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 1 Let
be continuous, compactly supported functions which are Lipschitz continuous. Show that the convolution
using Lebesgue measure on
obeys the
-type commutator estimate
for 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 2 Let
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 3 Let
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
is left-invariant (resp. right-invariant) if one has
(resp.
) for all
and Borel measurable sets
. A left-invariant Haar measure is a non-zero Radon measure which is left-invariant; a right-invariant Haar measure is defined similarly. A bi-invariant Haar measure is 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 1 A 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 2 If
is a countable discrete group, then counting measure is a bi-invariant Haar measure.
Example 3 If
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 4 If
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 not claim 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 5 Obtain 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 6 State 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 1 An important special case of the Haar measure construction arises for compact groups
. 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 2 The 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 7 Let
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 be unimodular if
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 8 Let
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 equation
for 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
-space for 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 action of
on
.
Exercise 10 If
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 11 Let
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 12 Show 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 14 Show 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 decomposition
of
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 15 Establish (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 3 The 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 decomposition
of
-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 16 Show that the hypothesis that
is Hausdorff can be omitted from Theorem 8. (Hint: use Exercise 3.)
Exercise 17 Show 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 the inverse limit
to 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 19 Let
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 20 Show 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 21 Let
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 a character to 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 10 Let
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 11 Let
be a locally compact Hausdorff abelian group with a cocompact finitely generated subgroup. Then
has a cocompact discrete finitely generated subgroup.
To prove this proposition, we need the following lemma.
Lemma 12 Let
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 23 Show 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 25 Show 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 the divisibility property that 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 4 Despite 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
Comments feed for this article
28 September, 2011 at 1:39 am
Marius Buliga
In 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 Tao
Thanks 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
Diego
Thank 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 Tao
I’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
Diego
Thanks 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 Strickland
In 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 Tao
Ah, yes, thanks for mentioning that method; I’ve added a remark referencing it to the notes.
30 September, 2011 at 4:49 am
Marius Buliga
Thank 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 zorin
Dear 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
Anonymous
Hi,
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
Anonymous
Oops, 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
alingalatan
In 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
Felix
Dear 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 Tao
If 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 Villanueva
Dear 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 Tao
If 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
Ignacio
Thank 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 Tsindlekht
I 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 Tsindlekht
Sorry, I have been wrong.
11 November, 2014 at 8:13 am
Anonymous
Dear 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
jcdiaz2015
Dear 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
VMT
Hi Terry,
stays far from 0 as
, as one sees from the identity
and Ex. 4 using a fixed nonnegative (nontrivial) f
.
– 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
– in the Hint for Ex. 7, I think you want the integrand
31 December, 2017 at 10:37 am
Terence Tao
Thanks 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
VMT
Ah 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.
…)