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of Bristol [2014]
TOPICS IN MODERN GEOMETRY
Lecture 1
1. Introduction: Motivation and Outline
Fundamental to much of modern mathematical research is the interplay between geometry
and algebra. It is difficult to characterise the ever-increasing number of ways such connections
have been exploited, but two of the most basic instances are the following.
(i) The use of algebraic tools to investigate geometric structures.
(ii) The use of geometric notions to yield novel insights about algebraic/arithmetic objects.
Approach (i) is exemplified in algebraic topology, a subject whose notable achievements include1:
The Brouwer fixed point theorem. Every continuous map from a closed ball in Rn to itself
has a fixed point.
The success of approach (ii) can be seen in the development of arithmetic geometry, which
studies problems in number theory using geometric ideas. A notable victory of this approach
is a theorem of Faltings2 which implies the following weak version of Fermat’s last theorem.
Weak FLT. For any n > 4 there are at most finitely many integer solutions to an + bn = cn
with hcf(a, b, c) = 1.
We prove neither of these results in this course3. Instead we focus on one of the most basic
objects endowed with both geometric and algebraic structure, a topological group. The definition
is exactly what one would guess: it is a group (one of the simplest algebraic objects) endowed
with the geometric structure of a topological space. The former encodes the symmetries of the
space, whilst the latter encodes how ‘connected’ the space is.
Although the concept of a topological group is simple, the first portion of our course is taken
up with building the necessary theory to enable us to give a precise definition. We first need
to recall the theory of topological spaces, continuous maps and the product topology. We then
investigate how the group structure forces some rigidity on the topology of the space - the
topology is much nicer than one would expect of an arbitrary topological space.
The next part of the course uses our basic knowledge of topological groups to develop the
beautiful theory of hyperbolic geometry. In particular we study how such spaces develop dynamically over time under the action of a group.
If time permits, we discuss how one can add further geometric structure to a topological
group. In particular, a structure that allows one to differentiate. Such Lie groups are ubiquitous
in modern mathematics, with applications as far a field as number theory and mathematical
physics. We have to put some work in merely to give their definition. Those interested can
investigate some of the further fascinating properties of Lie groups in their projects.
2. A Refresher in topology
Given a set X, we write 2X for its power set
2X := {Y : Y ⊂ X} .
Date: Autumn 2014.
Comments/corrections gratefully received at [email protected].
1
See http://en.wikipedia.org/wiki/Brouwer_fixed_point_theorem.
2See http://en.wikipedia.org/wiki/Faltings’_theorem.
3For courses developing some of the theory behind these results, see Algebraic Topology and Algebraic Number
Theory (respectively).
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of Bristol [2014]
Definition (Topological space). A topological space is a pair (X, τ ) where X is a set and τ ⊂ 2X
is a collection of subsets of X (the topology) such that the following three conditions are satisfied:
a. (Openness of ∅ and X). We have ∅ ∈ τ and X ∈ τ .
S
b. (Open sets are closed under arbitrary unions). If {Uα : α ∈ A} ⊂ τ then α∈A Uα ∈ τ .
c. (Open sets are closed under finite intersections). If U1 , . . . , Un ∈ τ then U1 ∩ · · · ∩ Un ∈ τ .
Definition (Open, closed, neighbourhood). Let (X, τ ) be a topological space.
a. (Open). An element U of the topology τ is called an open set.
b. (Closed). The complement F = X \ U of an open set U in X is called a closed set.
c. (Neighbourhood). Given a point x ∈ X, a neighbourhood of x is any set containing an open
set which contains x, i.e. N is a neighbourhood of x if there exists an open set U with
x ∈ U ⊂ N.
Exercise 2.1 (Homework, week 1). Let (X, τ ) be a topological space.
a. (Boolean properties of closed sets). Verify that closed sets are closed under arbitrary intersections and finite unions.
b. (Closure). Show that for any subset A of a topological space, the intersection of all closed
sets which contain A is itself a closed set containing A. This intersection is the smallest
closed set containing A, in the sense that it is contained in any closed set containing A. We
call it the closure of A and denote it A.
c. (Alternative characterisation of closure4). Verify that x ∈ A if and only if for any neighbourhood N of x we have N ∩ A 6= ∅.
Exercise 2.2 (Interior). Let (X, τ ) be a topological space. Given a subset E ⊂ X, define the
interior of E to be the union of all open sets contained in E, denoted int(E). Prove that the
interior of E equals the complement of the closure of its complement
int(E) = X \ (X \ E).
Examples 2.1 (Some topological spaces).
a. (Metric spaces). Recall that a set U in a metric space (X, d) is open if for any x ∈ U there
exists ε > 0 such that the open ball Bε (x) := {y ∈ X : d(x, y) < ε} is contained in U . If
(X, d) is a metric space, then the collection of open sets form a topology (prove it!).
n
b. (The Euclidean spaces).
pP The set R formsn a metric space with respect to the Euclidean
5
2
distance d(x, y) :=
i (xi − yi ) . Hence R inherits the topology induced by this metric .
n
2n
Notice that C with the usual norm can be viewed as R with the Euclidean norm. Hence
Cn also forms a topological space.
c. (The discrete topology). Any set forms a topological space with respect to the topology
2X = {E : E ⊂ X}. Notice that this is the largest possible topology on X, in that any
topology τ on X satisfies τ ⊂ 2X .
d. (The indiscrete or trivial topology). Any set forms a topological space with respect to the
trivial topology {∅, X}. Notice that this is the smallest possible topology on X in that
{∅, X} ⊂ τ for every topology τ on X.
e. (The co-finite topology). Given a set X, define U to be open if U = ∅ or X \ U is finite.
f. (The co-countable topology). Given a set X, define U to be open if U = ∅ or X \ U is (at
most) countable. Notice that this coincides with the discrete topology if X is countable, so
this is only really interesting if X is uncountable.
The latter three topologies often behave differently to the topological spaces we are familiar
with (metric spaces). They can therefore be a good source of counterexamples.
4This latter characterisation is often easier to work with. Notice that it captures the idea, found in metric
space theory, that the closure is the set of limit points of a set. Unfortunately limits are not always defined in
topological spaces, so this definition does not generalise directly.
5In fact, one can prove that any metric on Rn of the form d(x, y) = kx − yk where k·k is a norm (see http:
//en.wikipedia.org/wiki/Normed_vector_space) induces the same topology on Rn . An ambitious student
should have a go at proving this.
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University
of Bristol [2014]
Exercise 2.3. Verify that the spaces given in Examples 2.1 do indeed form topological spaces.
Definition (Continuous map). Given topological spaces X and Y , a function f : X →
Y is said to be continuous if for every open subset U ⊂ Y the inverse image f −1 (U ) :=
{x ∈ X : f (x) ∈ U } is open in X.
Contrast the notion of continuity with the following less-used notion.
Definition (Open map). Given topological spaces X and Y , a map f : X → Y is open if for
any open set U ⊂ X, the image f (U ) := {f (x) : x ∈ U } is open in Y .
Evidence that the notion of continuity is more useful is demonstrated in the following exercise.
Exercise 2.4. Prove that if (X, dX ) and (Y, dY ) are metric spaces, then the topological notion
of continuity coincides with the metric notion:
h
i
(∀x ∈ X)(∀ε > 0)(∃δ > 0)(∀x0 ∈ X) dX (x, x0 ) < δ =⇒ dY f (x), f (x0 ) < ε .
Definition (Homeomorphism). Given topological spaces X and Y , a homeomorphism is a
bijection f : X → Y such that both f and f −1 are continuous. If there exists a homeomorphism
between two topological spaces then we call them homeomorphic.
Alternatively, a homeomorphism is a bijection which is both open and continuous. Just
as two isomorphic groups are essentially the same group, possibly written in different ways,
homeomorphic topological spaces are essentially the same topological space (they have the
same topological properties).
Exercise 2.5. Give an example of a continuous bijection between topological spaces which does
not have a continuous inverse. (Hint6)
In checking that Examples 2.1 are indeed topological spaces, one should notice that it is often
laborious to check directly that the topology axioms are satisfied. It is useful to be able to build
a topology from a collection of sets which we would like to be open.
Lemma 2.1 (Topologies
Tare closed under intersection). Let τi (i ∈ I) be a collection of topologies
on a set X. Then τ := i τi is a topology on X whose sets are open in each of the topologies τi .
Exercise 2.6 (Homework, week 1). Prove Lemma 2.1.
Definition (Generated topology). Let σ ⊂ 2X denote a collection of subsets of a set X. The
topology generated by σ is the intersection of all topologies on X which contain σ. Notice that
this intersection is well-defined since 2X is a topology containing σ.
Under mild assumptions we can write down what the topology generated by σ actually looks
like.
Exercise 2.7 (Structure
of generated topology). Suppose that σ = {Eα ⊂ X : α ∈ A} covers
S
X, in that X ⊂ α Eα . Show that, under this covering assumption, the open sets in the
topology generated by σ coincide with those sets which can be
S written as a union of finite
intersections of elements in σ. i.e. U is open if and only if U = β Uβ where each Uβ takes the
form Eα1 ∩ · · · ∩ Eαn for some Eαi ∈ σ.
Notice that the topology τ generated by a collection of subsets σ ⊂ 2X is the smallest topology
on X for which each set E ∈ σ is open, for if τ 0 is a topology on X in which each E ∈ σ is open
then we have τ ⊂ τ 0 (why?).
Definition (Base of a topology). Given a topology τ on a set X, we say that a collection of
open subsets σ ⊂ τ forms a base (or basis) for τ if every element of τ can be written as a union
of elements of σ, i.e.
[
E∈τ
⇐⇒
E=
Bα for some Bα ∈ σ.
α∈A
6Take X and Y to be the same set (with at least two elements) where X has the discrete topology and Y the
trivial topology.
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University
of Bristol [2014]
Exercise 2.8. Let τ be a topology on a set X.
a. Show that if σ is a base for τ , then the topology generated by σ is equal to τ .
b. Show that if σ covers X (its union equals X), then the sets which can be written as finite
intersections of elements of σ form a base for the topology generated by σ.
c. Show that if σ covers X and the intersection of any two elements of σ equals a union of
elements of σ, then σ is itself a base for the topology generated by σ.
Exercise 2.9 (Homework, week 1).
a. Show that the open balls form a base for the topology induced by a metric. In particular,
the open intervals (a, b) generate and are a base for the usual topology on R.
b. By restricting to a subfamily of open balls, prove that the usual topology7 on Rn has a base
consisting of countably many open sets.
Definition (Second countable). A topological space X is said to be second countable if it has
a base consisting of countably many open sets (a countable base).
Exercise 2.10. Show that X is second countable if and only if
(∃ open sets {Bi }i∈N )(∀x ∈ X)(∀ open set U containing x)(∃i ∈ N)[x ∈ Bi ⊂ U ].
This leads one to wonder what a first countable space is. Rather like the difference between
uniform continuity and continuity, the order of the quantifiers is key.
Definition (First countable). A topological space is first countable if
(∀x ∈ X)(∃ open sets {Bi }i∈N )(∀ open set U containing x)(∃i ∈ N)[x ∈ Bi ⊂ U ].
Exercise 2.11. Show that a topological space X is first countable iff every x ∈ X has a
countable family of open neighbourhoods (Bi )i∈N such that any neighbourhood of x contains
some Bi .
Definition (Dense). A subset E of a topological space X is dense in X if for any non-empty
open subset U the intersection E ∩ U is also non-empty.
Exercise 2.12 (Homework, week 1). Prove that a subset E of a topological space X is dense
in X if and only if the closure of E is equal to the whole space X, i.e.
E = X.
Example 2.2. The rationals are dense in the reals.
Lecture 2
Definition (Separable). A topological space is separable if it has a countable dense subset.
Exercise 2.13. Prove that the Euclidean space Rn is separable.
Proposition 2.2 (2nd countable implies separable). If a topological space is second countable
then it is separable.
Proof. Let {Bi }i∈N denote a countable base for a topological space X. Deleting occurrences
of the empty set, we may assume each Bi is non-empty. For each i ∈ N let us choose some
xi ∈ Bi . We claim that the xi form our countable dense subset. It suffices to show that if U
is a non-empty open subset of X then there exists i ∈ N such that xi ∈ U . This follows since
Bi ⊂ U for some i.
Definition (Metrisable). A topological space (X, τ ) is metrisable if there exists a metric d on
X such that the topology on X inherited from this metric coincides with τ .
Proposition 2.3. Let X be a metrisable topological space. Then X is second countable if and
only if it is separable.
7The topology induced by the Euclidean metric.
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University
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Proof. Let X be a metrisable topological space with a countable dense subset {xi }i∈N . We must
show that X has a countable base. Consider the family of open balls
B := B1/n (xi ) : n, i ∈ N .
Since a countable union of countable sets is itself countable, the family B is countable.
To establish that B is a base, we must show that for any x ∈ X and N a neighbourhood of x,
there exists n, i ∈ N such that x ∈ B1/n (xi ) ⊂ N . Since N is a neighbourhood, there exists ε > 0
such that Bε (x) ⊂ N . The density of the xi ensures that there exists xi ∈ Bε (x). Choosing
1/n ≤ ε − d(x, xi ) the triangle inequality gives that B1/n (xi ) ⊂ Bε (x) ⊂ N , as required.
To motivate the next definition, let us call a subset S of a topological space X somewhere
dense if there exists a non-empty open set U such that S ∩ U is dense in the subspace topology8
on U (equivalently S ∩ U = U ).
Definition (Nowhere dense). We say that a subset S of a topological space X is nowhere dense
if it is not somewhere dense, so that for any non-empty open set U in X we have
S ∩ U 6= U .
Exercise 2.14.
a. Show that a subset S of a topological space X is nowhere dense if and only if the closure of
S has empty interior, i.e.
int(S) = ∅.
b. Given an example of a set S in a topological space X which is somewhere dense and whose
interior has empty closure, i.e.
int(S) = ∅.
3. The definition of a topological group
Definition (Topological Group). A topological group is a triple (G, τ, ·) where (G, τ ) is a topological space and (G, ·) is a group, such that both of the following conditions hold
a. The inverse operation x 7→ x−1 is continuous as a map G → G.
b. The group operation (x, y) 7→ xy is continuous as a map G × G → G when G × G is given
the product topology.
But what is the product topology? This definition is best seen in a broader context.
3.1. The product topology.
Definition (Pull-back topology). Given functions fi : X → Yi with each Yi a topological space,
the pull-back topology is the smallest topology on X such that
(i.e.
S all of the fi are continuous
the topology on X generated by the set of inverse images i fi−1 (U ) : U is open in Yi ).
Exercise 3.1. Prove that the pull-back
topology induced by a single function f : X → Y is
equal to the set of inverse images f −1 (U ) : U is open in Y .
Example 3.1 (Subspace topology). Given a topological space X and a subset Y ⊂ X, the
subspace topology is the pull-back topology of the inclusion function i : Y → X given by
x ∈ Y 7→ x ∈ X. This is the smallest topology on Y for which i is continuous. One can show
that E ⊂ Y is open if and only if E = U ∩ Y for some open subset U of X.
Definition (Product set). Given an indexed family of sets Xα (α ∈ A), the product set
Q
α∈A XαQis the set of indexed tuples (xα )α∈A such that
S xα ∈ Xα for all α ∈ A. One can
identify α∈A Xα with the set of functions f : A → α∈A Xα such that f (α) ∈ Xα for all
Q
α ∈ A. When Xα is the same set X for eachQα, we write X A for α∈A X. If A = {α1 , . . . , αn }
is finite, then we write Xα1 × · · · × Xαn for α∈A X.
8The subspace topology is defined in the next section.
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Exercise 3.2. Given a set A, show that the product set {0, 1}A is in bijective correspondence
with the power set 2A := {B : B ⊂ A}.
Definition (Product topology).
Given an indexed family of topological spaces Xα (α ∈ A),
Q
the
Q product topology on α∈A Xα is the pull-back of the collection of projection maps πβ :
α∈A Xα → Xβ given by (xα )α∈A 7→ xβ . In other words, the product topology is the smallest
topology such that all of the projections are continuous.
Exercise 3.3 (Homework, week 1).
Q
a. Show that U ⊂ α∈A Xα is an open set in the product topology if and only if for any x ∈ U
there exists a finite subset {α1 , . . . , αn } ⊂ A and open sets Ui ⊂ Xαi such that
(
)
Y
x∈ y∈
Xα : yαi ∈ Ui (1 ≤ i ≤ n) ⊂ U.
α∈A
Q
Writing πβ for the projection map α∈A Xα → Xβ , this exercise shows that a base for the
product topology consists of sets which can be written in the form
πβ−1
(U1 ) ∩ · · · ∩ πβ−1
(Un )
n
1
for some β1 , . . . , βn ∈ A and open sets Ui ⊂ Xβi
b. If X1 , . . . , Xn are topological spaces, show that sets of the form U1 × · · · × Un , where the Ui
are open in Xi , form a base for the product topology on X1 × · · · × Xn .
Exercise 3.4. Show that Rn equipped with the topology induced by the Euclidean metric
sX
d(x, y) :=
|xi − yi |2
i
is homeomorphic to the product topological space R × · · · × R (n copies of R).
The following is another topological space which is useful source of counterexamples.
Example 3.2 (The Cantor set). Let E0 := [0, 1], E1 := [0, 1/3] ∪ [2/3, 3/3], E2 := [0, 1/9] ∪
[2/9, 3/9] ∪ [6/9, 7/9] ∪ [8/9, 9/9], etc. In other words, Ei+1 is obtained from the union of
closed intervals Ei by removing the open middle third of every interval. We thereby obtain a
descending sequence of closed sets E0 ⊃ E1 ⊃ E2 ⊃ . . . . The intersection
E :=
∞
\
Ei
i=1
9
is called the Cantor set (after Georg Cantor ).
Exercise 3.5 (The Cantor set).
a. Justify why the Cantor set is a closed subset of R.
b. Show by induction that for any n ≥ 1 we have
" n
#
n
X
X
[
2
i 3−i , 2
i 3−i + 3−n .
En =
∈{0,1}n
i=1
i=1
c. Show that x is an element of the Cantor set if and only if there exists a sequence (i )i∈N with
i ∈ {0, 1} for each i such that
∞
X
x=2
i 3−i .
i=1
N
d. Show that {0, 1} equipped with the product topology is homeomorphic to the Cantor set
equipped with the subspace topology inherited from the usual topology on R.
9http://en.wikipedia.org/wiki/Cantor_set
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Exercise 3.6 (Homework, week 2). Given a topological space G, show that a function f :
G × G → G is continuous (with G × G under the product topology) if and only if for any
x1 , x2 ∈ G and any open neighbourhood U ⊂ G of f (x1 , x2 ), there exist open neighbourhoods
U1 , U2 ⊂ G of x1 and x2 (respectively) such that U1 × U2 ⊂ f −1 (U ).
Exercise 3.7 (Homework, week 2).
a. Show that a if (G, τ ) is a topological space and (G, ·) a group, then (G, τ, ·) forms a topological
group if and only if the (right) division map (x, y) 7→ xy −1 is continuous as a map G×G → G.
b. Hence show that a topological space (G, τ ) with a group structure (G, ·) forms a topological
group if and only if for any x, y ∈ G and any neighbourhood N ⊂ G of xy −1 there exist open
sets U1 , U2 such that U1 U2−1 ⊂ N .
Exercise 3.8. Given a topological group G, show that the squaring map x 7→ x2 is continuous
as a map G → G.
3.2. Examples of topological groups. At first sight, topological groups appear to be very
special kinds of group. However, every group is a topological group under both the discrete
topology and the trivial topology (exercise). This suggests that the same group can give rise to
different topological groups, but we must first define what we mean by ‘different’.
Definition (Topological homomorphism). Given topological groups G1 and G2 , a map f :
G1 → G2 is a topological homomorphism if it is both continuous and a group homomorphism.
Definition (Topologically isomorphic). We say that topological groups G1 and G2 are topologically isomorphic if there exists a map f : G1 → G2 which is both a group isomorphism and a
homeomorphism.
Topologically isomorphic groups are the same topological group (up to notation).
Example 3.3 (Isomorphic groups may not be topologically isomorphic). The additive group of
real numbers forms a topological group under the Euclidean topology. This is not topologically
isomorphic to the additive group of real numbers under the discrete topology. To see this note
that all singleton subsets {x} are open in the discrete topology, but no singleton set is open in
the usual topology on R (prove this). Since a homeomorphism maps open sets to open sets and
singleton sets to singleton sets, the two topologies cannot be homeomorphic.
Example 3.4 (Subgroups). Any subgroup of a topological group is itself a topological group
under the subspace topology (check this!).
Example 3.5 (Rn , +). Any subgroup of the additive group Rn is a topological group under
the subspace topology inherited from the usual (Euclidean) topology on Rn (prove this - by the
previous example, it suffices to check that Rn is a topological group).
Example 3.6 (Subgroups of GL(n, C)). One can regard the set GL(n, C) of n × n non-singular
2
complex matrices as a subset of Cn in the obvious manner. Equipping GL(n, C) with the
2
subspace topology inherited from the usual topology on Cn (the topology induced by the
2
Euclidean metric on R2n ), one can prove that GL(n, C) forms a topological group under matrix
multiplication (see the next exercise). Hence any subgroup of GL(n, C) forms a topological
2
group under the subspace topology inherited from R2n . In particular:
2
(1) GL(n, R) under the subspace topology inherited from Rn (check that the obvious embedding GL(n, R) → GL(n, C) gives a topological isomorphism with its image).
(2) The multiplicative group R× = GL(1, R) of non-zero real numbers.
(3) The special linear group SL(n, R) consisting of matrices in GL(n, R) with determinant
1.
(4) The special linear group SL(n, Z) consisting of matrices in SL(n, R) whose entries are
all integers.
(5) The orthogonal group O(n) of real matrices A = (aij ) ∈ GL(n, R) with AAT = I, where
AT = (aji ) is the transpose of A.
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(6) The circle group S 1 := {z ∈ C : |z| = 1} under complex multiplication and the subspace
topology inherited from C.
(7) The continuous Heisenberg group H3 (R) consisting of 3 × 3 upper triangular matrices
of the form


1 x y
0 1 z 
0 0 1
Exercise 3.9 (GL(n, C) is a topological group). Throughout this exercise we assume Rn has
the usual topology.
a. Show that a function f = (f1 , . . . , fm ) : Rn → Rm is continuous if and only if each of the
component functions fi : Rn → R is continuous.
b. Let P, Q ∈ R[x1 , . . . , xn ] be multivariable polynomials with real coefficients. Let EQ :=
{x ∈ Rn : Q(x) 6= 0} and endow this space with the subspace topology inherited from Rn .
Prove that the map f : EQ → R given by f (x) := P (x)/Q(x) is continuous.
c. Deduce, using both of the above, the continuity of the division map GL(n, C) × GL(n, C) →
GL(n, C) given by (A, B) 7→ AB −1 .
Exercise 3.10. Prove that the multiplicative group C× of non-zero complex numbers (under
the usual metric) is topologically isomorphic to the following subgroup of GL(2, R).
a −b
G=
: a, b ∈ R not both zero .
b a
Exercise 3.11. Construct a topology on the additive group Z/2Z under which Z/2Z is not a
topological group.
4. Homogeneity
The following definition is little used in what follows, but it adds context to our use of the
word homogeneity.
Definition (Homeo(X)). If X is a topological space, then we write Homeo(X) for the set of
all homeomorphisms from X to itself.
Exercise 4.1. Show that Homeo(X) forms a group (not necessarily topological).
One can endow Homeo(X) with the structure of a topological group, but we shall not require
that here.
Definition (Group action on a topological space). An action of a group G on a topological
space X is a map G → Homeo(X), g 7→ ρg . When the action is clear from the context, we often
write gx as a shorthand for ρg (x).
Definition (Homogeneous space). A topological space X equipped with a group action ρ :
G → Homeo(X) is termed a homogeneous space if the action is transitive, i.e. for any x, y ∈ X
there exists g ∈ G such that gx = ρg (x) = y.
Lemma 4.1. Let G be a topological group. For each g ∈ G define Lg , Rg : G → G by Lg (x) = gx
and Rg (x) = xg. Then g 7→ Lg and g 7→ Rg are both group actions G → Homeo(G) which give
G the structure of a homogeneous space.
The power of homogeneity is that it ensures that many (local) topological properties of the
group G are determined by the local structure of the group around the identity.
Definition (Continuity at a point). Given topological spaces X and Y , we say a map f : X → Y
is continuous at a point x ∈ X if for any open neighbourhood N of f (x) the inverse image f −1 (N )
is an open neighbourhood of x.
Exercise 4.2. Prove that a map f : X → Y between topological spaces is continuous if and
only if it is continuous at every point.
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Proposition 4.2. Let G1 and G2 be topological groups. A group homomorphism φ : G1 → G2
is continuous if it is continuous at a single point of G1 (for instance, at the identity).
Proof. Suppose that φ is continuous at x0 ∈ G1 . We must show that φ is continuous at every
x ∈ G1 . Let U be an open neighbourhood of φ(x). By homogeneity, there exists g ∈ G1 such
that x0 = gx. Since left multiplication by g is a homeomorphism, the set φ(g)U is an open
neighbourhood of φ(g)φ(x) = φ(gx) = φ(x0 ). As φ is continuous at x0 , there exists an open
neighbourhood V of x0 such that φ(V ) ⊂ φ(g)U . Then g −1 V is an open neighbourhood of x
satisfying φ(g −1 V ) = φ(g −1 )φ(V ) ⊂ φ(g −1 )φ(g)U = φ(g −1 g)U = U .
Definition (Compactness). A subset C of a topological space X is compact if each of its
open Scovers has a finite subcover. That is, if (Uα )α is a collection of open subsets of X with
C ⊂ α Uα , then there exist α1 , . . . , αn such that C ⊂ Uα1 ∪ · · · ∪ Uαn .
Definition (Locally Compact). A topological space is locally compact if every point has a
compact neighbourhood.
Proposition 4.3. A topological group is locally compact if and only if the identity has a compact
neighbourhood.
Proof. An exercise in using homogeneity.
Definition (Discrete). A topological space is discrete if it has the discrete topology. A subset
of a topological space is discrete if it is a discrete space under the subspace topology.
Exercise 4.3. Show that the integers form a discrete subgroup of the real numbers but that
the rationals do not.
Proposition 4.4. A topological group is discrete if {e} is open.
Proof. An exercise in homogeneity.
Definition (Locally Euclidean). A topological space is locally Euclidean if there exists n ∈ N
such that every point has a neighbourhood which is homeomorphic to Rn .
Exercise 4.4. Prove that an open ball Br (x) ⊂ Rn with the subspace topology is homeomorphic
to Rn . Deduce that a topological space is locally Euclidean if and only if there exists n ∈ N
such that every point has a neighbourhood homeomorphic to an open ball in Rn .
Proposition 4.5. A topological group is locally Euclidean if and only if there exists a neighbourhood of the identity which is homeomorphic to an open ball in Rn .
Proof. An exercise in homogeneity.
5. Quotient topological groups
Definition (Push-forward topology). Given functions fi : Xi → Y with each Xi a topological
space, the push-forward topology is the smallest topology
S on Y such that all of the fiare
continuous (i.e. the topology on Y generated by the set i E ⊂ Y : fi−1 (E) is open in Xi ).
Exercise 5.1.
topology induced by a single function f : X → Y
Prove that the push-forward
is equal to E : f −1 (E) is open in X .
Definition (Quotient topology). Given a set X with an equivalence relation ∼, we denote the
set of equivalence classes by X/ ∼. The quotient map is the function q : X → X/ ∼ which
maps x ∈ X to its equivalence class under ∼. If X is a topological space, the quotient topology
on X/ ∼ is the push-forward of the topology on X under the quotient map q : X → X/ ∼.
Equivalently, the smallest topology on the set of equivalence classes of X for which the quotient
map is continuous.
By the previous exercise, E ⊂ X/ ∼ is open in the quotient topology if and only if q −1 (E) is
open in X.
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Exercise 5.2. Prove that F ⊂ X/ ∼ is closed in the quotient topology if and only if q −1 (F ) is
closed in X.
Definition (Normal subgroup). Given a group G and a subgroup H, recall that H is said to
be normal in G if for every g ∈ G we have gH = Hg, or equivalently g −1 Hg ⊂ H.
Example 5.1. A subgroup H of a group G induces an equivalence relation on G by setting
x ∼ y iff xy −1 ∈ H (equivalently, x and y belong to the same coset of H). In this case, we write
G/ ∼ as G/H. Recall that this forms a group under multiplication of cosets xH · yH = xyH
provided that H is a normal subgroup of G.
The definition of the quotient topology ensures that the quotient map q : G → G/H is
continuous. Somewhat surprisingly it is also an open map. The group structure of G forces
some rigidity on its topological structure.
Lemma 5.1 (The quotient map is open). Let G be a topological group and H a subgroup of G.
Then the quotient map q : G → G/H is an open map of topological spaces.
Proof. Let U ⊂ G be open. By the definition of the quotient topology, our task is to show that
q −1 (q(U )) is open. Notice that x ∈ q −1 (q(U )) if and only if xH ∈ q(U ), which is true if and
only if there exists y ∈ U and h ∈ H with x = yh. Therefore
[
q −1 (q(U )) =
U h.
h∈H
By homogeneity, the above union is a union of open sets, hence itself open.
Exercise 5.3 (Quotient topological groups). Given a normal subgroup H of a topological group
G, prove that G/H is a topological group under the quotient topology.
Example 5.2 (The torus T). Since Z is a subgroup of the additive group R, the set T := R/Z
is a topological group under the quotient topology inherited from the usual topology on R.
This quotient group is a very good example to have in mind whenever you meet a new notion
in topological groups: one should repeatedly ask oneself ‘what does this mean for the torus?’
More generally, we define the n-dimensional torus Tn := Rn /Zn analogously.
Non-Example 5.3. Notice that if H is a normal subgroup of a topological group, then the set
of left-cosets
G/H := {gH : g ∈ G}
equals the set of right cosets
H\G := {Hg : g ∈ G} .
If H is a non-normal subgroup then these sets do not coincide. However, it can still be fruitful
to study the quotient topological spaces G/H and H\G, as well as the double quotient
H1 \G/H2 = {H1 gH2 : g ∈ G} .
Notice that G/H forms a homogeneous space under the left-action of G
(g, xH) 7→ gxH.
Similarly for H\G under the right action.
6. Compactness
Definition (Co-compact). A subgroup H of a topological group G is co-compact if the quotient
topological space G/H is compact.
Exercise 6.1. Prove that the torus T under addition is topologically isomorphic to the unit
circle S 1 under complex multiplication. Hence T is compact, and thus the integers are a cocompact subgroup of R.
Exercise 6.2. Show that Tn is homeomorphic to a compact subset of Rn and is therefore itself
compact.
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Compact topological groups are particularly nice to study. Their theory is only slightly more
complex than that of finite (discrete) groups, and is neater than the theory of non-compact
groups such as R. In fact, compact and non-compact topological groups are hard to compare,
as can be seen in the following result.
Proposition 6.1.
a. If G is a compact topological group then the only topological homomorphism f : G → R is
the trivial map x 7→ 0.
b. If G is a compact topological group, then any topological homomorphism f : G → R× satisfies
f (G) ⊂ {−1, 1}.
In order to prove Proposition 6.1, we must recall some standard properties of compactness.
Proposition 6.2 (The continuous image of a compact set is compact). If f : X → Y is a
continuous map and C is a compact subset of X, then f (C) is a compact subset of Y .
Proof. Let {Uα }α denote an open cover of f (C). Then f −1 (Uα ) α is an open cover of C.
Hence there exist α1 , . . . , αn such that
C ⊂ f −1 (Uα1 ) ∩ · · · ∩ f −1 (Uαn ).
It follows that
f (C) ⊂ Uα1 ∩ · · · ∩ Uαn .
Theorem 6.3 (Heine–Borel). A subset of Rn is compact if and only if it is both closed and
bounded.
Proof. See your notes from the Metric Spaces course, or visit http://en.wikipedia.org/wiki/
Heine-Borel_theorem.
Proposition 6.4 (Continuous real-valued functions achieve their maxima/minima on compact
sets). Let X be a compact topological space and f : X → R a continuous function. Then there
exists x, x ∈ X such that
f (x) = inf f (X)
and
f (y) = sup f (X).
(1)
Proof. The image f (X) is a compact subset of R, hence by Heine–Borel it is closed and bounded.
Boundedness implies that inf f (X) and sup f (X) are finite real numbers, so for any ε > 0 there
exist xε , yε ∈ X satisfying
inf f (X) ≤ f (xε ) ≤ inf f (X) + ε
and
sup f (X) − ε ≤ f (yε ) ≤ sup f (X).
It follows that inf f (X) and sup f (X) belong to the closure of f (X). Since f (X) is itself closed
we have inf f (X) ∈ f (X) and sup f (X) ∈ f (X), as required.
Exercise 6.3. Prove Proposition 6.1 using the above results.
7. Separation axioms
The topology and group structure of a topological group interact in non-trivial ways. In this
section we demonstrate how the group structure constrains the topological structure in a much
more rigid way than one would expect of an arbitrary topological space. In particular, we look
at how many of the different separation axioms for topological spaces collapse into the same
axiom for topological groups.
A separation axiom is an additional axiom one can add to the definition of a topological
space in order to allow one to use open sets to distinguish between different points in the space.
Abstract topological spaces are in some sense too general. In particular there are topologies,
like the trivial topology, whose open sets cannot distinguish between distinct points: there exist
distinct points whose neighbourhoods are exactly the same. A separation axiom ensures that
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the neighbourhoods of distinct points differ. The precise nature of this difference gives rise to
different separation axioms10.
Definition (Separation axioms). Let X be a topological space.
• X is T0 or Kolmogorov if for every distinct pair x, y ∈ X there exists an open set U
containing one but not the other, i.e. ∅ =
6 U ∩ {x, y} =
6 {x, y}.
• X is T1 if for every pair of distinct points there are open sets containing each point and
not the other.
• X is T2 or Hausdorff if for every pair of distinct points x, y ∈ X there exist disjoint
open sets U and V such that x ∈ U and y ∈ V .
• X is regular if for every x ∈ X and closed set F ⊂ X with x ∈
/ F there exist disjoint
open sets U and V such that x ∈ U and F ⊂ V .
• X is T3 if it is both T0 and regular.
Exercise 7.1. Prove that a topological space X is regular if and only if any x ∈ X and any
open neighbourhood U of x, there exists an open set U1 such that x ∈ U1 ⊂ U1 ⊂ U .
Proposition 7.1. We have the chain of implications
T3 =⇒ T2 =⇒ T1 =⇒ T0 .
Proof. Exercise.
Proposition 7.2. In a topological group T0 =⇒ T3 .
The following proof is somewhat lengthy as we try to motivate our argument. The reader is
encouraged to give their own streamlined one-paragraph proof.
Proof. The lemma follows from the surprising fact that any topological group is regular, hence
T3 if and only if T0 . From Exercise 7.1, it suffices to prove that for any x ∈ X and any
open neighbourhood U of x, there exists an open set U1 such that x ∈ U1 ⊂ U1 ⊂ U . By
homogeneity, we need only establish this when x = e (justify this), so we wish to show that any
open neighbourhood U of the identity contains the closure of another neighbourhood U1 of the
identity.
How does one proceed when G = (R, +) under the usual topology? In this case we may assume
that our open neighbourhood takes the form U = (−ε, ε) for some ε > 0 (justify why it suffices
to assume this). It is then clear that we can take U1 = (−ε/2, ε/2). This construction clearly
relies on the metric nature of R, but one can abstract what we have done (without any mention
of the distance ε) by noting that our choice of U1 is an open set satisfying U1 ⊂ U1 − U1 ⊂ U .
Given an open neighbourhood U of the identity in a general topological group, we hope to
model the above approach by finding another open neighbourhood of the identity U1 satisfying
U1 ⊂ U1 U1−1 ⊂ U . Recalling Exercise 3.7, continuity of the division map ensures that there exist
open neighbourhoods V1 , V2 of the identity such that V1 V2−1 ⊂ U (this is the point at which
we have used the key axiom linking the topological and group-theoretic structure). Setting
U1 := V1 ∩ V2 , we have obtained an open neighbourhood of the identity satisfying U1 U1−1 ⊂ U .
Finally we would like to show that U 1 ⊂ U1 U1−1 , i.e. for any x ∈ U 1 there exist u1 , u2 ∈ U1
such that x = u1 u−1
2 , or equivalently xu2 = u1 . This is equivalent to saying that xU1 ∩ U1 6= ∅.
Since left-multiplication by x is an open map G → G, the set xU1 is an open neighbourhood
of x · e = x. It follows from the alternative definition of closure (Exercise 2.1(c)) that any
neighbourhood N of x satisfies N ∩ U1 6= ∅. This completes our proof.
Exercise 7.2. Show that a topological group is Hausdorff if and only if the singleton set {e}
is closed.
It is very useful to assume that a topological group is Hausdorff. The above proposition
shows that we get this for free under the mild assumption that our space is T0 , and if fact we
10See http://en.wikipedia.org/wiki/Separation_axiom for more on separation axioms.
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get the even stronger property of regularity. If a topological group is not Hausdorff, one can
obtain a Hausdorff group by quotienting out by the closure of the identity {e}.
Lemma 7.3 (Closure preserves subgroups). Given a (normal) subgroup H of a topological group
G, its closure H is also a (normal) subgroup of G. In particular, {e} is a normal subgroup of
G.
Proof. To show that H is a subgroup, it suffices to show that if x1 , x2 ∈ H then x1 x−1
2 ∈ H. The
latter follows if, for any open neighbourhood U of x1 x−1
,
we
have
U
∩
H
=
6
∅.
By
continuity of
2
division, there exist open neighbourhoods U1 , U2 of x1 and x2 (respectively) such that U1 U2−1 ⊂
U . By the definition of closure, for i = 1, 2 there exist yi ∈ Ui ∩ H. Hence y1 y2−1 ∈ U ∩ H.
The proof that normality of H implies normality of H follows similarly and is left as an
exercise.
Proposition 7.4. If G is a topological group and H a closed normal subgroup, then G/H is a
Hausdorff topological group under the quotient topology.
Proof. Exercise 5.3 establishes that G/H is a topological group under the quotient topology.
By Exercise 7.2, the space G/H is Hausdorff if and only if {eH} is closed. By Exercise 5.2 this
is true if and only if the inverse image q −1 {eH} = H is closed in G, and we have assumed this
is true.
Corollary 7.5. Given any topological group G the quotient G/{e} is a Hausdorff topological
group.
Putting all the results of this section together, we see that given an arbitrary topological group
G, we can obtain a Hausdorff group by quotienting out by the (usually very small) subgroup
{e}. When G is already Hausdorff, we have lost nothing in this procedure since {e} = {e} and
G/ {e} is topologically isomorphic to G. The resulting group G/{e} is always a T3 space, which
is a stronger separation property than being Hausdorff (T2 ).
Notice that a metric space is always a T3 space. In fact, being a metric space is in some sense
the ultimate separation axiom, as it allows one to quantify the amount of separation between
two points.
Exercise 7.3. Show that a metrisable space is both first countable and Hausdorff.
Not every first countable Hausdorff space is metrisable
Example 7.1 (The Sorgenfrey line). We define the lower limit topology on the real line R as
the topology generated by the set of half-open intervals
{[a, b) : a, b ∈ R} .
The resulting topological space is called the Sorgenfrey line
Exercise 7.4 (The Sorgenfrey line is Hausdorff).
a. Show that every open interval (a, b) can be written as a countable union of half-open intervals
[c, d).
b. Hence show that the open sets of the usual topology on R are also open in the Sorgenfrey
line.
c. Deduce that the Sorgenfrey line is Hausdorff.
Exercise 7.5 (The Sorgenfrey line is first countable).
a. Verify that the half-open intervals form a base for the topology on the Sorgenfrey line.
b. Deduce that the Sorgenfrey line is first countable if any x ∈ R has a countable collection of
open neighbourhoods {Ui }i∈N such that for any half-open interval [a, b) containing x there
exists i ∈ N such that Ui ⊂ [a, b).
c. Complete the proof that the Sorgenfrey line is first countable by constructing such a set
{Ui }i∈N .
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Exercise 7.6. Prove that the rationals form a countable dense subset of the Sorgenfrey line.
Hence the Sorgenfrey line is separable.
Exercise 7.7. Prove that any separable metrisable space is second countable.
Proposition 7.6. The Sorgenfrey line is separable and not second countable, hence not metrisable.
Proof. One can check that the rationals form a countable dense subset of the Sorgenfrey line.
Suppose that {Ui }i∈N is a countable base for the topology on the Sorgenfrey line. For each
x ∈ R the half-open line [x, x + 1) is an open neighbourhood for x, hence there exists i(x) ∈ N
such that
x ∈ Ui(x) ⊂ [x, x + 1).
It follows that x = inf Ui(x) , so that the map x 7→ i(x) is an injective function from R to N.
This is impossible, since R is uncountable.
It is a remarkable fact that the rigidity imposed on the topology by the group structure of
a topological group ensures that the mild assumptions of first countability and the Hausdorff
axiom together guarantee metrisability.
Theorem 7.7 (Birkhoff–Kakutani). A topological group is metrisable if and only if it is first
countable and Hausdorff.
Proof. See the wonderful blog of Terry Tao http://terrytao.wordpress.com/2011/05/17/
the-birkhoff-kakutani-theorem/. The interested student could give an exposition of this
proof for their project.
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