MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
@36min just notifier - answered your quest on… – Alexander Chervov Nov 21 '12 at 11:10

Harish-Chandra isomorphism is defined for semisimple algebras. Let me try to give informal explanation for classical simple Lie algebras, take gl_n as main example.

We are interested in the center of U(gl_n) - the same as invariants of gl in that space. First let us look on S(gl_n) - symmetric algebra of gl_n - as a vector space and moreover as module of g, it is isomorphic to U(gl_n). So you need to describe the invariants of S(gl). Same as invariants of group GL on the space of all matrices where GL 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(gl)^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(gl) to U(gl) 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 ZU(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 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.


There is generalization of Harish-Chandra isomorphism A generalized Harish-Chandra isomorphism Sergey Khoroshkin, Maxim Nazarov, Ernest Vinberg (to confess S. Khoroshkin was my supervisor).

share|cite|improve this answer

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.

share|cite|improve this answer
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 $\mathfrak{h}^*$? – 36min Nov 21 '12 at 5:41
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. – Reladenine Vakalwe Nov 21 '12 at 5:54
In the comment above, by Cartan piece' I really mean purely Cartan piece'. I hope what I mean by this is clear? – Reladenine Vakalwe Nov 21 '12 at 5:56

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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