In the last few notes, we have been steadily reducing the amount of regularity needed on a topological group in order to be able to show that it is in fact a Lie group, in the spirit of Hilbert’s fifth problem. Now, we will work on Hilbert’s fifth problem from the other end, starting with the minimal assumption of local compactness on a topological group , and seeing what kind of structures one can build using this assumption. (For simplicity we shall mostly confine our discussion to global groups rather than local groups for now.) In view of the preceding notes, we would like to see two types of structures emerge in particular:

*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.

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).

## 31 comments

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28 September, 2011 at 1:39 am

Marius BuligaIn the uniqueness part of theorem 3, exactly which Fubini or Tonelli type theorem are you using? (related also to the interesting exercise 5 and comments inside).

Later, related to the comment which starts with: “In the case when {G} is metrisable…”, question: are there non metrisable examples where theorem 3 is true?

Typo: in exercise 10, “(Counterexample) Let {G} be the {ax+b} group from Example 6(vii)”, should be “exercise 6(viii)”

28 September, 2011 at 8:36 am

Terence TaoThanks for the correction.

In the proof of uniqueness in the post, one is working with continuous functions of compact support, so (by local finiteness) one has absolute integrability and finite measure and so any version of the Fubini-Tonelli theorem will work here. The more serious use of sigma-compactness in the uniqueness argument given above lies in the use of the Riesz representation theorem. Without that theorem, one would have to work with more general measurable functions in the Fubini-Tonelli argument, at which point sigma-compactness (and hence sigma-finiteness) will become important.

Thanks to the Birkhoff-Kakutani theorem (that I will discuss in the next set of notes), a group is metrisable iff it is first countable (and hence also second-countable, in the sigma-compact case), which is more or less the minimum needed for the Arzela-Ascoli argument to work, so metrisability is basically the limit of that approach. Of course, one could modify Arzela-Ascoli by replacing sequences with nets and ultrafilters to avoid the dependence on countability, but then one is basically back to one of the other compactness methods outlined in the post.

28 September, 2011 at 10:25 pm

DiegoThank you for this wonderful post Terry. Regarding further properties of the Haar measure, I know of an article in which the doubling property for the Haar measure on an *abelian* LCG is proved. Do you know whether this property (meaning, the existence of C > 0 s.t.

\mu(B(x,2r)) \leq C \mu(B(x,r), for all x \in G, r >0) holds true in general?

If not, how about the growth condition: there exists N > 0 such that

\mu(B(x,r)) \leq C r^N ?

Also, how about the “weak homogeneity property” (i.e. there exists N such that for all x \in G and r > 0, N balls of radius r/2 are enough to cover B(x, r))? Does it always hold true for a general LCG? This property is proved for some special groups in Coifman and Weiss’ seminal work on spaces of homogeneous type.

29 September, 2011 at 7:07 am

Terence TaoI’m not sure what results you are referring to, because the metric on the locally compact group G is not specified. There are certainly metrics on LCA groups that are not doubling at small scales, for instance take the sum of the circles for , which has increasingly large dimension at small scales and so is not doubling. At large scales the situation is better because one can show (after passing to an open subgroup, at least) that there is a cocompact discrete abelian subgroup of bounded rank (see Proposition 11 of the above post) which can be used to construct at least one metric on an LCA group that is doubling at large scales.

In the nonabelian case, though, one only expects doubling or polynomial growth at large scales if the group is essentially nilpotent at these scales; in the case of discrete finitely generated groups with the word metric, this is Gromov’s theorem, and (as we shall see in later notes) there are analogues for continuous groups also.

29 September, 2011 at 3:05 pm

DiegoThanks Terry. In principle I meant any given translation-invariant distance on GxG, but after your answer I realize that’s too weak a condition.

30 September, 2011 at 3:42 am

Neil StricklandIn the compact case I like this characterisation: $I(f)$ is the unique constant function in the closure of the convex hull of the translates of $f$. I don’t know if there is any way to adapt that to the noncompact case, though.

30 September, 2011 at 8:11 am

Terence TaoAh, yes, thanks for mentioning that method; I’ve added a remark referencing it to the notes.

30 September, 2011 at 4:49 am

Marius BuligaThank you for the reply. Re: Neil Strickland, the technique resembles the one used in the calculus of variations to show that the quasiconvexification of an integral functional defined on , with integrand depending only on the gradient, is an integral functional too, with integrand depending only on the gradient. Interestingly, by Morrey, quasiconvexity of the integrand is equivalent with the lower semicontinuity of the integral, and continuity of the integral (on functions with compact support, say) is equivalent with the integral being constant.

4 October, 2011 at 12:58 pm

254A, Notes 4: Building metrics on groups, and the Gleason-Yamabe theorem « What’s new[…] is the existence of a left-invariant Haar measure on any locally compact group; see Theorem 3 from Notes 3. Finally, we will also need the compact case of the Gleason-Yamabe theorem (Theorem 8 from Notes […]

8 October, 2011 at 12:57 pm

254A, Notes 5: The structure of locally compact groups, and Hilbert’s fifth problem « What’s new[…] the full group ; in particular, by arguing as in the treatment of the compact case (Exercise 19 of Notes 3), we conclude that any connected locally compact Hausdorff group is the inverse limit of Lie […]

9 October, 2011 at 2:27 pm

pavel zorinDear Prof. Tao,

the hint to Exercise 19 is misleading: one should rather index the linear groups by finite subsets of a neighborhood base of the identity, otherwise there are no obvious natural homomorphisms between them.

best regards,

pavel

[Corrected, thanks – T.]28 October, 2011 at 1:25 pm

AnonymousHi,

In Exercise 6.vi, by a nilpotent Lie group do you mean a Lie group whose Lie algebra is nilpotent? These are not the same since e.g., the group in Exercise 6.viii is nilpotent as a group. Also, Exercise 6.viii has a random equation in the middle of a sentence.

[Corrected, thanks. For connected Lie groups, nilpotency of the group is equivalent to nilpotency of the Lie algebra; this is a consequence of the Baker-Campbell-Hausdorff formula.]28 October, 2011 at 1:28 pm

AnonymousOops, never mind, that group is solvable, not nilpotent.

6 November, 2011 at 9:23 am

254A, Notes 8: The microstructure of approximate groups « What’s new[…] measure (or Lebesgue measure) on is a bi-invariant Haar measure on . (Recall from Exercise 6 of Notes 3 that connected nilpotent Lie groups are […]

6 November, 2011 at 5:46 pm

alingalatanIn lemma 10, shouldn’t be generated by S (and not K) ?

[Corrected, thanks – T.]6 December, 2011 at 10:39 am

254B, Notes 2: Cayley graphs and Kazhdan’s property (T) « What’s new[…] a proof that any locally compact group has a Haar measure, unique up to scalar multiplication), see this previous blog post of mine. Example 5 (Quasiregular representation) If is a measure space that acts on in a transitive […]

1 May, 2012 at 11:50 pm

FelixDear Prof. Tao,

I am very much interested in the question if (at least in the (sigma) compact case) the Haar measure is still unique (up to a constant factor), if one drops the regularity assumptions and just assumes it to be (left) translation invariant and locally finite.

Up to now I was unable to find any statement regarding that question in the literature. When I saw your exercise 5 above, I hoped that this could prove the uniqueness (because one cannot invoke the Riesz representation theorem without regularity), but I can only show that for every x in G there is some null set N_x (possibly depending upon x!) such that f equals its translation by x on the complement of N_x. Without further regularity assumptions I seem unable to show that there is one null set N which does the trick for all x in G.

Can you give me a hint or do you know whether the uniqueness still holds in the non-regular case?

Best regards,

Felix V.

3 May, 2012 at 6:31 pm

Terence TaoIf the group is sigma-compact and first-countable (hence metrisable), then all locally finite Borel measures are automatically Radon; see Exercise 12 of these notes of mine. Offhand, I don’t know what happens without the sigma-compact and first-countability hypotheses but suspect that some pathological counterexamples can arise in those cases (though maybe the translation invariance of the measure could prohibit this).

5 September, 2013 at 9:23 pm

Notes on simple groups of Lie type | What's new[…] of for any ). Indeed, letting be the compact form of , the Peter-Weyl theorem (as discussed in this previous blog post) we see that can be identified with a unitary Lie group (i.e. a real Lie subgroup of for some ); […]

24 August, 2014 at 4:16 am

Ignacio VillanuevaDear Prof. Tao,

is there any “bounded additive” version of the Haar measure?. That is, we consider a ($\sigma$-compact Hausdorff) locally compact group G. We know that there exists a unique left-invariant Radon measure \mu. If we now consider also bounded additive measures, is there another measure such measure \nu different from \mu that is also left-invariant? I guess this is essentially equivalent to prove the existence of a purely not countably additive measure which is left-invariant. Do you know if such objects exist?

Thank you very much,

Ignacio

24 August, 2014 at 7:57 am

Terence TaoIf the group G is amenable, then it has an invariant mean which will be a non-trivial finite left-invariant additive measure (I’m not sure what you mean by a bounded measure). If the group G is not amenable, then no such finite measure exists, but infinite measures would still exist. A trivial example would be to modify Haar measure by redefining the measure of any unbounded set to automatically be infinite.

26 August, 2014 at 3:56 am

IgnacioThank you very much for the very quick answer. Probably it is indeed amenability the relevant condition I am searching for. Thanks again.

25 September, 2014 at 5:42 am

Ilya TsindlekhtI think in the proof of uniqueness of Haar measure you need to show that behaves nicely as $epsilon$ goes to 0.

25 September, 2014 at 6:14 am

Ilya TsindlekhtSorry, I have been wrong.

11 November, 2014 at 8:13 am

AnonymousDear Prof. Tao,

do you know if a not connected locally compact group always can be inverse limit of second countable locally compact groups?

Thank you very much

Mauro

9 April, 2015 at 7:54 am

jcdiaz2015Dear Professor Tao,

I was glad to discover this nice post. I would just like to point out that Stroppel’s book ”Locally Compact Groups” contains an elementary proof of the uniqueness part of the Haar measure. It uses Urysohn’s Lemma and uniform convergence.

Best regards,

Juan.

25 July, 2015 at 8:58 pm

Lattices, the geometry of numbers, and adeles | the capacity to be alone[…] treatments of the (in my opinion, tedious) theory behind Haar measure readily available; I like this more algebraic treatment by Terry Tao more than some of the others, but you can look around if you’re interested in […]

31 December, 2017 at 5:55 am

VMTHi Terry,

– the first definition of convolution (little beyond (1)) seems inconsistent with the formula you use later on: they agree only when the group is unimodular

– in the proof of uniqueness of Haar measure, one may want to observe that stays far from 0 as , as one sees from the identity and Ex. 4 using a fixed nonnegative (nontrivial) f

– in the Hint for Ex. 7, I think you want the integrand .

31 December, 2017 at 10:37 am

Terence TaoThanks for the corrections! The lower bound on is true, but I don’t think it’s actually needed for the proof, as we don’t actually divide by this quantity.

1 January, 2018 at 2:30 pm

VMTAh yes, of course! (In my mind I was reaching the last estimate in a cumbersome way, rather than by direct substitution)

I would like to share a “variational” route for the last part of the proof of Theorem 6: is compact, being precompact and closed (as , and hence , are weakly compact and thus weakly closed). So there exists a vector maximising ; necessarily and . For any , the function

, defined for near 0, has a maximum at . Differentiation gives

for every , i.e. by self-adjointness . Hence at least one among and is an eigenvalue.

1 January, 2018 at 2:37 pm

VMT(I meant , i.e. …)