I have two matrices $U(\lambda, x,t)$ and $V(\lambda, x,t)$, where $\lambda$ is a parameter, which belong to the $sl(2)$ algebra, and satisfy the zero-curvature equation $$ \partial_t U - \partial_x V + [U,V] = 0 $$ And I would like to diagonalize $U$ using a local gauge transformation $$ U \to A = gUg^{-1}+ \left( \partial_x g \right) g^{-1} \tag{1} $$ Is there an algorithm or formula to achieve this? I tried to Google it, but couldn't find anything,.


For integrable field theories, the following quantities can be proven to be conserved in time $$ I_n(\lambda) = \mathrm{Tr} (T^n(t,\lambda)) $$

To answer fs137's question, there is a theorem which states that there exists a gauge transformation which can diagonalize both $U$ and $V$.

The reason why I want to diagonalize the matrix is because the quantities $$ Q_k(\lambda) = \int\limits_0^L A_k \mathrm{d} x $$ are the local conserved quantites of the field theory. Where $A$ is the diagonalized matrix (see equation (1)) expressed as formal series in $\lambda$ $$ A = \sum\limits_{r=-n}^\infty A_r \lambda^r $$

  • $\begingroup$ To be clear, you want the transformed $U$ to be diagonal for all $x$? What information do you have about $U$? (i.e. assumed continuity or differentiability, explicit formulas, analiticity, ... ?) What is the role of $t$? I also imagine the second term is $(\partial_x g)g^{-1}$, but the notation coud use some sprucing up. $\endgroup$ – Emilio Pisanty May 19 '16 at 17:18
  • $\begingroup$ Try to solve the PDE for $g$. Perhaps a cleaner starting question would be, "When is it possible to find a gauge transformation $g$ that diagonalizes $U(x,t)$". $\endgroup$ – TotallyRhombus May 19 '16 at 17:22
  • $\begingroup$ This seems to depend rather crucially on your gauge group. In any case, diagonalizing U or V seems a rather odd thing to do since you should think of them as abstract elements of the Lie algebra, not concrete matrices in a fixed basis. Why do you want to do this? $\endgroup$ – ACuriousMind May 19 '16 at 18:27
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    $\begingroup$ Well, as @fs137 points out, this is apparently 4 ODEs, but here is why they are only 3: the trace of your fundamental condition above (1) decouples in the commutator, and so decouples, contributing something already diagonal and proportional to the identity. The rest, for traceless 2x2 complex matrices can be recast in terms of Pauli vectors with the generic multiplication law $(\vec{a} \cdot \vec{\sigma})(\vec{b} \cdot \vec{\sigma}) = (\vec{a} \cdot \vec{b}) \, I + i ( \vec{a} \times \vec{b} )\cdot \vec{\sigma} ~.$ Solve for triplet g. $\endgroup$ – Cosmas Zachos May 31 '16 at 0:10
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    $\begingroup$ In case it is not self-explanatory, solve $( A_z \sigma_z)(\vec{g} \cdot \vec{\sigma})= (\vec{g} \cdot \vec{\sigma})(\vec{u} \cdot \vec{\sigma}) +(\partial_x\vec{g} \cdot \vec{\sigma})$. $\endgroup$ – Cosmas Zachos May 31 '16 at 18:39

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