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1 About this webpage

These lecture notes were created with Ximera, an interactive textbook platform hosted by Ohio State University. The Ximera Project is funded 2024-2026 (with no other external funding) by a $2,125,000 Open Textbooks Pilot Program grant from the federal Department of Education.

To load the most updated version, click the orange “update” button at the top of the page. If it is not there, then you are reading the most up-to-date version. The button looks like this:

Funding was provided by The London Mathematical Society and Lancaster University, University of Bristol, University of Edinburgh, Warwick University, Queen Mary University of London, and Ximera.

Acknowledgements: We thank Oleg Zaboronski for a careful reading of these notes and for correcting a few typos in [2].

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 braided Yang–Baxter equation \[ R_{1,2}R_{2,3}R_{1,2}=R_{2,3}R_{1,2}R_{2,3} \] 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 representation is defined by (fixing a typo in [2]) \[ \begin{align*} \rho _m(e)v_k &= \frac {q^{m-k}-q^{k-m}}{q-q^{-1}}v_{k+1}, \\ \rho _m(f)v_k &= \frac {q^k - q^{-k}}{q-q^{-1}}v_{k-1}, \\ \rho _m(h)v_k &= (2k-m)v_k. \end{align*} \] 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 (fixing a typo in [2]) \[ R(T \otimes 1)(1 \otimes T) = (T \otimes 1)(1 \otimes T)R. \] The relations can be explicitly written (details omitted) as \[ \begin{align*} t_{1,1} t_{1,2}=q^{-1} t_{1,2} t_{1,1}, \quad & t_{1,1} t_{2,1}=q^{-1} t_{2,1} t_{1,1}, \\ t_{1,2} t_{2,2}=q^{-1} t_{2,2} t_{1,2}, \quad &t_{2,1} t_{2,2}=q^{-1} t_{2,2} t_{2,1}, \\ t_{1,2} t_{2,1}=t_{2,1} t_{1,2}, \quad & t_{1,1} t_{2,2}-t_{2,2} t_{1,1}=\left (q^{-1}-q\right ) t_{1,2} t_{2,1}, \\ & t_{1,1} t_{2,2}-q^{-1} t_{1,2} t_{2,1}=1 \end{align*} \] 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 braided Yang–Baxter equation, \[ R_{1,2}R_{2,3}R_{1,2}=R_{2,3}R_{1,2}R_{2,3} \] then \(T(R)\) is a bialgebra with generators \(\{t_{i,j}\}\) and relations given by \[ R(T \otimes 1)(1 \otimes T)=(T \otimes 1)(1 \otimes T)R \] 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. We elaborate on this in the next section.

3 Dual algebras

Recall that the dual of a co–algebra is an algebra, and vice versa. The multiplication \(m\) in the algebra is defined by the co–product \(\Delta \) in the co–algebra. To describe this in more detail, we recall Sweedler’s notation. For any \(c\) in any co–algebra, we write \[ \Delta (c) = \sum _{(c)} c_{(1)} \otimes c_{(2)}. \] Sometimes the summation is dropped, and in that case it is called sumless Sweedler’s notation.

Now, let \(A\) be an algebra over a field \(k\) and let \(C=Hom_k(A,k)\). Then the co–product is defined by \[ \Delta (c)(a_1 \otimes a_2) = \sum _{(c)} c_1(a_1)c_2(a_2) = c(a_1a_2) \] In our situation, we are considering Hopf algebras, which are both algebras and co–algebras. So in this, it is more convenient to use the notation \[ \langle c,a\rangle = c(a). \]

It would perhaps be more helpful to illustrate this with the quantum group \(U_q(sl_2)\). More particularly, we have that for \( u \in U_q(sl_2) \), \[ \langle t_{ij}, u\rangle = [\rho _1(u)]_{ij}. \] In words, we say that \(t_{ij}\) is the \( (i,j)\) matrix entry in the fundamental representation. Suppose that we did not know the co–product applied to \(e\). We do know that since \(e\) increases weights by one (from classical Lie theory) that \[ \Delta (e) = k^{(1)} \otimes e + e\otimes k^{(2)} \] for some \( k^{(1)},k^{(2)} \in U^0\). Then \[ \langle k^{(1)},t_{i,j}\rangle \langle k^{(2)},t_{i',j'}\rangle =\langle \Delta (e), t_{i,j} \otimes t_{i',j'}\rangle = \langle e,t_{i,j}t_{i',j'}\rangle . \] We then need to solve for the eight variables \(\{\langle k^{(a)}, t_{i,j}\rangle : a,i,j=1,2 \}\). The relations among \(t_{i,j}\) give seven equations, and there is an additional equation because we know the co–product when \(q=1\). More explicitly, copying and pasting the relations, we have \begin{align*} \bracket {1}{t_{1,1}} \brackete {t_{1,2}} + \brackete {t_{1,1}} \bracket {2}{t_{1,2}}&=q^{-1} \left ( \bracket {1}{t_{1,2}} \brackete {t_{1,1}} + \brackete {t_{1,2}} \bracket {2}{t_{1,1}}\right ),\\ \bracket {1}{t_{1,1}} \brackete {t_{2,1}} + \brackete {t_{1,1}} \bracket {2}{t_{2,1}}&=q^{-1} \left ( \bracket {1}{t_{2,1}} \brackete {t_{1,1}} + \brackete {t_{2,1}} \bracket {2}{t_{1,1}}\right ), \\ \bracket {1}{t_{1,2}} \brackete {t_{2,2}} + \brackete {t_{1,2}} \bracket {2}{t_{2,2}}&=q^{-1} \left ( \bracket {1}{t_{2,2}} \brackete {t_{1,2}} + \brackete {t_{2,2}} \bracket {2}{t_{1,2}}\right ), \\ \bracket {1}{t_{2,1}} \brackete {t_{2,2}} + \brackete {t_{2,1}} \bracket {2}{t_{2,2}}&=q^{-1} \left ( \bracket {1}{t_{2,2}} \brackete {t_{2,1}} + \brackete {t_{2,2}} \bracket {2}{t_{2,1}}\right ), \\ \bracket {1}{t_{1,2}} \brackete {t_{2,1}} + \brackete {t_{1,2}} \bracket {2}{t_{2,1}}&= \bracket {1}{t_{2,1}} \brackete {t_{1,2}} + \brackete {t_{2,1}} \bracket {2}{t_{1,2}},\\ \left (\bracket {1}{t_{1,1}} \brackete {t_{2,2}} + \brackete {t_{1,1}} \bracket {2}{t_{2,2}}\right )-\left (\bracket {1}{t_{2,2}} \brackete {t_{1,1}} + \brackete {t_{2,2}} \bracket {2}{t_{1,1}}\right )&=\left (q^{-1}-q\right ) \left ( \bracket {1}{t_{1,2}} \brackete {t_{2,1}} + \brackete {t_{1,2}} \bracket {2}{t_{2,1}}\right ), \\ \left (\bracket {1}{t_{1,1}} \brackete {t_{2,2}} + \brackete {t_{1,1}} \bracket {2}{t_{2,2}}\right )-q^{-1} \left ( \brackete {t_{1,2}} \bracket {2}{t_{2,1}} + \bracket {1}{t_{1,2}} \brackete {t_{2,1}} \right )&=\langle k^{(1)},1\rangle \langle e,1\rangle + \langle e,1\rangle \langle k^{(2)},1\rangle \end{align*}

Using that \( \brackete {t_{1,2}}=1 \) and \( \brackete {t_{1,1}} = \brackete {t_{2,1}} = \brackete {t_{2,2}} = 0 \) we get that \begin{align*} \bracket {1}{t_{1,1}} &= q^{-1}\bracket {2}{t_{1,1}}\\ \bracket {2}{t_{2,2}} &= q^{-1}\bracket {1}{t_{2,2}}\\ \end{align*}

This does not have a unique solution, but once \(k^{(2)}\) is chosen then \(k^{(1)}\) is determined. One solution is \[ k^{(2)} = 1, \quad k^{(1)} = k^{-1}. \]

References