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

Hi All,

I am new to this (though I seem to be a latecomer); so forgive me if this is not your most favorite question:

I am trying to understand the structure (e.g., decomposition) of the unitary representation $L^2(N/M)$ where $N$ is a nilpotent Lie group acting by left translation on this Hilbert space (coming from the invariant measure on N/M). Surprisingly, I am unable to find any suitable references. Does anyone here know where one should look for an answer in the literature?

share|cite|improve this question
Appearently, it is clear what $M$ should be here? For a start with the Mackey Machine/Kirillov orbit method/Clifford theory (see F.Z.'s answer) I suggest Weintraub's book on representation theory for finite groups. This explains the representation theory of semidirect products/group extensions in terms of orbits, stabilizer and (proj.) reps of the stabilizer. – Marc Palm Nov 28 '12 at 9:06
up vote 8 down vote accepted

The representation you're looking at is $\mathrm{Ind}_M^N1$ and as such, its decomposition into irreducibles is very well understood using Kirillov's orbit method. (Essentially, the irreducibles that enter correspond to the coadjoint orbits in the image of the moment map $T^*(N/M)\to\mathfrak n^*$.)

I'd say the basic paper on the subject is this one by Corwin, Greenleaf, and Grélaud. It has references to the earlier work by Kirillov himself, and you'll find more in mathscinet's forward references to reviews citing it.

share|cite|improve this answer

Is $N$ or $M$ connected? simply connected? Is $N$ commutative? Ronald Lipsman has settled the decomposition of quasiregular representations of Lie groups in many settings. You will find his papers very readable. They are all available on google.

share|cite|improve this answer

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.