Question

In their great paper "Symplectic reflection algs. and Harish-Chandra hom." http://arxiv.org/abs/math/0011114 Etingof and Ginzburg write (page 9):

"In 1964, Harish-Chandra [HC] defined an algebra homomorphism : D(g)^g -> D(h)^W, that reduces to the restriction map: C[g]^g -> C[h]^W on zero order differential operators, and such that \Phi Laplacian_g = Laplacian_h "

Question:

Consider some higher order Laplacians - what are their image under this map ? Are there "nice explicit" formulas ? I am interested mostly in gl_n case.

Naive answer (probably incorrect): (ACTUALLY THIS CORRECT see answer) one may naively expect that e.g. cubic Laplacian on g will go to \partial_1 ^3 + \partial_2 ^3 + ... in the same way as it is stated above for quadratic Laplacian. But probably this is not correct.

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Further questions: 1) what will happen for "deformed Harish-Chandra" homomorphism ? (i.e. what about the algebra of quantum Calogero-Moser hamiltonians ? )

2) Consider the center of U(g) \in D(g)^g, the same question for it. In particular we can take Talalaev's "oper" det(d/dz - E_ij/z) - some special generating "function" for generators of the center of U(gl_n), might be some nice formula for its image under this map ... ?

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Let me translate the question in simple-minded terms, it order to try to be self-contained.

Setup: Consider Lie algebra g = gl_n, denote D(g)^g - invariant with respect to conjugation differential operators on it, denote C[g]^g - invariant functions. h \in gl_n - subspace of diagonal matrices, C[h]^W - symmetric function on h.

It is clear that D(g)^g acts on C[g]^g. It is well-known that C[g]^g isomorphic C[h]^W.

So we get map from \tilde Psi: D(g)^g -> D(h)^W.

Obviously zero-order differential operators (e.g. C(g)^g ) goes to C(h)^W. Obviously second order Laplacian on g will NOT go to second order Laplacian on h. However some educated guess is that we should conjugate it by the sqrt("volume of the orbit") (which is Vandermonde of h_i in gl_n case) i.e. consider delta * \Psi * delta^{-1} . Then some calculations see (page 45 loc. cit.) Claim that second order Laplacian on g will go to second order Laplacian on h.

Question: What happens with higher order Laplacians ?

Answer 1

I asked Pavel Etingof he answered that "naive answer is correct". Means that higher order Laplacians are mapped to their naive restrictions.

This is explicitly stated in Proposition 4.5 page 27 in

http://arxiv.org/abs/math/0606233

P. Etingof "Lectures on Calogero-Moser"

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Some steps of the proof - for quadratic Laplacian what makes explicit calculation. For higher Laplacians on should exploit that their images commute with quadratic and and their symbol is "naive restriction" one should look on the difference, by some arguments it is proved it must be zero.