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 $\Phi: D(\mathfrak{g})^{\mathfrak{g}} \rightarrow D(\mathfrak{h})^W$ that reduces to the restriction map: $\mathbb{C}[\mathfrak{g}]^{\mathfrak{g}} \rightarrow \mathbb{C}[\mathfrak{h}]^W$ on zero order differential operators, and such that $\Phi \Delta_{\mathfrak{g}} = \Delta_{\mathfrak{h}}$", where $\Delta_{\mathfrak{g}}$ is the second order Laplacian on $\mathfrak{g}$.


Consider some higher order Laplacians - what are their image under this map? Are there "nice explicit" formulas? I am interested mostly in $\mathfrak{gl}_n$ case.

Naive answer (probably incorrect): (ACTUALLY THIS IS CORRECT, see answer) one may naively expect that e.g. cubic Laplacian on $\mathfrak{g}$ 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.


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 $U(\mathfrak{g})$ in $D(\mathfrak{g})^{\mathfrak{g}}$, and 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 $U(\mathfrak{gl}_n)$, might be some nice formula for its image under this map ...?


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

Setup: Consider Lie algebra $\mathfrak{g} = \mathfrak{gl}_n$, denote $D(\mathfrak{g})^{\mathfrak{g}}$ - invariant with respect to conjugation differential operators on it, denote $\mathbb{C}[\mathfrak{g}]^{\mathfrak{g}}$ - invariant functions. $\mathfrak{h} \subseteq \mathfrak{gl}_n$ - subspace of diagonal matrices, $\mathbb{C}[\mathfrak{h}]^W$ - symmetric function on $\mathfrak{h}$.

It is clear that $D(\mathfrak{g})^{\mathfrak{g}}$ acts on $\mathbb{C}[\mathfrak{g}]^{\mathfrak{g}}$. It is well-known that $\mathbb{C}[\mathfrak{g}]^{\mathfrak{g}}$ isomorphic $\mathbb{C}[\mathfrak{h}]^W$.

So we get map from $\tilde \Psi: D(\mathfrak{g})^{\mathfrak{g}} \rightarrow D(\mathfrak{h})^W$.

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

Question: What happens with higher order Laplacians?

  • $\begingroup$ Perhaps the Duflo isomorphism is what you're looking for. $\endgroup$ Commented Oct 15, 2011 at 14:53
  • $\begingroup$ I do not understand why you think Duflo is related to this ? The image of Laplacians is some dif. opers. on C[h]^W. Duflo image is Z(U(gl)) I do not see any natural identification $\endgroup$ Commented Oct 15, 2011 at 17:10
  • 2
    $\begingroup$ Harish-Chandra's paper: justpasha.org/tmp/20111021 $\endgroup$ Commented Oct 21, 2011 at 5:10
  • $\begingroup$ Spasibo bolshoe ! $\endgroup$ Commented Oct 22, 2011 at 15:57

1 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


P. Etingof "Lectures on Calogero-Moser"


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.


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