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1 About this webpage

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2 Motivation

Given a solution \(R(\lambda )\) of the Yang–Baxter equation, the FRT construction of [1] produces a quasi–triangular Hopf algebra. To begin, suppose suppose the \(R\)–matrix \(R(\lambda )\) acts on the tensor power of vector spaces \(V \otimes V\). Let \(L(\lambda )\) be an operator acting on the tensor product \(V \otimes W\). Now consider the equation \[ R_{1,2}(\lambda -\mu )L_{1,3}(\lambda )L_{2,3}(\mu ) = L_{2,3}(\mu )L_{1,3}(\lambda )R_{1,2}(\lambda -\mu ) \] where both sides act on the tensor product \(V\otimes V \otimes W\). One solution occurs when the vector spaces \(V\) and \(W\) are the same and \(L\) is just equal to \(R.\) For the remainder of these notes, we consider the simpler case when \(V=W\).

Before launching into the most generalize case, it will be enlightening to consider the case of \(U_q(sl_2)\). Most of this portion of the exposition comes from the book [2]. Consider the \(4 \times 4\) matrix \[ R:= \left ( \begin{array}{cccc} q^{-2} & 0 & 0 & 0 \\ 0 & 0 & q^{-1} & 0 \\ 0 & q^{-1} & q^{-2}-1 & 0 \\ 0 & 0 & 0 & q^{-2} \end{array} \right ). \] We view this naturally as an operator on the tensor power \(\mathbb {C}^2 \otimes \mathbb {C}^2\). One can verify that this matrix solves the Yang–Baxter equation and \[ (R-q^{-2})(R+1)=0. \]

To relate this to \(U_q(sl_2)\), we first define a representation \(\rho _m\) on a \( (m+1)\)–dimensional vector space, with basis denoted \( \{v_0,\ldots ,v_{m}\}\). The represetation is defined by \[ \begin{align*} \rho _m(e)v_k &= \frac {q^{k-m}-q^{m-k}}{q-q^{-1}}v_{k+1}, \\ \rho _m(f)v_k &= \frac {q^k - q^{-k}}{q-q^{-1}}v_{k-1}, \\ \rho _m(h)v_k &= (2k-m)v_k. \end{align*} \] For \(i,j=1,2\) let \(t_{i,j}\) denote the matrix entries of the two–dimensional representation \(\rho _1\), and let \(T\) denote the \(2 \times 2\) matrix with entries \(t_{i,j}\). In other words, we consider \(t_{i,j}\) as abstract, non–commutating generators of the algebra \(\mathrm {Hom}(U_q(sl_2),\mathbb {C})\), defined by \[ t_{i,j}(u) = [\rho _1(u)]_{i,j}. \] The product in dual bi–algebras is defined by the co–product in the original algebra. In other words, \[ (t_{i,j}t_{k,l})(u) = (t_{i,j} \otimes t_{k,l})(\Delta (u)). \] Then one can verify that \[ R(T \otimes 1)(1 \otimes T) = (1 \otimes T)(T \otimes 1)R. \] The relations can be explicitly written (details omitted) as \[ \begin{align*} t_{11} t_{12}=q^{-1} t_{12} t_{11}, \quad & t_{11} t_{21}=q^{-1} t_{21} t_{11}, \\ t_{12} t_{22}=q^{-1} t_{22} t_{12}, \quad &t_{21} t_{22}=q^{-1} t_{22} t_{21}, \\ t_{12} t_{21}=t_{21} t_{12}, \quad & t_{11} t_{22}-t_{22} t_{11}=\left (q^{-1}-q\right ) t_{12} t_{21}, \\ & t_{11} t_{22}-q^{-1} t_{12} t_{21}=1 \end{align*} \] Of course, writing in the RTT form is much more concise. If the last relation is removed, then we obtain the dual to \(U_q(gl_2)\).

When \(q=1\), this reduces to the commutative algebra \(C[SL(2)]\), the usual algebra of functions on the group \(SL(2)\). This should not be surprising: the quantum group \(U_q(sl_2)\) is non–co–commutative except when \(q=1\), so its dual should be non–commutative except when \(q=1\). The algebra defined above is denoted \(C_q[SL(2)]\).

The co–product in a dual is defined from the product in the original algebra. In this case, this means that \[ \Delta (t_{i,j})(u_1 \otimes u_2) = t_{i,j}(u_1 u_2). \] The explicit formula is then \[ \Delta (t_{i,j}) = \sum _k t_{i,k}\otimes t_{k,j}. \] Note that there is no \(q\) in this formula, which should not surprise us. Since the quantization in \(U_q(sl_2)\) deforms the co–product and not the product, the corresponding deformation in the dual should not deform the co–product, but rather the product.

In the more general case, suppose that a matrix \(R\) satisfies the Yang–Baxter equation, then \(T(R)\) is a bialgebra with generators \(\{t_{i,j}\}\) and relations given by \[ R(1 \otimes T)(T \otimes 1) = (T \otimes 1)(1 \otimes T)R \] and co–product \[ \Delta (t_{i,j}) = \sum _k t_{i,k}\otimes t_{k,j}. \] To recover a quantum group analogous to a Drinfel’d–Jimbo quantum group, we would need to take the restricted dual, which is contained in \( \mathrm {Hom}(T,\mathbb {C})\). This is generated by all matrix entries of finite–dimensional modules. Since \(T\) is generally infinite–dimensional, the full dual is too large to reasonably consider.

References