Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. 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

There are many places, which describe the unitary irreducible representations of $GL(n, F)$ with $F = \mathbb{C}$ or $F =\mathbb{R}$. Basically, we obtain a bunch of parabolically induced representation indexed by irreducible reps of parabolic subgroups and complex parameters and construct various quotients.

However, I could not find how the restriction of a unitary irreducible representation to the maximal compact groups decomposes into irreducibles.

Does somebody have a reference for $GL(n, \mathbb{C})$?

For $SL(2, \mathbb{C})$, I found Barut-Raczka "The theory of group representations" pg.567 and for $GL(2, \mathbb{R})$, I found Knightly-Li "Hecke opeartors" and much more on $GL(n, \mathbb{R})$ in Goldfeld-Hundley's new book on automorphic forms.

share|cite|improve this question
Ask this on MO. – Mariano Suárez-Alvarez Apr 2 '12 at 16:06
up vote 1 down vote accepted

Let $\pi$ be an irreducible unitary representation of $G$, so $\pi$ is either in the discrete series or is parabolically induced. In the latter case, using the decomposition $G=MANK$, where $P=MAN$, $$\pi\simeq {\rm Ind}_{MAN}^G(\sigma\otimes\nu\otimes 1)$$ which is the completion of $$\lbrace {\rm continuous}\ F:G\rightarrow V^\sigma|F(mang)=a^\nu\sigma(m)F(g)\rbrace$$ where $\sigma$ is a representation of the compact group $M=P\cap K$ (with representation space $V^\sigma$), $\nu$ denotes the exponent of a representation of the split torus $A$, and $1$ denotes the trivial representation of $N$ (we use continuous functions so that we can make sense of pointwise values, we could have operated directly in the completion if we wrote everything in terms of the right regular representation).

If we restrict $\pi$ to $K$, it becomes the completion of $$\pi|_K=\lbrace {\rm continuous}\ F:K\rightarrow V^\sigma|F(mk)=\sigma(m)F(k)\rbrace$$ which is exactly $L^2(K,V^\sigma)$. So the question becomes about the decomposition of this space into irreducibles. When $\sigma$ is the trivial representation, this is the decomposition of $L^2(M\backslash K)$. I believe this decomposition is known (see section 3.3.1 here, or Helgason's books, "Groups and Geometric Analysis" and "Geometric Analysis on Symmetric Spaces"), but I'm not sure about the level of detail. This should be adaptable to non-trivial $\sigma$, but I don't know if anyone has bothered to write it out. When $M$ is abelian (e.g., the $GL_n(\mathbb C)$ case), you should be able to do this by hand.

For discrete series representations, see Theorem 9.20 of Knapp's "Representation Theory of Semisimple Groups [...]". In this case, there is a specific $K$-type with a certain highest weight (related to the parameter of the discrete series) and all other highest weights of $K$-types are translations of this one by non-negative multiples of positive roots.

share|cite|improve this answer
I am fairly confident that your idea is the correct one. This is essentially the Restriction Induction formula of Mackey, which reduces the problem to inducing in compact Lie groups, but I will check the references you gave. Thanks for the detailed answer. – Apr 3 '12 at 7:28
Sorry I wasn't able to add information you didn't already know! Maybe on MO someone knows a better reference or is willing to write out the complete answer . . . – B R Apr 3 '12 at 8:27
Your answer was helpful! Often hearing a rephrasing of what I think to understand, can clarify some stuff for me plus you gave some additional reference. Thanks – Apr 4 '12 at 9:59

For my own research I am intensively using the books by N. Ja. Vilenkin and A.U. Klimyk "Representation of Lie Groups and Special Functions". For $GL(n,\mathbb{C})$ I would recommend volumes 2 and 3.

share|cite|improve this answer
I will check the reference. Thanks. – Apr 4 '12 at 10:00

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.