# Application of Harish-Chandra theorem

Let $\mathfrak{g}$ be a semisimple finite dimensional Lie algebra and $V_\lambda$, resp. $V_\mu$ its finite dimensional highest weight modules with highest weights $\lambda$, resp. $\mu$. Let $\chi_\lambda, \chi_\mu : C(\mathcal{U}(\mathfrak{g})) \rightarrow \mathbb{C}$ be the corresponding central characters. Harish-Chandra theorem asserts that $\chi_\lambda = \chi_\mu$ if and only if $w(\lambda+\delta)-\delta = \lambda$ for some $w$ in the Weyl group, where $\delta$ is the Weyl vector, i.e. the sum of all fundamental weights.

Is it also true in this setting, that $V_\lambda \simeq V_\mu$ if and only if $\chi_\mu = \chi_\nu$ ?

How is this version of the theorem related to the fact that Harish-Chandra homomorphism is an isomorphism ?

Thank you very much for your answers. (I am studying a program in mathematical physics and trying to figure out how general is the procedure of labeling irreducible representations by values of Casimir operators, as physicists do so often)

-
 For $V_\lambda$ to be finite dimensional, we need $\lambda$ to be dominant. Then $\mu=w(\lambda+\delta)-\delta$ is not dominant unless $w=id$, $\mu=\lambda$. Therefore I don't quite understand the question. My recollection of enveloping algebras is a bit rusty, but isn't there stuff other than the Casimir operator in there as well? For example, isn't Casimir operator stable under the graph automorphism of the Dynkin diagram, so e.g. for type $A_n$ the Casimir operator would act as the same constant as on the dual of an irreducible rep? – Jyrki Lahtonen Jun 14 '12 at 19:25 Ah, thank you. I should have revised the theory of Lie algebras more before I asked. So it is clear from this argument with dominant weights that Harish-Chandra theorem implies directly what I asked, i.e. that we can distinguish irreps by their central characters. – user32817 Jun 14 '12 at 21:01 We can distinguish finite-dimensional irreps by their central characters. IIRC when you extend that to category $\mathcal{O}$ you get irreducible reps with non-dominant highest weights, but they are infinite dimensional, but there are indecomposable reps with composition factors (belonging to same central character), one f.d. the others not. That's what Kazhdan-Lusztig business was about. – Jyrki Lahtonen Jun 15 '12 at 6:03

If I interpret your question as asking: are irred. finite-dimensional rep's determined by their infinitesimal character (i.e. by the eigenvalues of the centre of the enveloping algebra on them) the answer is yes. As you essentially observe, if one has such an irrep. $V$, and you want to write it as $V_{\lambda}$, you can realize $\lambda$ as the unique dominant weight that $\chi_V$ (the central character of $V$) is equal to $\chi_{\lambda+\delta}$. (I am not sure what normalization you are using, but if you are using the normalized HC isomorphism, the one that identifies the centre of the enveloping algebra with $W$-invariants in the enveloping algebra of the Cartan, then $\chi_V$ will be the homomorphism corresponding to the character $\chi+\delta$ of the Cartan.)

The proof of this statement is closely tied up with the proof of the HC isomorphism (at least, with the proofs that I know). I learned the HC isomorphism from Knapp's overview by examples book, and I think I learned this fact, and its relationship to the HC isomorphism, from that book.

By the way, if you have a physics background, then you probably know one case of this: for the spherical harmonics, the Casimir (= spherical Laplacian) eigenvalue determines the irrep. of $SO(3)$ to which a given spherical harmonic belongs.

-

I am not an expert in Lie theory, but the following are some basic answers to your questions. All the material I use here comes from Humphrey's text 'Representations of semisimple Lie algebras in the BGG category $\mathcal{O}$'

First, there are two Harish-Chandra homomorphisms

1. defined by restricting the canonical projection $\mathrm{pr}:U(\mathfrak{g})\to U(\mathfrak{h})$ onto $Z(\mathfrak{g})$
2. 'twisted version' (which is the more useful one) which takes into account of the dot-action, and with a better range, denote $\psi:Z(\mathfrak{g})\to S(\mathfrak{h})$

Again I have to stress I am not expert in this field so in the literature, they may already assume we always talk about the twisted one; and this is the one I assume you are using here.

Harish-Chandra Theorem:

1. $\psi: Z(\mathfrak{g}) \simeq S(\mathfrak{g})^W$
2. For all $\lambda,\mu\in\mathfrak{h}^*$, the central characters $\chi_\lambda=\chi_\mu \Leftrightarrow \mu = w\cdot \lambda$ for some $w\in W$
3. Every central character $\chi:Z(\mathfrak{g})\to \mathbb{C}$ is of the form $\chi_\lambda$ for some $\lambda$.

Just to make sure our terminology matches up. When I read highest weight module (of weight $\lambda$), I assume you mean objects of the category $\mathcal{O}$ such that we have a maximal vector with weight $\lambda$ where $\mathfrak{n}$ acts as zero.

Let suppose $V_\lambda$ be highest weight module of highest weight $\lambda$, but NOT necessarily finite dimensional, then the answer to your first question ($V_\lambda \simeq V_\mu \Leftrightarrow \chi_\lambda=\chi_\mu$) is false. Some obvious examples are the Verma modules $M(\lambda) = U(\mathfrak{g})\otimes_{U(\mathfrak{b})}\mathbb{C}_\lambda$ where $\mathbb{C}_\lambda$ is the trivial character (representation) corresponding to weight $\lambda$ of the universal enveloping algebra of Borel subalgebra $\mathfrak{b}=\mathfrak{h}\oplus \mathfrak{n}$. Each $M(\lambda)$ is highest weight module of weight $\lambda$, and we can still have $\chi_\lambda=\chi_\mu \Rightarrow M(\lambda) \not\cong M(\mu)$. (cf Jyrki Lahtonen's comment under your question)

On the other hand, if we really restricts $V_\lambda$ to be finite dimensional (this actually impose a great restriction on the choice of $\lambda$ and $\mu$), then by my discussion with Matt below, who have pointed out there is at most one finite dimensional irreducible module in each block of category $\mathcal{O}$, we have $\chi_\lambda=\chi_\mu \Leftrightarrow V_\lambda = V_\mu$ is true IF $V_\lambda$ and $V_\mu$ are irreducibles; and in this case $\lambda =\mu$. Nevertheless, if you just need condition "finite dimensional" rather than "finite dimensional irreducible", then a counterexample is $V_\lambda = L(\lambda)\oplus L(\lambda)$ and $V_\mu = L(\mu) = L(\lambda)$. Both of these modules has central characters $\chi_\lambda$, but they are obviously not isomorphic.

My answer to your main question (application of Harish-Chandra Theorem) is the following. It breaks down weights into linkage classes, the amazing effect is the linkage class coincide with what we call a "block" in representation theory; the concept of blocks is well-defined for (abelian) category as well, if you think of a category as presented to you as a graph with vertices as (indecomposable) objects and directed arrows representing (indecomposable) morphisms, you can think of a block as the connected component of this graph. So the main corollary of Harish-Chandra theorem is $$\mathcal{O}=\bigoplus_{\chi \text{ central character}}\mathcal{O}_{\chi}=\bigoplus_{\lambda \in \mathfrak{h}^\times/W\cdot}\mathcal{O}_{\lambda}$$ where each indecomposable $U(\mathfrak{g})$-module lie in exactly one of $\mathcal{O}_{\lambda}$.

This really is a major reduction in studying category $\mathcal{O}$, because if we want to look at anything homological, we can concentrate back into each individual block. And we can study these blocks using finite dimensional algebras (due to the fact that each block has finitely many isomorphism classes of simple modules)! Hope this helps.

-
 Dear Aaron, There are some typos in your point 1. Also, I'm not sure I understand your second para: certainly, finite dim'l irreps are determined by their infinitesimal characters. Regards, – Matt E Jun 15 '12 at 20:49 Thanks for pointing out the typo. For the other thing, I assume here $\chi$ are the central characters, so they are not quite the same as characters of modules (which determine the module unique as you said). Also, I follows OP setting that $V_\lambda$ as (finite dimensional) highest weight module, rather than irreducible modules. So the idea is central characters determine blocks, characters determine composition (irreducible) factors. – Aaron Jun 16 '12 at 18:43 Dear Aaron, I guess I don't know what you mean by finite-dimensional highest weight module, other than finite-dimensional irreducible. Related to this, if by central character you mean infinitesimal character (the character through which the centre of the enveloping algebra acts on the representation) then the infinitesimal character of a finite-dimensional irreducible rep'n does determine it up to isomorphism. (I agree that it determines an entire block in category $\mathcal O$, but this block contains at most one finite-dimensional simple module.) Regards, – Matt E Jun 16 '12 at 23:27 My central character is the same as your infinitesimal character. I learn everything about category $\mathcal{O}$ from Humphrey's book, so I follow his terminology for central characters. – Aaron Jun 17 '12 at 14:14 I am sorry there are some confusion in terminology I should have clarify. When I wrote highest weight module I was acutally thinking about the Verma modules $M(\lambda)$, this is my bad. These are infinite dimensional unless the irreducible one corresponding to the minimal weight in a block of category O. I have re-edited the answer please check. – Aaron Jun 17 '12 at 15:00