How to understand the Harish-Chandra isomorphism? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T10:31:14Z http://mathoverflow.net/feeds/question/114020 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/114020/how-to-understand-the-harish-chandra-isomorphism How to understand the Harish-Chandra isomorphism? 36min 2012-11-21T05:12:08Z 2012-11-21T06:55:18Z <p>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 <code>$\mathfrak{h}^*$</code> for this isomorphism work.)</p> <p>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?</p> http://mathoverflow.net/questions/114020/how-to-understand-the-harish-chandra-isomorphism/114022#114022 Answer by Reladenine Vakalwe for How to understand the Harish-Chandra isomorphism? Reladenine Vakalwe 2012-11-21T05:33:52Z 2012-11-21T05:33:52Z <p>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.</p> http://mathoverflow.net/questions/114020/how-to-understand-the-harish-chandra-isomorphism/114029#114029 Answer by Alexander Chervov for How to understand the Harish-Chandra isomorphism? Alexander Chervov 2012-11-21T06:55:18Z 2012-11-21T06:55:18Z <p><a href="http://en.wikipedia.org/wiki/Harish-Chandra_isomorphism" rel="nofollow">Harish-Chandra isomorphism</a> is defined for semisimple algebras. Let me try to give informal explanation for classical simple Lie algebras, take gl_n as main example.</p> <p>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 </p> <p>Actually <strong>everybody knows the answer - invariants are coefficients of the characteristic polynomial</strong> ! 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).</p> <p><strong>Now, what is NON-obvious ?</strong> 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.</p> <p>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.</p> <p>This is what is called <a href="http://en.wikipedia.org/wiki/Duflo_isomorphism" rel="nofollow">Duflo isomorphism</a>. Moreover it was generalized by M. Kontsevich to quantization of arbitrary Poisson manifold.</p> <p>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.</p> <p><strong>Now what about why the Weyl group becomes shifted ?</strong> 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.</p> <p>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: <a href="http://mathoverflow.net/questions/93626/harish-chandra-isomorphism-for-loop-algebras-is-the-image-invariant-with-respect" rel="nofollow">http://mathoverflow.net/questions/93626/harish-chandra-isomorphism-for-loop-algebras-is-the-image-invariant-with-respect</a></p> <p>Similar reasons works for other classical semisimple Lie algebras, and may be not only for classical.</p> <p>PS</p> <p>There is generalization of Harish-Chandra isomorphism <a href="http://arxiv.org/abs/0912.1100" rel="nofollow">http://arxiv.org/abs/0912.1100</a> A generalized Harish-Chandra isomorphism Sergey Khoroshkin, Maxim Nazarov, Ernest Vinberg (to confess S. Khoroshkin was my supervisor).</p>