Timeline for infinitesimal character of Langlands quotient for GL(n,R)
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13 events
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| Jun 29, 2017 at 13:33 | comment | added | Vanya | In the comment before the previous comment.. " then Theorem 5.1.... ", I should write it as "Then Theorem 5.1 ( Langlands' classification) says that $J(Q_F,\sigma,\lambda)$ is unique irreducible quotient of $ind_{Q_F}^G(\sigma \otimes \lambda \otimes 1$ \textbf{with $\sigma$ irreducible tempered rep. of $M_F$}. | |
| Jun 29, 2017 at 13:25 | comment | added | Vanya | gives the infinitesimal character of $ind$ as $\Lambda_\sigma + \lambda$, where $\Lambda_\sigma$ is the I.C of unitary rep of $M_F$. So If I know that J and $ind$ have the same IC. then I am done....? | |
| Jun 29, 2017 at 13:24 | comment | added | Vanya | Thank you very much for the elaborate useful response. I will look into the references provided. Meanwhile, I was looking into the van den ban's paper you suggested. In it, the paragraph immediately after Theorem 4.5 says that irreducible tempered reps. are equivalent to unitary reps. (as H-C modules). Then Theorem 5.1 ( Langlands' classification) says that $J(Q_F, \sigma, \lambda)$ is unique irreducible quotient of $ind_{Q_F}^G(\sigma \otimes \lambda \otimes 1)$, So I am guessing that I may assume $\sigma$ in here to be unitary (by Theorem 4.5). Then Lemma 2.5 | |
| Jun 29, 2017 at 13:05 | comment | added | Vincent | Hmmm, I formulate a bit clumsily as a result of trying to seperate the 'more than one IC' from the 'generalized vs actual' IC issue, but that is unnecessary. The best thing to say is: a generalized IC is homomorphism $\mathfrak{Z} \to \mathbb{C}$ such that $\pi(Z)v = \nu(Z)v$ for some $v$ but all $Z$ and it is the unique infinitesimal character of $\pi$ when $\pi(Z)v = \nu(Z)v$ holds for all $v$ and all $Z$. If the latter situation occurs, then $(\pi, V)$ is quasi-simple; all irreducible admissible reps are quasi-simple by Schur's lemma, but there are others esp. among induced reps. | |
| Jun 29, 2017 at 12:55 | comment | added | Vincent | The point is just that elements of $\mathfrak{Z}$ might act as annoying jordan blocks. It they don't, and moreover there is only one eigenvalue around, then we are in the nice situation of quasi-simpleness. I believe this to be true representations induced from irreducible representations of parabolic subgroups, but please consult Knapp or other literature. | |
| Jun 29, 2017 at 12:53 | comment | added | Vincent | So fixing $Z \in \mathfrak{Z}$ we can say by looking at the fin. dim. space generated by one vector $v$ that a generalized infinitesimal character is just a eigenvalue of $\pi(Z)$, while it is called an actual infinitesimal character when the corresponding generalized eigenspace is actually an actual eigenspace. The nice thing about $\mathfrak{Z}$ being commutative is that the there is a joined decomposition of any fin dim space it acts on into generalized eigenspaces, so the choice of $Z \in \mathfrak{Z}$ in this comment is immaterial. | |
| Jun 29, 2017 at 12:50 | comment | added | Vincent | Ehm let me say just in words what I mean with generalized IC's vs ordinary ones. Write $\mathfrak{Z}$ for the center of $U(\mathfrak{g})$ and $(\pi, V)$ for the Harish-Chandra module. A generalized infinitesimal character is a character $\nu$ of $\mathfrak{Z}$ such that $\pi(Z)v = \nu(Z)v$ for all $Z \in \mathfrak{Z}$ but some $v$ in the representation space. It is an actual infinitesimal character if it holds for all $v$ in the finite dimensional space $\pi(\mathfrak{Z})v$. | |
| Jun 29, 2017 at 12:39 | comment | added | Vincent | 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) | |
| Jun 29, 2017 at 12:32 | comment | added | Vincent | 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. | |
| Jun 29, 2017 at 12:31 | comment | added | Vincent | 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. | |
| Jun 29, 2017 at 12:28 | comment | added | Vincent | 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) | |
| Jun 29, 2017 at 11:32 | comment | added | Vanya | 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? | |
| Jun 29, 2017 at 8:25 | history | answered | Vincent | CC BY-SA 3.0 |