Jack symmetric polynomials are known to be generalizations of Schur functions $\chi_\lambda$, for which powerful Weyl determinant formulas are known. Are there any generalizations of two determinant formulas for general Jack symmetric $P^\alpha_\lambda(x)$ functions?

The first determinant (Jacobi-Trudi) formula represents the character of the irrep GL(N) given by the partition $\lambda$ $$ \chi_\lambda(x)=\det_{i,j} s_{\lambda_i-i+j} $$ where $s_k$ are elementary Schur function and the second one gives the same function as determinant $$ \chi_\lambda(x)=\frac{\det_{i,j} x_i^{\lambda_j+N-j}}{\det_{i,j} x_i^{N-j}} $$ Jack symmetric polynomials are natural generalizations of Schur polynomials, and probably, to operate with them it would be useful to have as simple as possible analogs of Weyl formulas.


A natural analog would be a representation of the Jack polynomial as an alternating sum of eigenfunctions of the Calogero-Sutherland operator over an orbit of the symmetric group. This can be done even in the q-deformed case (i.e. for Macdonald polynomials and Macdonald operators). Such formula was conjectured by Felder and Varchenko and proved in the paper arXiv:q-alg/9603022, see formula (5-5).

| cite | improve this answer | |
  • 3
    $\begingroup$ Pavel, thank you very much for your answer. Unfortunatelly "Quantum" q-deformed polynomials are too involved for me, and, though in the classical limit relation you told me about could give one of the required formulas,I would prefer to have an "classical" explanation in terms like Calogero-Sutherland operator you mention. $\endgroup$ – Sasha Feb 8 '10 at 22:57
  • 2
    $\begingroup$ There is no difference whatsoever between the classical and quantum case. Just set q=1 and use usual Lie algebras instead of quantum groups. The arguments apply verbatim. $\endgroup$ – Pavel Etingof Feb 9 '10 at 1:30
  • 1
    $\begingroup$ At $\alpha=1$, can Jacobi-Trudi be explained as a result that follow from consideration of differential operators and eigenfunctions? $\endgroup$ – Student Nov 14 '19 at 4:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.