2
$\begingroup$

What is the relation between the definitions of Yangian $Y(sl_2)$ and quantum affine algebra $U_q(\widehat{sl_2})$?

There are two definitions of $U_q(\widehat{sl_2})$. The following is Jimbo presentation. The quantum affine algebra $U_q(\widehat{\mathfrak{sl}_2})$ is an associated algebra generated by $e_0^{\pm}, e_1^{\pm}, K_0^{\pm 1}, K_1^{\pm 1}$ subject to the relations: \begin{align} & K_0 K_1 = K_1 K_0, \\ & K_i K_i^{-1} = K_i^{-1} K_i = 1, \ i=0,1, \\ & K_i e_i^{\pm} = q^{\pm 2} e_i^{\pm} K_i, \ i=0,1, \\ & K_i e_j^{\pm} = q^{\mp 2}e_j^{\pm} K_i, \\ & e_0^{\pm} e_1^{\mp} = e_1^{\mp} e_0^{\pm} \\ & e_i^{+} e_i^{-} - e_i^{-}e_i^{+} = \frac{K_i-K_i^{-1}}{q-q^{-1}}, \ i=0,1, \end{align} and quantized Serre relations: \begin{align} (e_i^{\pm})^3 e_j^{\pm} - [3] (e_i^{\pm})^2 e_j^{\pm}e_i^{\pm} + [3]e_i^{\pm} e_j^{\pm} (e_i^{\pm})^{2} -e_j^{\pm}(e_i^{\pm})^3 = 0, \ i \neq j. \end{align}

The following is Drinfeld presentation. The quantum affine algebra $U_q(\widehat{\mathfrak{sl}_2})$ is an associated algebra generated by $x_m^{\pm}, h_r, K^{\pm 1}, c^{\pm 1}$, $m \in \mathbb{Z}, r \in \mathbb{Z}-\{0\}$, subject to the relations: \begin{align} & c^{\pm 1} \text{ are in the center}, \\ & K K^{-1} = K^{-1} K = 1, \\ & c c^{-1} = c^{-1} c = 1, \\ & [K, h_r] = 0, \\ & K x^{\pm} = q^{\pm 2} x_i^{\pm} K, \\ & [h_k, x_l^{\pm}] = \pm \frac{1}{k}[2k]c^{\mp |k|} x_{k+l}^{\pm}, \\ & x_{k+l}^{\pm} x_l^{\pm} - q^{\pm 2}x_l^{\pm} x_{k+l}^{\pm} = q^{\pm 2} x_k^{\pm} x_{l+1}^{\pm} - x_{l+1}^{\pm} x_k^{\pm}, \\ & [h_k, h_l] = \delta_{k,-l} \frac{1}{k} [2k] \frac{c^k-c^{-k}}{q-q^{-1}}, \\ & [x_k^+, x_l^-] = \frac{1}{q-q^{-1}} [c^{k-l}\psi_{k+l}-\phi_{k+l}], \end{align} where $\psi_k$, $\phi_k$ are given by: \begin{align} & \sum_{m=0}^{\infty} \psi_m z^m = K \exp\left( (q-q^{-1}) \sum_{s=1}^{\infty} h_s z^s \right), \\ & \sum_{m=0}^{\infty} \phi_{-m}z^{-m} = K^{-1} \exp\left( -(q-q^{-1}) \sum_{s=1}^{\infty} h_{-s} z^{-s} \right), \end{align} and $\psi_k=0$, $\phi_{-k}=0$, $k < 0$.

The definition of $Y(sl_2)$ is defined as follows.

Thank you very much.

$\endgroup$
1

1 Answer 1

2
$\begingroup$

This answer may be way too late to be relevant, so I'll keep it short. Both Yangians and quantum affine algebras are associative algebras that can be described in the Faddeev--Reshetikhin--Takhadjan presentation in terms of generators $T_{ij}(u)$, with $i,j\in\{1,\cdots,n\}$ for $\mathfrak{sl}_n$, subject to the relation

$$ R(u-v) \, (T(u)\otimes \mathbf{1}) \, (\mathbf{1}\otimes T(v)) = (T(v)\otimes \mathbf{1}) \, (\mathbf{1}\otimes T(u)) \, R(u-v) \qquad\qquad\qquad (*) $$

where $T(u)$ is an $n\times n$ matrix with entries $T_{ij}(u)$ and $\mathbf{1}$ is the $n\times n$ identity matrix. The R-matrix $R(u)$ has size $2n\times 2n$ with scalar entries (i.e. that commute with each other and the $T_{ij}$), and solves the (quantum) Yang--Baxter equation.


It is easy to see how this relates to the Yangian. Consider Yang's "rational" R-matrix

$$ R^\text{rat}(u) = u\,\mathbf{1}\otimes\mathbf{1} + P \, ,$$

with $P$ the permutation operator on $\mathbb{C}^n \times \mathbb{C}^n$; for $\mathfrak{sl}_2$ it is

$$ P = \begin{pmatrix} 1 & & & \\ & 0 & 1 & \\ & 1 & 0 & \\ & & & 1 \end{pmatrix} \, , $$

where I have omitted some entries that are zero. You can now rewite $(*)$ to get the defining relation for the Yangian that you quote.

The quantum-affine case instead comes from the "trigonometric" R-matrix, which for $\mathfrak{sl}_2$ reads

$$ R^\text{tri}(u) = \begin{pmatrix} \sin(u+\eta) & & & \\ & \sin(u) & \sin(\eta) & \\ & \sin(\eta) & \sin(u) & \\ & & & \sin(u+\eta) \end{pmatrix} \, . $$

[Note that $R^\text{tri}(\eta\,u)/\sin(\eta)\to R^\text{rat}(u)$ as $\eta\to0$.] It is a little bit more work to get Jimbo's presentation of $U_q(\widehat{\mathfrak{sl}_2})$ from $(*)$, but not very hard either. The details can be found in e.g. Section 2.3 of the book Quantum groups in two-dimensional physics by Gómez, Ruiz-Altaba and Sierra.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.