I'm trying to understand what the conditions are for the Lax pairs for the zero-curvature representation:


$$
\partial\_t U - \partial\_x V + [U,V]=0
$$


where $U=U(x,t,\lambda)$ and $V=V(x,t,\lambda)$ are matrix-valued functions and $\lambda$ is a parameter.


The motivation behind this question is that the Lax pairs for the KdV equation:


$$
u\_t + 6uu\_x - u\_{xxx} = 0
$$


is given by:


$$
U = \begin{pmatrix} 0 & 1 \\ \lambda + u & 0 \end{pmatrix} \text{ and } V = \begin{pmatrix} u\_x & 4 \lambda - 2u \\ 4 \lambda^2 + 2 \lambda u + u\_{xx} - 2u^2 & - u\_x \end{pmatrix}
$$


Now, it is not too difficult to verify that this indeed satisfies the zero-curvature representation, but I'm trying to figure out why we cannot use the Lax pairs:


$$
U = \begin{pmatrix} 0 & 0 \\ \lambda + u & 0 \end{pmatrix} \text{ and } V = \begin{pmatrix} 0 & 0 \\ \lambda + 3 u^2 - u\_{xx} & 0 \end{pmatrix}
$$


These matrices clearly satisfy the zero-curvature representation, but for some reason none of the notes I've been reading use them. What is the reason that they are not a valid Lax pair for the KdV equation?


I've also asked this question here (I hope that's ok): <http://www.physicsoverflow.org/26475/integrability-conditions-of-lax-pairs>



Integrability - conditions of lax pairs
One way to see this, is that you want the zero-curvature representation to be useful and tell you something you didn't know before. Your representation has the problem of being singular, in the sense that the Lax matrices have zero determinant. It would be more eveident if we were really speaking of the Lax representation:
$$L\_t=[M,L]$$
where all the usefulness comes from having the possibility to say that the eigenvalues (or the coefficients of the characteristic polynomial) of $L$ do not evolve (and hence are integrals of motion). You see that, if you take an $L$ for which the coefficients of the characteristic polynomials are identically zero, you don't achieve much.
This very same argument applies to the zero curvature representation if you remember how to pass from it to the Lax representation via the monodromy matrix. In any case, even if you ignore for the time being this problem and try to go through inverse scattering, you very soon hit the same wall.


