The trigonometric $E_8$ R-matrix

Abstract

An expression for the R-matrix associated with ${\mathcal {U}}_q({\widehat{\mathfrak e_8}})$ in its 249-dimensional representation is given using the diagrammatic calculus of ${\mathcal {U}}_q({\mathfrak {e}_8})$ invariants.

Appendices

Appendix A Some product rules

Appendix B The main formula

Appendix C Rational limit

The rational R-matrix is obtained from the trigonometric one in the correlated limit $q,z\rightarrow 1$:

$$\begin{aligned} q=e^{\epsilon \hbar /2} \qquad z=e^{\epsilon x} \qquad \epsilon \rightarrow 0 \end{aligned}$$

Representation theoretically, it corresponds to the limit from the quantized loop algebra ${\mathcal {U}}_q({\mathfrak {e}_8}[z^\pm ])$ (i.e., the quotient of ${\mathcal {U}}_q({\widehat{\mathfrak e_8}})$ by $\prod _{i=0}^8 k_i^{n_i}=1$) to the Yangian $\mathcal Y({\mathfrak {e}_8})\cong {\mathcal {U}}_q({\mathfrak {e}_8}[z])$. In this Appendix, we show briefly how to recover the results of [12].

The main simplification in the diagrammatic calculus (see the related discussion at the start of Sect. 3.3) is that undercrossings and overcrossings become indistinguishable:

and both become equal to the naive permutation of tensors of $V\otimes V$. The reader is invited to write the simplified relations that diagrams satisfy in this limit; we only provide one example:

The expression of the R-matrix in Appendix B becomes in the rational limit:

The alternative basis of Sect. 3.3 no longer makes sense because and are degenerate; we can however use an intermediate basis, in which the R-matrix becomes

the other terms (involving empty vertices) remaining the same.

It is not hard to see that the latter result coincides with that of [12] (see the equation between (3.9) and (3.10)), being careful that the expression there is $R(w)=P\check{R}_{\text {rat}}(x=-w)$ where P is permutation of factors, on condition that the following normalizing factor is used:

$$\begin{aligned} f(w)=\frac{w(w-5\hbar )(w-9\hbar )}{(w-\hbar )(w-6\hbar )(w-10\hbar )} \end{aligned}$$

with the substitution $i\hbar \rightarrow \hbar$ and with the free parameter $\alpha$ (which is related to our parameter $\kappa$) given the value

$$\begin{aligned} \alpha =\frac{1}{2\sqrt{15}} \end{aligned}$$