Contents

Chapter 1. Topological Groups 1 1. Topological spaces 1 2. Topological groups 2 3. Homogeneous spaces 5 4. Separation in topological groups 8 5. Examples of topological groups and homogeneous spaces 10 5.1. The general linear group GL(n, 𝔽) 10 5.2. The classical linear groups 12 5.3. The sphere S^{n} 15 5.4. The flag manifolds 15 5.5. The Euclidean motion group 16

CHAPTER 1 Topological Groups

In this chapter we discus the basic theory of topological groups, but only as far as we will need later on. A topological groups carries two different structure; the algebraic structure of a group, and the analytic/geometric structure of a topological space. We talk about a topological group if those two structures are compatible, i.e., the group operations are continuous. We will denote the identity element of a group G by 1, e, or e_{G} if we need to stress the group.

1. Topological spaces

We expect that the reader is familiar with the most basic notions in topology even if we recall some of them here. Let X be a topological space and N ⊂ X. Then the interior of N, denoted by N^{o}, is the biggest open set contained in N. Thus N^{o} = {x ∈ N | ∃U ∋ x open, U ⊂ N}. If N ⊂ X then N̄ denotes the closure of N, i.e., the smallest closed subset of X containing N. Thus

N̄ = ⋂_{N⊂C C closed}C = X \ (X \ N)^{o} .

Let x ∈ X. A subset N ⊂ X is called a neighborhood of x if x ∈ N^{o}. We denote by N (x) the collection of all neighborhoods of x. A map f : X → Y, where X and Y are topological spaces is continuous if f^{−1}(U) ⊂ X is open for every open subset U ⊂ Y.

Lemma 1.1. Let X and Y be topological spaces and f : X → Y. Then f is continuous if and only if for each x_{0} ∈ X and U ∈ N (x_{0}) there exists a V ∈ N (x_{0}) such that f(V) ⊂ U.

If X and Y are topological spaces, denote by C(X, Y) the set of continuous maps from X to Y. If Y = ℂ then we write simply C(X) for C(X, ℂ). Let C_{c}(X) be the set of compactly supported functions from X to ℂ. The map f is called a homeomorphism if f is bijective and both f and f^{−1} are continuous. In this case we say that X and Y are homeomorphic or just isomorphic. A map f : X → Y is open if f(U) is open in Y for each open set U ⊂ X. Thus f is a homeomorphism if and only if f is injective, surjective, continuous, and open. If X and Y are topological spaces then X × Y denotes the set {(x, y) | x ∈ X, y ∈ Y } with the topology generated by all the sets U × V, U open in X and V open in Y. This topology is called the product topology. There are several separation axioms, some of which we list here.

Definition 1. In the following X denotes a topological space. Further more x and y are two distinct points in X.

T_{0} [Kolmogoroff] The space X is called T_{0}-space if there exists an open set U in X containing exactly

one of the points x and y. T_{1} [Fréchet] The space X is called T_{1}-space if there is a U ∈ N (x) such that y ∉ U. T_{2} [Hausdorff] The space X is called T_{2} or Hausdorff if there exists U ∈ N (x) and V ∈ N (y) such

that U ∩ V = ∅. T_{3} [Vietoris] The space X is called T_{3} if for every closed set A ⊂ X and every x ∉ A there exists an

open sets U, and V ∈ N(x) such that A ⊂ U and U ∩ V = ∅. T_{3a} Let A and x be as in T_{3}. Then there exists a continuous function f : X → [0, 1] such that f(x) = 0

and f|A = 1. T_{4} [Tietze] The space X is called T_{4} space if for any two closed and disjunct sets A, B ⊂ X there

exists open sets U, V ⊂ X such that A ⊂ U, B ⊂ V and U ∩ V = ∅.

Finally we make the following definitions:

Definition 2. A topological space X is called regular if it is T_{0} and T_{3}. It is called normal if it is T_{1} and T_{4}. It is called completely regular if it is T_{1} and T_{3a}.

It is clear that X is T_{1} if and only if every finite subset of X is closed. The following implications are also obvious:

normal ⇒ regular ⇒ T_{2} ⇒ T_{1} ⇒ T_{0} .

Lemma 1.2. Let X, Y be two topological spaces and f : X → Y continuous. If Y is T_{1}, then {x ∈ X | f(x) = y} = f^{−1}(y) is closed for all y ∈ Y.

Theorem 1.3 (Urysohn). Let X be a T_{4} space, A, B ⊂ X non-empty, closed and disjunct. Then there exits a continuous map f : X → [0, 1] such that f|A = 0, and f|B = 1.

Definition 3. X is called locally compact if N (x) contains a compact set for each x ∈ X.

Lemma 1.4. Let X be a Hausdorff space and K ⊂ X compact. Then K is closed. If L is a closed subset of K then L is compact.

Lemma 1.5. Let X be a locally compact, Hausdorff, and T_{3a}. Then the following holds:

1. Let x ∈ X and U ∈ N (x). Then there exists a compact L ∈ N (x) such that L ⊂ U.

2. Let x ∈ X, U ∈ N (x) open, and K ⊂ U compact, then there exists a f ∈ C_{c}(X) such that 0 ≤ f ≤ 1, f|K = 1, and Supp(f) ⊂ U.

Proof. (1)Let K ∈ N (x) be compact. Let U ∈ N (x) be open. Let V = (K ∩ U)^{o}. Then V ∈ N (x) is open. Hence X \ V is closed. As X is T_{3a} there exists f ∈ C(X) such that 0 ≤ f ≤ 1, f(x) = 0 and f|X \V = 1. Let g = 1−f. Then g(x) = 1, 0 ≤ g ≤ 1, and L := {x | g(x) ≥ 1/2} ⊂ V ⊂ U. Furthermore L ⊂ K is closed and hence compact. As x ∈ {y ∈ X | g(x) > 1/2} ⊂ L is open, it follows that L ∈ N (x).

(2) For each y ∈ K choose g_{y} ∈ C(X) such that g_{y}(y) = 1 and Supp(g_{y}) ⊂ U as in the above construction. Let V_{y} = {z ∈ X | g_{y}(z) > 1/2}. Then V_{y} is open. Hence there exists finitely many y_{j}, j = 1, …, n such that with V_{j} = V_{y_{j}} we have K ⊂ ∪_{j=1}^{n}V_{k}. Let f = min{2 ∑_{j=1}^{n} g_{y_{j}}, 1}.

Theorem 1.6. Let X be a locally compact Hausdorff T_{3a} space topological space. Let ∅ ≠ K ⊂ X be comapct and U open such that K ⊂ U. Then there exists a compactly supported continuous function f such that f(x) = 1 for all x ∈ K and Supp(f) ⊂ U.

Proof. The set A = X \ U is closed as U is open. Hence for each x ∈ A there exists a function f ∈ C(X), 0 ≤ f ≤ 1 such that f_{x}(x) = 0 and f_{x}|A = 1. Let g_{x} = 1 − f_{x}. Then g_{x}(x) = 1 and g_{x}|A = 0.

2. Topological groups

In this section G denotes a group with identity element e.

Definition 4. Let G be a group and a topological space. We call G for a topological group if the maps 1. G × G ∋ (x, y) ↦ xy ∈ G

2. G ∋ x → x^{−1} ∈ G

are continuous.

Examples ℝ^{n} and the Heisenberg group. If G is a topological group, then the inversion ι : G → G, ι(x) = x^{−1} is a homeomorphism with inverse ι. In general our groups will be noncommutative. It is therefore necessary to distinguish between the left translation λ(a)(x) = ax and the right translation ρ(b)(x) = xa. For simplicity we will also use the notation λ_{a}(x) or λ(a)x and ρ_{a}(x) or ρ(a)x.

Lemma 1.7. Let G be a topological group and let a, b ∈ G. Define λ(a), ρ(b) : G → G by λ(a)(x) = ax and ρ(b)(x) = xb. Then λ(a) and ρ(b) are homeomorphisms with inverse λ(a^{−1}) respectively ρ(a^{−1}). Furthermore λ(e) = ρ(e) = id, λ(ab) = λ(a) ∘ λ(b) and ρ(ab) = ρ(a) ∘ ρ(b).

Proof. We can write λ(a) in the following way

\begin{matrix}G & → & G × G & → & G \\ x & ↦ & (a, x) & ↦ & ax \end{matrix} .

The first map is continuous by the definition of the product topology and the second map is continuous by part 1 in the definition of a topological groups. Hence λ(a) : G → G is continuous. We have

λ(ab)(x) = (ab)x = a(bx) = λ(a)(λ(b)(x)) = (λ(a) ∘ λ(b))(x).

Obviously λ(e)(x) = ex = x, so λ(e) is the identity map. Finally λ(a^{−1})λ(a) = λ(a^{−1}a) = id = λ(a) ∘ λ(a^{−1}). It follows that λ(a) is a homeomorphism with inverse λ(a^{−1}). The case ρ(a) is treated similarly.

Let A, B be non-empty subsets of G, and let x ∈ G. Define

AB = {ab | a ∈ A, b ∈ B}
A^{−1} = {a^{−1} | a ∈ A}
xA = {x}A = {xa | a ∈ A} = λ(x)(A)
Ax = A{x} = {ax | a ∈ A} = ρ(x)(A)
A^{2} = AA and A^{n+1} = AA^{n} , n ≥ 1 .

A set A is called symmetric if A = A^{−1}. Lemma 1.7 implies that if a ∈ G and U is an open neighborhood of the identity then aU and Ua are open neighborhoods of a.

Lemma 1.8. Assume that A, B ⊂ G are non-empty and that B is open. Then AB, BA, and B^{−1} are open.

Proof. We have AB = ⋃_{a∈A} aB and each aB is open according to Lemma 1.7. That BA is open follows in the same way. The last statement follows directly from the definition.

Corollary 1.9. Let U ∈ N (e), hen there exists a symmetric open set V ∈ N (e) such that V ⊂ U.

Proof. Let V := U^{o} ∩ U^{o−1}. Then V is open and e ∈ V because e = e^{−1} ∈ U^{o} ∩ U^{o−1}.

Corollary 1.10. Let a ∈ G. Then N (a) = aN (e) = N (e)a.

Proof. This follows immediately from the fact that λ_{a} and ρ_{a} are homeomorphism.

Theorem 1.11. Let G be a group and let U be a non-empty family of subsets of G such that the following holds:

1. e ∈ U for all U ∈ U;
2. If U, V ∈ U then U ∩ V ∈ U;
3. If U ∈ U then there is a V ∈ U such that V V^{−1} ⊂ U;
4. If U ∈ U and a ∈ G then aUa^{−1} ∈ U.

Call a set A in G open if for each a ∈ A there exits a U ∈ U such that aU ⊂ A. Let τ be the collection of all open sets. Then τ is a topology on G that makes G into a topological group.

Proof. It is clear that ∅ and G are both open. Let U_{1}, …, U_{r} be open sets. We have to show that U_{1}∩…∩U_{r} is open. If U_{1}∩…∩U_{r} is empty then there is nothing to prove. Otherwise let a ∈ U_{1}∩…∩U_{r}. Choose V_{j} ∈ U such that aV_{j} ⊂ U_{j}. Then by part 2 and induction it follows that V := V_{1} ∩ … ∩ V_{r} ∈ U. Finally aV ⊂ aV_{j} ⊂ U_{j} for j = 1, …, r. Hence aV ⊂ U_{1} ∩ … ∩ U_{r} and it follows that U_{1} ∩ … ∩ U_{r} is open. Let {U_{i}}_{i∈I} be a collection of open sets. Let a ∈ ∪_{i∈I}U_{i}. Then there exists a i_{0} such that a ∈ U_{i_{0}}. Let V ∈ U be such that aV ⊂ U_{i_{0}}, then aV ⊂ ∪_{i}U_{i} and it follows that ∪_{i}U_{i} is open. We have now shown that τ is a topology on G, but we still have to show that G is a topological group. According to exercise 2 it suffice to show that the map G×G ∋ (a, b) ↦^{ψ} ab^{−1} ∈ G is continuously a, b ∈ G and let U be an open set containing ab^{−1}. Let W ∈ U be such that ab^{−1}W ⊂ U. Let V be such that V V^{−1} ⊂ b^{−1}W b. For x ∈ aV and y ∈ bV we get:

xy^{−1} ∈ aV V^{−1}b^{−1} ⊂ ab^{−1}W ⊂ U .

It follows that ψ is continuous.

Let G and H be groups. A subgroup N ⊂ G is a normal subgroup if aNa^{−1} ⊂ N for all a ∈ G. A map φ : G → H is called a homomorphism if φ(ab) = φ(a)φ(b) for all a, b ∈ G. We notice that φ(G) is a subgroup of H and Ker(φ) = φ^{−1}(e) = {a ∈ G | φ(a) = e} is a normal subgroup of G. If G and H are topological groups, then G and H are isomorphic if there exists a homomorphism φ : G → H which is also a homeomorphism. In that case φ^{−1} : H → G is also an isomorphism.

Lemma 1.12. Let G and H be topological groups, and φ : G → H a homomorphism of groups. Then the following holds:

1. φ is continuous if and only if φ is continuous at e ∈ G.

2. If φ is continuous and H is T_{1} then Ker(φ) is a closed normal subgroup of G.

Proof. 1) The claim follows from u ↦ φ(au) = λ_{φ(a)}φ(u) and φ(u) = λ_{φ(a)^{−2}}φ(au) and the fact that λ_{φ(a)} is a homeomorphism.

2) This follows from Lemma 1.2.

We will see several examples of topological groups in the next section, but here are two simple examples.

Example 1.1 (Banach spaces). Let B any topological vector space. Then B is a topological group with respect to addition. In particular this holds for B an Banach space. As B is abelian it follows that the maps λ(x) and ρ(x) agree and are just the usual translation by x, λ_{x}(y) = y + x. Furthermore

AB = {x + y | x ∈ A, y ∈ B} = A + B
A^{−1} = −A
A^{n} = nA .

If continuous homomorphism in this case is a ℝ-linear map. As a special example this shows that all the spaces ℝ^{n} and ℂ^{n} are topological groups.

Example 1.2 (ℂ^{∗n} and the torus 𝕋^{n}). Let ℂ^{∗} = {z ∈ ℂ | z ≠ 0}. Then ℂ^{∗n} is a group under componentwise multiplication

(z_{1}, …, z_{n}) · (w_{1}, …, w_{n}) = (z_{1}w_{1}, …, z_{n}w_{n}) .

The inverse is

(z_{1}, …, z_{n})^{−1} = (z_{1}^{−1}, …, z_{n}^{−1}) .

Consider the relative topology (with respect to ℂ^{n}) on ℂ^{∗n}. Thus the topology is given by the norm ||z|| = \sqrt{∑ |z_{j}|^{2}}. Let a, b, s, t ∈ ℂ^{∗}^{n}. Then

||st − ab|| = ||st − sb + sb − ab||
≤ ||s|| ||t − b|| + ||b|| ||s − a||
≤ max{||s||, ||b||}(||t − b|| + ||s − a||) .

Let 0 < ϵ < 1 and let R = max{||a||, ||b||} > 0. Let ϵ′ = ϵ/2(R + 1). If ||s − a||, ||t − b|| < ϵ′ then ||s|| < R + 1 and

||st − ab|| < (R + 1)2ϵ′ = ϵ .

Hence the multiplication is continuous. We also have

||z^{−1} − a^{−1}|| = \sqrt{∑ |\frac{1}{z_{j}} − \frac{1}{a_{j}} |^{2}}

= \sqrt{∑ |\frac{a_{j} − z_{j}}{z_{j}a_{j}} |^{2}}
≤ C||z − a||

where C = max{1/|z_{j}a_{j}|} < ∞. It follows that ℂ^{∗n} is a locally compact Hausdorff topological group. Let 𝕋^{n} be the n-dimensional torus:

𝕋^{n} = {z ∈ ℂ^{n} | z = (z_{1}, …, z_{n}) ∀j : |z_{j}| = 1} .

Then 𝕋^{n} is a closed subgroup of ℂ^{∗n} and hence also a locally compact, Hausdorff topological group.

Example 1.3 (The Heisenberg group). A further example is the Heisenberg group ℂ^{n} + ℝ with multiplication

(z, t) · (w, s) = (z + w, s + t + \frac{1}{2}ω(z, w))

where ω is a symplectic form on ℝ^{2n} = ℂ^{n}.

3. Homogeneous spaces

The importance of many groups comes from their connection to geometry as a symmetry group of a geometric structure and their action on sets or manifolds. We recall therefore the main definitions. Let G be a group and X a set. A (left) action of G on X is a map μ : G × X → X, μ(g, x) = g · x, such that the following holds:

1. e · x = x for all x ∈ X, i.e., μ(e, ·) is the identity map on X;

2. (ab) · x = a · (b · x) for all a, b ∈ G and all x ∈ X. Similarly we define a right-action as a map G × X → X, (g, x) ↦ x · g such that x · e = x for all x ∈ X and x · (ab) = (x · a) · b. If μ is a left action then we can define a right action μ_{R} by μ_{R}(a, x) = a^{−1} · x. If x ∈ X and a ∈ G let

μ_{x} : G → X, g ↦ g · x

and

λ_{a} : X → X, x ↦ a · x .

We will also use the notation λ(a) = λ_{a}. If G is a topological group and X is a topological space, then the action is continuous if μ is continuous. In this case we say that X is a (continuous) G-space. The action is separately continuous if for each x ∈ X and a ∈ G the maps μ_{x} and λ_{a} are continuous. Let x ∈ X, Y ⊂ X, and B ⊂ G be non-empty. Then

Y^{B} : = {x ∈ Y | ∀g ∈ B : g · x = x} ,
B^{Y} : = {b ∈ B | ∀x ∈ Y : b · x = x} ,
B^{x} : = B^{{x}} = {g ∈ G | g · x = x} .

The space X is homogeneous if there exists a x ∈ X such that X = G · x. We phrase this also by saying that G acts transitively on X. In that case X = G · y for every y ∈ X. If X and Y are G-spaces and φ : X → Y then we call φ a G-map if φ(a · x) = a · φ(x) for all a ∈ G and all x ∈ X. If G acts separately continuously on X and Y then we will always require φ to be continuous. Denote by M_{G}(X, Y) the set of all G-maps X → Y. If X = Y then M_{G}(X) := M_{G}(X, X). If M_{G}(X, Y) contains an isomorphism then we say that X and Y are isomorphic, X ≃_{G} Y.

Lemma 1.13. Assume that G acts separately continuously on X. Then the following holds: 1. Let a ∈ G then the map λ(a) : X → X, x ↦ a · x is a homeomorphism with inverse λ(a^{−1}).

2. Assume that X is T_{2}. If Y ⊂ X is closed then Y^{G} is closed in X.

3. If B ⊂ G is closed in G and Y ⊂ X is non-empty, then B^{Y} ⊂ G is closed in G.

4. G^{Y} is a subgroup of G.

Proof. 1) By assumption the map λ(a) is continuous for each a ∈ G. Hence λ(a^{−1}) : X → X is continuous and λ(a)λ(a^{−1}) = λ(aa^{−1}) = λ(e) = id. In the same way it follows that λ(a)^{−1}λ(a) = id.

2) Let g ∈ G and x ∈ X such that g · x ≠ x. Then there exists open sets U ∈ N (x), V ∈ N (g · x) such U ∩ V = ∅. As λ(g) : X → X is continuous it follows that there exists a open U_{1} ∈ N (x) such that λ(g)(U_{1}) ⊂ V. If y ∈ U_{1} is such that g · y = y then that would imply that y ∈ U_{1} ∩ V ⊂ U ∩ V = ∅, which is a contradiction. Hence V ⊂ {x ∈ X | g · x ≠ x} and it follows that {x ∈ X | g · x ≠ x} is open. Hence {x ∈ X | g · x = x} is closed. As Y is closed in X and

Y^{B} = {x ∈ Y | ∀g ∈ B : g · x = x} = ( ⋂_{g∈B}{x ∈ X | g · x ≠ x}) ∩ Y

it follows that Y^{B} is closed in X

3) Assume that b · x ≠ x. Let U ∈ N (x), W ∈ N (b · x) be open and such that U ∩ W = ∅. As

G ∋ a ↦ a · x ∈ X

is continuous it follows that we can find a neighborhood V ∋ b such that V · x ⊂ W. Assume that a ∈ V and a · x = x. Then a · x ∈ W ∩ U = ∅, a contradiction. Hence {g ∈ G | g · x ≠ x} is open in G. It follows that {g ∈ G | g · x = x} is closed in G. As B is closed it G if follows that

B^{Y} = ⋂_{x∈Y} {b ∈ G | b · x = x} ∩ B

is closed in G.

4) Obviously e ∈ G^{Y}. Let x ∈ Y and a, b ∈ G^{Y}. Then (ab) · x = a · (b · x) = a · x = x and

Define a equivalence relation ∼ on X by

x ∼ y ⟺ ∃g ∈ G : g · x = y .

If x ∈ X let [x] denote the equivalence class containing x. Then G\X = {[x] | x ∈ X}. We define X/G in the same way if G acts from the right. Let H ⊂ G be a subgroup. Then H acts on G by g · h = gh, g ∈ G, h ∈ H. Then G/H is the space of all subset of G of the form {aH | a ∈ G}. The group G acts on G/H by a · (bH) = (ab)H. We denote this action sometimes by L_{a}, i.e., L_{a}(bH) = (ab)H. Let κ : G → G/H be the canonical map a ↦ aH. Assume that G is a topological group. Define a topology on G/H by requiring a set U ⊂ G/H to be open if and only if κ^{−1}(U) ⊂ G is open (see exercise 7). We call this topology the quotient topology on G/H.

Lemma 1.14. Let G be a topological group and H a subgroup. Then the following holds:

1. The canonical map κ : G → G/H is continuous and open.

2. The action on G on G/H is continuous.

3. G/H is Hausdorff if and only if H is closed in G.

Proof. 1) The quotient map is continuous by definition. Let U ⊂ G be open. Then

κ^{−1}(κ(U)) = UH = ⋃_{h∈H} Uh

which is open as Uh is open in G for all h ∈ H. Hence κ(U) ⊂ G/H is open by the definition of the topology on G/H.

2) Consider the commutative diagram

\begin{matrix}G × G & → & G & → & G/H \\ ↓ & & ↗ \\ G × G/H \end{matrix} \begin{matrix}(a, b) & → & ab & → & abH \\ ↓ & & ↗ \\ (a, bH) \end{matrix} .

The horizontal and vertical maps are continuous. Hence the map (a, bH) ↦ a(bH) is also continuous.

3) Assume that H is Hausdorff. Then κ(e) = {H} ∈ G/H is closed in G/H. Therefore

κ^{−1}(G/H \ {H}) = G \ H

is open in G. It follows that H is closed. Suppose that H is closed in G. Let μ : G × G → G be the map (a, b) ↦ a^{−1}b. Then μ^{−1}(H) is closed in G. Assume that aH ≠ bH. Then μ(a, b) = a^{−1}b ∉ H. As H is closed in G there exists U, V ⊂ G open, such that a ∈ U, b ∈ V, and U × V ∩ m^{−1}(H) = ∅. Furthermore κ(U) is an open neighborhood of aH and κ(V) is an open neighborhood of bH. Assume that x ∈ κ(U) ∩ κ(V). Then we can write

x = cH = dH

with c ∈ U and d ∈ V. It follows that c^{−1}d = μ(c, d) ∈ H or (c, d) ∈ μ^{−1}(H)∩U ×V = ∅, a contradiction. Hence G/H is Hausdorff.

It is now a natural question to ask: Which G-spaces are of the form G/H for a subgroup H of G? We will not attempt to answer this question in its most generality. First of all it is clear that X need to be homogeneous. Assume that G acts transitively and separately continuously on X and that x ∈ X. Let H = G^{x} = {a ∈ G | a · x = x}. Then we have a map

G → X, a ↦^{κ_{x}} a · x

which factors to a map

G/H ∋ aH ↦^{π_{x}} a · x ∈ X,

i.e., κ_{x} = π_{x} ∘ κ. As the action of G is continuous and because of the definition of the topology on G/H it follows that π_{x} is continuous and bijective. Thus G/H ≃ X if and only if π_{x} is open. This is in general not the case. Notice that π_{x} is open if and only if κ_{x} is open. We prove that π_{x} is a homeomorphism under the assumptions that G is locally compact Lindelöff space and X is a locally compact.

Definition 5. Let X be a topological space. Then X is locally compact if N (x) contains a compact set for each x ∈ X. The space X is called a Lindelöff space if every open covering {U_{i}}_{i∈I} of X has a countable subcoverging, i.e., there exists a subset J ⊂ I such that J is isomorphic to a subset of ℕ and X = ∪_{j∈J}U_{j}.

Lemma 1.15. Let X and Y be topological space and K ⊂ X compact. Then the following holds: 1. If L ⊂ K is closed, then L is compact.

2. If f : X → Y is continuous then f(K) is compact in Y.

3. If X is Hausdorff then K is closed.

Lemma 1.16. Let G be a locally compact topological group. Let a ∈ G and U ∈ N (a). Then there exists a compact K ∈ N (a) such that K ⊂ U.

Proof. We can assume that a = e because λ_{a} is a homeomorphism. We can also assume that U is open. Choose V ∈ N (a) symmetric and such that V^{2} ⊂ U ∩ K^{o}. If V̄ ⊂ V^{2} then V̄ ⊂ K. As each closed subset of a compact set is compact it follows that V̄ is compact. Furthermore V̄ ⊂ U and the Lemma would follows. We now show that V̄ ⊂ V^{2}. Let x ∈ V̄. Then V^{−1}x ∈ N (x). Hence V^{−1}x ∩ V ≠ ∅. But then x ∈ V^{2}.

For the proof of our next theorem we need the Baire’s category theorem.

Definition 6. Let X be a topological space. A set N ⊂ X is nowhere dense if N̄^{o} = ∅. The set N is of the first category in X if N is a countable union of nowhere dense sets. A set N is of the second category if N is not for the first category.

A obvious consequence of the definition is that a closed set A is of the first category if and only if A^{o} = ∅. Furthermore a countable union of set of the first category is of the first category. But notice that a set of the first category can be ”big”. As a example ℚ = ∪_{q∈ℚ}\{q\} is of the first category, but \overline{ℚ}^{o} = ℝ!

Theorem 1.17 (Baire’s Theorem). Let X be a locally compact Hausdorff space or a complete metric space. Then the intersection of every countable collection of dense open subsets of X is dense in X.

Corollary 1.18. Let X be a locally compact Hausdorff space or a complete metric space. Let {A_{j}} be a collection of closed subsets of X. If ∪A_{j} = X then there exists a j such that A_{j}^{o} ≠ ∅.

Proof. Let E_{j} = X \ A_{j}. Then E_{j} is open. We have

⋂_{j} E_{j} = X \ ⋃_{j} A_{j} = ∅ .

Hence at least on E_{j} is not dense in E. It follows that A_{j} contains an open set, or A_{j}^{o} ≠ ∅.

Theorem 1.19. Let G be a Hausdorff locally compact topological group and X a Hausdorff locally compact space. Assume that G acts separately continuously and transitively on X and that G is a Lindelöff space. Then π_{x} : G/G^{x} → X is a homeomorphism for all x ∈ X and G acts continuously on X.

Proof. We only have to show that κ_{x} is open for one x ∈ X (see exercise 9). Let H = G^{x}. Let U ⊂ G be open. We have to show that κ_{x}(U) is open. Let g ∈ U and y = κ_{x}(g) = gH ∈ κ_{x}(U). Let V ∈ N (e) be symmetric, compact and such that gV^{2} ⊂ U. We have G = ∪_{a∈G}aV^{o}. As G is a Lindelöff space there is a countable set J, such that G = ∪_{j}a_{j}V^{o}. As the map κ_{x} is surjective it follows that

X = ⋃_{j∈J} κ_{x}(a_{j}V) = ⋃_{j∈J} a_{j} · κ_{x}(V).

Using that κ_{x} is continuous it follows that κ_{x}(a_{j}V) is compact and hence closed. By Corollary 1.18 there is a j such that κ_{x}(a_{j}V)^{o} ≠ ∅. As λ(a_{j}) : X → X is a homeomorphism it follows that κ_{x}(V)^{o} ≠ ∅. Let z ∈ κ_{x}(V)^{o}. Then z = v·x for some ν ∈ V. But then x ∈ v^{−1}·z is an interior point in V ·κ_{x}(V) = κ_{x}(V^{2}). As y = g · x and λ_{g} is a homeomorphism we finally get

y = gv^{−1} · z ∈ (g · κ_{x}(V^{2}))^{o} = κ_{x}(gV^{2})^{o} ⊂ κ_{x}(U) .

Thus y is an interior point in κ_{x}(U). As y was an arbitrary point in κ_{x}(U) if follows that κ_{x}(U) is open and hence κ_{x} is an open map.

4. Separation in topological groups

Our aim in this section is to show that any T_{0} topological group is completely regular. We start with two simple consequences of Lemma 1.14.

Lemma 1.20. Let G be a topological group and N a closed normal subgroup. Then G/H is a Hausdorff topological group.

Lemma 1.21. Let G be a topological group. Then the following are equivalent:

1. G is Hausdorff;

2. G is T_{0}

3. G is T_{1}.

4. The set {e} is closed in G,

5. The set {a} is closed in G for every a ∈ G, i.e., G is T_{1}.

Let U = U_{0} ∈ N (e) be open. If U_{n} is defined choose U_{n+1} ∈ N (e) open, symmetric and such that U_{n+1}^{2} ⊂ U_{n}. Let 𝔹 ⊂]0, 1] be the set of rational numbers of the form r2^{−n}, r, n ∈ ℕ_{0}, 1 ≤ r < 2^{n}.

Lemma 1.22. There exists a sequence of open sets U(r) ∈ N (e) such that U(2^{−n}) is symmetric for all n and

1. U(2^{−n}) ⊂ U_{n} for all n ∈ ℕ_{0};
2. U(2^{−n})U(k2^{−n}) ⊂ U((k + 1)2^{−n}) for all n ∈ ℕ_{0} and 1 ≤ k < 2^{n};
3. If r, s ∈ 𝔹 and r ≤ s then U(r) ⊂ U(s).

Proof. Let U_{n}, n ∈ ℕ_{0}, be as above. Let U(1) = U(2^{0}) := U_{0}. Let n ≥ 1 and assume that we have defined U(k2^{−j}) for all j < n and all k ≤ 2^{j}. Let 2 ≤ k < 2^{n}. Then k2^{−n} = (\frac{k−1}{2}) 2^{−n+1} and U(\frac{k−1}{2}2^{−n+1}) is defined. We can therefore define

U(2^{−n}) : = U_{n} ,
U(k2^{−n}) : = \{ \begin{matrix}U (\frac{k}{2} 2^{−n+1}) & if k is even \\ U(2^{−n})U (\frac{k−1}{2}2^{−n+1}) & if k is odd \end{matrix} .

Obviously U(2^{−n}) ⊂ U_{n} is symmetric. Suppose that r = k2^{−n} = l2^{−m}, with m < n. Then k = l2^{n−m} is even. Hence

U(k2^{−n}) = U(l2^{n−m−1}2^{−n+1}) = U(\frac{l}{2}2^{−m+1}) .

Hence U(r) is well defined.

(2) If n = 1 then k = 1 and (2) holds. Let n ≥ 2. If k = 1 then U(2^{−n})U(2^{−n}) = U_{n}^{2} ⊂ U_{n−1} = U(2^{−n+1}) = U(2 · 2^{−n}). Hence (2) holds for k = 1. Let 2^{n} > k ≥ 2 and assume that (2) is holds for all l < k. If k is even then k + 1 is odd. Thus U((k + 1)2^{−n}) = U(2^{−n})U (\frac{k}{2} 2^{−n+1}). It follows that

U(2^{−n})U(k2^{−n}) = U(2^{−n})U (\frac{k}{2} 2^{−n+1}) = U ((k + 1)2^{−n}) . (4.1)

If k is odd, then k − 1 is even and by definition and (4.1) we get

U(2^{−n})U(k2^{−n}) = U(2^{−n})U(2^{−n})U (\frac{k − 1}{2}2^{−n+1})
⊂ U(2^{−n+1})U (\frac{k − 1}{2}2^{−n+1})
⊂ U ((\frac{k − 1}{2} + 1) 2^{−n+1}) = U((k + 1)2^{−n}) .

(3) Let r, s ∈ 𝔹. Write r = k2^{−n} and s = h2^{−m} with 1 ≤ r < 2^{n} and 1 ≤ h < 2^{m}. Assume that n < m. Then k2^{m−n} < 2^{m} and r = (k2^{m−n})2^{−m}. Hence we can assume that m = n. Suppose that r < s. Then

U(r) = U(k2^{−n}) ⊂ U((k + 1)2^{−n}) ⊂ … ⊂ U(h2^{−n}) = U(s) .

Definition 7. A function f : G → ℂ is called uniformly continuous from the right if for all ϵ > 0 there is a U ∈ N (e) such that for all x, y ∈ G with xy^{−1} ∈ U we have

|f(x) − f(y)| < ϵ .

f is called uniformly continuous from the left if for all ϵ > 0 there is a U ∈ N (e) such that for all x, y ∈ G with x^{−1}y ∈ U we have

|f(x) − f(y)| < ϵ .

f is called uniformly continuous if it is uniformly continuous from left and right.

Notice that a function that is uniformly continuous from the left or the right is obviously continuous.

Lemma 1.23. Let U_{n}, n ∈ ℕ_{0}, be a sequence of symmetric open sets in N (e) with U_{n+1}^{2} ⊂ U_{n}. Let H be a subgroup of G. Then there exists a f ∈ C(G) such that:

1. f(G) ⊂ [0, 1].

2. f(e) = 0;

3. If a^{−1}b ∈ H then f(a) = f(b).

4. If n ∈ ℕ_{0} and f(x) < 2^{−n} then x ∈ U_{n}H.

5. f is uniformly continuous from the right.

Proof. For x ∈ G let F_{x} := {r ∈ 𝕌 | x ∈ U(r)H}. If F_{x} = ∅ let f(x) = 1. Otherwise let

f(x) := inf F_{x} ≥ 0.

Then (1) holds. As e ∈ U(r) for all r it follows that f(1) = 0. Thus (2) holds. If aH = bH then F_{a} = F_{b} and (3) holds. Suppose now that f(x) < 2^{−n}. Then there exists a s < 2^{−n} such that x ∈ U(s)H. By Lemma 1.22, 3, it follows that x ∈ U(2^{−n})H = U_{n}H.

(5) Let ϵ > 0. Choose n ∈ ℕ_{0} such that 2^{−n} < ϵ/2. Assume that x, y ∈ G are such that xy^{−1} ∈ U_{n}. If F_{y} = ∅ then f(y) = 1. Then f(x) ≤ 1 = f(y). Hence |f(y) − f(x)| = f(y) − f(x) = 1 − f(x). If 1 − f(x) ≥ ϵ then f(x) ≤ 1 − ϵ < 1 − 2^{−n+1} = 2^{−n+1}(2^{n−1} − 1). This implies in particular that n ≥ 2 and x ∈ V ((2^{n−1} − 1)2^{−n+1})H ⊂ U_{n−1}H. As U_{k} is symmetric for all k and U_{n} ⊂ U_{n−1} it follows that

y ∈ U_{n}x ⊂ U_{n−1}^{2}H .

As n ≥ 2 this would contradict F_{y} = ∅. Hence 1 − f(x) < ϵ. Assume now that F_{y} ≠ ∅. Let k ≤ 2^{n} be minimal such that y ∈ U(k2^{−n})H. Then y ∉ U((k − 1)2^{−n})H. Thus (k − 1)2^{−n} ≤ f(y) ≤ k2^{−n}. If k = 2^{n}. Then f(x) ≤ 1 and

f(x) − f(y) < 1 − (2^{n} − 1)2^{−n} = 2^{−n} < ϵ .

In case k < 2^{n} we get

x = (xy^{−1})y ∈ U_{n}U(k2^{−n})H ⊂ U((k + 1)2^{−n}) H.

Hence f(x) ≤ (k + 1)2^{−n}. Hence

f(x) − f(y) < (k + 1)2^{−n} − (k − 1)2^{−n} = 2 · 2^{−n} < ϵ .

As U_{n} is symmetric it follows in the same way that f(y) − f(x) < ϵ. Hence |f(x) − f(y)| < ϵ for xy^{−1} ∈ U_{n}.

Theorem 1.24. Let H be a subgroup of G. Then G/H is a T_{3a} space. If H is closed in G then G/H is completely regular.

Proof. Let A ⊂ G/H be closed and let x = aH ∈ G/H. We have to show that there is a continuous function g : G/H → [0, 1] such that g|A = 1 and g(x) = 0. Let U := a^{−1}(G \ κ^{−1}(A)). Then U ∈ N (e) is open. Choose the sequence U_{n}, n ∈ ℕ_{0}, with U_{0} = U and let f : G → [0, 1] be as in Lemma 1.23. Let g(bH) = f(a^{−1}b). Then g is well defined and continuous. We have

g(aH) = f(e) = 0 .

Let bH ∈ A. Then a^{−1}b ∉ U. Hence f(a^{−1}b) = 1. It follows that g|A = 1. If H is closed then G/H is T_{2}. In particular G/H is T_{1} and T_{3a}, which is completely regular.

Theorem 1.25. Let G be a topological group. Then G is T_{3a}. Furthermore G is completely regular if and only if {e} is close in G.

Let U and U(r) be as above. Then f(x) = 1 for all x ∈ G \ U. Assume that r < 1 and x ∈ U(r). Then f(x) < 1. In particular if f(x) < 1/2 then x ∈ U_{1}. Let g = 1 − f. Then it follows that Supp(g) ⊂ Ū and g(x) ≥ 1/2 for x ∈ U_{1}.

Lemma 1.26. Assume that G is T_{2} and locally compact. Let K ⊂ G, K ≠ ∅, be compact, and U open such that K ⊂ U. Then there exists a f ∈ C_{c}(G) such that f(G) ⊂ [0, 1], f|K = 1, and Supp(f) ⊂ U.

Proof. This follows from Theorem 1.25 and Lemma 1.4.

Exercise

1. Let X be a topological space. Show the following:

(a) T_{4} and T_{1} ⇒ T_{3};

(b) T_{3} and T_{0} ⇒ T_{2};

(c) T_{3a} ⇒ T_{3};

(d) Completely regular ⇒ regular;

(e) Normal ⇒ regular.

2. Let G be a group with topology τ. Show that G is a topological group if and only if the map

G × G ∋ (x, y) ↦ xy^{−1} ∈ G

is continuous.

3. Prove Lemma 1.1.

4. Show that ρ(a) : G → G is a homeomorphism.

5. Let G be a topological group. Let U ⊂ G be open and let a ∈ U. Then there exists an open set V ∋ e such that aV ⊂ U.

6. Let G be a topological group and H ⊂ G an open subgroup. Then H is also closed.

7. Suppose that G is a topological group and H a subgroup. Let τ be the collection of all sets U ⊂ G/H such that κ^{−1}(U) ⊂ G is open in G. Show that τ is a topology on τ.

8. Let B_{1}, B_{2} be Banach spaces over ℝ viewed as topological groups. Show that a continuous map φ : B_{1} → B_{2} is a group homomorphism if and only if φ is linear.

9. Let G be a topological group, H ⊂ G a subgroup and X a topological space.

(a) Show that a function f : G/H → X is continuous if and only if the map f ∘ κ : G → X is continuous.

(b) Assume that G acts separately continuously on X. Show that the following are equivalent: (i) There exists a x ∈ X such that κ_{x} is open; (ii) κ_{x} is open for all x ∈ X; (iii) There exists a x ∈ G such that π_{x} : G/G^{x} → X is a homeomorphism; (iv) π_{x} : G/G^{x} → X is a homeomorphism for all x ∈ X.

10. Let G = ℝ, and let τ ∈ ℝ \ ℚ. Show that G acts continuously on 𝕋^{2} by

x · (e^{iu}, e^{iv}) := (e^{i(u+x)}, e^{i(v+τx)}) , x, u, v ∈ ℝ .

Let z ∈ 𝕋^{2}, then G^{z} = {e}, G · z is dense in 𝕋^{2}, and κ_{z} : x ↦ x · z is not open.

11. Prove Lemma 1.15.

12. Let G be a topological group and X and Y be G spaces. Assume that φ : X → Y is an Gisomorphism. Show that φ^{−1} : Y → X is also a G-isomorphism.

5. Examples of topological groups and homogeneous spaces

In this section we give some examples of topological groups. We have meet some of them before, others will play an important role later on.

5.1. The general linear group GL(n, 𝔽). Let 𝔽 be the field of real or complex numbers. Let M(n × m, 𝔽) = {(x_{ij})_{i,j=1}^{n,m} | x_{ij} ∈ 𝔽} be the space of n by m matrices. If n = m then we simply write M(n, 𝔽). Let x_{1}, …, x_{m} ∈ 𝔽^{n} be the column vectors. Then

M(n × m, 𝔽) ∋ (x_{ij})_{i,j=1}^{n,m} ↦ ( x_{1} ; ⋮ ; x_{m} ) ∈ 𝔽^{nm}

or

M(n × m, 𝔽) ∋ (x_{ij})_{i,j=1}^{n,m} ↦ (x_{11}, …, x_{n1}, x_{12}, …, x_{n2}, …, x_{m1}, …, x_{mn})^{t} ∈ 𝔽^{nm}

is a linear isomorphism. In particular the usual norm on 𝔽^{nm} defines a norm and metric on M(n, ℝ). ||(x_{ij})|| = \sqrt{∑_{i,j=1}^{nm} |x_{ij}|^{2}} defines a metric on M(n, 𝔽). The determinant function det : M(n, 𝔽) → 𝔽,

X ↦ det(X) = ∑_{σ∈γ_{n}} sign(σ)x_{1σ(1)}x_{2σ(2}) · . . . · x_{nσ(n)}

is a polynomial in the coordinates and hence continuous. It follows that

GL(n, 𝔽) := {x ∈ M(n, 𝔽) | det(x) ≠ 0}

is open and dense in M(n, 𝔽). As x ∈ M(n, 𝔽) is invertible if and only if det(x) ≠ 0, det(xy) = det(x) det(y) and det(x^{−1}) = det(x)^{−1} it follows that GL(n, 𝔽) is a group, the general linear group. Let x = (x_{ij}) and y = (y_{ij}). Then

(xy)_{ij} = ∑_{ν=1}^{n} x_{iν}y_{νj}

and as the inverse is a rational function in the coordinates it follows that both multiplication and inverse are continuous. Hence GL(n, 𝔽) is a topological group. The above argument shows that det : GL(n, 𝔽) → 𝔽 is a continuous homomorphism. The kernel

SL(n, 𝔽) := {a ∈ GL(n, 𝔽) | det(a) = 1}

is a closed normal subgroup of GL(n, 𝔽) and hence a locally compact Hausdorff topological group. It is the special linear group. The group SL(2, 𝔽) acts naturally on 𝔽^{2} by (A, x) = Ax. The set {0} is obviously an orbit. Let x = (x_{1}, x_{2})^{t} ∈ 𝔽^{2} \ {0}. If A = (a_{1}, a_{2}) with column vectors a_{1}, a_{x} ∈ 𝔽^{2} it follows that Ae_{1} = a_{1}. If x_{1} ≠ 0 let

A = ( \begin{matrix}x_{1} & 0 \\ x_{2} & 1/x_{1} \end{matrix} ) ∈ SL(2, 𝔽) .

If x_{1} = 0 then x_{2} ≠ 0. Let

A = ( \begin{matrix}0 & −1/x_{2} \\ x_{2} & 0 \end{matrix} ) ∈ SL(2, 𝔽)

Then Ae_{1} = x. It follows that 𝔽^{2}\ {0} = SL(2, 𝔽)e_{1} is one orbit. Hence 𝔽^{2} decomposes into exactly two orbits. Assume that Ae_{1} = e_{1}. Then A must have the form

A = ( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) , x ∈ N .

It follows that

G^{e_{1}} = \{( \begin{matrix}1 & x \\ 0 & 1 \end{matrix} ) | x ∈ ℝ\} =: N

and 𝔽^{2} \ {0} ≃ SL(2, 𝔽)/N.

Assume now that 𝔽 = ℝ. For A = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ GL(2, ℂ) and z ∈ ℂ let

A · z = \frac{az + b}{cz + d}

if az + d ≠ 0. Let

H_{+} = {z ∈ ℂ | Im(z) > 0}

Lemma 1.27. Let A ∈ SL(2, ℝ) and z ∈ H_{+}. Then A · z is well defined, and A · z ∈ H_{+}. Furthermore SL(2, ℝ) × H_{+} ∋ (A, z) ↦ A · z ∈ H_{+} defines a continuous action of SL(2, ℝ) on H_{+}. The action is

transitive and SL(2, ℝ)^{i} = SO(2) = \{( \begin{matrix}cos(θ) & sin(θ) \\ − sin(θ) & cos(θ) \end{matrix} ) | θ ∈ ℝ\}.

Proof. Write z = x + iy ∈ H_{+}. Let A = ( \begin{matrix}a & b \\ c & d \end{matrix} ) ∈ SL(2, ℝ). Then cz + d = (cx + d) + icy. If

cy = 0 then c = 0 as y > 0. But then d ≠ 0 and cz + d = d ≠ 0. Furthermore

Im A · z = \frac{y}{(cx + d)^{2} + c^{2}y^{2}} > 0 .

Hence A · z ∈ H_{+}. As (A, z) ↦ A · z is a rational function in a, b, c, d, and z it follows that (A, z) ↦ A · z is continuous. We refer to the exercises to show that (AB) · z = A · (B · z).

Let x + iy ∈ H_{+}. Then y > 0. Let A = ( \begin{matrix}\sqrt{y} & \frac{x}{\sqrt{y}} \\ 0 & \frac{1}{\sqrt{y}} \end{matrix} ) ∈ SL(2, ℝ). Then

A · i = \frac{i\sqrt{y} + x/\sqrt{y}}{1/\sqrt{y}} = x + iy .

Hence the action is transitive. Assume that

( \begin{matrix}a & b \\ c & d \end{matrix} ) · i = \frac{ai + b}{ci + d} = i .

Then

ai + b = di − c .

Thus a = d and b = −c. As det A = ad − bc = a^{2} + b^{2} = 1 it follows that we can write a = cos(θ) and b = sin(θ) for some θ ∈ ℝ. Now the lemma follows.

Corollary 1.28. Let g ∈ SL(2, ℝ). Then there exists unique k ∈ SO(2, ℝ), a ∈ A, and n ∈ N such that g = kan.

Proof. Let z = g · i. Then there exits unique a ∈ A and n ∈ N such that z = an · i. Hence g(an)^{−1} = k ∈ SO(2, ℝ).

5.2. The classical linear groups. We will later on be looking at other closed subgroups of the general linear group. The most important of those are the classical linear groups some of which we list here. Let 𝔽 = ℝ or 𝔽 = ℂ. Let V be a vector space over 𝔽.

Definition 8. A map β : V × V → 𝔽 is called 1. bilinear if for each fixed u ∈ V the maps

V ∋ x ↦ β(x, u) ∈ 𝔽 and V ∋ x ↦ β(u, x) ∈ 𝔽

are linear;

2. symmetric if β(u, v) = β(v, u) for all u, v ∈ V;

3. skew-symmetric if β(u, v) = −β(v, u) for all u, v ∈ V;

4. Hermitian if β is linear in the first variable and β(u, v) = \overline{β(v, u)};

5. non-degenerate if β is bilinear or Hermitian and for each u ≠ 0 there exists a v ∈ V such that β(u, v) ≠ 0.

6. positive if β(u, v) > 0 for all v ∈ V, v ≠ 0.

7. inner product on V if β is Hermitian, non-degenerate and positive.

If 𝔽 = ℝ and β is Hermitian, then β is actually symmetric, but we prefer to use Hermitian in both cases, so that we do not have to make the difference between ℝ and ℂ. Notice also that if β is Hermitian, then

β(u, λx + y) = \overline{β(λx + y, u)}
= λ̄\overline{β(x, u)} + \overline{β(y, u)}
= λ̄β(u, x) + β(u, y) .

Hence β is conjugate linear in the second variable. If V is a Hilbert space and β is Hermitian define

||β|| := sup_{u,v∈V, ||u||=||v||=1} |β(u, v)| .

We say that β is bounded if ||β|| < ∞. In that case we call ||β|| for the norm of β. Notice that Lemma 1.29. Assume that V is a Hilbert space (over ℝ or ℂ) with inner product (·, ·) and corresponding norm || · ||. Suppose that β is a bounded Hermitian form on V. Then there exists a bounded linear map T : V → V such that

β(u, v) = (T u, v)

for all u, v ∈ V. Furthermore ||β|| = ||T ||. T is injective with dense range if and only if β is nondegenerate. We have β(u, v) = \overline{β(v, u)}, respectively β(u, v) = −\overline{β(v, u)}, for all u, v ∈ V, if and only if T is symmetric, i.e., T^{∗} = T, respectively skew-symmetric, T^{∗} = −T.

Proof. Let v ∈ V. Define f = f_{β,v} : V → ℂ by u ↦ β(u, v). Then f is linear and
|f(u)| ≤ ||β|| ||v|| ||u|| .

Hence f is continuous with ||f_{β,v}|| ≤ ||β|| ||v||. By the Riez representation theorem there exists a unique w ∈ V such that

f(u) = (u, w) .

Define S(v) := w. Then S is linear, continuous, and ||S|| = ||f||. Define T := S^{∗}. Then T is continuous with norm ||T || = ||S|| = ||f||. By definition we have β(u, v) = (T u, v). Hence |β(u, v)| ≤ ||T || ||u|| ||v||. It follows that ||β|| ≤ ||T ||. On the other hand we also have

|(T u, v)| = |β(u, v)| ≤ ||β|| ||u|| ||v|| .

Hence ||T || = ||β||. We have β(u, v) = \overline{β(v, u)} ⟺ (T u, v) = \overline{(T v, u)} = (u, T v) ⟺ T^{∗} = T. The last statement follows in the same way.

Definition 9. Let V be a (complex) Hilbert space. A map j : V → V is called a conjugation if j is conjugate linear, i.e., j(λu + v) = λ̄j(u) + j(v).

Let j be a conjugation on V. Then V_{ℝ} := {v ∈ V | j(v) = v} is a real Hilbert space such that V_{ℝ}⊕iV_{ℝ} = V and iV_{ℝ} = {v ∈ V | j(v) = −v}. On the other hand any decomposition V = V_{ℝ} ⊕ iV_{ℝ} with V_{ℝ} a real sub-Hilbert space gives rise to a conjugation by defining j to be the identity on V_{ℝ} and −id on iV_{ℝ}.

Lemma 1.30. Let V be a Hilbert space. Then there exists a conjugation j : V → V.

Proof. Let {e_{j}}_{j∈I} be a orthonormal basis for V. Define j : V → V by

j (∑ λ_{j}e_{j}) := ∑ λ̄_{j}e_{j} .

Then j is a conjugation.

In case dim V = n < ∞ we choose e_{1}, …, e_{n} to be the standard basis e_{1} = (1, 0, …, 0)^{t}, …, e_{n} = (0, …, 0, 1)^{t}. Then j((z_{1}, …, z_{n})^{t}) = (z̄_{1}, …, z̄_{n}). the usual conjugation. Let j be a conjugation on V. Then we have a canonical bilinear form β_{j} give by β_{j}(u, v) = (u, j(v)). Define A^{t} := jA^{∗}j. Then

β_{j}(Au, v) = (Au, jv) = (u, A^{∗}jv) = (u, j(jA^{∗}j)v) = β_{j}(u, A^{t}v) .

Lemma 1.31. Assume that V is a Hilbert space with inner product (·, ·) and conjugation j. Suppose that β is a bounded bilinear form on V. Then there exists a linear map A : V → V such that ||A|| = ||β|| and β(u, v) = β_{j}(Au, v) for all u, v ∈ V. β is non-degenerate if and only if A is injective with dense range. β is symmetric if and only if A = A^{t} and skew symmetric if and only if A = −A^{t}.

Here are the standard example for 𝕍 = 𝔽^{n}.

1. (Need to be rewritten) Let Symb_{n}(𝔽) be the space of symmetric bilinear forms on V = 𝔽^{n}. Let Symb_{n}^{∗}(𝔽) be the subset of non-degenerate symmetric bilinear forms. If a ∈ GL(n, 𝔽) and β ∈ Symb_{n}(𝔽) define a · β ∈ Symb_{n}(𝔽) by a · β(u, v) := β(a^{t}u, a^{t}v). Then

(ab) · β(u, v) = β((ab)^{t}u, (ab)^{t}v)
= β(b^{t}(a^{t}u), b^{t}(a^{t}v))
= (b · β)(a^{t}u, a^{t}v)
= a · (b · β)(u, v) .

Thus GL(n, 𝔽) acts on Symb_{n}(𝔽). If β is non-degenerate then a · β is also non-degenerate. Thus GL(n, 𝔽) leaves Symb_{n}^{∗}(𝔽) invariant. Let

O(β, 𝔽) := {a ∈ M(n, 𝔽) | a · β = β} .

Then O(β, 𝔽) is a group (see exercise 4). We let SO(β, 𝔽) = {a ∈ O(β, 𝔽) | det(a) = 1}. Then SO(β, 𝔽) is a closed subgroup of O(β, 𝔽). Let β(z, w) = ∑_{j=1}^{n} z_{j}w_{j} = z · w. Then β = β_{j} with j the standard conjugation. In this case we have

O(β, 𝔽) = {A ∈ M(n, 𝔽) | A^{t} = A^{−1}} .

If 𝔽 = ℂ, then we write O(n, ℂ) for O(β, 𝔽), and if 𝔽 = ℝ we write O(n).

Let Sym_{n}(𝔽) be the space of symmetric n × n-matrices. Let Sym_{n}^{∗}(𝔽) = Sym_{n}(𝔽) ∩ GL(n, 𝔽). We make Sym_{n}(𝔽) into a topological space using the relative topology as a subset of M(n, 𝔽). Then GL(n, 𝔽) acts continuously on Sym_{n}(𝔽) by a · A := aAa^{t}. Let A ∈ Sym_{n}(𝔽). Notice that this action leaves Sym_{n}^{∗}(𝔽) invariant. Define β_{A}(u, v) = (Au) · v. Then β_{A} is symmetric.

Lemma 1.32. The map Sym_{n}(𝔽) ∋ A ↦ β_{A} ∈ Symb_{n}(𝔽) is an isomorphism of GL(n, 𝔽)- spaces.

Proof. That the map is bijective follows from Lemma 1.15. Let a ∈ GL(n, 𝔽) and A ∈ Sym_{n}(𝔽). Then

β_{a·A}(u, v) = β((aAa^{t})u, v) = β(A(a^{t}u), a^{t}v) = a · β_{A}(u, v) .

If 𝔽 = ℂ then every symmetric matrix in Sym_{n}^{∗}(ℂ) can be written in the form aa^{t}. This implies that the group GL(n, ℂ) acts transitively on Sym_{n}^{∗}(ℂ) and Symb_{n}(ℂ). The stabilizer of I = id is the group O(n, ℂ). It follows that

Sym_{n}^{∗}(ℂ) ≃ Symb_{n}^{∗}(ℂ) ≃ GL(n, ℂ)/O(n, ℂ) .

Let now 𝔽 = ℝ. Then the situation is more complicated. Let 0 ≤ p, q ≤ n such that n = p+q. Let

I_{p,q} = ( \begin{matrix}I_{p} & 0 \\ 0 & −I_{p} \end{matrix} ) .

and

β_{p,q}(u, v) = β_{I_{p,q}}(u, v) = ∑_{j=1}^{p} u_{j}v_{j} − ∑_{j=p+1}^{n} u_{j}v_{j} .

Let A be a symmetric matrix with det A ≠ 0. Then there exists a g ∈ O(n) such that gAg^{t} is diagonal, i.e.,

gAg^{t} = d(λ_{1}, …, λ_{n}).

We can choose g such that λ_{1} ≥ … ≥ λ_{n}. Assume that λ_{1} ≥ … ≥ λ_{p} > 0 > λ_{p+1} ≥ … ≥ λ_{n}. Let

h = d (1/\sqrt{λ_{1}}, …, 1/\sqrt{λ_{p}}, −1/\sqrt{|λ_{p+1}|}, …, −1/\sqrt{|λ_{n}|}) g ∈ GL(n, ℝ) .

Then

hAh^{t} = I_{p,q}

Hence if A corresponds to the symmetric matrx A then h · β = β_{p,1} and say that β has signature (p, q). Let Sym_{p,q}(ℝ) be the set of matrices conjugate to I_{p,q} and Symb_{p,q}(ℝ) bet the set of symmetric bilinear forms with signature (p, q). Then Sym_{p,q}(ℝ) ≃ Symb_{p,q}(ℝ) as GL(n, 𝔽)-spaces.

O(p, q) = {A ∈ GL(n, ℝ) | A · β_{p,q} = β_{p,q}}
= {A ∈ GL(n, ℝ) | AI_{p,q}A^{t} = I_{p,q}} .

We let SO(p, q) = {A ∈ O(p, q) | det A = 1}. Then SO(p, q) is a closed subgroup. From the above discussion it follows that

Sym_{p,q}(ℝ) ≃ Symb_{p,q}(ℝ) ≃ GL(n, ℝ)/O(p, q) .

2. Let 𝔽 = ℝ. Define a skew-symmetric bilinear form ω on ℝ^{2n} by

ω(u, v) := ∑_{j=1}^{n} u_{j}v_{n+j} − ∑_{j=1}^{n} u_{n+j}v_{j} = J_{n}u · v

where

J_{n} = ( \begin{matrix}0 & I_{n} \\ −I_{n} & 0 \end{matrix} ) .

We let

Sp(n, ℝ) : = {A ∈ GL(n, ℝ) | A · ω = ω}
= {A ∈ GL(n, ℝ) | AJ_{n}A^{t} = J_{n}} .

The group Sp(n, ℝ) is the symplectic group. We let Sp(n) = Sp(n, ℝ) ∩ SO(n). Then Sp(n) is a compact subgroup of Sp(n, ℝ).

5.3. The sphere S^{n}. The group O(n + 1) and SO(n + 1) act continuously on S^{n} = {x ∈ ℝ^{n+1} | |x| = 1}. Let x = e_{1}. Then

(a_{ij})e_{1} = ( \begin{matrix}a_{11} \\ . \\ . \\ . \\ a_{n+1,1} \end{matrix} ) . (5.1)

Let x = (x_{1}, …, x_{n})^{t} ∈ S^{n}. Extend x to an orthonormal basis x = x_{1}, …, x_{n+1} of ℝ^{n+1}. Let A be the matrix with column vectors x_{1}, …, x_{n+1}. Thus A = (x_{1}, …, x_{n+1}). Then det A = ±1. If det A = −1 replace x_{n+1} by −x_{n+1}. We can therefore assume that det A = 1. But then A ∈ SO(n + 1) and by (5.1) we have Ae_{1} = x. It follows that SO(n + 1) (and hence O(n + 1) too) acts transitively on S^{n}. Assume that Ae_{1} = e_{1}. By (5.1) we have that a_{11} = 1 and a_{21} = … = a_{n+1,1} = 0. Hence a_{12} = … = a_{1,n+1} = 0. It follows that A has the form

A = ( \begin{matrix}1 & 0 \\ 0 & B \end{matrix} )

with B ∈ SO(n). The result is:

1. Lemma 1.33. S^{n} ≃ SO(n + 1)/SO(n).

5.4. The flag manifolds. Let V be an n-dimensional vector space over 𝔽. Let 1 ≤ n_{1} < n_{2} < … < n_{k} < n with n_{j} ∈ ℕ. Let n = (n_{1}, …, n_{k}). Let Flag_{k}(𝔽) = Flag_{k} = Flag(n_{1}, …, n_{k}) be the set of nested chains V_{1} ⊂ … ⊂ V_{k} of subspaces with dim V_{j} = n_{j}. Then GL(n, 𝔽) acts on Flag_{k} by

A · (V_{1}, …, V_{k}) = (AV_{1}, …, AV_{k}) .

Let e_{1}, …, e_{n} be the standard basis. Let E_{j} = ∑_{i=1}^{n_{j}} 𝔽e_{i}. Let E := (E_{1}, …, E_{k}) ∈ Flag_{n}. Let V = (V_{1}, …, V_{k}) ∈ Flag_{n}. Choose f_{1}, …, f_{n} a basis of V such that f_{1}, …, f_{n_{j}} is a basis of V_{j}. Then there exists a A ∈ GL(n, 𝔽) such that Ae_{j} = f_{j}. It follows that A · E = V. Hence GL(n, ℝ) acts transitively on Flag_{n}. We can even say more: If we choose f_{1}, …, f_{n} orthonormal with the same orientation as the standard basis we see that we can even choose A ∈ SO(n) if 𝔽 = ℂ and A ∈ SU(n) if 𝔽 = ℂ. Hence SO(n) (respectively SU(n)) acts transitively on Flag_{n}Assume that A · E = E. Then AV_{1} = V_{1}. It follows that A has the form

A = ( \begin{matrix}A_{1} & B \\ 0 & C \end{matrix} )

with A ∈ GL(n_{1}, 𝔽), C ∈ GL(n − n_{1}, 𝔽) and B an arbitrary n_{1} × (n − n_{1})-matrix. Iterating this, using that AV_{j} ⊂ V_{j} it follows that A has the form

A = ( \begin{matrix}A_{1} & ∗ & ∗ & ∗ & ∗ \\ 0 & A_{2} & ∗ & ∗ & ∗ \\ 0 & 0 & ... & ∗ & ∗ \\ 0 & 0 & 0 & A_{k−1} & ∗ \\ 0 & 0 & 0 & 0 & A_{k} \end{matrix} )

with A_{j} ∈ GL(n_{j} − n_{j−1}, 𝔽), 1 ≤ j ≤ k. Here ∗ stands for an arbitrary matrix of the correct size. Let P (n) be the group GL(n, 𝔽)^{E}. Then P (n) is a parabolic subgroup of GL(n, 𝔽). We can now introduce a topology on Flag_{n} such that Flag_{n} ≃ GL(n, 𝔽)/P (n). Let G_{j} ⊂ GL(n_{j}, 𝔽) be a subgroup. Denote by S(G_{1} × ⋯ × G_{k}) the subgroup of GL(n, 𝔽) given by

S(G_{1} × ⋯ × G_{k}) = \{( \begin{matrix}A_{1} & 0 & 0 & 0 & 0 \\ 0 & A_{2} & 0 & 0 & 0 \\ 0 & 0 & ... & 0 & 0 \\ 0 & 0 & 0 & A_{k−1} & 0 \\ 0 & 0 & 0 & 0 & A_{k} \end{matrix} ) | A_{j} ∈ G_{j}, det(A_{1}) · . . . · det(A_{k}) = 1\}

= G_{1} × ⋯ × G_{k} ∩ SL(n, 𝔽) .

Theorem 1.34. The space Flag_{n} is compact. In fact we have:

1. Flag_{n}(ℝ) ≃ SO(n)/S(O(n_{1}) × ⋯ × O(n_{k}));

2. Flag_{n}(ℂ) ≃ SU(n)/S(U(n_{1}) × ⋯ × U(n_{k}))

5.5. The Euclidean motion group. Let G and H be topological groups. Let G×H be the product group. Denote the group of homomorphisms H → H by Aut(H). Assume that α : G → Aut(H), g ↦ α_{g}, is a homomorphism. Define g · h = α_{g}h. This defines an action of G on G such that (g · a)(g · b) = g · (ab). We let G ×_{α} H be G × H with the multiplication

(a, b) · (c, d) = (ab, bα_{a}(d)) .

Then G ×_{α} H is a group. In particular we can take G to be a closed subgroup of GL(n, ℝ) and H = ℝ^{n}. The map α is then the canonical map α_{A}(x) = Ax. Then G ×_{α} ℝ^{n} is a group with multiplication (A, x)(B, y) = (AB, Ay +x). The group G×_{α} ℝ^{n} acts on ℝ^{n} by affine transformations (A, x)·y = Ay +x. This action is obviously transitive and we have (G ×_{α} ℝ^{n})^{0} = G. Hence

ℝ^{n} ≃ G ×_{α} ℝ^{n}/G .

The group O(n)×_{α} ℝ^{n} is called the Euclidean motion group and denoted by E(n). It is exactly the group of affine transformation of ℝ^{n} such that ||T x − T y|| = ||x − y|| for all x, y ∈ ℝ^{n}. In the case n = 1 this is the so-called ax + b-group. If G = SO(1, n − 1) then G ×_{α} ℝ^{n} is the Poincare-group.

Exercise

1. Show that the standard norm on M(n × m, 𝔽) is given by

||X|| = \sqrt{Tr(XX^{∗})}, X ∈ M(n × m, 𝔽)

where X^{∗} = (\overline{x_{ji}}) = X̄^{t}.

2. Prove lemma 1.31.

3. Let 𝔽 = ℂ and let V be a complex vector space. Let β be a Hermitian bilinear form on V. Show that β(u, u) ∈ ℝ for all u ∈ V.

4. Show that O(β, 𝔽) is a group.

5. Let β_{A}(u, v) = β(Au, v) where β(u, v) = ∑_{j=1}^{n} u_{i}u_{j}. Show the following:

(a) Assume that A and B are symmetric and B = aAa^{t} for sone a ∈ GL(n, 𝔽). Then

O(β_{B}, 𝔽) = aO(β_{A}, 𝔽)a^{t}

(b) Assume that A is in GL(n, 𝔽). Then O(β_{A}, 𝔽) := {a ∈ GL(n, 𝔽) | a^{t}Aa = a}, where a^{t}

stands for the transposed matrix a_{ij}^{t} = a_{ji}.

6. Let A ∈ O(p, q). Then det(A) = ±1. If A ∈ Sp(n, ℝ) then det(A) = 1.

7. Let n = (n_{1}, …, n_{k}) ∈ ℕ^{j} be such that n_{1} < n_{2} < … < n_{k}. Let G_{j} be a subgroup of GL(n_{j} − n_{j−1}, ℝ). Define

S(G_{1} × … × G_{k}) := \{( \begin{matrix}A_{1} & 0 & 0 \\ 0 & ... & 0 \\ 0 & 0 & A_{k} \end{matrix} ) | A_{j} ∈ G_{j}, det(A_{1}) · . . . · det(A_{k}) = 1\} .

Show that SO(n) acts transitively on Flag_{n} and that Flag_{n} ≃ SO(n)/S(O(n_{1}) × … × O(n_{k} − n_{k−1}))).

5. EXAMPLES OF TOPOLOGICAL GROUPS AND HOMOGENEOUS SPACES 17

8. Show that E(n) is the group of distance preserving affine transformation of ℝ^{n}.

9. Let Sym^{+}(n, ℝ) be the set of positive definite symmetric matrices. Show that Sym^{+}(n, ℝ) ≃ GL(n, 𝔽)/O(n).

10. Show that the map ( \begin{matrix}a & b \\ c & d \end{matrix} ) · z = \frac{az+b}{cz+d} satisfies AB · z = A · (B · z) where ever defined.