3
$\begingroup$

Let $G = GL(n,\mathbb{R})$. Consider a Langlands data $(Q_F, \sigma, \lambda)$ with $F \subset \Delta$ (the set of simple roots), $Q_F$ the associated standard parabolic subgroup, $\sigma$ an irreducible tempered representation of $M_F$, and $\lambda$ an element of $(\mathfrak{a}_F)^*_\mathbb{C} = Hom(\mathfrak{a},\mathbb{C})$ satisfying $\langle Re \lambda, \alpha \rangle \ge 0 $ for $\alpha \in \Delta \setminus F$

Here $\mathfrak{a}_F$ is the Lie algbera of $A_F$ appearing in the Langlands decompostion $Q_F = M_F A_F N_F$.

Then the Langlands classification of irreducible admissible $(\mathfrak{g},K)$-modules is essentially given as the module $J(Q_F, \sigma, \lambda)$ occuring in the parabolically induced module $Ind_{Q_F}^G(\sigma \otimes \lambda \otimes 1)_K$ (the subscript $K$ denotes $K-$finite).

Is it the case that the infinitesimal character of Langland's quotient same as that of the $Ind_{Q_F}^G(\sigma \otimes \lambda \otimes 1)$ (with out $K$ subscript?)

(I am a novice in this subject.)

Improve this question
$\endgroup$

1 Answer 1

Reset to default
3
$\begingroup$

It is quite some years ago that I used to know this, but as far as I remember the ansewr is YES.

$J(Q_F, \sigma, \lambda)$ is the unique irreducible quotient of $Ind_{Q_F}^G(\sigma \otimes \lambda \otimes 1)$. In general, if $W$ is a quotient of $V$ then the set of infinitesimal characters of $W$ is a subset of that of $V$, and in this case we know (by irreducibility) that $W$ only has a single infinitesimal character.

What remains to be shown (but is implicit in your question) is that $Ind_{Q_F}^G(\sigma \otimes \lambda \otimes 1)$ only has a single infinitesimal character. This is true and in fact this infinitesimal character can be computed from $\lambda$ and the infinitesimal character of $\sigma$ in a way that I believe is described here: https://dspace.library.uu.nl/handle/1874/1625 (Induced representations and the Langlands classification by Erik van den Ban, lecture notes from a summer school in Edinburgh in the '90s.)

With regards to the $K$-subscript. The space indicated by $K$-subscript is a) dense in the space without the $K$-subscript en b) closed under the action of the Lie-algebra (and so in particular the center of the universal enveloping algebra) on the bigger (no-$K$-subscript) space. These two facts should be enough to see that you can read off the infinitesimal character of the bigger space from the smaller space.

Improve this answer
$\endgroup$
12
  • $\begingroup$ Thank you very much for the response. The reference provided contains the computation only when $\sigma$ is unitary (Lemma 2.5). Moreover can you offer a reference for the statement "What remains to be shown ... is that .. only has a single infinitesimal character" - the beginning sentence of the second paragraph? $\endgroup$
    – Vanya
    Commented Jun 29, 2017 at 11:32
  • $\begingroup$ Well, whether or not 'it remains to be shown' that this rep has a single infinitesimal character depends on how one interprets your question, I was more referring to the general fact that generically representations come with more than one 'generalized infinitesimal characters', where both them being actual infinitesimal characters and there being only one of them is kind of special but also the most famous and most discussed situation because irreducible (Harish-Chandra would say 'simple') are of this type (one single, simple infinitesimal character) $\endgroup$
    – Vincent
    Commented Jun 29, 2017 at 12:28
  • $\begingroup$ The converse does not hold: some non-irreducible representations do have a well defined single simple infinitesimal character. The oldest source I know discussing this is Harish-Chandra, who used the term quasi-simple for reps with one infinitesimal character, the reference is: Harish-Chandra. Representations of semisimple Lie groups on a Banch space. Proc. Nat. Acad. Sci. U. S. A., 37:170–173, 1951, text preceeding Thm 2. $\endgroup$
    – Vincent
    Commented Jun 29, 2017 at 12:31
  • $\begingroup$ However if you are looking for a reference of the statement that the induced modules you discuss are quasi-simple, I'm not sure where to look. As I said, I believed the proof was in the notes of Van den Ban. $\endgroup$
    – Vincent
    Commented Jun 29, 2017 at 12:32
  • $\begingroup$ Here are some more references for te distinction between generalized infinitesimal characters and actual infinitesimal characters: Knapp: representation theory of semi-simple groups, an overview based on examples, sections VIII.6 and X.9 and Vogan, Representations of real reductive groups, Def. 0.3.18. (I don't actually remember what is in these texts at these particular places, I just copied the references from my thesis.) I would imagine that the book by Knapp also has a proof of the quasi-simpleness of parabolically induced representations, although perhaps for SL(n) rather than GL(n) $\endgroup$
    – Vincent
    Commented Jun 29, 2017 at 12:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .