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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, $$ r(u^2_1 \otimes I^{12}_{22}) = \sum_{r,s,x} R^{21}_{rs}R^{2r}_{x2}R^{xs}_{12} - \sum_{r,s,x} R^{21}_{xr}R^{x1}_{1s}R^{sr}_{22} = R^{21}_{21}R^{22}_{22}R^{21}_{12} + R^{21}_{12}R^{21}_{12}R^{12}_{12} - R^{21}_{12}R^{12}_{12}R^{22}_{22} = (q-q^{-1})^2 \neq 0. $$

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Kassel's book, if I recall correctly, has all details. –  Mariano Suárez-Alvarez Aug 17 '10 at 3:02
    
Clearly, we do have $\sum_{r,s,x} R^{ji}_{rs}R^{kr}_{xn}R^{xs}_{lm} - \sum_{r,s,x} R^{ki}_{xr}R^{xj}_{ls}R^{sr}_{nm}= (R_{23}R_{12}R_{13}-R_{13}R_{12}R_{23})^{kij}_{lnm}$ 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). –  DamienC Aug 17 '10 at 9:11
    
Ok, I agree 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} = (R_{23}R_{12}R_{13} - R_{13}R_{12}R_{23})^{kij}_{lnm}$, but I don't see why this vanishes. In particular, I don't see why this resolves the specific problem of $R(u^2_1 \otimes I^{12}_{22})$ seeming to not equal zero.. –  Abtan Massini Aug 17 '10 at 12:26
    
Applying the flip $\tau_{2,3}$ to $$ R_{23}R_{12}R_{13}-R_{13}R_{12}R_{23} $$ we get $$ R_{32}R_{13}R_{12}-R_{12}R_{13}R_{32} $$ which vanishes. –  DamienC Aug 17 '10 at 21:25
    
I have made some small formatting changes. –  Theo Johnson-Freyd Aug 18 '10 at 3:05
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1 Answer

Let me try to rephrase everything in more modern terms.

First of all $R\in\mathrm{End}(V\otimes V)$. Then let me denote by $U$ the $\mathbb{C}\langle u^i_j \rangle$-valued matrix with entries being $u^i_j$'s ($U\in{\rm End}(V)\langle u^i_j \rangle$).

Now the two-sided ideal $\mathcal J(R)$ is generated by the entries of the matrix $$ M(U):=U_1U_2R-RU_2U_1. $$ This expression lies in ${\rm End}(V\otimes V)\langle u^i_j \rangle\cong{\rm End}(V)\otimes{\rm End}(V)\otimes \mathbb{C}\langle u^i_j \rangle$. Viewing elements of this space as tensors with $3$ components some people rewrite it as follows: $$ U_{1,3}U_{2,3}R_{1,2}-R_{1,2}U_{2,3}U_{1,3}. $$

REMARK: it seems that Klimyk and Schmudgen chose a similar but different $M(U)$, but it is not very important.

Now we want to define $r:\mathbb{C}\langle u^i_j \rangle\otimes \mathbb{C}\langle u^i_j \rangle\to \mathbb{C}$. It is sufficient to define it on generators, and the best way to organize the corresponding coefficients is to give an expression for $$ r(U\otimes U)\in{\rm End}(V)\otimes{\rm End}(V)\cong{\rm End}(V\otimes V). $$ We define naively $r(U\otimes U):=R$.

We then need to check that the elements $r(M(U)\otimes U)$ and $r(U\otimes M(U))$, lying in ${\rm End}(V^{\otimes3})$, vanish. Let me try with the second one.

First part: $$ r(U\otimes U_1U_2R)=r(U\otimes U_1U_2)R_{2,3}=r(U\otimes U_1)r(U\otimes U_2)R_{2,3}=R_{1,2}R_{1,3}R_{2,3} $$

Second part: $$ r(U\otimes RU_2U_1)=R_{2,3}r(U\otimes U_2U_1)=R_{2,3}r(U\otimes U_2)r(U\otimes U_1)=R_{2,3}R_{1,3}R_{1,2} $$

So there is no problem here since we find Yang-Baxter.

Let me try now with the first one.

First part: $$ r(U_1U_2R\otimes U)=r(U_1U_2\otimes U)R_{1,2}=r(U_1\otimes U)r(U_2\otimes U)R_{1,2}=R_{1,3}R_{2,3}R_{1,2} $$

Second part: $$ r(RU_2U_1\otimes U)=R_{1,2}r(U_2U_1\otimes U)=R_{1,2}r(U_2\otimes U)r(U_1\otimes U)=R_{1,2}R_{2,3}R_{1,3} $$

There seems to be a problem here since we find an expression which is not Yang-Baxter: $$ R_{1,3}R_{2,3}R_{1,2}-R_{1,2}R_{2,3}R_{1,3} $$ But there is no: applying the flip $\tau_{1,2}$ we get $$ R_{2,3}R_{1,3}R_{2,1}-R_{2,1}R_{1,3}R_{2,3} $$ which is an avatar of Yang-Baxter (as far as $R^{op}=R^{-1}$).

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