• Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.


PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback


(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,064 questions , 2,215 unanswered
5,347 answers , 22,731 comments
1,470 users with positive rep
818 active unimported users
More ...

  Algebra of operators for Heisenberg XXZ spin chain model

+ 2 like - 0 dislike

Recently I've been reading Reshetikhin's lecture note https://arxiv.org/pdf/1010.5031.pdf on integrability of the 6-vertex model. The author defines a complex algebra $C_q(\widehat{\text{SL}}_2)$ as the following.

Consider the $R$-matrix

$$R=\begin{bmatrix} 1 & 0 & 0& 0\\ 0 & f(z) & z^{-1}g(z) & 0\\ 0 & zg(z) & f(z) & 0 \\ 0 & 0 & 0 & 1 \end{bmatrix}$$


$$ f(z)=\frac{z-z^{-1}}{zq-z^{-1}q^{-1}},\quad g(z)=\frac{q-q^{-1}}{zq-z^{-1}q^{-1}}. $$

Consider the matrix $\mathcal{T}(z)$ which is the generating function for the elements $T^{(k)}_{ij}$

$$\mathcal{T}(z)=\sum^{\infty}_{k=1}T^{(k)}z^{2k}+\begin{bmatrix} T^{(0)}_{11} & T^{(0)}_{12}\\ 0 & T^{(0)}_{22} \end{bmatrix}, $$

where $T^{(k)}_{ij}$ is a matrix element of $T^{(k)}$ for $k\ge 1$. Then the defining relations of $C_q(\widehat{\text{SL}}_2)$ can be written as the following matrix identities with entries in $C_q(\widehat{\text{SL}}_2)$:

$$ R(z)\mathcal{T}(zw)\otimes \mathcal{T}(w)=(1\otimes \mathcal{T}(w))(\mathcal{T}(zw)\otimes 1)R(z)$$


$$ \mathcal{T}(qz)_{11}\mathcal{T}(z)_{22}- \mathcal{T}(qz)_{12}\mathcal{T}(z)_{21}=1.$$

I have a feeling that this definition of $C_q(\widehat{\text{SL}_2})$ might have something to do with the affine quantum group $U_q(\widehat{\text{SL}_2})$, but I am not sure at this moment. Are they the same, or if not how are they related to each other?

asked Feb 3, 2023 in Theoretical Physics by anonymous [ no revision ]

1 Answer

+ 1 like - 0 dislike

The complex algebra Cq(SL2) defined by Reshetikhin is related to the affine quantum group Uq(SL2) as follows:

Cq(SL2) is a specialization of the quantized enveloping algebra Uq(sl2), which is the quantization of the Lie algebra sl2. This means that Cq(SL2) can be obtained from Uq(sl2) by setting certain parameters to specific values.

Uq(SL2) is the quantum group associated with the Lie algebra sl2 and can be thought of as a deformation of the Lie group SL2. It is a Hopf algebra and has several important structures and properties, such as the R-matrix, the coproduct, and the antipode.

In the case of Cq(SL2), the R-matrix is defined in terms of a function f(z) and g(z) and the matrix elements T(k)ij. The defining relations of Cq(SL2) are then given by the matrix identities involving T(z) and R(z).

In summary, Cq(SL2) is a specialization of Uq(sl2) and is related to the affine quantum group Uq(SL2) as a subalgebra or a quotient algebra.

answered Feb 6, 2023 by anonymous [ no revision ]

Hi, thanks for the answer. Do you have a reference for that? I would like to check the precise statement.

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification

user contributions licensed under cc by-sa 3.0 with attribution required

Your rights