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I have some basic question regarding the Lax formalism for 2-dimensional integrable systems.

The monodromy matrix is build from the Lax component $U(\lambda)$ $$T(\lambda)=\mathcal P \exp \int_0^{2\pi}dx U(\lambda), $$ on can compute that $$\partial_t T(\lambda)=\int_0^{2\pi}dx \partial_x \left(\mathcal P e^{\int_x^{2\pi}dyU}V(\lambda)\mathcal P e^{\int_0^{x}dyU}\right), $$ where $v$ is the other Lax component. How does one arrive at the conclusion that this is equivalent to $$\partial_t T(\lambda)=[V(2\pi,t,\lambda),T(\lambda)].$$ And how can one conclude that the eigenvalues of $T(\lambda)$ generate an infinite set of integrals of motion upon expansion in the spectral parameter $\lambda$.

(I'm referring to notes )

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The reason is in fact very simple: $$\partial_t T(\lambda)=\int_0^{2\pi}dx \partial_x \left(\mathcal P e^{\int_x^{2\pi}dyU}V(\lambda)\mathcal P e^{\int_0^{x}dyU}\right)\\= \mathcal P e^{\int_{2\pi}^{2\pi}dyU}V(\lambda)e^{\int_0^{{2\pi}}dyU} -\mathcal P e^{\int_{2\pi}^{0}dyU}V(\lambda)e^{\int_0^{{0}}dyU}\\ =\mathcal P V(\lambda)e^{\int_0^{{2\pi}}dyU} -\mathcal P e^{\int_{2\pi}^{0}dyU}V(\lambda)\\ =[V(\lambda),T(\lambda)].$$ This answers the first question. For the second one: the trace of a matrix is nothing but the sum of the eigenvalues, and thus, assuming $T$ is diagonalizable, the sum of the diagonal elements in the basis determined by the eigenvectors. But any diagonal matrix sits in the centre of the matrix group and the commutant has to vanish.

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