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Let $\mathfrak{gl}_n(x)= \mathfrak{gl}_n \otimes_\mathbb{C}\mathbb{C}(x)$ be the algebra of matrices taking values in rational functions.

Definition: Two matrices $A, B \in \mathfrak{gl}_n(x)$ are said to be gauge equivalent iff there exists an invertible matrix $P \in {GL}_n(\mathbb{C}(x))$ satisfying: $$B=PAP^{-1}+P'P^{-1}$$

The above is easily seen to be an equivalence relation on the matrices in $\mathfrak{gl}_n(x)$.

The classical Jordan decomposition for complex scalar matrices gives a canonical (modulo transposition of blocks) representative for every equivalence class of matrices for the equivalence relation of similarity. My question is basically "is there a similar thing for this case and if there is what is it?".

Question: Can one define a "canonical form" for matrices in $\mathfrak{gl}_n(x)$ modulo gauge equivalence? That is a sensible and computable assignment of a canonical representative matrix for every gauge equivalence class in $\mathfrak{gl}_n(x)$

If the above is some kind of extremely difficult untractable super-problem I'd appreciate an answer which explains why is this the case and additionally:

If the above is hopless can this be done at least for $\mathfrak{gl}_n \otimes_\mathbb{C}\mathbb{C}((x))$? With $P$ above lying in $GL_n(\mathbb{C}((x)))$ of course.

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  • $\begingroup$ Are you sure that this is the relation you want? It should be $$ A = P^{-1}BP + P^{-1}\,P', $$ shouldn't it? $\endgroup$
    – Robert Bryant
    Commented Jun 17, 2017 at 16:34
  • $\begingroup$ @RobertBryant Previous version was indeed wrong. Now I set it up so that one gets the expression for $B$ by putting $P(\partial - A)P^{-1}=(\partial - B)$. The expression you mention probably comes from a different convention. $\endgroup$
    – Saal Hardali
    Commented Jun 17, 2017 at 20:41
  • $\begingroup$ Rewrite the equation as $P' = BP - PA$. Use the usual ODE theory to solve for $P$. Since the equation is linear, the solution exists globally, on any simply connected domain. If your domain has closed loops, then the holonomy of $\partial -A$ and $\partial -B$, seen as flat connections, must agree, otherwise your solution will be at most multivalued. $\endgroup$
    – Igor Khavkine
    Commented Jun 18, 2017 at 7:53
  • $\begingroup$ @Igor Khavkine What does this do? Takes two given matrices A and B which are locally gauge equivalent and finds a gauge transformation between them? That's not really the topic of the question. $\endgroup$
    – Saal Hardali
    Commented Jun 18, 2017 at 7:56
  • $\begingroup$ @SaalHardali, If the solution exists for $A=0$ (modulo holonomy constraints) then zero is your canonical form. $\endgroup$
    – Igor Khavkine
    Commented Jun 18, 2017 at 8:17

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For your second question, you may want to look at "Jordan decomposition for a class of singular differential operators" by A. H. M. Levelt, https://projecteuclid.org/download/pdf_1/euclid.afm/1485896417

The only difference with your problem is that he allows gauge transformations by an extension of the Laurent series field.

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