Quantized universal enveloping algebras, the Yang-Baxter equation and invariants of links, I

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 2 June 2014

Notes and References

This is a typed copy of the LOMI preprint Quantized universal enveloping algebras, the Yang-Baxter equation and invariants of links, I by N.Yu. Reshetikhin.

Recommended for publication by the Scientific Council of Steklov Mathematical Institute, Leningrad Department (LOMI) 6, June, 1987.

$𝔤=G_2$

The irreducible finite dimensional representation of
$U_q\left(G_2\right)$ are parametrized by h.w. $\lambda ={\lambda }_1{\omega }_1+{\lambda }_2{\omega }_2,$ where ${\omega }_1$ and ${\omega }_2$ are the fundamental weights (see [Bou1981]) and ${\lambda }_i\in {\mathbb{Z}}_{+}\text{.}$ If $e_{\lambda }\in V^{\lambda }$ the h.w. vector, $H_1{e}_{\lambda }={\lambda }_1{e}_{\lambda },$ $H_2{e}_{\lambda }=3{\lambda }_2{e}_{\lambda }\text{.}$

The basic representation of $U_q\left(G_2\right)$ have h.w. $\lambda ={\omega }_1\text{.}$ Let us consider the basics in $V^{{\omega }_1}$ formed by weight vectors,
 $e_1=e_{{\omega }_1},$ $e_2=e_{{\alpha }_1+{\alpha }_2},$ $e_3=e_{{\alpha }_1},$ $e_4=e_0,$ $e_5=e_1,$ $e_6=e-{\alpha }_1-{\alpha }_2,$ $e_7=e-{\omega }_1\text{.}$ The roots are shown on the figure 1. $$\begin{array}{c} α1 α2 ω1 ω2 \\ \text{Figure 1.}\end{array}$$ In this basis the elements $H_i,X_i^{\pm }$ are represented by the matrices: $$\begin{array}{cc}\pi \left(X_1^{+}\right)=\left(\begin{array}{ccccccc}0& b& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& c& 0& 0& 0\\ 0& 0& 0& 0& c& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& b\\ 0& 0& 0& 0& 0& 0& 0\end{array}\right),\pi \left(X_2^{+}\right)=\left(\begin{array}{ccccccc}0& 0& 0& 0& 0& 0& 0\\ 0& 0& 1& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 1& 0\\ 0& 0& 0& 0& 0& 0& 0\\ 0& 0& 0& 0& 0& 0& 0\end{array}\right)& \text{(7.1)}\\ \pi \left(H_1\right)=\left(\begin{array}{ccccccc}1& & & & & & \\ & -1& & & & 0& \\ & & 2& & & & \\ & & & 0& & & \\ & & & & -2& & \\ & 0& & & & 1& \\ & & & & & & -1\end{array}\right),\pi \left(H_2\right)=\left(\begin{array}{ccccccc}0& & & & & & \\ & 1& & & & 0& \\ & & -1& & & & \\ & & & 0& & & \\ & & & & 1& & \\ & 0& & & & -1& \\ & & & & & & 0\end{array}\right)& \text{(7.2)}\end{array}$$ $$\pi \left(X_i^{-}\right)=\pi {\left(X_i^{+}\right)}^t\text{.}$$ Here $b=\sqrt{\frac{\text{sh}\left(\frac{h}{2}\right)}{\text{sh}\left(\frac{3h}{2}\right)}},$ $c=\sqrt{\frac{\text{sh}\left(h\right)}{\text{sh}\left(\frac{3h}{2}\right)}}\text{.}$

In this normalization the commutation relations between the generators $H_i,X_i^{\pm }$ have the following form: $$\begin{array}{c}[H_1,X_1^{\pm }]=\pm 2X_1^{\pm },[H_2,X_1^{\pm }]=\mp 3X_1^{\pm }\\ [H_1,X_2^{\pm }]=\mp 3X_2^{\pm },[H_2,X_2^{\pm }]=\pm 6X_2^{\pm }\\ [X_i^{+},X_j^{-}]=\frac{\text{sh}\left(\frac{h}{2}{H}_i\right)}{\text{sh}\left(\frac{3h}{2}\right)},q=e^h\text{.}\end{array}$$

The representation $V^{{\omega }_1}$ is selfdual: $V^{{\omega }_1^{*}}=V^{{\omega }_1}\text{.}$

A tensorial product of the two basic representations of $U_q\left(G_2\right)$ has the following decomposition: $$\begin{array}{cc}{V}^{{\omega }_1}\otimes V^{{\omega }_1}=V^{2{\omega }_1}\oplus V^{{\omega }_2}\oplus V^{{\omega }_1}\oplus V^0\text{.}& \text{(7.3)}\end{array}$$ In this decomposition $\overline{2{\omega }_1}=0,$ $\overline{{\omega }_2}=\overline{{\omega }_1}=1,$ $\bar{0}=0\text{.}$ It is not difficult to calculate the matrix $K_{{\omega }_1}^{{\omega }_1{\omega }_1}\text{:}$ $$\begin{array}{cc}\begin{array}{ccc}{e}_1& =& g(c{q}^{\frac{3}{4}}{e}_2\otimes e_3-b{q}^{\frac{3}{2}}{e}_1\otimes e_4+b{q}^{-\frac{3}{2}}{e}_4\otimes e_1-c{q}^{-\frac{1}{4}}{e}_3\otimes e_2),\\ e_2& =& g(b{q}^{\frac{1}{2}}{e}_2\otimes e_4-c{q}^{\frac{5}{4}}{e}_1\otimes e_5+c{q}^{-\frac{5}{4}}{e}_5\otimes e_1-b{q}^{-\frac{1}{2}}{e}_4\otimes e_2),\\ e_3& =& g(b{q}^{\frac{1}{2}}{e}_3\otimes e_4-c{q}^{\frac{5}{4}}{e}_1\otimes e_6+c{q}^{-\frac{5}{4}}{e}_6\otimes e_1-b{q}^{-\frac{1}{2}}{e}_4\otimes e_3),\\ e_4& =& gb(-q{e}_1\otimes e_7-q^{\frac{3}{2}}{e}_2\otimes e_6+e_3\otimes e_5+(q^{\frac{1}{2}}-q^{-\frac{1}{2}})e_4\otimes e_4-e_5\otimes e_3+q^{-\frac{3}{2}}{e}_6\otimes e_2+q^{-1}{e}_7\otimes e_1),\\ e_5& =& g(b{q}^{\frac{1}{2}}{e}_4\otimes e_5-c{q}^{\frac{5}{4}}{e}_2\otimes e_7+c{q}^{-\frac{5}{4}}{e}_7\otimes e_2-b{q}^{-\frac{3}{2}}{e}_5\otimes e_4),\\ e_6& =& g(b{q}^{\frac{1}{2}}{e}_4\otimes e_6-c{q}^{\frac{5}{4}}{e}_3\otimes e_7+c{q}^{-\frac{5}{4}}{e}_7\otimes e_3-b{q}^{-\frac{1}{2}}{e}_6\otimes e_4),\\ e_7& =& g(c{q}^{\frac{3}{4}}{e}_5\otimes e_6-b{q}^{\frac{3}{2}}{e}_4\otimes e_7+b{q}^{-\frac{3}{2}}{e}_7\otimes e_4-c{q}^{-\frac{3}{4}}{e}_6\otimes e_5),\\ g& =& \frac{1}{\sqrt{q^2{q}^{-2}}},\end{array}& \text{(7.4)}\end{array}$$ So the matrix ${\left(K_{{\omega }_1}^{{\omega }_1{\omega }_1}\right)}_i^{jk}$ $$\begin{array}{cc}{\displaystyle e_i=\sum _{j,k=1}^7{\left(K_{{\omega }_1}^{{\omega }_1{\omega }_1}\right)}_i^{jk}{e}_j\otimes e_k}& \text{(7.5)}\end{array}$$ is given explicitly.

Theorem 7.1. The matrix $R^{{\omega }_1{\omega }_1}$ satisfies the following relations: $$\begin{array}{cc}\begin{array}{c} \end{array}-q\begin{array}{c} \end{array}=(q-1)\{\begin{array}{c} \end{array}-\alpha \begin{array}{c} \end{array}+\beta \begin{array}{c} \end{array}\}& \text{(7.6)}\\ \begin{array}{c} \end{array}-q\begin{array}{c} \end{array}=(q-1)\{\begin{array}{c} \end{array}-\alpha \begin{array}{c} \end{array}+\beta \begin{array}{c} \end{array}\}\text{.}& \text{(7.7)}\end{array}$$ Here $\alpha =q+q^{-1},$ $\beta =\frac{q^{-3}(q^3-1)(q^4+1)}{q-1}\text{.}$

		Proof.
	From the theorem 1.5 we have the spectral decomposition of $R^{{\omega }_1{\omega }_1}\text{:}$ $$\begin{array}{cc}\begin{array}{c} ω1 ω1 \end{array}=q\begin{array}{c} ω1 ω1 2ω1 ω1 ω1 \end{array}-\begin{array}{c} ω1 ω1 ω2 ω1 ω1 \end{array}-q^{-3}\begin{array}{c} ω1 ω1 ω1 ω1 ω1 \end{array}+q^{-6}\begin{array}{c} ω1 ω1 0 ω1 ω1 \end{array}\text{.}& \text{(7.8)}\end{array}$$ One-dimensional projector is defined in section 1 (1.47): $$\begin{array}{cc}\begin{array}{c} ω1 ω1 0 ω1 ω1 \end{array}=\frac{1}{{\text{tr}}_{V^{{\omega }_1}}\left(q^{-\rho }\right)}\begin{array}{c} \end{array}=\frac{q^2+1}{q}\frac{q-1}{(q^7-1)(q^{-6}+1)}\begin{array}{c} \end{array}\text{.}& \text{(7.9)}\end{array}$$ Substituting (7.8), (7.9) in the l.h.s. of 7.6 and using the decomposition of units in $V^{{\omega }_1\otimes 2}\text{:}$ $$\begin{array}{c} \end{array}=\begin{array}{c} 2ω1 \end{array}+\begin{array}{c} ω2 \end{array}+\begin{array}{c} ω1 \end{array}+\begin{array}{c} 0 \end{array}$$ we obtain (7.6). The equality (7.7) is obtained by rotating the pictures (7.6) by ${90}^{\circ }$ (making crossing-transformation). $\square$

Considering the equalities (7.6) and (7.7) as a system of linear equations for $R^{{\omega }_1{\omega }_1}$ and ${\left(R^{{\omega }_1{\omega }_1}\right)}^{-1}$ we obtain an expression for these matrices in terms of the matrix $K_{{\omega }_1}^{{\omega }_1{\omega }_1}$ (7.4):

Proposition 7.1. $$\begin{array}{cc}\begin{array}{c} \end{array}=\frac{1}{q+1}[q^2\begin{array}{c} \end{array}+q^{-3}\frac{(q^3-1)(q^4+1)}{(q-1)}(\begin{array}{c} \end{array}+q\begin{array}{c} \end{array})+q^{-1}\begin{array}{c} \end{array}]& \text{(7.10)}\\ \begin{array}{c} \end{array}=\frac{1}{q+1}[q^2\begin{array}{c} \end{array}+q^{-3}\frac{(q^3-1)(q^4+1)}{(q-1)}(q\begin{array}{c} \end{array}+\begin{array}{c} \end{array})+q^{-1}\begin{array}{c} \end{array}]\text{.}& \text{(7.11)}\end{array}$$

Let us consider now the restriction of $G_2$ to the $𝔰𝔩\left(2\right)$ triple formed by $X_1^{\pm },H_1\text{.}$ The representation $V^{{\omega }_1}$ is the sum of three $𝔰𝔩\left(2\right)\text{-irreducible}$ components. $$\begin{array}{cc}{V}^{{\omega }_1}{|}_{𝔰𝔩\left(2\right)}=V^{{\omega }_1}\otimes V^2\otimes {\stackrel{\sim }{V}}^1\text{.}& \text{(7.12)}\end{array}$$ Here $\text{dim}\hspace{0.17em}{V}^1=\text{dim}\hspace{0.17em}{\stackrel{\sim }{V}}^1=2,$ $\text{dim}\hspace{0.17em}{V}^2=3\text{.}$ The spaces $V^1,V^2$ and ${\stackrel{\sim }{V}}^1$ are formed by the vectors $(e_1,e_2),$ $(e_3,e_4,e_5)$ and $(e_6,e_7)$ correspondingly.

For $𝔰𝔩\left(2\right)$ CGC we have: $$V^2\subset V^1\otimes V^1$$ $$\begin{array}{cc}\begin{array}{ccc}{e}^2& =& e^1\otimes e^1,\\ e^0& =& \frac{1}{\sqrt{q^{1/2}+q^{-1/2}}}(q^{\frac{1}{4}}{e}^{-1}\otimes e^1+q^{-\frac{1}{4}}{e}^1\otimes e^{-1}),\\ e^{-2}& =& e^{-1}\otimes e^{-1},\end{array}& \text{(7.12)}\end{array}$$ where $e^{\pm 1}$ are the bases in $2\text{-dimensional}$ representation of $U_q\left(𝔰𝔩\left(2\right)\right)$ and $e^2,e^0,e^{-2}$ are the bases of a $3\text{-dimensional}$ one. $$V^2\subset V^2\otimes V^2$$ $$\begin{array}{cc}\begin{array}{ccc}{e}^2& =& \frac{1}{\sqrt{q+q^{-1}}}(q^{\frac{1}{2}}{e}^2\otimes e^0-q^{-\frac{1}{2}}{e}^0\otimes e^2),\\ e^0& =& \frac{1}{\sqrt{q+q^{-1}}}(e^2\otimes e^{-2}+(q^{\frac{1}{2}}-q^{-\frac{1}{2}})e^0\otimes e^0-e^{-2}\otimes e^2),\\ e^{-2}& =& \frac{1}{\sqrt{q+q^{-1}}}(q^{\frac{1}{2}}{e}^0\otimes e^{-2}-q^{-\frac{1}{2}}{e}^{-2}\otimes e^0)\text{.}\end{array}& \text{(7.13)}\end{array}$$ In accordance with the graphical representation of CGC given in section 2 we can associate tho following picture with the formulae (7.12) and (7.13) $$\begin{array}{cc}\begin{array}{c} 1 1 2 \end{array}⟷\text{(7.12),}\begin{array}{c} 2 2 2 \end{array}⟷\text{(7.13).}& \text{(7.14)}\end{array}$$ An index $1$ corresponds to a $2\text{-dimensional}$ representation. An index $2$ corresponds to a $3\text{-dimensional}$ representation.

From (7.4) and (7.12)-(7.14) we obtain the $𝔰𝔩\left(2\right)\text{-block}$ form of the matrix $K_{{\omega }_1}^{{\omega }_1{\omega }_1}\text{:}$ $$K_{{\omega }_1}^{{\omega }_1{\omega }_1}=\begin{array}{c} \end{array}=\frac{1}{\sqrt{q^2+q^{-2}}}$$ $$\begin{array}{cc}\begin{array}{|cccccccc|}\hline 0& -q\begin{array}{c} 1 2 2 \end{array}& 0& q^{-1}\begin{array}{c} 1 2 1 \end{array}& 0& 0& 0& 0\\ 0& 0& x{q}^{-\frac{5}{4}}\begin{array}{c} 2 1 1 \end{array}& 0& y\begin{array}{c} 2 2 2 \end{array}& 0& -x{q}^{\frac{5}{4}}\begin{array}{c} 2 1 1 \end{array}& 0\\ 0& 0& 0& 0& 0& q\begin{array}{c} 1 2 1 \end{array}& 0& -q^{-1}\begin{array}{c} 1 1 2 \end{array}\\ \hline\end{array}& \text{(7.15)}\end{array}$$ where $$x={\left(\frac{1-q^2}{1-q^3}\right)}^{\frac{1}{2}}{q}^{\frac{1}{4}},y={\left(\frac{(1-q)(1+q^2)}{(1-q^3)}\right)}^{\frac{1}{2}}\text{.}$$

From (7.15) we obtain the following expression for projector $P_{{\omega }_1}^{{\omega }_1{\omega }_1}\text{:}$ $$\begin{array}{cc}\begin{array}{c} ω1 ω1 ω1 ω1 ω1 \end{array}=\frac{1}{q^2-q^{-2}}\begin{array}{|ccc|}\hline \begin{array}{ccc}0& 0& 0\\ 0& q^2\begin{array}{c} 1 2 1 1 2 \end{array}& 0\\ 0& 0& x^2{q}^{-\frac{3}{2}}\begin{array}{c} 1 1 2 1 1 \end{array}\end{array}& \begin{array}{ccc}0& 0& 0\\ -\begin{array}{c} 1 1 2 1 1 \end{array}& 0& 0\\ 0& xy{q}^{-\frac{5}{4}}\begin{array}{c} 1 1 2 2 1 \end{array}& 0\end{array}& \begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ -x^2\begin{array}{c} 1 1 2 1 1 \end{array}& 0& 0\end{array}\\ \begin{array}{ccc}0& -\begin{array}{c} 2 1 1 1 2 \end{array}& 0\\ 0& 0& xy{q}^{-\frac{5}{4}}\begin{array}{c} 2 2 2 1 1 \end{array}\\ 0& 0& 0\end{array}& \begin{array}{ccc}{q}^{-2}\begin{array}{c} 2 1 1 2 1 \end{array}& 0& 0\\ 0& y^2\begin{array}{c} 2 2 2 2 2 \end{array}& 0\\ 0& 0& q^2\begin{array}{c} 2 1 1 2 1 \end{array}\end{array}& \begin{array}{ccc}0& 0& 0\\ -yx{q}^{\frac{5}{4}}\begin{array}{c} 2 2 2 1 1 \end{array}& 0& 0\\ 0& -\begin{array}{c} 2 1 1 1 2 \end{array}& 0\end{array}\\ \begin{array}{ccc}0& 0& -x^2\begin{array}{c} 1 1 2 1 1 \end{array}\\ 0& 0& 0\\ 0& 0& 0\end{array}& \begin{array}{ccc}0& -yx{q}^{\frac{5}{4}}\begin{array}{c} 1 1 2 2 2 \end{array}& 0\\ 0& 0& -\begin{array}{c} 1 2 1 2 1 \end{array}\\ 0& 0& 0\end{array}& \begin{array}{ccc}{x}^2{q}^{\frac{5}{2}}\begin{array}{c} 1 1 2 1 1 \end{array}& 0& 0\\ 0& q^{-4}\begin{array}{c} 1 2 1 1 2 \end{array}& 0\\ 0& 0& 0\end{array}\\ \hline\end{array}& \text{(7.16)}\end{array}$$ and the similar expression for crossing-conjugate matrix. For one dimensional projector there is a similar representation: $$\begin{array}{cc}\begin{array}{c} ω1 ω1 ω1 ω1 \end{array}=\begin{array}{|ccc|}\hline \begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& 0& q^{\frac{9}{2}}\begin{array}{c} 1 1 \end{array}\end{array}& \begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ 0& q^{\frac{9}{4}}\begin{array}{c} 1 2 \end{array}& 0\end{array}& \begin{array}{ccc}0& 0& 0\\ 0& 0& 0\\ \begin{array}{c} 1 1 \end{array}& 0& 0\end{array}\\ \begin{array}{ccc}0& 0& 0\\ 0& 0& q^{\frac{9}{4}}\begin{array}{c} 2 1 \end{array}\\ 0& 0& 0\end{array}& \begin{array}{ccc}0& 0& 0\\ 0& \begin{array}{c} 2 2 \end{array}& 0\\ 0& 0& 0\end{array}& \begin{array}{ccc}0& 0& 0\\ \begin{array}{c} 2 1 \end{array}{q}^{-\frac{9}{4}}& 0& 0\\ 0& 0& 0\end{array}\\ \begin{array}{ccc}0& 0& \begin{array}{c} 1 1 \end{array}\\ 0& 0& 0\\ 0& 0& 0\end{array}& \begin{array}{ccc}0& q^{-\frac{9}{4}}\begin{array}{c} 1 2 \end{array}& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}& \begin{array}{ccc}\begin{array}{c} 1 1 \end{array}{q}^{-\frac{9}{2}}& 0& 0\\ 0& 0& 0\\ 0& 0& 0\end{array}\\ \hline\end{array}& \text{(7.17)}\end{array}$$ where $\begin{array}{c} 1 -1 1 \end{array}=q^{-\frac{1}{4}},$ $\begin{array}{c} 1 1 -1 \end{array}=q^{\frac{1}{4}},$ $\begin{array}{c} 2 -2 2 \end{array}=q,$ $\begin{array}{c} 2 0 0 \end{array}=1,$ $\begin{array}{c} 2 2 -2 \end{array}=q^{-1}\text{.}$

Substituting (7.15)-(7.17) in (7.10) and (7.11) we obtain an explicit expression for $R^{{\omega }_1{\omega }_1}$ though $𝔰𝔩\left(2\right)\text{-CGC.}$

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