Solution of a $q$-difference Noether problem and the quantum Gelfand–Kirillov conjecture for $\mathfrak gl _N$

Abstract

It is shown that the $q$-difference Noether problem for all classical Weyl groups has a positive solution, simultaneously generalizing well known results on multisymmetric functions of Mattuck (Proc Am Math Soc 19:764–765, 1968) and Miyata (Nagoya Math J 41:69–73, 1971) in the case $q=1$, and $q$-deforming the noncommutative Noether problem for the symmetric group (Futorny et al. in Adv Math 223:773–796, 2010). It is also shown that the quantum Gelfand–Kirillov conjecture for $\mathfrak gl _N$ (for a generic $q$) follows from the positive solution of the $q$-difference Noether problem for the Weyl group of type $D_n$. The proof is based on the theory of Galois rings (Futorny and Ovsienko in J Algebra 324:598–630, 2010). From here we obtain a proof of the quantum Gelfand–Kirillov conjecture for $\mathfrak gl _N$, and for a certain extension of $\mathfrak sl _N$. Previously, the case of $\mathfrak sl _N$ was shown by Fauquant-Millet (J Algebra 218:93–116, 1999) and by Alev and Dumas (J Algebra 170:229–265, 1994) (for $N=2,3$). Moreover, we give an explicit description of the skew fields of fractions for $U_q(\mathfrak gl _N)$ and $U_q^\mathrm{ext}(\mathfrak sl _N)$ which generalizes the results of Alev and Dumas (J Algebra 170:229–265, 1994).

Appendix

1.1 Proof of Proposition 3.1

The statement is equivalent to proving that

$$\begin{aligned}{}[P(X),P(Y)]=0 \end{aligned}$$

(8.1)

in $C_n^q[X,Y]$. Put

$$\begin{aligned} Q_j(X)= \left( \prod _{k\in \{1,\ldots ,n\}\setminus \{j\}} \frac{X-x_k}{x_j-x_k}\right) y_j. \end{aligned}$$

(8.2)

so that $P(X)=\sum _{i=1}^n Q_i(X)$. Observe that

$$\begin{aligned} w\big (Q_j(X)\big ) = Q_{w(j)}(X),\qquad w\in S_n. \end{aligned}$$

(8.3)

Thus, to prove (8.1), it is enough to show the following two identities:

$$\begin{aligned}{}[Q_1(X),Q_1(Y)]&= 0,\end{aligned}$$

(8.4)

$$\begin{aligned}{}[Q_1(X),Q_2(Y)]+[Q_2(X),Q_1(Y)]&= 0. \end{aligned}$$

(8.5)

Since $y_1 x_i = q^{\delta _{i1}} x_i y_1$ we have

$$\begin{aligned} Q_1(X)Q_1(Y)&= \prod _{k=2}^n \frac{X-x_k}{x_1-x_k}y_1 \prod _{k=2}^n\frac{Y-x_k}{x_1-x_k}y_1\\&= \prod _{2\le k\le n} \frac{(X-x_k)(Y-x_k)}{(x_1-x_k)(q x_1-x_k)}y_1^2 \end{aligned}$$

which is symmetric in $X,Y$. This proves (8.4).

Next we prove (8.5). Let

$$\begin{aligned} R_j(X)=\prod _{k=3}^n \frac{X-x_k}{x_j-x_k},\quad j=1,2. \end{aligned}$$

Then

$$\begin{aligned} Q_1(X)&= \frac{X-x_2}{x_1-x_2} R_1(X) y_1,\qquad Q_2(X)=\frac{X-x_1}{x_2-x_1} R_2(X) y_2,\\ {[}y_1, R_2(X)]&= [y_2,R_1(X)]=0, \end{aligned}$$

and

$$\begin{aligned} R_1(X)R_2(Y)=R_1(Y)R_2(X)=R_2(X)R_1(Y)=R_2(Y)R_1(X). \end{aligned}$$

We have

$$\begin{aligned}&[Q_1(X),Q_2(Y)]+[Q_2(X),Q_1(Y)]= Q_1(X)Q_2(Y)-Q_1(Y)Q_2(X)\\&\qquad + Q_2(X)Q_1(Y) - Q_2(Y)Q_1(X) \\&\quad =\frac{X-x_2}{x_1-x_2} \cdot \frac{Y-qx_1}{x_2-qx_1} R_1(X)R_2(Y) y_1y_2 -\frac{Y-x_2}{x_1-x_2} \cdot \frac{X-qx_1}{x_2-qx_1} R_1(Y)R_2(X) y_1y_2\\&\qquad + \frac{X-x_1}{x_2-x_1} \cdot \frac{Y-qx_2}{x_1-qx_2} R_2(X)R_1(Y) y_1y_2 - \frac{Y-x_1}{x_2-x_1} \cdot \frac{X-qx_2}{x_1-qx_2} R_2(Y)R_1(X) y_1y_2\\&\quad =\Bigg (\frac{(X-x_2)(Y-qx_1)-(Y-x_2)(X-qx_1)}{(x_1-x_2)(x_2-qx_1)}\\&\qquad \qquad + \frac{(X-x_1)(Y-qx_2)-(Y-x_1)(X-qx_2)}{(x_2-x_1)(x_1-qx_2)} \Bigg ) R_1(X)R_2(Y) y_1y_2 \\&\quad =\Bigg (\frac{(x_2^2-q^2x_1^2)X - (x_2-qx_1)Y}{(x_1-x_2)(x_2-qx_1)} + \frac{(x_1-qx_2)X-(x_1-qx_2)Y}{(x_2-x_1)(x_1-qx_2)}\Bigg ) R_1(X)R_2(Y)y_1y_2\\&\quad =0 \end{aligned}$$

This shows (8.5) and completes the proof that $[t_i,t_j]=0$ for all $i,j$.

1.2 Proof of Proposition 3.7

The relation (3.15) holds by Proposition 3.1, while (3.16) holds by the definition, (3.11), of $e_d$. Relation (3.17) is trivial for $k=0$.

Using (3.12) and that $w(e_k)=e_k$ for any $w\in S_n$ we have, for any $j,k\in \{1,\ldots ,n\}$,

$$\begin{aligned} (-1)^{j-1}\Delta \cdot t_je_k=\sum _{w\in S_n} {{\mathrm{sgn}}}(w)w\Big (x_1^{n-2}x_2^{n-3}\cdots x_{n-2} e_{n-j}^{\prime }e_k(x_1,\ldots ,x_{n-1},qx_n)y_n\Big ) \end{aligned}$$

Substituting $y_n=t_1+x_nt_2+\cdots +x_n^{n-1}t_n$ and using that $w(t_i)=t_i$ for all $w\in S_n$ we get

$$\begin{aligned}&(-1)^{j-1}\Delta \cdot t_je_k \\&\quad =\sum _{i=1}^n\sum _{w\in S_n} {{\mathrm{sgn}}}(w)w\Big (x_1^{n-2}x_2^{n-3}\cdots x_{n-2} \cdot x_n^{i-1}e_{n-j}^{\prime }e_k(x_1,\ldots ,x_{n-1},qx_n)\Big )t_i. \end{aligned}$$

Write $e_{n-j}^{\prime }$ as a sum of monomials $x_{i_1}\cdots x_{i_{n-j}}$ and $1\le i_1<\cdots <i_{n-j}\le n-1$. We claim that the only way to get a nonzero contribution is when $i_r=r$ for all $r$. Indeed, suppose $i_r>r$ for some $r$ chosen minimal. Then the product

$$\begin{aligned} x_1^{n-2}x_2^{n-3}\cdots x_{n-2} \cdot x_{i_1}\cdots x_{i_{n-j}}\cdot x_n^{i-1}e_k(x_1,\ldots ,x_{n-1},qx_n) \end{aligned}$$

will be fixed by the transposition $(i_r-1\;\;i_r)$. Therefore, after anti-symmetrization, the term will cancel out. In other words, the substitution $w\mapsto w\cdot (i_r-1\;\; i_r)$ in the sum

$$\begin{aligned} \sum _{w\in S_n} {{\mathrm{sgn}}}(w)w\Big (x_1^{n-2}x_2^{n-3}\cdots x_{n-2} \cdot x_{i_1}\cdots x_{i_{n-j}}\cdot x_n^{i-1}e_k(x_1,\ldots ,x_{n-1},qx_n)\Big )t_i \end{aligned}$$

gives the same expression with opposite sign, proving it is zero. Thus, noting also that

$$\begin{aligned} e_k(x_1,\ldots ,x_{n-1},qx_n)=e_k^{\prime } + qx_n e_{k-1}^{\prime }, \end{aligned}$$

we have

$$\begin{aligned}&(-1)^{j-1}\Delta \cdot t_je_k\nonumber \\&\quad =\sum _{i=1}^n\sum _{w\in S_n} {{\mathrm{sgn}}}(w)w\Big (x_1^{n-1}\cdots x_{n-j}^j\cdot x_{n-j+1}^{j-2}\cdots x_{n-2} x_n^{i-1} (e_k^{\prime }+qx_ne_{k-1}^{\prime })\Big )t_i.\qquad \end{aligned}$$

(8.6)

The term $i=j$: Write $e_k^{\prime }= \sum _{1\le i_1<\cdots <i_k\le n-1} x_{i_1}\cdots x_{i_k}$. Consider

$$\begin{aligned} x_1^{n-1}\cdots x_{n-j}^j x_{n-j+1}^{j-2}\cdots x_{n-2} x_n^{j-1} \cdot x_{i_1}\cdots x_{i_k} \end{aligned}$$

An expression like this containing factors $(x_rx_{r^{\prime }})^s$ ($r\ne r^{\prime }$) will become zero after anti-symmetrization. If $n-j\ge k$ there is a unique way to get a nonzero result, namely to choose $(i_1,\ldots ,i_k)=(1,2,\ldots ,k)$. If $n-j<k$ there is no way to get nonzero result. Thus

$$\begin{aligned}&\sum _{w\in S_n} {{\mathrm{sgn}}}(w)w\Big (x_1^{n-1}\cdots x_{n-j}^j\cdot x_{n-j+1}^{j-2}\cdots x_{n-2} x_n^{j-1}e_k^{\prime }\Big ) \\&\quad ={\left\{ \begin{array}{ll} a(n,n-1,\ldots ,n-k+1,n-k-1,\ldots ,j,j-2,\ldots ,1,0,j-1),&{} j+k\le n\\ 0,&{} j+k>n \end{array}\right. } \end{aligned}$$

where $a(i_1,\ldots ,i_n):=\sum _{w\in S_n}{{\mathrm{sgn}}}(w)w(x_1^{i_1}\cdots x_n^{i_n})$. Use that $w(a(i_1,\ldots ,i_n))={{\mathrm{sgn}}}(w)a$ $(i_1,\ldots ,i_n)$ with

$$\begin{aligned} w=(n-j+1\quad n-j+2\quad \cdots \quad n), \end{aligned}$$

which is a cycle of length $j$, to get

$$\begin{aligned}&a(n,n-1,\ldots ,n-k+1,n-k-1,\ldots ,j,j-2,\ldots ,1,0,j-1)\\&\quad =(-1)^{j-1}a(n,n-1,\ldots ,n-k+1,n-k-1,\ldots ,0). \end{aligned}$$

Using that the Schur function

$$\begin{aligned} s_\lambda = a(\lambda _1+n-1,\lambda _2+n-2,\ldots ,\lambda _n)/a(n-1,n-2,\ldots ,0), \end{aligned}$$

defined for a partition $\lambda =(\lambda _1,\ldots ,\lambda _n)$, $\lambda _1\ge \cdots \ge \lambda _n\ge 0$, satisfies $s_{1^k0^{n-k}}=e_k$ and that $\Delta =a(n-1,n-2,\ldots ,0)$ we get that

$$\begin{aligned}&\sum _{w\in S_n} {{\mathrm{sgn}}}(w)w\Big (x_1^{n-1}\cdots x_{n-j}^j\cdot x_{n-j+1}^{j-2}\cdots x_{n-2} x_n^{j-1}e_k^{\prime }\Big ) \nonumber \\&\quad = {\left\{ \begin{array}{ll} (-1)^{j-1} \Delta \cdot e_k, &{} j+k\le n,\\ 0,&{} j+k>n. \end{array}\right. } \end{aligned}$$

(8.7)

Similarly, if we look at the term containing $qx_ne_{k-1}^{\prime }$, there is at most one tuple $(i_1,\ldots ,i_{k-1})$, $1\le i_1<\cdots <i_{k-1}\le n-1$ such that the antisymmetrization of

$$\begin{aligned} qx_1^{n-1}\cdots x_{n-j}^j x_{n-j+1}^{j-2}\cdots x_{n-2}x_n^j x_{i_1}\cdots x_{i_{k-1}} \end{aligned}$$

is nonzero, namely $(i_1,\ldots ,i_{k-1})=(1,\ldots ,k-1)$ and this time, due to the presence of $x_n^j$, it gives nonzero result if and only if $k-1\ge n-j$ i.e. $j+k>n$. Thus

$$\begin{aligned}&\sum _{w\in S_n} {{\mathrm{sgn}}}(w)w\Big (x_1^{n-1}\cdots x_{n-j}^j\cdot x_{n-j+1}^{j-2}\cdots x_{n-2} \cdot qx_n^je_{k-1}^{\prime }\Big ) \\&\quad ={\left\{ \begin{array}{ll} 0,&{} j+k\le n\\ qa(n,n-1,\ldots ,j+1,j-1,\ldots ,n-k+1,n-k-1,\ldots ,1,0,j),&{} j+k>n \end{array}\right. } \end{aligned}$$

To get a descending sequence inside the parenthesis we apply the cyclic permutation which places $j$ between $j-1$ and $j+1$. This cycle has length $j$, giving a factor $(-1)^{j-1}$. As before, this gives

$$\begin{aligned}&\sum _{w\in S_n} {{\mathrm{sgn}}}(w)w\Big (x_1^{n-1}\cdots x_{n-j}^j\cdot x_{n-j+1}^{j-2}\cdots x_{n-2} \cdot qx_n^je_{k-1}^{\prime }\Big ) \nonumber \\&\quad = {\left\{ \begin{array}{ll} 0,&{} j+k\le n\\ (-1)^{j-1}q \Delta \cdot e_k, &{} j+k> n,\\ \end{array}\right. } \end{aligned}$$

(8.8)

Combining (8.7) and (8.8) yields

$$\begin{aligned}&(-1)^{j-1}\Delta \cdot (t_je_k - q^{\delta _{j+k>n}} e_kt_j) \nonumber \\&\quad = \sum _{i\in \{1,\ldots ,n\}\setminus \{j\}}\sum _{w\in S_n} {{\mathrm{sgn}}}(w)w\Big (x_1^{n-1}\cdots x_{n-j}^j\cdot x_{n-j+1}^{j-2}\cdots x_{n-2} x_n^{i-1} (e_k^{\prime }+qx_ne_{k-1}^{\prime })\Big )t_i. \nonumber \\ \end{aligned}$$

(8.9)

The terms where $i>j$: We first look at the $e_k^{\prime }$ term in (8.9). That $i>j$ means the exponent $i-1$ of $x_n$ occurs in one of the exponents in $x_1^{n-1}x_2^{n-2}\cdots x_{n-j}^j$, namely in $x_{n-(i-1)}^{i-1}$. Therefore $x_{i_1}\cdots x_{i_k}$ must contain $x_1x_2\cdots x_{n-(i-1)}$. In particular $k\ge n-(i-1)$. The remaining factors must be $x_{n-j+1}x_{n-j+2}\cdots$ and they cannot continue beyond $x_{n-1}$ meaning that $k-(n-i+1) + (n-j) \le n-1$. Thus the following inequalities are necessary conditions in order to avoid having two variables with the same exponent:

$$\begin{aligned} k\ge n-i+1,\quad \text{ and } \quad k+i-j-1\le n-1, \end{aligned}$$

i.e.

$$\begin{aligned} n-k+1\le i\le n-k+j. \end{aligned}$$

If these inequalities hold there is a unique tuple

$$\begin{aligned} (i_1,\ldots ,i_k)=(1,2,\ldots ,n-i+1, n-j+1,n-j+2,\ldots ,k+i-j-1) \end{aligned}$$

with $1\le i_1<\cdots <i_k\le n-1$ such that

$$\begin{aligned} \sum _{w\in S_n} {{\mathrm{sgn}}}(w)w\Big (x_1^{n-1}\cdots x_{n-j}^j\cdot x_{n-j+1}^{j-2}\cdots x_{n-2}^2 x_n^{i-1} \cdot x_{i_1}\cdots x_{i_k}\Big ). \end{aligned}$$

is nonzero. With this choice we get

$$\begin{aligned}&\sum _{w\in S_n} {{\mathrm{sgn}}}(w)w\Big (x_1^{n-1}\cdots x_{n-j}^j\cdot x_{n-j+1}^{j-2}\cdots x_{n-2} x_n^{i-1} \cdot x_{i_1}\cdots x_{i_k}\Big )\nonumber \\&\quad =a(n,\ldots ,i,i-2,\ldots ,n-(k+i-j)+1,n-(k+i-j)-1,\ldots ,0,i-1)\nonumber \\&\quad = (-1)^i a(n,n-1,\ldots ,n-(k+i-j)+1,n-(k+i-j)-1,\ldots ,0) \nonumber \\&\quad = (-1)^i \Delta \cdot e_{k+i-j} \end{aligned}$$

(8.10)

where we applied the cyclic permutation $(n-i+2\;\; n-i+3\;\; \cdots \;\; n-1\;\; n)$ of length $i-1$ in the second equality.

The argument for the term containing $qx_1e_{k-1}^{\prime }$ is analogous, but gives an extra minus sign. Together with (8.10) one obtains that for $i>j$ we have

$$\begin{aligned}&\sum _{w\in S_n} {{\mathrm{sgn}}}(w)w\Big (x_1^{n-1}\cdots x_{n-j}^j\cdot x_{n-j+1}^{j-2}\cdots x_{n-2} x_n^{i-1} (e_k^{\prime }+qx_ne_{k-1}^{\prime })\Big )t_i\nonumber \\&\quad ={\left\{ \begin{array}{ll} (-1)^{i+1}(q-1)\Delta \cdot e_{k+i-j}t_i,&{} n-k+1\le i\le n-k+j,\\ 0,&{}\text{ otherwise }. \end{array}\right. } \end{aligned}$$

(8.11)

The terms where $i<j$: We look at the $e_k^{\prime }$ term in (8.9). Necessary conditions for nonzero contribution are $k\ge j-i$ and $k-(j-i)\le n-j$, i.e.

$$\begin{aligned} j-k\le i \le n-k. \end{aligned}$$

(8.12)

After a similar computation as the $i>j$ case we obtain

$$\begin{aligned}&\sum _{w\in S_n} {{\mathrm{sgn}}}(w)w\Big (x_1^{n-1}\cdots x_{n-j}^j\cdot x_{n-j+1}^{j-2}\cdots x_{n-2} x_n^{i-1} (e_k^{\prime }+qx_ne_{k-1}^{\prime })\Big )t_i\nonumber \\&\quad ={\left\{ \begin{array}{ll} (-1)^i(q-1)\Delta \cdot e_{k+i-j}t_i , &{} j-k\le i\le n-k\\ 0,&{} \text{ otherwise }. \end{array}\right. } \end{aligned}$$

(8.13)

Combining (8.13), (8.11) and (8.9) we obtain

$$\begin{aligned}&t_je_k - q^{\delta _{j+k>n}}e_kt_j =(q-1)\sum _{\begin{array}{c} i>j\\ n-k+1\le i\le n-k+j \end{array}} (-1)^{j-1+i+1} e_{k+i-j} t_i \nonumber \\&\quad + (q-1)\sum _{\begin{array}{c} i<j\\ j-k\le i\le n-k \end{array}} (-1)^{j-1+i} e_{k+i-j}t_i \end{aligned}$$

(8.14)

Making the change of summation variables $i\mapsto i+j$ we get

$$\begin{aligned}&t_je_k - q^{\delta _{j+k>n}}e_kt_j =(q-1)\sum _{\begin{array}{c} i>0\\ n-(j+k)+1\le i\le n-k \end{array}} (-1)^{i+\delta _{i<0}} e_{k+i} t_{j+i} \nonumber \\&\quad + (q-1)\sum _{\begin{array}{c} i<0\\ -k\le i\le n-(j+k) \end{array}} (-1)^{i+\delta _{i<0}} e_{k+i}t_{j+i}. \end{aligned}$$

(8.15)

In the first sum, the condition $i\le n-k$ is redundant since, by the notational convention, $e_{k+i}=0$ for $i>n-k$. Similarly, $-k\le i$ is superfluous in the second sum. Thus we obtain (3.17).

1.3 Proof of Proposition 3.9

First note that (3.24) implies that

$$\begin{aligned}{}[T_j,E_0]&= 0,\quad \forall j\in [\![{1},{n}]\!],\end{aligned}$$

(8.16)

$$\begin{aligned}{}[T_j, E_n]_q&= 0,\quad \forall j\in [\![{1},{n}]\!]. \end{aligned}$$

(8.17)

We now prove (3.29). Let $j\in [\![{1},{n-1}]\!]$ and $k\in [\![{0},{n-1}]\!]$. Then the left hand side of (3.29) equals

$$\begin{aligned}&[\widetilde{E}_k, \widetilde{T}_j]_{q^{\delta _{j+k>n-1}}} = [T_{k+1}, E_jT_1T_n-(-1)^jE_0T_{n-j}T_1-(-1)^{n-j}E_nT_{n+1-j}T_n]_{q^{\delta _{j+1+k>n}}} \nonumber \\ \end{aligned}$$

(8.18)

$$\begin{aligned}&\quad =(q-1)\sum _{i\in \mathbb{Z }\setminus I(n-1-j-k)} (-1)^{i+\delta _{i<0}} E_{j+i} T_{k+1+i} T_1T_n \end{aligned}$$

(8.19)

$$\begin{aligned}&\qquad -(1-q^{\delta _{j+k+1>n}})(-1)^jE_0T_{k+1}T_{n-j}T_1\end{aligned}$$

(8.20)

$$\begin{aligned}&\qquad -(q-q^{\delta _{j+k+1>n}})(-1)^{n-j}E_nT_{k+1}T_{n+1-j}T_n. \end{aligned}$$

(8.21)

By (3.19) with $(j,k)$ replaced by $(k+1,n-j)$, the term (8.20) equals

$$\begin{aligned} -(q-1)\sum _{i\in \mathbb{Z }\setminus I(n-1-j-k)}(-1)^{j+\delta _{i<0}}E_0T_{k+1+i}T_{n-j-i}T_1. \end{aligned}$$

(8.22)

Similarly, applying (3.20) with $(j,k)$ replaced by $(k+1,n+1-j)$ shows that (8.21) is equal to

$$\begin{aligned} -(q-1)\sum _{i\in \mathbb{Z }\setminus I(n-1-j-k)}(-1)^{n-j+\delta _{i<0}}E_nT_{k+1+i}T_{n+1-j-i}T_n. \end{aligned}$$

(8.23)

Adding together (8.22), (8.23) and (8.19) gives the right hand side of (3.29). This proves (3.29).

In particular, taking $k=0$ and $k=n-1$ in (3.29) we get

$$\begin{aligned} T_1\widetilde{T}_j-\widetilde{T}_jT_1&= 0,\end{aligned}$$

(8.24)

$$\begin{aligned} T_n\widetilde{T}_j-q\widetilde{T}_jT_n&= 0, \end{aligned}$$

(8.25)

for all $j\in [\![{1},{n-1}]\!]$. Using these identities, together with $[T_j,E_0]=0$ and $[T_j,E_n]_q=0$ which follow from (3.24), one can check that

$$\begin{aligned} q\widetilde{T}_j=T_1T_n\widetilde{T}_j(T_1T_n)^{-1}= T_nT_1E_j-(-1)^{j}T_1T_{n-j}E_0-(-1)^{n-j}T_nT_{n+1-j}E_n, \end{aligned}$$

proving (3.30).

That (3.28) holds is trivial from the assumption (3.22).

We now prove (3.27). Let $j,k\in [\![{1},{n-1}]\!]$. We will bring $\widetilde{T}_j\widetilde{T}_k$ to the normal form where all the $E$’s are to the left of all the $T$’s and prove that the resulting expression is symmetric in $j,k$. We may assume $j\ne k$. Using (8.24), (8.25) and (3.29), we have

$$\begin{aligned} \widetilde{T}_j\widetilde{T}_k&= (E_jT_1T_n-(-1)^jE_0T_{n-j}T_1-(-1)^{n-j}E_nT_{n+1-j}T_n)\widetilde{T}_k\nonumber \\&= qE_j\widetilde{T}_kT_1T_n-(-1)^jE_0T_{n-j}\widetilde{T}_kT_1-(-1)^{n-j}qE_nT_{n+1-j}\widetilde{T}_kT_n\nonumber \\&= qE_j\widetilde{T}_kT_1T_n\nonumber \\&\quad -(-1)^jE_0\Big (q^{\delta _{-j+k>0}}\widetilde{T}_k T_{n-j}+(q-1)\sum _{i\in \mathbb{Z }\setminus I(j-k)}(-1)^{i+\delta _{i<0}}\widetilde{T}_{k+i}T_{n-j+i}\Big )T_1\nonumber \\&\quad -(-1)^{n-j}qE_n\Big (q^{\delta _{1-j+k>0}}\widetilde{T}_kT_{n+1-j} \nonumber \\&\qquad +(q-1)\sum _{i\in \mathbb{Z }\setminus I(-1+j-k)} (-1)^{i+\delta _{i<0}} \widetilde{T}_{k+i} T_{n+1-j+i}\Big )T_n\nonumber \\&= qE_jE_kT_1^2T_n^2 - (-1)^kqE_jE_0T_{n-k}T_1^2T_n -(-1)^{n-k}qE_jE_nT_{n+1-k}T_1T_n^2\nonumber \\&\quad -(-1)^jq^{\delta _{k>j}} E_0E_kT_{n-j}T_1^2T_n +(-1)^{j+k}q^{\delta _{k>j}}E_0^2T_1^2 T_{n-k}T_{n-j} \nonumber \\&\qquad \qquad +(-1)^{n-k+j}q^{\delta _{k>j}} E_0E_nT_{n-j}T_{n+1-k}T_1T_n\nonumber \\&\quad -(-1)^j(q-1)\sum _{i\in \mathbb{Z }\setminus I(j-k)}(-1)^{i+\delta _{i<0}} E_0\Big (E_{k+i}T_1T_n \nonumber \\&\qquad \qquad -(-1)^{k+i}E_0T_{n-k-i}T_1-(-1)^{n-k-i}E_nT_{n+1-k-i}T_n\Big )T_{n-j+i}T_1\nonumber \\&\quad -(-1)^{n-j}q^{1+\delta _{k\ge j}}E_nE_k T_1T_n T_{n+1-j}T_n + (-1)^{n-j+k}q^{1+\delta _{k\ge j}}E_nE_0T_{n-k}T_1T_{n+1-j}T_n\nonumber \\&\qquad \qquad + (-1)^{2n-j-k}q^{1+\delta _{k\ge j}} E_n^2T_{n+1-k}T_n^2T_{n+1-j}\nonumber \\&\quad -(-1)^{n-j}q(q-1)\sum _{i\in \mathbb{Z }\setminus I(-1+j-k)} (-1)^{i+\delta _{i<0}}E_n\Big (E_{k+i}T_1T_n\nonumber \\&\qquad \qquad -(-1)^{k+i}E_0T_{n-k-i}T_1-(-1)^{n-k-i}E_nT_{n+1-k-i}T_n\Big )T_{n+1-j+i}T_n. \end{aligned}$$

(8.26)

We prove that all parts of this expression are symmetric in $j,k$. The first term, containing $E_jE_k$, is trivially symmetric.

The terms containing $E_0^2T_1^2$. There are two terms in (8.26) containing $E_0^2T_1^2$:

$$\begin{aligned} (-1)^{j+k}q^{\delta _{k>j}}E_0^2T_{n-k}T_{n-j}T_1^2 + (-1)^{j+k}(q-1)\sum _{i\in \mathbb{Z }\setminus I(j-k)} (-1)^{\delta _{i<0}} E_0^2T_1^2T_{n-k-i}T_{n-j+i}. \nonumber \\ \end{aligned}$$

(8.27)

Applying (3.19) with $(j,k)$ replaced by $(n-j,n-k)$ we get that (8.27) equals

$$\begin{aligned} (-1)^{j+k}E_0^2T_{n-j}T_{n-k}T_1^2 \end{aligned}$$

which is symmetric in $j,k$.

The terms containing $E_n^2T_n^2$.

$$\begin{aligned}&(-1)^{2n-j-k}q^{1+\delta _{k\ge j}} E_n^2T_n^2T_{n+1-k}T_{n+1-j} \nonumber \\&\quad +(-1)^{2n-j-k}q(q-1)\sum _{i\in \mathbb{Z }\setminus I(-1+j-k)} (-1)^{\delta _{i<0}} E_n^2T_n^2T_{n+1-k-i}T_{n+1-j+i} \end{aligned}$$

(8.28)

Here we can apply (3.20) with $(j,k)$ replaced by $(n+1-j,n+1-k)$ to see that (8.28) equals

$$\begin{aligned} (-1)^{j+k}q^2 E_n^2T_n^2T_{n+1-j}T_{n+1-k} \end{aligned}$$

which is symmetric in $j,k$.

The terms containing $E_0T_1^2T_n$.

$$\begin{aligned}&-E_0\Big ((-1)^kqE_jT_{n-k} +(-1)^jq^{\delta _{k>j}} E_0E_k T_{n-j}\nonumber \\&\quad +(-1)^j(q-1)\sum _{i\in \mathbb{Z }\setminus I(j-k)} (-1)^{i+\delta _{i<0}} E_0E_{k+i}T_{n-j+i}\Big )T_1^2T_n \end{aligned}$$

(8.29)

The parenthesis equals

$$\begin{aligned}&(-1)^kE_j T_{n-k} + (-1)^j E_kT_{n-j} \nonumber \\&\quad +(-1)^k(q-1)E_jT_{n-k} + (-1)^j(q^{\delta _{k>j}}-1)E_kT_{n-j}\nonumber \\&\quad +(-1)^j(q-1)\sum _{i\in \mathbb{Z }\setminus I(j-k)} (-1)^{i+\delta _{i<0}} E_{k+i} T_{n-j+i}. \end{aligned}$$

(8.30)

If $j\ge k$, we can include $(-1)^k(q-1)E_jT_{n-k}$ as the term $i=j-k$ in the sum. If $j<k$, the term $(-1)^k(q-1)E_jT_{n-k}$ cancels the term $i=j-k$ in the sum, and $(-1)^j(q^{\delta _{k>j}}-1)E_kT_{n-j}$ may be included in the sum as $i=0$. Thus (8.30) can be written

$$\begin{aligned}&(-1)^kE_jT_{n-k}+(-1)^jE_kT_{n-j}\nonumber \\&\quad +(-1)^j(q-1)\sum \limits _{{i\in {\left\{ \begin{array}{ll} \mathbb{Z }\setminus [\![{0},{j-k-1}]\!],&{} j\ge k\\ \mathbb{Z }\setminus [\![{j-k},{-1}]\!],&{} j<k \end{array}\right. }}} (-1)^{i+\delta _{i<0}} E_{k+i}T_{n-j+i}. \end{aligned}$$

(8.31)

Making the change of variables $i\mapsto i+j-k$ in this sum gives the same expression but with $j$ and $k$ interchanged. Thus it is symmetric in $j$ and $k$.

The terms containing $E_nT_1T_n^2$.

$$\begin{aligned}&-qE_n\Big ((-1)^{n-k}E_jT_{n+1-k} +(-1)^{n-j}q^{\delta _{k\ge j}} E_k T_{n+1-j} \nonumber \\&\quad \quad +(-1)^{n-j}(q-1)\sum _{i\in \mathbb{Z }\setminus I(-1+j-k)} (-1)^{i+\delta _{i< 0}} E_{k+i} T_{n+1-j+i} \Big )T_1T_2^2 \end{aligned}$$

(8.32)

Similarly to the previous case, the expression inside the parenthesis can be written as

$$\begin{aligned}&(-1)^{n-k}q E_jT_{n+1-k} + (-1)^{n-j} q E_kT_{n+1-j}\nonumber \\&\quad +(-1)^{n-j}(q-1)\sum _{\begin{array}{c} i\in {\left\{ \begin{array}{ll} \mathbb{Z }\setminus [\![{1},{j-k}]\!],&{}j>k\\ \mathbb{Z }\setminus [\![{j-k+1},{0}]\!],&{}j\le k \end{array}\right. } \end{array}} (-1)^{i+\delta _{i\le 0}} E_{k+i}T_{n+1-j+i}.\qquad \end{aligned}$$

(8.33)

Substituting $i\mapsto i-k+j$ one checks this is symmetric in $j$ and $k$.

The terms containing $E_0E_nT_1T_n$. Finally, there are four terms in (8.26) containing $E_0E_nT_1T_n$:

$$\begin{aligned}&E_0E_n\Big ((-1)^{n-k+j}q^{\delta _{k> j}} T_{n+1-k}T_{n-j} + (-1)^{n-j+k}q^{1+\delta _{k\ge j}} T_{n-k}T_{n+1-j}\nonumber \\&\quad + (-1)^{n-k+j}(q-1)\sum _{i\in \mathbb{Z }\setminus I(j-k)} (-1)^{\delta _{i<0}} T_{n+1-k-i}T_{n-j+i}\nonumber \\&\quad + (-1)^{n-j+k}q(q-1)\sum _{i\in \mathbb{Z }\setminus I(-1+j-k) }(-1)^{\delta _{i<0}} T_{n-k-i} T_{n+1-j+i} \Big ) T_1T_n \end{aligned}$$

(8.34)

Applying (3.19) with $(j,k)$ replaced by $(n-j+1,n-k)$ and (3.20) with $(j,k)$ replaced by $(n-j,n-k+1)$ we obtain that the parenthesis in (8.34) equals

$$\begin{aligned} (-1)^{n+j-k}qT_{n+1-k}T_{n-j} + (-1)^{n+k-j}qT_{n+1-j}T_{n-k} \end{aligned}$$

which is symmetric in $j$ and $k$. This completes the proof that (8.26) is symmetric in $j$ and $k$. Thus (3.27) holds.

The last statement about generators follows from the fact that (3.25) and (3.26) can be used to express $E_j$ for $j\in [\![{1},{n-1}]\!]$ and $T_k$ for $k\in [\![{1},{n}]\!]$, in terms of the new generators $\{E_0,E_n\}\cup \{\widetilde{T}_j\}_{j=1}^{n-1}\cup \{\widetilde{E}_k\}_{k=0}^{n-1}$.

1.4 Example: the case $n=2$

If $n=2$ then (3.3) becomes

$$\begin{aligned} y_1=t_1+x_1 t_2 ,\qquad y_2=t_1+x_2 t_2 \end{aligned}$$

and from this, or using (3.12), we get

$$\begin{aligned} t_1&= (x_1-x_2)^{-1}(x_1y_2-x_2y_1),\\ t_2&= -(x_1-x_2)^{-1}(y_2-y_1). \end{aligned}$$

By definition (3.11), we have

$$\begin{aligned} e_0=1,\quad e_1=x_1+x_2, \quad e_2=x_1x_2. \end{aligned}$$

By Corollary 3.5, $\mathbb{k }_q(\bar{x},\bar{y})^{S_2}$ is generated as a skew field over $\mathbb{k }$ by $e_1,e_2, t_1, t_2$. By Proposition 3.7 we have the following relations:

$$\begin{aligned} t_1 t_2&= t_2 t_1, \\ e_1 e_2&= e_2 e_1, \\ t_1 e_2&= q e_2 t_1, \\ t_2 e_2&= q e_2 t_2, \\ t_1 e_1&= e_1 t_1 + (1-q)e_2 t_2, \\ t_2 e_1&= q e_1 t_2 + (q-1)t_1. \end{aligned}$$

Using the notation in (3.32) and (3.31) we have

$$\begin{aligned} X_1&= e_2^{(0)} = e_2, \\ X_2&= e_1^{(1)} = t_2, \\ Y_1&= e_0^{(1)} = t_1, \\ Y_2&= e_0^{(2)} = e_1^{(0)}e_0^{(1)}e_1^{(1)}+e_0^{(0)}e_0^{(1)}e_0^{(1)}+e_2^{(0)}e_1^{(1)}e_1^{(1)}\\&= e_1t_1t_2+e_0t_1^2+e_2t_2^2. \end{aligned}$$

By (3.48) or direct computations,

$$\begin{aligned} {[}Y_1,Y_2{]}&= 0,\qquad [X_2,X_1]_q =0,\\ {[}Y_1,X_2{]}&= 0,\qquad [Y_2,X_1]_{q^2} =0,\\ {[}Y_1,X_1{]}_q&= 0,\qquad [Y_2,X_2]_{q^{-1}} =0. \end{aligned}$$

Thus, $(Z_1,Z_2,Z_3,Z_4)=(X_1,Y_1,X_2,Y_2)$ satisfy $Z_iZ_j=q^{s_{ij}}Z_jZ_i$ with

$$\begin{aligned} (s_{ij})=\left[ \begin{array}{lccc} 0 &{}\quad -1 &{}\quad -1 &{}\quad -2 \\ 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 1 &{}\quad 0 &{}\quad 0 &{}\quad 1 \\ 2 &{}\quad 0 &{}\quad -1 &{}\quad 0 \end{array}\right] . \end{aligned}$$

Using the definition (3.33),

$$\begin{aligned} \widehat{X}_1=X_1,\quad \widehat{X}_2=Y_1X_2^{-1}, \quad \widehat{Y}_1=Y_1,\quad \widehat{Y}_2=Y_1^{-2}Y_2. \end{aligned}$$

As in the proof of Theorem 3.10, $\widehat{X}_1, \widehat{X}_2, \widehat{Y}_1,\widehat{Y}_2$ generate $\mathbb{k }(\bar{x},\bar{y})^{S_2}$ as a skew field, the following relations hold:

$$\begin{aligned}&[\widehat{X}_1,\widehat{X}_2]=0,\qquad [\widehat{Y}_1,\widehat{Y}_2]=0,\\&\widehat{Y}_i\widehat{X}_j = q^{\delta _{ij}} \widehat{X}_j\widehat{Y}_i,\qquad \forall i,j\in \{1,2\}. \end{aligned}$$

and we have an isomorphism $\mathbb{k }_q(x_1,x_2,y_1,y_2)^{S_2}\simeq \mathbb{k }_q(x_1,x_2,y_1,y_2)$.

1.5 Example: the case $n=3$

The elementary symmetric polynomials $e_d$ are

$$\begin{aligned} e_0&= 1,\\ e_1&= x_1+x_2+x_3,\\ e_2&= x_1x_2+x_2x_3+x_3x_1,\\ e_3&= x_1x_2x_3. \end{aligned}$$

By (3.12) we have

$$\begin{aligned} t_1&= \Delta ^{-1}\cdot \big ( (x_2^2x_3-x_3^2x_2)y_1+(x_3^2x_1-x_1^2x_3)y_2+(x_1^2x_2-x_2^2x_1)y_3\big ),\\ t_2&= \Delta ^{-1}\cdot \big ( (x_2^2-x_3^2)y_1+(x_3^2-x_1^2)y_2+(x_1^2-x_2^2)y_3\big ),\\ t_3&= \Delta ^{-1}\cdot \big ( (x_2-x_3)y_1+(x_3-x_1)y_2+(x_1-x_2)y_3\big ), \end{aligned}$$

where

$$\begin{aligned} \Delta =(x_1-x_2)(x_1-x_3)(x_2-x_3). \end{aligned}$$

By Corollary  3.5, $\mathbb{k }_q(\bar{x},\bar{y})^{S_3}$ is generated as a skew field over $\mathbb{k }$ by $e_1,e_2,e_3,t_1,t_2,t_3$ and by Proposition 3.7 or direct computations, we have the following relations:

$$\begin{aligned} {[}t_i,t_j{]}&= 0, \quad \forall i,j\in \{1,2,3\},\\ {[}e_i,e_j{]}&= 0,\quad \forall i,j\in \{1,2,3\},\\ {[}t_i,e_3{]}_q&= 0,\quad \forall i\in \{1,2,3\},\\ {[}t_1,e_1{]}&= (q-1)e_3 t_3,\\ {[}t_2,e_1{]}&= (q-1)(t_1 -e_2 t_3), \\ {[}t_3,e_1{]}_q&= (q-1)t_2,\\ {[}t_1,e_2{]}&= (1-q)e_3 t_2, \\ {[}t_2,e_2{]}_q&= (1-q)(e_3t_3-e_1 t_1),\\ {[}t_3,e_2{]}_q&= (1-q)t_1. \end{aligned}$$

By (3.32) and (3.31),

$$\begin{aligned} X_1&= e_3^{(0)} = e_3=x_1x_2x_3,\\ X_2&= e_2^{(1)} = t_3,\\ X_3&= e_1^{(2)} = e_2^{(0)}e_0^{(1)}e_2^{(1)}-e_0^{(0)}e_0^{(1)}e_0^{(1)}+e_3^{(0)}e_1^{(1)}e_2^{(1)}\\&= e_2t_1t_3-e_0t_1^2+e_3t_2t_3,\\ Y_1&= e_0^{(1)}=t_1,\\ Y_2&= e_0^{(2)}=e_1^{(0)}e_0^{(1)}e_2^{(1)}+e_0^{(0)}e_1^{(1)}e_0^{(1)}-e_3^{(0)}e_2^{(1)}e_2^{(1)}\\&\quad =e_1t_1t_3+e_0t_2t_1-e_3t_3^2,\\ Y_3&= e_0^{(3)}=e_1^{(1)}e_0^{(2)}e_1^{(2)}+e_0^{(1)}e_0^{(2)}e_0^{(2)}+e_2^{(1)}e_1^{(2)}e_1^{(2)}\\&\quad = t_2Y_2X_3+t_1Y_2^2+t_3X_3^2. \end{aligned}$$

By (3.48),

$$\begin{aligned} {[}Y_k,Y_i]&= 0,\quad \forall k,i\in \{1,2,3\},\\ {[}X_2,X_1]_q&= 0,\quad [X_3,X_1]_{q^2}=0,\quad [X_3,X_2]_{q^{-1}}=0,\\ {[}Y_1,X_2]&= 0,\quad [Y_1,X_3]=0,\quad [Y_2,X_3]=0,\\ {[}Y_1,X_1]_q&= 0,\quad [Y_2,X_1]_{q^2}=0,\quad [Y_3,X_1]_{q^5}=0,\\ {[}Y_2,X_2]_{q^{-1}}&= 0,\quad [Y_3,X_2]_{q^{-2}}=0,\quad \\ {[}Y_3,X_3]_q&= 0. \end{aligned}$$

Thus, if we let $(Z_1,Z_2,\ldots ,Z_6)=(X_1,Y_1,X_2,Y_2,X_3,Y_3)$, then $Z_iZ_j=q^{s_{ij}}Z_jZ_i$ with

$$\begin{aligned} (s_{ij})=\left[ \begin{array}{lccccc} 0 &{}\quad -1 &{}\quad -1 &{}\quad -2 &{}\quad -2 &{}\quad -5 \\ 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 1 &{}\quad 0 &{}\quad 0 &{}\quad 1 &{}\quad 1 &{}\quad 2 \\ 2 &{}\quad 0 &{}\quad -1 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 2 &{}\quad 0 &{}\quad -1 &{}\quad 0 &{}\quad 0 &{}\quad -1 \\ 5 &{}\quad 0 &{}\quad -2 &{}\quad 0 &{}\quad 1 &{}\quad 0 \end{array}\right] . \end{aligned}$$

(8.35)

By performing simultaneous elementary row and column transformations, this matrix can be brought to the skew normal form

$$\begin{aligned} \left[ \begin{array}{cccccc} 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ -1&{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad -1 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 1 &{}\quad 0 &{}\quad 0 &{}\quad 0 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad 1 \\ 0 &{}\quad 0 &{}\quad 0 &{}\quad 0 &{}\quad -1 &{}\quad 0 \end{array}\right] . \end{aligned}$$

(8.36)

As in (3.33), changing generators to

$$\begin{aligned} \widehat{X}_1&= X_1,\quad \widehat{X}_2=Y_1X_2^{-1},\quad \widehat{X}_3=Y_2^{-1}X_3,\\ \widehat{Y}_1&= Y_1,\quad \widehat{Y}_2=Y_1^{-2}Y_2,\quad \widehat{Y}_3=Y_1^{-1}Y_2^{-2}Y_3. \end{aligned}$$

one can also verify directly that

$$\begin{aligned}{}[\widehat{X}_i,\widehat{X}_j]&= [\widehat{Y}_i,\widehat{Y}_j]=0,\quad \forall i,j\in \{1,2,3\},\\ \widehat{Y}_i\widehat{X}_j&= q^{\delta _{ij}} \widehat{X}_j \widehat{Y}_i,\quad \forall i,j\in \{1,2,3\}, \end{aligned}$$

which means that there is an isomorphism of skew fields

$$\begin{aligned}&\mathbb{k }_q(\bar{x},\bar{y}) \overset{\sim }{\longrightarrow } \mathbb{k }_q(\bar{x},\bar{y})^{S_3}\\&x_i \longmapsto \widehat{X}_i,\qquad \forall i\in \{1,2,3\},\\&y_i \longmapsto \widehat{Y}_i,\qquad \forall i\in \{1,2,3\}.\\ \end{aligned}$$