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The Yang-Baxter equation, or rather one given solution of the Yang-Baxter equation corresponding to some model is often described as the fundamental relation defining an integrable model. However in the construction of the transfer matrix of the XXZ spin chain or equivalently of the 6-vertex model, which can be found e.g. here or here, the Yang Baxter equation for the $R$-matrix is deduced as a compatibility condition for two reordering of a product of three monodromy matrices to be equivalent (see for instance (B.39) in Franchini 2017), and is not even a necessary condition for the $RTT$ relation to hold, and thus for the transfer matrix to commute.

Thus, I fail to see, in this elementary example, why is the Yang-Baxter equation (understood as the relation between two products of three $R$ matrices) so fundamental in integrability.

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The quantum inverse scattering method (QISM) revolves around 'quantum operators' (acting on the spin-chain Hilbert space), which are usually denoted by $A(x),\dots$. They are conveniently grouped together as the matrix entries (in 'auxiliary space') of the monodromy matrix $T_a(x)$. As you've probably seen, the exchange relations encoded in $$R_{ab}(x-y)\,T_a(x)\,T_b(y)=T_b(y)\,T_a(x)\,R_{ab}(x-y) \,,$$ guarantee that the corresponding transfer matrices $t(x) = \mathrm{tr}_a \, T_a(x)$ and $t(y) = \mathrm{tr}_b \, T_b(y)$ commute and can be simultaneously diagonalised by the algebraic Bethe ansatz. But for these 'RTT relations' to be consistent -- i.e. define an (associative) algebra -- we need the $R$-matrix, which similarly groups together the structure constants, to obey the YBE -- which ensures the associativity of that algebra. This is the two ways triple products of monodromy matrices can be reordered, encoding cubic relations for the quantum operators.

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