The purpose of this post is to describe more of Segal’s paper. Actually, I won’t be covering any legitimate K-theory in this post; that’ll have to wait for a third. I’ll mostly be describing various classical constructions for vector bundles in the equivariant setting.

In the classical theory of (ordinary) vector bundles on compact spaces, a basic result is the Serre-Swan theorem, which identifies the category of (complex) vector bundles with the category of projective modules over the ring of complex-valued continuous functions on . This is essentially a reflection of the fact that any vector bundle on , say , can be obtained as a retract of some trivial bundle . Taking retracts corresponds to choosing idempotents in the ring of -by- matrices in , and this description via idempotents applies as well to projective modules over (or, in fact, any commutative ring).

The crucial statement here, that any vector bundle is a retract of a trivial one, fails in the equivariant case, simply because a vector bundle on which acts nontrivially can’t be a retract of a vector bundle with trivial action. But we have something reasonably close to it.

Definition 1Given a -representation and a -space , we can form a vector bundle , which his naturally -equivariant.

This bundle is, equivalently, formed by taking the equivariant map . -vector bundles on are identified with -representations, so we just have to pull back.

Anyway, the claim is:

Theorem 2 (Equivariant Serre-Swan (Segal))Any -vector bundle is a direct summand of a bundle for some -representation .

The proof of this result turns out to be significantly harder than the non-equivariant case. There, one has a plain vector bundle over a plain space , and one just needs to find a finite number of global sections of that generate . Then, that gives a surjection from a trivial bundle to , which necessarily splits.

In order to handle the equivariant case, however, one needs more work. Let be a -vector bundle over the -space . We can always find a finite-dimensional space of sections such that generates all the fibers; that is, the image of in each fiber is all of . This has nothing to do with equivariance. Now if were a -invariant subspace of the -vector space , then we could just think of as a -representation and consider a map

which would be the desired surjection.

Unfortunately, this is not the case. It is not even true necessarily that is *contained* in a finite-dimensional -invariant vector space of sections (which would also be enough). We’ll actually need something topological.

**1. The Peter-Weyl theorem**

For a compact group , the classical Peter-Weyl theorem gives a decomposition of as a -module: namely, if the are the finite-dimensional irreducible representations of , then we have:

Theorem 3 (Peter-Weyl)The Hilbert space (where is given the Haar measure) decomposes as an orthogonal sum .

This is similar to (and a generalization of) the decomposition for a *finite* group , as the range over the irreducible representations of . The difference, of course, is that the Peter-Weyl theorem gives a topological decomposition of a Hilbert space, rather than a vector space decomposition.

Anyway, we won’t actually need just the Peter-Weyl theorem, but a generalization thereof.

Theorem 4 (Mostow)If is a -Banach space, then the elements such that is finite-dimensional are dense.

The Peter-Weyl theorem gives a (Hilbert space) decomposition of into a collection of finite-dimensional -invariant subspaces, so this is a generalization. To prove Mostow’s theorem, one argues as follows. One consequence of the proof of the Peter-Weyl theorem is that the statement of Mostow’s theorem is true not only for , but for with the sup norm. That is, any continuous function on can be uniformly approximated by a continuous function whose translates by elements of span a finite-dimensional space.

Now, because is a Banach space, one may *integrate* an -valued continuous function over any nice probability space. (This is probably not the most general case under which one may do this.) So, this means we can integrate an -valued function over .

We thus define a *pairing* as follows. Given , , we define

When , this is just averaging, for instance.

Let us fix , and try to approximate by elements of whose -translates span a finite-dimensional space. To do so, we take

Now, note that if is such that , then

In particular, if is chosen such that is supported in a small neighborhood of the identity , then is very close to . So we can think of these “generalized averages” as good approximations to . Note moreover that is well-behaved with respect to translation; that is,

if the action of on is appropriately defined.

With these facts, we can easily prove the theorem. Fix , and fix . If we choose the continuous function whose support is very near and whose total integral is one, then

Next, let be a function, approximating by in -norm, say such that , and we can also assume that the -translates of span a finite-dimensional subspace of . In this case, we have:

where . In particular, we find

though by construction of , we have also seen that the translates of span a finite-dimensional subspace of . So we have gotten the desired approximation to .

**2. Proof of the equivariant Serre-Swan theorem**

With Mostow’s theorem proved, we can now give the proof of the equivariant Serre-Swan theorem. Let be a -vector bundle. As we’ve seen, to prove the result (i.e. to express as a retract of a bundle associated to a -representation), it suffices to find a finite-dimensional -invariant subspace

such that the image of in at each is all of .

In order to do this, we start by choosing any subspace (not necessarily -invariant) with the property that the image of in fills for all . By the Peter-Weyl theorem, we can find a subspace which is “close” to (i.e., by approximating each element of a basis of very closely) such that is contained in a finite-dimensional invariant -representation . But if is “very close” to , then generates at each , since does, and since is compact. Thus, we’re done, and the result we wanted is proved.

**3. The equivariant Grassmannian**

One of the classical facts about vector bundles in topology is the following. If is a compact space (or, more generally, a suitably nice space, e.g. a CW complex), then isomorphism classes of -dimensional vector bundles on are in bijection with homotopy classes of maps from into the Grassmannian of -planes in the vector space . Since isomorphism classes of -dimensional vector bundles are the same as isomorphism classes of principal -bundles, it follows that the infinite Grassmannian is a model for the classifying space .

One consequence of all this discussion is that we can get an analogous result for -spaces. Let be a compact -space. Let be the -space defined as follows. For each (finite-dimensional) -representation , we let be the -space of -dimensional subspaces of . For an imbedding , there is an obvious map . Now since the category of finite-dimensional -representations is filtered, we define

This is a -space (not compact!) and it is the analog of the infinite Grassmannian in the equivariant case.

The claim is that there is a “universal” -dimensional -vector bundle on . To see this, note that there is an -dimensional -vector bundle on for each (contained in ); this is the “tautological” one, defined in the usual way. These are compatible with the morphisms. Let be the -vector bundle on given by the colimit.

Theorem 5There is a natural isomorphism between isomorphism classes of -dimensional -vector bundles on and (that is, -homotopy classes of -maps).

Note that . One can prove this as in the real case. Namely, if is a -vector bundle, then we can write for some finite-dimensional -representation . Then we get a map sending to the image of in . This defines a map from to the infinite Grassmannian, and it’s easy to see that becomes the pull-back of the universal bundle. The uniqueness is similar as in the non-equivariant case.

]]>-
*representations*of into some more structured group, such as a matrix group ; and -
*metrics*on that capture the escape and commutator structure of (i.e. Gleason metrics).

To build either of these structures, a fundamentally useful tool is that of (left-) Haar measure – a left-invariant Radon measure on . (One can of course also consider right-Haar measures; in many cases (such as for compact or abelian groups), the two concepts are the same, but this is not always the case.) This concept generalises the concept of Lebesgue measure on Euclidean spaces , which is of course fundamental in analysis on those spaces.

Haar measures will help us build useful representations and useful metrics on locally compact groups . For instance, a Haar measure gives rise to the regular representation that maps each element of to the unitary translation operator on the Hilbert space of square-integrable measurable functions on with respect to this Haar measure by the formula

(The presence of the inverse is convenient in order to obtain the homomorphism property without a reversal in the group multiplication.) In general, this is an infinite-dimensional representation; but in many cases (and in particular, in the case when is compact) we can decompose this representation into a useful collection of finite-dimensional representations, leading to the Peter-Weyl theorem, which is a fundamental tool for understanding the structure of compact groups. This theorem is particularly simple in the compact abelian case, where it turns out that the representations can be decomposed into one-dimensional representations , better known as characters, leading to the theory of Fourier analysis on general compact abelian groups. With this and some additional (largely combinatorial) arguments, we will also be able to obtain satisfactory structural control on locally compact abelian groups as well.

The link between Haar measure and useful metrics on is a little more complicated. Firstly, once one has the regular representation , and given a suitable “test” function , one can then embed into (or into other function spaces on , such as or ) by mapping a group element to the translate of in that function space. (This map might not actually be an embedding if enjoys a non-trivial translation symmetry , but let us ignore this possibility for now.) One can then pull the metric structure on the function space back to a metric on , for instance defining an -based metric

if is square-integrable, or perhaps a -based metric

if is continuous and compactly supported (with denoting the supremum norm). These metrics tend to have several nice properties (for instance, they are automatically left-invariant), particularly if the test function is chosen to be sufficiently “smooth”. For instance, if we introduce the differentiation (or more precisely, finite difference) operators

(so that ) and use the metric (1), then a short computation (relying on the translation-invariance of the norm) shows that

for all . This suggests that commutator estimates, such as those appearing in the definition of a Gleason metric in Notes 2, might be available if one can control “second derivatives” of ; informally, we would like our test functions to have a “” type regularity.

If was already a Lie group (or something similar, such as a local group) then it would not be too difficult to concoct such a function by using local coordinates. But of course the whole point of Hilbert’s fifth problem is to do without such regularity hypotheses, and so we need to build test functions by other means. And here is where the Haar measure comes in: it provides the fundamental tool of convolution

between two suitable functions , which can be used to build smoother functions out of rougher ones. For instance:

Exercise 1Let be continuous, compactly supported functions which are Lipschitz continuous. Show that the convolution using Lebesgue measure on obeys the -type commutator estimatefor all and some finite quantity depending only on .

This exercise suggests a strategy to build Gleason metrics by convolving together some “Lipschitz” test functions and then using the resulting convolution as a test function to define a metric. This strategy may seem somewhat circular because one needs a notion of metric in order to define Lipschitz continuity in the first place, but it turns out that the properties required on that metric are weaker than those that the Gleason metric will satisfy, and so one will be able to break the circularity by using a “bootstrap” or “induction” argument.

We will discuss this strategy – which is due to Gleason, and is fundamental to all currently known solutions to Hilbert’s fifth problem – in later posts. In this post, we will construct Haar measure on general locally compact groups, and then establish the Peter-Weyl theorem, which in turn can be used to obtain a reasonably satisfactory structural classification of both compact groups and locally compact abelian groups.

** — 1. Haar measure — **

For technical reasons, it is convenient to not work with an absolutely general locally compact group, but to restrict attention to those groups that are both -compact and Hausdorff, in order to access measure-theoretic tools such as the Fubini-Tonelli theorem and the Riesz representation theorem without bumping into unwanted technical difficulties. Intuitively, -compact groups are those groups that do not have enormously “large” scales – scales are too coarse to be “seen” by any compact set. Similarly, Hausdorff groups are those groups that do not have enormously “small” scales – scales that are too small to be “seen” by any open set. A simple example of a locally compact group that fails to be -compact is the real line with the discrete topology; conversely, a simple example of a locally compact group that fails to be Hausdorff is the real line with the trivial topology.

As the two exercises below show, one can reduce to the -compact Hausdorff case without much difficulty, either by restricting to an open subgroup to eliminate the largest scales and recover -compactness, or to quotient out by a compact normal subgroup to eliminate the smallest scales and recover the Hausdorff property.

Exercise 2Let be a locally compact group. Show that there exists an open subgroup which is locally compact and -compact. (Hint:take the group generated by a compact neighbourhood of the identity.)

Exercise 3Let be a locally compact group. Let be the topological closure of the identity element.

- (i) Show that given any open neighbourhood of a point in , there exists a neighbourhood of whose closure lies in . (
Hint:translate to the identity and select so that .) In other words, is a regular space.- (ii) Show that for any group element , that the sets and are either equal or disjoint.
- (iii) Show that is a compact normal subgroup of .
- (iv) Show that the quotient group (equipped with the quotient topology) is a locally compact Hausdorff group.
- (v) Show that a subset of is open if and only if it is the preimage of an open set in .

Now that we have restricted attention to the -compact Hausdorff case, we can now define the notion of a Haar measure.

Definition 1 (Radon measure)Let be a -compact locally compact Hausdorff topological space. The Borel -algebra on is the -algebra generated by the open subsets of . A Borel measure is a countably additive non-negative measure on the Borel -algebra. A Radon measure is a Borel measure obeying three additional axioms:

- (i) (Local finiteness) One has for every compact set .
- (ii) (Inner regularity) One has for every Borel measurable set .
- (iii) (Outer regularity) One has for every Borel measurable set .

Definition 2 (Haar measure)Let be a -compact locally compact Hausdorff group. A Radon measure isleft-invariant(resp.right-invariant) if one has (resp. ) for all and Borel measurable sets . Aleft-invariant Haar measureis a non-zero Radon measure which is left-invariant; a right-invariant Haar measure is defined similarly. Abi-invariant Haar measureis a Haar measure which is both left-invariant and right-invariant.

Note that we do not consider the zero measure to be a Haar measure.

Example 1A large part of the foundations of Lebesgue measure theory (e.g. most of these lecture notes of mine) can be summed up in the single statement that Lebesgue measure is a (bi-invariant) Haar measure on Euclidean spaces .

Example 2If is a countable discrete group, then counting measure is a bi-invariant Haar measure.

Example 3If is a left-invariant Haar measure on a -compact locally compact Hausdorff group , then the reflection defined by is a right-invariant Haar measure on , and the scalar multiple is a left-invariant Haar measure on for any .

Exercise 4If is a left-invariant Haar measure on a -compact locally compact Hausdorff group , show that for any non-empty open set .

Let be a left-invariant Haar measure on a -compact locally compact Hausdorff group. Let be the space of all continuous, compactly supported complex-valued functions ; then is absolutely integrable with respect to (thanks to local finiteness), and one has

for all (thanks to left-invariance). Similarly for right-invariant Haar measures (but now replacing by ).

The fundamental theorem regarding Haar measures is:

Theorem 3 (Existence and uniqueness of Haar measure)Let be a -compact locally compact Hausdorff group. Then there exists a left-invariant Haar measure on . Furthermore, this measure is unique up to scalars: if are two left-invariant Haar measures on , then for some scalar .Similarly if “left-invariant” is replaced by “right-invariant” throughout. (However, we do

notclaim that every left-invariant Haar measure is automatically right-invariant, or vice versa.)

To prove this theorem, we will rely on the Riesz representation theorem:

Theorem 4 (Riesz representation theorem)Let be a -compact locally compact Hausdorff space. Then to every linear functional which is non-negative (thus whenever ), one can associate a unique Radon measure such that for all . Conversely, for each Radon measure , the functional is a non-negative linear functional on .

We now establish the uniqueness component of Theorem 3. We shall just prove the uniqueness of left-invariant Haar measure, as the right-invariant case is similar (and also follows from the left-invariant case by Example 3). Let be two left-invariant Haar measures on . We need to prove that is a scalar multiple of . From the Riesz representation theorem, it suffices to show that is a scalar multiple of . Equivalently, it suffices to show that

for all .

To show this, the idea is to approximate both and by superpositions of translates of the same function . More precisely, fix , and let . As the functions and are continuous and compactly supported, they are uniformly continuous, in the sense that we can find an open neighbourhood of the identity such that and for all and ; we may also assume that the are contained in a compact set that is uniform in . By Exercise 4 and Urysohn’s lemma, we can then find an “approximation to the identity” supported in such that . Since

for all in the support of , we conclude that

uniformly in ; also, the left-hand side has uniformly compact support in . If we integrate against , we conclude that

where the implied constant in the notation can depend on but not on . But by the left-invariance of , the left-hand side is also

which by the Fubini-Tonelli theorem is

which by the left-invariance of is

which simplifies to . We conclude that

and similarly

which implies that

Sending we obtain the claim.

Exercise 5Obtain another proof of uniqueness of Haar measure by investigating the translation-invariance properties of the Radon-Nikodym derivative of with respect to .

Now we show existence of Haar measure. Again, we restrict attention to the left-invariant case (using Example 3 if desired). By the Riesz representation theorem, it suffices to find a functional from the space of non-negative continuous compactly supported functions to the non-negative reals obeying the following axioms:

- (Homogeneity) for all and .
- (Additivity) for all .
- (Left-invariance) for all and .
- (Non-degeneracy) for at least one .

Here, is the translation operation as discussed in the introduction.

We will construct this functional by an approximation argument. Specifically, we fix a non-zero . We will show that given any finite number of functions and any , one can find a functional that obeys the following axioms:

- (Homogeneity) for all and .
- (Approximate additivity) for all .
- (Left-invariance) for all and .
- (Uniform bound) For each , we have , where does not depend on or .
- (Normalisation) .

Once one has established the existence of these approximately additive functionals , one can then construct the genuinely additive functional (and thus a left-invariant Haar measure) by a number of standard compactness arguments. For instance:

- One can observe (from Tychonoff’s theorem) that the space of all functionals obeying the uniform bound is a compact subset of the product space ; in particular, any collection of closed sets in this space obeying the finite intersection property has non-empty intersection. Applying this fact to the closed sets of functionals obeying the homogeneity, approximate additivity, left-invariance, uniform bound, and normalisation axioms for various , we conclude that there is a functional that lies in all such sets, giving the claim.
- If one lets be the space of all tuples , one can use the Hahn-Banach theorem to construct a bounded real linear functional that maps the constant sequence to . If one then applies this functional to the one can obtain a functional with the required properties.
- One can also adopt a nonstandard analysis approach, taking an ultralimit of all the and then taking a standard part to recover .
- A closely related method is to obtain from the by using the compactness theorem in logic.
- In the case when is metrisable (and hence separable, by -compactness), then becomes separable, and one can also use the Arzelá-Ascoli theorem in this case. (One can also try in this case to directly ensure that the converge pointwise, without needing to pass to a further subsequence, although this requires more effort than the compactness-based methods.)

These approaches are more or less equivalent to each other, and the choice of which approach to use is largely a matter of personal taste.

It remains to obtain the approximate functionals for a given and . As with the uniqueness claim, the basic idea is to approximate all the functions by translates of a given function . More precisely, let be a small quantity (depending on and ) to be chosen later. By uniform continuity, we may find a neighbourhood of the identity such that for all and . Let be a function, not identically zero, which is supported in .

To motivate the argument that follows, pretend temporarily that we have a left-invariant Haar measure available, and let be the integral of with respect to this measure. Then , and by left-invariance one has

and thus

for any scalars and . In particular, if we introduce the *covering number*

of a given function by , we have

This suggests using a scalar multiple of as the approximate linear functional (noting that can be defined without reference to any existing Haar measure); in view of the normalisation , it is then natural to introduce the functional

(This functional is analogous in some ways to the concept of outer measure or the upper Darboux integral in measure theory.) Note from compactness that is finite for every , and from the non-triviality of we see that , so is well-defined as a map from to . It is also easy to verify that obeys the homogeneity, left-invariance, and normalisation axioms. From the easy inequality

we also obtain the uniform bound axiom, and from the infimal nature of we also easily obtain the subadditivity property

To finish the construction, it thus suffices to show that

for each , if is chosen sufficiently small depending on .

Fix . By definition, we have the pointwise bound

and

if is small enough. Indeed, we have

If is non-zero, then by the construction of and , one has and , which implies that

Using (3) we thus have

which gives (5); a similar argument gives (6). From the subadditivity (and monotonicity) of , we conclude that

and

where equals on the support of . Summing and using (4), we conclude that

and the claim follows by taking small enough. This concludes the proof of Theorem 3.

Exercise 6State and prove a generalisation of Theorem 3 in which the hypothesis that is Hausdorff and -compact are dropped. (This requires extending concepts such as “Borel -algebra”, “Radon measure”, and “Haar measure” to the non-Hausdorff or non--compact setting. Note that different texts sometimes have inequivalent definitions of these concepts in such settings; because of this (and also because of the potential breakdown of some basic measure-theoretic tools such as the Fubini-Tonelli theorem), it is usually best to avoid working with Haar measure in the non-Hausdorff or non--compact case unless one is very careful.)

Remark 1An important special case of the Haar measure construction arises forcompactgroups . Here, we can normalise the Haar measure by requiring that (i.e. is a probability measure), and so there is now a unique (left-invariant) Haar probability measure on such a group. In Exercise 7 we will see that this measure is in fact bi-invariant.

Remark 2The above construction, based on the Riesz representation theorem, is not the only way to construct Haar measure. Another approach that is common in the literature is to first build a left-invariant outer measure and then use the Carathéodory extension theorem. Roughly speaking, the main difference between that approach and the one given here is that it is based on covering compact or open sets by other compact or open sets, rather than covering continuous, compactly supported functions by other continuous, compactly supported functions. In the compact case, one can also construct Haar probability measure by defining to be the mean of , or more precisely the unique constant function that is an average of translates of . See Exercise 6 of these notes for further discussion (the post there focuses on the abelian case, but the argument extends to the nonabelian setting).

The following exercise explores the distinction between left-invariance and right-invariance.

Exercise 7Let be a -compact locally compact Hausdorff group, and let be a left-invariant Haar measure on .

- (i) Show that for each , there exists a unique positive real (independent of the choice of ) such that for all Borel measurable sets and for all absolutely integrable . In particular, a left-invariant Haar measure is right-invariant if and only if for all .
- (ii) Show that the map is a continuous homomorphism from to the multiplicative group . (This homomorphism is known as the
modular function, and is said to beunimodularif is identically equal to .)- Show that for any , one has . (
Hint:take another function and evaluate in two different ways, one of which involves replacing by .) In particular, in a unimodular group one has and for any Borel set and any .- (iii) Show that is unimodular if it is compact.
- (iv) If is a Lie group with Lie algebra , show that , where is the adjoint representation of , defined by requiring for all (cf. Lemma 13 of Notes 1).
- (v) If is a connected Lie group with Lie algebra , show that is unimodular if and only if for all , where is the adjoint representation of .
- (vi) Show that is unimodular if it is a connected nilpotent Lie group.
- (vii) Let be a connected Lie group whose Lie algebra is such that (where is the linear span of the commutators with ). (This condition is in particular obeyed when the Lie algebra is semisimple.) Show that is unimodular.
- (viii) Let be the group of pairs with the composition law . (One can interpret as the group of orientation-preserving affine transformations on the real line.) Show that is a connected Lie group that is not unimodular.

In the case of a Lie group, one can also build Haar measures by starting with a non-invariant smooth measure, and then correcting it. Given a smooth manifold , define a *smooth measure* on to be a Radon measure which is a smooth multiple of Lebesgue measure when viewed in coordinates, thus for any smooth coordinate chart , the pushforward measure takes the form for some smooth function , thus

for all . We say that the smooth measure is *nonvanishing* if is non-zero on for every coordinate chart .

Exercise 8Let be a Lie group, and let be a nonvanishing smooth measure on .

- Show that for every , there exists a unique smooth function such that
- Verify the
cocycle equationfor all .- Show that the measure defined by
is a left-invariant Haar measure on .

There are a number of ways to generalise the Haar measure construction. For instance, one can define a local Haar measure on a local group . If is a neighbourhood of the identity in a -compact locally compact Hausdorff local group , we define a *local left-invariant Haar measure* on to be a non-zero Radon measure on with the property that whenever and is a Borel set such that is well-defined and also in .

Exercise 9 (Local Haar measure)Let be a -compact locally compact Hausdorff local group, and let be an open neighbourhood of the identity in such that is symmetric (i.e. is well-defined and equal to ) and is well-defined in . By adapting the arguments above, show that there is a local left-invariant Haar measure on , and that it is unique up to scalar multiplication. (Hint:a new technical difficulty is that there are now multiple covering numbers of interest, namely the covering numbers associated to various small powers of . However, as long as one keeps track of which covering number to use at various junctures, this will not cause difficulty.)

One can also sometimes generalise the Haar measure construction from groups to spaces that acts transitively on.

Definition 5 (Group actions)Given a topological group and a topological space , define a (left) continuous action of on to be a continuous map from to such that and for all and .This action is said to be transitive if for any , there exists such that , and in this case is called a homogeneous space with structure group , or

homogenous -spacefor short.For any , we call the stabiliser of ; this is a closed subgroup of .

If are smooth manifolds (so that is a Lie group) and the action is a smooth map, then we say that we have a

smooth actionof on .

Exercise 10If acts transitively on a space , show that all the stabilisers are conjugate to each other, and is homeomorphic to the quotient spaces after weakening the topology of the quotient space (or strengthening the topology of the space .

If and are -compact, locally compact, and Hausdorff, a (left) *Haar measure* is a non-zero Radon measure on such that for all Borel and .

Exercise 11Let be a -compact, locally compact, and Hausdorff group (left) acting continuously and transitively on a -compact, locally compact, and Hausdorff space .

- (i) (Uniqueness up to scalars) Show that if are (left) Haar measures on , then for some .
- (ii) (Compact case) Show that if is compact, then is compact too, and a Haar measure on exists.
- (iii) (Smooth unipotent case) Suppose that the action is smooth (so that is a Lie group and is a smooth manifold). Let be a point of . Suppose that for each , the derivative map of the map at is unimodular (i.e. it has determinant ). Show that a Haar measure on exists.
- (iv) (Smooth case) Suppose that the action is smooth. Show that any Haar measure on is necessarily smooth. Conclude that a Haar measure exists if and only if the derivative maps are unimodular.
- (v) (Counterexample) Let be the group from Example 7(viii), acting on by the action . Show that there is no Haar measure on . (This can be done either through (iv), or by an elementary direct argument.)

** — 2. The Peter-Weyl theorem — **

We now restrict attention to compact groups , which we will take to be Hausdorff for simplicity (although the results in this section will easily extend to the non-Hausdorff case using Exercise 3). By the previous discussion, there is a unique bi-invariant Haar probability measure on , which gives rise in particular to the Hilbert space of square-integrable functions on (quotiented out by almost everywhere equivalence, as usual), with norm

and inner product

For every group element , the translation operator is defined by

One easily verifies that is both the inverse and the adjoint of , and so is a unitary operator. The map is then a continuous homomorphism from to the unitary group of (where we give the latter group the strong operator topology), and is known as the regular representation of .

For our purposes, the regular representation is too “big” of a representation to work with because the underlying Hilbert space is usually infinite-dimensional. However, we can find smaller representations by locating *left-invariant* closed subspaces of , i.e. closed linear subspaces of with the property that for all . Then the restriction of to becomes a representation to the unitary group of . In particular, if has some finite dimension , this gives a representation of by a unitary group after expressing in coordinates.

We can build invariant subspaces from applying spectral theory to an invariant operator, and more specifically to a *convolution operator*. If , we define the convolution by the formula

Exercise 12Show that if , then is well-defined and lies in , and in particular also lies in .

For , let denote the right-convolution operator . This is easily seen to be a bounded linear operator on . Using the properties of Haar measure, we also observe that will be self-adjoint if obeys the condition

and it also commutes with left-translations:

In particular, for any , the *eigenspace*

will be a closed invariant subspace of . Thus we see that we can generate a large number of representations of by using the eigenspace of a convolution operator.

Another important fact about these operators, is that the are compact, i.e. they map bounded sets to precompact sets. This is a consequence of the following more general fact:

Exercise 13 (Compactness of integral operators)Let and be -finite measure spaces, and let . Define an integral operator by the formula

- Show that is a bounded linear operator, with operator norm bounded by . (
Hint:use duality.)- Show that is a compact linear operator. (
Hint:approximate by a linear combination of functions of the form for and , plus an error which is small in norm, so that becomes approximated by the sum of a finite rank operator and an operator of small operator norm.)

Note that is an integral operator with kernel ; from the invariance properties of Haar measure we see that if (note here that we crucially use the fact that is compact, so that ). Thus we conclude that the convolution operator is compact when is compact.

Exercise 14Show that if is non-zero, then is not compact on . This example demonstrates that compactness of is needed in order to ensure compactness of .

We can describe self-adjoint compact operators in terms of their eigenspaces:

Theorem 6 (Spectral theorem)Let be a compact self-adjoint operator on a complex Hilbert space . Then there exists an at most countable sequence of non-zero reals that converge to zero and an orthogonal decompositionof into the eigenspace (or kernel) of , and the -eigenspaces , which are all finite-dimensional.

*Proof:* From self-adjointness we see that all the eigenspaces are orthogonal to each other, and only non-trivial for real. If , then has an orthonormal basis of eigenfunctions , each of which is enlarged by a factor of at least by . In particular, this basis cannot be infinite, because otherwise the image of this basis by would have no convergent subsequence, contradicting compactness. Thus is finite-dimensional for any , which implies that is finite-dimensional for every non-zero , and those non-zero with non-trivial can be enumerated to either be finite, or countable and go to zero.

Let be the orthogonal complement of . If is trivial, then we are done, so suppose for sake of contradiction that is non-trivial. As all of the are invariant, and is self-adjoint, is also invariant, with being self-adjoint on . As is orthogonal to the kernel of , has trivial kernel in . More generally, has no eigenvectors in .

Let be the unit ball in . As has trivial kernel and is non-trivial, . Using the identity

valid for all self-adjoint operators (see Exercise 15 below). Thus, we may find a sequence of vectors of norm at most such that

for some . Since , we conclude that

By compactness of , we may pass to a subsequence so that converges to a limit , and thus . As has no eigenvectors, must be trivial; but then converges to zero, a contradiction.

Exercise 15Establish (9) whenever is a bounded self-adjoint operator on . (Hint:Bound by the right-hand side of (8) whenever are vectors of norm at most , by playing with for various choices of scalars , in the spirit of the proof of the Cauchy-Schwarz inequality.)

This leads to the consequence that we can find non-trivial finite-dimensional representations on at least a single non-identity element:

Theorem 7 (Baby Peter-Weyl theorem)Let be a compact Hausdorff group with Haar measure , and let be a non-identity element of . Then there exists a finite-dimensional invariant subspace of on which is not the identity.

*Proof:* Suppose for contradiction that is the identity on every finite-dimensional invariant subspace of , thus annihilates every such subspace. By Theorem 6, we conclude that has range in the kernel of every convolution operator with , thus for any with obeying (7), i.e.

for any such . But one may easily construct such that is non-zero at the identity and vanishing at (e.g. one can set where is an open symmetric neighbourhood of the identity, small enough that lies outside ). This gives the desired contradiction.

Remark 3The full Peter-Weyl theorem describes rather precisely all the invariant subspaces of . Roughly speaking, the theorem asserts that for each irreducible finite-dimensional representation of , different copies of (viewed as an invariant -space) appear in , and that they are all orthogonal and make up all of ; thus, one has an orthogonal decompositionof -spaces. Actually, this is not the sharpest form of the theorem, as it only describes the left -action and not the right -action; see this previous blog post for a precise statement and proof of the Peter-Weyl theorem in its strongest form. This form is of importance in Fourier analysis and representation theory, but in this course we will only need the baby form of the theorem (Theorem 7), which is an easy consequence of the full Peter-Weyl theorem (since, if is not the identity, then is clearly non-trivial on and hence on at least one of the factors).

The Peter-Weyl theorem leads to the following structural theorem for compact groups:

Theorem 8 (Gleason-Yamabe theorem for compact groups)Let be a compact Hausdorff group, and let be a neighbourhood of the identity. Then there exists a compact normal subgroup of contained in such that is isomorphic to a linear group (i.e. a closed subgroup of a general linear group ).

Note from Cartan’s theorem (Theorem 2 from Notes 2) that every linear group is Lie; thus, compact Hausdorff groups are “almost Lie” in some sense.

*Proof:* Let be an element of . By the baby Peter-Weyl theorem, we can find a finite-dimensional invariant subspace of on which is non-trivial. Identifying such a subspace with for some finite , we thus have a continuous homomorphism such that is non-trivial. By continuity, will also be non-trivial for some open neighbourhood of . Using the compactness of , one can then find a finite number of such continuous homomorphisms such that for each , at least one of is non-trivial. If we then form the direct sum

then is still a continuous homomorphism, which is now non-trivial for any ; thus the kernel of is a compact normal subgroup of contained in . There is thus a continuous bijection from the compact space to the Hausdorff space , and so the two spaces are homeomorphic. As is a compact (hence closed) subgroup of , the claim follows.

Exercise 16Show that the hypothesis that is Hausdorff can be omitted from Theorem 8. (Hint:use Exercise 3.)

Exercise 17Show that any compact Lie group is isomorphic to a linear group. (Hint:first find a neighbourhood of the identity that is so small that it does not contain any non-trivial subgroups.) The property of having no small subgroups will be an important one in later notes.

One can rephrase the Gleason-Yamabe theorem for compact groups in terms of the machinery of inverse limits (also known as *projective limits*).

Definition 9 (Inverse limits of groups)Let be a family of groups indexed by a partially ordered set . Suppose that for each in , there is a surjective homomorphism which obeys the composition law for all . (If one wishes, one can take a category-theoretic perspective and view these surjections as describing a functor from the partially ordered set to the category of groups.) We then define theinverse limitto be the set of all tuples in the product set such that for all ; one easily verifies that this is also a group. We let denote the coordinate projection maps .If the are topological groups and the are continuous, we can give the topology induced from ; one easily verifies that this makes a topological group, and that the are continuous homomorphisms.

Exercise 18 (Universal description of inverse limit)Let be a family of groups with the surjective homomorphisms as in Definition 9. Let be the inverse limit, and let be another group. Suppose that one has homomorphisms for each such that for all . Show that there exists a unique homomorphism such that for all .Establish the same claim with “group” and “homomorphism” replaced by “topological group” and “continuous homomorphism” throughout.

Exercise 19Let be a prime. Show that is isomorphic to the inverse limit of the cyclic groups with (with the usual ordering), using the obvious projection homomorphisms from to for .

Exercise 20Show that every compact Hausdorff group is isomorphic (as a topological group) to an inverse limit of linear groups. (Hint:take the index set to be the set of all non-empty finite collections of open neighbourhoods of the identity, indexed by inclusion.) If the compact Hausdorff group is metrisable, show that one can take the inverse limit to be indexed instead by the natural numbers with the usual ordering.

Exercise 21Let be an abelian group with a homomorphism into the unitary group of a finite-dimensional space . Show that can be decomposed as the vector space sum of one-dimensional -invariant spaces. (Hint:By the spectral theorem for unitary matrices, any unitary operator on decomposes into eigenspaces, and any operator commuting with must preserve each of these eigenspaces. Now induct on the dimension of .)

Exercise 22 (Fourier analysis on compact abelian groups)Let be a compact abelian Hausdorff group with Haar probability measure . Define acharacterto be a continuous homomorphism to the unit circle , and let be the collection of all such characters.

- (i) Show that for every not equal to the identity, there exists a character such that . (
Hint:combine the baby Peter-Weyl theorem with the preceding exercise.)- (ii) Show that every function in is the limit in the uniform topology of finite linear combinations of characters. (
Hint:use the Stone-Weierstrass theorem.)- (iii) Show that the characters for form an orthonormal basis of .

** — 3. The structure of locally compact abelian groups — **

We now use the above machinery to analyse locally compact abelian groups. We follow some combinatorial arguments of Pontryagin, as presented in the text of Montgomery and Zippin.

We first make a general observation that locally compact groups contain open subgroups that are “finitely generated modulo a compact set”. Call a subgroup of a topological group *cocompact* if the quotient space is compact.

Lemma 10Let be a locally compact group. Then there exists an open subgroup of which has a cocompact finitely generated subgroup .

*Proof:* Let be a compact neighbourhood of the identity. Then is also compact and can thus be covered by finitely many copies of , thus

for some finite set , which we may assume without loss of generality to be contained in . In particular, if is the group generated by , then

Multiplying this on the left by powers of and inducting, we conclude that

for all . If we then let be the group generated by , then lies in and . Thus is the image of the compact set under the quotient map, and the claim follows.

In the abelian case, we can improve this lemma by combining it with the following proposition:

Proposition 11Let be a locally compact Hausdorff abelian group with a cocompact finitely generated subgroup. Then has a cocompactdiscretefinitely generated subgroup.

To prove this proposition, we need the following lemma.

Lemma 12Let be a locally compact Hausdorff group, and let . Then the group generated by is either precompact or discrete (or both).

*Proof:* By replacing with the closed subgroup we may assume without loss of generality that is dense in .

We may assume of course that is not discrete. This implies that the identity element is not an isolated point in , and thus for any neighbourhood of the identity , there exist arbitrarily large such that ; since we may take these to be large and positive rather than large and negative.

Let be a precompact symmetric neighbourhood of the identity, then (say) is covered by a finite number of left-translates of . As is dense, we conclude that is covered by a finite number of translates of left-translates of by powers of . Using the fact that there are arbitrarily large with , we may thus cover by a finite number of translates of with . In particular, if , then there exists an such that . Iterating this, we see that the set is left-syndetic, in that it has bounded gaps as one goes to . Similarly one can argue that this set is right-syndentic and thus syndetic. This implies that the entire group is covered by a bounded number of translates of and is thus precompact as required.

Now we can prove Proposition 11.

*Proof:* Let us say that a locally compact Hausdorff abelian group has *rank at most * if it has a cocompact subgroup generated by at most generators. We will induct on the rank . If has rank , then the cocompact subgroup is trivial, and the claim is obvious; so suppose that has some rank , and the claim has already been proven for all smaller ranks.

By hypothesis, has a cocompact subgroup generated by generators . By Lemma 12, the group is either precompact or discrete. If it is discrete, then we can quotient out by that group to obtain a locally compact Hausdorff abelian group of rank at most ; by induction hypothesis, has a cocompact discrete subgroup, and so does also. Hence we may assume that is precompact, and more generally that is precompact for each . But as we are in an abelian group, is the product of all the , and is thus also precompact, so is compact. But is a quotient of and is also compact, and so itself is compact, and the claim follows in this case.

We can then combine this with the Gleason-Yamabe theorem for compact groups to obtain

Theorem 13 (Gleason-Yamabe theorem for abelian groups)Let be a locally compact abelian Hausdorff group, and let be a neighbourhood of the identity. Then there exists a compact normal subgroup of contained in such that is isomorphic to a Lie group.

*Proof:* By Lemma 10 and Proposition 11, we can find an open subgroup of and discrete cocompact subgroup of . By shrinking as necessary, we may assume that is symmetric and only intersects at the identity. Let be the projection to the compact abelian group , then is a neighbourhood of the identity in . By Theorem 8, one can find a compact normal subgroup of in such that is isomorphic to a linear group, and thus to a Lie group. If we set , it is not difficult to verify that is also a compact normal subgroup of . If is the quotient map, then is a discrete subgroup of and from abstract nonsense one sees that is isomorphic to the Lie group . Thus is locally Lie. Since is an open subgroup of the abelian group , is locally Lie also, and is thus is isomorphic to a Lie group by Exercise 15 of Notes 1.

Exercise 23Show that the Hausdorff hypothesis can be dropped from the above theorem.

Exercise 24 (Characters separate points)Let be a locally compact Hausdorff abelian group, and let be not equal to the identity. Show that there exists a character (see Exercise 22) such that . This result can be used as the foundation of the theory of Pontryagin duality in abstract harmonic analysis, but we will not pursue this here; see for instance this text of Rudin.

Exercise 25Show that every locally compact abelian Hausdorff group is isomorphic to the inverse limit of abelian Lie groups.

Thus, in principle at least, the study of locally compact abelian group is reduced to that of abelian Lie groups, which are more or less easy to classify:

Exercise 26

- Show that every discrete subgroup of is isomorphic to for some .
- Show that every connected abelian Lie group is isomorphic to for some natural numbers . (
Hint:first show that the kernel of the exponential map is a discrete subgroup of the Lie algebra.) Conclude in particular thedivisibility propertythat if and then there exists with .- Show that every compact abelian Lie group is isomorphic to for some natural number and a which is a finite product of finite cyclic groups. (You may need the classification of finitely generated abelian groups, and will also need the divisibility property to lift a certain finite group from a certain quotient space back to .)
- Show that every abelian Lie group contains an open subgroup that is isomorphic to for some natural numbers and a finite product of finite cyclic groups.

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Remark 4Despite the quite explicit description of (most) abelian Lie groups, some interesting behaviour can still occur in locally compact abelian groups after taking inverse limits; consider for instance the solenoid example (Exercise 6 from Notes 0).

We call two unitary representations and *isomorphic* if one has for some unitary transformation , in which case we write .

Given two unitary representations and , one can form their direct sum in the obvious manner: . Conversely, if a unitary representation has a closed invariant subspace of (thus for all ), then the orthogonal complement is also invariant, leading to a decomposition of into the *subrepresentations* , . Accordingly, we will call a unitary representation *irreducible* if is nontrivial (i.e. ) and there are no nontrivial invariant subspaces (i.e. no invariant subspaces other than and ); the irreducible representations play a role in the subject analogous to those of prime numbers in multiplicative number theory. By the principle of infinite descent, every finite-dimensional unitary representation is then expressible (perhaps non-uniquely) as the direct sum of irreducible representations.

The Peter-Weyl theorem asserts, among other things, that the same claim is true for the regular representation:

Theorem 1 (Peter-Weyl theorem)Let be a compact group. Then the regular representation is isomorphic to the direct sum of irreducible representations. In fact, one has , where is an enumeration of the irreducible finite-dimensional unitary representations of (up to isomorphism). (It is not difficult to see that such an enumeration exists.)

In the case when is abelian, the Peter-Weyl theorem is a consequence of the Plancherel theorem; in that case, the irreducible representations are all one dimensional, and are thus indexed by the space of *characters* (i.e. continuous homomorphisms into the unit circle ), known as the Pontryagin dual of . (See for instance my lecture notes on the Fourier transform.) Conversely, the Peter-Weyl theorem can be used to deduce the Plancherel theorem for compact groups, as well as other basic results in Fourier analysis on these groups, such as the Fourier inversion formula.

Because the regular representation is faithful (i.e. injective), a corollary of the Peter-Weyl theorem (and a classical theorem of Cartan) is that every compact group can be expressed as the inverse limit of Lie groups, leading to a solution to Hilbert’s fifth problem in the compact case. Furthermore, the compact case is then an important building block in the more general theory surrounding Hilbert’s fifth problem, and in particular a result of Yamabe that any *locally* compact group contains an open subgroup that is the inverse limit of Lie groups.

I’ve recently become interested in the theory around Hilbert’s fifth problem, due to the existence of a *correspondence principle* between locally compact groups and *approximate groups*, which play a fundamental role in arithmetic combinatorics. I hope to elaborate upon this correspondence in a subsequent post, but I will mention that versions of this principle play a crucial role in Gromov’s proof of his theorem on groups of polynomial growth (discussed previously on this blog), and in a more recent paper of Hrushovski on approximate groups (also discussed previously). It is also analogous in many ways to the more well-known *Furstenberg correspondence principle* between ergodic theory and combinatorics (also discussed previously).

Because of the above motivation, I have decided to write some notes on how the Peter-Weyl theorem is proven. This is utterly standard stuff in abstract harmonic analysis; these notes are primarily for my own benefit, but perhaps they may be of interest to some readers also.

** — 1. Proof of the Peter-Weyl theorem — **

Throughout these notes, is a fixed compact group.

Let and be unitary representations. An (linear) equivariant map is defined to be a continuous linear transformation such that for all .

A fundamental fact in representation theory, known as Schur’s lemma, asserts (roughly speaking) that equivariant maps cannot mix irreducible representations together unless they are isomorphic. More precisely:

Lemma 2 (Schur’s lemma for unitary representations)Suppose that and are irreducible unitary representations, and let be an equivariant map. Then is either the zero transformation, or a constant multiple of an isomorphism. In particular, if , then there are no non-trivial equivariant maps between and .

*Proof:* The adjoint map of the equivariant map is also equivariant, and thus so is . As is also a bounded self-adjoint operator, we can apply the spectral theorem to it. Observe that any closed invariant subspace of is -invariant, and is thus either or . By the spectral theorem, this forces to be a constant multiple of the identity. Similarly for . This forces to either be zero or a constant multiple of a unitary map, and the claim follows. (Thanks to Frederick Goodman for this proof.)

Schur’s lemma has many foundational applications in the subject. For instance, we have the following generalisation of the well-known fact that eigenvectors of a unitary operator with distinct eigenvalues are necessarily orthogonal:

Corollary 3Let and be two irreducible subrepresentations of a unitary representation . Then one either has or .

*Proof:* Apply Schur’s lemma to the orthogonal projection from to .

Another application shows that finite-dimensional *linear* representations can be canonically identified (up to constants) with finite-dimensional unitary representations:

Corollary 4Let be a linear representation on a finite-dimensional space . Then there exists a Hermitian inner product on that makes this representation unitary. Furthermore, if is irreducible, then this inner product is unique up to constants.

*Proof:* To show existence of the Hermitian inner product that unitarises , take an arbitrary Hermitian inner product and then form the average

(this is the “Weyl averaging trick”, which crucially exploits compactness of ). Then one easily checks (using the fact that is finite dimensional and thus locally compact) that is also Hermitian, and that is unitary with respect to this inner product, as desired. (This part of the argument does not use finite dimensionality.)

To show uniqueness up to constants, assume that one has two such inner products , on , and apply Schur’s lemma to the identity map between the two Hilbert spaces and . (Here, finite dimensionality is used to establish

A third application of Schur’s lemma allows us to express the trace of a linear operator as an average:

Corollary 5Let be an irreducible unitary representation on a non-trivial finite-dimensional space , and let be a linear transformation. Thenwhere is the identity operator.

*Proof:* The right-hand side is equivariant, and hence by Schur’s lemma is a multiple of the identity. Taking traces, we see that the right-hand side also has the same trace as . The claim follows.

Let us now consider the irreducible subrepresentations of the left-regular representation . From Corollary 3, we know that those subrepresentations coming from different isomorphism classes in are orthogonal, so we now focus attention on those subrepresentations coming from a single class . Define the *-isotypic component* of the regular representation to be the finite-dimensional subspace of spanned by the functions of the form

where are arbitrary vectors in . This is clearly a left-invariant subspace of (in fact, it is bi-invariant, a point which we will return to later), and thus induces a subrepresentation of the left-regular representation. In fact, it captures precisely all the subrepresentations of the left-regular representation that are isomorphic to :

Proposition 6Let . Then every irreducible subrepresentation of the left-regular representation that is isomorphic to is a subrepresentation of . Conversely, is isomorphic to the direct sum of copies of . (In particular, has dimension ).

*Proof:* Let be a subrepresentation of the left-regular representation that is isomorphic to . Thus, we have an equivariant isometry whose image is ; it has an adjoint .

Let and . The convolution

can be re-arranged as

where

In particular, we see that for every . Letting be a sequence (or net) of approximations to the identity, we conclude that as well, and so , which is the first claim.

To prove the converse claim, write , and let be an orthonormal basis for . Observe that we may then decompose as the direct sum of the spaces

for . The claim follows.

From Corollary 3, the -isotypic components for are pairwise orthogonal, and so we can form the direct sum , which is an invariant subspace of that contains all the finite-dimensional irreducible subrepresentations (and hence also all the finite-dimensional representations, period). The essence of the Peter-Weyl theorem is then the assertion that this direct sum in fact occupies all of :

Proposition 7We have .

*Proof:* Suppose this is not the case. Taking orthogonal complements, we conclude that there exists a non-trivial which is orthogonal to all , and is in particular orthogonal to all finite-dimensional subrepresentations of .

Now let be an arbitrary self-adjoint kernel, thus for all . The convolution operator is then a self-adjoint Hilbert-Schmidt operator and is thus compact. (Here, we have crucially used the compactness of .) By the spectral theorem, the cokernel of this operator then splits as the direct sum of finite-dimensional eigenspaces. As is equivariant, all these eigenspaces are invariant, and thus orthogonal to ; thus must lie in the kernel of , and thus vanishes for all self-adjoint . Using a sequence (or net) of approximations to the identity, we conclude that vanishes also, a contradiction.

Theorem 1 follows by combining this proposition with 6.

** — 2. Nonabelian Fourier analysis — **

Given , the space of linear transformations from to is a finite-dimensional Hilbert space, with the Hilbert-Schmidt inner product ; it has a unitary action of as defined by . For any , the function can be easily seen to lie in , giving rise to a map . It is easy to see that this map is equivariant.

*Proof:* As and are finite-dimensional spaces with the same dimension , it suffices to show that this map is an isometry, thus we need to show that

for all . By bilinearity, we may reduce to the case when are rank one operators

for some , where is the dual vector to , and similarly for . Then we have

and

The latter expression can be rewritten as

Applying Fubini’s theorem, followed by Corollary 5, this simplifies to

which simplifies to , and the claim follows.

As a corollary of the above proposition, the orthogonal projection of a function to can be expressed as

We call

the *Fourier coefficient* of at , thus the projection of to is the function

which has an norm of . From the Peter-Weyl theorem we thus obtain the *Fourier inversion formula*

and the *Plancherel identity*

We can write these identities more compactly as an isomorphism

where the dilation of a Hilbert space is formed by using the inner product . This is an isomorphism not only of Hilbert spaces, but of the left-action of . Indeed, it is an isomorphism of the bi-action of on both the left and right of both and , defined by

and

It is easy to see that each of the are irreducible with respect to the action. Indeed, first observe from Proposition 8 that is surjective, and thus must span all of . Thus, any bi-invariant subspace of must also be invariant with respect to left and right multiplication by arbitrary elements of , and in particular by rank one operators; from this one easily sees that there are no non-trivial bi-invariant subspaces. Thus we can view the Peter-Weyl theorem as also describing the irreducible decomposition of into -irreducible components.

Remark 1In view of (1), it is natural to view as being the “spectrum” of , with each “frequency” occuring with “multiplicity” .

In the abelian case, any eigenspace of one unitary operator is automatically an invariant subspace of all other , which quickly implies (from the spectral theorem) that all irreducible finite-dimensional unitary representations must be one-dimensional, in which case we see that the above formulae collapse to the usual Fourier inversion and Plancherel theorems for compact abelian groups.

In the case of a finite group , we can take dimensions in (1) to obtain the identity

In the finite abelian case, we see in particular that and have the same cardinality.

Direct computation also shows other basic Fourier identities, such as the convolution identity

for , thus partially diagonalising convolution into multiplication of linear operators on finite-dimensional vector spaces . (Of course, one cannot expect complete diagonalisation in the non-abelian case, since convolution would then also be non-abelian, whereas diagonalised operators must always commute with each other.)

Call a function a class function if it is conjugation-invariant, thus for all . It is easy to see that this is equivalent to each of the Fourier coefficients also being conjugation-invariant: . By Lemma 5, this is in turn equivalent to being equal to a multiple of the identity:

where the character of the representation is given by the formula

The Plancherel identity then simplifies to

thus the form an orthonormal basis for the space of class functions. Analogously to (1), we have

(In particular, in the case of finite groups , has the same cardinality as the space of conjugacy classes of .)

Characters are a fundamentally important tool in analysing finite-dimensional representations of that are not necessarily irreducible; indeed, if decomposes into irreducibles as , then the character then similarly splits as

and so the multiplicities of each component in can be given by the formula

In particular, these multiplicities are unique: all decompositions of into irreducibles have the same multiplicities.

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Remark 2Representation theory becomes much more complicated once one leaves the compact case; convolution operators are no longer compact, and can now admit continuous spectrum in addition to pure point spectrum. Furthermore, even when one has pure point spectrum, the eigenspaces can now be infinite dimensional. Thus, one must now grapple with infinite-dimensional irreducible representations, as well as continuous combinations of representations that cannot be readily resolved into irreducible components. Nevertheless, in the important case oflocally compact groups, it is still the case that there are “enough” irreducible unitary representations to recover a significant portion of the above theory. The fundamental theorem here is theGelfand-Raikov theorem, which asserts that given any non-trivial group element in a locally compact group, there exists a irreducible unitary representation (possibly infinite-dimensional) on which acts non-trivially. Very roughly speaking, this theorem is first proven by observing that acts non-trivially on the regular representation, which (by the Gelfand-Naimark-Segal (GNS) construction) gives a state on the *-algebra of measures on that distinguishes the Dirac mass at from the Dirac mass from the origin. Applying the Krein-Milman theorem, one then finds anextremestate with this property; applying the GNS construction, one then obtains the desired irreducible representation.