Download TOPOLOGICAL GROUPS 1. Introduction Topological groups are

TOPOLOGICAL GROUPS
MAX PITZ
1. Introduction
Topological groups are objects that combine two separate structures—the structure of
a topological space and the algebraic structure of a group—linked by the requirement that
the group operations are continuous with respect to the underlying topology.
Many of the natural infinite groups one encounters in mathematics are in fact topological groups: The additive and multiplicative groups of fields Q, R and C, their subgroups
such as the multiplicative group S 1 of complex numbers of absolute value 1. Also, the
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invertible n × n matrices over R and C (equipped with the topology from Cn ) with composition form natural topological groups. In fact, this last example is a special case of
a large class of interesting examples: group of autohomeomorphisms of some topological
space.
In this course we introduce the relevant concepts of topological groups and develop
their basic theory.
Some initial questions one might ask are the following:
• Given a topological space X, can can we find a group structure ◦ on X making
(X, ◦) into a topological group?
• Given a group G, can we find a non-discrete T0 -topology on G making it into a
topological group?
A remark: There is a high chance that there are several typos and other mistakes in
this script. If you find one, please email me under [email protected].
2. Homogeneous spaces
A topological space X is called homogeneous if for all x, y ∈ X there is a homeo
morphism f : X → X such that f (x) = y. The spheres S n = x ∈ Rn+1 : ||x|| = 1 are
homogeneous, and so are R, Q and P = R \ Q. The topological product of homogeneous
spaces is also homogeneous. The unit interval I = [0, 1] is not homogeneous, and neither
is I n for n ∈ N. (why?). However, the Hilbert Cube H = I N is homogeneous—this is a
surprising result due to Keller. Thus, homogeneity can improve when taking products,
but can never be lost.
Recall that an action of a group G on a space X is called transitive if the action has
only one orbit, i.e. X = G·x for some x ∈ X; and it is called free if the mapping x 7→ g · · · x
has no fixed-points apart from g = e ∈ G being the identity. For example, every group
acts on itself by left- or right-multiplication, and this action is transitive and free.
Homogeneity can be phrased in terms of group actions as follows. For a topological
space X, denote by Aut(X) the group of self-homeomorphisms of X. Then the group
Aut(X) acts naturally on X, via (f, x) 7→ f (x) for f ∈ Aut(X) and x ∈ X. Then X is
homogeneous if and only if Aut(X) acts transitively on X. We also write [x]∼ for the
orbit of x under Aut(X).
Let us see a less trivial class of homogeneous spaces: all connected Hausdorff manifolds
are homogeneous.
Date: Summer term 2016.
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Lemma 2.1 (Pasting Lemma). Let f : X → Y be a function between two top. spaces.
(1) If U is an open cover of X, then f is continuous if and only if f U is continuous
for all U ∈ U .
(2) If F is a locally finite closed cover of X, then f is continuous if and only if f F
is continuous for all F ∈ F .
Proof. Exercise/revision.
A space X is called strongly locally homogeneous provided that every point x ∈ X has a
neighbourhood base of open sets U 3 x such that for all y ∈ U there exists an f ∈ Aut(X)
with f (x) = y and f (z) = z for all z ∈
/ U.
Lemma 2.2. Let E be a normed vector space. For every basic open U = Bε (x) ⊂ E and
y ∈ U there is an autohomeomorphism h : E → E such that h(x) = y and h E \ U = id.
In particular, E is strongly locally homogeneous.
Proof. It suffices to show the claim U = B1 (0) and x = 0. By pasting lemma, we only
need to find an autohomeomorphism h : U → U that leaves the boundary pointwise fixed.
Indeed, in this case that mappings h : U → U and id : E \ U → E \ U , each defined on a
closed subset of E, agree on the intersection of their domains U \ U .
As for the proof, the idea is that if f : U → E is a ‘nice’ homeomorphism, we might
be able to reduce homogeneity of U to the homogeneity of E as follows: First apply f ,
then translate f (0) to f (y) in E, and then apply f −1 . We then have to check that we can
make this approach work such that we fix the boundary of U .
Our candidate for such a map f (compare to the real case (−1, 1) → R) is the map
x
.
f : U → E, x 7→
1 − kxk
This is a continuous bijection between U and E with, with (continuous) inverse
y
g : E → U, y 7→
.
1 + kyk
Now, the translation will be Ty : E → E, x 7→ x + f (y).
We claim that the following function works.
(
gTy f (x) if x ∈ U,
hy : U → U , x 7→
x
if kxk = 1.
This map is a bijection with inverse h−1
y = h−y . It is also continuous on U . So it remains
to show that it is continuous at points of the boundary. So suppose xn → d ∈ ∂U .
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This entails
kxn k →1 (clear) and xn / kxn k → d (in say R , this means angles converge.
xn 1
Formally, d − kxn k ≤ kd − xn k + (1 − kxn k ) kxn k → 0).
Now, to establish continuity of hy at d we must show that kd − hy (xn )k → 0. Observe
f (x)+f (y)
(x)
x
and that kxk
= kff (x)k
(think: haven’t changes angles). Then
that hy (x) = 1+kf
(x)+f (y)k
xn + f (xn ) − hy (xn ) → 0,
kd − hy (xn )k ≤ d
−
kxn k
kf (xn )k
because the first term goes to 0 be the above, and the last term grows like kf (x1n )k → 0 as
kxn k → 1.
(Quick check: after expanding, the denominator of the last expression grows like
kf (xn )k (1 + kf (xn ) + f (y)k) ∼ kf (xn )k2 , whereas the numerator is bounded by
k(f (xn ) + f (xn ) kf (xn ) + f (y)k − f (xn ) kf (xn )k − f (y) kf (xn )k)k
≤ kf (xn )k (1 + kf (y)k) + kf (xn )k (kf (xn ) + f (y)k − kf (xn )k)
≤ kf (xn )k (1 + kf (y)k) + kf (xn )k (kf (y)k) ∼ kf (xn )k ,
where kf (xn ) + f (y)k−kf (xn )k ≤ kf (y)k is simply the triangle-inequality in disguise.)
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Lemma 2.3. Assume that X is strongly locally homogeneous. Then every orbit under the
action of Aut(X) is open. In particular, if X is connected, then X is homogeneous.
Proof. Let y ∈ [x]∼ . Then take U a neighbourhood of y witnessing that X is strongly
locally homogeneous. Then U ⊂ [x]∼ : indeed, if f (x) = y and g(y) = u then u ∈ [x]∼ .
But now the orbits form a partition of X into open equivalent classes. Hence every orbit
is also closed, being the complement of the remaining orbits. Hence, if X is connected,
then [x]∼ = X.
A topological manifold (without boundary) is a topological space X that has a basis
of open sets all homeomorphic to Rn .
Corollary 2.4. All connected Hausdorff manifolds are (strongly locally) homogeneous.
Proof. Let X be a connected Hausdorff manifold. To show that X is homogeneous, it
suffices to show, by Lemma 2.3, that X is strongly locally homogeneous.
Let x ∈ X be arbitrary. By assumption, there is a neighbourhood V of x and a
homeomorphism g : Rn → V with g(0) = x. Now, g restricts to a homeomorphism on
B1 (0) ⊂ Rn . Since B1 (0) is compact and X is Hausdorff, g(B1 (0)) is closed in X.
Next, we claim that U = g(B1 (0)) ⊂ V is as desired. Indeed, pick u ∈ U . By Lemma 2.2
there is an autohomeomorphism h of B1 (0) mapping 0 7→ g −1 (u) and fixing the boundary
of B1 (0) pointwise. Then g ◦ h ◦ g −1 : g(B1 (0)) → g(B1 (0)) is an autohomeomorphism
sending x to u and fixing U \ U pointwise. Now paste together with the identity on
X \ U . Both functions agree on their intersection, so this gives a continuous function by
the pasting lemma. The inverse is also continuous.
The following example shows that restricting ourselves to Hausdorff manifolds is necessary. (Where does the above proof break down?)
Example 2.5. There is a connected non-homogeneous non-Hausdorff manifold.
Construction. This is the standard example of a non-Hausdorff manifold. Consider the
quotient (R × {0, 1})/∼ induced by the relation (x, 0) ∼ (x, 1) for all x ∈ R \ {0}. This
space has precisely two orbits: {(0, 0), (0, 1)} and its complement.
In fact, the above techniques, especially Lemma 2.2, can be generalised to prove that
Rn is countable dense homogeneous (or cdh for short) for all n ∈ N, i.e. that for every
two countably dense sets A, B ⊂ Rn there is an autohomeomorphism f of Rn such that
f (A) = B. At this point, we simply state the corresponding theorem for R as an exercise.
We will come back to the general situation later in this course when investigating groups
of autohomeomorphisms in more detail.
Exercise 2.6. Prove that the real line is countable dense homogeneous.
Clearly, the notion of countable dense homogeneity requires that the space is separable,
i.e. has a countable dense subset. Countable dense homogeneity can be thought of as a
strengthening of homogeneity—however, we have to be careful: S 1 ⊕ S 2 is countable dense
homogeneous, but not homogeneous. However:
Lemma 2.7. Let X be a connected separable metrizable space. If X is cdh then it is
homogeneous.
Proof. As in Lemma 2.3 it suffices to show that every orbit [x]∼ is open in X.
Claim: [x]∼ is open if and only if it has non-empty interior. The direct implication is
trivially true. Conversely, suppose U ⊂ [x]∼ is a non-empty open subset, and pick y ∈ U .
Let z ∈ [x]∼ be arbitrary. Then there is an autohomeomorphism h of X with h(y) = z.
But then it is clear that z ∈ h(U ) ⊂ [x]∼ with h(U ) open. This proves the claim.
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So we may suppose for a contradiction some orbit [x]∼ has empty interior, i.e. that
X \ [x]∼ is dense in X. Then let A ⊂ X \ [x]∼ be countable, dense in X.1 Then there is an
autohomeomorphism h of X such that h(A) = A ∪ {x}. Pick a ∈ A such that h(a) = x.
But then a ∈ A ∩ [x]∼ = ∅, a contradiction.
3. Basic properties of topological groups
Product topology. Recall Q
that if {Xs : s ∈ S} is a (multi-)set of topological spaces, then
the Cartesian product X = s∈S Xs can be endowed it withQ
the Tchonoff product topology.
The product topology is generated by all sets of the form s∈S Us where Us ⊂ Xs is an
open subset of Xs for all s ∈ S, and further Us = Xs for all but finitely many s ∈ S. Let
πs : X → Xs , hxs : s ∈ Si 7→ xs
be the projection map from T
X onto the sth coordinate space Xs . The basic open subsets
of X can also be written as s∈F πs−1 (Us ) where F ⊂ S is a finite set of coordinates, and
Us ⊂ Xs open for all s ∈ F .
The most important properties of the product topology are the following:
• all projection maps πs : X → Xs are open, continuous maps, and further
• a function f : Y → X from some topological space Y into the product X is
continuous if and only if fs = πs ◦ f is continuous for all s ∈ S.
Definition 3.1. (Topological groups)
(1) A structure G = (G, τ, ◦, (·)−1 ) is called a topological group provided that (G, τ )
is a topological space, (G, ◦, (·)−1 ) is a group, and
• composition ◦ : G × G → G is continuous (where G × G carries the product
topology), and
• inversion (·)−1 : G → G is continuous.
(2) A topological group G acts topologically on a topological space X if the group
action G×X → X, sending (g, x) 7→ g ·x is continuous w.r.t. the product topology.
We usually use the multiplicative notation for our groups. For
of the
A, B ⊂ G subsets
group we write as usual AB = {ab : a ∈ A, b ∈ B} and A−1 = a−1 : a ∈ A . Instead of
{x}A we also simply write xA. A subset A ⊂ G is called symmetric if A = A−1 .
Condition (1) asserts that for every neighbourhood U 3 xy there are neighbourhoods
V and W of x and y respectively such that V W ⊂ U . Condition (2) asserts that for every
neighbourhood U of x−1 there is a neighbourhood V of x such that V −1 ⊂ U .
Lemma 3.2. Let G be a topological group. Then left- and right-translations (and hence
inner automorphisms), and inversion are autohomeomorphisms of G.
In particular, the underlying topological space of every topological group is homogeneous.
Proof. Let G be a topological group. For g ∈ G, we write tg : G → G for the lefttranslation x 7→ gx. This is of course a bijection with inverse t−1
= tg−1 . It remains to
g
show that any tg is continuous. First, consider the map
ιg : G ,→ G × G, x 7→ (x, g).
This map is a continuous map into the product G × G, because the individual coordinate
maps are continuous, being the identity and the constant function respectively. By assumption, also multiplication · : G × G → G is continuous. Hence, it follows that tg can
be written as the composition
(·) ◦ ιg : G ,→ G × G → G, x 7→ (x, g) 7→ xg,
which is continuous.
The other cases are similar.
1This argument uses the result that every subspace of a separable metrizable space is again separable
and that ‘being dense’ is upwards transitive.
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The fact the every topological group is homogeneous is a useful criterion to see that
spaces like the half line [0, ∞) or the closed unit interval [0, 1] cannot be turned into
a topological group. However, the converse is far from true, and we will need more
sophisticated methods to decide this question. For example, we remarked earlier that all
spheres S n are homogeneous topological spaces. However, only S 0 , S 1 and S 3 can be
turned into topological groups. Similarly, the Hilbert cube H = I N cannot be turned into
a topological group, because it has the fixed-point property: every autohomeomorphism
H → H has a fixed point, which clashes with any possibility to assign a continuous right
translation.
Definition 3.3. Let (X, τ ) be a topological space, and B, U ⊂ τ .
• B is a base of X if for every open V ⊂ X and x ∈ V there is B ∈ B such that
x ∈ B ⊂ V , i.e. if every open subset of X is the union of elements of B.
• U = {Us : s ∈ S} is a (open) neighbourhood base for x ∈ X if for every open
V ⊂ X there is U ∈ U such that x ∈ U ⊂ V .
Lemma 3.4. Let G be a topological group and U an open neighbourhood base at e. Then
both families {xU : x ∈ G, U ∈ U} and {U x : x ∈ G, U ∈ U} form a basis for the topology
of G.
Proof. Exercise.
Lemma 3.5. In a topological group G, the identity element e ∈ G has a neighbourhood
base consisting of symmetric sets.
Proof. Suppose U is an open set containing e. Then U −1 is also an open neighbourhood of
e by continuity, so V = U ∩U −1 is open. Then V is clearly symmetric, and e ∈ V ⊂ U . Lemma 3.6. If V is a symmetric neighbourhood of the identity, then V ⊂ V 2 .
Proof. Let x ∈ V . We have to show that x ∈ V 2 . Since xV is a neighbourhood of x,
it follows that xV ∩ V 6= ∅. So there are v1 , v2 ∈ V such that xv1 = v2 . But then
x = v2 v1−1 ∈ V V −1 = V 2 by symmetry. Hence, V ⊂ V 2 .
Theorem 3.7. For a topological group G the following are equivalent.
(1) G is T0 ,
(2) G is T1 ,
(3) G is T2 (Hausdorff ), and
(4) G is T3 (regular).
Proof. Since G is homogeneous (Lemma 3.2), it suffices to prove: if G is a T0 -topological
group, then it satisfies the axiom of regularity at e. So let U be an open neighbourhood
of e. For regularity, we have to find an open set V such that e ∈ V ⊂ V ⊂ U . But since
e · e = e, continuity of the composition and Lemma 3.5 imply that there is a symmetric
open neighbourhood V of e such that V 2 ⊂ U .
It follows from Lemma 3.6 that e ∈ V ⊂ V ⊂ V 2 ⊂ U as desired.
(T3 is usually defined as T1 plus the axiom of regularity. But T0 + regularity already
implies T2 , as is easily seen: let x 6= y ∈ G. By T0 , there is an open set U containing one
point and not the other. Say x ∈ U 63 y. By regularity, there is an open set V such that
x ∈ V ⊂ V ⊂ U . But then the open sets V and G \ V witness that G is Hausdorff.)
Exercise 3.8. Let G be a topological group and A and B subsets of G. Then
(1) if A is open and B arbitrary then AB and BA are open,
(2) if A and B are compact then AB and BA are compact,
(3) (A)(B) ⊂ AB,
(4) (A)−1 = A−1 , and
(5) xAy = xAy.
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Moreover, if G is a T0 -topological group, then we also have
(6) if ab = ba for all a ∈ A and b ∈ B, then ab = ba for all a ∈ A and b ∈ B.
Corollary 3.9. If H is an [abelian] subgroup / normal subgroup of a [T0 ] topological group
G then H is also an [abelian] subgroup / normal subgroup of G.
Proof. If H is a subgroup of G then H 2 ⊂ H and H −1 ⊂ H. Hence, by the previous
2
exercise, we also have H ⊂ H 2 ⊂ H and (H)−1 = H −1 ⊂ H. Hence, H is a subgroup.
The remaining assertions follow in similar manner from the previous exercise.
Lemma 3.10 (First Closure Lemma). Let G be a topological group, and let U be a neighbourhood basis at e. Then for any subset A ⊂ G we have
\
\
A=
AU =
AU .
U ∈U
U ∈U
Proof. Without loss of generality we may assume that U consists of symmetric sets. We
begin with the following observation: If V is a symmetric neighbourhood of e ∈ G, and
x, y ∈ G are arbitrary group elements, then we have the following symmetry:
x ∈ yV ⇔ y ∈ xV.
(compare to ε-balls in metric spaces, where we also have x ∈ Bε (y) ⇔ y ∈ Bε (x)). Indeed,
x ∈ yV means there is v ∈ V such that x = yv. But then y = xv −1 ∈ xV by symmetry of
V . And conversely. T
Step 1. Show A ⊂ U ∈U AU . So let x ∈ A, and consider V ∈ U. The fact that x ∈ A
means that every neighbourhood of x intersects A, so xV ∩ A 6= ∅. So there exists a ∈ A
such that a ∈ xV T
. By symmetry
T this means x ∈ aV . It follows that x ∈ AV for all V ∈ U .
Step 2. Show U ∈U AU ⊂ U ∈U AU . This is trivial.
T
Step 3. Show U ∈U AU ⊂ A. We will show the contrapositive. So assume x ∈
/ A.
Then there is open U 3 e such that xU ∩ A = ∅. Find a symmetric V ∈ U such that
x ∈ V ⊂ V 2 ⊂ U . We claim that xV ∩ AV = ∅ (and hence x ∈
/ AV ). Suppose otherwise.
Then there are v1 , v2 ∈ V and a ∈ A such that xv1 = av2 . But then
xv1 v2−1 = a ∈ xV 2 ∩ A ⊂ xU ∩ A = ∅,
a contradiction.
Lemma 3.11 (Second Closure Lemma). Let G be a topological group, A ⊂ G closed and
K ⊂ G compact. Then AK is closed.
Proof. Let x ∈ AK. Have to show that x ∈ AK. First note that A closed implies that both
A−1 are closed (as inversion is a homeomorphism) and so is A−1 x (since right translation
by x is a homeomorphism).
Now x ∈ AK means xU ∩ AK 6= ∅, and hence also A−1 xU ∩ K 6= ∅ for all U ∈ U . Then
n
o
C = K ∩ A−1 xU : U ∈ U
is a collection of closed subset of K with the finite intersection property, as
K ∩ A−1 xU1 ∩ · · · ∩ A−1 xUn ⊃ K ∩ A−1 x(U1 ∩ · · · ∩ Un ) 6= ∅.
By
T compactness of K, the collection C has non-empty intersection, so we may pick
k ∈ C ⊂ A−1 x = A−1 x by the First Closure Lemma and the remarks at the beginning
of the proof. So k = a−1 x for some a ∈ A, hence x = ak ∈ AK as desired.
The compactness assumption on K cannot be weakened to simply being closed, as the
next example shows.
Exercise 3.12. If A, B ⊂ G are closed then AB need not be closed. Indeed, consider
(R, +) with the Euclidean topology, and the two closed subsets Z and αZ for α irrational.
Their sum is a countable dense subset of R—so not closed.
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Corollary 3.13. Then following assertions hold in any topological group G with identity
element e.
T
(1) {e} = U is a closed normal subgroup of G, which is contained in every open
and in every closed set meeting {e},
(2) {e} = G if and only if G is indiscrete, and
(3) for every compact subset K of G, we have {e}K = K
(4) the factor group G/{e} (endowed with the quotient topology) is a Hausdorff topological group.
T
Proof. The fact that {e} = U is immediate from the First Closure Lemma. That {e} is
a normal subgroup of G follows from Corollary 3.9, because {e} itself is a normal subgroup
of G. Next, suppose that U ⊂ G is open such that U ∩ {e} =
6 ∅. Then e ∈ U , so U ∈ U
shows that {e} ⊂ U .
Further, suppose that A ⊂ G is a closed subset with A ∩ {e} 6= ∅. Then {e} 6⊂ G \ A,
which is open, so {e} ∩ (G \ A) = ∅ by the previous argument, implying that {e} ⊂ A.
For (2), suppose that {e} = G. If U ⊂ G is non-empty open, then (1) shows that
G = {e} ⊂ U .
For (3), we note that the set {e}K is closed by the Second Closure Lemma 3.11, so
K ⊂ {e}K. However, we also have {e}K ⊂ {e}K = K by Exercise 3.8
For (4), note that we saw in an homework exercise that G/N for N E G is a topological
group with respect to the quotient topology. And if N ⊂ G is closed then G/N is T1 . It
follows from Theorem 3.7 that G/N is automatically regular Hausdorff.
4. Quotients, open maps and the first isomorphism theorem
The quotient topology. In the following, we will be using quotient topologies for quotients of topological groups modulo subgroups. So let G be a topological group, and
H ≤ G a subgroup. The space of (left-)cosets is defined as G/H = {xH : x ∈ G}, and
π : G → G/H, x 7→ xH denotes the quotient map.
As a quotient of a topological space, G/H carries a natural topology, namely the
quotient topology. It is defined as follows: a subset V ⊂ G/H is declared to be open if
and only if π −1 (V ) ⊂ G is open.
Lemma 4.1. Let G be a topological group with subgroup H. Then V ⊂ G/H is open if
and only if there is U ⊂ G such that V = {xH : x ∈ U }.
Proof. For ‘⇒’, let U = π −1 (V ). By definition of the product topology, U ⊂ G is open,
and V = {xH : x ∈ U }. For ‘⇐’, let U ⊂ G open. Then
π(U ) = {xH : x ∈ U } = {xH : x ∈ U H}.
To see whether this is an open subset of G/H,Swe have to check whether its preimage
under π is open in G. But π −1 (π(U )) = U H = x∈H U x is open.
In particular, it follows that π : G → G/H is an open, surjective continuous map
(Warning: quotient maps of arbitrary topological quotients don’t need to be open maps–
but they are for coset spaces).
Let G be a topological group with closed subgroup H. We let G act on G/H 2 by
G × G/H → G/H, (g, xH) 7→ gxH.
This action is also called the natural action of G on G/H. This is a continuous and
transitive group action, as the following (commutative) diagram shows:
2This might not be a group, but it is certainly a topological space.
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G×G
(g, x) 7→ gx
G
π
id × π
G × G/H
(g, xH) 7→ gxH
G/H
Theorem 4.2. Let G be a topological group with closed subgroup H. The natural action of
G on G/H is continuous and transitive. In particular, G/H is a homogeneous topological
space.
Proof. The assertion follows from the fact that the above diagram commutes and the fact
that id × π is an open map. Indeed, to see that every preimage of an open set V ⊂ G/H
under the map (g, xH) 7→ gxH is open, it suffices to observe that this preimage equals
(id × π) ◦ (·)−1 ◦ π −1 (V ).
Lemma 4.3. If H ≤ G is a subgroup of a topological group G, then the quotient G/H is
Hausdorff iff H ⊂ G is closed.
Proof. Since the quotient map π : G → G/H is continuous, it follows that if G/H is
Hausdorff, then H = π −1 ({H}) is closed as preimage of closed H under the continuous π.
Conversely, suppose that H ⊂ G is closed. Since G/H is homogeneous by Theorem 4.2,
it suffices to separate the identity H = eH in G/H from elements yH where y ∈
/ H. By
Lemma 4.1, it suffices to find an identity neighbourhood V such that
(1)
V H ∩ V yH = ∅.
Since H is closed, and y ∈
/ H there is an open identity neighbourhood U such that
H ∩ U y = ∅. Then
H ∩ U yH = ∅.
For if h1 = uyh2 , then h1 h−1
2 = uy ∈ H ∩ U y = ∅, a contradiction.
Now pick a symmetric open identity neighbourhood V of e such that V 2 ⊂ U . We claim
that V satisfies (1). If not, then there are element vi ∈ V and hi ∈ H with v1 h1 = v2 yh2 ,
so by symmetry
h1 = v1−1 v2 yh2 ∈ H ∩ V 2 yH ⊂ H ∩ U yH = ∅,
a contradiction.
Definition 4.4. (Filter) Let X be a set and F ⊂ P(X) \ {∅} be a collection of non-empty
subsets of X. Consider the following properties that F might have:
(F1) for all A, B ∈ F there is C ∈ F such that C ⊂ A ∩ B,
(F2) whenever A ∈ F and B ⊂ X such that A ⊂ B then B ∈ F ,
(F3) for all A ⊂ X, either A ∈ F or X \ A ∈ F .
A (non-empty) collection F with (F1) is called filterbase, a collection with (F1) and
(F2) is a filter and a collection F with (F1)–(F3) is called an ultrafilter.
For example, P(x) = {Y ⊂ X : x ∈ Y } is a filter, the so-called principal filter at x.
In fact, it is easy to see that P(x) is an ultrafilter. If X is infinite, then the collection
{A ⊂ X : |X \ A| < ∞} is a filter, called the co-finite or the Fréchet-filter.
If X is a topological space, and x ∈ X, then the neighbourhood filter N (x) =
{U ⊂ X : x ∈ int(U )} is a filter, as is quickly checked. In a topological group G, we
denote by NG the filter of all identity neighbourhoods in G.
Given a filter F on a space X, and a map f : X → Y , then it is easy to check that
{f (F ) : F ∈ F} is a filterbase. We define
f (F) = {W ⊂ Y : ∃F ∈ F s.t. f (F ) ⊂ W }
to be the filter generated by the filterbase {f (F ) : F ∈ F}.
TOPOLOGICAL GROUPS
9
Definition 4.5. A morphism of topological groups f : G → H is a continuous map between
the underlying spaces which is a group homomorphism at the same time.
Theorem 4.6.
(a) A homomorphism between topological groups f : G → H is a morphism if and only if it is continuous at e.
(b) The following are equivalent:
(1) f is open,
(2) there is a basis U at e such that for each U ∈ U, f (U ) has non-empty interior,
(3) there is a basis U at e such that for each U ∈ U, f (U ) is an identity neighbourhood in H.
(c) Let G and H be topological groups with nhoodfilters NG and NH for the identity
elements. Then a morphism f : G → H is continuous and open iff f (NG ) = NH .
Proof. (a). The forwards direction is clear. Conversely, we have to show that f is continuous at every point of G. So let y ∈ G and f (y) = x. Using Ex. 3.4, consider a basic open
subset xU ⊂ H. We need to find an identity neighbourhood V such that f (yV ) ⊂ xU .
By assumption, V = f −1 (U ) contains e in its interior. Then yV contains y in its
interior, and
f (yV ) ⊂ xU.
Indeed, if yv ∈ yV then f (v) ∈ U and hence f (yv) = f (y)f (v) ∈ xU , using that f is a
homomorphism. Hence, f is continuous at every point.
(b). The implications (1) ⇒ (3) ⇒ (2) are clear, since f (e) = e by the homomorphism
property.
For (2) ⇒ (3), let U be an identity neighbourhood of eG such that eH is not yet in the
interior of f (U ). Using the halving property, find a symmetric identity neighbourhood V
such that V 2 ⊂ U . By assumption, f (V ) has non-empty interior. If eH is in the interior,
we are done. Otherwise, we know that there is v ∈ V such that f (v) ∈ int(f (V )). Consider
the identity neighbourhood v −1 V 3 eG . Then v −1 V ⊂ V 2 ⊂ U , and
eH = f (eG ) = f (v −1 v) = f (v)−1 f (v) ∈ int f (v)−1 f (V ) = int f (v −1 V ) .
And (3) ⇒ (1) is now easy. Let W ⊂ G open. We have to show f (W ) is open, or
equivalently, that for every w ∈ W , the image f (w) lies in the interior of f (W ). Using
Ex. 3.4, find an identity neighbourhood that such wV ⊂ W and eH ∈ int(f (V )). Since f
is a homomorphism, we have
f (w) ∈ f (w)f (V ) ⊂ f (W )
and since eH ∈ int(f (V )) it follows that f (w) lies in the interior of f (w)f (V ).
(c). If V ∈ NG , then by openness, f (V ) is a neighbourhood for eH , and hence f (V ) ∈
NH . This shows {f (V ) : V ∈ NG } ⊂ NH , and hence the same inclusion holds for the
generated filter, i.e. f (NG ) ⊆ NH .
Conversely, let W ∈ NH . To show that W lies in the filter generated by {f (V ) : V ∈ NG }
we have to find some V ∈ NG such that f (V ) ⊂ W . Buy V = f −1 (W ) ∈ NG by continuity,
and f (V ) ⊆ W as required. It follows that f (NG ) ⊇ NH .
Theorem 4.7 (First Isomorphism Theorem for Topological Groups). A morphism f : G →
H between topological groups G and H decomposes canonically as in the following commutative diagram
f
G
H
π
G/ ker f
ι
f¯
im f
10
MAX PITZ
where f¯: gN 7→ f (g) is a bijective morphism of topological groups.
Moreover, f¯ is an isomorphism of topological groups (so a homeomorphism + homomorphism) iff f : G → H is open onto its image.
Proof. Without the topological content, this is simply the statement of the First Isomorphism Theorem for Groups. Now, that f¯ is continuous follows is a direct consequence of
the definition of the quotient topology: If U ⊂ H is open, then f¯−1 (U ) ⊂ G/ ker f is open
(by definition) iff π −1 (f¯−1 (U )) = f −1 (U ) is open in G. But the latter statement holds
true since f was continuous.
We now want to understand when f¯ is a homeomorphism. Since f¯ is a morphism of
topological groups, we get an isomorphism iff f¯ is an open map.
Suppose first that f¯ is open. Since π is an open map, it follows that f¯ ◦ π is an open
map from G → im f . Since the diagram commutes, it follows that f : G → im f is open.
Conversely, assume that f is open onto its image. Then by commutativity, it follows
that f¯ ◦ π : G → im f is an open map. To see that f¯ is open, let V ⊂ G/ ker f be an open
subset. Then U = π −1 (V ) ⊂ G is open by continuity, and V = π(U ) by surjectivity of π.
So f¯(V ) = f¯ ◦ π(U ) is open.
Some remarks: As an important special case, note that if G is compact, and H is
Hausdorff, then the continuous bijection f¯ is automatically a homeomorphism.
What does “open onto its image” mean? Consider the embedding S 1 ,→ C× as a
topological subgroup. This embedding not an open map, as the circle is not open in C.
But the standard inclusion is open onto its image.
The first isomorphism theorem for topological groups raises the question under which
conditions a morphism between topological groups is open. This is definitely not always
the case: Let id : (Rd , +) → (R, +) be the identity from the additive group Rd of the reals
endowed with the discrete topology to the usual additive groups of the reals. Then id is
a bijective morphism of topological groups which is not an isomorphism.
However, in Functional Analysis, we have learned about the following theorem:
Theorem 4.8 (Open Mapping Theorem, Banach & Schauder). Every surjective continuous linear operator X → Y between Banach spaces X and Y is an open map.
So in some cases, we have something like “automatic openness”. Later in this course
we will see that such a behaviour of automatic openness also also holds for morphisms
between many (complete, metric) groups.
5. Connectedness
Recall that a topological space is connected if it cannot be written as the disjoint union
of two non-trivial open subsets. Equivalently, X is connected if the only closed-and-open
(clopen) subsets of X are ∅ and X itself.
Let X be a topological space and x ∈ X. The connected component C(x) of x in X is
the largest connected set in X containing x. Formally, one can write
[
C(x) =
{C : x ∈ C, C ⊂ X connected}.
Since the closure of a connected set is again connected, maximality implies that every
component is a closed subset of X.
By Q(x) we denote the quasi-component of X, namely the set
\
Q(x) =
{D : x ∈ D, D ⊂ X clopen in X}.
It is easy to see that C(x) ⊂ Q(x) always. Indeed, since C(x) is connected, it follows
that whenever C(x) ∩ D 6= ∅ for some clopen set D of X, then C(x) ⊂ D, as otherwise
D ∩ C(x) and (X \ D) ∩ C(x) would disconnect the connected set C(x).
In general, one has C(x) ( Q(x). But see Lemma 5.7 below for more. A space is
called totally disconnected if C(x) = {x} for all x ∈ X; in other words, a space is totally
TOPOLOGICAL GROUPS
11
disconnected if there are no non-trivial connected subspaces. Note that such a space is
then T1 . Further, a space X is zero-dimensional if X has a basis consisting of closed
an open subsets. Every zero-dimensional T1 -space is totally disconnected. We start with
investigating connectedness properties in topological groups.
Example 5.1. A subset X ⊂ R is zero-dimensional if and only if it does not contain a
non-trivial interval.
Lemma 5.2. Let G be a topological group and C = C(e) be the component of the identity.
Then C is a closed normal subgroup of G.
Proof. The set C −1 is homeomorphic to C, and therefore a connected set, containing e.
By maximality of C, we have C −1 ⊂ C. In particular, for any a ∈ C we also have a−1 ∈ C,
and hence aC 3 e is again connected containing e, and hence aC ⊂ C by maximality. Thus
C 2 ⊂ C, i.e. C is a subgroup. Also, e ∈ xCx−1 ⊂ C, so C is normal. Finally, C is closed,
just as every connected component is in a topological space.
In fact, by the same arguments it is easy to see that C is invariant under any automorphism ψ of G, as e ∈ ψ(C) witnesses that ψ(C) ⊂ C.
Exercise 5.3.
(1) A subgroup H of a topological group is open if and only if it contains a non-empty open subset.
(2) If H is an open subgroup of a topological group G then H is closed and C(e) ⊂ H.
(3) If G contains a connected subset with non-empty interior then C(e) is open and
G/C(e) is discrete.
(4) if G is a locally connected topological group and f : G → H is an open morphism of
topological groups, then the identity component C of G is mapped onto the identity
component of H.
(5) G/C(e) is a totally disconnected Hausdorff group.
Theorem 5.4. Let GSbe a topological group, C the component of e and U any
S neighbourhood of e. Then C ⊂ n∈N U n . In particular, if G is connected, then G = n∈N U n .
S
Proof. First, pick a symmetric neighbourhood e ∈ V ⊂ U . We show that W = n∈N V n
is a subgroup. It clearly contains e, and if x, y ∈ W then say x ∈ V k and y ∈ V l , so
xy −1 ∈ V k V −l = V k V l = V k+l ⊂ W . Hence W W −1 ⊂ W , i.e. W is indeed a subgroup.
Next, W contains a non-trivial open
S set, so it is open, and hence closed by Exercise 5.3.
Since C is connected, C ⊂ W ⊂ n∈N U n .
Theorem 5.5. Let G be a topological group, and let U be a compact clopen neighbourhood
of the identity e. Then U contains a subgroup H of G that is compact clopen.
Proof. The first step of the proof is to claim that there is a neighbourhood V 3 e such
that U V = U . For this, it suffices find, for the complement A = G \ U , a symmetric
identity neighbourhood V of e such that U ∩ AV = ∅. Because then also U V ∩ A = ∅,
as any element uv = a ∈ V U ∩ A gives rise to an element u = av −1 ∈ U ∩ AV = ∅,
contradiction. And U V ∩ A = ∅ means U V ⊂ U .
So suppose that for all identity neighbourhoods W that U ∩ AW 6= ∅. Then
U ∩ AW : W 3 e open
has the finite intersection property, so compactness gives
\
\
u∈
U ∩ AW ⊂ U ∩
AW = U ∩ A = U ∩ A = ∅
W
W
where the first inclusion follows from the property of the closure operator, and the middle
equality follows from the First Closure Lemma, and that both A and U are closed, a
contradiction. This establishes the claim.
12
MAX PITZ
S
In the second step, note that V n ⊂ U V n ⊂ U V n−1 ⊂ · · · ⊂ U . So set H = n V n ⊂ U .
As in the previous result, this is the desired clopen subgroup. H inherits its compactness
from U .
Before we come to the main result about connectedness in topological groups, we
need an important result from general topology, the so-called Šura-Bura Lemma. First,
a warning: whether a subset is clopen depends on the surrounding space: Let X =
[0, 1] ∪ [2, 3] ⊂ R. Then [0, 1] is clopen in X, but of course not clopen in R. What we can
say, however, is that since X ⊂ R is closed, every clopen subset of X must be closed in R.
Similarly, if we take an open subspace Y ⊂ R then every clopen subset of Y is open in R.
We will make use of these tricks in the following results.
Lemma 5.6.
T If X is a compact space, C a family of closed subsets of X,Tand U ⊂ X open
such that C ⊂ U , then there is a finite subcollection C 0 ⊂ C such that C 0 ⊂ U .
T
Proof. Since C ⊂ U it follows that {U } ∪ {X \ C : C ∈ C} is an open cover of X.
T By
compactness, it has a finite subcover of the form {U } ∪ {X \ C : C ∈ C 0 }. But then C 0 ⊂
U as desired.
Lemma 5.7 (Šura-Bura Lemma). In a compact Hausdorff space, components and quasicomponents agree, i.e. Q(x) = C(x) holds for all x ∈ X.
Proof. The inclusion C(x) ⊂ Q(x) is always true. Hence, it remains to show that Q(x) ⊂
C(x). This will follow from the fact that Q(x) is connected.
Suppose for a contradiction that Q(x) was disconnected. Then Q = A ⊕ B for nontrivial disjoint Q-closed subsets. Let’s say x ∈ A. Since Q ⊂ X is closed (as intersection
of closed subsets), it follows that A and B are disjoint closed subsets of X. Since compact
Hausdorff implies normal, there are disjoint open subsets U and V of X such that A ⊂ U
and B ⊂ V . Observe also that U ∩ V = ∅ = U ∩ V .
But now Q ⊂ U ∪ V , so by Lemma 5.6,T there is a finite collection {D0 , . . . , Dn } of
clopen subsets of X such that Q(x) ⊂ D = i≤n Di ⊂ U ∪ V .
Claim that U ∩ D is clopen. Indeed, both D and U are X-open, hence so is their
intersection. To see that U ∩ D is X-closed, we compute
U ∩ D ⊂ U ∩ D = U ∩ D = U ∩ ((U ∪ V ) ∩ D) = U ∩ D
since U ∩ V = ∅.
But now, x ∈ A ⊂ U ∩D is clopen, so Q(x) ⊂ U ∩D. Hence B = ∅, a contradiction.
Phrased differently, the Šura-Bura Lemma says that in a compact Hausdorff space,
between any distinct connected components C1 and C2 , there is a clopen set separating
them, i.e. a clopen D such that C1 ⊂ D and C2 ∩ D = ∅.
A natural weakening of compactness is local compactness. A Hausdorff space X is said
to be locally compact if for every open set U ⊂ X and every x ∈ U there is an open set
V such that x ∈ V ⊂ V ⊂ U and V is compact. For example, Euclidean spaces Rn are
not compact, but they are locally compact, as the neighbourhoods Bε (x) are (closed and
bounded, and hence) compact.
Example 5.8. There is a locally compact Hausdorff space X containing a point x such that
C(x) ( Q(x). Hence, the conclusion of the Šura-Bura Lemma can fail for non-compact
spaces.
Construction. Consider the compact
Hausdorff space X = [0, 1] × ({0} ∪ {1/n : n ∈ N}) ⊂
R2 , and consider Y = X \ ( 12 , 0) . Then Y is a locally compact space, is which
the point x = (0, 0) has component C(x) = [0, 12 ) × {0}, but quasicomponent Q(x) =
[0, 12 ) ∪ ( 21 , 1] × {0}.
Corollary 5.9. A totally disconnected, locally compact Hausdorff space is zero-dimensional.
TOPOLOGICAL GROUPS
13
Proof. Consider a point x ∈ U for U ⊂ X open. We need to find a clopen set D ⊂ X
such that x ∈ D ⊂ U .
First, by local compactness, find a neighbourhood V of x such that V is compact and
x ∈ V ⊂ V ⊂ U . Then V is totally disconnected, too, so in V , the Šura-Bura Lemma
gives us C(x) = Q(x) ⊂ V . By Lemma 5.6, there is a finite collection {D0 , . . . , Dn } of
T
clopen subsets of V such that x ∈ Q(x) ⊂ D = i≤n Di ⊂ V .
We claim that D is clopen in X. First, D ⊂ V is closed, which is in turn closed in X.
So D ⊂ X is closed. Next, D is an open subset of V , which is in turn open in X. So
D ⊂ X is open.
So the bottom line is that the Šura-Bura Lemma may fail for locally compact spaces,
but it holds again if our locally compact space is totally disconnected.
Theorem 5.10. Let G be a compact group and let U be a clopen neighbourhood of the
identity e. Then U contains an open and closed normal subgroup N of G. The group G/N
is finite and discrete.
Proof. By Theorem 5.5, U contains a compact, open and closed subgroup H. So our task
is to argue that we can shrink H further to get a normal subgroup. We define
\
N=
xHx−1 .
x∈G
Step 1: N is indeed a normal subgroup: Every conjugate xHx−1 is a subgroup of G, so
N , as an intersection of subgroups, is itself a subgroup of G. To see that N is normal,
consider any conjugation map fy : G → G, g 7→ ygy −1 . We need to show that fy (N ) ⊂ N .
Note that since f is injective, we have
\
\
fy (
xHx−1 ) =
fy (xHx−1 ).
x∈G
x∈G
And since
\
x∈G
fy (xHx−1 ) =
\
(yx)H(yx)−1 = N
x∈G
the result follows.
Step 2. We have forced N to be normal, but is it clopen? For this, it suffices to show
that N contains a non-empty open subset, by Exercise 5.3(1) and (2). For later use, we
will state this result as a lemma.
Lemma 5.11. Let G be a compact topological group and H ⊂ G an open identity neighbourhood. Then there is open V 3 e such that x−1 V x ⊂ U for all x ∈ G.
For any such neighbourhood V , we have V ⊂ N , so the lemma implies Step 2 as desired.
Proof of Lemma 5.11. Consider the continuous map α : G × G → G, (g, x) 7→ x−1 gx.
Then α({e}×G) = {e} ⊂ H, so by continuity, α−1 (H) is an open neighbourhood of {e}×G.
By definition of the product topology, for every g ∈ G there exists open neighbourhood
Vg 3 e and Wg 3 g such that
(e, g) ∈ Vg × Wg ⊂ α−1 (H).
By compactness of {e}×G, there is a finite subcover
T indexed by elements
S g1 , . . . gn covering
{e}×G. Define open subsets of G as follows: V = i≤n Vi , and W = i≤n Wi . We observe
that
V × G ⊂ V × W ⊂ α−1 (H).
But α(V × G) ⊂ H translates to x−1 V x ⊂ H for all x ∈ G.
Finally, since N is an open subgroup of G, also all co-sets of N in G are open subsets
of G. Thus, G/N is discrete (in particular: Hausdorff), and compact (it’s the image of
compact G under continuous quotient map). So it must be finite.
14
MAX PITZ
Theorem 5.12. In a totally disconnected group G that is locally compact [compact], every
identity neighbourhood contains a clopen compact [normal] subgroup.
Proof. Since components are closed, a totally disconnected group is automatically T1 , and
hence Hausdorff (Theorem 3.7). It follows from Corollary 5.9 that G is zero-dimensional.
So let U be an identity neighbourhood of e in G. Then there is a compact clopen subset
D ⊂ G such that e ∈ D ⊂ U . By Theorem 5.5, there is a compact clopen subgroup H of
G such that e ∈ H ⊂ D. This proves the first assertion of the theorem.
Now assume that G is compact. It suffices to show that the H from above contains a
clopen normal subgroup N . But this is the assertion of Theorem 5.10.
We express the assertion of Theorem 5.12 also as G has arbitrarily small clopen compact
[normal] subgroups.
The assumption of local compactness is necessary as demonstrated by the group (Q, +).
The rationals are zero-dimensional, but contain no bounded open subgroups. An interesting example where the theorem applies is the product group 2N (where 2 denotes the
group on two elements with the discrete topology).
Definition 5.13. Let J be a directed set, i.e. a set with a reflexive, transitive and antisymmetric relation ≤ such that every finite non-empty subset of J as an upper bound.
An inverse system of compact T0 -topological groups over J is a family of morphisms
{fkj : Gk → Gj : j ≤ k ∈ J}
between compact topological groups Gj (j ∈ J) such that
(1) fjj = idGj for all j ∈ J, and
(2) fkl ◦ fjk = fjl for all j ≤ k ≤ l ∈ J.
The maps fkj are also called bonding maps.
Theorem 5.14 (Existence of the inverse limit). Let I = {fjk : Gk → Gj : j ≤ k ∈ J} be
an inverse system of compact T0 -topological groups. Then
(
)
Y
lim I = g ∈
Gj : j ≤ k ∈ J ⇒ fkj (gk ) = gj
←
j∈J
is a closed compact (non-empty) subgroup of the product group P =
Q
j∈J
Gj .
Proof. Assume that j ≤ k ∈ J and define Gkj = {g ∈ P : fkj (gk ) = gj }. Since fkj is a
homomorphism between topological groups, the set Gkj is a subgroup of P : Indeed, if
g, h ∈ Gkj then gh−1 ∈ Gkj as
−1
−1
fkj ((gh−1 )k ) = fkl (gk h−1
= (gh−1 )l .
k ) = fkl (gk )fkl (hk ) = gl hl
And since fkj is continuous, and the graph of a continuous map into a Hausdorff space
is closed, it follows
T that Gkj is closed.
So lim← I = j≤k∈J Gkj is a closed subgroup of P . Since P is compact by Tychonoff’s
theorem, so is lim← I. It is non-empty, because it contains the identity of P .
An example: assume we have a sequence
f
f
f
f
1
2
3
4
G0 ←−
G1 ←−
G2 ←−
G3 ←−
···
Then we can obtain an inverse system by simply defining fji as the composition fi+1 ◦
fi+2 ◦ · · · ◦ fj . Then the inverse limit is simply given by
{hgn : n ∈ Ni : fn+1 (gn+1 ) = gn for all n ∈ N}.
Choose a natural number p and let Gn = Z(pn ) = Z/pn Z, and define
fn+1 : Gn+1 → Gn , z + pn+1 Z 7→ z + pn Z.
TOPOLOGICAL GROUPS
15
The inverse limit of the system
f
f
f
f
1
2
3
4
Z(p0 ) ←−
Z(p1 ) ←−
Z(p2 ) ←−
Z(p3 ) ←−
···
is called the group Zp of p-adic integers. Picture to be added in lectures. Zp is a
compact totally disconnected Hausdorff group with a basis of identity neighbourhoods
{pn Zp : n ∈ N}. The embedding Z ,→ Zp , z 7→ (z + pn Z)n∈N is an injective homomorphism, and its image is dense in Zp . Elements are close to zero if they are divisible by
large powers of p.
Theorem 5.15. Assume that G is a compact Hausdorff group such that the collection
N of compact normal subgroups forms a neighbourhood basis for the identity of G (this
happens in Theorem 5.12). For M ⊃ N ∈ N let
fN M : G/N → G/M, gN 7→ gM
denote the natural (quotient) morphism. Then
I = {fN M : G/N → G/M : M ⊃ N ∈ N }
forms a inverse system of finite discrete topological groups such that G and lim← I are
isomorphic as topological groups.
Proof. N is directed be reverse inclusion: Whenever N1 , N2 ∈ N are compact normal
subgroups, then N1 ∩ N2 is an identity neighbourhood of e, so from our assumption that
N is a basis it follows that there exists M ∈ N such that M ⊂ N1 ∩ N2 .
We claim that
ϕ : G → lim I, g 7→ hgN : N ∈ N i
←
is an isomorphism of topological groups. We verify the claim in several steps.
Q
• By definition, we have ϕ(g) is an element of the product space N G/N . To see
that ϕ(g) is indeed an element of the inverse limit, we have to show that it is
compatible with all bonding maps fN M (where M ⊃ N ). But by definition:
fN M (ϕ(g)N ) = fN M (gN ) = gM = ϕ(g)M .
• Moreover, ϕ is a morphism, because its “coordinates” are morphisms.
• To see that ϕ is injective,
we consider the kernel. But if ϕ(g) = e then gN = N
T
for all N ∈ N so g ∈ N = {e}.
• To see that ϕ is surjective, assume that γ = hgN N : N ∈ N i is an element of the
inverse limit. Then
Γ = {gN N : N ∈ N }
is a collection of closed subsets of G with the finite intersection property. For if
−1
N ⊂ M then fN M (gN N ) = gM M means gM
gN ∈ M and hence gN ∈ gM M ∩
gN N . Now for any finite such collection gMi , by directedness, we can find a N
such that Mi ⊃ N for all i. So by compactness, there is an element
\
g∈
Γ.
And g ∈ gN N is equivalent to gN = gN N . Thus, ϕ(g) = γ.
• All in all, ϕ is a bijective continuous morphism between compact Hausdorff groups.
And compactness of G implies that ϕ is a homeomorphism.
Thus, we can think of the inverse limit as the group G approximated by factor groups
G/N of smaller and smaller subgroups. For example, we may think of the co-sets in G/N
as a tiling of G, where the quotient structure tells us roughly where element of G go
under composition. The smaller the subgroup, the finer the tiling, and the better the
approximation.
Corollary 5.16. Let G be a totally disconnected compact group. Then G is isomorphic
to an inverse limit of finite discrete topological groups.
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MAX PITZ
Proof. Theorems 5.12 and 5.15.
6. Metrizability
A topological space X is said to be completely regular if for every open set U and point
x ∈ U there is a continuous function f : X → [0, 1] such that f (x) = 1 and f (X \U ) ⊂ {0}.
A T1 space that is completely regular is usually called T3.5 .
In this section we show that every topological group is completely regular (and hence
that every T0 group is T3.5 ); and further, that first-countable T0 topological groups are in
fact metrizable by a left-invariant metric (a metric d : G × G → G is called left-invariant
if for all g, h ∈ G we have d(xg, xh) = d(g, h) for all x ∈ G. Right-invariance is defined
accordingly, and we say the metric is bi-invariant if it is both left- and right-invariant).
Recall that a pseudo-metric σ : G × G → [0, ∞) is a function that satisfies all properties
of a metric, apart from possibly the positive-definite requirement: σ(x, y) = 0 does not
necessarily imply that x = y. It is an easy exercise that for every (pseudo)metric σ, the
map σ 0 = min {σ, 1} is also a (pseudo)metric which generates the same topology. Thus, it
is never any true loss of generality to consider bounded (pseudo)metrics only.
Theorem 6.1. Let {Uk : k ∈ N} be a sequence of symmetric
T neighbourhoods of e in a
2
topological group G such that Uk ⊇ Uk+1
for all k. Let H = k∈N Uk .
Then there is a left-invariant pseudometric σ on G such that
(1)
(2)
(3)
(4)
σ : G × G → [0, 1] is continuous (w.r.t. product topology on G × G),
σ(x, y) = 0 iff y −1 x ∈ H,
σ(x, y) ≤ 2−k+2 whenever y −1 x ∈ Uk ,
σ(x, y) ≥ 2−k whenever y −1 x 6∈ Uk .
If in addition xUk x−1 for all x ∈ G and k ∈ N, then σ is also right-invariant and
σ(x−1 , y −1 ) = σ(x, y) for all x, y ∈ G.
Proof. The idea is to pretend that the given sets Uk form open balls of radius 2−k around
the identity e ∈ G. From this knowledge, how would we construct all ε-balls around e?
For example, the ball with radius 3/4 should be U1 U2 . This idea can be used to build Br
for all dyadic rationals r. And this is exactly what we will be doing.
Let us rename Uk = V2−k , and we will use either name, whatever is more convenient.
For a dyadic rational
r = 2−l1 + 2−l2 + · · · + 2−ln , 0 < l1 < · · · < ln ,
we define
Vr = Ul1 Ul2 · · · Uln .
Claim 1: For dyadic rationals r < s we have Vr ⊆ Vs . Indeed, for
s = 2−m1 + 2−m2 + · · · + 2−mp , 0 < m1 < · · · < mp ,
the fact that r < s implies that there for the first index k where mk 6= lk , for which we
have mk < lk . Define
W = Ul1 Ul2 · · · Ulk−1 = Um1 Um2 · · · Umk−1 ,
and note that
Umk ⊇ Ulk −1 .
TOPOLOGICAL GROUPS
17
Then it follows that
Vr
=
Ul1 Ul2 · · · Uln = W Ulk Ulk+1 · · · Uln
=
W Ulk Ulk +1 · · · Uln −1 Uln (fill in intermediate terms for missing subscripts)
⊆
W Ulk Ulk +1 · · · Uln −1 Ul2n (add an extra term Uln )
⊆
W Ulk Ulk +1 · · · Ul2n −1 (combine last two factors)
⊆
···
⊆
W Ulk −1 ⊂ W Umk ⊂ Vs .
Claim 2: For dyadic rationals r and l ≥ 1 we have Vr Ul = Vr V2−l ⊆ Vr+2−l+2 .
Let r be given in its dyadic representation as above. Define Vr = G for all dyadic rationals
r ≥ 1. If l > ln then Claim 2 is immediate from Claim 1. So suppose that l ≤ ln , then
lk−1 < l ≤ lk for some positive integer k.
Define r1 = 2−l+1 − 2−lk − 2−lk+1 − · · · − 2−ln > 0 and let r2 = r + r1 then r < r2 <
r + 2−l+1 . In defining r2 , we essentially considered the negative of a tail of r, so
r2 = 2−l1 + 2−l2 + · · · + 2−lk−1 + 2−(l−1) .
Hence, by Claim 1,
Vr Ul ⊆ Vr2 Ul = Vr2 +2−l ⊆ Vr+2−(l−1) +2−l ⊆ Vr+2−(l−2) .
Claim 3: If we define
ϕ : G → [0, 1], x 7→ inf {r : x ∈ Vr },
then
σ : G × G → [0, 1], σ(x, y) = sup {|ϕ(zx) − ϕ(zy)| : z ∈ G}.
gives rise to a pseudometric as desired.
We complete the proof in several steps.
Observation 1: The ∆-inquality of σ follows from the one for | · |. Indeed,
σ(x, w) = sup {|ϕ(zx) − ϕ(zw)| : z ∈ G}
≤ sup {|ϕ(zx) − ϕ(zy)| + |ϕ(zy) − ϕ(zw)| : z ∈ G}
≤ sup {|ϕ(zx) − ϕ(zy)| : z ∈ G} + sup {|ϕ(zy) − ϕ(zw)| : z ∈ G}
= σ(x, y) + σ(y, w).
Observation 2: The pseudo-metric σ is left-invariant. Indeed,
σ(gx, gy) = sup {|ϕ(zgx) − ϕ(zgy)| : z ∈ G}
= sup {|ϕ(zx) − ϕ(zy)| : z ∈ G} = σ(x, y).
Observation 3: Property (3) holds. Let u = y −1 x ∈ Uk . By left-invariance, we have
σ(x, y) = σ(u, e). We want to show that
σ(u, e) = sup {|ϕ(zu) − ϕ(z)| : z ∈ G} ≤ 2−k+2 .
So let z ∈ G be arbitrary. If z ∈ Vr then zu ∈ Vr Uk ⊂ Vr+2−k+2 . Hence ϕ(zu) ≤
ϕ(z) + 2−k+2 . If zu ∈ Vr then zuu−1 ∈ Vr Uk ⊂ Vr+2−k+2 . Hence ϕ(z) ≤ ϕ(zu) + 2−k+2 .
The claim follows.
Observation 4: Continuity. Let (x, y) and (x1 , y1 ) be elements of G × G. Using leftinvariance of σ and the ∆-inequality, we see
−1
−1
−1
|σ(x, y) − σ(x1 , y1 )| = σ(x−1
x1 ) − σ(y −1 x1 , y −1 y1 )
1 x, x1 y) − σ(x1 y, e) + σ(e, y
−1
−1
−1
≤ σ(x−1
x1 ) − σ(y −1 x1 , y −1 y1 )
1 x, x1 y) − σ(x1 y, e) + σ(e, y
−1
≤ σ(x−1
y1 ),
1 x, e) + σ(e, y
where the last line follows from the reverse ∆-inequality.
18
MAX PITZ
Together with observation 3, this shows that |σ(x, y) − σ(x1 , y1 )| ≤ 2 · 2−k+2 whenever
(x1 , y1 ) ∈ (xUk , yUk ). (since Uk is symmetric, x1 ∈ xUk implies x−1
1 x ∈ Uk ) Hence, σ is
continuous.
Observation 5: Property (4) holds. Indeed, assume y −1 x ∈
/ Ul . Then ϕ(y −1 x) ≥ 2−l
and hence
σ(x, y) = σ(y −1 x, e) ≥ sup · · · ≥ ϕ(y −1 x) ≥ 2−l .
Observation 6: Property (5) holds. Suppose that all Uk are invariant under inner
automorphisms. Then
xVr x−1 = xUl1 x−1 xUl2 x−1 · · · xUln x−1 = Vr .
Hence, ϕ(xyx−1 ) = ϕ(y) for all x, y ∈ G. As a consequence, using that ϕ(zxa) =
ϕ(a(zxa)a−1 ) = ϕ(azx) we get right-invariance as follows:
σ(xa, ya) = sup {|ϕ(zxa) − ϕ(zya)| : z ∈ G} = sup {|ϕ(azx) − ϕ(azy)| : z ∈ G} = σ(x, y).
And also, applying, in order, right-invariance, left-invariance and symmetry, we get
σ(x−1 , y −1 ) = σ(e, y −1 x) = σ(y, x) = σ(x, y).
Theorem 6.2 (Birkhoff-Kakutani). Let G be a T0 topological group. Then G is metrizable
if and only if G is first countable at the identity. In this case, the metric can be taken to
be left-invariant.
Proof. For the non-trivial direction, suppose that G has a countable open neighbourhood
base {Vn : n ∈ N} at e. Aiming to apply the previous theorem, it suffices to find a sequence
2
of symmetric neighbourhoods {Un : n ∈ N} of e with Un+1
⊂ Un ∩ Vn for all n ∈ N, so
that the corresponding H = {e}. We will build this family recursively. Set U0 = V0 , and
suppose that U0 , . . . , Un have been constructed as desired. We use Lemma 3.6 to pick a
2
symmetric neighbourhood Un+1 of e such that Un+1
⊂ Vn ∩ Un .
Now let σ be the left-invariant pseudometric
T for the family
T {Un : n ∈ N} as in Theorem 6.1. Since G is T1 , we then have H = n∈N Un ⊂ n∈N Vn = {e}. Hence, the
left-invariant pseudometric is in fact a metric, as σ(x, y) = 0 iff y −1 x ∈ H = {e} iff x = y.
It remains to show that σ generates the topology of G. This follows from Theorem 6.1(3)&(4), as
B2−k (e) ⊂ Uk ⊂ B2−k+2 (e).
Indeed, for the first inclusion, note that if x ∈
/ Uk then σ(x, e) ≥ 2−k by (4). And if x ∈ Uk
then σ(x, e) ≤ 2−k+2 by (3).
We remark that this theorem is far from true for general topological spaces: the Sorgenfrey line, or the (compact Hausdorff) Double Arrow Space are both first countable,
but not metrizable.
Exercise 6.3. Every topological group is completely regular.
Exercise 6.4. Prove that every T0 topological abelian group admits a continuous isomorphism into a product of metrizable abelian groups.
Theorem 6.5. A first countable compact T0 group G has a bi-invariant metric.
Proof. By Theorem 6.2, G admits a left-invariant metric σ compatible with its topology.
Define
d : G × G → [0, ∞), d(x, y) = sup {σ(xz, yz) : z ∈ G}.
Since G × G is compact and σ continuous, it follows that σ is bounded, and hence d is
well-defined for all x, y ∈ G.
Claim 1: d is a bi-invariant metric. We made d right invariant by brute force, and
it inherits being left-invariant from σ. The ∆-inequality follows as in the proof for σ.
Finally, d(x, y) ≥ σ(x, y) easily shows that d(x, y) = 0 iff σ(x, y) = 0 iff x = y.
Claim 2: The topologies defined by d and σ coincide.
TOPOLOGICAL GROUPS
19
Let ε > 0 be arbitrary. By Lemma 5.11, there is an identity neighbourhood Bδσ (e) such
that
z −1 Bδσ (e)z ⊂ Bεσ (e)
for all z ∈ G. The implies that
Bδσ (e) ⊂ Bεd (e),
σ
−1
since if g ∈ Bδ (e), then σ(z gz, e) < ε for all z ∈ G. It follows σ(gz, ez) < ε for all z ∈ G
by left-invariance, and hence d(g, e) = sup · · · < ε.
Together with the obvious inclusion Bεd (e) ⊂ Bεσ (e) (if x ∈ Bεd (e), then d(x, e) < ε, so
in particular σ(x, e) ≤ d(x, e) < ε, so x ∈ Bεσ (e)), this proves the claim.
7. Open mapping and closed graph theorems
Recall from the First Isomorphism Theorem for Topological Groups (Theorem 4.7) that
is f : G → H is a surjective open morphism of topological groups, then
G/ ker f ∼
=H
as topological groups. For this theorem, it is crucial that the map f is open; hence, we are
interested in finding good conditions which morphisms (=
ˆ continuous homomorphism) are
open. It turns out that in several cases, surjective morphisms between topological groups
are automatically open. Such results are commonly gathered under the heading Open
Mapping Theorems. The extra conditions usually involve assuming something like (local)
compactness or (local) completeness.
Recall that a metric space is complete if and only if every Cauchy sequence converges. A
topological space X is completely metrizable if for some metric d inducing the topology on
X, the space (X, d) is complete. So completeness is a property of metric spaces, whereas
being completely metrizable is a property of topological spaces.
Deciding whether a given metric space is complete is often not so hard. But there are
some surprises which topogical spaces are completely metrizable: For example, (0, 1) is not
complete in its usual metric, but it is homeomorphic to R, so completely metrizable. For
a more interesting example, it can be shown that the irrationals R \ Q are homeomorphic
to ZN with the product topology; and as a countable product of the completely metrizable
spaces Z, it follows that the irrationals are completely metrizable (exercise). The rationals,
however, are not completely metrizable—can you prove this?
Banach spaces are examples of complete topological groups.
Definition 7.1. (i) A filter F on a topological space X converges to a point x ∈ X (write
F → x) if N (x) ⊂ F , where N (x) denotes the neighbourhood filter of x, see Definition 4.4.
Equivalently, for each U 3 x open there is an F ∈ F such that F ⊂ U .
(ii) A filter F on a topological group G is a Cauchy filter if for each U ∈ N (e) there is
an F ∈ F such that F F −1 ⊂ U .
(iii) A topological group is complete if every Cauchy filter converges.
Exercise 7.2 (Equivalence of the two definitions of completeness in metric spaces). Let
X be a metric space. For a sequence S = {xn : n ∈ N} ⊂ X let F be the filter on X
generated by the Frechet filter on S (see Def 4.4).
(i) xn → x iff F → x.
(ii) Let G be a metric group endowed with a left-invariant metric. Show that S is a
Cauchy sequence iff F is a Cauchy filter.
(iii) Show that a left-invariant metric group is complete in the sense of metric spaces
iff it is complete in the sense of topological groups.
Theorem 7.3. Consider the following properties that a morphism f : G → H between
topological groups G and H might have:
(1) f is open,
(2) for all neighbourhoods U 3 eG we have that f (U ) has non-empty interior,
20
MAX PITZ
(3) for all neighbourhoods U 3 eG we have that f (U ) is a neighbourhood of eH ,
(4) for all open sets U ⊂ G there is an open setV ⊂ H
such that f (U ) ⊂ V ⊂ f (U ),
(5) for all open sets U ⊂ G we have f (U ) ⊂ int f (U ) .
Then (2) − (5) are equivalent and implied by (1). If G is a first countable complete T0
topological group and H is Hausdorff, then (1) − (5) are all equivalent.
Proof. Compare also to Theorem 4.6. Indeed, from there we know that openness of f
is equivalent to the condition that for all identity neighbourhoods U ⊂ G, f (U ) is a
neighbourhood of eH . Hence, (1) ⇒ (2).
For (2) ⇒ (3), let U ∈ N (e
a symmetric identity neighbourhood V such that
G ). Find
V 2 ⊂ U . By (2), find v ∈ int f (V ) . Then, using Exercise 3.8,
−1
⊂ f (V )f (V )−1 = f (V V −1 ) ⊂ f (U ).
eH = vv −1 ∈ int f (V ) int f (V )
For (3) ⇒ (4), let U ⊂ G open. We now want to apply the previous case pointwise. So
for every f (x) ∈ f (U ) we want to find an open neighbourhood Vx such that
f (x) ∈ Vx ⊂ f (U ).
S
Then clearly, V = x∈U Vx does the job.
But note that x−1 U is a neighbourhood of eG , and
therefore
by (3), we have that
f (x−1 U ) is a neighbourhood of eH . Put Vx = f (x)int f (x−1 U ) . This works, as
Vx ⊂ f (x)f (x−1 U ) ⊂ f (x) f (x−1 U ) ⊂ f (x)f (x−1 U ) = f (U )
where for the third inclusion we have used again Exercise 3.8).
(4) ⇒ (5) ⇒ (2) are trivial. Thus, it remains to show that under the additional
assumptions we have say (3) ⇒ (1). We will do this separately in the next theorem. Theorem 7.4. Suppose (3) above is satisfied, i.e. for every identity neighbourhood U 3 eG
there is open V 3 eH such that V ⊂ f (U ). Suppose further that G is first countable, T0
and complete, and H is Hausdorff.
Pick a left-invariant metric for G, and let Un = B1/4n (e) be the neighbourhood base
consisting of 1/4n -balls. Then Vn ⊂ f (Un−1 ) for all n ∈ N. In particular, f is open by
Theorem 4.6.
Proof. Suppose h ∈ Vn . We want to show that h ∈ f (Un−1 ).
The idea is as follows. We will construct a Cauchy sequence {xk : k ∈ N} ⊂ Un2 ⊂ G
such that f (xk ) → h. Since G is complete, there is x ∈ Un2 ⊂ Un−1 with x = lim xk , and
by Hausdorffness and continuity, we will have f (x) = h.
Step 1. The family {Vn : n ∈ N} is a countable open neighbourhood base for eH ∈ H.
Pick an arbitrary open W 3 eH . We need to find n ∈ N such that Vn ⊂ W . To do this,
use regularity of H (Theorem 3.7) to find W 0 3 eH open such that W 0 ⊂ W . Since f
is continuous at eG and {Uk : k ∈ N} is a neighbourhood base at eG , there is n ∈ N such
that f (Un ) ⊂ W 0 . Hence, it follows that
Vn ⊂ f (Un ) ⊂ W 0 ⊂ W
as desired.
By moving to a subsequence if necessary, we may assume that {Vn : n ∈ N} is nested,
i.e. Vn+1 ⊂ Vn .
Step 2. We are planning to find a Cauchy sequence hxk : k ∈ Ni ⊂ Un2 such that
f (xk ) → h. In order to achieve the last item, it suffices to arrange, in view of Step 1, for
f (xk ) ∈ hVn+k+1 for all k ∈ N.
TOPOLOGICAL GROUPS
21
Step 3. We recursively pick elements g0 , g1 , . . . in G with gk ∈ Un+k and define xk =
g0 g1 · · · gk such that f (xk ) ∈ hVn+k+1 (equivalently: h−1 f (xk ) ∈ Vn+k+1 ) for all k ∈ N.
By Step 4, xk is a Cauchy sequence converging to say x. By Step 2, f (x) = lim f (xk ) = h.
To do this, note first that since h ∈ f (Un ), we have
hVn+1 ∩ f (Un ) 6= ∅.
So there exists g0 ∈ Un such that f (g0 ) ∈ hVn+1 .
Next, suppose that we have found g0 , . . . gk with gi ∈ Un+i such that h−1 f (xk ) ∈
Vn+k+1 .
Since h−1 f (xk ) ∈ f (Un+k+1 ), we have
−1
Vn+k+1
h−1 f (xk ) ∩ f (Un+k+1 ) 6= ∅.
Find gk+1 ∈ Un+k+1 such that
−1
−1
f (gk+1
) ∈ Vn+k+1
h−1 f (xk )
−1
(note that by symmetry, Un+k+1
= Un+k+1 ). It follows
−1
−1
h−1
)f −1 (xk ) ∈ Vn+k+1
f (gk+1
and hence
f (xk+1 ) = f (xk )f (gk+1 ) ∈ hVn+k+1
as desired.
Step 4. xk ∈ Un2 for all k ∈ N. This follows from the fact that xk = g0 g1 · · · gk ∈
2
Un Un+1 · · · Un+k ⊂ Un Un+1 · · · Un+k
⊂ · · · ⊂ Un2 .
Step 5. Finally, it remains to argue that xk is indeed a Cauchy sequence. But since
d(e, gk ) < 1/4n+k+1 it easily follows that xk is Cauchy: By left-invariance, for k > m,
d(xk , xm ) = d(g0 · · · gk , g0 , · · · gm ) = d(gm+1 · · · gk , e)
(left-invariance)
≤ d(gm+1 · · · gn , gm+1 · · · gk−1 ) + · · · + d(gm+1 gm+2 , gm+1 ) + d(gm+1 , e)
(∆-inequality)
≤ d(gk , e) + d(gk−1 , e) + · · · + d(gm+1 , e)
(left-invariance)
m+n
≤ 1/4
.
This completes the proof.
A topological space X is called Baire space if the intersection of countably many dense
open subsets of X is again dense. Or, equivalently, if the union of any countable family
of closed subsets with empty interior in X has again empty interior. In particular, a
Baire space is inexhaustible, i.e.
S whenever we have a countable family {An : n ∈ N} of
closed subsets of X with X = An , then some int(An ) 6= ∅ (so if X is not the union of
countably many nowhere dense closed subsets).
Theorem 7.5 (The Baire Category Theorem). Every locally compact Hausdorff space
and every locally completely metrizable space is a Baire space.
Proof. Exercise.
S
For example, this implies now easily that Q is not completely metrizable: Q = q∈Q {q}
is a countable union of closed sets with empty interior, so not Baire.
A separable (=
ˆ having a countable dense subset) completely metrizable space X is
called Polish. Observe that a closed subspace of a topologically complete space is topologically complete as well. It can be shown that every Gδ -subset (a countable intersection
of open sets) of a Polish space is Polish. So P = R \ Q ∼
= ZN is a Polish group, and so is
N
the Cantor set C ∼
= 2 . Conversely, it can also be shown that every Polish subspace A of
a space X is a Gδ -subset of X.
22
MAX PITZ
Theorem 7.6 (Open Mapping Theorem I). Let f : G → H be a surjective morphism of
topological groups. If G is a complete, separable metric group, and H is inexhaustible and
Hausdorff, then f is open.
Proof. By Theorem 7.3(2) it suffices to show that for each identity neighbourhood U 3 eG ,
closure of the image f (U ) has non-empty interior in H. Let D be a countable dense subset
of G. Then G = D ⊂ DU be the First Closure Lemma. Thus, as f is surjective, we have
[
[
H = f (G) =
f (d)f (U ) ⊂
f (d)f (U ).
d∈D
d∈D
We have written H as the countable union of closed subsets. Since H is inexhaustible, it
follows that some f (d)f (U ) has non-empty interior, and hence so does f (U ).
Corollary 7.7. Every surjective morphism between complete separable metric groups is
open.
Proof. By the Baire Category Theorem.
Warning: In an earlier version of these notes it said that every surjective morphism
between Polish groups is open. However, the above proof does not show this stronger
claim: We know that every Polish group has a left-invariant metric by Birkhoff-Kakutani;
and that it has a complete metric by assumption. However, it is not clear why there should
be a metric which is both left-invariant and complete at the same time. And indeed, not
all Polish groups are complete groups in this stricter sense: e.g. Hom([0, 1]), the group of
all autohomeomorphisms of the unit interval under composition with the sup norm.3
A space X is called σ-compact if it is the countable union of compact subspaces.
Theorem 7.8 (Open Mapping Theorem II). Let f : G → H be a surjective morphism of
topological groups. If G is σ-compact, and H is inexhaustible and Hausdorff, then f is
open.
Proof. If G is σ-compact, then so is the quotient G/ ker f . Hence, without loss of generality, we may assume,
by the First Isomorphism Theorem 4.7, that f is bijective.
S
S
Write G = n∈N Kn as a union of compact subsets Kn ⊂ G. Then H = f (Kn ) is
a union of compact subsets, each of which is closed in H by Hausdorffness. Since H is
inexhaustible, there is n ∈ N such that f (Kn ) has non-empty interior, say V .
Then U = f −1 (V ) is an open subset contained in Kn (since f is bijective). We may
assume that e ∈ U , as otherwise we could pick g ∈ U and shift U by g −1 .
To see that f is open, we now verify that the conditions of Theorem 4.6 are satisfied, i.e.
that e ∈ G has a neighbourhood base of open sets whose images have non-empty interior
in H. Indeed, since f Kn : Kn → f (Kn ) is a continuous bijection from a compact space
to a Hausdorff space, it is a homeomorphism. In particular, for any open neighbourhood
W of e such that W ⊂ U , we have that f (W ) is open in f (Kn ), and contained in V ; so
f (W ) is open in V , which in turn is open in H, so f (W ) is open in H.
Since all such sets W form a neighbourhood bases at e in G, the result follows.
Corollary 7.9. Every surjective morphism between locally compact, σ-compact Hausdorff
groups is open.
Proof. By the Baire Category Theorem.
Theorem 7.10 (Open Mapping Theorem III, Banach & Schauder). Every surjective
continuous linear operator X → Y between Banach spaces X and Y is an open map.
Proof. Exercise.
3See also M. Malicki: On polish groups admitting a compatible complete left-invariant metric, J.
Symbolic Logic, 2011.
TOPOLOGICAL GROUPS
23
We now come to the closed graph theorems. Roughly speaking, these theorems provide
us with an often easy-to-check criterium when an algebraic homomorphism between topological groups is continuous. They are, in some sense, consequences of the open mapping
theorems above.
Let f : X → Y be a function between topological spaces. The graph of f is the subspace
Gr(f ) = {(x, f (x)) : x ∈ X} ⊂ X × Y.
If X and Y are groups, then Gr(f ) is a subgroup of X × Y if and only if f is a homomorphism.
Exercise 7.11. (1) Let f : X → Y be a continuous map from a topological space X into
a Hausdorff space Y . Show that G(f ) is a closed subset of the product X × Y .
(2) Now assume that Y is compact Hausdorff. Show conversely that if Gf ⊂ X × Y is
closed, then f is continuous.
The following lemma gives us the key to applying the open mapping theorems to
conclude something about continuity of maps.
Lemma 7.12. f : X → Y is continuous if and only if π1 : Gr(f ) → X is open.
Proof. Note that fˆ: X → Gr(f ), x 7→ (x, f (x)) and π1 are inverses of each other. And
f is continuous if and only if fˆ is continuous (recall that by the product topology, a map
into a product is continuous if and only if all its coordinate maps are continuous) if and
only if its inverse π1 is open.
In particular, if f is continuous, then fˆ is a homeomorphism between X ∼
= Gr(f ).
Theorem 7.13 (Closed graph theorem). For a homomorphism f : G → H of topological
groups, assume one of the following hypothesis:
(1) G and H are Hausdorff, locally compact and σ-compact;
(2) G and H are complete, separable metric groups;
(3) G and H are Banach spaces (and f linear).
Then f is continuous if and only if Gr(f ) is closed.
Proof. In all cases, we have that G × H, and hence Gr(f ), are Hausdorff spaces. So by
Exercise 7.11 we only have to prove the non-trivial direction that closed graph implies
continuity.
(1) Under these assumptions, Gr(f ) is a locally compact, σ-compact topological Hausdorff group, and π1 : Gr(f ) → G is a surjective morphism between locally compact groups
with σ-compact domain. Hence, by the Second Open Mapping Theorem 7.8, the projection π1 is open. So f is continuous by Lemma 7.12.
(2) If G and H are complete, separable metric groups, then π1 : Gr(f ) → G is a
surjective morphism of complete, separable metric groups, and hence open by the First
Open Mapping Theorem 7.6 and its corollary. So f is continuous by Lemma 7.12.
(3) If Gr(f ) is closed in the Banach space G × H, then it is itself a Banach space, so
we can apply the Third Open Mapping Theorem 7.10 to see that π1 : Gr(f ) → G is open.
So again, it follows that f is continuous by Lemma 7.12.
8. Haar measure
In this section we will see that every locally compact group G has a Haar measure: a
left-invariant, not identically vanishing measure on the Borel σ-algebra of G. We will give
a proof of this result in the metrizable case, which gives a good impression of the ideas
involved in constructing Haar measure.
Definition 8.1. Let G be a topological group. A measure µ on the Borel σ-algebra of G
is called a left-invariant Haar measure if
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MAX PITZ
• for all compact subsets K ⊂ G we have µ(G) < ∞,
• for all non-empty open subsets U ⊂ G we have µ(U ) > 0,
• for every Borel set B ⊂ G and g ∈ G we have µ(gB) = µ(B).
Note that the third property is well-defined: If B ⊂ G is Borel, then so is gB, because
left-translation by g is a homeomorphism. Haar measure is a generalisation of the Lebesgue
measure on the topological group (R, +). Indeed, recall that on Rn the Lebesgue measure
λ is the unique translation invariant measure such that λ([0, 1]n ) = 1.
Note also that what we call Haar measure should more precisely be called “left Haar
measure”. However, there is no crucial difference between the existence part of a leftor a right Haar measure, in the sense that if µ is a left Haar measure, then µ0 : E →
[0, ∞], B → µ(B −1 ) is a right Haar measure. We leave it as an exercise to check the
details.
Example 8.2. Consider G = GL(R, 2), the topological group of invertible 2 × 2 matrices,
regarded as an (open) subset of R4 . Specifically, a vector (x1 , x2 , x3 , x4 ) ∈ R4 corresponds
to the matrix
x1 x2
.
x3 x4
Then both left and right Haar measure are given by
Z
1
µ(U ) =
dX, dX = dx1 dx2 dx3 dx4 .
|
det
X|2
U
Solution. Let U ⊂ GL(R, 2) be a measurable subset and let B = bb13 bb24 ∈ GL(R, 2).
We have to show that our measure µ is invariant under left-translating by B, i.e. that
µ(BU ) = µ(U ).
For this, we first have to understand to which operation on R4 left-multiplication by a
matrix corresponds to. Write
ϕB : G → G, X 7→ BX.
Written out, this means
x1 x2
b1
7→
x3 x4
b3
b2
b4
x1
x3
x2
x4
=
b1 x1 + b2 x3
b3 x1 + b4 x3
b1 x2 + b2 x4
.
b3 x2 + b4 x4
On R4 , this corresponds to the map


 
x1
x1
 
x2 
4
4 x2 

ϕ̂B : R → R ,   7→ B̂  
,
x3
x3 
x4
x4
where

b1
0
B̂ = 
b3
0
0
b1
0
b3
b2
0
b4
0

0
b2 
.
0
b4
Observe that | det B̂| = | det B|2 : when exchanging the order of
we have to multiply the determinant by -1, so
b 1 b 2
b 1 0 b 2 0 b 1 0 b 2 0 0 b1 0 b2 b 3 0 b 4 0 = (−1)2 b3 b4
= −1 0
b 3 0 b 4 0 0 b1 0 b2 0
0
0 b3 0 b4 0 b3 0 b4 0
two rows or two colums,
0
0
b1
b3
0 0 b1
=
b2 b3
b4 2
b2 b4 TOPOLOGICAL GROUPS
25
Also, the linear mapping ϕ̂B : R4 → R4 has Jacobi matrix Dϕ̂B = B̂. Now observe that
by the substitution rule for multi-variable functions, we have
Z
Z
1
1
µ(BU ) = µ(ϕ̂(U )) =
dX
=
· | det Dϕ̂B |dX
2
2
ϕ̂(U ) | det X|
U | det (BX)|
Z
Z
1
1
2
=
·
|
det
B|
dX
=
dX = µ(U ).
2 | det X|2
|
det
B|
|
det
X|2
U
U
Hence, the integral is left-invariant under multiplication with B. The argument for right
multiplication by C is similar, where the corresponding Ĉ has automatically the form


c1 c2 0
0
c3 c4 0
0
.

0
0 c1 c2 
0
0 c3 c4
So in this example, the left- and right Haar measures agree, but this need not be the
case.
Exercise 8.3. Consider the group G of 2 × 2 non-singular matrices X ∈ GL(2, R) with
second row (0, 1), i.e. all matrices
x y
: x, y ∈ R, x 6= 0 .
0 1
Thus G can be identified with (the locally compact metrizable space) R2 \ {(x, y) : x = 0},
turning G into a topological group. Consider the measures
Z
Z
1
1
µ(U ) =
dX
=
dxdy,
2
2
|
det
X|
|x|
U
U
and
Z
Z
1
1
ν(U ) =
dX =
dxdy.
|
det
X|
|x|
U
U
Then µ is a left- and ν is a right Haar measure. So the left and right Haar measures of a
locally compact group can be different.
In the remainder of this chapter, we will see that every locally compact group always
has a Haar measure.
Theorem 8.4. Every locally compact Hausdorff topological group has a left-invariant
Haar measure which is unique up to scalar multiples.
We will give the existence proof in the case of locally compact metrizable topological
groups.
8.1. Recap. This section is for reference purposes.
Let X be a set. An algebra A ⊂ P(X) is a collection of subsets of X which contains
the empty set, is closed under taking complements and binary (⇒ finite) unions (⇒ finite
intersections). Sometimes, an algebra is also called ‘field of sets’.
A σ-algebra (or ‘σ-field’) Σ ⊂ P(X) is an algebra which is also closed under countable
unions (⇒ countable intersections).
Let X be a set, and Σ a σ-algebra on X. A function µ : Σ → [0, ∞] is a measure
provided µ(∅) = 0, and µ is σ-additive, i.e. for all countable collections {En : n ∈ N} ⊂ Σ
of pairwise disjoint sets we have
!
[
X
µ
En =
µ(En ).
n
n
One checks that every measure is monotone, i.e. E1 ⊂ E2 ∈ Σ implies µ(E1 )S≤ µ(E
2 ),
and
σ-subadditive,
i.e.
for
countable
families
{E
≤
E
n : n ∈ N} ⊂ Σ we have µ
n
n
P
µ(E
).
n
n
26
MAX PITZ
Let F be an arbitrary family of subsets of X. Then the intersection of all σ-algebras
containing F is the unique smallest σ-algebra which contains F . This σ-algebra is denoted
σ(F ) and is called the σ-algebra generated by F .
Let (X, τ ) be a topological space. The Borel σ-algebra on X is σ(τ ), the σ-algebra
generated by all open sets of X. It is the smallest σ-algebra containing all open and all
closed subsets of X.
8.2. Intuition. On the topological group (R, +), we already know that a Haar measure
exists, namely the usual Lebesgue measure λ. How could we describe the length of an
interval [a, b] without explicitly referencing subtraction in R?
Here is an idea. Consider the open set B0 = (− 12 , 12 ), and assume for a moment that we
know that this open set has measure 1. Now, all translates {xU : x ∈ [a, b]} form an open
cover of [a, b], so by compactness, finitely many of these translates already cover [a, b]. Let
n0 be the minimum number of translates that cover [a, b]. Then it is not hard to see that
n0 − 1 ≤ λ([a, b]) < n0 .
(Indeed, if the length λ([a, b]) lies in the interval [k, k + 1) for some integer k ∈ N then
precisely n0 = k + 1 many translates are needed.) For a better approximation, let n1 be
the minimum number of translates of U1 = (− 14 , 41 ) that cover [a, b]. Again, it can be
shown that
n1 − 1
n1
≤ λ([a, b]) <
.
2
2
We could continue in this fashion to get better an better approximations. This works,
but the problem is that it uses the measures of the sets Bn , which we somehow know in
advance. In general, we have to define λ for all sets, without any knowledge to start with.
To work around this, choose a “reference interval” Z ⊂ R and assign to it value 1. For
the Lebesgue measure, we usually take Z = [0, 1]. Now fix any nested neighbourhood base
{Bn : n ∈ N} of 0 ∈ R such that diam(Bn ) → 0. For every k ∈ N, let nk be the number of
translates of Bk needed to cover [a, b], and mk be the number of translates of Bk needed
nk
.
to cover Z = [0, 1]. Then, as the kth approximation, the measure of [a, b] is roughly m
k
And indeed, for (R, +) it can be shown that limk→∞ nk /mk = λ([a, b]). Had we chosen
a different reference interval, we would have gotten some positive scalar multiple of the
Lebesgue measure—but it would still be a left-invariant Haar measure.
We can extend bits of this approach to any metrizable locally compact topological
group. We can pick a decreasing neighbourhood base {Bn : n ∈ N } of the identity of
sets with compact closure (by local compactness) and decreasing diameter, and choose
as our reference any non-empty open set Z with compact closure. As before, for any
open set U with compact closure, we look at nk /mk , where nk and mk are the minimum
number of translates of Bk needed to cover U and Z respectively. We can show that
{nk /mk : k ∈ N} ⊂ R is a bounded sequence—but will it converge? Indeed, it might not.
To work around this, we have to generalise our notion of convergence.
8.3. Convergence along ultrafilters. Recall that a non-empty collection of subsets
V ⊂ P(N) \ {∅} is called an ultrafilter if
(F1) for all A, B ∈ V there is C ∈ V such that C ⊂ A ∩ B,
(F2) whenever A ∈ V and B ⊂ N such that A ⊂ B then B ∈ V,
(F3) for all A ⊂ N, either A ∈ V or N \ A ∈ V.
Moreover, F is called free if for all n ∈ N, we have {n} ∈
/ N (or equivalently, N \ {n} ∈ F).
A free ultrafilter contains all co-finite sets as it contains all complements of singletons and
is closed under finite intersection. Thus, any element of a free ultrafilter is infinite.
TOPOLOGICAL GROUPS
27
Lemma 8.5. Let X be a compact Hausdorff space and suppose hxn : n ∈ Ni is a sequence
of elements in X. Let V be a free ultrafilter on N. Then
\
{xn : n ∈ V }
V ∈V
consists of exactly one element x, which we denote by x = V-lim xn .
We also say the the sequence converges to x along the ultrafilter V, and write xn −
→
V
x. Note that in fact the above lemma holds also for non-free ultrafilters—the so-called
principal ultrafilters. Indeed, if P(k) is the principal ultrafilter at k ∈ N (see Def. 4.4)
then P(k)-lim xn is simply xk .
n
o
Proof. By (F 2), the collection {xn : n ∈ V } : V ∈ V has the finite intersection property.
By compactness, it has non-empty intersection.
It remains to show that the intersection consists of one element only. So suppose for
T
a contradiction that x 6= y ∈ V ∈V {xn : n ∈ V }. Using Hausdorffness, find disjoint open
neighbourhoods U 3 x and V 3 y.
Let A = {n ∈ N : xn ∈ U }. Then either A ∈ V or N \ A ∈ V. In the first case,
V ∩ {xn : n ∈ A} = ∅, so y ∈
/ {xn : n ∈ A}, a contradiction. In the second case, U ∩
{xn : n ∈ N \ A} = ∅, so x ∈
/ {xn : n ∈ N \ A}, a contradiction.
Here is an indication how convergence along ultrafilters compares to the usual convergence in analysis.
Exercise 8.6. Suppose hxn : n ∈ Ni is a bounded sequence of reals, and x ∈ R. Let cof
denote the co-finite filter on N.
(1) hxn : n ∈ Ni converges to x in the usual sense if and only if for every ε > 0 there
is A ∈ cof such that {xn : n ∈ A} ⊂ Bε (x), and
(2) hxn : n ∈ Ni converges to x in along some free ultrafilter V if and only if for every
ε > 0 there is V ∈ V such that {xn : n ∈ V } ⊂ Bε (x).
Consider the sequence h1, 0, 1, 0, 1, . . .i. This sequence does not converge in the usual
sense, but it does converge along a free ultrafilter V. Indeed, either the set of even numbers,
or the set of odd numbers belongs to V. So along the ultrafilter V, the sequence converges
either to 1 or 0.
Any sequence of reals converges along some ultrafilter in the extended real line [−∞, ∞].
If a sequence of real numbers hxn : n ∈ Ni is bounded (i.e. satisfies kxn k ≤ M for some
M for all n) then V-lim xn ∈ [−M, M ] as well.
Exercise 8.7 (Some properties of ultrafilter limits). Let V be a free ultrafilter on N,
xn , yn , x, y ∈ R.
(1) hxn : n ∈ Ni ≤ hyn : n ∈ Ni pointwise then V − lim xn ≤ V − lim yn ,
(2) Let hxn : n ∈ Ni ≤ hyn : n ∈ Ni be sequences of real numbers. Then V − lim xn ≤
V − lim yn if and only if {n ∈ N : xn ≤ yn } ∈ V.
(3) if xn −
→ x and yn −
→ y then xn yn −
→ xy and (xn + yn ) −
→ (x + y).
V
V
V
V
Theorem 8.8. There exists a free ultrafilter on N.
Proof. Consider all filters extending the co-finite filter. This family is partially ordered
by inclusion. It is easy to check that the conditions of Zorn’s lemma are satisfied (since
the filter axioms are finitary), so let V be a maximal (free) filter with respect to inclusion.
We have to show that V satisfies the complement property (F 3).
Towards this, let A ⊂ N be arbitrary. We show that N \ A ∈
/ V implies A ∈ V. Let
V ∈ V be arbitrary. If V ∩ A = ∅, then V ⊂ N \ A, and hence N \ A ∈ V by (F 2), a
contradiction. Thus, we have V ∩ A 6= ∅ for all V ∈ V. But then the filter V 0 generated
28
MAX PITZ
by V ∪ {V ∩ A : V ∈ V} (i.e. take all supersets, c.f. Def. 4.4) is a filter with V 0 ⊃ V. By
maximality, V 0 = V. But now we have A = A ∩ N ∈ V 0 = V as desired.
8.4. Existence of Haar measure—proof outline. In this section, let (G, d) be a
metrizable, locally compact topological group, and d a left-invariant metric (Theorem 6.2).
By
U0 = ∅ =
6 U ⊂ G open : U compact
we denote the collection of non-empty open sets with compact closure. By local compactness, there are many such sets in G; indeed, G has a basis of open sets with compact
closure.
Definition 8.9. For U, V ∈ U0 define VU ∈ N as the least number of left-translates of V
that cover U . Since V is open and U compact, this is well defined.
Now let B = Bn = B1/n (e) : n ∈ N be a neighbourhood basis of the identity e ∈ G,
given by the (1/n)-balls around e. Fix a non-empty reference set Z ∈ U0 , and a free
ultrafilter V on N.
Step I: Define maps λn : U0 → [0, ∞) (n ∈ N) and λ : U0 → [0, ∞) as follows:
U . Z
λn (U ) =
, and
Bn
Bn
λ(U ) = V − lim λn (U ).
Here, the limit for λ is taken along our chosen ultrafilter V, cf. Lemma 8.5. Note that we
need hλn (U ) : n ∈ Ni to be bounded for this to be well-defined, but this will turn out to
be true. The map λ is left-invariant.
Next, put U = U0 ∪ {∅, G} and extend λ to a function λ : U0 → [0, ∞] by defining
λ(∅) = 0 and, if G ∈
/ U0 (i.e. if G is non-compact), λ(G) = ∞.
Step II: We extend λ to all of P(G) in the sense of an “outer measure”. Indeed, every
subset E ⊂ G can be covered by a countable collection S of elements of U, as {G} is
always such a collection. As measure for E, we take the infimum value of all such covers
S. More precisely, define a map µ∗ : P(G) → [0, ∞] by
(
)
X
µ∗ (E) = inf
λ(S) : S ⊂ U a countable cover of E .
S∈S
It is this µ∗ restricted to the Borel σ-algebra of G that will be the Haar measure we are
looking for. Note that µ∗ (U ) ≤ λ(U ) for all U ∈ U; and indeed, µ∗ (U ) < λ(U ) can be
possible.
We will establish the following properties, showing that µ∗ is indeed an outer measure:
(II.1) µ∗ (∅) = 0,
(II.2) µ∗ is monotonically increasing,
(II.3) µ∗ is countably subadditive,
(II.4) µ∗ is left-invariant, and
(II.5) µ∗ is additive on sets E1 , E2 ⊂ G that are ‘far apart’ (if there exists some ε > 0
such that Bε (E1 ) ∩ Bε (E2 ) = ∅).
Step III: So we now have a map µ∗ defined on P(G) but we know from (R, +) that
we cannot hope for µ∗ to be countably additive on all of P(G). Hence, we have to find a
smaller σ-algebra A ⊂ P(G) such that µ = µ∗ A becomes countably additive, and such
that A contains all the Borel sets of G.
More precisely, we take A to consist of all subsets E ⊂ G that split all other subsets
A ⊂ G correctly w.r.t. µ∗ , i.e.
A = {E ⊂ G : ∀A ⊂ G we have µ∗ (A) = µ∗ (A ∩ E) + µ∗ (A \ E)}.
We will show that
(III.1) A is a σ-algebra,
TOPOLOGICAL GROUPS
29
(III.2) µ is countably additive on A, and
(III.3) A contains all open subsets of G, and hence is has the Borel σ-algebra of G as a
subcollection.
Step IV: Finally, we show that µ : A → [0, ∞] has also all the other properties we
required of Haar measure, namely
(IV.1) µ(C) < ∞ for all compact subsets C ⊂ G (compact subsets are closed, hence
Borel, so C ∈ A is well-defined),
(IV.2) µ(U ) > 0 for all non-empty open subsets U ⊂ G, and
(IV.3) µ(gB) = µ(B) for all Borel sets B ⊂ G.
8.5. Haar measure—Step I.
U W Claim (I.1). ∀U, V, W ∈ U0 : 1 ≤ VU ≤ W
· V .
U Proof. 1 ≤ V holds because U is non-empty, so at least one translate of V is required
U
to cover U . To prove the second inequality, let k = W
an ` = W
. Then there are
V
g1 , . . . , gk ∈ G and h1 , . . . , h` ∈ G such that
U ⊂ g1 W ∪ · · · ∪ gk W and W ⊂ h1 V ∪ · · · ∪ h` V.
We claim that {gi hj V : i ≤ k, j ≤ `} covers U . Let u ∈ U . The first cover gives us gi ∈ G
such that u ∈ gi W , to u = gi w for some w ∈ W . The second cover gives us hj ∈ G such
that w ∈ hj V , so there is v ∈ V such that w = hj v. But then u = gi hj v so u ∈ gi hj V as
claimed.
Z U Claim (I.2). For all n ∈ N and for all U ∈ U0 : 1 U ≤ λn (U ) ≤ Z .
h i h i
· BZn . This implies
Proof. By the previous claim, we have for all n ∈ N that BUn ≤ U
Z
U . Z
U
λn (U ) :=
≤
Bn
Bn
Z
which gives us the right inequality of the claim.
h i h i
Z
Similarly, the previous claim implies for all n ∈ N that BZn ≤ U
· BUn . This implies
Z
Z . U
≤
1 λn (U ) =
Bn
Bn
U
which gives us the left inequality of the claim.
Z Claim (I.3). For all U ∈ U0 : 1 U
≤ λ(U ) ≤ U
.
Z
Proof.
Z This
follows since hλn (U ) : n ∈ Ni is a sequence in the closed and bounded interval
1 U
, U
. Thus, every accumulation point of this sequence, and in particular the
Z
V-limit, lies also within that interval.
Claim (I.4). The extended function λ : U → [0, ∞] is sub-additive, i.e. for all U1 , U2 ∈ U
we have λ(U1 ∪ U2 ) ≤ λ(U1 ) + λ(U2 ).
Proof. If one of U1 and U2 is G or ∅, then the claim is trivially true. So we may assume
U1 , U2 ∈ U0 . First, it is easy to see that for all n ∈ N,
U1 ∪ U2
U1
U2
≤
+
.
Bn
Bn
Bn
(If one can cover U1 and U2 separately
by left-translates of Bn , then together one has
h i
Z
covered U1 ∪ U2 .) Dividing by Bn one both sides gives
λn (U1 ∪ U2 ) ≤ λn (U1 ) + λn (U2 ).
All that remains is to observe that the V-limit preserves such inequalities.
30
MAX PITZ
Claim (I.5). The map λ is additive for far-apart sets, i.e. if U1 , U2 ∈ U are far apart,
then λ(U1 ∪ U2 ) = λ(U1 ) + λ(U2 ).
Recall that ‘far apart’ means that there is ε > 0 such that Bε (U1 ) ∩ Bε (U2 ) = ∅.
Proof. Recall that ‘far apart’ means that there is ε > 0 such that Bε (U1 ) ∩ Bε (U2 ) = ∅.
First, observe that if 1/N < ε then no translate of Bn (for n ≥ N ) can intersect both
U1 and U2 . This holds, because our metric was left-invariant, so diam(Bn ) = diam(gBn )
for all g ∈ G. This observation implies
U1 ∪ U2
U1
U2
=
+
,
Bn
Bn
Bn
because “≤” is always true by sub-additivity, and “≥” follows by observing that it we
cover U1 ∪ U2 by a minimal number of left-translates g1hBn , i. . . gk Bn , then every gi Bn
intersects either U1 or U2 (provided n ≥ N ). Dividing by BZn one both sides gives
λn (U1 ∪ U2 ) ≤ λn (U1 ) + λn (U2 )
for all n ≥ N.
Now observe that the V-limit preserves this equality, as free ultrafilters only care about
co-finite sets.
Claim (I.6). The map λ is left-invariant.
h i h i
gU
Proof. The claim is trivial for G and ∅. So let U ∈ U0 , and observe that B
= BUn ,
n
as any minimal cover of Bn -translates of U can be further translated by g to give a cover
of gU , and conversely, any minimal cover of Bn -translates of gU can be further translated
by g −1 to give a cover of U .
This readily gives λn (gU ) = λn (U ) for all g ∈ G and n ∈ N, and hence λ(gU ) = λ(U )
for all g ∈ G.
8.6. Haar measure—Step II. Recall that we defined for all subsets E ⊂ G a map
(
)
X
∗
µ (E) = inf
λ(S) : S ⊂ U a countable cover of E .
S∈S
∗
Claim (II.1). µ (∅) = 0.
Proof. Observe that more generally, we have µ(U ) ≤ λ(U ) for all U ∈ U. In particular,
µ∗ (∅) ≤ λ(∅) = 0.
Claim (II.2). µ∗ is monotonically increasing.
Proof. If E1 ⊂ E2 then any cover S of E2 is also a cover of E1 . Thus, the infimum for E1 is
taken over a larger set of values that the infimum for E2 is. If follows µ∗ (E1 ) ≤ µ∗ (E2 ). Claim (II.3). µ∗ is countably sub-additive.
Proof. Recall that countable sub-additivity says that for any countable family {En : n ∈ N}
of subsets of G, we have
!
[
X ∗
∗
µ
En ≤
µ (En ).
n∈N
n∈N
Let ε > 0 be arbitrary. For every n ∈ N choose a cover Sn of En that gets ε/2n+1 -close
within the infimum for µ∗ (En ), i.e. such that
X
µ∗ (En ) ≤
λ(S) ≤ µ∗ (En ) + ε/2n+1 .
s∈Sn
TOPOLOGICAL GROUPS
Define S =
S
31
Sn . Clearly, S is an admissible countable cover of E =
X
X X
µ∗ (E) ≤
λ(S) ≤
λ(S)
S∈S
=
X
S
En , and we have
n∈N S∈Sn
∗
µ (En ) +
n∈N
n∈N
∗
As this holds for all ε > 0, we have µ (
S
ε
X
2n+1
En ) ≤
P
=
X
µ∗ (En ) + ε.
n∈N
∗
µ (En ) as required.
Claim (II.4). µ∗ is left-invariant.
Proof. To show µ∗ (E) = µ∗ (gE) for all g ∈ G, note that whenever S covers E then
gS = {gS : S ∈ S} covers gE, and conversely. Hence, the sets over which we take the
infima are the same, and hence so are their infima.
Claim (II.5). µ∗ is additive on sets E1 , E2 ⊂ G that are far apart.
Proof. Let E1 , E2 ⊂ G be far apart subsets of G. By sub-additivity of µ∗ (Claim II.3) we
only need to show that
µ∗ (E1 ∪ E2 ) ≥ µ∗ (E1 ) + µ∗ (E2 ).
Now if µ∗ (E1 ∪ E2 ) = ∞ there is nothing to do. So assume µ∗ (E1 ∪ E2 ) < ∞. Let ε > 0
arbitrary and find a countable cover S ⊂ U0 within ε of the infimum, i.e. such that
X
µ∗ (E1 ∪ E2 ) ≤
λ(S) ≤ µ∗ (E1 ∪ E2 ) + ε.
S∈S
Since E1 and E2 are far apart, there is some δ > 0 such that Bδ (E1 ) ∩ Bδ (E2 ) = ∅.
Consider V1 = Bδ/2 (E1 ) and V2 = Bδ/2 (E2 ). Then V1 and V2 are still far apart.
Now consider the open covers
S1 = {S ∩ V1 : S ∈ S} and S2 = {S ∩ V2 : S ∈ S}.
Note that S∩Vi is an open subset of G such that S ∩ Vi is compact (as a closed subset of the
compact S). Further, both Si are countable, and they cover Ei respectively. Hence, they
are admissible open covers for E1 and E2 , so their λ-value counts towards the infimum.
Further, note that for any S1 , S2 ∈ S the sets S1 ∩ V1 and S2 ∩ V2 are far apart (as subsets
of the far-apart sets V1 and V2 ). Hence, that far-apart property for λ (Claim I.5) gives us
X
X
µ∗ (E1 ∪ E2 ) ≤
λ(S) +
λ(S)
S∈S1
=
X
S∈S2
λ(S ∩ V1 ) + λ(S ∩ V2 )
S∈S
=
X
λ(S ∩ (V1 ∪ V2 ))
(far-apart property of λ)
λ(S)
(µ∗ monotonically increasing)
S∈S
≤
X
S∈S
≤ µ∗ (E1 ∪ E2 ) + ε.
And µ∗ (E1 ∪ E2 ) ≤ µ∗ (E1 ∪ E2 ) + ε for all ε > 0 gives µ∗ (E1 ∪ E2 ) ≤ µ∗ (E1 ∪ E2 ).
8.7. Haar measure—Step III. Recall that we put
A = {E ⊂ G : ∀A ⊂ G we have µ∗ (A) = µ∗ (A ∩ E) + µ∗ (A \ E)}.
We wanted to show that this is a σ-algebra which a) contains the Borel-σ-algebra of G
and on which b) the function µ∗ is countably additive.
32
MAX PITZ
Before we begin, we remark that since µ∗ is subadditive, the inequality µ∗ (A) =
µ ((A ∩ E) ∪ (A \ E) ≤ µ∗ (A ∩ E) + µ∗ (A \ E) is always true, and that we therefore
could have also defined
∗
A = {E ⊂ G : ∀A ⊂ G we have µ∗ (A) ≥ µ∗ (A ∩ E) + µ∗ (A \ E)}.
Claim (III.1). A is a σ-algebra.
Proof. A is closed under taking complements. If E ∈ A then for all A ⊂ G we have
µ∗ (A) = µ∗ (A ∩ E) + µ∗ (A \ E) = µ∗ (A \ (G \ E)) + µ∗ (A ∩ (G \ E)),
so G \ E ∈ A.
We have {∅, G} ⊂ A. Indeed, ∅ ∈ A as µ∗ (A) = µ∗ (A \ ∅) + µ∗ (A ∩ ∅), and the last
term is 0 by Claim II.1. But if ∅ ∈ A then so is its complement G.
A is closed under finite unions. Let E1 , E2 ∈ A. Need to show that E1 ∪ E2 ∈ A. Let
A ⊂ G be arbitrary. Then
µ∗ (A) = µ∗ (A ∩ E1 ) + µ∗ (A \ E1 )
(since E1 ∈ A)
= µ∗ (A ∩ E1 ) + [µ∗ ((A \ E1 ) ∩ E2 ) + µ∗ (A \ (E1 ∪ E2 ))]
(since E2 ∈ A)
= [µ∗ (A ∩ E1 ) + µ∗ ((A \ E1 ) ∩ E2 )] + µ∗ (A \ (E1 ∪ E2 ))
≥ µ∗ ((A ∩ E1 ) ∪ ((A \ E1 ) ∩ E2 )) + µ∗ (A \ (E1 ∪ E2 ))
(by subadditivity)
= µ∗ (A ∩ (E1 ∪ E2 )) + µ∗ (A \ (E1 ∪ E2 )).
Since A was arbitrary, it follows that E1 ∪ E2 ∈ A.
ASis closed under countable unions. Let {En : n ∈ N} ⊂ A. We need to show that
E = En ∈ A.
Since A is closed under finite unions and complements, we may assume without loss
of generality
pairwise disjoint. Indeed, consider E00 = E0 and
S that the {En : n0 ∈ N} are S
0
En = En \ i<n Ei . Then En ∈ A and En0 = E.
∗
Now let A ⊂ G be arbitrary. We need to show that µS
(A) ≥ µ∗ (A ∩ E) + µ∗ (A \ E).
Note that since A is closed under finite unions, we have i≤n Ei ∈ A for all n ∈ N, so
µ∗ (A) = µ∗ (A ∩
n
[
Ei ) + µ∗ (A \
i=0
≥ µ∗ (A ∩
n
[
n
[
Ei )
(
i=0
[
Ei ∈ A)
i≤n
Ei ) + µ∗ (A \ E)
(monotonicity)
i=0
=µ
∗
(A ∩
n
[
!
Ei ) ∩ E0
+µ
∗
(A ∩
= µ (A ∩ E0 ) + µ
!
Ei ) \ E0
+ µ∗ (A \ E)
(E0 ∈ A)
i=0
i=0
∗
n
[
∗
A∩
n
[
!
Ei
+ µ∗ (A \ E)
({Ei : i ≤ n} disjoint)
i=1
= ···
=
n
X
µ∗ (A ∩ Ei ) + µ∗ (A \ E).
i=0
As this inequality holds for all n ∈ N, we conclude
µ∗ (A) ≥
∞
X
µ∗ (A ∩ Ei ) + µ∗ (A \ E)
i=0
≥ µ∗ (
∞
[
A ∩ Ei ) + µ∗ (A \ E)
(countable sub-additivity, Claim II.3)
i=0
= µ∗ (A ∩ E) + µ∗ (A \ E).
TOPOLOGICAL GROUPS
33
Claim (III.2). µ∗ is σ-additive on A.
Proof. Let us first observe that µ∗ is finitely additive on A. Indeed, let E1 , E2 ∈ A be
disjoint. Choose A = E1 ∪ E2 . Since E1 ∈ A we get
µ∗ (E1 ∪ E2 ) = µ∗ (A) = µ∗ (A ∩ E1 ) + µ∗ (A \ E1 ) = µ∗ (E1 ) + µ∗ (E2 ),
as desired.
For σ-additivity, let again {En : n ∈ N} ⊂ A be a countable disjoint collection of sets
in A. From sub-additivity we know that
X ∗
µ∗ (E) ≤
µ (En ).
n∈N
For the reverse inequality, use
µ∗ (E) = µ∗ (E ∩
n
[
Sn
i=0
Ei ∈ A and take A = E to get
Ei ) + µ∗ (E \
i=0
≥ µ∗ (
n
[
n
[
Ei )
i=0
(µ∗ positive)
Ei )
i=0
=
n
X
µ∗ (Ei )
(finite additivity).
i=0
This holds for all n ∈ N, and hence we finally obtain
µ∗ (E) ≥
∞
X
µ∗ (Ei ).
i=0
So far we have shown that A is a σ-algebra and µ = µ∗ A is a measure on it. However,
so far we still don’t know whether A contains any interesting subsets of G apart from ∅
and G itself. The next claim shows that in fact, A contains the Borel σ-algebra of G.
Claim (III.3). U ⊂ G open implies U ∈ A.
Proof. Let U ⊂ G be a non-empty open subset. Fix A ⊂ G. We need to show that
µ∗ (A) ≥ µ∗ (A ∩ U ) + µ∗ (A \ U ).
For all n ∈ N define
An = x ∈ A ∩ U : B1/n (x) ⊂ U ⊂ A ∩ U.
S
Then An ⊂ An+1 and An = A∩U , as U is open. Now An ∪(A \ U ) ⊂ A, so monotonicity
gives us
µ∗ (A) ≥ µ∗ (An ∪ (A \ U )) = µ∗ (An ) + µ∗ (A \ U )
where the last equality holds because An and A\U are far apart (their δ-Balls for δ = 1/2n
are disjoint).
Next, we will show that µ∗ (An ) % µ∗ (A ∩ U ) as n → ∞. Then since the above
inequality
µ∗ (A) ≥ µ∗ (An ) + µ∗ (A \ U )
holds for all n ∈ N, this will imply
µ∗ (A) ≥ µ∗ (A ∩ U ) + µ∗ (A \ U ).
By monotonicity, hµ∗ (An ) : n ∈ Ni is an increasing sequence bounded by µ∗ (A ∩ U ).
To show that this bound if the least upper bound requires a tricky argument. Consider
the sequence of ‘onion rings’ B0 = A0 , and Bn = An \ An−1 for all n > 0. All elements in
this sequences are clearly disjoint; but successive elements might touch, so they are not
34
MAX PITZ
far apart. However, the distance between two elements that are not adjacent is strictly
positive: if n − 1 > m then for x ∈ Bn and y ∈ Bm we find for any z ∈ G \ U that
1
1
≤ d(x, z) <
.
n
n−1
Hence
d(y, z) ≤ d(y, x) + d(x, z)
implies that
1
1
−
> 0.
m
n−1
Since this lower bound holds uniformly for all x and y, we see that An+1 \An and Am+1 \Am
are indeed far apart.
In order to use additivity for
S far-apart sets, we now add up all even and all odd onion
rings separately. Since An = i≤n Bn , we have
!
n
[
∗
∗
(monotone)
µ (A2n ) ≥ µ
B2i
d(x, y) ≥ d(y, z) − d(x, z) >
i=0
=
n
X
µ∗ (B2i )
(far apart),
i=0
and similarly,
µ∗ (A2n+1 ) ≥
n
X
µ∗ (B2i+1 ).
i=0
Now if any of these two sums converge to ∞, then µ∗ (A ∩ U ) ≥ µ∗ (An ) → ∞ implies
that
µ∗ (A ∩ U ) = ∞ = lim µ∗ (An )
n→∞
as desired.
Otherwise, both sums converge to finite numbers, so their tail sums converge to 0. It
follows that
!
∞
[
∗
∗
Bi
µ (A ∩ U ) = µ A2n ∪
i=2n+1
≤ µ∗ (A2n ) +
∞
X
µ∗ (Bi )
(sub-additivity)
i=2n+1
= µ∗ (A2n ) +
∞
X
i=n
µ∗ (B2i+1 ) +
∞
X
µ∗ (B2i+2 )
(rearranging abs. conv. sum)
i=n
where the last two terms converge to 0 as n → ∞. It follows that
µ∗ (A ∩ U ) ≤ lim µ∗ (A2n ) = lim µ∗ (An ).
n→∞
n→∞
Hence, we have shown that τ ⊂ A and hence that the Borel σ-algebra of G is a subalgebra
of A.
8.8. Haar measure—Step IV. Finally, it remains to establish that our measure µ has
indeed the properties we want, namely that it assigns finite measure to every compact set,
and that it assigns positive measure to every non-empty open set.
Claim (IV.1). Let C ⊂ G be compact. Then µ(C) < ∞.
TOPOLOGICAL GROUPS
35
Proof. Since C ⊂ G is compact, finitely many open sets {U1 , . . . , Un } with compact closure
cover C. It follows by monotonicity that
µ(C) ≤
n
X
µ(U ) ≤
i=1
n
X
λ(U ) < ∞
i=1
as λ(·) is always finite (for Ui 6= G).
Claim (IV.2). Let U ⊂ G be non-empty open. Then µ(U ) > 0.
Proof. Let U ⊂ G be non-empty open. By local compactness, there is V ∈ U0 such that
V ⊂ U . For calculating µ(V ) is suffices to consider only finite covers S of V , as every
open cover has a finite subcover which gives a smaller λ-value. So let ε > 0 and find a
finite cover S ⊂ U0 of V such that
X
λ(Si ) ≤ µ(V ) + ε.
Then
[
λ(V ) ≤ λ( S)
X
λ(S)
≤
(monotonicity of λ)
(finite subadditivity, I.4)
S∈S
≤ µ(V ) + ε
≤ µ(U ) + ε
(monotonicity of µ).
Since this inequality holds for all ε > 0, it follows that µ(U ) ≥ λ(V ) > 0 (Claim I.3).
8.9. Uniqueness of Haar measure.
Exercise 8.10. Let G be a locally compact metrizable
groupRwith left Haar measure µ. Let
R
f : G → R be an integrable function. Then G f (x)dµ(x) = G f (yx)dµ(x) for all y ∈ G.
A Borel measure µ on G is called
• inner regular if µ(A) = sup {µ(K) : K ⊂ A compact} for all A measurable,
• outer regular if µ(A) = inf {µ(U ) : U ⊃ A open} for all A measurable, and
• regular of µ is both inner and outer regular.
Note that the Haar measure we constructed above was outer regular. We will see in
an exercise that under some mild assumption (like having a countable base), every Haar
measure is automatically regular.
The support of a continuous function f : G → R is the set f −1 (R \ {0}).
Theorem 8.11 (Fubini for locally compact spaces). Let X and Y be locally compact
Hausdorff spaces, let µ and ν be regular Borel measures on X and Y respectively, and let
f : X × Y → R be a continuous function with compact support. Then
Z Z
Z Z
f (x, y)dν(y)dµ(x) =
f (x, y)dµ(x)dν(y) < ∞.
X
Y
Proof. See e.g. [1, 7.6.4].
Y
X
The crucial point is that the standard Fubini theorem is usually proven only for σ-finite
measure spaces (i.e. spaces that are the countable union of subsets of finite measure), an
assumption that we don’t yet want to make for our topological groups.
Theorem 8.12 (Uniqueness of Haar measure). Let G be a locally compact metrizable
group, and let µ and ν be left-invariant regular Haar measures. Then there is a real
number α > 0 such that µ = α · ν.
36
MAX PITZ
Proof. Since G is locally compact and T3.5 , we can find an open U with compact closure,
a point x ∈ U , and a continuous function ϕ : G → [0, 1] such that ϕ(x) = 1 and ϕ(G\U ) ⊂
{0}. Then ϕ is a non-vanishing continuous function with compact support.
Since ϕ−1 ( 12 , 1]) is non-empty and open, it is assigned a positive measure by any Haar
measure. Hence,
Z
Z
ϕdµ > 0 and
ϕdν > 0.
By scaling the measures appropriately, we may assume that
Z
Z
ϕdµ = 1 = ϕdν.
Consider
Z
∆ : G → [0, ∞), g 7→
ϕ(xg −1 )dµ(x).
G
Claim: ∆ is a well-definedRcontinuous function. To see that ∆ is well-defined, we need
to argue that the integral ϕ(xg −1 )dµ(x) is finite for all g ∈ G. But if x 7→ ϕ(x) has
(compact!) support K, then fg : x 7→ ϕ(xg −1 ) has support
R Kg, which is also compact.
Hence, µ(Kg) < ∞.4 Since ϕ is bounded by 1, we have ϕ(xg −1 )dµ(x) ≤ µ(Kg), and
the result follows.
To see that ∆ is continuous, consider a convergent sequence gn → g ∈ G. We need
to show that ∆(gn ) → ∆(g). Let X = {g} ∪ {gn : n ∈ N}, a compact subset of G. By
S
Exercise 3.8(2), the set K̃ = KX = Kg ∪ n Kgn is compact. Then all functions fgn
are bounded by the integrable function 1K̃ , and fgn → fg pointwise (by continuity of ϕ).
Hence, by the dominated convergence theorem, we obtain
Z
Z
∆(gn ) = fgn (x)dµ(x) → fg (x)dµ(x) = ∆(g).
To complete the proof, we need the following Lemma.
Lemma 8.13. For all continuous f : G → R with compact support, the following statements hold with ρ either one of µ and ν:
R
R
(1) R f (x)dµ(x) = R f (x−1 )∆(x−1 )dρ(x), and
(2) f (x)dµ(x) = f (x)∆(x)∆(x−1 )dρ(x).
Proof. To prove (1),
Z
Z
f (x)dµ(x) =
Z
=
Z
=
Z
=
Z
=
Z
=
Z
=
let ρ be either one of µ and ν. Then
Z
Z
f (x)dµ(x) ϕ(y)dρ(y)
since
ϕ(y)dρ(y) = 1
Z
f (x)dµ(x) ϕ(y)dρ(y)
Z
f (y −1 x)dµ(x) ϕ(y)dρ(y)
by Exercise 8.10 with x 7→ y −1 x
Z
f (y −1 x)ϕ(y)dρ(y)dµ(x)
Fubini, 8.11
Z
f (y)−1 ϕ(xy)dρ(y)dµ(x)
by Exercise 8.10 with y 7→ xy
Z
f (y −1 )
ϕ(xy)dµ(x) dρ(y)
Fubini, 8.11
f (y −1 )∆(y −1 )dρ(y).
4Note though that µ(K) 6= µ(Kg) is possible, as µ is not necessarily right-invariant.
TOPOLOGICAL GROUPS
37
R
To see that (1) implies (2), consider f (x)∆(x)∆(x−1 )dρ(x) and apply (1) twice as
follows:
Z
Z
[f (x)∆(x)]∆(x−1 )dρ(x) =
f (x−1 )∆(x−1 ) dµ(x)
Z
= f (x)dµ(x).
This completes the proof of the lemma.
LetRus now see howR to put these results together to prove the theorem. From (2)%=µ , we
know f (x)dµ(x) = f (x)∆(x)∆(x−1 )dµ(x) for all continuous f with compact support,
implying ∆(x)∆(x−1 ) ≡ 1. So together with (2)%=ν we see that
Z
Z
Z
f (x)dµ(x) = f (x)∆(x)∆(x−1 )dν(x) = f (x)dν(x)
(†)
for all continuous f : G → R with compact support.
Claim: µ(K) = ν(K) for all compact K ⊂ G. For the proof, note that for every
x ∈ K there is open V such that V ⊂ B1/n (x) is compact. By compactness, there is a
finite subcover, whose union provides us with an open set Vn such that Vn is compact and
V n ⊂ B1/n (K). Note that K and G \ Vn are disjoint closed sets. Therefore, there is a
continuous function ϕn : G → [0, 1] such that ϕ(K) ⊂ {1} and ϕ(G \ Vn ) ⊂ {0}. In other
words,
we
T
T have continuous function ϕn with compact support such that ϕn K = 1. Since
Vn ⊂ B1/n (K) = K it follows that ϕn → 1K pointwise. Without loss of generality,
we may assume that ϕn are decreasing, and hence |ϕn (x)| ≤ ϕ1 (x), so we are bounded by
an integrable function. So by the dominated convergence theorem, it follows that
Z
Z
Z
Z
(†)
ϕn dν = 1K dν = ν(K).
ϕn dµ = lim
µ(K) = 1K dµ = lim
n→∞
n→∞
So µ and ν agree on all compact sets, and by inner regularity, it follows µ = ν.
9. The autohomeomorphism group of a topological space
9.1. The compact open topology. If X and Y are topological spaces, then by C(X, Y )
we denote the family of all continuous functions from X to Y . The compact open topology
on C(X, Y ) is defined as follows. For K ⊂ X and U ⊂ Y we set
[K, U ] = {f ∈ C(X, Y ) : f (K) ⊂ U }.
The collection
{[K, U ] : K ⊂ X compact, U ⊂ Y open}
serves as a subbase for a topology on C(X, Y ), the so-called compact-open topology on
C(X, Y ). The resulting topological spaces is also denoted by Ck (X, Y ).
It is instructive to compare this with the topology of pointwise convergence on C(X, Y ),
which is the subspace topology C(X, Y ) ⊂ Y X . Indeed, subbasic open subsets for the
product topology are of the form
[{x}, U ] = {f ∈ C(X, Y ) : f (x) ∈ U }.
So for the compact-open topology, we allow ourselves to specify more information in our
(sub)-basic open subsets; in particular, there are more open subsets under the compactopen topology then under the topology of pointwise convergence.
Exercise 9.1. Let X and Y be topological spaces.
(1) If S is a subbase for Y , then {[K, S] : K ⊂ X compact, S ∈ S} is a subbase for
Ck (X, Y ).
(2) Let F be a collection of compact subsets of X such that for every
S K ⊂V ⊂X
with K compact and V open, there is a finite G ⊂ F with A ⊂ G ⊂ U . Then
{[F, S] : F ∈ F , S ∈ S} is a subbase for Ck (X, Y ).
38
MAX PITZ
Exercise 9.2. Let X and Y be topological spaces. If B ⊂ Y is closed, then [A, B] ⊂
Ck (X, Y ) is closed.
Lemma 9.3. Let X and Y be topological spaces. If Y is T1 (T3 ) then Ck (X, Y ) is T1
(T3 ). Moreover, if X and Y are separable metrizable spaces, and X is locally compact,
then Ck (X, Y ) is separable metrizable.
Proof. For T1 , let f 6= g be distinct elements in Ck (X, Y ). Then there is x ∈ X such that
f (x) 6= g(x). Then [{x}, Y \ {f (x)}] is an open neighbourhood of g not containing f . So
Ck (X, Y ) is T1 .5
For T3 , let f ∈ [K, U ]. (Note that it suffices to show regularity for subbasic opens). By
continuity of f , the set f (K) is a compact subset of U , so by compactness and regularity,
there is an open W ⊂ Y such that f (K) ⊂ W ⊂ W ⊂ U . Then f ∈ [K, W ] ⊂ [K, W ] ⊂
[K, W ] ⊂ [K, U ] as desired (where [K, W ] is closed by Ex. 9.2).
To prove metrizability, it suffices to show by Urysohn’s metrization theorem, that in
addition to being T3 and T1 , Ck [X, Y ] has a countable base. Let A and B be countable
bases for X and Y (here we use separable metrizable), respectively, both closed under finite
unions, such that A is compact for all A ∈ A (here we use that X is locally compact). We
claim that
[A, B] : A ∈ A, B ∈ B
is a countable basis for Ck (X, Y ). It is clearly countable. To see that it is a basis, consider
f ∈ [K, U ]. Find B ∈ B such that f (K) ⊂ B ⊂ U (by compactness of f (K) and the fact
that B is closed under finite unions). Then find A ∈ A such that K ⊂ A ⊂ A ⊂ f −1 (B)
(by the same argument). Then f ∈ [A, B] ⊂ [K, U ].
We now consider composition of continuous functions. To be more precise, assume that
X, Y and Z are topological spaces. Consider the composition operator
T : C(X, Y ) × C(Y, Z) → C(X, Z), (f, g) 7→ g ◦ f.
Lemma 9.4. Let X, Y, Z be topological spaces such that Y is locally compact and regular.
Then T is continuous.
Proof. Let T (f, g) = g ◦f ∈ [K, U ] where K ⊂ X compact and U ⊂ Y open. By continuity
of f and g it follows that f (K) ⊂ Y is compact, and g −1 (U ) ⊂ Y is open. Moreover,
g ◦ f ∈ [K, U ] implies f (K) ⊂ g −1 (U ). Using that Y is locally compact and regular, find
an open V ⊂ Y with compact closure such that
f (K) ⊂ V ⊂ V ⊂ G−1 (U ).
It is now easy to verify that f ∈ [K, V ] and g ∈ [V , U ] and [K, V ] ◦ [V , U ] ⊂ [K, U ], so
composition is continuous.
9.2. Groups of homeomorphisms. We now pick up the thread from the beginning of
this script and consider groups of autohomeomophisms of topological spaces X. So let X
be a topological space. By Aut(X) we denote the subspace
{h ∈ C(X, X) : h is a homeomorphism}
as a subspace of Ck (X, X). As usual, we call Aut(X) the (auto-)homeomorphism group
of X. Recall that Aut(X) has a standard group structure, as for every f, g ∈ Aut(X),
also f ◦ g −1 ∈ Aut(X).
The group Aut(X) has the following useful subgroups: For A ⊂ X we denote by
Aut(X A) the subgroup
h ∈ Aut(X) : h X \ A = idX\A
of Aut(X).
5Alternatively, we could argue that the compact-open topology is finer than the product topology,
which we already know is T1 .
TOPOLOGICAL GROUPS
39
Theorem 9.5. Let X be a locally compact separable metrizable space.
(1) The composition function Aut(X) × Aut(X) → Aut(X) defined by (f, g) 7→ g ◦ f
is continuous.
(2) If X is compact, then inversion on Aut(X) is continuous.
(3) If every point of X has a compact connected neighbourhood, then Aut(X) =
Aut(ωX X) is a topological group.6
Hence, Aut(X) with the standard group operations becomes a topological group under
the compact-open topology, provided that X is compact, or X is locally compact and
locally connected.
Proof. (1) follows from Lemma 9.4. For (2), simply observe that f ∈ [K, U ] if and only
if f −1 ∈ [X \ U, X \ K] (this is, because x ∈ K ⇒ f (x) ∈ U is logically equivalent to
f (x) ∈
/U ⇒x∈
/ K).
The proof of (3) is more interesting (due to Arens, and van Mill in the presented form).
First note that Aut(ωX X) = {f ∈ Aut(ωX) : f (∞) = ∞}. Hence, the topology that
Aut(X) = Aut(ωX X) inherits from Aut(ωX) is generated by the subbasis
{[K, O] : K ⊂ X compact, O ⊂ X open} ∪ {[F, X \ K] : F ⊂ X closed, K ⊂ X compact}.
This is because the second type of neighbourhoods are the traces of functions that send
∞ 7→ ∞. Now clearly, the first type of neighbourhoods are open in Aut(X). Hence, the
topology inherited from Aut(ωX) is finer than the compact open topology on Aut(X).
To show that both topologies coincide, it therefore remains to show that open sets of
the second type [F, X \ K] for F ⊂ X closed and K ⊂ X compact are also open in the
compact-open topology of Aut(X).
For this, pick f ∈ [F, X \ K] arbitrary. Since f is a homeomorphisms, f −1 (K) is
compact. For all x ∈ X pick a compact connected neighbourhood Cx of x. Then
int(Cx ) : x ∈ f −1 (K)
is an open cover of the compact set f −1 (K), so there is a finite subset A ⊂ f −1 (K) such
that
[
f −1 (K) ⊂
int(Ca ).
a∈A
Then C =
S
0
a∈A Ca is compact. Similarly, find an even bigger compact set C such that
f −1 (K) ⊂ int(C) ⊂ C ⊂ int C 0 ⊂ C 0 .
Now consider the open subset
U = [C 0 ∩ F, X \ K] ∩ [∂C 0 , f (X \ C)] ∩
\
[{a}, f (int(Ca )].
a∈A
We claim that this set is as desired. First, we check that f ∈ U . First f (C 0 ∩ F ) ⊂
f (F ) ⊂ X \ K. Next, ∂C 0 ⊂ X \ C gives f (∂C 0 ) ⊂ f (X \ C). And lastly, a ∈ int(Ca )
implies f (a) ∈ f (int(Ca )) for all a ∈ A.
It remains to verify that U ⊂ [F, X \ K]. Suppose not. Then there is a homeomorphism
h ∈ U \[F, X \K]. And h ∈
/ [F, X \K] means that there is some x ∈ F such that h(x) ∈ K.
Where can this x lie? Since h ∈ U ⊂ [C 0 ∩ F, X \ K] we see that x ∈
/ C 0 . Also note that
x ∈ h−1 (K) ⊂ h−1 (f (C)),
−1
(since f (K) ⊂ C ⇒ K ⊂ f (C)) so x ∈ h−1 (f (Ca )) for some a ∈ A.
However, since h ∈ U ⊂ [{a}, f (int(Ca )], we also have a ∈ h−1 (f (Ca ). So now we are
in the situation where a ∈ C ⊂ C 0 and x ∈
/ C 0 are contained in the same continuum7
−1
−1
h (f (Ca ) (this is a continuum since h ◦ f is a homeomorphism). Hence, there is
y ∈ h−1 (f (Ca )) ∩ ∂C 0 6= ∅. Let us check where this y has to be mapped to.
6Here, ωX denotes the 1-point compactification of X.
7Continuum is shorthand for compact connected.
40
MAX PITZ
First, y ∈ h−1 (f (Ca )) ⊂ h−1 (f (C)) means that h(y) ∈ f (C). However, h ∈ U ⊂
[∂C 0 , f (X \ C)] means h(y) ∈ f (X \ C). This is a contradiction, since h(y) ∈ f (C) ∩ X \
f (C) = ∅, since f is a bijection.
We remark that Aut(X) need not be a topological group if X is just locally compact,
a counterexample is given by Aut(C × N). Therefore, in the following we will only deal
with groups Aut(X) for spaces separable metrizable spaces that are either compact, or
locally compact plus locally connected (such as X = Rn or X = N, where in the latter
case we have already seen in an exercise that this is a topological groups).
9.3. A metric for Aut(X). Let X be a compact metric space, and Y a separable metric
space with metric %. In the next theorem we will see that under these assumptions,
%̂(f, g) = sup {%(f (x), g(x)) : x ∈ X}
is an admissible metric on Ck (X, Y ). This metric is the metric of uniform convergence.
Indeed, recall that for f, fn : X → Y we say that fn converges to f uniformly if for all
ε > 0 there is a natural number N such that |fn (x) − f (x)| < ε for all x ∈ X and n ≥ N .
And indeed, it is easily checked that this happens if and only if fn ∈ Bε (f ) for all n ≥ N
where we now measure the distance with respect to %̂.
Theorem 9.6. Let X and (Y, %) be metric spaces, and X compact. Then %̂ is a metric
for Ck (X, Y ) (so compatible with the compact-open topology).
Proof. We leave the proof that %̂ is a well-defined metric as an exercise.
To show that %̂ induces the compact open topology, we have to show two directions.
(a) that for every function f inside a subbasic open subset f ∈ [K, U ] there is some ε > 0
such that f ∈TBε (f ) ⊂ [K, U ]. And (b), that whenever f ∈ Bε (f ), we can find a basic
open set U = i≤n [Ki , Ui ] such that f ∈ U ⊂ Bε (f ).
So for (a), let f ∈ [K, U ] be given. We claim that there is some ε > 0 such that
f (K) ⊂ Bε (f (K)) ⊂ U . Indeed, since f (K) is compact and U is open, this follows
as in the proof of Theorem 5.5: suppose for a contradiction that no ε works, i.e. that
B1/n (f (K)) ∩ X \ U 6= ∅. Then the collection
n
o
B1/n (f (K)) ∩ (X \ U ) : n ∈ N
of closed subsets of X \ U has the finite intersection property. Hence, by compactness of
T
X \ U , there is x ∈ (X \ U ) ∩ n∈N B1/n (f (K)). Now since the distance of x to f (K) is
0, it follows that x ∈ f (K) = f (K). This, however, contradicts x ∈ X \ U .
Now claim that f ∈ Bε (f ) ⊂ [K, U ]. Indeed, if g ∈ Bε (f ) and x ∈ K then g(x) ∈
Bε (f (K)) ⊂ U .
Conversely, for (b), let f ∈ Bε (f ) be given. Find a finite subcover of the open cover
−1
f (Bε/4 (y)) : y ∈ Y
S
of the compact space X. So there are {y1 , . . . , yn } ⊂ Y such that X = i≤n Ui where
Ui = f −1 (Bε/4 (yi )). Then Ui is compact, and f (Ui ) ⊂ f (Ui ) ⊂ Bε/4 (xi ) ⊂ Bε/3 (yi ) by
continuity of f . We claim that
\
U=
[Ui , Bε/3 (yi )]
i≤n
S
is as desired. By the above, we have f ∈ U . Now let g ∈ U . Since Ui covers X, and
f (Ui ) ∪ g(Ui ) ⊂ Bε/3 (yi ) it follows that for all x ∈ X, %(f (x), g(x)) < 2
, so
3
%̂(f, g) = sup {%(f (x), g(x)) : x ∈ X} ≤
as desired.
2
<ε
3
TOPOLOGICAL GROUPS
41
We now observe that under certain assumptions on the completeness of Y , also Ck (X, Y )
is complete.
Theorem 9.7. Let X, Y be separable metric spaces with X compact and % a complete
metric for Y . Then %̂ is complete.
Proof. Exercise.
Now we would like to say that for a compact metrizable space X, our metric %̂ on
Ck (X, X) is also a complete metric on Aut(X). However, while we have shown that the
limit of a Cauchy sequence of continuous functions under uniform convergence is still
continuous, we have no control over the fact whether the inverse is continuous (or whether
it even exists in the first place).
Lemma 9.8. Let (X, ρ) be compact metric. Then %̂ is a right-invariant admissible metric
for Aut(X). In addition, the function σ defined by
σ(f, g) = %̂(f, g) + %̂(f −1 , g −1 )
is a complete admissible metric for Aut(X) (but not necessarily left- or right-invariant).
Proof. Right-invariant. Let h ∈ Aut(G). We have to show that %̂(h · f, h · g) = %̂(f, g) for
all h ∈ Aut(X). And indeed,
%̂(f, g) = sup {%(f (x), g(x)) : x ∈ X}
= sup {%(f (h(x)), g(h(x))) : x ∈ X} = %̂(f ◦ h, g ◦ h)
= %̂(h · f, h · g).
To show that σ is admissible, note first that Bεσ (f ) ⊂ Bε%̂ (f ) for all ε > 0. Therefore,
the topology induced by σ is as least as fine as the topology induced by %̂. Conversely,
we have to show that for all ε > 0 there is δ > 0 such that Bδ%̂ (f ) ⊂ Bεσ (f ). Consider
%̂
U = Bε/2
(f −1 ). Since inversion is continuous, f ∈ U −1 is a point contained in an open
set, so there is δ > 0 with δ < ε/2 such that
Bδ%̂ (f ) ⊂ U −1 .
To see that this δ is as desired, consider g ∈ Bδ%̂ (f ). Since g ∈ U −1 it follows that g −1 ∈ U
and therefore %̂(f −1 , g −1 ) < ε/2. Thus,
σ(f, g) = %̂(f, g) + %̂(f −1 , g −1 ) ≤ δ + ε/2 ≤ ε/2 + ε/2 = ε.
To establish completeness, assume that hfn : n ∈ Ni is Cauchy w.r.t. σ. Then by the
definition of σ it follows that both fn and fn−1 are Cauchy w.r.t. %̂, so by Theorem 9.7,
fn → f converges uniformly to a continuous function, and fn−1 → g converges also to
a continuous function. It remains to show that g = f −1 ∈ Aut(X). However, since
composition is continuous, it follows from fn → f and fn−1 → g that id = fn fn−1 → f g. 9.4. The Inductive Convergence Criterion. A useful application of the fact that
Aut(X) and Ck (X, X) are complete for compact metrizable spaces is the so-called Inductive Convergence Criterium.
Theorem 9.9. Let X be compact metrizable. If hhn : n ∈ Ni ⊂ Aut(X) is a sequence
such that for all n ∈ N we have
(1) %̂(hn+1 , hn ) < 2−n , and
(2) %̂(hn+1 , hn ) < 3−n · min {min {%(hi (x), hi (y)) : %(x, y) ≥ 1/n} : 1 ≤ i ≤ n},
then the limit h = limn→∞ hn is a homeomorphism of X.
42
MAX PITZ
Proof. Since by (1), the sequence hhn : n ∈ Ni is Cauchy in the complete space Aut(X)
(Theorem 9.7), the function h = limn→∞ hn exists and is continuous. By compactness of
X, it suffices to show that h is a bijection.
Surjectivity: By compactness, h(X) ⊂ X is a closed subset. Suppose h(X) ( X. Let
y ∈ X \ h(X). Then there is ε > 0 such that Bε (y) ⊂ X \ h(X). Find n ∈ N such
that %̂(h, hn ) < ε and pick x ∈ X such that hn (x) = y. Then %(h(x), hn (x)) ≥ ε, a
contradiction.
Injectivity: Take x 6= y ∈ X and take N ≥ 2 large enough such that %(x, y) ≥ 1/N .
Put δ = %(hN (x), hN (y)). Then for all n ≥ N we have
%̂(hn+1 , hn ) < 3−n · δ.
Using the triangle inequality, we obtain
%(hn (x), hN (x)) <
n
X
3−k · δ,
k=N
so taking the limit n → ∞ we see
%(h(x), hN (x)) ≤
∞
X
k=N
Likewise, we get %(h(y), hN (y)) <
would imply
1
δ.
2
3−k · δ ≤
∞
X
k=2
3−k · δ <
1
δ.
2
Now if h(x) = h(y), then the triangle inequality
δ = %(hN (x), hN (y)) ≤ %(h(x), hN (x)) + %(h(y), hN (y)) <
1
1
δ + δ = δ,
2
2
a contradiction.
Let X be a compact metrizable space, and let hhn : n ∈ Ni be a sequence in Aut(X).
Then for each n ∈ N, the function fn = hn ◦ · · · ◦ h1 belongs to Aut(X). If f = limn→∞ fn
exists, then it will also be denoted by
lim hn ◦ · · · ◦ h1 .
n→∞
Note that since the metric %̂ is left-invariant, we have
%̂(hn+1 ◦ · · · ◦ h1 , hn ◦ · · · ◦ h1 ) = %̂(hn+1 , idX ).
This means by the Inductive Convergence Criterium that if the sequence %̂(hn+1 , idX )
converges rapidly to 0 (subject to conditions (1) and (2)) then the infinite left-product of
the hn exists and is a homeomorphism of X.
Let us see as a first application that the Euclidean spaces Rn are countably dense
homogeneous.
Theorem 9.10. Let X be locally compact, separable, metrizable, and strongly locally
homogeneous. Then for all countable dense subsets A and B of X, and for every closed
subset C ⊂ X which misses A ∪ B, there is a homeomorphism h ∈ Aut(X) such that
h A : A → B is a homeomorphism and h C = id.
Proof. Enumerate A = {a1 , a2 , . . .} and B = {b1 , b2 , . . .}. The hypothesis of strongly
locally homogeneous implies that for every point x ∈ X and each open neighbourhood U
of x there is a homeomorphism of ωX which is supported on U and maps x into A (or B).
Now using the Inductive Convergence Criterium, we construct a sequence hhn : n ∈ Ni
of homeomorphisms of ωX 8 such that its infinite left-product is a homeomorphism, and
the following conditions are satisfied:
(1) hn ◦ · · · ◦ h1 (ai ) = h2i ◦ · · · ◦ h1 (ai ) ∈ B for each i and n ≥ 2i,
(2) (hn ◦ · · · ◦ h1 )−1 (bi ) = (h2i+1 ◦ · · · ◦ h1 )−1 (bi ) ∈ A for each i and n ≥ 2i + 1, and
(3) for all n ∈ N, hn restricts to the identity on C ∪ {∞}.
8Note that for separable metrizable X, ωX is also metrizable by Urysohn’s metrization theorem.
TOPOLOGICAL GROUPS
43
Conditions (1) − (3) ensure that h(A) = B, h−1 (B) = A and h(∞) = ∞, so h X ∈
Aut(X) as desired.
Assume that h1 , . . . , h2i−1 have been defined for certain i. We will now show how to
define h2i and h2i+1 satisfying the above conditions.
If h2i−1 ◦ · · · ◦ h1 (ai ) ∈ B take h2i = idωX . Otherwise, choose a small neighbourhood
U2i of h2i−1 ◦ · · · ◦ h1 (ai ) disjoint from the closed set
C ∪ {∞} ∪ {b1 , . . . , bi−1 } ∪ h2i−1 ◦ · · · ◦ h1 ({a1 , . . . , ai−1 }).
The idea is that these points have already been taken care of, and if we choose our h2i
such that we don’t move points in this set, then the left product h2i ◦ · · · ◦ h1 will still
take care of all the old points, so we don’t destroy what we have built earlier. Take h2i
be a homeomorphism of ωX supported on U2i sending the point h2i−1 ◦ · · · ◦ h1 (ai ) into
B ∩ U2i . This completes the definition of h2i .
Next, if (h2i ◦ · · · ◦ h1 )−1 (bi ) ∈ A then take h2i+1 = idωX . Otherwise, choose a small
neighbourhood U2i+1 of bi disjoint from the closed set
C ∪ {∞} ∪ {b1 , . . . , bi−1 } ∪ h2i ◦ · · · ◦ h1 ({a1 , . . . , ai }).
Then take h2i+1 a homeomorphism of ωX supported on U2i+1 such that h−1
2i+1 sends the
point bi into h2i ◦ · · · ◦ h1 (A) ∩ U2i+1 .
This completes the recursive construction. If the neighbourhoods U2i and U2i+1 are
chosen small enough (note that %̂(h2i , idX ) ≤ diam(U2i )) then the conditions of the Inductive Convergence Criterion are satisfied, and we are done.
So by taking C = ∅ in the above result we obtain:
Corollary 9.11. The Euclidean spaces Rn (n ∈ N), and more generally every compact
Hausdorff manifold are countable dense homogeneous.
As a second application, we will show that the Hilbert cube [0, 1]N is topologically
homogeneous. Note that [0, 1]N , the space of all sequences with values in the unit interval
has the standard metric
X |xn − yn |
%(x, y) =
.
2−n
n∈N
Theorem 9.12 (Keller). The Hilbert cube is topologically homogeneous.
For the proof, we shall need two lemmas.
Lemma 9.13. Consider the subspace P = (0, 1)N of Q consisting of all sequences with
values strictly between 0 and 1. Then for any two points x, y ∈ P there is f ∈ Aut(Q)
such that f (x) = y.
Proof. For each coordinate n ∈ N, use the technique of Lemma 2.4 to find a homeomorphism fn : I → I such that fn (0) = 0, fn (1) = 1 and fn (xn ) = yn . Then
f : Q → Q, z 7→ (fn (zn ))n∈N
is as required.
Hence, to show that Q is homogeneous, it suffices to show that for every x ∈ Q there
is f ∈ Aut(Q) such that f (x)n ∈ (0, 1) for all n ∈ N. We will construct such a homeomorphisms using the Inductive Convergence Lemma. The following lemma encompasses the
crucial construction step.
Lemma 9.14. Suppose x ∈ Q, m ∈ N and ε > 0. Then there is an element hm ∈ Aut(Q)
such that
(1) %̂(hm , idQ ) < ε,
(2) hm (x)m ∈ (0, 1), and
(3) hm (y)i = yi for all i < m and y ∈ Q.
44
MAX PITZ
Proof. If xm ∈ (0, 1) then let hm = idQ . So without loss of generality, we may assume
that xm = 1. LetQn > m be such that 2−(n−1) < ε.
Consider the k∈{m,n} [0, 1] consisting of the product of the two coordinates m, n of
Q. It is geometrically obvious that there is a homeomorphism f : I 2 → I 2 such that
(4) f [0, 1 − 2−n ] × I = id, and
(5) f ({1} × I) ⊂ (1 − 2−n , 1) × {1}.
Q
Q
Think of Q = k∈{m,n} [0, 1] × k∈{m,n}
[0, 1] and extend the map f to h = f × id. So h
/
only affects the m and nth coordinate of any point. So h satisfies (2) and (3) alright. Let
us check that also (1) is satisfied. But note that for every z ∈ Q,
%(h(z), z)
=
2−m · |h(z)m − zm | + 2−n |h(z)n − zn |
≤
2−m · 2−n + 2−n · 1 ≤ 2−(n−1) < ε.
Thus, it follows that %̂(hm , idQ ) = sup {%(h(z), z) : z ∈ Q} < ε as required.
Proof of Theorem 9.12. Let x ∈ Q. We use the Inductive Convergence Criterium to find
a h ∈ Aut(Q) such that h(x) ∈ P . Together with Lemma 9.13 this clearly implies the
result.
Inductively, find smaller and smaller hn such that fn = hn ◦· · ·◦h1 (x) satisfies fn (x)i ∈
(0, 1) for all i ≤ n. By Lemma 9.14 we can always find a next homeomorphism hn+1
that also takes care of the coordinate n + 1 without changing anything about the earlier
coordinates. The proof is complete.
The fact the Q is homogeneous raises the question whether there is an admissible
group structure on Q turning it into a topological group. It turns out, however, that
this is not possible: Recall that every non-trivial topological group has fixed-point free
autohomeomorphisms—namely the translations.
Theorem 9.15. The Hilbert cube Q has the fixed-point property: for every h ∈ Aut(Q)
there is x ∈ Q such that h(x) = x.
Proof. Exercise. Use the Brouwer Fixed Point Theorem: Every continuous map f : I n →
I n has a fixed point.
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[1]
[2]
[3]
[4]
[5]
D.L. Cohn, Measure Theory, Birkäuser Boston, 1980.
D. Dikranjan, Introduction to topological groups, Lecture script, 2013.
E. Hewitt, K.A. Ross, Abstract Harmonic Analysis, Band 1 (2nd ed.), Springer, 1979.
K.H. Hofmann, S.A. Morris, The Structure of Compact Groups, (3rd ed.), DeGruyter, 2013.
J. van Mill, The Infinite-Dimensional Topology of Function Spaces, Elsevier, 2001.