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This is supposed to be a one-line derivation, but how to deduce the commutativity of the transfer matrix (in algebraic bethe ansatz) from the RTT relation?

Thank you!

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  • $\begingroup$ Just apply (R)^(-1) at each side then take the trace, then just by cyclicity of the trace you have your result $\endgroup$ – Giuseppe May 2 '18 at 21:30
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Let me flesh out what @Giuseppe more or less said.

Let me suppress spectral parameters for now, since nothing happens to those. Multiply by the inverse of the R-matrix on one side, so that you get e.g. $T_a\,T_b = R_{ab}^{-1} \, T_b\,T_a\, R_{ab}$. Now trace over both auxiliary spaces:

$$\text{tr}_{ab}[T_a\,T_b] = \text{tr}_{ab}[R_{ab}^{-1} \, T_b\,T_a\, R_{ab}] = \text{tr}_{ab}[T_b\,T_a\, R_{ab} \, R_{ab}^{-1}] = \text{tr}_{ab}[T_b\,T_a] \ ,$$

where the middle equality uses cyclicity of the trace. (Note that we cannot use this cyclicity to reverse the two monodromy matrices since, unlike the R-matrix, the $T$s also act on the 'physical' space, aka 'quantum' space, where they don't commute and we don't take a trace.) Finally use that $\text{tr}_{ab} = \text{tr}_a \otimes \text{tr}_b$ to get the commutativity of the transfer matrices $t(u) = \text{tr}_a\, T_a(u)$ and $t(v) = \text{tr}_b\, T_b(v)$, where I put back the spectral parameters.

To turn this one-line computation into a complete argument we have to add one more paragraph. What it really shows is that transfer matrices commute as long as the R-matrix is invertible. For example, in the six-vertex model (the XXZ spin chain) this is the case for almost all values of its argument $u-v$, as one checks by computing $\det R(u-v)$. It follows that $t(u)$ commutes with almost all $t(v)$. But typically, for physical spaces of the form $W=V^{\otimes L}$, the entries of the transfer matrix are polynomial in the vertex weights (= entries of the Lax operator/R-matrix used to define $T$ and thus $t$), and the vertex weights are analytic in the spectral parameter. By analyticity of the transfer matrix the commutativity thus extends to all values of $u,v$.

For a graphical version you can check out e.g. Section 4.2 of my lecture notes or any of the references I give in its introduction.

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    $\begingroup$ @WeichengYe You're welcome! If you are satisfied with my answer you can accept it by clicking the check-mark on the right. $\endgroup$ – Jules Lamers May 3 '18 at 15:13

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