Let $G$ be a locally compact group and let $\mu$ be a left Haar measure. We know that $\mu$ is unique up to a scalar in $\mathbf{R}_{>0}$. I don't know so much about unitary representations of groups but for the sake of convenience let us make the following definition:

Let $(V,\langle\ ,\ \rangle)$ be an Hilbert space over $\mathbf{C}$ of countable (finite or infinite) orthonormal Schauder basis. We let Let $GL_{cont}(V)$ be the set of bounded (with respect to the operator norm) linear operators on $V$. We may view $GL_{cont}(V)$ as a topological group via the discrete topology. Now let $M$ be a $\mathbf{C}$ vector space with a linear $G$-action. We will say that $M$ is a unitary irreducible representation of $G$ if there exists an abstract isomorphism of $\mathbf{C}$ vector spaces $f:M\rightarrow V$ (where $V$ is chosen as above) such that the natural map $GL(M)\rightarrow GL(V)$

(1) factors through $GL_{cont}(V)$

(2) $V$ is irreducible as a $G$-module

(3) For all $g\in G$ and all $v,w\in V$ one has that $\langle\rho(g)v,\rho(g)w\rangle=\langle v,w\rangle$.

Now let us consider the space $L^2(G)$ of all functions $f:G\rightarrow\mathbf{C}$ where $f$ is measurable and square integrable with respect to the Haar measure. Note that this space has a natural structure of a $G$-module through left action.

Now in the special case where $G$ is a compact Lie group ($G$ is not necessarily connected so in particular this covers all finite groups) then all irreducible representation are unitary (the average trick) and finite dimensional (this I think is non-trivial and follows from Peter-Weyl, actually I never looked at the proof of this result). Moreover, if $\widehat{G}$ denote a complete set irreducible $\mathbf{C}$ representations of $G$ (up to isomorphisms as (unitary) $G$-modules) then one has that

$L^2(G)=\bigoplus_{\phi\in\widehat{G}}\oplus_{i,j}\sqrt{n_{\phi}}\phi_{ij}$ where $n_{\phi}=dim(\phi)$ and $\phi_{ij}$ is the $(i,j)$-th entry of $\phi:G\rightarrow GL(V_{\phi})$. In other words all irreducible unitary representations (say $\phi$ is one of them) of $G$ occur in $L^2(G)$ with multiplicities $n_{\phi}$. The direct sum here should be understood in the sense of Schauder basis with respect to the topology induced by $\langle\ ,\ \rangle$. Note that $\lbrace\sqrt{n_{\phi}}\phi_{i,j}\rbrace$ gives an orthonormal basis of $L^2(G)$.

Now here is a set of natural questions:

(1) Do all the irreducible unitary representations of a semi-semiple (reductive) algebraic group over $\mathbf{R}$ occur in $L^2(G)$?

(2) On the other side of the spectrum, what about algebraic solvable groups?

(3) What is the minimal example of a locally compact topological group $G$ (with an non artificial tailor made topology, in particular $G$ has to be infinite) for which one can find an irreducible unitary representation which does not occur in $L^2(G)$?

  • $\begingroup$ I think you have some sort of implicit assumption that "occur" will be as simple as the compact case, i.e. as direct summand. In this type of question, that is not how it works. But see en.wikipedia.org/wiki/Noncommutative_harmonic_analysis fpr what to expect. $\endgroup$ Commented Dec 22, 2010 at 22:05
  • $\begingroup$ Hi @Charles, thanks a lot for your comment. So by "occur" I simply mean that if $M$ is an irreducible unitary $\mathbf{C}$-representation of $G$ as defined above then there exists a continuous inclusion map of $G$-module $\iota:M\hookrightarrow L^2(G)$ which respects the inner products. And then because of the inner product one may define an orthogonal complement which "seems" to imply indeed that $M$ may be viewed as a direct summand! $\endgroup$ Commented Dec 22, 2010 at 23:56
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    $\begingroup$ Complementing Charles's comment: take $G=\mathbf{R}$ and the trivial 1-dimensional representation. It is not included in $L_2(\mathbf{R})$ since non-zero constant functions are not Lebesgue integrable. $\endgroup$
    – algori
    Commented Dec 23, 2010 at 0:05
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    $\begingroup$ ... by the way, there is no need to take $G$ to be $\mathbf{R}$, one can take any other non-compact group instead. $\endgroup$
    – algori
    Commented Dec 23, 2010 at 0:09
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    $\begingroup$ Dear Hugo, See my answer here: mathoverflow.net/questions/37021/… . It provide a summary of unitary representation theory of reductive Lie groups, which is somewhat more complicated than what you have anticipated. (Your concept of "occur" corresponds essentially to discrete series representations. Not all Lie groups admit discrete series, and even for those that do, most unitary irreps. are not discrete series.) $\endgroup$
    – Emerton
    Commented Dec 23, 2010 at 1:20

2 Answers 2


I'm not quite sure if this is the answer that you looking for but anyway he we go. For a locally compact group you are going to generally want to look at strongly continuous representation. By this is mean endow $B(H)$, the bounded operators on a hilbert space $H$ with the topology of point-wise norm convergence. And only consider reps $\pi:G\rightarrow B(H)$ that are continuous with this topology. Now such a rep is unitary if, for every $g\in G$, $\pi(g)$ is a unitary operator.

Now the notion of "occur in" that you mention seems to be the notion of strong containment. We say that $\rho:G\rightarrow B(K)$ is strongly contained in $\pi:G\rightarrow B(H)$ if there is a $G$-equivarient unitary operator from $K$ to a closed subspace of $H$.

So it now seems that you are asking when does the left regular rep ($L^2(G)$) strongly contain all irreducibles. So yes for compact Lie groups this follows from Peter-weyl this is true.

However as soon as you go to something non-compact this might not be true.

In fact, there is a much weaker notion known as weak containment of representation. and it is known that $L^2(G)$ weakly contains all irreducible reps if and only if $G$ is amenable.

Non-compact Lie groups are in general not amenable, (any groups which contains $\mathbb{F}_2$ the free group on 2 generators is non-amenable)

There is much more to be said about this but I think that this should suffice for now

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    $\begingroup$ There's something I don't understand here. One can embed the free group on two generators in the group $SO(3)$ of rotations in three dimensions, which is as compact as it gets. So according to what you are saying, $SO(3)$ would not be amenable, whence $L^2(SO(3))$ would not contain all irreps (even weakly), contradicting Peter-Weyl. Not being an expert, I'm surely confused... but where? $\endgroup$ Commented Dec 23, 2010 at 0:59
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    $\begingroup$ Right, sorry. The statement is the the embedding must be closed. The embedding into $SO(3)$ is not closed. $\endgroup$ Commented Dec 23, 2010 at 2:03

For (1) and (3), there are plenty of irreducible unitary representations of SL2(R) that do not occur in the regular representation, such as complementary series, limits of discrete series, and for that matter the trivial representation.

  • $\begingroup$ Yes indeed, thanks @Richard. Do you have a good reference where the author constructs all irreducible (no more unitary) representations of $SL_2(\mathbf{R})$? I know that there is the book from Lang on $SL_2(\mathbf{R})$ but there might be better references... $\endgroup$ Commented Dec 23, 2010 at 3:53
  • $\begingroup$ Hugo, you should take a look at Godement's Seminarie Bourbaki Expose "Introduction aux travaux de Selberg" (1957) where a proof of the Plancherel theorem for $SL_2(\mathbf R)$ is beautifully explained. For the construction of representations, I like Howe and Tan in "Nonabelian Harmonic Analysis". $\endgroup$ Commented Dec 23, 2010 at 4:55
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    $\begingroup$ Welcome, Amri ! $\endgroup$ Commented Dec 23, 2010 at 11:59

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