MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

Let $G$ is a semisimple algebraic group over characteristic 0. I'm curious about the finite set of differential operators $D_i$ on algebra $\Bbbk [V]$ ($V$ is a representation of $G$), this differential operators are satisfying the following conditions. For every irreducible $G$-representation $W$ (denote it highest weight as $\omega$), there exists an isomorphism of $W$ with the set of $f\in\Bbbk[V]$, s.t $D_i f=n_i(\omega) f$ (where $n_i(\omega)$ are the functions from the set of all highest weights to $\Bbbk$).

Example: $xp_x+yp_y=np$ - Euler equation on $k[x,y]$ defines an irreducible representation of $SL_2$.

So I want to know if there exist: $V$, $D_i$ and $n_i$.

share|cite|improve this question

If I understand the intent of the question: partly iterating what @Alexander Chervov's answer notes, ... For example, a theorem of Harish-Chandra asserts that the collection of irreducible admissible representations of (some nice class of reductive or semi-simple) real Lie groups giving the same repn of the center of the universal enveloping algebra is finite.

(My own thinking about this has been mostly not about finite-dimensional repns of non-compact groups... but maybe this is not an essential obstacle to a helpful remark.)

In line with what @Alexander C. notes, for SL(2,R), the Casimir eigenvalues do distinguish irreducibles, at least in two important subfamilies: finite-dimensional, and unitary.

In fact, by Casselman's subrepn theorem, most interesting repns are subrepns (and quotients) of principal series, but an important technical note is that some unitary repns occur as subrepns of non-unitary (and not-unitarizable) principal series. This happens already for SL(2,R), where the holomorphic discrete series are subs of principal series far out of the unitary range. And so on.

Perhaps it is my own failing, but I do not know offhand a good reference for discussion of the possible extent of the finitely-many non-isomorphic irreducibles with the same character for the center of the enveloping algebra.

share|cite|improve this answer
Thanks for nice answer ! Let me also mention Verma modules. It is almost obvious that ALL Verma modules with the highest weights W(L) has same value for Casimirs. Where W - means the Weyl, its action on highest weights defined by "dotted rule" l-> w( l - rho) - rho. See discussion… This observation plays role in BGG resolution - it shows that only Verma modules with weights W(l) can be non-trivial components of resolution of the finite dimensional representation irrep L. – Alexander Chervov Nov 17 '11 at 6:02

Any element of the center of the universal envelopping will satisfy this property.

More precisely if you really request that it should happen for EVERY W, you should also request that "V" is sufficiently big, means that k[V] contains all irreps (otherwise question does not make sense). For example if G= GL_n V=C^n is not big enough. You can take C^n \otimes (C*) ^n - regular representation, or bigger spaces: C^n \otimes (C*)^m for m> n. than its Okay.

More detailed: since G acts on V , Lie algebra acts by vector fields on V and so universal enveloping acts by differential operators, center of it by Schur lemma acts by scalars on any irrep W - thar is what you required.

Example of sl2 which you mention is a part of this picture. But we should take gl2 not sl2 then Euler's operator comes from the central element of gl2 (identity matrix is center of gl2).

The centers of universal envelopings (sometimes elements of centers are called Casimir elements) are relatively well understood. For example for glen there are so-called Capelli elements written quite explicitly (see Wikipedia).

share|cite|improve this answer
Is this sufficient, if we will take only central elements from universal enveloping algebra? – zroslav Nov 16 '11 at 20:06
I mean: will these equalities for universal enveloping algebra define the required irreducible representation? – zroslav Nov 16 '11 at 20:08

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.