Let $V$ be a finite dimensional vector space and let $R$ be a linear invertible mapping from $V \otimes V$ to itself. If we fix a basis $\{e_i\}_{i=1,2, \cdots ,n}$ then we have $N^4$ complex numbers $R^{mn}_{ij}$ such that $$ R(e_i \otimes e_j) = \sum_{m,n=1}^N R_{ij}^{mn} e_m \otimes e_n.$$

As is well-known, we can use the numbers $R^{ij}_{mn}$ to construct a bialgebra $A(R)$, called the *FTR-bialgebra associated to* $R$. We will describe here just the algebra structure of $A(R)$: Let $\mathbb{C}\langle u^i_j \rangle$ denote the free algebra over $\mathbb{C}$ generated by $u^i_j$ and let ${\cal J}(R)$ be the bi-ideal generated by the elements
$$
I^{ij}_{mn} = \sum R^{ji}_{kl} u^k_m u^l_n - \sum u^i_k u^j_l R^{lk}_{mn}, \qquad i,j,m,n = 1, \ldots ,N.
$$
Then we define
$$
A(R) = \mathbb{C}\langle u^i_j \rangle/{\cal J}(R).
$$

Now in Section 10.1.2 of their book *Quantum Groups and their Representations,* Klimyk and Schmudgen claim that if $R$ is a solution of the Quantum Yang Baxter Equation
$$
R_{12}R_{13}R_{23} = R_{23}R_{13}R_{12},
$$
then the bialgebra is coquasi-triangular with a unique universal r-form for which
$$
r(u^i_j \otimes u^k_l) = R^{ik}_{jl}.
$$
The crucial step in establishing this is to show that $r({\cal J}(R) \otimes u^k_l) = r(u^k_l \otimes {\cal J}(R)) = 0$. Using the elementary properties of $r$, they prove that
$$r(I^{ij}_{nm} \otimes u^k_l) = \sum_{r,s,x}R^{ji}_{rs}R^{rk}_{nx}R^{sx}_{ml} - \sum_{r,s,x} R^{ik}_{rx}R^{jx}_{sl}R^{sr}_{nm}.$$
The fact that $R$ satisfies the Quantum Yang Baxter equation then easily implies that this sum vanishes.

The proof that $r(u^k_l \otimes {\cal J}(R)) = 0$ is left as an excercise. Now it routine to show that $$ r(u^k_l \otimes I^{ij}_{mn}) = \sum_{r,s,x} R^{ji}_{rs}R^{kr}_{xn}R^{xs}_{lm} - \sum_{r,s,x} R^{ki}_{xr}R^{xj}_{ls}R^{sr}_{nm}. $$ However, I cannot see why the Yang-Baxter Equation implies that this vanishes.

I had hoped to use the proof of this fact to resolve the difficulty in this question. Indeed if we have $$ R^{ij}_{mn} = q^{-\frac{1}{2}}(q^{\delta_{ij}}\delta_{im}\delta_{jn} + (q-q^{-1})\theta(i-j)\delta_{in}\delta_{jm}), \qquad i,j,m,n = 1,2, $$ we get the $2 \times 2$ quantum matrices. Since $I^{12}_{22}$ is easily shown to give $ab - qba$, the proof would imply that $P_{u^2_1}(ab-qba)$ vanishes (where we are using the $P$ notation of the other question). However, $$ r(u^2_1 \otimes I^{12}_{22}) = \sum_{r,s,x} R^{21}_{rs}R^{2r}_{x2}R^{xs}_{12} - \sum_{r,s,x} R^{21}_{xr}R^{x1}_{1s}R^{sr}_{22} = R^{21}_{21}R^{22}_{22}R^{21}_{12} + R^{21}_{12}R^{21}_{12}R^{12}_{12} - R^{21}_{12}R^{12}_{12}R^{22}_{22} = (q-q^{-1})^2 \neq 0. $$

`$\sum_{r,s,x} R^{ji}_{rs}R^{kr}_{xn}R^{xs}_{lm} - \sum_{r,s,x} R^{ki}_{xr}R^{xj}_{ls}R^{sr}_{nm}= (R_{23}R_{12}R_{13}-R_{13}R_{12}R_{23})^{kij}_{lnm}$`

and $R_{23}R_{12}R_{13}-R_{13}R_{12}R_{23}$ is not yang-Baxter (and it does not have any meaning in terms of braids). $\endgroup$ – DamienC Aug 17 '10 at 9:11