Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

I am interested in finding non-constant solutions to the following Yang Baxter equation

$$R_{12}(x/y) R_{13}(x/z) R_{23}(y/z) = R_{23}(y/z) R_{13}(x/z) R_{12}(x/y)$$

where $R(x)$ is an endomorphism of $V\otimes V$. Personally I happen to be interested in finding explicit solutions in the case where $V$ is of dimension 2.

I would like to know what families of solutions are known. For example, there is an example at the end of Ch 12 of Chari and Pressley related to quantum affine sl2. I would prefer answers that are as explicit as possible (eg writing $R$ as 4x4 matrix) and some context as to where they come from.

share|improve this question

5 Answers 5

The form of the Yang Baxter equation you are studying developed to a large extent in the context of solvable lattice modes in statistical mechanics. There the R matrix is used to construct a transition function in a certain statistical system. At least in some cases, the R matrix coming from statistical mechanics agrees with the R matrix coming from a quantum affine algebra. The standard example is the relationship between the ``six-vertex model" and $U_q(\widehat{sl}_2)$. A good reference is Jimbo and Miwa's book "algebraic analysis of solvable lattice models" (mathscinet review). At least for the R matrix coming from $U_q(\widehat{sl}_2)$, they do give explicit formulas. Unfortunately, I don't know a good online reference.

share|improve this answer
2  
Jimbo and Miwa's book can be found online if you look in the right place (gigapedia). –  Peter McNamara Nov 21 '09 at 18:15

Maybe I should point out the paper by Hietarinta "Solving the two-dimensional constant quantum Yang--Baxter equation", which can be found on the web. There, he completely classifies constant solutions in 2 dimensions. Of course, this is not quite what you need (since in his paper there is no spectral parameters), but still maybe it's useful for something.

share|improve this answer

I think this is a very well-understood problem; it might be helpful for you to know (which you may already?) that Yang-Baxter equation with a parameter like that are known as "QYBE with spectral parameter". Well that phrase usually refers to the additive version of what you wrote with arguments (a-b), (a-c) and (b-c) respectively.

What you are interested in is called the "QYBE with multiplicative spectral parameter", also called the "trigonometric QYBE" (because your x,y,z can express as e^a, e^b, and e^c of the additive situation; it's just a naming convention).

A paper about the constructions is here (but having those keywords you will find many papers online):

http://www.google.com/url?sa=t&source=web&ct=res&cd=4&ved=0CBkQFjAD&url=http%3A%2F%2Fwww.ma.utexas.edu%2Fmp_arc%2Fc%2F94%2F94-127.ps.gz&ei=gET7StLOLdDdlAfcsPW5Aw&usg=AFQjCNEP59Q-x84lijdXHJkV95Oxgei3Cg&sig2=8JebQcUBnWYKgQPfgO0hVw.

I'm not an expert by any means, but I hoped the keywords and link i provided might help your search. If you're interested in solutions where V is 2-dimensional, then I think all of them you'll get from quantum affine Lie algebras are going to come, as you said from affine sl_2.

Are you able to decipher from Chari and Pressley the explicit form of the R-matrix in the quantum affine sl_2 case that interests you?

For constant QYBE there is a theorem (which for sl2) one can prove more or less by hand, that if you require the R-matrix to be Hecke (i.e. having only two eigenvalues which you can normalize to be q, -q^{-1}) then the only R-matrix you can find is the R-matrix for quantum sl2 (not affine since there aren't parameters). I would guess that whatever context motivates your search imposes some restriction on the QYBEwMSP you want to find, so that the quantum affine sl_2 R-matrices are the only examples. But I am just guessing as to your motives when I say that.

share|improve this answer

Both

$R(\lambda)=\left(\begin{array}{cccc} \lambda + \alpha & 0 & 0 & 0 \\\\ 0 & \lambda & \alpha & 0 \\\\ 0 & \alpha & \lambda & 0 \\\\ 0 & 0 & 0 & \lambda + \alpha \end{array}\right)$

and

$R(\lambda) = \left(\begin{array}{cccc} \sin(\lambda + \alpha) & 0 & 0 & 0 \\\\ 0 & \sin(\lambda) & \sin(\alpha) & 0 \\\\ 0 & \sin(\alpha) & \sin(\lambda) & 0 \\\\ 0 & 0 & 0 & \sin(\lambda + \alpha) \end{array}\right)$

satisfy

$R_{12}(\lambda_1-\lambda_2)R_{13}(\lambda_1-\lambda_3)R_{23}(\lambda_2-\lambda_3)=R_{23}(\lambda_2-\lambda_3)R_{13}(\lambda_1-\lambda_3)R_{12}(\lambda_1-\lambda_2)$

One good introduction is this paper of Faddeev.

share|improve this answer

The general theory (due to Jimbo) is that each irreducible finite dimensional representation of the quantised enveloping algebra of a Kac-Moody algebra (not of finite type) gives a trigonometric R-matrix. There is substantial information on these representations but the R-matrices are not explicit. There is a special case which is explicit and is given by the "tensor product graph" method (this was worked out by Niall MacKay and Gustav Delius).

I used this in my paper: R-matrices and the magic square. J. Phys. A, 36(7):1947–1959, 2003. and you can find the references there.

If you want to go beyond this special case and be explicit then you can use "cabling" a.k.a "fusion".

The only papers which deal with R-matrices not covered by the tensor product graph method that I know of are

Vyjayanthi Chari and Andrew Pressley. Fundamental representations of Yangians and singularities of R-matrices. J. Reine Angew. Math., 417:87–128, 1991.

G´abor Tak´acs. The R-matrix of the Uq(d(3)4 ) algebra and g(1)2 affine Toda field theory. Nuclear Phys. B, 501(3):711–727, 1997.

Bruce W. Westbury. An R-matrix for D(3) 4 . J. Phys. A, 38(2):L31–L34, 2005

Deepak Parashar, Bruce W. Westbury R-matrices for the adjoint representations of Uq(so(n)) arXiv:0906.3419

The Chari & Pressley paper deals with rational R-matrices. The last preprint was an incomplete attempt to try and find the trigonometric analogues of these R-matrices.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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