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Consider group algebra $C[S_n]$. Take any of its representation $(\pi, V)$ (for example regular). Take some complex numbers $z_i$ i=2...n. Denote as usually by $(1i)\in S_n$ the transpositions of (1) and (i).

Question solve the following ODE (it is known that its solution will be polynom p(z) in $z$) :

$ p'(z) = m\sum_{i=2...n} \frac{ Id + \pi( (1i) )}{z-z_i} p (z) $

Here "m" is integer, $p(z)$ is $V$-valued polynom.

The hope is of course for some "nice" formula, although I do not think it is easy question.

Well, here should the background... [EDIT - see background below.] May I kindly apologize and hide the background from you, at least for some hours, I will insert it later, I think that if I put the background you will say "ah famous equation people studied it for many years and everything is known", however I think only 5 people thought on this particular question, and I do not think these thoughts are exhaustive. So I am afraid background will be demotivating...

There are two simple cases where it is easy to get the solution - trivial representation and sign representation.

For sign representation solution is constant and can be embeded in C[S_n] as p(z) = antisymmetrizer.

For trivial representation the solution is $(symmetrizer) * \prod_i (z-z_i)^2 $.

Even for vector representation in $C^n$ I do not know the nice solution.

[EDIT] BACKGROUND

This equation is basically Knizhnik-Zamolodchikov equation (for gl_n (by Schur-Weyl for S_n)). Much is known about it and about its solutions. However main interest is focused on the case when m=1/integer - in this case the solutions are NON-polynomial.

[Some theory] For the Fuchsian ODE $f'(z) = \sum_i \frac{A_i}{z-z_i} f(z)$, the obvious necessary condition to have rational (respectively polynomial) solutions is that eigenvalues of A_i are integers (respectively naturals). This is clear since near z=z_i solutions have the form $(z-z_i)^{(A_i)}$, which is up to change of basis $z^{(\lambda_i)}$, where $\lambda_i$ are eigenvalues of $A_i$.

One can easily see that for our particular equation these conditions are satisfied. (Since (1i) is transposition so its eigs are {-1,1}).

Also considering z=infinity we should impose same condition on $\sum_i A_i$. This sum is actually a Jacus-Murphy element and it is known to have integer eigs, see Why are Jucys-Murphy elements' eigenvalues whole numbers?

But these conditions are not sufficient. So the first surprise is that for this particular equation it is nevertheless true, that solutions are polynomial.

This actually can be seen as a corollary of Drinfeld-Kohno theorem, which predicts that monodromy of KZ is given by the R-matrix of corresponding quantum group with q=e^(2\pi i m) =1 for m-integer and R-matrix is degenerated to identity matrix. However proofs of Drinfeld-Kohno theorem assumes "m" is in general position, and so this case should be considered specially.

The conditions for Fuchsian equation to be have rational solutions has been analysed:

http://arxiv.org/abs/math/0607555

Meromorphic Solutions of Linear Differential Systems, Painleve type functions

Lev Sakhnovich

[Related works] http://arxiv.org/abs/math/0610383 Polynomial solutions of the Knizhnik-Zamolodchikov equations and Schur-Weyl duality Giovanni Felder, Alexander P. Veselov

Moreover they have constructed the polynomial solutions. however in terms of some integrals - which may not be the end of the story..

Lev Sakhnovich studied the question in a series of papers:

Explicit calculations are contained in http://arxiv.org/abs/math-ph/0609067 Rational solutions of Knizhnik-Zamolodchikov system

http://arxiv.org/abs/0903.3873 http://arxiv.org/abs/0805.2662

http://arxiv.org/abs/math/0702404 Rational solutions of KZ equation

We came to this question from the point of view of the Gaudin integrable system. It appears to be that KZ solutions for m=1 gives rise to the solutions the so-called GL-oper=BaxterTQequation=quantum spectral curve equation for the Gaudin model. Which encode the information about the Bethe ansatz spectrum of quantum Gaudin Hamiltonians. See e.g. http://arxiv.org/abs/0711.2236 page 15 section "4.1.1 Application to the Knizhnik-Zamolodchikov equation" and references their in.

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    $\begingroup$ I loved the «background will be demotivating» :) $\endgroup$
    – Mariano Suárez-Álvarez
    Apr 25, 2012 at 19:31
  • $\begingroup$ more we learn less we know :) $\endgroup$
    – Alexander Chervov
    Apr 25, 2012 at 19:38
  • $\begingroup$ Sasha, do you know the solution for the group $S_4$, or at least $S_3$ (regular presentation)? $\endgroup$
    – Dmitri Panov
    Apr 26, 2012 at 10:50
  • $\begingroup$ @Dima, I tried to calculate, as far as I remember for S_3 it was managable, however for S_4 I used Mathematica... Let me write the background and give the references.. Here is reference with explicit calculations for S_4, but the "nice" answer is not yet found arxiv.org/abs/math/0702404 Rational solutions of KZ equation, case $S_4$ Lev Sakhnovich $\endgroup$
    – Alexander Chervov
    Apr 26, 2012 at 11:28
  • $\begingroup$ Background - this equation is basically Knizhnik-Zamolodchikov equation for the group S_n and can be rewritten as gl_n KZ in vector representation. arxiv.org/abs/math/0610383 Polynomial solutions of the Knizhnik-Zamolodchikov equations and Schur-Weyl duality Giovanni Felder, Alexander P. Veselov gives solution however in terms of some integrals - which may not be the end of the story... Explicit calculations are contained in arxiv.org/abs/math-ph/0609067 Rational solutions of Knizhnik-Zamolodchikov system L Sakhnovich arxiv.org/abs/0903.3873 arxiv.org/abs/0805.2662 $\endgroup$
    – Alexander Chervov
    Apr 26, 2012 at 11:28

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