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The Harish-Chandra isomorphism describe the center $Z(\mathfrak{g})$ of $U(\mathfrak{g})$ as invariants of $\text{Sym}^*\mathfrak{h}$ under the action of the Weyl group. (One need to twist the action or make a change of coordinate on the affine space $\mathfrak{h}^*$ for this isomorphism work.)

My question is, how does one understand this isomorphism, and what's the geometric context of it? Why should one expect something like this might be true?

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Harish-Chandra isomorphism is defined for semisimple algebras. Let me try to give an informal explanation for classical simple Lie algebras, take $\mathfrak{gl}_n$ as main example.

We are interested in the center of $U(\mathfrak{gl}_n)$ - the same as invariants of gl in that space. First let us look on $S(\mathfrak{gl}_n)$ - symmetric algebra of $\mathfrak{gl}_n$ - as a vector space and moreover as module of g, it is isomorphic to $U(\mathfrak{gl}_n)$. So you need to describe the invariants of S(gl). Same as invariants of group $GL_n$ on the space of all matrices where $GL_n$ acts by conjugation

Actually everybody knows the answer - invariants are coefficients of the characteristic polynomial ! Coefficients of the char.pol are symmetric functions of eigenvalues - so you get the $S(\mathfrak{gl_n})^{GL} = Sym(h^*)^W$. (As for me this simple fact is main "motivation" for HC-isomorphism).

Now, what is NON-obvious ? We need to pass from $S(\mathfrak{gl_n})$ to $U(\mathfrak{gl_n})$ and this passing is actually related to twisting the action Weyl group by shift on $\rho/2$.

I do not think there is simple explanation of this, but there is some general context which explains at least why $Z(U(g))$ is isomorphic as commutative algebra to $S(g)^g$. Moreover for any Lie algebra "g" not just semisimple one.

This is what is called Duflo isomorphism. Moreover it was generalized by M. Kontsevich to quantization of arbitrary Poisson manifold.

So roughly speaking the "general context" is the following - "quantization is GOOD" i.e. it many algebraic structures which we can see on the classical level of Poisson algebras can be transfered to quantum level isomorphically. This is formalized by "formality theorems" like Kontsevich one, and later ones.

Now what about why the Weyl group becomes shifted ? I do not know the answer - it is clear how one can guess this - just take Casimir for $\mathfrak{sl}(2)$ and you will see it. But conceptual meaning is not clear for me.

If it is clear for someone - then please would you be so kind to comment what happens in the case of affine gl ? See my question: Harish-Chandra isomorphism for loop algebras. Is the image invariant with respect to affine Weyl group ? Is there relation - Miuru transformation - affine Weyl group ?

Similar reasons works for other classical semisimple Lie algebras, and may be not only for classical.

PS

There is generalization of Harish-Chandra isomorphism https://arxiv.org/abs/0912.1100 A generalized Harish-Chandra isomorphism Sergey Khoroshkin, Maxim Nazarov, Ernest Vinberg (to confess S. Khoroshkin was my supervisor).

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Here is one algebraic/representation theory perspective on why such a morphism might exist, although I don't think historically this is the way it went at all. Let's admit for a moment that one might be interested in Verma modules. It is easy to see that the center acts via scalars on these. Verma modules are parametrized by $\mathfrak{h}^*$ and in this way one gets an algebra morphism $Z(\mathfrak{g}) \to Sym(\mathfrak{h})$. Now it is also relatively straightforward to see (if I remember correctly) that for each simple reflection $s$, the Verma module $M(s\cdot \lambda)$ occurs as a submodule of the Verma module $M(\lambda)$, for $\lambda$ integral. It follows that the algebra morphism constructed earlier lands in $W$-invariants.

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  • $\begingroup$ Verma modules are parametrized by $\mathfrak{h}^∗$ and in this way one gets an algebra morphism $Z(\mathfrak{g})\to \text{Sym}(\mathfrak{h})$... How do you see one gets a regular function on $\mathfrak{h}^*$? $\endgroup$
    – 36min
    Nov 21, 2012 at 5:41
  • $\begingroup$ Hmmm, one needs a tiny bit more knowledge about Verma modules plus the triangular decomposition: the triangular decomposition for the enveloping algebra shows that on a highest weight vector an element in the center acts as the weight that the Verma corresponds to evaluated on the Cartan piece appearing in the triangular decomposition of the element. Thus, it's a regular function. $\endgroup$ Nov 21, 2012 at 5:54
  • $\begingroup$ In the comment above, by Cartan piece' I really mean purely Cartan piece'. I hope what I mean by this is clear? $\endgroup$ Nov 21, 2012 at 5:56
  • $\begingroup$ Do you really mean that $M(s\cdot\lambda)$ occurs as a submodule of $M(\lambda)$ for $\lambda$ merely integral, rather than just for $\lambda$ dominant? $\endgroup$
    – LSpice
    Jan 20, 2023 at 13:31
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Depending what one means by "intuition", the computation of the behavior of the center $\mathfrak z$ of $U\mathfrak g$ on unramified principal series repns not only suggests the form of the Harish-Chandra isomorphism, but also suggests the $W$-invariance, since the intertwinings (generically isomorphisms) among principal series are exactly given by Weyl group elements (with the shift by the half-sum $\rho$ of positive roots as a result of a change-of-measure at the group level).

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  • $\begingroup$ does it not also prove the HC isomorphism? $\endgroup$ Mar 3 at 15:52
  • $\begingroup$ @Venkataramana, maybe so! I'd have to think about it... Is it immediately clear to you? $\endgroup$ Mar 3 at 18:25
  • $\begingroup$ Professor Garrett: It seems to me that it is enough to prove this in the linear ( even split real ) case; using finite dimensional representations it seems clear that an element of the enveloping algebra is completely determined by its action on principal series. Using the intertwiners on principal series it seems to follow that the HC homomorphism is injective. I don't see why it is surjective onto the algebra of twisted Weyl group invariants. $\endgroup$ Mar 4 at 2:18
  • $\begingroup$ I now think the subjectivity follows from the fact that $G$ invariant polynomials on the Lie algebra $\mathfrak g$ map isomorphically $onto$ $W$ invariant polynomials on the Carton subalgebra $\mathfrak h$ (and by considering the graded algebras associated the enveloping algebra $U(\mathfrak g)$ and $U(\mathfrak h)$. $\endgroup$ Mar 4 at 4:23
  • $\begingroup$ somehow, my computer "corrected" surjectivity into subjectivity. $\endgroup$ Mar 4 at 6:40

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