every $\displaystyle
f\in\mathcal{Z}_{0}({\mathcal{O}}_{\epsilon})$,
$\displaystyle\alpha\in\mathcal{O}_{\epsilon}$. We have

$\displaystyle\displaystyle\textstyle(f\star g)\lhd t_{i}=\sum_{(f\star
g)}t_{i}\left((f\star g)_{(1)}\right)(f\star g)_{(2)}$
$\displaystyle\displaystyle=\sum_{(f),(g)}t_{i}\left(f_{(1)}\star
g_{(1)}\right)f_{(2)}\star g_{(2)}$
$\displaystyle\displaystyle=\sum_{(f),(g)}t\left(\nu_{i}(f_{(1)})\nu_{i}(g_{(1)})\right)f_{(2)}\star
g_{(2)}$ (72)
$\displaystyle\displaystyle=\sum_{(f),(g)}t\left(\nu_{i}(f_{(1)})\right)t\left(\nu_{i}(g_{(1)})\right)f_{(2)}\star
g_{(2)}$

using that $\displaystyle\nu_{i}$ is a homomorphism in the third equality, and
(71) in the last one. The result is just $\displaystyle(f\lhd t_{i})(g\lhd
t_{i})$. $\displaystyle\Box$

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