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\author{Jeffrey Kuan}
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\title{The FRT Construction}
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\begin{abstract}
The Yang--Baxter equation is cool.
\end{abstract}
\maketitle
%\part{Introduction}
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Accessibility statement: \href{https://ximera.osu.edu/firststeps24html/aFirstStepInXimera/basics/FRT_Construction_WCAG}{A WCAG2.1AA compliant version of these notes} will eventually be available, once I finish writing them.
You can also \href{https://ximera.osu.edu/firststeps24html/aFirstStepInXimera/basics/FRT_Construction.tex}{download the TeX source file.}
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\section{Motivation}
Given a solution \(R(\lambda)\) of the Yang--Baxter equation, the FRT construction of
\cite{FRT88} 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 \cite{KS98}. 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.
\begin{thebibliography}{10}
\bibitem{FRT88} L. D. Faddeev, N. Yu. Reshetikhin, L. A. Takhtajan, Quantization
of Lie groups and Lie algebras, Algebraic analysis, Vol. I, Academic
Press, Boston, MA, (1988), 129-139.
\bibitem{KS98} Leonid I. Korogodski, Yan S. Soibelman, Algebras of Functions
on Quantum Groups: Part I, American Mathematical Soc., 1998
\end{thebibliography}
\end{document}