# Why all irreducible representations of compact groups are finite-dimensional ? [EDIT: Subtleties: AC,etc]

About 20 years ago I read in textbook that "all irreducible representations of compact groups are finite-dimensional", but me and the proof of this fact never met each other :)

May I ask dear MO colleagues, is there (simple?) argument to prove it ?

As far as I heard this result can be generalized in the realm of non-commutative geometry, Woronowicz compact quantum groups (?).

So the "bonus" question - what is appropriate "compactness" condition for some algebra (and/or Hopf algebra) such that it will guarantee the same property (i.e. all irreps are finite-dim.) ?

[EDIT] Thanks very much for excellent answers ! Let me ask about some more details, to finally clarify.

1) What is maximal possible relaxation of the requirement on vector space V ? Is it enough to require arbitrary linear topological space or we need to restrict to Hausdorff (?) Banach (?) Hilbert (?), whatever spaces ? (It seems restrictions on the space may come from the Schur lemma, it is not clear for me what is appropriate generality it holds).

2) Do we need axiom of choice here ? (Probably not, we need existence of Haar measure, but Wikipedia writes that "Henri Cartan furnished a proof of existence of Haar measure which avoided AC use."

3) Informally: what is the hardest tool one uses in the proof ? (May be existence of Haar measure ?)

[END EDIT].

[EDIT]

Let me add sketch of arguments by Aakumadula, as I understand it. It might be helpful to clarify new questions.

1) Tool: Continuous functions on the group can be mapped to operators on V. (Need measure here). (Group algebra acts on V).

2) Fact: Continuous function will be mapped to COMPACT operators. (In R^n I know how to prove it, in general no).

3) Observe: Conjugation invariant function are mapped to operators which commute with action of group.

4) Schur Lemma: operators commuting with group in irrep are Lamda*Id. (What do we need from the space V for this to be true ? )

5) Corollary: If we find invariant continuous function which is mapped in NON-zero in V, then we are done, because by (2) it is compact operator and by (4) it is Lambda*Id.

So we need to find invariant function which will be non-zero in V.

6) Take arbitrary "approximate identity" i.e. sequence of continuous (non-invariant) functions f_n which converge as functionals to delta-function in identity of the group. (It is local fact. But how to prove it ? Do we need Axiom of choice here ? )

7) Make averaging over the group of f_n - get sequence of INVARIANT continuous functions which again converge to detla(e), since delta(e) is invariant.

8) Operators T(f_n) converge to identity operator, hence for some N they are NON-ZERO. WE ARE DONE by (5) ! Because T(f_N) is compact and Lambda*Id and Lambda is NON-ZERO.

[End EDIT].

• I don't know about the quantum groups aspect of this question, so I won't try to give an answer, just a comment about the first part: The "complete reducibility to finite dimensional representations" result you read about is a consequence of (or part of, depending on how you phrase it) the Peter-Weyl Theorem. I'm not aware of any simpler proof than the standard one, which can be found in any text book on compact groups and their representations, so it's not really a research question. – Robert Bryant Jan 20 '13 at 15:45
• If you have new questions that are at an appropriate level, please ask a new question, instead of turning the existing question into a mess. – S. Carnahan Jan 22 '13 at 13:57
• To add on what S. Carnahan, seeing this thread for the first time now I have no idea how to read it. – Asaf Karagila Jan 22 '13 at 14:36
• I completely agree with S.Carnaghan – Venkataramana Jan 22 '13 at 16:02
• I agree with Scott, Adaf and Aakumadula, this should not be treated like some kind of blog post with running commentary. I suggest reverting to an earlier version of the question and asking a new question – Yemon Choi Jan 22 '13 at 16:39

[First proof] I will address only the first part. Suppose $G$ is a compact group and $\pi : G \rightarrow U(H)$ a unitary irreducible representation on a Hilbert space $H$. Suppose $\phi$ a continuous function on $G$. Then it is easy to show that $\pi (\phi)$ is a compact operator on $H$.

Since the OP has asked for clarification let me state: convolution by continuous functions on $L^2(G)$ where $G$ a COMPACT group, is a compact operator. This fact is used in all proofs of Peter-Weyl. The proof is by showing that the convolution is an $L^2$ kernel, and any $L^2$ function on $G\times G$ may be approximated in $L^2$ by simple functions (linear combination of char functions of the form $E\times F$ where $E,F$ ae measurable subsets of $G$), each of these simple kernels have finite dimensional image and are hence compact operators. This is standard material in any functional analysis book (in fact Kirillov's book).

What I have used is that for a compact group, any irreducible unitary representation is a sub of the regular representation (see the comments), and hence $\pi (\phi)$ ,which is a convolution by $\phi$ is a compact operator.

If $(\phi _{\epsilon} )$ is an approximate identity on $G$ consisting of continuous conjugation invariant real valued continuous functions on $G$, then $\pi (\phi _{\epsilon})$ is a sequence of compact operators (which are scalars because of irreducibility) converging weakly to the identity, and hence the identity operator is also compact. Therefore, $H$ is finite dimensional.

I mention this because this does not use the Peter-Weyl theorem (but uses ideas of the proof).

[Second Proof] If you use the Peter-Weyl theorem, then the proof of finite dimensionality is easy. Peter weyl says that $L^2(G)$ is a Hilbert space direct sum of irreducible FINITE dimensional representations of $G$. The algebraic direct sum $X$ is then a dense subspace of $L^2(G)$. If an infinite dimensional irrep $H$ existed, it would still be in $L^2(G)$ but by Schur orthogonality (another use of Schur's lemma) functions in $H$ would be orthogonal to functions in $X$, a contradiction, since $X$ is dense.

[Remark] if $G$ is already a closed subgroup of $U(n)$, and $V$ is the standard rep of $U(n)$ then by the Stone-Weierstrass theorem, $V$, $V^*$ and irreducible subs of their tensor powers, will all be in $L^2(G)$ and will be dense in the space of continuous functions on $G$. And hence these representation functions $Y$ will be dense in $L^2(G)$. So vectors in $H$ cannot be orthogonal to the elements of $Y$. All this is standard material in any book on the Peter-Weyl theorem (actually worked out in Kirillov's book) and I suggest that the OP looks at them

[Remark] The proof given by Francois Ziegler is the simplest, in my opinion. .

• Cool! Thank you very much! What about semigroups? – Alexander Chervov Jan 20 '13 at 18:11
• I don't know what happens with semi-groups. The above proof uses Haar measure, conjugation action etc. – Venkataramana Jan 20 '13 at 18:35
• Let me wrap my mind on that proof. 1) "it is easy to show that π(ϕ) is a compact operator on H." how to see it ? What is the generality it is true ? If group is locally compact but function $\phi$ is with local support is it true ? 2) Why do you need to choose conjugation invariant real valued functions ? I see you want to have "(which are scalars because of irreducibility) ", but why do you need this ? – Alexander Chervov Jan 20 '13 at 19:04
• By the way, you can work with Banach space representations too. The existence of approximate identities is easy to show for any locally compact group. We consider a sequence of continuous functions, whose support decreases to the identity, whose total integral is one and the associated convolution operators, converge pointwise (not in the operator norm) to the identity. – Venkataramana Jan 21 '13 at 9:41
• @Venkataramana I'm not understanding the step of your proof via Peter-Weyl where you claim that the infinite-dimensional irreducible representations also appear in $L^2(G)$. Is this obviously the case? The proof that the Peter-Weyl map from $V^* \otimes V \to L^2(G)$ is injective for finite-dimensional $V$ seems to rely on the fact that $V^* \otimes V$ has a canonical inner product (the Hilbert-Schmidt inner product), and this doesn't seem to work in the infinite-dimensional case. – Ashvin Swaminathan Nov 24 '17 at 23:03

(Addressing only the title question.) There is a short proof avoiding Peter-Weyl and the theory of compact operators. It is due to Nachbin and is reproduced in Hewitt-Ross, Abstract Harmonic Analysis 1, p. 344. A slight simplification of it runs as follows: Pick a unit vector $v$ in your representation space $V$. Schur's lemma gives $$\int_G gv(gv,\cdot)dg = \lambda1 \tag 1$$ where $\lambda = \int_G|(v,gv)|^2dg>0$ (sandwiching (1) with $(v,\cdot v)$). Now let $W\subset V$ be finite-dimensional, and write $E=E^2$ for the orthogonal projection $V\to W$. We get $$\int_GEgv(gv,E\cdot)dg = \lambda E, \tag 2$$ whence, taking traces in (2), $\lambda\dim(W)=\int_G\|Egv\|^2dg\leqslant\operatorname{vol}(G)$. Thus, the dimension of any finite-dimensional subspace is bounded, as was to be shown.

• Thank you very much ! I'll think and may be ask questions tomorrow. What about semi-groups ? – Alexander Chervov Jan 20 '13 at 20:36
• Very nice! Makes the finite dimensionality elementary. – Venkataramana Jan 21 '13 at 1:42
• I should correct myself: this version of Schur's Lemma uses spectral theory of bounded operators, which has a bit more "muscle" than that for compact operators. – Venkataramana Jan 21 '13 at 3:01
• @Aakumadula: You are quite right about Schur's Lemma. (A detailed reference for it would be Hewitt-Ross, p. 324.) So I'd call this proof short and simple, but not quite "elementary". – Francois Ziegler Jan 21 '13 at 3:54
• May I respectfully disagree with both of you (V. and F.Z.) about non-elementariness? In fact the LHS of (1) is clearly a compact operator (because by compactness of the $G$-orbit of $v$, the integral is a norm-limit of finite rank operators). Hence the result by compact spectral theory and Riesz (because $\lambda 1$ is compact). – BS. Nov 15 '15 at 10:50

In comments the question is asked about semigroups. The answer is that a compact semigroup can have no non-trivial finite dimensional reps. Take the unit interval with min. Then since its connected its image under a rep is connected. But it is idempotent and commutative and hence simultaneously diagonalizable. But there are only finitely many diagonal idempotents so the image is a point. More generally you can separate points of a compact inverse semigroup iff the idempotents are totally disconnected.

• That's actually pretty cute... – Todd Trimble Jan 20 '13 at 21:32
• This actually just shows you can't separate points with finite dimensional representations, which means no Peter-Weyl theorem. For a compact idempotent commutative semigroup there are no infinite dimensional irreps. This is because the only central idempotents can be 0 or 1 in an irreducible rep of a compact semigroup because the image of a central idempotent is invariant. – Benjamin Steinberg Jan 20 '13 at 22:03
• Thank you very much ! By the way if semi-group is finite, is it also false ? (Your argument breaks, since set is discrete and it can be mapped to discrete set of idempotents). – Alexander Chervov Jan 21 '13 at 6:37
• If it's finite, any vector in any representation generates a finite-dimensional subrepresentation, so any irreducible representation must be finite-dimensional. – Eric Wofsey Jan 21 '13 at 13:25
• Or said differently every simple module for a finite dimensional algebra is cyclic and hence a quotient of the regular module and hence finite dimensional. – Benjamin Steinberg Jan 21 '13 at 13:38

Here's a pretty easy direct argument. Let $X$ be a unitary representation of a compact group $G$. We note that for any finite-rank operator $T$ on $X$, $g\mapsto gTg^{-1}$ is a norm-continuous map $G\to B(X)$. This is because $T$ is a sum of operators of the form $\langle -, u\rangle v$, which conjugate to $\langle -, gu\rangle gv$, and $g\mapsto gv$ is norm-continuous for any fixed $v$ (the map $G\to U(X)$ is strong operator continuous).

Now let $T$ be any finite-rank positive operator on $X$. By averaging the conjugates of $T$ over $G$, we obtain an invariant positive operator $\tilde{T}$. By the continuity noted above and compactness of $G$, $\tilde{T}$ can be approximated in norm by finite "Riemann sums" of conjugates of $T$, and is thus compact.

If $X$ is irreducible, $\tilde{T}$ has to be a multiple of the identity. Since $\tilde{T}$ is compact, it follows that $X$ is finite-dimensional. More generally, eigenspaces of $\tilde{T}$ give finite-dimensional subrepresentations of any representation, and it follows easily that every unitary representation is a sum of irreducible representations (which are finite-dimensional).

• (This assumes the space is Hilbert, though) – Mariano Suárez-Álvarez Jan 20 '13 at 20:04
• True. The important thing that's being used from that is that we have a nice notion of positivity, which allows us to be sure that $\tilde{T}$ is nonzero. – Eric Wofsey Jan 20 '13 at 20:13
• Thank you very much ! I'll think and may be ask questions tomorrow. What about semi-groups ? – Alexander Chervov Jan 20 '13 at 20:36