Involutive solutions to the Yang-Baxter equation (and triangular Hopf algebras)

Question

I'm interested in solutions to the Yang-Baxter equation $$R_{12}R_{23}R_{12}=R_{23}R_{12}R_{23},$$ that are involutive $R^2_{12}=1$. Or put it another way, I'm interested in representations of the symmetric group $S_N$ in the tensor product vector space $V^{\otimes N}$, where the element $P_{j,j+1}$ acts like $I^{\otimes (j-1)}\otimes R\otimes I^{\otimes (N-j-1)}$.

Some trivial solutions are $R_{12}=\pm I_{12}$ (where $I_{12}=I\otimes I$ is the identity matrix), $R_{12}=\Pi_{12}\phi_{12}$, where $\Pi_{12}$ flips the first two factor spaces, i.e. $\Pi(a\otimes b)=b\otimes a, \forall a,b\in V$, and $\phi_{12}$ is a diagonal matrix with diagonal values $\pm 1$.

A very well studied family of solutions are the so called set-theoretical solutions, see this post. However, it seems that most (or all) of the involutive solutions in this family satisfy the "non-degenerate" condition, and, as shown in arXiv:math/9801047, in the resulting representation of $S_N$, $P_{j,j+1}$ is conjugate to the standard flip, i.e. $P_{j,j+1}=J_N \Pi_{j,j+1} J^{-1}_N$ for some invertible map $J_N$.

Does there exist an involutive solution to YBE that is not conjugate (when acting on $V^{\otimes N}$) to any of the trivial solutions?

Also, it seems to me that a systematic way of generating new solutions to YBE is by an algebraic structure called quasi-triangular Hopf algebras, and involutive solutions correspond to triangular Hopf algebras. Triangular Hopf algebras have been classified in arXiv:math/0202258. But that paper is too formidable for me to read, and it seems that a lot of triangular Hopf algebras, e.g. Sweedler's four dimensional $H_4$, gives a trivial $R$ matrix, something like $R=\Pi$. So, may I ask, which families of triangular Hopf algebras give us "non-trivial" solutions to the YBE? ("non-trivial" means not conjugate to any of the trivial solutions when acting on $V^{\otimes N}$)

Answer 1

I am not sure about the conjugation property that you are asking, but I know some solutions to the set theoretical YB equation, which are degenerate, nevertheless interesting. The simplest one comes from the models treated in solv-int/9712008. So if your set is X={1,2,3} then the map is R(1,2)=(2,1) R(2,1)=(1,2) R(1,3)=(3,1) R(3,1)=(1,3) and on all other pairs it acts as the identity. It can be checked that this satisfies set. th. YB and also it is degenerate, so it is not discussed in the paper you mention. I did not think about the conjugation property you mentioned... maybe it should be checked. In any case many other degenerate solutions can be found, and they are interesting from a physical point of view.

Update: I don't know what is the quickest proof, there should be a simple one. But I know this one. Take my map I wrote, and construct on a chain of length 4 the following combination: $\tilde R=R_{14}R_{23}R_{34}R_{12}$. If the nearest neighbour maps $R_{j,j+1}$ are conjugate to the elementary exchanges $\Pi_{j,j+1}$, then the map $R_{14}$ is also conjugate to the exchange $\Pi_{1,4}$. If we look at the permutation group $S_4$, then the permutation corresponding to the element $\tilde R$ has order 2. So if $R$ is conjugate to the permutations, then we should see $(\tilde R)^2=1$. But this is not true, as direct computations shows, you just apply these combinations twice and you see that you don't come back to the original configuration. For example $(\tilde R)^2(1,2,3,3)=(1,3,3,2)$.

I guess there should be a simpler proof.

Update The identity map is degenerate. My example is also degenerate, but at least it ,,does something''. It is actually a simple union (in the sense of the Etingof et al paper) of the identity map on the states 2,3 and the separate state 1.

Answer 2

Here is a family of examples indexed by an integer $m\geq 3$. Let \begin{eqnarray} R^{ij}_{kl}=\lambda_{ij}c_{kl}-\delta_{ik}\delta_{jl},\tag{1} \end{eqnarray} where $\lambda, c$ are $m\times m$ matrices satisfying \begin{equation}\label{eq:lambdac} \lambda c \lambda^T c^T=I_{m}, ~~~ \mathrm{Tr}(\lambda c^T)=2, \end{equation} where $I_m$ denotes the $m\times m$ identity matrix. [For example, for $m=2k$, we can take $\lambda=\mathrm{diag}\{I_{2},\sigma^z,\sigma^z,\ldots,\sigma^z\}, c=I_m$, and for $m=2k+1$, we can take $\lambda=\mathrm{diag}\{-1,(\frac{\sqrt{5}+3}{2})^{\sigma^y},\sigma^z,\ldots,\sigma^z\}, c=I_m$. Here $\sigma^{x},\sigma^{y},\sigma^{z}$ are the Pauli matrices.]

To show they are not conjugate to any of the trivial solutions, we use the tool of Hilbert series: for any involutive $R$-matrix, let $V_N(R)$ denote the representation of $S_N$ generated by $\{R_{12},R_{23},\ldots,R_{N-1,N}\}$ [as a vector space , $V_N(R)\cong V^{\otimes N}$]. Let $d_N$ denote the number of trivial representations contained in $V_N(R)$ [or, equivalently, $d_N$ is the dimension of common eigenspace of $\{R_{12},R_{23},\ldots,R_{N-1,N}\}$ with eigenvalue $+1$.]. Then the Hilbert series of $R$ is defined as: $$h_R(x)=\sum_{n\geq 0} d_n x^n.$$ Obviously, if two $R$-matrices $R_1, R_2$ are conjugate [i.e. $V_N(R_1)\cong V_N(R_2), \forall N\geq 0$], then their Hilbert series must be the same. The following table lists the Hilbert series for all the trivial $R$-matrices along with the $R$ in Eq.(1), which shows that this family of $R$ matrices are not conjugate to any of the trivial ones [and also not their direct sums, as $h_{R_1\oplus R_2}(x)=h_{R_1}(x)h_{R_2}(x)$]. $$\begin{array}{|c|c|c|} \hline R^{ij}_{kl} & h_R(x) & h_{-R}(x) \\ \hline -\delta_{il}\delta_{jk} & (1+x)^m & 1/(1-x)^m \\ \hline \delta_{il}\delta_{jk}(-1)^{\delta_{ij}} & (1+x)^m & 1/(1-x)^m \\ \hline -\delta_{ik}\delta_{jl} & 1+mx & 1/(1-mx) \\ \hline \lambda_{ij}c_{kl}-\delta_{ik}\delta_{jl} & 1+mx+x^2 & 1/(1-mx+x^2) \\ \hline \end{array}$$