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For any Lie algebra symmetrization maps Poisson center of S(g) to center of U(g). Consider g=gl_n, the Poisson center of S(gl_n) is isomorphic to algebra of symmetric polynomials in eigenvalues. Consider some element c \in S(gl_n), which corresponds to some symmetric polynomial P(l1,...ln).

Question: is there something known about value of symmetrization(c) in irrep of highest weight (w1,...wn), i.e. what is the Harish-Chandra image of symmetrization(c) ?

We should expect that this value is P(w1,...,wn)+correction, but it is quite difficult to calculate this "correction".


For sl(2) the answer is known - see Kirillov's paper "Merits and demerits of orbit method" section 3.5 (http://www.ams.org/bull/1999-36-04/S0273-0979-99-00849-6/S0273-0979-99-00849-6.pdf). Answer and proof are rather non-trivial and obtained with the help of the Duflo map...

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Well, long time passed and no answer, I also asked by the e-mail M. Duflo, A. Kirillov, E. Vinberg, G. Olshanski, so I got the impression that it seems indeed uknown.

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  • $\begingroup$ You should not be too presumptious based on the lack of response at MO: I, for one, missed your question the first time around. Some things are certainly known, e.g. didn't Okounkov and Olshanskii investigate this question ("shifted Schur functions")? $\endgroup$ Dec 12, 2011 at 16:23
  • $\begingroup$ @Victor Okounkov Olshanskii is not related to this question. They consider "special" symmetrization, in opposite to the "standard" one I am asking about. The "special" symmetrization introduced by Olshaskii can be seen in two ways: 1) consider representation of GL_n by right-invariant vector fields E=XD^t and sym(E...E)= WickOrdering(XD) i.e. put "X" to the left "D" to the right - that is why it is related to Capelli identities 2) see second comment $\endgroup$ Dec 12, 2011 at 17:20
  • $\begingroup$ 2) "general" symmetrization (in the sense of Olshanki) comes from ANY local (nearby origins) identification of g and G - then U(g) as delta functions at "e" gots identified with S(g). Standard symmetrization comes from exponential mapping. "Special" comes from X-> 1+X for gl_n. For other semisimple it appears to be more tricky with "special" one. $\endgroup$ Dec 12, 2011 at 17:24
  • $\begingroup$ Ok, I think that I now understand your question (it wasn't clear from the original post). Let $$\sigma: S(\mathfrak{g}) \to U(\mathfrak{g})$$ be the standard symmetrization map, $$HC: S(\mathfrak{h})^W \to Z(\mathfrak{g})$$ be the Harish-Chandra homomorphism and $$HC_{qc}: S(\mathfrak{h})^W \to S(\mathfrak{g})^{\mathfrak{g}}$$ be its quasi-classical version. Your question, then, concerns an explicit form of the composite map $$HC^{-1}\circ\sigma\circ HC_{qc}: S(\mathfrak{h})^W \to S(\mathfrak{h})^W.$$ I am busy this week (exams), but if it's what you want, I'll think about it later. $\endgroup$ Dec 12, 2011 at 21:47
  • $\begingroup$ @Victor. Yes absolutely correct. You are welcome to edit my post to make it more readable. As a small remark, I think it is necessary to emphasize that there are two actions of W on S(h) - standard one and shifted by \rho - the image of HC is invariant with respect to shifted action. If you have some interest in the question I can share some calculations and the letter by M. Duflo how to approach this. But I am NOT sure this is "good" question with "nice" answer... May be Kirillov's "nice" answer only for sl_2. $\endgroup$ Dec 13, 2011 at 5:59

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