$V$ be a 3 dimensional space with basis $\{e_1,e_2,e_3\}$ over $\mathbb{C}$. Let $q\in \mathbb{C}^\ast$.
$\beta: V\otimes V \to V\otimes V$ defined by \begin{equation} \beta(e_i\otimes e_j) = \begin{cases} e_i\otimes e_i & \text{if } i = j\\ q(e_j\otimes e_i) & \text{if } i< j\\ q(e_j\otimes e_i) + (1-q^2)(e_i\otimes e_j)& \text{if } i> j \end{cases} \end{equation}
Is there a smart way of verifying that this operator $\beta$ satisfies Yang Baxter Equation?
Let $\beta_1=id\otimes \beta :V\otimes V\otimes V\to V\otimes V\otimes V$ and $\beta_2=\beta\otimes id:V\otimes V\otimes V\to V\otimes V\otimes V$
I need to verify $\beta_1\circ\beta_2\circ\beta_1 = \beta_2\circ\beta_1\circ\beta_2$.
checking on every elements or computing matrices and then performing matrix multiplication are really very painful. Is there any logical reasoning why this operator is a solution to Yang Baxter equation. I found that this operator is called Jimbo's operator of type $A^{(1)}_2$. I am not able to figure what that is?
Any motivational comments will also be helpful. Thanks.
You should look into $R$-matrix. There are a structure designed to automatically solve Yang Baxter. So maybe the "smart way" you are looking for is simply checking whether or not this can be written into an $R$-matrix.
But of course, I might have not fully understood the subtleties of your question. I am myself pretty new to Quantum Groups...