Topology: 2020, SEU Yi Li

Topology: 2020, SEU

Yi Li

SCHOOLOF MATHEMATICS AND SHING-TUNG YAU CENTER,SOUTHEAST UNIVERSITY,NANJING,CHINA 210096 E-mail address: [email protected]; [email protected]; [email protected]

Contents

Chapter 1. Basic topology 1 1.1. Metric spaces 1 1.2. Topological spaces 11 1.3. Constructions 37 1.4. Topological groups and matrix Lie groups 50 1.5. Connectedness and compactness 67

Chapter 2. Analysis on topological groups 81 2.1. Haar measures 81 2.2. Iwasawa’s decomposition 87 2.3. Harish, Mellin and spherical transforms 92

Chapter 3. Fundamental groups 109

v CHAPTER 1

Basic topology

Main references: • Kelley, John L. General topology, Reprint of the 1955 edition [Van Nostrand, Toronto, Ont.], Graduate Texts in Mathe- matics, No. 27, Springer-Verlag, New York-Berlin, 1975. xiv+298 pp. MR0370454 (51# 6681) • Bredon, Glen E. Topology and geometry, Graduate Texts in Mathematics, No. 139, Springer-Verlag, New York, 1993. xiv+557 pp. MR1224675 (94d: 55001) • Lee, John M. Introduction to topological manifolds, Second edition, Graduate Texts in Mathematics, No. 202, Springer, New York, 2011. xviii+433 pp. ISBN: 978-1-4419-7939-1 MR2766102 (2011i: 57001) • Munkres, James R. Topology: a ﬁrst course, Prentice-Hall, Inc., Englewood Cliffs, N. J., 1975. xvi+413 pp. MR0464128 (57 # 4063) • Burago, Dmitri; Burago, Yuri; Ivanov, Sergei. A course in metric geometry, Graduate Studies in Mathematics, 33, Amer- ican Mathematical Society, Providence, RI, 2001. xiv+415 pp. ISBN: 0-8218-2129-6 MR1835418 (2002e: 53053)

1.1. Metric spaces 1.1.1. Metric spaces. A metric on a set X is a map d : X × X → R := R ∪ {∞} satisfying, for any x, y, z ∈ X, (a) (positiveness) d(x, y) ≥ 0 and d(x, y) = 0 if and only if x = y; (b) (symmetry) d(x, y) = d(y, x); (c) (triangle inequality) d(x, z) ≤ d(x, y) + d(y, z). A metric space is a pair (X, d) where d is a metric on X.

Given a metric space (X, d) we deﬁne a relation ∼ on X by x ∼ y ⇐⇒ d(x, y) < ∞. Then ∼ is an equivalence relation on X, and the equivalence class [x] of x can be endowed with a natural metric (still denoted by d).

EXERCISE 1.1.1. For any x ∈ X, show that ([x], d) is a ﬁnite metric space. → A map f : (X, dX) (Y, dY) between metric spaces is called distance-preserving if dY( f (x1), f (x2)) = dX(x1, x2)

1 2 1. BASIC TOPOLOGY

∈ for all x1, x2 X. A distance-preserving bijection is called an isometry.

A semi-metric on a set X is a map d : X × X → R := R ∪ {∞} satisfying, for any x, y, z ∈ X, (a) (nonnegativity) d(x, y) ≥ 0; (b) (symmetry) d(x, y) = d(y, x); (c) (triangle inequality) d(x, z) ≤ d(x, y) + d(y, z). A semi-metric space is a pair (X, d) where d is a semi-metric on X.

Given a semi-metric space (X, d) we deﬁne a relation ∼ on X by x ∼ y ⇐⇒ d(x, y) = 0. Then ∼ is an equivalence relation on X. Deﬁne Xˆ := X/ ∼= {[x] : x ∈ X}, dˆ([x], [y]) := d(x, y). Then (X/d, d) := (Xˆ , dˆ) is a metric space.

EXERCISE 1.1.2. Show that dˆ is well-deﬁned and (X/d, d) is a metric space.

EXERCISE 1.1.3. Let X = R2 and deﬁne ′ ′ ′ ′ d((x, y), (x , y )) := |(x − x ) + (y − y )|. Show that d is a semi-metric on X. Deﬁne f : R2/d → R by f ([(x, y)]) := x + y. Show that f is an isometry.

EXERCISE 1.1.4. Consider the set X of all continuous real-valued functions on [0, 1]. Show that ∫ 1 d( f , g) := | f (x) − g(x)|dx 0 deﬁnes a metric on X. Is this still the case if continuity is weakened to integrabil- ity?

EXERCISE 1.1.5. Let (X, dX) and (Y, dY) be two metric spaces. We deﬁne √

dX×Y((x1, y1), (x2, y2)) := dX(x1, x2) + dY(y1, y2). × Show that (X Y, dX×Y) is a metric space.

EXERCISE 1.1.6. As a subspace of R2, the unit circle S1 carries the restricted 2 Euclidean metric from R . We can deﬁne another (intrinsic) metric dint by

dint(x, y) := the length of the shorter arc between them. ∈ π ∈ 1 Note that dint(x, y) [0, ] for any points x, y S . (a) Show that any circle arc of length less than or equal to π, equipped with the intrinsic metric, is isometric to a straight line segment. (b) The whole circle with the intrinsic metric is not isometric to any subset of R2. 1.1. METRIC SPACES 3

Let (X, d) be a metric space. Set B(x, r) := {y ∈ X : d(x, y) < r}, B(x, r) := {y ∈ X : d(x, y) ≤ r}. We deﬁne the metric topology on X by saying that U ⊂ X is open if and only for any x ∈ U there exists ϵ > 0 such that B(x, ϵ) ⊂ U. Let (1.1.1.1) T := {all open subsets in X} ∪ {∅, X}. The set T satisﬁes the following (1) U ∩ V ∈ T for any U, V ∈ T ; ∪ ∈ T { } ⊂ T (2) i∈IUi for any collection Ui i∈I ; (3) ∅, X ∈ T . In general, we can introduce a topology on any set X.A topology on X is a subset T of 2X satisfying (1) U ∩ V ∈ T for any U, V ∈ T ; ∪ ∈ T { } ⊂ T (2) i∈IUi for any collection Ui i∈I ; (3) ∅, X ∈ T . The pair (X, T ) is called the topological space, and any element in T is said to be open; a closed set is a subset of X whose complement in X is open. In particular, we see that any metric space must be a topological space.

Given a metric space (X, d). • (X, d) is complete if any Cauchy sequence1 in X has a limit in X. • A diameter of S ⊂ X is deﬁned by (1.1.1.2) diam(S) := sup d(x, y). x,y∈S • (X, d) is compact if any sequence in X has a converging subsequence. • ⊂ ϵ { ∈ ≤ ϵ} S X is an -net if Sϵ := x X : d(x, S) = infy∈X d(x, y) = X. • X is totally bounded if for any ϵ > 0 there exists a ﬁnite ϵ-net in X. The following theorem is a basic result.

THEOREM 1.1.7. Let (X, d) is a metric space. Then X is compact if and only if X is complete and totally bounded.

1.1.2. Length spaces. Let (X, T ) be a topological space. We say X is Haus- dorff if for any different points x, y ∈ X, there exist open sets Ux, Uy ⊂ X such that x ∈ Ux, y ∈ Uy, and Ux ∩ Uy = ∅. Clearly that any metric space must be Hausdorff. A path γ in X is a continuous2 map γ : I → X. where I ⊂ R is an interval (i.e., connected subset) of R.

A length structure on X is a pair (A , L), where A is a collection of paths (called admissible paths) and L : A → R+ is a map on A (called the length of paths), satisfying

1 { } ϵ ∈ A sequence xn n∈N in X is said to be Cauchy if for any > 0 there exists an integer n0 N ϵ ≥ such that d(xn, xm) < whenever n, m n0. − 2That is, γ 1(U) is open in I for any open set U in X. 4 1. BASIC TOPOLOGY

A γ → A γ| ∈ A (a) ( is closed under restrictions) If : [a, b] X in , then [c,d] for any subinterval [c, d] ⊂ [a, b]; A γ → γ (b) ( is closed under concatenations of paths) If 1 : [a, c] X and 2 : → γ γ [c, d] X are admissible with 1(c) = 2(c), then the obvious product γ γ · γ → := 1 2 : [a, d] X is also admissible; (c) (A is closed under reparameterizations) If γ : [a, b] → X is admissible and φ : [c, d] → [a, b] is a homeomorphism, then γ ◦ φ : [c, d] → x is also admissible; γ| γ| γ| (d) (Length of paths is additive) L( [a,b]) = L( [a,c]) + L( [c,b]) for any c ∈ [a, b] and γ : [a, b] → X in A ; (e) (The length of a piece of a path continuously depends on the piece) If γ : [a, b] → X is a admissible path with L(γ) < +∞, then L(γ, a, t) := γ L( [a,t]) is continuous in t; (f) (L(·) is invariant under reparametrizations) L(γ ◦ φ) = L(γ) for any admissible path γ and any homeomorphism φ as in (c); (g) (Length structures agree with the topology of X) For any x ∈ X and open set Ux ∋ x, one has inf {L(γ) : γ(a) = x, γ(b) ∈ X \ U } > 0. γ∈A x Let (X, T ) is a Hausdorff space with a length structure (A , L). Deﬁne (1.1.2.1) d (x, y) := inf {L(γ)|γ : [a, b] → X, γ(a) = x, γ(b) = y} . L γ∈A ∞ If there is no path between x and y, we set dL(x, y) = + . We say a length struc- A ∈ ∞ ture ( , L) is complete if for any x, y X with dL(x, y) < + , there exists a path γ ∈ A γ joining x and y such that dL(x, y) = L( ).

EXERCISE 1.1.8. Show that (X, dL) is a metric space, where dL is deﬁned in (1.1.2.1).

Note that dL is not necessarily a ﬁnite metric.

DEFINITION 1.1.9. A length space is a metric space (X, d) such that d = dL for some length structure (A , L). In this case, we call d as an intrinsic metric on X. In the following we shall construct a length structure on a given metric space γ → { } (X, d). If : [a, b] X is a path in X, for any partition yi 0≤i≤N of [a, b] with ≤ ≤ ≤ · · · ≤ a = y0 y1 y2 yN = b, we define Σ γ { } γ γ ( , yi 0≤i≤N) := ∑ d ( (yi−1), (yi)) . 1≤i≤N The length of γ is given by γ Σ γ { } (1.1.2.2) Ld( ) := sup ( , yi 0≤i≤N) . { } yi 0≤i≤N γ γ ∞ A The path is said to be rectifiable if Ld( ) < + . The length structure ( , L) now is defined by A { } (1.1.2.3) := all paths parametrized by closed intervals , L := Ld.

THEOREM 1.1.10. Let (X, d) be a metric space, for the length structure (A , L) de- ﬁned in (1.1.2.3), we have the following properties: 1.1. METRIC SPACES 5

(a) L(γ) ≥ d(γ(a), γ(b)) for any path γ : [a, b] → X; γ| γ| γ γ → (b) L( [a,c]) + L( [c,b]) = L( ) for any path : [a, b] X and a < c < b; γ γ γ| (c) If is rectiﬁable, then L( , c, d) := L( [c,d]) is continuous in c and d; (d) L is a lower semi-continuous functional on C([a, b], X) with respect to the point- {γ } wise convergence; that is, if i i∈N is a sequence of rectiﬁable paths in X with γ γ ∈ the same domain and i(t) converges to (t) for every given t [a, b], then γ ≤ γ L( ) lim infi→∞ L( i).

{ } PROOF. (a) For any partition yi 0≤i≤N of [a, b] we have Σ γ { } γ γ ≥ γ γ γ γ (b , yi 0≤i≤N) = ∑ d ( (yi−1), (yi)) d( (y0), (yN)) = d( (a), (b)) 1≤i≤M by the triangle inequality, so that L(γ) ≥ d(γ(a), γ(b)). ′ ′′ { } ′ { } ′′ (b) Given partitions yi 0≤i≤N and yj 0≤j≤N of [a, c] and [c, b], respectively. ′ ′′ { } ′ ∪ { } ′′ Then yi 0≤i≤N yj 0≤j≤N forms a partition of [a, b] and ( ) ′ ′′ γ ≥ Σ γ { } ′ ∪ { } ′′ L( ) , yi 0≤i≤N yj 0≤j≤N ( ) ( ) γ ′ γ ′ γ ′′ γ ′′ = ∑ d (yi−1), (yi) + ∑ d (yj−1), (yj ) ≤ ≤ ≤ ≤ ′′ 1 (i N ) 1( j N ) ′ ′′ Σ γ| { } ′ Σ γ| { } ′′ = [a,c], yi 0≤i≤N + [c,b], yj 0≤j≤N . γ ≥ γ| γ| Hence L( ) L( [a,c]) + L( [c,b]). { } { } ∪ { } Conversely, for any partition yi 0≤i≤N, we see that yi 0≤i≤N c is also a partition of [a, b]. Consequently, we get γ| γ| ≥ Σ γ { } ∪ { } ≥ Σ γ { } L( [a,c]) + L( [c,b]) ( , yi 0≤i≤N c ) ( , yi 0≤i≤N) γ| γ| ≥ γ and then L( [a,c]) + L( [c,b]) L( ). (c) We only prove that L(γ, c, d) is continuous in d ∈ (a, b]. For any ϵ > 0, γ ∞ { } ≤ γ − since L( ) < + ), there exists a partition yi 0≤i≤N of [a, b] such that 0 L( ) Σ γ { } ϵ ( , yi 0≤i≤N) < . We may assume that yj−1 < d = yj for some j, otherwise we { } ∪ { } can use the partition yi 0≤i≤N d . Then

L(γ| ) − d(γ(y − ), γ(d)) ≤ L(γ) − Σ(γ, (y ) ≤ ≤ ) < ϵ [yj−1,d] j 1 i 0 i N ′ ∈ by (a) and (b). For any d [yj−1, d], one has ( ) ( ) ( ) ( ) γ| − γ| ′ ≤ ϵ γ γ − γ| ′ L [y − ,d] L [y − ,d ] + d (yj−1), (d) L [y − ,d ] j 1 j 1 ( ) ( j 1 ) ≤ ϵ γ γ − γ γ ′ + d (yj−1), (d) d (yj−1), (d ) ′ ≤ ϵ + d(γ(d ), γ(d)). ′ ′ Since γ is continuous, it follows that L(γ, c, d ) − L(γ, c, d) = L(γ, d , d) ≤ 2ϵ when ′ d is very close to d. Thus L(γ, c, d) is continuous in d. (d) We ﬁrst assume that L(γ) < +∞. For any ϵ > 0 there exists a partition { } γ − Σ γ { } ϵ ≫ yi 0≤i≤N of [a, b] such that L( ) ( , yi 0≤i≤N) < . Choose j 1 so that 6 1. BASIC TOPOLOGY

γ γ ϵ ≤ ≤ d( j(yi), (yi)) < /N for each 0 i N. Compute γ ≤ Σ γ { } ϵ L( ) ( , yi 0≤i≤N) + ϵ ≤ Σ(γ , {y } ≤ ≤ ) + ϵ + · 2 · N j i 0 i N N ≤ γ ϵ L( j) + 3 where we used ( ) ( ) γ γ ≤ γ γ γ γ d ( (yi−1), (yi)) d (yi−1), (yi−1) + d (yi−1), (yi) ( j ) j j γ γ + d (yi), j(yi) ϵ ( ) ≤ · 2 + d γ (y − ), γ (y ) . N j i 1 j i γ ≤ γ Therefore L( ) lim infj→∞ L( j). γ ∞ { } Σ γ { } When L( ) = + , we choose a partition yi 0≤i≤N such that ( , yi 0≤i≤N) > 1/ϵ. Then 1 ϵ < Σ(γ, {y } ≤ ≤ ) ≤ Σ(γ , {y } ≤ ≤ ) + · 2 · N ≤ L(γ ) + 2ϵ ϵ i 0 i N j i 0 i N N j ≫ → ∞ γ → ∞ for j 1. Letting j yields L( j) . A For a metric space (X, d), we have a length structure ( , L), where L := Ld, and then the induced intrinsic metric ∧ (1.1.2.4) d := dL. ∧ ∧ However, d may be equal to d. As a direct consequence of Theorem 1.1.10, d ≤ d always holds.

EXAMPLE 1.1.11. Consider[ the subset( )] ∪ 1 X := (0, 1), , 0 ∪ [(0, 1), (0, 0)] ⊂ R2. n n∈N | → For the standard metric d := d 2 X we see that d((1/n, 0), (0, 0)) = 1/n 0 as R ∧ n → ∞. However the limit limn→∞ d ((1/n, 0), (0, 0)) does not exist; in fact, for any path γ connecting (1/n, 0) with (1/m, 0), we have (( ) ) (( ) ) 1 1 L(γ) = L (γ) ≥ d , 0 , (0, 0) + d , 0 , (0, 0) d n m √ √ 1 1 = 1 + + 1 + > 2. n2 m2 Given a metric space (X, d), we have the following diagram: (X, d) −−−−→ (A , L = L )  d  y

∧ (X, d = d )  L  y ∧ ∧ b ( ( ) = ) ←−−−− (A = ∧ ) X, d dbL , L Ld ∧ ∧ by Theorem 1.1.10. A natural question is whether d = (d ) . 1.1. METRIC SPACES 7 b THEOREM 1.1.12. Let (X, d) be a metric space and d the intrinsic metric induced by d. γ ∞ γ γ (a) If Ld( ) < + , then Ld( ) = L ∧ ( ). ∧d ∧ ∧ ∧ ∧ (b) The intrinsic metric induced by d coincides with d , i.e., (d ) = d .

PROOF. (a) For any two points x, y ∈ X, choose a path γ : [a, b] → X from x to y. Then γ ≥ γ γ Ld( ) d( (a), (b)) = d(x, y) ≤ ∧ γ ≤ γ which implies d(x, y) d (x, y) and hence Ld( ) Ld∧ ( ). γ → { } Conversely, for any path : [a, b] X and any partition yi 0≤i≤N of the interval [a, b], we have ( ) ∧ ∑ (γ, {y } ≤ ≤ ) ≤ ∑ d (γ(y − ), γ(y )) ≤ ∑ L γ| = L (γ) i 0 i N i 1 i d [yi−1,yi] d b ≤ ≤ ≤ ≤ d 1 i N 1 i N according to Theorem 1.1.10 (b). Part (b) follows from an obvious way.

EXAMPLE 1.1.13. Let (M, g) be a connected Riemannian manifold. Deﬁne ∫ b γ |γ | (1.1.2.5) L(x, y) = Ld( ) := ˙ (t) gdt, a where { } 1 (1.1.2.6) d(x, y) = dg(x, y) := inf Lg(γ)|γ is a piecewise C -path joining x to y . Then (M, d) is a metric space.

1.1.3. Gromov-Hausdorff distance. In this subsection, all proofs can be found in BBI’s book “A course in metric geometry”.

ϵ ϵ Let (Z, dZ) be a metric space. Given an > 0, the -neighborhood of a subset S ⊂ Z is given by { } ∈ ≤ ϵ (1.1.3.1) Sϵ := z Z : dZ(z, S) = inf dZ(z, x) . x∈S We say a subset S ⊂ X is an ϵ-net if Sϵ = X. The Hausdorff distance of A, B ⊂ Z is deﬁned to be the quantity Z {ϵ ⊂ ⊂ } (1.1.3.2) dH(A, B) := inf > 0 : A Bϵ and B Aϵ . ϵ Z ∞ If no such , we put dH(A, B) = + . A basic fact is that the set of all compact Z subsets of Z, equipped with dH, forms a metric space.

For any metric spaces (X, dX) and (Y, dY), we deﬁne Gromov-Hausdorff distance between them by

(1.1.3.3) dGH ((X,dX), (Y, dY))  →  f : (X, dX) (Z, dZ)   →  Z g : (Y, dY) (Z, dZ) := inf dH ( f (X), g(Y)) :  . (Z,dZ), f,g  are isometric embeddings  into a metric space (Z, dZ) 8 1. BASIC TOPOLOGY

One can show that if (X, dX), (Y, dY) are compact metric spaces with zero Gromov- Hausdorff distance, then (X, dX) and (Y, dY) are isometric. Therefore, the set of all compact metric space, modulo the isometric equivalence, is a metric space with respect to dGH. → A distortion of a map f : (X, dX) (Y, dY) between metric spaces is deﬁned to be | − | (1.1.3.4) dis( f ) := sup dX(x1, x2) dY( f (x1), f (x2)) . ∈ x1,x2 X Note that dis( f ) = 0 if and only if f is distance-preserving. Such a f is called an ϵ-isometry if dis( f ) ≤ ϵ and f (X) is an ϵ-net in Y. The following result characterizes the Gromov-Hausdorff distances with ϵ- isometries. ϵ LEMMA 1.1.14. Let (X, dX) and (Y, dY) be metric spaces and > 0. ϵ ϵ → (a) If dGH((X, dX), (Y, dY)) < , then there exist -isometries f : (X, dX) → (Y, dY) and g : (Y, dY) (X, dX). → ϵ ϵ (b) If f : (X, dX) (Y, dY) is an -isometry, then dGH((X, dX), (Y, dY)) < 2 .

A pointed metric space is a triple (X, dX, x0), where (X, dX) is a metric space → and x0 is a ﬁxed base-point. A pointed map f : (X, dX, x0) (Y, dY, y0) is a map → f : (X, dX) (Y, dY) which preserves the base-points (that is, f (x0) = y0). An ϵ → -pointed-isometry f : (X, dX, x0) (Y, dY, y0) is a pointed map f satisfying | − | ϵ dX(x1, x2) dY( f (x1), f (x2)) < ∈ ϵ−1 ϵ−1 ⊂ ϵ−1 for any x1, x2 BX(x0, ), and BY(y0, ) ( f (BX(x0, )))ϵ. For two pointed metric spaces (X, dX, x0) and (Y, dY, y0) deﬁne the pointed-Gromov-Hausdorff distance as pt (( ) ( )) dGH x, dX, x0 , Y, dY, y0   there exist ϵ-pointed-isometries  ϵ → (1.1.3.5) := inf > 0 : f : (X, dX, x0) (Y, dY, y0) and .  →  g : (Y, dY, y0) (X, dX, x0)

ϵ pt ϵ LEMMA 1.1.15. Given > 0. If dGH((X, dX, x0), (Y, dY, y0)) < , then there exist ϵ ϵ−1 → ϵ−1 ϵ 2 -isometries f : (BX(x0, ), dX) (BY(y0, + ), dY) with f (x0) = y0 and ϵ−1 → ϵ−1 ϵ g : (BY(y0, ), dY) (BX(x0, + ), dX) with g(y0) = x0.

ϵ → PROOF. By deﬁnition, there exist 2 -pointed-isometries f : (X, dX, x0) (Y, dY, y0) → ∈ ϵ−1 | − and g : (Y, dY, y0) (X, dX, x0). For any x BX(x0, ) we have dX(x, x0) | ϵ ϵ−1 ⊂ ϵ−1 ϵ dY( f (x), y0) < and then f (BX(x0, )) BY(y0, + ). ( ) We say that a sequence of pointed metric spaces Xi, dXi , xi converges to a pointed metric space (Y, dY, y0) in the sense of pointed Gromov-Hausdorff distance, if ( ) pt lim d (Xi, dX , xi), (Y, dY, y0) = 0. i→∞ GH i A subset S in a topological space X is precompact if any sequence in S has a subsequence that converges to a point in X. 1.1. METRIC SPACES 9

THEOREM 1.1.16. (Gromov’s precompactness theorem) Let M := a collection of compact metric spaces, Mpt := a collection of pointed metric spaces. (1) If M is uniformly totally bounded, that is, (i) there exists D < ∞ such that diam(X) ≤ D for any X ∈ M, and (ii) for any ϵ > 0 there exists N(ϵ) ∈ N such that any X ∈ M contains an ϵ-net S ⊂ X consisting of at most N(ϵ) points, then M is precompact in the collection of metric spaces with respect to the Gromov- Hausdorff distance. (2) If Mpt has the property that ϵ ρ ϵ ρ ∈ ∈ pt for any , > 0, there exists N( , ) N such that any (X, x0) M ϵ ⊂ ρ ⊂ ϵ ρ contains an -net S B(x0, ) X consisting of at most N( , ) points, then Mpt is precompact in the collection of pointed metric spaces with respect to the pointed Gromov-Hausdorff distance. As a corollary of volume comparison theorem, we have

THEOREM 1.1.17. (Gromov) Given K ∈ R, m ≥ 2, let { } pt pointed complete Riemannian m-manifolds (M, g, O) Mm,K := . with Ricg ≥ (m − 1)K g and O ∈ M pt Then Mm,K is precompact in the collection of pointed metric spaces with respect to pointed Gromov-Hausdorff distance. The tangent cone of a metric space (X, d) at x ∈ X is given by ( ) = ( α ) (1.1.3.6) TxX, dx, 0x : αlim→∞ X, kd, x k provided this limit (in the sense of pointed Gromov-Hausdorff distance) exists for α → ∞ {α } any sequence k and is independent of k k∈N (up to isometry). (1) The tangent cone is a metric space. (2) The tangent cone of a Riemannian manifold (M, g) at p ∈ M is isometric to (T M, g(x), 0 ). x x ∼ (3) For any c > 0, we have an isometry (TxX, cdx, 0x) = (TxX, dx, 0x). The Gromov-Hausdorff asymptotic cone of a metric space (X, d) at x ∈ X is given by (1.1.3.7) (AX, d , 0) := lim (X, ω d, x) AX ω → i i 0 provided this limit exists and is independent (up to isometry) of the sequence {ω } i i∈N. ∼ (1) For any c > 0, we have an isometry (AX, cdAX, 0) = (AX, dAX, 0). (2) If (M, g) is a complete noncompact Riemannian manifold with nonnegative sectional curvature, then the asymptotic cone exists and is isometric to a Euclidean metric cone. (For a proof of this result, read Theorem I.26 in The Ricci Flow: Techniques and Applications. Part III: Geometric-Analytic Aspects.) 10 1. BASIC TOPOLOGY

The topological cone of a topological space X is deﬁned by (1.1.3.8) Cone(X) := (X × [0, ∞))/(X × {0}) = {[(x, r)] : x ∈ X, r ∈ [0, ∞)}. − ∼ We call the point X × {0} the vertex of Cone(X). It is clear that Cone(Sm 1) = Rm.

We deﬁne the Euclidean metric cone of a metric space (X, d) with diam(X) ≤ π as follows. It is a topological cone Cone(X) equipped with the metric dCone(X) deﬁned by √ 2 2 − (1.1.3.9) dCone(X) ([(x1, r1)], [(x2, r2)]) := r1 + r2 2r1r2 cos (d(x1, x2)).

One can show that (Cone(X), dCone(X)) is a metric space, and for any c > 0

dCone(X) ([(x1, cr1)], [(x2, cr2)]) = cdCone(X) ([(x1, r1)], [(x2, r2)]) . For Riemannian manifolds, we have the following equivalent description of Euclidean metric cones. Given a Riemannian manifold (M, g) with diameter ≤ π, deﬁne 2 ⊗ (1.1.3.10) gCone := r g + dr dr on M × (0, ∞). We can show that

(1.1.3.11) dgCone = dCone(M) M × ∞ ⊂ M ∈ M ∈ ∞ on (0, ) Cone( ). Indeed, for any x1, x2 and r1, r2 (0, ), one has

(1.1.3.12) dgCone ((x1, r1), (x2, r2)) = dCone(M) ([(x1, r1)], [(x2, r2)]) . ∈ M γ → M In fact, given two points x1, x2 , let : [0, d(x1, x2)] be a unit speed ∈ ∞ minimal geodesic joining x1 to x2, where d := dg. Given r1, r2 (0, ), join the ∈ M × ∞ two points (x1, r1), (x2, r2) (0, ) by paths γ −→ M × ∞ 7−→ γ : [0, d(x1, x2)] (0, ), u ( (u), r(u)) → ∞ where r : [0, d(x1, x2)] (0, ) satisﬁes r(0) = r1 and r(d(x1, x2)) = r2. The γ length of with respect to gCone is given by ∫ √ d(x1,x2) (γ) = ( )2 + ( )2 LgCone r u r˙ u du. 0 a For any variation r = s with s(0) = s(d(x1, x2)) = 0, we have ∫ [ ] d(x ,x ) −1/2 a γ 1 2 2 2 LgCone ( ) = r(u) + r˙(u) [r(u)s(u) + r˙(u)s˙(u)] du 0 ∫ [ ] d(x1,x2) −1/2 = r(u)2 + r˙(u)2 r(u)s(u) { 0 [ ] [ ] ∫ −3/2 d(x1,x2) − − r(u)2 + r˙(u)2 r(u)r˙(u) + r˙(u)r¨(u) r˙(u) 0 [ ] } −1/2 + r(u)2 + r˙(u)2 r¨(u) s(u)du ∫ [ ] [ ] d(x1,x2) −3/2 = r(u)2 + r˙(u)2 −r(u)r¨(u) + 2r˙(u)2 + r(u)2 r(u)s(u)du. 0 1.2. TOPOLOGICAL SPACES 11

The Euler-Lagrange equation is now of the form 1 r(u) = a cos u + b sin u ∈ α ∈ π for some a, b R. Let := d(x1, x2) [0, ] and consider the Euclidean planar α triangle with side-angle-side equal to r1- -r2 and vertices with polar coordinates α (r1, 0), (0, 0), and (r2, ). By the law of sines, we obtain −1 − −1 α r1 − cos α 1 r2 r1 cos r2 a = , b = α = α r1 sin r1 sin and ∫ α √ 2 2 dgCone ((x1, r1), (x2, r2)) = r(u) + r˙(u) du 0 √ ∫ √ α α du − cos u = 2 + 2 = 2 + 2 a b 2 a b 0 (a cos u + b sin u) b(b sin u + a cos u) 0 √ √ ab2 sin α = a2 + b2 = r2 + r2 − 2r r cos α b sin α + a cos α 1 2 1 2 = dCone(Mm) ([(x1, r1)], [(x2, r2)]) .

1.2. Topological spaces A topological space is a pair (X, T ), where X is a set and T is a set of 2X, satisfying (1) U ∩ V ∈ T for any U, V ∈ T ; ∪ ∈ T { } ⊂ T (2) i∈IUi for any collection Ui i∈I ; (3) ∅, X ∈ T . An element of T is called an open subset of X (or open set if both X and T are understood), while a subset in X is called closed it its complement in open.

Suppose that (X, T ) is a topological space and A ⊆ X. Deﬁne T { ∩ | ∈ T } A := A U I . T Then clearly (A, A) is also a topological space.

If (X, T ) is a topological space, then, by deﬁnition, any intersection of ﬁnitely many open subsets of X is still an open subsets of X.

We have seen in Section 1.1 that any metric space is always a topological space. Hence the Euclidean space Rm is a topological space, whose topology is called the Euclidean topology. The following are some standard subsets of Rm: • The unit interval: I = [0, 1] = {x ∈ R|0 ≤ x ≤ 1}. • The (open) unit ball in Rm:   ( )  1/2  m 1 ··· m ∈ m | | i 2 B := x = (x , , x ) R x = ∑ (x ) < 1 . 1≤i≤m When m = 2, we call D := B2 the (open) unit disk. 12 1. BASIC TOPOLOGY

• The (closed) unit ball in Rm:

Bm ≡ Bm := {x ∈ Rm||x| ≤ 1}.

When m = 2, we call D := B2 the closed unit disk. • The (unit) circle:

S1 := {x ∈ R2||x| = 1} = {z ∈ C||z| = 1}.

• The (unit) m-sphere:

Sm := {x ∈ Rm+1||x| = 1}.

Let (X, T ) be a topological space. Then • X and ∅ are closed subsets of X. • Any union of ﬁnitely many closed subsets of X is a closed subset of X. • Any intersection of arbitrarily many closed subsets of X is a closed subset of X. For example, let X = {1, 2, 3} and

T := {∅, {1}, {1, 2}, {1, 2, 3}} .

Then (X, T ) is a topological space. Clearly ∅, {3}, {2, 3}, and {1, 2, 3} are all closed subsets of X. However, the subset {2} is neither open nor closed of X. To describe {2}, introduce the following concepts.

Suppose that (X, T ) is a topological space and A is any subset of X. • The closure of A in X is ∩ Cle(A) ≡ A := {B ⊂ X|B ⊃ A and B is closed in X} .

Then A is closed in X. • The interior of A is ∪ Int(A) ≡ A˚ := {C ⊂ X|C ⊂ A and C is open in X} .

Then A˚ is open in X. • The exterior of A is

Ext(A) := X \ A.

• The boundary of A is

Bdy(A) ≡ ∂A := X \ (Int(A) ∪ Ext(A)).

By deﬁnition, we obtain

A˚ ⊆ A ⊆ A, X = Int(A) ⨿ Ext(A) ⨿ Bdy(A). 1.2. TOPOLOGICAL SPACES 13

1.2.1. Continuous maps and bases. Let (X, T ) be a topological space. A neighborhood of x ∈ X is a set N ⊆ X containing an open subset U ∈ T with the property x ∈ U ⊂ N. (1) An neighborhood may not be open. (2) An neighborhood can be the entire space X. (3) The intersection of two neighborhoods of x is also a neighborhood of x. X A neighborhood basis at x ∈ X is a set Bx ⊂ 2 of neighborhoods of x with the property that every neighborhood N of x in X contains some B ∈ Bx. EXERCISE 1.2.1. Let (X, T ) be a topological space and A ⊆ X be any subset. It is easy to prove the following statements: (1) A point is in Int(A) if and only if it has a neighborhood contained in A. (2) A point is in Ext(A) if and only if it has a neighborhood contained in X \ A. (3) A point is in Bdy(A) if and only if every neighborhood of it contains both a point of A and a point if X \ A. (4) A point is in A if and only if every neighborhood of it contain a point of A. (5) A = A ∪ ∂A = A˚ ∪ ∂A. (6) Int(A) and Ext(A) are open in X, while A and ∂A are closed in X. (7) The following are equivalent: – A is open in X. – A = Int(A). – A contains none of its boundary points. – Every point of A has a neighborhood contained in A. (8) The following are equivalent: – A is closed in X. – A = A. – A contains all of its boundary points. – Every point of X \ A has a neighborhood contained in X \ A. ◦ (9) X \ A = X \ A˚ and (X \ A) = X \ A. (10) Let A be a collection of subsets of X. Show that ∩ ∩ ∪ ∪ ⊆ A, A ⊇ A A∈A A∈A A∈A A∈A

and ( )◦ ( )◦ ∩ ∩ ∪ ∪ A ⊆ A˚, A ⊇ A˚. A∈A A∈A A∈A A∈A Suppose that (X, T ) is a topological space and A ⊆ X is a subset. • We say that x ∈ X is a limit point (or accumulation point, cluster point) of A if every neighborhood of x contains a point of A other than x (note that x itself may not be in A). • We say that x ∈ X is an isolated point of A if x has a neighborhood U in X such that U ∩ X = {x}. • A is said to be dense in X if A = X. Clearly ever point of A is either a limit point or an isolated point, but not both.

EXERCISE 1.2.2. Let (X, T ) be a topological space and A ⊆ X. Show that 14 1. BASIC TOPOLOGY

• A is closed if and only if it contains all of its limit points. • A is dense if and only if every nonempty open subset of X contains a point of A.

THEOREM 1.2.3. Let (X, d) be a metric space. If A ⊆ X, then { ∈ { } ⊂ } (1.2.1.1) A = x X : lim xn = x for some sequence xn n∈N A . n→∞

∈ { } ⊂ PROOF. If x X with limn→∞ xn = x for some sequence xn n∈N A, any open set around x contains at least one point in A. Hence x ∈ A. Conversely, for −1 −1 a point x ∈ A, consider the ball B(x, n ). Then x ∈ B(x, n ) ∩ A for some xn. −1 Thus d(x, xn) < n . T → T A continuous map f : (X, X) (Y, Y) between topological spaces is a → −1 ∈ T ∈ T map f : X Y such that f (U) Y whenever U Y. T → T We say that a map f : (X, X) (Y, Y) is continuous at x if for any neighborhood N of f (x) in Y there exists a neighborhood M of x in X with f (M) ⊂ N. From the deﬁnition of neighborhoods, f is continuous at x if and only if for any − neighborhood N of f (x) in Y the inverse image f 1(N) is a neighborhood of x in X. T → T THEOREM 1.2.4. A map f : (X, X) (Y, Y) is continuous if and only if f is continuous at each point in X.

PROOF. Suppose ﬁrst that f is continuous. Let N be a neighborhood of f (x) in − − Y and choose an open set U in Y with f (x) ∈ U ⊂ N. Then x ∈ f 1(U) ⊂ f 1(N) − and f 1(N) is a neighborhood of x in X. Hence f is continuous at x. − − Conversely, given an open set U ⊂ Y. For any x ∈ f 1(U), we see that f 1(U) −1 is a neighborhood of x so that x ∈ Vx ⊂ f (U) for some open set Vx in X. Then ∪ −1 ⊂ T f (U) = Vx X − x∈ f 1(U) and f is continuous.

T → T EXAMPLE 1.2.5. (1) Every constant map f : (X, X) (Y, Y) is continuous. T → T (2) The identity map 1X : (X, X) (X, X) is continuous. T → T (3) If f : (X, X) (Y, Y) is continuous, so is the restriction of f to any open subset U ⊆ X. T → T T → T (4) For continuous maps f : (X, X) (Y, Y) and g : (Y, Y) (Z, Z), the ◦ T → T composite map g f : (X, X) (Z, Z) is also continuous. We also observe that for T → T ◦ ◦ any continuous map f : (X, X) (Y, Y) we have f 1X = f and 1Y f = f. T → T We say a map f : (X, X) (Y, Y) is a homeomorphism if f is bijective and −1 T T f, f are continuous. In this case, we say that (X, X) and (Y, Y) are homeomorphic or topologically equivalent, and denote by X ≈ Y. T → T EXERCISE 1.2.6. (1) Let f : (X, X) (Y, Y) be a bijection between topo- T T logical spaces. Show that f is homeomorphic if and only if f ( X) = Y, that is, ∈ T ∈ T U X if and only if f (U) Y. 1.2. TOPOLOGICAL SPACES 15

T → T ∈ T (2) Let f : (X, X) (Y, Y) be a homeomorphism and U X. Show that ∈ T | T → T f (U) Y and f U is a homeomorphism from (U, U) ( f (U), f (U)). (3) Any translation m −→ m 7−→ R R , x x + x0 and dilation Rm −→ Rm, x 7−→ λx are homeomorphic. (4) The map F : Bm → Rm given by x F(x) := . 1 − |x| (5) Let C := {(x, y, z) ∈ R3| max{|x|, |y|, |z|} = 1} be the cubical surface of side 2 centered at (0, 0, 0). Deﬁne (x, y, z) F : C −→ S2, (x, y, z) 7−→ √ . x2 + y2 + z2 Show that F is homeomorphic. (6) Let X := [0, 1) ⊆ R and Y := S1 ⊆ C. Deﬁne √ √ π − e : X −→ Y, e(s) := e2 1s = cos(2πs) + −1 sin(2πs). Show that e is continuous and bijective, but is not homeomorphic. T → T ⊆ T ∈ T A map f : (X, X) (Y, Y) is open if f (U) Y for all U X; f is closed if f (C) is closed in Y for all closed set C ⊆ X. T → T EXERCISE 1.2.7. Let f : (X, X) (Y, Y) be a bijective continuous map. Show that the following are equivalent: • f is a homeomorphism. • f is open. • f is closed.

T → T PROPOSITION 1.2.8. Let f : (X, X) (Y, Y) be a map between topological spaces. Show that (1) f is continuous if and only if f (A) ⊆ f (A) for all A ⊆ X. (2) f is closed if and only if f (A) ⊇ f (A) for all A ⊆ X. − − ◦ (3) f is continuous if and only if f 1(B˚) ⊆ ( f 1(B)) for all B ⊆ Y. − − ◦ (4) f is open if and only if f 1(B˚) ⊇ ( f 1(B)) for all B ⊆ Y.

PROOF. We only give proves of (1) and (2). (1) Let f be a continuous map and A be a given subset of X. For any point x ∈ A, we shall prove f (x) ∈ f (A). According to Exercise 1.2.1 (4), it suffices to verify that every neighborhood N of f (x) contains a point in f (A). By definition, we have an open subset V of f (x) such that V ⊆ N. Then, by continuity, U := −1 ∈ T ∈ f (V) X and x U. Hence by Exercise 1.2.1 (4) again, we can find a point ′ ′ ′ x ∈ A ∩ U, so that y := f (x ) ∈ f (A) ∩ V and f (x) ∈ f (A). 16 1. BASIC TOPOLOGY

Conversely, assume f (A) ⊆ f (A) holds for all A ⊆ X. We ﬁrst show that for − − any closed subset B of Y, f 1(B) is closed in X. Let A := f 1(B) ⊆ X. To prove − f 1(B) is closed in X, we verify that A = A by Exercise 1.2.1 (8). If x ∈ A, then − f (x) ∈ f (A) ⊆ f (A) = f ( f 1(B)) ⊂ B = B ∈ ∈ T so that x A. Hence A = A. Next we show the continuity of f. Let V Y and set B := Y \ V. Then − − − − f 1(B) = f 1(Y) \ f 1(V) = X \ f 1(V). − − Since f 1(B) is closed, it follows that f 1(V) is open. (2) Let f be a closed map. Then for any A ⊆ X, we have that f (A) is closed and then f (A) ⊆ f (A), f (A) ⊆ f (A) = f (A). Conversely, assume that f (A) ⊇ f (A) holds for all A ⊆ X. For a closed subset C of X, we get f (C) ⊆ f (C) = f (C) so that f (C) is closed in Y.

In Proposition 1.2.8, together with Theorem 1.2.4, we actually proved that for T → T a map f : (X, X) (Y, Y) between topological spaces, the following are equivalent: • f is continuous, • for any x ∈ X, f is continuous at x, − • for any closed subset B of Y, f 1(B) is closed in X, • for any A ⊆ X, f (A) ⊆ f (A).

T → T A map f : (X, X) (Y, Y) between topological spaces is said to be locally homeomorphic if every point x ∈ X has a neighborhood U ⊆ X such that f (U) ∈ T | → Y and f U : U f (U) is a homeomorphism.

EXERCISE 1.2.9. Show that • Every homeomorphism is a local homeomorphism. • Every local homeomorphism is continuous and open. • Every bijective local homeomorphism is a homeomorphism.

Given a topological space (X, T ).A basis for T is a subset B ⊆ T such that any open set U of X can be written as an union of some members of B. (1) Let B ⊆ T be a basis for T . Then clearly ∪ – X = B∈B B, and ∈ B ∈ ∩ ∈ B – if B1, B2 and x B1 B2, there exists an element B3 such ∈ ⊆ ∩ that x B3 B1 B2. 1.2. TOPOLOGICAL SPACES 17

(2) Consider a subset C ⊆ T satisfying

for any U ∈ T and x ∈ U there exists Cx ∈ C such that x ∈ Cx ⊆ U. Then C is a basis for T . In fact, one has ∪ U = Cx, Cx ∈ C . x∈U

PROPOSITION 1.2.10. Let X be a set and suppose that B is a collection of subsets of X. Then B is a basis for some topology on X if and only if it satisﬁes the following two conditions: ∪ (1) X = B∈B B. ∈ B ∈ ∩ ∈ B (2) If B1, B2 and x B1 B2, there exists an element B3 such that ∈ ⊆ ∩ x B3 B1 B2. If so, there is a unique topology on X for which B is a basis, called the topology generated by B.

PROOF. One direction is obvious. Suppose now that B satisfies (1) and (2). Let TB be the collection of all unions of elements of B. We shall verify that TB is a topology on X. • X, ∅ ∈ TB. By condition (1), X ∈ TB, and ∅ ∈ TB as the union of the empty collection of elements of B. • T { } ⊆ T B is closed under arbitrary unions. For any collection Ui i∈I B, write each Ui as ∪ Ui = Bij. ∈ j Ji Then ∪ ∪ ∪ ∈ T Ui = Bij B. ∈ ∈ ∈ i I i I j Ji • TB is closed under finite intersections. It is enough to prove that whenever U, V ∈ TB one has U ∩ V ∈ TB. Given x ∈ U ∩ V. Then there exist ∈ B ∈ ⊆ ∈ ⊆ B1, B2 such that x B1 U and x B2 V. By condition (2), there ∈ B ∈ ⊆ ∩ ⊆ ∩ exists Bx such that x Bx B1 B2 U V. Consequently, ∪ U ∩ V = Bx x∈U∩V so that U ∩ V ∈ TB. Next we show that B is a basis for TB. By definition of TB, it is clear. ′ If T is another topology on X for which B is a basis, then B ⊆ T and, ∈ T ′ ∪ { } B for every U , we have U = i∈I Bi, for a collection Bi i∈I of . Hence ′ ′ ′ U ∈ TB and then T ⊆ TB. Since B ⊆ T , it follows that TB ⊆ T . Therefore ′ T = TB.

Given a topological space (X, T ).A subbasis for T is a set S ⊂ 2X such that B is a basis for T , where B is the collection of all ﬁnite intersections of members of S (so ∅, X ∈ B). 18 1. BASIC TOPOLOGY

Given any collection S of subsets of any set X, we can construct a topology T on X such that S is a subbasis for T . Indeed, we deﬁne T to be the collection of arbitrary unions of the ﬁnite intersections of members of C , assuming the convenience that ∅ is the union of an empty collection of sets and X is the intersection of an empty collection of sets. This topology is called the topology generated by S .

A topology space (X, T ) is said to be first countable if for any x ∈ X there exists a countable neighborhood basis at x (that is, a collection Bx of neighborhoods of x such that every neighborhood of contains some B ∈ Bx). (X, T ) is said to be second countable if T has a countable basis B. (1) Any metric space is first countable. If (X, d) is a metric space, then d in- T ∈ { } duces the metric topology . For each x X, the collection B(x, r) r∈Q+ is a neighborhood basis at x. (2) Euclidean spaces are second countable. Consider the countable base B = {Bm(x, r)|x ∈ Qm, r ∈ Q+} of Rm. (3) Not every metric space is second countable, for example X = [0, 1] with the trivial metric. (4) Any second countable space is first countable.

THEOREM 1.2.11. Let (X, d) be a metric space. X is second countable if and only if X contains a countable dense subset.

B { } PROOF. If X is second countable, then X has a countable basis = Bi i∈N. ∈ ∈ { } Choose xi Bi for each i N and let A := xi i∈N. Then A is a countable dense set. { } B Conversely, let A = xi i∈N be a countable dense set in X. If we take = ( ( )) ∈ ∈ B B xi, r i N,r Q>0 , then is a countable basis. A topological space is said to be separable if it contains a countable dense subset.

Let (X, T ) be a topological space and (Y, d) is a metric space. We say a se- { } quence of maps f n n∈N between X and Y converges uniformly to a map f : X → Y if ( ) ( ( ) ( )) = lim→∞ sup d f i x , f x 0. n x∈X ⇒ In this case, we write it as f n f. { } T THEOREM 1.2.12. Let f n n∈N, f be maps between a topological space (X, ) and ⇒ a metric space (Y, d). If f n are continuous and f n f , then f is continuous.

ϵ ∈ ϵ PROOF. Given > 0 there is an integer n0 N such that d( f (x), f (x0)) < ∈ ≥ ∈ 3 for any x X and n n0. Given x0 X there exists a neighborhood N of x0 in X such that ϵ ∈ d( f (x), f (x0)) < , x N. m0 n0 3 1.2. TOPOLOGICAL SPACES 19

Hence d( f (x), f (x )) ≤ d( f (x), f (x)) + d( f (x), f (x )) + d( f (x ), f (x )) 0 n0 n0 n0 0 n0 0 n0 0 ϵ ϵ ϵ < + + = ϵ. 3 3 3 Thus f is continuous at x0. By Theorem 1.2.3, we see that f is continuous.

Let X be a set and P some condition given on subsets of X. Consider a topology T on X whose open sets satisfy P. (1) T is the smallest topology or coarsest topology satisfying P, if for any ′ topology T on X whose open sets satisfying P, then T -open sets are ′ ′ T -open. In this case we use the notion T ≺ T . (2) T is the largest topology or ﬁnest topology satisfying P, if for any topol- ′ ′ ogy T on X whose open sets satisfying P, then T -open sets are T - ′ open. In this case, we use the notion T ≺ T .

EXAMPLE 1.2.13. (1) (Trivial topology) Any set X can be equipped with a topology Tt = {∅, X}. This is the smallest topology on X. T (2) (Discrete topology) Any set X can also be equipped with another topology d = 2X. This is the largest topology on X. (3) (Finite complement topology) Given an infinite set X. The collection of sub- T sets whose complement are finite, together with the empty set, forms a topology fc. (4) (Countable complement topology) Given an infinite set X. The collection of subsets whose complement are countable, together with the empty set, forms a topology Tcc. (5) (Particular point topology) Given an infinite set X and x ∈ X. Show that

Tpp := {U ⊆ X|U = ∅ or x ∈ U} is a topology on X. (6) (Excluded point topology) Given an inﬁnite set X and x ∈ X. Show that

Tep := {U ⊆ X|U = X or x ∈/ U} is a topology on X. (7) Show that (R, Tpp) is first countable and separable, but not second countable. (8) Show that (R, Tep) is first countable, but not second countable or separable. T (9) Show that (R, fc) is separable, but not first or second countable.

1.2.2. Categories: an introduction. A category C is a triple

(Ob(C), HomC(·, ·), ◦) , where Ob(C) is a class of objects, HomC(X, Y) is a set for every pair (X, Y) of objects (its element, called morphism, is usually written as f : X → Y; we call X is the domain of f and Y the range of f ), and ◦ is the composition on sets (that is, given X, Y, Z ∈ Ob(C), we deﬁne ◦ : HomC(X, Y) × HomC(Y, Z) → HomC(X, Z) by ◦( f , g) := g ◦ f ), satisfying the following axioms (1) (Associativity) h ◦ (g ◦ f ) = (h ◦ g) ◦ f for any morphisms

f g h X −−−−→ Y −−−−→ Z −−−−→ W 20 1. BASIC TOPOLOGY

∈ ∈ (2) (Identity) For any X Ob(C) there exists a morphism 1X HomC(X, X), called the identity morphism of X, such that ◦ ◦ 1X f = f , g 1X = g ∈ ∈ for any f HomC(Y, X) and g HomC(X, Y). Note that 1X is unique. C is called small if Ob(C) is a set.

∈ ∈ ◦ If morphisms f HomC(X, Y) and g HomC(Y, X) satisfying g f = 1X, we say that g is a left inverse of f and f is a right inverse of g respectively. A two-sided inverse or an inverse of a morphism f is a morphism which is both a left inverse of f and a right inverse of f . A morphism f : X → Y is said to be an equivalence if there exists a morphism g : Y → X which is a two-sided inverse of f . (1) If f : X → Y has a left inverse and a right inverse, then they are equal ′ and f is an equivalence. Indeed, if g : Y → X is a left inverse and ′′ → ′ ◦ ◦ ′′ g : Y X is a right inverse, then g f = 1X and f g = 1Y so that ′ ′ ◦ ′ ◦ ◦ ′′ ◦ ′′ ′′ g = g 1Y = g ( f g ) = 1X g = g . − (2) If f : X → Y is an equivalence, then by (1) it has a unique inverse f 1 : − Y → X and f 1 is also an equivalence. (3) We say two objects X and Y are equivalent, written as X ≈ Y, if there exists an equivalence f : X → Y. (4) It is clear that ≈ is an equivalence relation in any set of objects of C. A morphism f : X → Y is an isomorphism if there exists g : Y → X such ◦ ◦ that f g = 1Y and g f = 1X. Such a g, which is unique, is called the inverse − of f and is denoted by f 1. Clearly that an isomorphism is exact an equivalence. An endomorphism is a morphism f : X → X with same domain and range. An automorphism is an endomorphism which is an isomorphism. Two morphisms f and g are parallel if they have same domain and same range, i.e., f , g : X ⇒ Y, or

f X −−−−→ Y. g A morphism f : X → Y is a monomorphism if for any pair of parallel morphisms ⇒ ◦ ◦ g1, g2 : Z X, f g1 = f g2 implies g1 = g2.

g f Z −−−−→1 X −−−−→ Y g2 A morphism f : X → Y is an epimorphism if for any pair of parallel morphisms ⇒ ◦ ◦ g1, g2 : Y Z, g1 f = g2 f implies g1 = g2.

f g X −−−−→ Y −−−−→1 Z. g2

Note that f : X → Y is a monomorphism if and only if the map f ◦ : HomC(Z, X) → HomC(Z, Y) is injective for any object Z ∈ Ob(C), and f : X → Y is an epimorphism if and only if the map ◦ f : HomC(Y, Z) → HomC(X, Z) is injective for any object Z ∈ Ob(C). If f g X −−−−→ Y −−−−→ Z 1.2. TOPOLOGICAL SPACES 21 are morphisms and if f and g are monomorphisms (resp. epimorphisms, resp. isomorphisms), then g ◦ f is a monomorphism (resp. epimorphism, resp. isomorphism).

We sometimes write f : X Y or else f : X ,→ Y to denote a monomorphism and f : X Y to denote an epimorphism.

EXAMPLE 1.2.14. (1) Set is the category of sets and set maps. (2) Group is the category of groups and group homomorphisms. (3) AGroup is the category of Abelian groups and group homomorphisms. (4) Ring is the category of rings and ring homomorphisms. (5) CRing is the category of commutative rings and ring homomorphisms. (6) ModR is the category of R-modules and R-module homomorphisms. (7) Vect(R) is the category of real vector spaces and R-linear maps. (8) Vect(C) is the category of complex vector spaces and C-linear maps. (9) Top is the the category of topological spaces and continuous maps. (10) Top∗ is the category of pointed topological space and pointed continuous maps. ∈ A pointed topological space is a pair (X, x0) where X is a topological space and x0 X; a → → pointed continuous map f : (X, x0) (Y, y0) is a continuous map f : X Y such that f (x0) = y0. π The fundamental groups 1 roughly speaking is a map from Top∗ to Group; π namely associated to a pointed topological space (X, x0) a group 1(X, x0). We π also can deﬁne higher homotopy groups i(X, x0) which turns out to be Abelian for any i ≥ 2.

′ Let C be a category. A subcategory C of C is itself a category and satisﬁes ′ • any object in C is also object in C, • ∈ ′ ⊆ For any X, Y Ob(C ), we have HomC′ (X, Y) HomC(X, Y), and • ∈ ∈ ′ 1X HomC′ (X, X) for any X Ob(C ). ′ A subcategory C is called a full subcategory if moreover

HomC′ (X, Y) = HomC(X, Y) ′ for any X, Y ∈ Ob(C ). Top is a subcategory, but not a full subcategory, of Set.

◦ The opposite category C of a category C is deﬁned by ◦ Ob(C ) := Ob(C), HomC◦ (X, Y) := HomC(Y, X).

EXERCISE 1.2.15. Let C be a category. An object P ∈ Ob(C) is initial if for any Y ∈ Ob(C), HomC(P, Y) has exactly one element. An object Q ∈ Ob(C) is ﬁnal if for any X ∈ Ob(C), HomC(X, Q) has exactly one element. Show that two initial (resp. ﬁnal) objects are isomorphic. ′ A (covariant) functor F : C → C between two categories consists of ′ (1) a map F : Ob(C) → Ob(C ), and (2) for any two objects X, Y ∈ Ob(C), a map → ′ F : HomC(X, Y) HomC′ (F(X), F(X )). 22 1. BASIC TOPOLOGY

These data satisfy ◦ ◦ (1.2.2.1) F(1X) = 1F(X), F( f g) = F( f ) F(g). ′ ◦ ′ A contravariant functor G : C → C is a covariant functor from C to C .

EXAMPLE 1.2.16. (1) Let C be a category and take an object X ∈ Ob(C). Deﬁne

(1.2.2.2) HomC(X, ·) : C −→ Set, Z 7−→ HomC(X, Z), (1.2.2.3) HomC(·, X) : C −→ Set, Z 7−→ HomC(Z, X).

Then HomC(X, ·) is a functor, while HomC(·, X) is a contravariant functor. (2) The forgetful functor F : Top → Set. π → 7→ (3) The fundamental group functor 1 : Top∗ Group, given by (X, x) π 1(X, x). → ′ Let F1, F2 : C C be two functors. A morphism or natural transformation Θ from F1 to F2 consists of Θ ∈ ∈ (X) HomC′ (F1(X), F2(X)) whenever X Ob(C). These data satisfy the following commutative diagram: Θ(X) F (X) −−−−→ F (X) 1 2   ◦ Θ Θ ◦ (1.2.2.4) F1( f )y yF2( f ) F2( f ) (X) = (Y) F1( f ) Θ −−−−→(Y) F1(Y) F2(Y) for any X, Y ∈ Ob(C) and any f ∈ HomC(X, Y). ′ ′ DEFINITION 1.2.17. For two categories C and C , deﬁne a new category Func(C, C ) as follows: ′ ′ Ob(Func(C, C )) := {functors from C to C }, { } HomFunc(C,C′)(F1, F2) := morphisms from F1 to F2 . Let C be a category. A functor F : C → Set is said to be representable if there exists an object X ∈ Ob(C) such that F is isomorphic to HomC(X, ·) in the category Func(C, Set). ′ EXERCISE 1.2.18. (1) Show that Func(C, C ) is a category. (2) If F : C → Set is representable for some X ∈ Ob(C), then X is unique to isomorphism and is called a representative of F. ′ Let F : C → C be a functor. (1) F is fully faithful if for any objects X, Y ∈ Ob(C), the map −→ HomC(X, Y) HomC′ (F(X), F(Y)) is bijective. (2) F is an equivalence of categories if F is fully faithful and for any object ′ ′ ′ X ∈ Ob(C ) there exists X ∈ Ob(C) and an isomorphism X → F(X). ′ EXERCISE 1.2.19. F ∈ Ob(Func(C, C )) is an equivalence of categories if and ′ ′ only if there exists F ∈ Ob(Func(C , C)) and isomorphisms ◦ ′ → ′ ′ ′ ◦ → F F 1C′ in Func(C , C ) and F F 1C in Func(C, C). 1.2. TOPOLOGICAL SPACES 23

Let C be a category.Deﬁne ∨ ◦ (1.2.2.5) C := Func(C , Set) and ∨ (1.2.2.6) HomC : C −→ C , X 7−→ HomC(·, X).

THEOREM 1.2.20. (Yoneda’s lemma) (1) For any X ∈ Ob(C) and any F ∈ ∨ Ob(C ), one has

(1.2.2.7) HomC∨ (HomC(X), F) is isomorphic to F(X) in Set.

(2) HomC is a fully faithful functor.

∈ · ϕ ∈ PROOF. (1) To f HomC∨ (HomC( , X), F), we associate ( f ) F(X) as follows: −→ 7−→ ϕ f (X) : HomC(X, X) F(X), 1X ( f ) := f (X)(1X). ∈ ψ ∈ · Conversely, to s F(X), we can associate (s) HomC∨ (HomC( , X), F) as follows: For Y ∈ Ob(C), −−−−→F −−−−→s HomC(Y, X) HomSet(F(X), F(Y)) F(Y) we deﬁne ψ(s)(Y) := s ◦ F, where ∈ s( f ) := f (s), f HomSet(F(X), F(Y)). Then ϕ and ψ are inverse to each other. (2) For any X, Y ∈ Ob(C),

HomC ∨ (HomC(·, X), HomC(·, Y)) → HomC(·, Y)(X) = HomC(X, Y) is bijective by (1). 1.2.3. Subspaces. Let (X, T ) be a topological space and S ⊆ X is a subset. ⊆ T T { ∩ | ∈ T } S X is a subspace if (S, S) is a topological space, where S := U S U is the relative topology or subspace topology. The inclusion map ι T −→ T (1.2.3.1) S : (S, S) (X, ) is continuous.

EXAMPLE 1.2.21. Consider the subspaces ∪ { } S1 := [0, 1] (2, 3), S2 := 1/n n≥1 of R. Note that [0, 1] is not an open interval of R, but is an open subset of S1, since [0, 1] = ∩ − { } S1 ( 1, 2). In S2, the one-point sets 1/n are all open so the subspace topology on S2 is discrete. Let S be a subspace of (X, T ). We sat a subset U ⊆ S is relatively open or T relatively closed in S if U is open or closed in the subspace topology S. PROPOSITION 1.2.22. Suppose that S is a subspace of a topological space (X, T ). (1) If U ⊆ S ⊆ X, U is open in S and S is open in X, then U is open in X. The same is true with “closed” in place of “open”. (2) If U is a subset of S that is either open or closed in X, then it is also open or closed in S, respectively. (3) If U ⊆ S ⊆ X, then the closure of U in S is equal to U ∩ S. 24 1. BASIC TOPOLOGY

(4) If U ⊆ S ⊆ X, then the interior of U in S contains U˚ ∩ S.

PROOF. (1) U is open in S implies U = S ∩ V for some V ∈ T . Because S ∈ T , we must have U ∈ T . (2) Assume that U is open in X. Then U = S ∩ U is open in S. (3) Write US as the closure of U in S. We shall prove that US = U ∩ S. Suppose ′ x ∈ US and N is any neighborhood of x in X. Then N := N ∩ S is a neighborhood ′ of x in S, so N ∩ U ̸= ∅. Because U ⊂ S, we get N ∩ U ̸= ∅ and x ∈ U ∩ S. ′ ′ Conversely, let x ∈ U ∩ S and N is any neighborhood of x in S. By definition, N ′ ∈ T ′ ∋ ′ ∩ contains an open subset V S satisfying V x. Write V = N S for some N ∈ T . Then N itself is a neighborhood of x in X, so N ∩ U ̸= ∅. Consequently, ′ ′ V ∩ U ̸= ∅ and N ∩ U ̸= ∅, so x ∈ US. ∈ ∩ (4) Write IntS(U) as the interior of U in S. Let x U˚ S. Then we can find ′ a neighborhood N ⊆ X containing x. Set N := N ∩ S a neighborhood of x in S. ∈ Hence x IntS(U). ∩ ( In Proposition 1.2.22 (4), we can find an example so that U˚ S IntS(U). ∪ Indeed, consider X = R, S = [0, 1] (2, 3) and U = [0, 1]. Then IntS(U) = [0, 1] but U˚ ∩ S = (0, 1).

THEOREM 1.2.23. (1) (Characteristic property of the subspace topology) Sup- pose that (X, T ) is a topological space and S ⊆ X is a subspace. For any topological space T T → T (Y, Y), a map f : (Y, Y) (S, S) is continuous if and only if the composite map ι ◦ T → T S f : (Y, Y) (X, ) is continuous:

? XO ι ◦ f S ι S / Y S f

(2) (Uniqueness of the subspace topology) Suppose that S is a subset of a topological space (X, T ). The subspace topology on S is the unique topology for which the characteristic property (1) holds.

ι ◦ T → T ∈ T PROOF. (1) Suppose that S f : (Y, Y) (X, ) is continuous. If U S, ∩ ι−1 ∈ T then U = V S = S (V) for some V . Thus −1 −1 ι−1 ι ◦ −1 ∈ T f (U) = f ( S (V)) = ( S f ) (V) Y. Conversely, suppose that f is continuous. For any V ∈ T , we have ι ◦ −1 −1 ι−1 −1 ∩ ∈ T ( S f ) (V) = f ( S (V)) = f (S V) Y. ι ◦ Hence S f is continuous. T ′ (2) Suppose S is equipped with some topology S that satisﬁes the character- T istic property (1). For convenience, let Ss denote the topological space (S, S) and T ′ T T ′ Ss′ denote the topological space (S, ). To prove S = we should verify that A → S (by Exercise 1.2.6) the identity map 1S : S S is a homeomorphism between Ss and Ss′ . 1.2. TOPOLOGICAL SPACES 25

ι → T Clearly that the inclusion map s : Ss (X, ) is continuous. Since Ss′ satis- ι → T ﬁes the characteristic property, it follows that the inclusion map s′ : Ss′ (X, ) is also continuous. Consider the following two diagrams: > XO > XO }} }} ι ′ ◦1 ′ } ιs◦1 ′ } s ss } ι ′ s s } ι }} s }} s }} }} / / S S ′ S ′ S s ′ s s ′ s 1ss 1s s Here −→ −→ 1ss′ : Ss Ss′ , 1s′s : Ss′ Ss. ι ◦ ι ι ◦ ι Because s′ 1ss′ = s and s 1s′s = s′ , we, using again the characteristic property, → obtain the continuities of both 1ss′ and 1s′s. Thus 1S : S S is a homeomorphism between Ss and Ss′ .

T → T COROLLARY 1.2.24. Let f : (X, X) (Y, Y) be a continuous map between topological spaces. (1) (Restricting the domain) The restriction of f to any subspace S ⊆ X is continuous. (2) (Restricting the codomain) If T is a subspace of Y that contains f (X), then f : X → T is continuous. (3) (Expanding the codomain) If Y is a subspace of Z, then f : X → Z is continuous.

| ◦ ι PROOF. (1) Because f S = f S. (2) By Theorem 1.2.23. (3) Because the map X → Z is the composite of f : X → Y with the inclusion Y ,→ Z.

PROPOSITION 1.2.25. Suppose S is a subspace of a topological space (X, T ). (1) If R ⊆ S is a subspace of S, then R is a subspace of X. B T B { ∩ | ∈ B} T (2) If is a basis for , then S := B S B is a basis for S. (3) Every subspace of a ﬁrst countable space is ﬁrst countable. (4) Every subspace of a second countable space if second countable.

T T ∈ T ∩ PROOF. (1) It sufﬁces to prove ( S)R = R. Let U ( S)R. Then U = R V ∈ T ∩ ∈ T for some V S. But V can be written as V = S W for some W , we obtain ∩ ∩ ∩ ∈ T U = R (S W) = R W R. ∈ T ∩ ∈ T Conversely, Let U R. Then U = R W for some W and ∩ ∩ ∈ T U = R V, V := S W S. ∈ T Hence U ( S)R. B ⊆ T ∈ T (2) By deﬁnition of subspace topology, S S. For any U S, we have U = S ∩ W for some W ∈ T . Since B is a basis for T , it follows that ∪ ∈ B W = Bi, Bi . i∈I ∪ ∩ ∩ ∈ B B T Therefore U = i∈I Bi S with Bi S S. Thus S is a basis for S. 26 1. BASIC TOPOLOGY

(3) Assume that S is a subspace of a first countable topological space (X, T ). ∈ B { } For any x S, we can find a countable neighborhood basis x = Bi i∈N at x B { ∩ } in X. Then S,x := Bi S i∈N is a countable neighborhood basis at x in S, so T (S, S) is first countable. (4) The proof is similar to (3).

Let (X, T ) be a topological space and S be closed in X. Clearly that if A is closed in S, then it is also closed in X.

THEOREM 1.2.26. (Gluing lemma) Let (X, T ) be a topological space and X = A ∪ B, where A, B are closed in X. → → | | (1) If f : A Y and g : B Y are continuous and f A∩B = g A∩B, then f, g combine to give a continuous function h : X → Y given by { f (x), x ∈ A, h(x) = g(x), x ∈ B. → | | (2) If f : X Y is a map, then f is continuous provided f A, f B are continuous.

PROOF. Evidently (2) directly follows from (1). To prove (1), we use criterion T → T of continuity below Proposition 1.2.8, that is, f : (X, X) (Y, Y) is continuous − if and only if f 1(B) is closed in X for any closed subset B of Y. For any closed subset C of Y, one has − − − h 1(C) = f 1(C) ∪ g 1(C). − − Since f and g are continuous, we have that f 1(C) and g 1(C) are closed in A and − B respectively. Hence h 1(C) is closed in X.

1.2.4. Separation axioms. Let (X, T ) be a topological space.

(1) T0-space: for any x ̸= y ∈ X, there exists U ∈ T containing x but not y, or V ∈ T containing y but not x. (2) T1-space: for any x ̸= y ∈ X, there exist U, V ∈ T such that U contains x but not y and V contains y but not x. (3) T2-space or Hausdorff space: for any x ̸= y ∈ X, there exist U, V ∈ T such that x ∈ U, y ∈ V, and U ∩ V = ∅. ∈ (4) T3-space or regular space: it is T1-space, and for any x X, any closed set F ⊆ X with x ∈/ F, there exist U, V ∈ T such that x ∈ U, F ⊆ V, and U ∩ V = ∅. ⊆ (5) T4-space or normal space: it is T1-space, and for any closed sets F, G X with F ∩ G = ∅, there exist U, V ∈ T such that F ⊆ U, G ⊆ V, and U ∩ V = ∅. By deﬁnition, it is clear that {normal spaces} ( {regular spaces} ( {Hausdorff spaces}.

Indeed, we shall prove that any Hausdorff space must be T1. By Theorem 1.2.27 below, we shall prove that any one-point set {x} in a Hausdorff space (X, T ) is closed. For any y ∈ X and y ̸= x, we can ﬁnd disjoint open subsets U, V ∈ T such that y ∈ U and x ∈ V respectively. But x ∈/ U, we have y ∈/ {x}. Thus {x} = {x}. We now give examples on strict inclusions. 1.2. TOPOLOGICAL SPACES 27

(1) RK is Hausdorff but not regular. Recall that RK denotes the real line with the topology generated by the basis consisting of all open intervals (a, b) \ { } and all subsets of the form (a, b) K, where K = 1/n n∈N. Observe that K is closed in RK and does not contain 0. Suppose that there exist disjoint open sets U and V containing 0 and K, respectively. We shall ﬁnd a contradiction. Choose a basis element, which must be of the form (a, b) \ K, containing 0 and lying in U. Take n sufﬁciently large so that 1/n ∈ (a, b) and then choose a basis element, which must be of the form (c, d), about 1/n contained in B. Choose z satisfying max(c, 1/(n + 1)) < ∈ ∩ z < 1/n; then z U V! Therefore RK is not regular. (2) Rℓ is regular. Recall that Rℓ denotes the real line with the topology generated by the basis consisting of all half-open intervals of the form [a, b) with a < b. Clearly one-point sets are closed in Rℓ so that, according to Theorem 1.2.27 below, Rℓ is T1. To prove normality, we suppose that A and B are disjoint closed subsets in Rℓ. For a ∈ A choose a basis element [a, xa) not intersecting B and similarly for any b ∈ B choose a basis element [b, x ) not intersecting A. Then open subsets b ∪ ∪ U := [a, xa), V := [b, xb) a∈A b∈B are disjoint open subsets about A and B respectively. 2 × (3) The Sorgenfrey plane Rℓ := Rℓ Rℓ is regular but not normal.

THEOREM 1.2.27. (1) A topological space is T1 if and only if one-point sets are closed sets. (2) Any subspaces of a Ti-space is Ti, where i = 2, 3. ∈ (3) AT2-space is T3 if and only if for any x X and any open subset U of x, there exists an open subset V of x such that V ⊆ U. ⊆ (4) AT2-space is T4 if and only if for any open set U and closed set A with A U, there exists an open set V such that A ⊆ V ⊆ V ⊆ U. (5) Any metric space is normal.

T ∈ ∈ PROOF. (1) Let (X, ) be a T1-space. Given any point x X. For any y X \{x} we have an open set Uy ∈ T such that y ∈ Uy but x ∈/ Uy. Then X \{x} = ∪ { } y∈X\{x}Uy is open and x is closed in X. Conversely, assume that a topological space (X, T ) satisfies the properties that any one-point set is closed. Consider any two distinct points , y ∈ X. We have that Uy := X \{x}, Vx := X \{y} are open. Therefore Uu ∈ T is an open set such ∈ ∈ T that y Yy but x / Uy. The similar result holds for Vx. Hence (X, ) is T1. T T ̸ (2) Assume first that (X, ) is T2 and S is a subspace of (X, ). For any x = y in S, we can find two disjoint open subsets U, V ∈ T of x, y respectively, satisfying ∩ ∅ ∩ ∩ ∩ ∅ T U V = . Then (U S) (V S) = so (S, S) is Hausdorff. T T T Next assume that (X, ) is T3 and S is subspace of (X, ). Clearly (S, S) is ∈ T1 space. Let x S and B be a closed subset of S disjoint from x. Then B ∩ S = BS = B by Proposition 1.2.22 (3), so x ∈/ B. Using the regularity of X, there exist disjoint open subsets U ∋ x and V ⊇ B. Then U ∩ S and V ∩ S are disjoint open subsets in T Y containing x and B respectively. Therefore (S, S) is regular. 28 1. BASIC TOPOLOGY

(3) Let (X, T ) be a Hausdorff space. Suppose first that (X, T ) is regular. Take any x ∈ X and any open subset U of x, and let B := X \ U. By regularity, there exist disjoint open subsets V ∋ x and W ⊃ B. We claim that V ∩ B = ∅. Otherwise, we can find y ∈ B ∩ V so that W is a neighborhood of y but disjoint from V, contradicting y ∈ V. Hence V ⊆ U. Conversely, suppose that x ∈ X and a closed subset B not containing x. Let U := X \ B/ By hypothesis, there exists an open subset V of x such that V ⊂ U. The open subsets V and X \ V are disjoint open subsets containing x and B respectively. (4) The proof is essentially similar to that of (3). (5) Let (X, d) be a metric space with the induced metric topology T , and A, B be disjoint closed subsets of X. For any a ∈ A, there exists ϵa > 0 so that ϵ ∩ ∅ ∈ ϵ ϵ ∩ B(a, a) B = . Similarly, for any b B there exists b > 0 so that B(b, b) A = ∅. Define ∪ ∪ ϵ ϵ U := B(a, a/2), V := B(b, b/2). a∈A b∈B Then U and V both are open and contain A and B respectively. We claim that ∩ ∅ ∈ ∩ ∈ ϵ ∩ ϵ U V = . Otherwise, for z U V, we have z B(a, a/2) B(b, b/2) for some a ∈ A and b ∈ B. Then ϵ + ϵ d(a, b) ≤ d(a, z) + d(z, b) < a b . 2 ϵ ≤ ϵ ϵ ∈ ϵ ϵ ≤ ϵ ∈ ϵ If a b, then d(a, b) < b so that a B(b, b). If b a, then b B(a, a. Both give a contradiction.

EXERCISE 1.2.28. Prove (4) in Theorem 1.2.27. Remark that a subspace of a normal space need not be normal. For example, consider the product space (for its deﬁnition see Section 1.3) RJ, where J is uncountable. It can be proved that RJ is regular but not normal. Moreover RJ is homeomorphic to the subspace (0, 1)J of [0, 1]J, which is normal.

THEOREM 1.2.29. Any regular space with a countable basis is normal.

PROOF. Let (X, T ) be a regular space with a countable basis B. Let A and B be disjoint closed subsets of X. For any x ∈ A there exists an open subset Ux not intersecting B. By Theorem 1.2.27 (3), we can ﬁnd an open subset Vx of x such that Vx ⊆ Ux. Choose an element of B containing x and contained in Vx; therefore we { } ⊆ B obtain a countable open subsets Un n∈N such that ∪ A ⊆ Un =: U, Un ∩ B = ∅. n∈N { } Similarly, we can ﬁnd another countable open subsets Vn n∈N such that ∪ B ⊆ Vn =: V, Vn ∩ A = ∅. n∈N Then U and V are open subsets containing A and B respectively, but they may have nonempty intersection. Given n ∈ N, set ∪ ∪ ′ \ ′ \ Un := Un Vi, Vn := Vn Ui. 1≤i≤n 1≤i≤n 1.2. TOPOLOGICAL SPACES 29

′ ′ Then Un and Vn are open, and ∪ ∪ ⊆ ′ ⊆ ′ A Un, B Vn. n∈N n∈N

Finally, deﬁne ∪ ∪ ′ ′ ′ ′ U := Un, V := Vn. n∈N n∈N ′ ∩ ′ ∅ ∈ ′ ∩ ′ ∈ ′ ∩ ′ ∈ Then U V = . Indeed, if x U V , then x Uj Vk for some j, k N. ≤ ∈ ∈ ∪ ≤ ∈ ∪ Assume j k. Then x Uj but x / 1≤i≤jVi; since j k, x / 1≤i≤kUk implies ∈ ′ ∩ ′ ∅ x / Uj. Hence U V = .

Consider a metric space (X, d), and A, B are disjoint closed subsets of X. De- ﬁne a map d(x, A) f : X −→ [0, 1], x 7−→ d(x, A) + d(x, B) where d(x, A) := inf{d(x, a)|a ∈ A}. Clearly that | ≡ | ≡ f A 0, f B 1. ∈ { } ⊂ We claim that f is continuous. Consider a given point x X and xn n∈N X with d(xn, x) → 0 as n → +∞. We may assume that x ∈/ A ∪ B, since A and B are closed. In this case, d(xn, A) > 0 and d(xn, B) > 0 for sufﬁciently large n. Compute

d(xn, A) d(x, A) f (xn) − f (x) = − d(xn, A) + d(xn, B) d(x, A) + d(x, B) d(x , A)d(x, B) − d(x, A)d(x , B) = n n [d(xn, A) + d(xn, B)][d(x, A) + d(x, B)] [d(x , A) − d(x, A)]d(x, B) + d(x, A)[d(x, B) − d(x , B)] = n n . [d(xn, A) + d(xn, B)][d(x, A) + d(x, B)] From the triangle inequalities

According to Theorem 1.2.27 (5), any metric spaces are normal. We ask whether the above construction holds for normal spaces.

THEOREM 1.2.30. (Urysohn’s lemma) Let (X, T ) be a normal space and A, B be disjoint closed subsets of X. For any closed interval [a, b] ⊆ R, there exists a continuous map f : X → [a, b] such that | ≡ | ≡ f A a and f B b. 30 1. BASIC TOPOLOGY

PROOF. Since the map [0, 1] → [a, b], t 7→ (1 − t)a + tb, is continuous, we may assume that [a, b] = [0, 1].

Step 1. Let P := [0, 1] ∩ Q. We shall deﬁne for each p ∈ P an open set Up ⊆ X, in such a way that Up ⊆ Uq whenever p < q. Moreover ⊆ ⊆ ⊆ \ A U0 U0 U1, U1 := X B. Step 2. We deﬁne { ∅, if p < 0 and p ∈ Q, U := p X, if p > 1 and p ∈ Q.

Clearly for any p, q ∈ Q with p < q, we still have Up ⊆ Uq.

Step 3. Given x ∈ X, deﬁne

Q(x) := {p ∈ Q|Up ∋ x} ⊆ Q. Then Q ∩ (1, +∞) ⊂ Q(x). Thus Q(x) is bounded below and its greatest lower bound lies in [0, 1]. Deﬁne

f (x) := inf Q(x) = inf{p ∈ Q|x ∈ Up}, x ∈ X, which is well-deﬁned.

Step 4. If x ∈ A, then x ∈ Up for all p ∈ Q ∩ [0, +∞), so that f (x) = 0. If x ∈ B, then x ∈ Up for all p ∈ Q ∩ (1, +∞), so that f (x) = 1.

Step 5. We claim

x ∈ Ur =⇒ f (x) ≤ r and x ∈/ Ur =⇒ f (x) ≥ r.

Indeed, if x ∈ Ur, then x ∈ Us for all rational numbers s > r; therefore f (x) ≤ r. If x ∈/ Ur, then x ∈/ Us for any rational numbers s < r; so f (x) ≥ r. ∈ ⊆ Step 6. f is continuous. For any x0 X and any open interval (c, d) R ∋ ⊂ containing f (x0), we want to ﬁnd an open subset U x0 such that f (U) (c, d). Choose p, q ∈ Q such that

c < p < f (x0 < q < d. ∈ ∈ By Step 5, f (x0) < q implies x0 Uq, and f (x0) > p implies x0 / Up. Hence ∈ \ ⊂ ∈ x0 U := Uq Up which is open in X. To verify f (U) (c, d), take x U. Then x ∈ Uq ⊂ Uq so that f (x) ≤ q by Step 5. From x ∈/ Up we get f (x) ≥ p again by Step 5. Thus p ≤ f (x) ≤ q and f (U) ⊂ (c, d).

Return back to Step 1. Because P is countable, we can arrange the elements of P in an infinite sequence, denote by P, so that 1 and 0 are the first two elements of the sequence. Define \ ⊇ U1 := X B A.

By normality (see Theorem 1.2.27 (4)), there exists an open subset U0 such that ⊂ ⊆ ⊆ A U0 U0 U1. 1.2. TOPOLOGICAL SPACES 31

Let Pn (n ≥ 2) denote the set of the ﬁrst n rational numbers in P. Assume that Up is deﬁned for all p ∈ Pn satisfying the condition

(1.2.4.1) p < q =⇒ Up ⊂ Uq. Let r denote the next rational number in P. Consider ∪ { } Pn+1 := Pn r . ∈ It is a finite subset of [0, 1], and we can find p, q Pn+1 such that p < r < q in the usual order relation and in such a way that there are no other elements of Pn+1 in [p, r] and [r, q]. Such p and q are called immediate predecessor and immediate successor respectively. Since p, q actually lie in Pn, we have Up ⊂ Uq. Applying the normality of X to the pair (Up, X \ Uq) of disjoints closed subsets, there exists an open subset Ur such that Up ⊆ Ur ⊆ Ur ⊆ Uq. We claim that (1.2.4.1) holds for every pair of elements of Pn+1. If both of them lie in Pn,(1.2.4.1) holds trivially. If one is r and another is s ∈ Pn, then either s ≤ p so that Us ⊂ Up ⊆ Ur, or s ≥ q so ⊂ ⊆ that Ur Uq Us. Hence (1.2.4.1) also holds in Pn+1. By induction on n, we have defined the open subset Up for each p ∈ P.

If P in Step 1 is the following sequence { } 1 1 2 1 3 1 2 3 4 1 P = 1, 0, , , , , , , , , , , ··· , 2 3 3 4 4 5 5 5 5 6 then ∪ { } ⇒ 1 ⇒ ⊆ ⊆ P3 = P2 1/2 = 0 < < 1 = U0 U 1 U1, 2 2 ∪ { } ⇒ 1 1 ⇒ ⊆ ⊆ P4 = P3 1/3 = 0 < < = U0 U 1 U 1 , 3 2 3 2 ∪ { } ⇒ 1 2 ⇒ ⊆ ⊆ P5 = P4 2/3 = < < 1 = U 1 U 2 U1, 2 3 2 3 ∪ { } ⇒ 1 1 ⇒ ⊆ ⊆ P6 = P5 1/4 = 0 < < = U0 U 1 U 1 , 4 3 4 3 ∪ { } ⇒ 2 3 ⇒ ⊆ ⊆ P7 = P6 3/4 = < < 1 = U 2 U 3 U1, 3 4 3 4 ∪ { } ⇒ 1 1 ⇒ ⊆ ⊆ P8 = P7 1/5 = 0 < < = U0 U 1 U 1 . 5 4 5 4

DEFINITION 1.2.31. Let A, B be two subsets of a topological space (X, T ). We say that A and B can be separated by a continuous function, if there exists a → | ≡ | ≡ continuous function f : X [0, 1] such that f A 0 and f B 1. Observe that the following are equivalent for a topological space (X, T ): (1) (X, T ) is normal. (2) Every pair of disjoint closed subsets in X can be separated by disjoint open subsets. (3) Every pair of disjoint closed subsets can be separated by a continuous function. 32 1. BASIC TOPOLOGY

Let (X, T ) be a topological space and f : X → R be a continuous function. The support of f is

(1.2.4.2) supp( f ) := {x ∈ X| f (x) ̸= 0}. If A is a closed subset of X and U is an open subset containing A, a continuous → | ≡ ⊆ function f : X [0, 1] such that f A 1 and supp( f ) U is called a bump function for A supported in U.

COROLLARY 1.2.32. (Existence of bump functions) Let (X, T ) be a normal space. If A is a closed subset of X and U is an open subset containing A, there exists a bump function for A supported in U.

PROOF. By normality, we can ﬁnd an open subset V such that A ⊆ V ⊆ V ⊆ U. Apply Theorem 1.2.30 to the pair (A, B := X \ V).

THEOREM 1.2.33. (Tietze’s extension theorem) Let (X, T ) be a normal space and A ⊆ X be closed subset. (1) Any continuous function of A into [a, b] ⊂ R may be extended to a continuous map of X into [a, b]. (2) Any continuous function of A into R may be extended to a continuous map of X into R.

PROOF. Given a continuous function f : A → [−r, r], we construct a continuous function g : X → R such that ¯ 1 2 |g(x)| ≤ r, x ∈ X and |g(a) − f (a)| ≤ r, a ∈ A. 3 3 Deﬁne − − − −1 −1 I1 := [ r, r/3], I2 := [ r/3, r/3], I3 := [r/3, r], B := f (I1), C := f (I3). Since f is continuous, it follows that B, C are closed disjoint subsets in A and then B, C are closed disjoint subset of X, by Proposition 1.2.22 (1). According to Theo- rem 1.2.30, there exists a continuous function g : X → [−r/3, r/3] such that 1 1 g| ≡ − r, g| = r. B 3 C 3 Then |g(x)| ≤ r/3 for all x ∈ X. Given a ∈ A. If a ∈ B, then |g(a) − f (a)| = | − r/3 − f (a)| ≤ 2r/3; if a ∈ C, then |g(a) − f (a)| = |r/3 − f (a)| ≤ 2r/3; if ∈ ∪ ∈ | − | ≤ a / B C, then f (a), g(a) I2 and g(a) f (a) 2r/3.

(1) Since [a, b] is homeomorphic to [−1, 1], we may assume [a, b] = [−1, 1]. Let f : A → [−1, 1] be a continuous map. Then there exists a continuous function → g1 : X R such that 1 2 |g (x)| ≤ , x ∈ X and |g (a) − f (a)| ≤ , a ∈ A. 1 3 1 3 1.2. TOPOLOGICAL SPACES 33

− − The function f g1 maps A into [ 2/3, 2/3]. We can ﬁnd a continuous function → g2 : X R such that ( ) ( ) 1 2 2 2 |g (x)| ≤ , x ∈ X and | f (a) − g (a) − g (a)| ≤ , a ∈ A. 2 3 3 1 2 3

In general, for any n ≥ 2, we can construct a continuous function bgn defined on X such that ( ) − ( ) n 1 n | | ≤ 1 2 ∈ − ≤ 2 ∈ gn(x) , x X and f (a) ∑ gi(a) , a A. 3 3 1≤i≤n 3 Define g(x) := ∑ gn(x), x ∈ X. n≥1 Since ( ) − 1 2 n 1 ∑ |gn(x)| ≤ ∑ = 1, 1≤n≤N 3 1≤n≤N 3 it follow that the above series is absolutely and uniformly convergent and then g is continuous on X. Clearly that g(a) = f (a) for all a ∈ A, and g maps X into [−1, 1]. ∼ (2) We may, without loss of generality, assume that f : A → (−1, 1) = R is a continuous function. By (1), f can be extended to a continuous map g : X → [−1, 1]. Define − − D := g 1({−1}) ∪ g 1({1}) | ⊆ − ∩ ∅ which is closed in X. from g A = f (A) ( 1, 1), we see that A D = . By Theorem 1.2.30, there exists a continuous function ϕ : X → [0, 1] such that ϕ| ϕ| D = 0, A = 1. Define h(x) := ϕ(x)g(x). Then g is continuous on X and is an extension of f : h(a) = ϕ(a)g(a) = g(a) = f (a), a ∈ A. For any x ∈ X, if x ∈ D then h(x) = 0 · g(x) = 0; if x ∈/ D then h(x) = ϕ(x)g(x) ∈ ∼ (−1, 1). Thus h is a continuous from X into (−1, 1) = R.

1.2.5. Topological manifolds. Let (X, T ) be a topological space. (1) A cover of X is a collection U of subsets of X such that any point x ∈ X is contained in some U ∈ U . If each of the sets in U is open (resp. closed), then we say U is an open cover (resp. a closed cover). ′ (2) Given a cover U , a subcover of U is a subcollection U ⊆ U that is still a cover of X.

THEOREM 1.2.34. Any open cover of a second countable topological space has a countable subcover. 34 1. BASIC TOPOLOGY

PROOF. Let B be a countable basis for X and let U be any open cover of X. Deﬁne ′ B := {B ∈ B : B ⊂ U for some U ∈ U } ̸= ∅. B′ ∈ B′ ∈ U ⊃ It is clear that is countable. For any B let UB be such that UB B as above. Let U ′ { ∈ U ∈ B′} := UB : B . This is a countable subcover of X. A topological space (X, T ) is said to be a Lindel¨ofspace if every open cover of X has a countable subcover.

EXERCISE 1.2.35. Recall Example 1.2.13. Show that (R, Tpp) is not Lindelof,¨ T T (R, ep) is Lindelof,¨ and (R, fc) is Lindelof.¨

A topological space M = (M, T ) is local Euclidean of dimension m if for any p ∈ M there is an open set U ⊂ M containing p, called coordinate domain, that is homeomorphic to an open subset U of Rm. The homeomorphism is denoted by φ : U → φ(U) = U, and is called a coordinate map on U. The pair (U,φ) is called a coordinate chart for M. Write (1.2.5.1) φ(q) := (x1(q), ··· , xm(q)), q ∈ U. We see that xi can be viewed as a function on U and φ = (x1, ··· , xm). Equivalently, M is local Euclidean of dimension m if and only if any point p ∈ M has an open set U ⊂ M of p that is homeomorphic to an open ball in Rm. In this case we usually say U is a coordinate ball in M. We may also assume that φ(p) = 0 ∈ Rm.

An m-dimensional topological manifold is a second countable Hausdorff space M that is local Euclidean of dimension m. Write dim(M) = m and call the dimension of M. A typical example of m-dimensional topological manifold is Rm.

PROPOSITION 1.2.36. Every open subset of an m-dimensional topological manifold is still an m-dimensional topological manifold.

PROOF. Let M be an m-dimensional topological manifold and V be an open subset of M. (1) V is locally Euclidean of dimension m. For any p ∈ V ⊆ M, we have an open subset U in M containing p, that is homeomorphic to an open subset U of Rm. Then U ∩ V still contains p, is open, and is homeomorphic to an open subset of Rm. (2) V is second countable. By Proposition 1.2.25 (4). (3) V is Hausdorff. By Theorem 1.2.27 (2).

The (closed) m-dimensional upper half-space Hm ⊆ Rm is deﬁned by (1.2.5.2) Hm := {(x1, ··· , xm) ∈ Rm|xm ≥ 0}. 1.2. TOPOLOGICAL SPACES 35

The boundary and interior of Hm are ∂Hm = {(x1, ··· , xm) ∈ Hm|xm = 0}, Int(Hm) = {(x1, ··· , xm) ∈ Hm|xm > 0}. When m = 0, we have H0 = R0 = {0}, so Int(H0) = H0 and ∂H0 = ∅.

When m = 2, we obtain the upper half-plane (1.2.5.3) H ≡ H2 := {τ ∈ C|Im(τ) > 0}. Deﬁne the modular group {[ ] }

a b (1.2.5.4) SL (Z) := a, b, c, d ∈ Z, ad − bc = 1 2 c d and the associated map ( [ ] ) a b aτ + b (1.2.5.5) SL (Z) × C −→ C, γ = , τ 7−→ γ(τ) := . 2 c d cτ + d Here C := C ∪ {∞} can be considered as the Riemann sphere S2.

EXERCISE 1.2.37. (1) Let Γ be the subgroup of SL (Z) generated by [ ] [ ] 2 1 1 0 −1 , . 0 1 1 0 Γ Prove = SL2(Z). γ γ′ ∈ τ ∈ (2) For any , SL2(Z) and any H, one has ′ ′ (γ · γ )(z) = γ(γ (z)). (3) Prove [ ] Im(τ) d 1 a b Im(γ(τ)) = , γ(τ) = , γ = ∈ SL (Z). |cτ + d|2 dτ (cτ + d)2 c d 2

Since Hm ⊆ Rm we can give it the subspace topology of Rm. An m-dimensional topological manifold with boundary is a second countable Hausdorff space M in which every point p ∈ M has an open subset U containing p, which is homeomorphic, under the map φ, either to an open subset of Rm or to an open subset of Hm. The set U is called a coordinate domain of p, φ is called a coordinate map, and the pair (U,φ) is called a coordinate chart for M. We say (U,φ) is an interior chart if φ(U) is an open subset of Rm, and a boundary chart if φ(U) is an open subset of Hm with φ(U) ∩ ∂Hm ̸= ∅.

Let M be an m-dimensional topological manifold with boundary. A point p ∈ M is called an interior point of M if it is in the domain of an interior chart, and it is called a boundary point of M if it is in the domain of a boundary chart that takes p to ∂Hm. The boundary of M, denoted by ∂M, is the set of all its boundary points, and its interior, denoted by Int(M) or M˚ , is the set of all its interior points.

PROPOSITION 1.2.38. If M is an m-dimensional topological manifold with boundary, then Int(M) is an open subset of M, which is itself an m-dimensional topological manifold. 36 1. BASIC TOPOLOGY

EXERCISE 1.2.39. Prove Proposition 1.2.38. When M is Hm, an m-dimensional topological manifold with boundary, we have Hm = Int(Hm) ⨿ ∂Hm, a disjoint union. Actually this fact holds for any m- dimensional topological manifold with boundary M, i.e., M = Int(M) ⨿ ∂M. PROPOSITION 1.2.40. If M is a nonempty m-dimensional topological manifold with boundary, then ∂M is closed in M, and M is an m-dimensional topological manifold if and only if ∂M = ∅.

PROOF. According to Proposition 1.2.38 and the above mentioned fact, ∂M = M\ Int(M) is closed. If M is an m-dimensional topological manifold, then M = Int(M) so that ∂M = ∅. If ∂M = ∅, then M = Int(M) which is an m-dimensional topological manifold by Proposition 1.2.38. U { } T Let := Ui 1≤i≤n be an open cover of a topological space (X, ).A partition of unity subordinate to U is a finite family of continuous functions ϕ → i : X R, with the following properties: • ≤ ϕ ≤ ∈ 0 i(x) 1 for all i and all x X; • ϕ ⊆ Supp( i) Ui; • ϕ ∈ ∑1≤i≤n i(x) = x for all x X. U { } THEOREM 1.2.41. (Existence of finite partitions of unity) If = Ui 1≤i≤n is a finite open cover of a normal space (X, T ), then there exists a partition of unity subordinate to U .

W { } PROOF. Basic idea is to construct two ﬁnite open covers = Wi 1≤i≤n and V { } = Vi 1≤i≤n of X so that ⊆ ⊆ ⊆ Wi Vi Vi Ui. {ϕ } Then, by Theorem 1.2.30 we can ﬁnd i 1≤i≤n. U { } V { } Step 1: We can shrink = Ui 1≤i≤n to an open cover = Vi 1≤i≤n of X ⊆ such that Vi Ui for each i. \ ∪ Let A := X 2≤i≤nUi. Then A is closed in X and is contained in U1. By normality, Theorem 1.2.27 (4), there exists an open subset V1 of X such that ⊆ ⊆ ⊆ A V1 V1 U1. { ··· } So V1, U2, , Un is still an open cover of X. In general, given open subsets ··· { ··· ··· } V1, , Vk−1 such that V1, , Vk−1, Uk, Uk+1, , Un is an open cover of X, let ( )   ∪ ∪ \  ∪   A := X Vi Uj . 1≤i≤k−1 1≤j≤n

Then A is closed in X and is contained in Uk. By normality again, there exists an ⊆ ⊆ ⊆ { ··· ··· } open subset Vk such that A Vk Vk Uk. Then V1, , Vk−1, Vk, Uk+1, , Un is an open cover of X. After ﬁnite steps, we prove Step 1.

Step 2: Complete the proof. Using Step 1 again, we can choose an open cover W { } = Wi 1≤i≤n of X such that ⊆ ⊆ ⊆ Wi Vi Vi Ui. 1.3. CONSTRUCTIONS 37

ψ → By Theorem 1.2.30, for each i, there exists a continuous function i : X [0, 1] ψ ψ \ ψ−1 \{ } ⊆ such that i(Wi) = 1 and i(X Vi) = 0. Because i (R 0 ) Vi, we have ψ ⊆ ⊆ W ψ Supp( i) Vi Ui. Since covers X, the finite sum ∑1≤i≤n i(x) > 0 for all x ∈ X. Define ψ ϕ i(x) ∈ i(x) := ψ , x X. ∑1≤j≤n j(x) {ϕ } U Then i 1≤i≤n is a partition of unity subordinate to . 1.3. Constructions We in this section discuss two important constructions of new topological spaces from old one. At first we introduce topological embeddings.

An injective continuous map that is a homeomorphism onto its image (in the subspace topology) is called a topological embedding.

PROPOSITION 1.3.1. (1) A continuous injective map that is either open or closed is a topological embeddings. (2) A surjective topological embedding is a homeomorphism.

T → T PROOF. (1) Let f : (X, X) (Y, Y) be a continuous injective map. Then f → T → deﬁnes a bijective map f : X f (X) that is continuous. To prove f : (X, X) T ( f (X), f (X)) is homeomorphic, we need to check that it is open or closed by Ex- ercise 1.2.7. T → T ∈ T If f : (X, X) (Y, Y) is open (resp. closed) and A X, then f (A) is open (resp. closed) in Y. By Proposition 1.2.22 (2), f (A) is also open (resp. closed) in T → T f (X). Thus f : (X, X) ( f (X), f (X)) is open (resp. closed). (2) Clearly. T ι If S is a subspace of a topological space (X, ), then the inclusion map S : T → T (S, S) (X, ) is a topological embedding.

EXAMPLE 1.3.2. (Examples of topological embeddings) (1) Let F : R → R2 be the injective continuous map F(s) := (s, s2). Its image is the parabola P := {(x, y) ∈ R2|y = x2}. Hence F is an injective continuous π 2 → π π| map. Denote : R R by (x, y) := x. Then P is continuous and gives the inverse of F : R → P. Therefore F is a topological embedding. (2) Let f : [0, 1) → R2 be the map f (s) := (cos(2πs), sin(2πs)). It is clear that f is not a homeomorphisms onto its image in the subspace topology. However the restriction of f to any interval [0, b) for b ∈ (0, 1) is an topological embedding. (3) Let U ⊆ Rn be an open subset and : U → Rk be any continuous map. The graph Γ ⊆ n+k of f is the subset f R deﬁned by Γ { ∈ × k| } f := (x, y) U R y = f (x) , with the subspace topology inherited from Rn+k. Deﬁne two maps Φ −→ n+k 7−→ f : U R , x (x, f (x)) and π : Rn+k −→ Rn, (x1, ··· , xn, xn+1, ··· , xn+k) 7−→ (x1, ··· , xn). 38 1. BASIC TOPOLOGY

Φ Γ π|Γ Then f is a continuous bijection from U onto f and f is a continuous inverse for Φ Φ Γ Γ f . Hence f is a topological embedding and f is homeomorphic to U, so that f is a topological manifold. (4) Consider the n-dimensional unit sphere or unit n-sphere { } Sn = x = (x1, ··· , xn+1) ∈ Rn+1| ∑ (xi)2 = 1 . 1≤i≤n+1 For example, S0 = {−1, 1}, S1 is the unit circle in R2, and S2 is the spherical surface of radius 1 in R3. For each i ∈ {1, ··· , n + 1}, deﬁne ± { ∈ n| ± i } Ui := x S x > 0 . Then ∪ ∪ n + ∪ − S = Ui Ui . 1≤i≤n+1 1≤i≤n+1 ± ± On each U , we see that x ∈ Sn ∩ U if and only if i √i xi = ± 1 − ∑ (xj)2. 1≤j̸=i≤n+1

n ∩ ± Hence S Ui is the graph of a continuous function, is therefore locally Euclidean of dimension n. Thus Sn is an n-dimensional topological manifold. Using the stereographic projection we can give an explicit coordinate map. Let N := (0, ··· , 0, 1) ∈ Sn and consider the map ( ) x1 xn φ : Sn \{N} −→ Rn, (x1, ··· , xn, xn+1) 7−→ , ··· , . 1 − xn+1 1 − xn+1 Evidently, ( ) 1 n 2 − 2u 2u |u| − 1 φ 1(u1, ··· , un) = , ··· , , , u = (u1, ··· , un). 1 + |u|2 1 + |u|2 1 + |u|2 Similarly, we let S := {0, ··· , 0, −1} and consider the the map ( ) x1 xn ψ : Sn \{S} −→ Rn, (x1, ··· , xn, xn+1) 7−→ , ··· , . 1 + xn+1 1 + xn+1 Evidently, ( ) 1 n 2 − 2u 2u 1 − |u| ψ 1(u1, ··· , un) = , ··· , , , u = (u1, ··· , un). 1 + |u|2 1 + |u|2 1 + |u|2 Consequently, {(Sn \ N,φ), (Sn \ S, ψ)} covers Sn. (5) Suppose that M is a topological manifold and S ⊂ M. Under the relative topology, S is itself a topological space and furthermore a topological manifold. When U is open, we say that U is an open submanifold of M. Let us denote GL(n, R) the set of all n × n singular real matrices. Since any n × n 2 matrix can be viewed as a point in Rn , the “determent function” given by 2 2 det : Rn ⊃ GL(n, R) −→ Rn , A 7−→ det(A) 2 − shows that GL(n, R) = Rn \ det 1(0) is a topological manifold (actually it is an open submanifold). 1.3. CONSTRUCTIONS 39

→ ··· If fi : Xi Yi are maps for i = 1, , n, their product map is × · · · × × · · · × −→ × · · · × f1 fi : X1 Xn Y1 Yn given by × · · · × ··· ··· f1 fn(x1, , xn) := ( f1(x1), , fn(xn)) . T ≤ ≤ 1.3.1. Product spaces. Let (Xi, i), 1 i n be topological spaces. Deﬁne B { × · · · × | ∈ T ≤ ≤ } := U1 Un Ui i, 1 i n . According to Proposition 1.2.10, B is a basis for some topology. We call such a T ×···× T ⊗ · · · ⊗ T × · · · × topology the product topology X1 Xn or 1 n on X1 Xn. The × · · · × T ⊗ · · · ⊗ T topological space (X1 Xn, 1 n) is called the product space. The × · · · × basis subsets of the form U1 Un are called product open subsets. For example, on R2 = R × R, the product topology is generated by sets of the form I × J, where I and J are open subsets of R. Hence the product topology on R2 is generated by open rectangle, so that it is the same as the metric topology induced by the Euclidean distance function.

× · · · × T ⊗ · · · ⊗ T Consider the above product space (X1 Xn, 1 n). The canon- ical projection π × · · · × −→ (1.3.1.1) i : X1 Xn Xi ∈ T is actually continuous. Indeed, for any Ui i we have − π 1( ) = × · · · × − × × × · · · × ∈ T ×···× Ui X1 Xi 1 Ui Xi+1 Xn X1 Xn π so that i is continuous. × THEOREM 1.3.3. (1) (Characteristic property of product topologies) Let X1 · · · × → × · · · × Xn be a product space. For any topological space B, a map f : B X1 Xn π ◦ is continuous if and only if each of its component functions f i := i f is continuous: × · · · × X1 9 Xn rr f rr rr π rr i rr r / B Xi f i ··· (2) (Uniqueness of the product topology) Let X1, , Xn be topological spaces. × · · · × The product topology on X1 Xn is the unique topology that satisﬁes the characteristic property (1). ··· (3) Let X1, , Xn be topological spaces. π × · · · × → (a) The projection maps i : X1 Xn Xi are all continuous and open. (b) The product topology is “associative” in the sense that the three product topolo- × × × × × × × gies X1 X2 X3, (X1 X2) X3, and X1 (X2 X3) on the set X1 × X2 X3 are all equal. ∈ ̸ → × · · · × (c) For any i and points xj Xj, j = i, the map f i : Xi X1 Xn given by ··· ··· f i(x) := (x1, , xi−1, x, xi+1, , xn)

is a topological embedding of Xi into the product space. B { × · · · × (d) If for each i, i is a basis for the topology of Xi, then the set B1 Bn : ∈ B } × · · · × Bi i is a basis for the product topology on X1 Xn. 40 1. BASIC TOPOLOGY

··· (e) If Ai is the subspace of Xi for i = 1, , n, the product topology and the sub- × · · · × ⊆ × · · · × space topology on A1 An X1 Xn are equal. × · · · × (f) If each Xi is Hausdorff so is X1 Xn. × · · · × (g) If each Xi is second countable so is X1 Xn. (4) A product of continuous maps is continuous, and a product of homeomorphisms is a homeomorphism. M ··· M ··· (5) If 1, , n are topological manifolds of dimensions m1, , mn, respec- M × · · · × M tively, the product spade 1 n is a topological manifold of dimension m1 + ··· + mn.

PROOF. (1) Suppose ﬁrst that all f i are continuous. To prove f is continuous, −1 × · · · × it sufﬁces to prove that f (U1 Un) in open in B. According tyo −1 × · · · × −1 ∩ · · · ∩ −1 f (U1 Un) = f 1 (U1) f n (Un), we see that f is continuous. ∈ T Conversely, if f is continuous, then for any Ui i we have −1 −1 × · · · × × × · · · × f i (Ui) = f (X1 Xi−1 U Xi+1 Xn) that is open in B. Hence f i is continuous. (2) − (5): as an exercise.

EXERCISE 1.3.4. Prove (2) − (5) in Theorem 1.3.3. From Theorem 1.3.3, we have the m-torus m 1 × · · · × 1 (1.3.1.2) T := |S {z S} m which is an m-dimensional topological manifold. When m = 2, we call T2 the torus.

For any indexed family (Xα)α∈ of sets, deﬁne its Cartesian product by  J 

 ∪  → α ∈ ∏ Xα := x : J Xα xα := x( ) Xα . α∈J α∈J α ∈ J When Xα = X for all J, we write X := ∏α∈A Xα. T T Given a collection (Xα, α)α∈J of topological spaces, deﬁne the topology to be the topology generated by the basis { }

∏ Uα Uα ∈ Tα and Uα = Xα for all but a ﬁnite number of α . α∈J T We call the Tychonoff topology or product topology on ∏α∈J Xα. The box topology is generalized by the basis { }

∏ Uα Uα ∈ Tα for all α ∈ A . α∈J Most theorems about ﬁnite products will also hold for arbitrary products if we use the product topology. Whenever we consider the product ∏α∈J Xα we shall assume it is given by the product topology. 1.3. CONSTRUCTIONS 41

THEOREM 1.3.5. (1) Suppose that the topology Tα on each Xα is generated by a basis B α ∈ α, J. Then the basis for the box topology on ∏α∈J Xα is { }

∏ Bα Bα ∈ Bα , α∈J and the basis for the product topology on ∏α∈J Xα is { }

Bα ∈ Bα for ﬁnitely many indices α and ∏ Bα . Bα = Xα for all the remaining indices α∈J π → π (2) Let β : ∏α∈J Xα Xβ be the natural projection, that is, β(x) = xβ. Deﬁne ∪ { } −1 S := Sβ, Sβ := πβ (Uβ)|Uβ ∈ Tβ . β∈J Then S is a subbasis for the product topology. (3) Let Aα, α ∈ J, be subspace of Xα.

(3.1) ∏α∈J Aα is a subspace of ∏α∈J Xα if both products are given by the box product, or if both products are given by the product product. (3.2) If ∏α∈J Xα is given either the box or the product topology, then

∏ Aα = ∏ Aα. α∈J α∈J α ∈ (4) If each Xα, J, is Hausdorff, then ∏α∈J Xα is Hausdorff in both the box and product topologies. T (5) (Characteristic property of inﬁnite product spaces) Let (Xα, α)α∈J be an T → indexed family of topological spaces. For any topological space (B, B), a map f : B ∏α∈J Xα (with the product topology) ia continuous if and only if its component functions π ◦ f α = α f is continuous. The product topology is the unique topology on ∏α∈J Xα that satisﬁes this property.

PROOF. (1) is clear. B (2) Let S be the basis on ∏α∈A Xα consisting of ﬁnite intersections of elements of S . Then for any B ∈ BS we have ∩ −1 B = πβ (Uβ ) i i 1≤i≤n

β ··· β β ∈ Tβ α = α α ̸= β ··· β where 1, , n are distinct and U i i . Deﬁning U X if 1, , n, we obtain B = ∏α∈J Uα which is a basis element for the product topology. (3) − (4): as an exercise. (5) Consider the diagram

f / B F ∏α∈ Xα FF J FF FF π FF β f β F" Xβ 42 1. BASIC TOPOLOGY

−1 Assume ﬁrst that f is continuous. For any Uβ ∈ Tβ, πβ (Uβ) is a subbasis element π for the product topology on ∏α∈J Xα, we see that β is continuous and then f β = πβ ◦ f is also continuous. Conversely, assume that each f α is continuous. For any subbasis element π−1 ∈ T β (Uβ) for the product topology on ∏α∈J Xα, where Uβ β, one has −1 π−1 −1 ∈ T f ( β (Uβ)) = f β (Uβ) B. Hence f is continuous. The uniqueness is similar to that of ﬁnite product topology.

EXERCISE 1.3.6. Prove Theorem 1.3.5 (3) − (4).

EXAMPLE 1.3.7. This example shows that Theorem 1.3.5 (5) may not hold if the ω ≡ product topology is replaced by the box topology. Consider R := ∏n≥1 Xn with Xn R, and deﬁne f / ω R G R GG GG GG π GG n f n # R = Xn ω where f (t) := (t, t, t, ··· ) and fn(t) := t. If R is given the box topology, then f is not continuous. For example, let ( ) 1 1 B := ∏ − , n≥1 n n ω − be an open subset of R relative to the box topology. We prove that f 1(B) is not open in − R. Otherwise, f 1(B) contains some interval (−δ, δ) about 0 ∈ R. So f ((−δ, δ)) ⊆ B and ( ) 1 1 f ((−δ, δ)) = (−δ, δ) ⊆ − , n n n for all n ≥ 1, a contradiction!

T ∆ THEOREM 1.3.8. A topological space (X, ) is Hausdorff if and only if X := { ∈ × } ⊂ × × T (x, y) X X : x = y X X is closed in (X X, X×X). ∆ × PROOF. Suppose ﬁrst that X is closed in X X with respect to the product × \ ∆ ̸ ∈ × ⊂ T ∩ topology. Then X X X is open. If x = y in X, then (x, y) U V X×X × \ ∆ (X X X). Hence X is Hausdorff. ∈ × \ ∆ ∈ ∈ T ∈ ∈ T For any (x, y) X X X, we have x U X and y V Y for some ∩ ∅ ∈ × × ⊂ ∆ open sets U, V with U V = . Hence (x, y) U V. If U V X, then ∈ ∩ × ⊂ × \ ∆ ∆ z U V, a contradiction. Therefore U V X X X and X is closed. T 1.3.2. Quotient spaces. Let (X, X) be a topological space, Y be a set, and q : X → Y a surjective map. T (1) Deﬁne a topology Y on Y by T { ⊆ −1 ∈ T } (1.3.2.1) Y := V Y : q (V) X . This is called the quotient topology induced by q. In this topology, q : T → T (X, X) (Y, Y) is continuous (called a quotient map). 1.3. CONSTRUCTIONS 43

T T (2) (Y, Y) is a topological space and Y is the largest topology on Y which makes q continuous. T ∼ Let (X, X) be a topological space and be an equivalence relation on X. (1) Let Y := X/ ∼:= {[x] : x ∈ X} be the set of equivalence classes, and deﬁne π : X −→ Y, x 7−→ [x], be the natural projection. (2) X/ ∼ together with the topology determined by π is called the quotient space of X by the given equivalence relation. The quotient map π is called the projection. π T → T More generally, a surjective map : (X, X) (Y, Y) between topological spaces is called a quotient map if the following condition holds: ∈ T ⇐⇒ π−1 ∈ T V Y (V) X. It is clear that a quotient map is continuous.

EXAMPLE 1.3.9. (1) (Quotient spaces may not be Hausdorff) Let X = R and ∼ − ∈ ∼ T T T x y if and only x y Q. Then X/ is noncountable. If X := R, then X/∼ is the trivial topology on X/ ∼. (2) Let Pn := (Rn+1 \{0})/ ∼ be the quotient space of Rn+1 \{0} where the equivalence relation is defined by x ∼ y ⇐⇒ y = λx for some λ ̸= 0. Equivalently, Pn is the set of 1-dimensional linear subspaces in Rn+1. Denote the quotient map by (1.3.2.2) π : Rn+1 \{0} −→ Pn, x 7−→ [x] = [x1, ··· , xn+1]. U If i is given by { } U ∈ n i ̸ ≤ ≤ i := [x] P : x = 0 , 1 i n + 1, U n then ( i)1≤i≤n+1 is an open cover of P with respect to the quotient topology induced by π. Define ( − ) x1 xi 1 xi+1 xn+1 φ ([x]) := , ··· , , , ··· , ; i xi xi xi xi φ n then i is homeomorphic and hence P is locally Euclidean. Lemma 1.3.11 below implies that Pn is second countable. The Hausdorff property of Pn will be proved in Section 1.5. (3) Let (X, T ) be any topological space and A be any subset of X. Define a relation ∼ on X by ∼ ⇐⇒ ∈ × \ × \ x1 x2 (x1, x2) A A or (X A) (X A). Clearly that ∼ is an equivalence relation on X. The quotient space X \ A := X\ ∼ is said to be obtained by collapsing A to a point. − − For example, Bn \ Sn 1 is homeomorphic to Sn 1. (4) If (X, T ) is any topological space, the quotient (1.3.2.3) C(X) := (X × [0, 1])/(X × {0}) obtained from X × [0, 1] by collapsing one end to a point is called the cone on X. For example, C(Sn) is homeomorphic to Bn. 44 1. BASIC TOPOLOGY

T ∈ (5) Let (Xi, i)1≤i≤n be a finite family of nonempty topological spaces. For each i { ··· } ∈ ∨ ··· 1, , n , let xi Xi be a fixed point. The wedge product 1≤i≤nXi of X1, , Xn is the quotient space obtained from the disjoint union ⨿1≤i≤n Xi by collapsing the set { ··· } x1, , xn to a point. For example, the wedge sum R ∨ R is homeomorphic to the union of the x-axis and the y-axis in the plane, and the wedge sum S1 ∨ S1 is homeomorphic to the figure-eight space consisting of the union of the two circles of radius 1 centered at (0, 1) and (0, −1) in the plane. (6) Quotient maps may not be open or closed. For example, consider π × −→ 7−→ 1 : R R R, (x, y) x and A := {(x, y) ∈ R × R|x ≥ 0 or y = 0}. Let q : A → R be obtained by restricting π 1 on A. Then qis a quotient map but is neither open or closed. (7) Consider X := [0, 1] ∪ [2, 3] ⊆ R with the subspace topology, Y := [0, 2] ⊆ R with the subspace topology. Define a map { x, x ∈ [0, 1], p : X −→ Y, x 7−→ x − 1, x ∈ [2, 3]. Then p is surjective, continuous, closed, quotient, but is not open. For A := [0, 1) ∪ [2, 3] ⊆ X, define q : A → Y to be the restriction of p on A; then q is continuous, surjective, but is not quotient. π × → 7→ π (8) Consider the projection 1 : R R R, (x, y) x. Then 1 is continuous, surjective, open, but is not closed (consider C := {(x, y) ∈ R × R|xy = 1}). Let ∪ { } π | A := C (0, 0) and q := 1 A; then q is continuous, surjective, but is not a quotient map. (9) Define   a, x > 0, −→ { } 7−→ q : R A := a, b, c , x  b, x < 0, c, x = 0. T {{ } { } { } { } ∅} Then the quotient topology on A induced by q is A = a , b , a, b , a, b, c , . T → T ⊆ If q : (X, X) (Y, Y) is a quotient map, a subset U X is said to be − saturated (with respect to q) if U = q 1(V) for some subset V ⊆ Y. − (1) U is saturated if and only if U = q 1(q(U)). − (2) A subset q 1(y) ⊆ X for y ∈ Y is called a fiber of q; a saturated set is one that is a union of fibers.

THEOREM 1.3.10. (1) (Characteristic property of quotient topologies) Let q : T → T T (X, X) (Y, Y) be a quotient map. For any topological space (B, B), a map f : Y → B is continuous if and only if the composite map f ◦ q is continuous: X ? ?? ??f ◦q q ?? ? / Y B f T T (2) (Characterization of quotient maps) Let (X, X) and (Y, Y) be topological spaces, and let q : X → Y be any surjective map. Then q is a quotient map if and only if the characteristic property holds. 1.3. CONSTRUCTIONS 45

T → T (3) (Passing to the quotient) Suppose that q : (X, X) (Y, Y) is a quotient T T → T map. (B, B) is a topological space, and f : (X, X) (B, B) is any continuous map ′ ′ that is constant on the ﬁbers of q (i.e., if q(x) = q(x ), then f (x) = f (x )). Then there e T → T e ◦ exists a unique continuous map f : (Y, Y) (B, B) such that f = f q: X ? ?? ??f q ?? ? / Y e B f In this situation, we say that the map f passes to the quotient or descends to the quotients. ( ) ( T ) → ( T ) 4 (Uniqueness of quotient spaces) Suppose q1 : X, X Y1, Y1 and T → T q : (X, X) (Y2, Y ) are quotient maps that make the some identiﬁcations (i.e., 2 ′ 2 ′ q1(x) = q1(x ) if and only if q2(x) = q2(x )). Then there is a unique homeomorphism φ ( T ) → ( T ) φ ◦ = : Y1, Y1 Y2, Y2 such that q1 q2. ( ) ( T ) → ( T ) 5 (Composition property of quotient maps) Suppose q1 : X, X Y, Y1 T → T ◦ and q2 : (Y, Y) (Z, Z) are quotient maps. Then their composition q2 q1 : T → T (X, X) (Z, Z) is also a quotient map. T → T (6) A continuous surjective map q : (X, X) (Y, Y) is a quotient map if and only if it takes saturated open sets to open sets, or saturated closed sets to closed set. T → T (7) Suppose that q : (X, X) (Y, Y) is a quotient map. The restriction of q to any saturated open or closed set is a quotient map. T → T (8) If q : (X, X) (Y, Y) is a surjective continuous map that is also an open or closed map, then it is a quotient map.

∈ T −1 −1 −1 PROOF. (1) For any U B, f (U) is open in Y if and only if q ( f (U)) = − ( f ◦ q) 1(U) is open in X. (2): as an exercise. (3) For any y ∈ Y, we choose some x ∈ X with q(x) = y (because q is surjective). Define e f (y) := f (x). e Since f is constant on the fibers of q, it follows that f is well-defined. The continu- e ity of f follows from (1). ′ e T → T If there exists another continuous map f : (Y, Y) (B, B) such that f = e′ f ◦ q. Then e′ e′ e e f (y) = f (q(x)) = f (x) = ( f ◦ q)(x) = f (y). (4) Consider three diagrams

X AA X AA X AA AA AA AA AAq2 AAq1 AAq1 q1 AA q2 AA q1 AA 1 / / Y1 / Y1 ∃e Y2 Y2 ∃e Y1 Y1 e ◦e Y2 q2 q1 q1 q2 e e By (3), there exist q2 and q1. Since e ◦ e ◦ e ◦ q1 q2 q1 = q1 q2 = q1, 46 1. BASIC TOPOLOGY

( ) e ◦ e = e ◦ e = we obtain from 2 that q1 q2 1Y1 . Similarly we have q2 q1 1Y2 . (5) − (6): as an exercise. π | → ⊆ (7) Let A be a saturated open set and let = q A : A q(A). Given V − q(A) and assume π 1(V) is open in A, we shall show that V is open in q(A). We ﬁrst claim − − π 1(V) = q 1(V). − Indeed, if V ⊆ q(A) and A is saturated, then q 1(V) ⊆ A (why?) and hence − − π 1(V) = q 1(V) (why?) − − Now π 1(V) is open in A and A is open in X, so that π 1(V) is open in X. − − − From q 1(V) = π 1(V), we see that q 1(V) is open in X. Since q is a quotient map, it follows that V is open in Y and in particular open in q(A). − (8) Assume that q is open. Given V ⊆ Y and assume that q 1(V) is open in X, − we shall show that V is open in Y. Because q is surjective, we get q(q 1(V)) = V. − Hence V = q(q 1(V)) is open in Y (because q is open).

EXERCISE 1.3.11. Prove Theorem 1.3.10 (2), (5), and (6). Note that the product of two quotient maps need not be a quotient map. If T → T T T q : (X, X) (Y, Y) is a quotient map and X is Hausdorff, then Y need not be Hausdorff. π T → T LEMMA 1.3.12. Suppose : (X, X) (Y, Y) is a quotient map and X is second countable. If Y is locally Euclidean, it is second countable.

PROOF. Since Y is locally Euclidean, it follows that Y has an open cover U = U U π−1 U {π−1 U ( α)α∈A with α is homeomorphic to some open ball. Then ( ) := ( α) : Uα ∈ U } is an open cover of X. The second countability of X shows that (by The- − ′ orem 1.2.34) there exists a countable subcover of π 1(U ). Let U ⊆ U denote a − ′ ′ subcover such that π 1(U ) is a countable subcover of X and U is a countable cover of Y by Euclidean balls. Each such ball has a countable basis, and the union of all these bases is a countable basis for Y.

1.3.3. The de Rham cohomology groups on Rn. Let U be an open set in Rn and Ωp(U) denote the space of smooth p-forms on U generated by dxI, where

I i1 ∧ · · · ∧ ip ≤ ··· ≤ (1.3.3.1) dx := dx dx , I = (1 i1 < < ip n). ∞ over C (U). Any element of Ωp(U) can be written as ω ω I = ∑ I dx . |I|=p Then ⊕ • (1.3.3.2) Ω (U) := Ωp(U) 0≤p≤n is a graded algebra. Deﬁne a differential operator d (exterior differentiation) by (1.3.3.3) d : Ωp(U) −→ Ωp+1(U) as follows: 1.3. CONSTRUCTIONS 47

(i) if f ∈ Ω0(U), then ∂ f d f = ∑ dxi = d f ; ∂ i 1≤i≤n x ω ∑ ω I ∈ Ωp (ii) if = |I|=p I dx (U), then ω ω ∧ I ω ∧ I d := ∑ d I dx = ∑ d I dx . |I|=p |I|=p

It is easy to show that d2 = d ◦ d = 0 and ω (1.3.3.4) d(ω ∧ η) = dω ∧ η + (−1)deg( )ω ∧ dη, where deg(ω) = p if ω ∈ Ωp(U).

• The pair (Ω (U), d) is called the de Rham cohomological complex on U. We denote by Zp(U) and Bp(U) the set of closed p-forms and exact p-forms respectively. Note that exact forms are always closed, however the closed forms may not be exact3. The p-th de Rham cohomology group of U is p p Z (U) (1.3.3.5) H (U) := , 0 ≤ p ≤ n. DR Bp(U) Set ⊕ • p (1.3.3.6) HDR(U) := HDR(U). 0≤p≤n ω η ∈ • For any [ ], [ ] HDR(U) we can deﬁne (1.3.3.7) [ω] · [η] := [ω ∧ η] which is independent of the choice of representatives.

If F : U → V is a smooth map from an open subset U ⊂ Rn to another open ∗ subset V ⊂ Rm, and if ω ∈ Ωp(V), the pull-back F ω is the smooth p-form on U obtained by making the substitution y = F(x) = (F1(x), ··· , Fm(x) in ω ∗ ω ω i1 ∧ · · · ∧ ip (1.3.3.8) (F )(x) = ∑ I (F(x))dF dF . |I|=p If we have a second smooth map G : V → W and if η is a smooth form on W, then ∗ ∗ F (G η) is obtained by substituting z = G(y) and y = F(x), thus ∗ ∗ ∗ (1.3.3.9) F (G η) = (G ◦ F) η. Moreover, we have ∗ ∗ (1.3.3.10) d(F ω) = F (dω) for any smooth form ω on V. Consequently, the smooth map F : U → V induces a morphism on cohomology groups: ∗ p −→ p ω 7−→ ∗ω (1.3.3.11) F : HDR(V) HDR(U), [ ] [F ].

3Consider U = R2 \{(0, 0)} and ω = (xdy − ydx)/(x2 + y2) 48 1. BASIC TOPOLOGY

Let ω be a smooth form on [0, 1] × U. For (t, x) ∈ [0, 1] × U we write ω ω I ωe ∧ J (1.3.3.12) (t, x) = ∑ I (t, x)dx + ∑ J(t, x)dt dx . |I|=p |J|=p−1 Deﬁne (∫ ) 1 Ωp × −→ Ωp−1 ω 7−→ ωe J (1.3.3.13) K : ([0, 1] U) (U), ∑ J(t, x)dt dx . |J|=p−1 0 We then have the following diagram

d Ωp([0, 1] × U) −−−−→ Ωp+1([0, 1] × U)     yK yK

− d Ωp 1(U) −−−−→ Ωp(U) Observe that (1.3.3.14) (∫ ) 1 ∂ω (Kd + dK) ω = ∑ I (t, x)dt dxI = ∑ [ω (1, x) − ω (0, x)] dxI ∂t I I |I|=p 0 |I|=p where ω is given in (1.3.3.12). Thus ω ι∗ω − ι∗ω (1.3.3.15) (Kd + dK) = 1 0 where ιt : U → [0, 1] × U is the injection x 7→ (t, x).

THEOREM 1.3.13. (Poincar´e’s lemma) Let U ⊆ Rn be a star-shaped open set with respect to 0 (that is, 0 ∈ U and any line segment 0x is entirely contained in U). If ω ∈ p ≥ ω η η ∈ Ωp−1 p { } Z (U) and p 1, then = d for some (U). Consequently, HDR(U) = 0 ≥ 0 for any star-shaped open set (with respect to some point in U) and p 1, and HDR(U) = R for any star-shaped open set.

PROOF. Let H : [0, 1] × U → U be deﬁned by H(t, x) := tx and set F(x) := x, G(x) := 0, x ∈ Ω. Then ◦ ι ◦ ι F = H 1, G = H 0, so that { ∗ ∗ ∗ ∗ ∗ ∗ ω − ω(0), p = 0, d (K(H ω)) = ι∗ H ω − ι H ω = F ω − G ω = 1 0 ω, p ≥ 1. ∗ When p ≥ 1, we get η = K(H ω).

A direct sum of vector spaces ⊕ • C := Cp p∈Z is called a differential complex if there are homomorphisms − − dp 1 dp dp+1 · · · −−−−→ Cp 1 −−−−→ Cp −−−−→ Cp+1 −−−−→· · · 1.3. CONSTRUCTIONS 49 such that dp+1 ◦ dp = 0 for any p ∈ Z. d is the differential operator of the complex • • C . The cohomology of C is the direct sum of vector spaces ⊕ • • • (1.3.3.16) H (C ) := Hp(C ) p∈Z where p ∩ p p • Ker(d ) C H (C ) := − . Im(dp 1) ∩ Cp • • • A map f : A → B between two differential complexes is a chain map if it • • • ◦ • • ◦ • commutes with the differential operators of A and B : f dA• = dB• f . That is, p−1 p p+1 − d • d • d • · · · −−−−→ Ap 1 −−−−→A Ap −−−−→A Ap+1 −−−−→·A · ·     −   y f p 1 y f p y f p+1

p−1 p p+1 − d • d • d • · · · −−−−→ Bp 1 −−−−→B Bp −−−−→B Bp+1 −−−−→·B · · for any p ∈ Z, p ◦ p−1 p−1 ◦ p−1 f dA• = dB• f . A sequence of vector spaces − − f i 1 f i · · · −−−−→ Vi 1 −−−−→ Vi −−−−→ Vi+1 −−−−→· · · − is said to be exact if for all i ∈ Z, Ker( f i) = Im( f i 1). An exact sequence of vector spaces 0 → U → V → W → 0 is called a short exact sequence.

Given a short exact sequence of differential complexes • • • f • g • 0 −−−−→ A −−−−→ B −−−−→ C −−−−→ 0 • • in which the maps f and g are chain maps, there is a long exact sequence of cohomology groups: • • • Hp( f ) • Hp(g ) • δp • · · · −−−−→ Hp(A ) −−−−→ Hp(B ) −−−−→ Hp(C ) −−−−→ Hp+1(A ) −−−−→ • • • Indeed, the map Hp( f ) : Hp(A ) → Hp(B ) is deﬁned by • Hp( f )([a]) := [ f p(a)]. ∈ p If a Ker(dA• ), then p+1 ◦ p p ◦ p 0 = f dA• (a) = dB• f (a) p ∈ p ∈ p • which implies f (a) Ker(dB• ). If [a1] = [a2] H (A ), then p−1 a2 = a1 + dA• (a) − for some a ∈ Ap 1. Then p p p ◦ p−1 p p−1 ◦ p−1 f (a2) = f (a1) + f dA• (a) = f (a1) + dB• f (a). p p Thus [ f (a1)] = [ f (a2)]. • • • Similarly, we can deﬁne Hp(g ) : Hp(B ) → Hp(C ). 50 1. BASIC TOPOLOGY

• • Next we deﬁne δp : Hp(C ) → Hp+1(A ) as follows...... x x x    p+1 p+1 p+1  •  •  • dA dB dC

p+1 p+1 p+1 f p+1 p g p+1 0 −−−−→ A ∋ a −−−−→ B ∋ d • (b) −−−−→ C −−−−→ 0 x x B x  p  p  p  •  •  • dA dB dC f p gp 0 −−−−→ Ap −−−−→ Bp ∋ b −−−−→ Cp ∋ c −−−−→ 0 x x x    p−1 p−1 p−1  •  •  • dA dB dC ...... ∈ p p p For any c Ker(dC• ), since g is surjective, it follows that g (b) = c for some b ∈ Bp. Because p ◦ p p+1 ◦ p 0 = dC• g (b) = g dB• (b), p ∈ p+1 p+1 p+1 p ∈ we get dB• (b) Ker(g ) = Im( f ). Thus f (a) = dB• (b) for some a Ap+1. From p+1 ◦ p p+1 ◦ p+1 p+2 ◦ p+1 0 = dB• dB• (b) = dB• f (a) = f dA• (a) p+2 p+1 and the injectivity of , we see that dA• (a) = 0 and then a is closed. Deﬁne δp([c]) := [a]. It is clear that the map δp is independent of the choice of a, b, c.

1.4. Topological groups and matrix Lie groups A topological group is a group G endowed with a topology such that µ × −→ 7−→ : G G G, (g1, g2) g1g2 and − ι : G −→ G, g 7−→ g 1 are both continuous. If the topology is discrete, G is called a discrete group.

For example, R, together with additive group structure and Euclidean topol- ∗ ogy, is a topological group; R := R \{0}, together with multiplication and the relative topology, is a topological group; the general linear group GL(n, R) := {n × n real nonsingular matrices}, together with matrix multiplication and the rel- 2 ative topology inherited from Rn , is a topologiocal group.

1.4.1. Subgroups and left/right translations. Let G be a topological group. A subgroup H of G is a subset which is also a subgroup in the algebraic sense. Under the relative topology, H is also a topological group. If A, B are subsets of G we let − − AB := {ab : a ∈ A, b ∈ B}, A 1 := {a 1 : a ∈ A}. − We say a subset A of G is symmetric if A = A 1. 1.4. TOPOLOGICAL GROUPS AND MATRIX LIE GROUPS 51

PROPOSITION 1.4.1. (1) Any ﬁnite product of topological groups is a topological group. (2) Let G be a topological group with unity element e. (a) The symmetric neighborhoods of e form a neighborhood basis at e. (b) If g ∈ G and U is any neighborhood of g, then there is a symmetric neighborhood − V of e such that VgV 1 ⊆ U. (c) If U is any neighborhood of e and n is any positive integer, then there exists a symmetric neighborhood V of e such that Vn ⊆ U. (3) If H is any subgroup of G, then H is also a subgroup of G. If H is a normal subgroup then so is H.

PROOF. (1): as an exercise. − − (2) If U is a neighborhood of e then so is U 1, and hence U ∩ U 1 is a symmetric neighborhood of e. This implies (a). (3) ????

By Proposition 1.4.1, Rn, S1, Tn are topological groups. Since the orthogonal group O(n) = {A ∈ GL(m, R) : AAT = I} is a subgroup of GL(m, R), O(m) is a topological group.

Let G be a topological group and g ∈ G. (1) The left translation is defined by ′ ′ (1.4.1.1) Lg : G −→ G, g 7−→ gg . − 1 − ◦ Then Lg = Lg 1 and Lg1 Lg2 = Lg1g2 . (2) The right translation is defined by ′ ′ −1 (1.4.1.2) Rg : G −→ G, g 7−→ g g . − 1 − ◦ Then Rg = Rg 1 and Rg1 Rg2 = Rg1g2 . (3) The adjoint homomorphism is defined by ′ ′ −1 (1.4.1.3) adg : G −→ G, g 7−→ gg g .

Then adg = Lg ◦ Rg = Rg ◦ Lg. 1.4.2. Group actions. Let X be a topological space and G a topological group. (1) The left action of G on X is a continuous map (1.4.2.1) G × X −→ X, (g, x) 7−→ g · x with the properties that · · · · g1 (g2 x) = (g1g2) x, 1 x = x ∈ ∈ for any g1, g2 G and x X. (2) The right action of G on X is a continuous map (1.4.2.2) X × G −→ X, (x, g) 7−→ x · g with the properties that · · · · (x g1) g2 = x (g1g2), x 1 = x 52 1. BASIC TOPOLOGY

∈ ∈ for any g1, g2 G and x X.

Observe that Lg induces a left action of G on itself by setting

g · x := Lgx.

Similarly, Rg induces a right action of G on itself by setting · x g := Rg−1 x.

We now always assume that G X is a left action. For any x ∈ x, we define the orbit of x by (1.4.2.3) G · x := {g · x : g ∈ G} ⊆ X. If G · x ∩ G · y ̸= ∅, then g · g = y for some g ∈ G and G · x = G · y. We can define an equivalence relation ∼ on X by (1.4.2.4) x ∼ y ⇐⇒ g · x = y for some g ∈ G. Then the equivalence class [x] of x is precisely the orbit G · x of x. Define the orbit space by ∪ (1.4.2.5) G \ X := X/ ∼= {[x] : x ∈ X} = G · x. x∈X

EXAMPLE 1.4.2. (1) The left action GL(n, R) × Rn → Rn given by (A, x) 7→ Ax has orbits that are {0} and Rn \{0}. (2) The left action O(n) × Rn → Rn given by (A, x) 7→ Ax has orbits that are {0} and spheres at 0. ∗ (3) The left action R × (Rn \{0}) → Rn \{0} given by (r, x) 7→ rx has orbits that are lines through 0.

THEOREM 1.4.3. Let G be a topological group acted from left on a topological space X. Then (i) π : X → G \ X is open, and (ii) G \ X is Hausdorff if and only if {(x, y) ∈ X × X : y = g · x for some g ∈ G} is closed in X × X.

− PROOF. (i) Let U ⊂ X be open and consider π 1(π(U)). To prove π is open, −1 −1 we shall verify π (π(U)) is open in X. For any x ∈ π (π(U)), set Ux := {g · x : g ∈ G}. The continuity implies that Ux is a neighborhood of x and Ux ⊂ − π 1(π(U)). ∆ { ∈ \ } (ii) G\X = ([x], [x]) : [x] G X implies that π × π −1∆ { ∈ × · ∈ } ( ) G\X = (x, y) X X : g x = y for some g G . π × π ∆ Since is open, surjective, and continuous, it follows that G\X is closed if π × π −1 ∆ and only if ( ) ( G\X) is closed. 1.4. TOPOLOGICAL GROUPS AND MATRIX LIE GROUPS 53

In general, we can show that if f : X → Y is open, surjective, and continuous, − then a subset S ⊆ X is closed if and only if f 1(S) ⊆ X is closed. Indeed, if S is closed, then − − − − f 1(Y \ S) = f 1(Y) \ f 1(S) = X \ f 1(S) − − is open in X. If f 1(S) is closed, then f (X \ f 1(S)) = Y \ S is open in Y.

1.4.3. Matrix Lie groups. Let K = R or C. The general linear map over K is the group

2 (1.4.3.1) GL(n, K) := {n × n invertible matrices with K-entries} ⊆ Kn . Then GL(n, K) is a topological group. Deﬁne ∼ 2 (1.4.3.2) Max(n, K) := {n × n matrices with K-entries} = Kn . { } We say a sequence Am m≥1 of complex matrices in Max(n, C) converges to ∈ → → ∞ ≤ ≤ a matrix A Max(n, C) if (Am)ij Aij as m for all 1 i, j n.

DEFINITION 1.4.4. A matrix Lie group is any subgroup G of GL(n, C) with { } the following property: if Am m≥1 is any sequence of matrices in G and Am converges to some matrix A, then either A ∈ G or A is not invertible. Equivalently G ⊆ GL(n, C) is a matrix Lie group if and only if it is a closed subgroup of GL(n, C).

EXAMPLE 1.4.5. (Counterexamples of matrix Lie groups) (1) The set of all n × n invertible matrices all of whose entries are rational. (2) Let a be irrational and consider {[ √ ] } −1t e √0 ∈ G := − t R . 0 e 1ta Clearly −I ∈/ G. However we can ﬁnd a sequence of matrices in G which converges to −I, so G is not a matrix Lie group. Actually {[ √ ] } −1t e √0 ∈ ∼ 1 × 1 2 G = − s, t R = S S = T . 0 e 1s We introduce some examples of matrix Lie groups. (1) The general linear groups GL(n, R) and GL(n, C). Moreover GL(n, R) is a subgroup of GL(n, C). (2) The special linear groups SL(n, K): (1.4.3.3) SL(n, C) := {A ∈ GL(n, K)| det A = 1}. (3) The orthogonal and special orthogonal groups. We say A ∈ Max(n, R) is orthogonal if δ ≤ ≤ ∑ Aki Akj = ij, 1 i, j n. 1≤k≤n The usual inner product on Rn is ⟨x, y⟩ := ∑ xkyk, x = (x1, ··· , xn), y = (y1, ··· , yn) ∈ Rn. 1≤k≤n 54 1. BASIC TOPOLOGY

Then A is orthogonal is equivalent to ⟨x, y⟩ = ⟨Ax, Ay⟩ for any x, y ∈ Rn. Hence A is orthogonal if and only if AT A = I, where AT denotes the T transpose of A, i.e., (A )ij = Aji. If A is orthogonal, then 1 = det(AT A) = (det A)2 and det A = ±1. Let (1.4.3.4) O(n) := {A ∈ GL(n, R)|A orthogonal} and (1.4.3.5) SO(n) := {A ∈ O(n)| det A = 1}. For n = 2, SO(2) is the group of rotations on the plane, i.e., {[ ] }

cos θ − sin θ (1.4.3.6) SO(2) = θ ∈ R . sin θ cos θ (4) The unitary and special unitary groups. We say A ∈ Max(n, C) is unitary if δ ≤ ≤ ∑ Aki Akj = ij, 1 i, j n. 1≤k≤n The usual inner product on Cn is ⟨z, w⟩ := ∑ zkwk, z = (z1, ··· , zn), w = (w1, ··· , wn) ∈ Cn. 1≤k≤n Then A is unitary is equivalent to ⟨z, w⟩ = ⟨Az, Aw⟩ for any z, w ∈ Cn. ∗ ∗ Hence A is unitary if and only if A A = I, where A denotes the adjoint ∗ of A, i.e., (Aij = Aji. ∗ If A is unitary, then 1 = det(A A) = | det A|2 and det A ∈ S1. Let (1.4.3.7) U(n) := {A ∈ L(n, C)|A unitary} and (1.4.3.8) SU(n) := {A ∈ U(n)| det A = 1}. (5) The complex orthogonal groups. Instead the inner product on Cn, we consider the bilinear form (·, ·) on Cn, (z, w) := ∑ zkwk, z = (z1, ··· , zn), w = (w1, ··· , wn) ∈ Cn. 1≤k≤n We say A ∈ Max(n, C) is (complex) orthogonal of (z, w) = (Az, Aw) for all z, w ∈ Cn. If A is orthogonal, then 1 = det A · det A and det A = ±1. Let (1.4.3.9) O(n, C) := {A ∈ Max(n, C)|A orthogonal} and (1.4.3.10) SO(n, C) := {A ∈ O(n, C)| det A = 1}. (6) The generalized orthogonal and Lorentz groups. For any positive inte- · · n+k gers n and k, deﬁne a symmetric bilinear form [ , ]n,k on R : i i − j j (1.4.3.11) [x, y]n,k := ∑ x y ∑ x y . 1≤i≤n n+1≤j≤n+k 1.4. TOPOLOGICAL GROUPS AND MATRIX LIE GROUPS 55

Let { } ∈ | (1.4.3.12) O(n, k) := A Max(n + k, R) [Ax, Ay]n,k = [x, y]n,k . For A ∈ Max(n + k, R), let   A  1,i  (i)  .  ≤ ≤ A := . , 1 i n + k. An+k,i Then   (i) (j) ̸  [A , A ]n,k = 0, i = j, ∈ ( ) ⇐⇒ (i) (i) ≤ ≤ A O n, k  [A , A ]n,k = 1, 1 i n  (i) (i) − ≤ ≤ [A , A ]n,k = 1, n + 1 i n + k. Let [ ] In 0 g := − 0 Ik Then (1.4.3.13) AO(n, k) ⇐⇒ ATgA = g. Hence det A = ±1. When (n, k) = (3, 1), we obtain the Lorentz group O(3, 1). (7) The symplectic groups. Define the following skew-symmetric bilinear form B on R2n: (1.4.3.14) B[x, y] := ∑ (xkyn+k − xn+kyk). 1≤k≤n Let us define the real symplectic group by (1.4.3.15) Sp(n, R) := {A ∈ Max(2n, R)|B[Ax, Ay] = B[x, y], ∀ x, y ∈ R2n}. Clearly Sp(1, R) = SL(2, R). If define [ ] 0 I J := n −In 0 then B[x, y] = ⟨x, Jy⟩ and (1.4.3.16) A ∈ Sp(n, R) =⇒ AT JA = J. Hence det A = ±1 for all A ∈ Sp(n, R). Actually, we can prove that det A = 1 for all A ∈ Sp(n, R). Similarly we can define the complex symplectic group (1.4.3.17) Sp(n, C) := {A ∈ Max(2n, C)|B[Ax, Ay] = B[x, y], ∀ x, y ∈ C2n}. ¯ Then (1.4.3.18) A ∈ Sp(n, C) ⇐⇒ AT JA = J. Hence det A = ±1 for all A ∈ Sp(n, C). Actually, we can prove that det A = 1 for all A ∈ Sp(n, C). Clearly Sp(1, C) = SL(2, C). The compact symplectic group4 is defined to be (1.4.3.19) Sp(n) := Sp(n, C) ∩ U(2n).

4Some books used Sp(n) for our Sp(n, C) and USp(n) for our Sp(n). 56 1. BASIC TOPOLOGY

Clearly Sp(1) = SU(2). The special and general linear groups, the orthogonal and unitary groups, and the symplectic groups make up the classical groups. (8) The Heisenberg group H is    

 1 a b    ∈ ∈ ⊆ (1.4.3.20) H := A = 0 1 c Max(3, R) a, b, c R SL(3, R). 0 0 1 ∗ ∼ ∗ ∼ (9) R = {a ∈ R|a ̸= 0} = GL(1, R), C = {a ∈ C|a ̸= 0} = GL(1, C), ∗ ∼ and S1 = {z ∈ C ||z| = 1} = U(1). The additive group R is isomorphic to GL+(1, R) the group of 1 × 1 matrices with positive determinant, via the mapx 7→ [ex]. Moreover, the additive group Rn is isomorphic to the group of diagonl real matrices with positive diagonal entries, i.e., ∼ Rn = GL+(1, R) × · · · × GL+(1, R)(n times). (10) The Euclidean and Poincar´egroups. Let (1.4.3.21) E(n) := {distance-preserving bijections on Rn} that is, f ∈ E(n) if and only if f : Rn → Rn is bijective and |x − y| = | f (x) − f (y)| for all x, y ∈ Rn. Note that f need not to be linear. Then O(n) is a subgroup of E(n) and is the group of all linear distance- preserving maps on Rn. For x ∈ Rn deﬁne the translation by x by n n Tx : R −→ R , y 7−→ x + y. A basic fact about R(n) is any T ∈ E(n) can be written uniquely as an orthogonal linear transformation followed by a translation, i.e., T = Tx ◦ R for some x ∈ Rn and R ∈ O(n). Write

{x, R} := Tx ◦ R. Then for y ∈ Rn, one has {x, R}y = Ry + x and { }{ } x1, R1 x2, R2 y = R1(R2y + x2) + x1 = R1R2y + (x1 + R1x2). The product on E(n) is { }{ } { ◦ } (1.4.3.22) x1, R1 x2, R2 = x1 + R1x2, R1 R2 and the inverse of {x, R} is − − − {x, R} 1 = {−R 1x, R 1}. Because translations are not linear, we conclude that E(n)is not a subgroup of GL(n, R). However, E(n) is isomorphic to a subgroup of GL(n + 1, R) via the following map [ ] R xT {x, R} 7−→ ∈ GL(n + 1, R). 0 1 The Poincare´ group is deﬁned to be n+1 (1.4.3.23) P(n, 1) := {T = Tx ◦ A|x ∈ R , A ∈ O(n, 1)}. 1.4. TOPOLOGICAL GROUPS AND MATRIX LIE GROUPS 57

This is the group of afﬁne transformations of Rn+1 which preserves the Lorentz distance i i 2 n+1 n+1 2 n+1 dL(x, y) := ∑ (x − y ) − (x − y ) , x, y ∈ R . 1≤i≤n Here an afﬁne transformation is one of the form x 7→ Ax + b, where A is a linear transformation and b is a constant. Similarly, we can prove that P(n, 1) is isomorphic to the group of (n + 2) × (n + 2) matrices of the form [ ] A xT A ∈ O(n, 1), x ∈ Rn+1 0 1

EXERCISE 1.4.6. (1) Show that O(n) is a group, a subgroup of GL(n, C), and is a matrix Lie group. (2) Show that SO(n) is a group, a subgroup of O(n), and is a matrix Lie group. (3) Deduce (1.4.3.6) from (1.4.3.5). (4) Show that U(n) is a group, a subgroup of GL(n, C), and is a matrix Lie group. (5) Show that SU(n) is a group, a subgroup of U(n), and is a matrix Lie group. (6) Show that O(n, k) is a group, a subgroup of GL(n + k, R), and a matrix Lie group. (7) Determine elements in O(1, 1). (8) Verify (1.4.3.13). (9) Show that Sp(n, R) is a group, a subgroup of GL(2n, R), and a matrix Lie group. (10) Show that the Heisenberg H is a group, a subgroup of GL(3, R), and a matrix Lie group. Actually, every A ∈ SU(2) can be expressed in the form [ ] α −β α β ∈ |α|2 |β|2 A = β α , , C, + = 1.

Hence we can view SU(2) as the three-dimensional sphere S3 inside C2 = R4.

DEFINITION 1.4.7. A matrix Lie group G is said to be compact if the following conditions holds: (1) if Am is any sequence of matrices in G and Am converges to a matrix A, then A ∈ G, and ∈ || || | | ≤ (2) there exists a constant C such that for all A G, A ∞ = max1≤i,j≤n Aij C. 2 If we view Max(n, C) as Cn , then G is compact in the sense of Deﬁnition 1.4.7 2 if and only if G is a closed and bounded subset of Cn .

EXAMPLE 1.4.8. (1) Compact matrix Lie groups: O(n), SO(n), U(n), SU(n), Sp(n). (2) Noncompact matrix Lie groups: GL(n, R), GL(n, C), SL(n, R), SL(n, C), O(n, C), SO(n, C), O(n, k), SO(n, k), the Heisenberg group H, Sp(n, R), Sp(n, C), ∗ ∗ E(n), P(n, 1), R, R , C, C .

EXERCISE 1.4.9. Verify Example 1.4.8. 58 1. BASIC TOPOLOGY

DEFINITION 1.4.10. A matrix Lie group G is path-connected if given any two matrices A and B in G, there exists a continuous path A(t) in G, 0 ≤ t ≤ 1, such that A(0) = A and A(1) = B.

2 Since any matrix Lie group G can be viewed as a subset of Cn that is simply- connected, it follows that any matrix Lie group is path-connected if and only if it is path-connected.

A matrix Lie group which is not path-connected can be decomposed uniquely as a union of path-connected components, such that two elements in the same path-connected component can be joined by a continuous path, but two elements in the different path-connected components cannot.

PROPOSITION 1.4.11. (1) If G is a matrix Lie group, then the path-connected component of G containing the identity is a subgroup of G. (2) For any n ≥ 1, the group GL(n, C) is path-connected. (3) For any n ≥ 1, the group SL(n, C) is path-connected. (4) For any n ≥ 1, the groups U(n) and SU(n) are path-connected.

PROOF. (1) Let H be the path-connected component of G containing the identity e. For A, B ∈ H ⊆ G, there exists a continuous paths A(t), B(t), such that A(0) = B(0) = e, A(1) = A, B(1) = B. Then (A(t)B(t) is a continuous path connecting e and AB. Hence the product AB is still in H. Similarly the inverse of A is again in H. Hence H is a subgroup. ∼ ∗ (2) When n = 1, GL(1, C) = C that is obviously path-connected. When n ≥ 2, we claim that any element A ∈ GL(n, C) can be connected to the identity I by a continuous path lying in GL(n, C). From linear algebra, we know that any matrix is similar to an upper triangular matrix. That is, for any A ∈ Max(n, C) there exists C ∈ GL(n, C) such that   λ ∗ · · · ∗ 1    .. .  −  0 . ∗ .  A = CBC 1, B =    . . .  . .. .. ∗ 0 ··· 0 λn If A ∈ GL(n, C), then all λ are nonzero. Set  i  λ − ∗ · · · − ∗ 1 (1 t) (1 t)    . .   0 .. (1 − t)∗ .  B(t) =   (0 ≤ t ≤ 1), D = diag(λ , ··· , λn).  . . .  1 . .. .. (1 − t)∗ 0 ··· 0 λn − − Then A(t) := CB(t)C 1 is a continuous path in GL(n, C) from A to CDC 1 and ∈ ∈ { ··· } λ ∈ A(t) GL(n, C). Next, for each i 1, , n , choose a continuous path i(t) ∗ ≤ ≤ λ λ λ e C , 1 t 2, such that i(1) = i and i(2) = 1. Then A(t) is a continuous path −1 in GL(n, C) from CDC to In, where e · λ ··· λ · −1 ≤ ≤ A(t) = C diag( 1(t), , n(t)) C , 1 t 2.

Hence we ﬁnd a continuous path in GL(n, C) from A to In. 1.4. TOPOLOGICAL GROUPS AND MATRIX LIE GROUPS 59

λ ≤ ≤ − (3) In the proof of (2), deﬁne i(t), 1 i n 1 as before and λ 1 n(t) = λ ··· λ . 1(t) n−1(t) (4) From linear algebra, any unitary matrix U ∈ U(n) can be written as √ √ − θ − θ − · 1 1 ··· 1 n · 1 ∈ θ ··· θ ∈ U = U1 diag(e , , e ) U1 , U1 U(n), 1, , n R. If we deﬁne √ √ − − θ − − θ − · 1(1 t) 1 ··· 1(1 t) n · 1 ≤ ≤ U(t) = U1 diag(e , , e ) U1 , 0 t 1, then U(t) ∈ U(n) and connects U to In.

Group Connected? Components GL(n, C) Y 1 SL(n, C) Y 1 GL(n, R) N 2 SL(n, R) Y 1 O(n) N 2 SO(n) Y 1 U(n) Y 1 SU(n) Y 1 O(n, 1) N 4 SO(n, 1) N 2 Heisenberg Y 1 E(n) N 2 P(n, 1) N 4

DEFINITION 1.4.12. A matrix Lie group G is simply-connected if it is path- connected and any loop in G can be shrunk continuously to a point in G. Precisely, assume that G is path-connected. Then G is simply-connected if for any given continuous path A(t), 0 ≤ t ≤ 1, lying in G with A(0) = A(1), there exists a continuous function A(s, t), 0 ≤ s, t ≤ 1, such that A(s, 0) = A(s, 1)(0 ≤ s ≤ 1), A(0, t) = A(t), A(1, t) = A(1, 0)(0 ≤ t ≤ 1).

Deﬁne As(t) := A(s, t). Then A(s, 0) = A(s, 1) means that As is a loop; A(0, t) = A(t) means A0 = A; A(1, t) = A(1, 0) means A1 is the point A(0) = A(1).

PROPOSITION 1.4.13. The group SU(2) is simply-connected.

PROOF. Any A ∈ SU(2) can be written as [ ] α −β α β ∈ |α|2 |β|2 A = β α , , C with + = 1.

Deﬁne [ ] α −β −→ 3 ⊆ 2 ∼ 4 7−→ α β SU(2) S C = R , β α ( , ). The above map is homeomorphic so that SU(2) is simply-connected. 60 1. BASIC TOPOLOGY

Group Simply-connected? Fundamental group SO(2) N Z SO(n)(n ≥ 3) N Z/2Z U(n) X Z SU(n) Y {1} Sp(n) Y {1} GL+(n, R) N Z (n = 2), Z/2Z (n ≥ 3) GL(n, C) N Z SL(n, R)(n ≥ 2) N Z (n = 2), Z/2Z (n ≥ 3) SL(n, C) Y {1} SO(n, C) N Z (n = 2), Z/2Z (n ≥ 3) Sp(n, R) N Z Sp(n, C) Y {1} Let X ∈ Max(n, K). Deﬁne formally Xm (1.4.3.24) eX ≡ exp X := ∑ . m≥0 m!

2 Regarding Max(n, C) as Cn , we can deﬁne the Hilbert-Schmidt norm by ( ) 1/2 || || | |2 (1.4.3.25) X := ∑ Xij . 1≤i,j≤n n2 { } Then we can pullback all analysis on C to Max(n, C). A sequence Xm m≥1 in || − || → → ∞ { } Max(n, C) converges to X, if Xm X 0 as m . A sequence Xm m≥1 in Max(n, C) is a Cauchy sequence if ||Xm − Xℓ|| → 0 as m, ℓ → ∞. Therefore, any Cauchy sequence in Max(n, C) converges.

PROPOSITION 1.4.14. For any X ∈ Max(n, K), the series (1.4.3.24) converges. Moreover eX is continuous in X.

PROOF. Since m m X ||X|| || || ∑ ≤ ∑ ≤ e X , 0≤m≤M m! 0≤m≤M m! || m || { i } it follows that the series ∑m≥0 X /m! is convergent. Hence ∑0≤i≤m X /i! m≥0 is a Cauchy sequence and then is a convergent sequence. Thus eX is well-deﬁned. Because the series (1.4.3.24) uniformly converges on {X ∈ Max(n, C)||X|| ≤ R} and each Xm/m! is continuous, we can conclude that eX is continuous.

Consider the following 2 × 2 matrices [ ] [ ] 0 −a b c X = , Y = a 0 0 b Clearly [ ] [ ] [ ] −a2 0 0 a3 a4 0 X2 = , X3 = , X4 = , ··· . 0 −a2 −a3 0 0 a4 1.4. TOPOLOGICAL GROUPS AND MATRIX LIE GROUPS 61

Then [ ( )] 2 4 3 [ ] 1 − a + a + · · · − a − a + ··· cos a − sin a eX = 2! 4! 3! = − a3 ··· − a2 a4 ··· sin a cos a a 3! + 1 2! + 4! + On the other hand [ ] eb ebc XY ̸= YX, eY = , eXeY ̸= eYeX. 0 eb To compute eX for more general matrices X, we ﬁrst prove the following

PROPOSITION 1.4.15. Let X, Y ∈ Max(n, C). Then 0 (1) e = In. ∗ ∗ (2) (eX) = eX . − − (3) eX is invertible and (eX) 1 = e X. α β α β (4) e( + )X = e Xe X for all α, β ∈ C. (5) If XY = YX, then eX+Y = eXeY = eYeX. −1 − (6) If C is invertible, then eCXC = CeXC 1. || || (7) ||eX|| ≤ e X . (8) etX is a smooth curve in Max(n, C) and

d d etX = XetX = etX X, etX = X. dt dt t=0

PROOF. (1) − (2) and (6) − (8) are clearly. Also (3) − (4) follow from (5). Since XY = YX, it follows that for any m ≥ 1 we have ( ) m − (X + Y)m = ∑ XkYm k. 0≤k≤m k Hence ( )( ) ( ) − Xm Ym Xk Ym k eXeY = ∑ ∑ = ∑ ∑ ( − ) m≥0 m! m≥0 m! m≥0 0≤k≤m k! m k ! ( ( ) ) 1 m − 1 = ∑ ∑ XkYm k = ∑ (X + Y)m = eX+Y. m≥0 m! 0≤k≤m k m≥0 m! Now we have eXeY = eX+Y = eY+X = eYeX.

EXERCISE 1.4.16. Prove (1) − (2) and (6) − (8).

Return back to the above matrix [ ] 0 −a X = a 0 Because the eigenvectors of X are [ ] [√ ] 1 −1 √ , −1 1 √ √ with eigenvalues − −1a and −1a, respectively, we get [ √ ] [ √ ] − 1 −1 − −1a 0 X = CDC 1, C := √ , D = √ −1 1 0 −1a 62 1. BASIC TOPOLOGY

From Proposition 1.4.15 (6), we have [ √ ] [ √ ][ √ ] − − −1a 1 − −1 X D −1 √1 1 e √0 √2 2 e = Ce C = − − −1 1 0 e 1a − 1 1 [ ] 2 2 − = cos a sin a sin a cos a For any X, Y ∈ Max(n, C), we know from Proposition 1.4.15 (8) that etX is smooth and then ( ) ( ) ( ) 2 2 d − d X X etXYe tX = I + tX + t2 + ··· Y I − tX + t2 + ··· dt dt n 2 n 2 t=0 t=0 ( ) d = Y + t(XY − YX) + t2(··· ) = XY − YX. dt t=0

DEFINITION 1.4.17. For any X ∈ Max(n, K) deﬁne −→ 7−→ − (1.4.3.26) adX : Max(n, K) Max(n, K), Y [X, Y] := XY YX. We also call [X, Y] the Lie bracket of X and Y. m If g(z) = ∑ ≥ am(z − a) is a formal series, then deﬁne m 0 ( ) − m −→ g(adX) := ∑ am adX a1Max(n,K) : Max(n, K) Max(n, K). m≥0 In particular, we obtain the operator eadX for each X ∈ Max(n, K). Moreover we can prove − (1.4.3.27) eXYe X = eadX (Y).

Indeed, ( ) m k m−k (ad ) − X Y (−X) eadX (Y) = ∑ X Y, eXYe X = ∑ ∑ . ( − ) m≥0 m! m≥0 0≤k≤m k! m k ! To prove (1.4.3.27) we need to verify ( ) m m k − m−k (adX) Y = ∑ X Y( X) . 0≤k≤m k m ∑ m k − m−k When m = 1, it is trivial. Assume that (adX) Y = 0≤k≤m ( k )X Y( X) . Then [ ( ) ] m+1 m m k − m−k (adX) Y = adX ((adX) Y) = X, ∑ X Y( X) 0≤k≤m k ( ) ( ) m − − = ∑ Xk+1Y(−X)m k − XkY(−X)m+1 k k 0≤(k≤m ) ( ) m − m − = ∑ XkY(−X)m+1 k + ∑ XkY(−X)m+1 k k − 1 k 1≤k≤m+1 ( ) 0≤k≤m m + 1 − = ∑ XkY(−X)m+1 k. 0≤k≤m+1 k A ﬁnite-dimensional K-Lie algebra is a ﬁnite-dimensional K-vector space g, together with a map [·, ·] : g × g into g, with the following properties 1.4. TOPOLOGICAL GROUPS AND MATRIX LIE GROUPS 63

(1) [·, ·] is bilinear, (2) [X, Y] = −[Y, X] for all X, Y ∈ g, (3) (Jacobi identity) [X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0 for all X, Y, Z ∈ g. It is clear that Max(n, K) together with [·, ·] deﬁned in (1.4.3.26) is a ﬁnite-dimensional K-Lie algebra.

THEOREM 1.4.18. Every invertible n × n matrix can be expressed as eX for some X ∈ Max(n, C). Recall that the function (z − 1)m ln z = ∑ (−1)m+1 m≥1 m is defined and holomorphic in {z ∈ C||z − 1| < 1}. In particular, when |u| < ln 2, ln(eu) = u, when |z − 1| < 1, eln z = z. DEFINITION 1.4.19. For any A ∈ Max(n, C), define (A − I )m (1.4.3.28) ln A := ∑ (−1)m+1 n . m≥1 m As in the proof of Proposition 1.4.14, we can prove that the series (1.4.3.28) is defined and continuous in {A ∈ Max(n, C)|||A − In|| < 1}. Moreover, when X ln A ||X|| < ln 2, ln e = X, when ||A − In|| < 1, e = A.

To extend Proposition 1.4.15 (5) to the case that [X, Y] ̸= 0, we need the following deﬁnition: ln z z ln z (−1)m+1 (1.4.3.29) g(z) = = = 1 + ∑ (z − 1)m, |z − 1| < 1. − 1 z − 1 m(m + 1) 1 z m≥1

EXERCISE 1.4.20. Verify (1.4.3.29). For any X, Y ∈ Max(n, C) with sufﬁciently small ||X|| and ||Y||, we have ( ) X Y 1 2 1 2 e e = In + (X + Y) + X + XY + Y ( 2 2 ) 1 1 1 1 + X3 + X2Y + XY2 + Y3 + ··· 6 2 2 6 Then ( ) 1 1 eXeY − I = (X + Y) + X2 + XY + Y2 + ··· n 2 2 and ( X Y 2 2 3 3 (e e − In) = (X + Y) + X + Y + XYX + YXY ) 1 1 3 2 + Y2X + YX2 + X2Y + XY2 + ··· 2 2 2 3 Therefore − m+1 X Y ( 1) X Y m X Y 1 X Y 2 ln(e e ) = ∑ (e e − 1) = (e e − In) − (e e − In) + ··· m≥1 m 2 64 1. BASIC TOPOLOGY

1 1 1 = X + Y + X2 + XY + Y2 − (X2 + XY + YX + Y2) + ··· 2 2 2 XY − YX 1 = X + Y + + ··· = X + Y + [X, Y] + ··· . 2 2 The higher order terms are given by the famous Baker-Campbell-Hausdorff formula.

THEOREM 1.4.21. (Baker-Campbell-Hausdorff) For all X, Y ∈ Max(n, C) with ||X|| and ||Y|| sufﬁciently small, we have ∫ 1 (1.4.3.30) ln(eXeY) = X + g(eadX ◦ etadY )(Y)dt. 0 The above formula gives the next terms in the series of ln(eXeY).

EXERCISE 1.4.22. Show by (1.4.3.30) that 1 1 1 (1.4.3.31) ln(eXeY) = X + Y + [X, Y] + [X, [X, Y]] − [Y, [X, Y]] + ··· 2 12 12 for sufﬁciently small X and Y. We will give a proof of Theorem 1.4.21 in a special case that X and Y commute with [X, Y] (we now do not assume further that ||X|| and ||Y|| are sufﬁciently small).

THEOREM 1.4.23. If X, Y ∈ Max(n, C) satisﬁes (1.4.3.32) [X, [X, Y]] = [Y, [X, Y]] = 0, then X Y X+Y+ 1 [X,Y] (1.4.3.33) e e = e 2 .

PROOF. For any t ∈ R we shall prove

2 tX tY tX+tY+ t [X,Y] e e = e 2 which is equivalent, under the assumption (1.4.3.32),

2 t(X+Y) tX tY − t [X,Y] e = e e e 2 . Set 2 t(X+Y) tX tY − t [X,Y] A(t) := e , B(t) := e e e 2 . Clearly that d A(t) = A(t)(X + Y) dt by Proposition 1.4.15 (8). For B(t) one has ( ) 2 d d tX tY − t [X,Y] B(t) = e e e 2 dt dt ( ) ( ) ( ) 2 2 2 d tX tY − t [X,Y] tX d tY − t [X,Y] tX tY d − t [X,Y] = e e e 2 + e e e 2 + e e e 2 dt dt dt 2 2 2 tX tY − t [X,Y] tX tY − t [X,Y] tX tY − t [X,Y] = e Xe e 2 + e e Ye 2 + e e e 2 (−t[X, Y]). 1.4. TOPOLOGICAL GROUPS AND MATRIX LIE GROUPS 65

Since Y commutes with [X, Y], it follows that 2 2 tX tY − t [X,Y] tX tY − t [X,Y] e e Ye 2 = e e e 2 Y. Using (1.4.3.27) and [Y, [X, Y]] = 0 yields − − XetY = etYe tY XetY = etYe adY (X) = X − t[Y, X] = X + t[X, Y]. Consequently,

2 2 2 d tX − t [X,Y] tX tY − t [X,Y] tX tY − t [X,Y] B(t) = e (X + t[X, Y])e 2 + e e e 2 Y + e e e 2 (−t[X, Y]) dt 2 tX tY − t [X,Y] = e e e 2 (X + t[X, Y] + Y − t[X, Y]) = B(t)(X + Y). Because both A(t) and B(t) satisfy the same differential equation, and A(0) = B(0) = In, we can conclude that A(t) ≡ B(t).

EXERCISE 1.4.24. (1) Show that for any two elements X, Y of the Heisenberg group H, we always have (1.4.3.32). (2) Show by direct computation that for any X, Y ∈ h, we have (1.4.3.33), where   

 0 α β   γ α β γ ∈ (1.4.3.34) h :=  0 0 , , R . 0 0 0 Actually, h deﬁned in (1.4.3.34) is the Lie algebra of the Heisenberg group H in the sense of Deﬁnition 1.4.25.

DEFINITION 1.4.25. Let G be a matrix Lie group. The Lie algebra of G, denoted g, is the set of all matrices X such that etX is in G for all t ∈ R. tX If we write A(t) := e , then A(0) = In, A(t) is continuous in t, and A(t + s) = A(t)A(s) for all t, s ∈ R. Such a function A : R → GKL(n, C) is called a one- parameter subgroup of GL(n, C). Conversely, we can prove that for a given one- parameter subgroup A(t) of GL(n, C) there exists a unique X ∈ Max(n, C) such that A(t) = etX.

THEOREM 1.4.26. For any X ∈ Max(n, C) we have (1.4.3.35) det(eX) = etr(X).

PROOF. From linear algebra, we know that X can be uniquely written as X = S + N, where S, N ∈ Mat(n, C), SN = NS, S is diagonalizable and N is nilpotent. From SN = NS, we obtain eX = eS+N = eSeN, det(eX) = det(eS) det(eN). λ ··· λ −1 ∈ Write S = Cdiag( 1, , n)C for some C GL(n, C). Then λ λ − eS = Cdiag(e 1 , ··· , e n )C 1 and λ ∑ λ det(eS) = ∏ e i = e 1≤i≤n i = etr(S). 1≤i≤n 66 1. BASIC TOPOLOGY

Since N is nilpotent, from linear algebra, we can ﬁnd some D ∈ GL(n, C) such that   0 · · · ∗    . . .  −1 N = D . .. . D 0 ··· 0 Hence   1 · · · ∗   N  . . .  −1 e = D . .. . D 0 ··· 1 So det(eN) = 1 and tr(N) = 0. Consequently det(eX) = etr(S) = etr(X).

We now can determine Lie algebras of classical groups. (1) G = GL(n, C): If X ∈ Max(n, C), we have etX ∈ GL(n, C). Hence the Lie algebra gl(n, C) of GL(n, C) is Max(n, C). (2) G = GL(n, R): The Lie algebra gl(n, R) of GL(n, R) is Max(n, R). (3) G = SL(n, C): If etX ∈ SL(n, C), then from (1.4.3.35) we have · 1 = det(etX) = etr(tX) = et tr(X) ⇐⇒ tr(X) = 0. Hence the Lie algebra sl(n, C) of SL(n, C) is given by (1.4.3.36) sl(n, C) = {X ∈ Max(n, C)|tr(X) = 0}.

(4) G = SL(n, R): The Lie algebra sl(n, R) of SL(n, R) is given by (1.4.3.37) sl(n, R) = {X ∈ Max(n, R)|tr(X) = 0}.

∗ − − ∗ (5) G = U(n): etX ∈ U(n) if and only if (etX) = (etX) 1 = e tX, or etX = − e tX. Thus the Lie algebra u(n) of U(n) is ∗ (1.4.3.38) u(n) = {X ∈ Max(n, C)|X = −X}.

(6) G = SU(n): The Lie algebra su(n) of SU(n) is ∗ (1.4.3.39) su(n) = {X ∈ Max(n, C)|X = −X, tr(X) = 0}.

− T − (7) G = O(n): etX ∈ O(n) if and only if (etX)T = (etX) 1 or etX = e tX. Thus the Lie algebra o(n) of O(n) is (1.4.3.40) o(n) = {X ∈ Max(n, R)|XT = −X}. If X ∈ o(n), then tr(X) = 0 and det(eX) = 0. (8) G = SO(n): The Lie algebra so(n) of SO(n) is the same as o(n). (9) G = O(n, k): The Lie algebra of O(n, k) is given by { [ ]}

∈ T − In 0 (1.4.3.41) o(n, k) = X Max(n + k, R) gX g = X, g = − . 0 Ik

(10) G = SO(n, k): the Lie algebra so(n, k) is the same as o(n, k). 1.5. CONNECTEDNESS AND COMPACTNESS 67

(11) G = Sp(n, C): The Lie algebra of Sp(n, C) is given by {[ ] }

AB (1.4.3.42) sp(n, C) = A ∈ Max(n, C) and B, C symmetric . C −AT

(12) G = Sp(n, R): The Lie algebra of Sp(n, R) is given by { [ ]} 0 I (1.4.3.43) sp(n, R) = X ∈ Max(2n, R)|JXT J = X, J = n . −In 0

(13) G = Sp(n): The Lie algebra of Sp(n) is given by sp(n) = sp(n, C) ∩ u(2n). (14) G = Heisenberg group H: The Lie algebra of H is given by (1.4.3.34). (15) G = E(n): The Lie algebra of E(n) is given by {[ ] } T Y y T (1.4.3.44) e(n) = Y ∈ Max(n, R), Y = −Y . 0 1

(16) G = P(n, 1): The Lie algebra of P(n, 1) is given by {[ ] }

Y yT (1.4.3.45) p(n, 1) = Y ∈ so(n, 1) . 0 0

1.5. Connectedness and compactness In this section we give general deﬁnitions of connectednes and compactness.

1.5.1. Connectedness. Let (X, T ) be a topological space. (1) A separation of X is a pair (U, V) such that ∅ ̸= U, V ⊆ T , U ∩ V = ∅, X = U ∪ V. (2) X is called disconnected if there exists a separation of X. If there does not, X is called connected. Thus, a topological space X is called connected if it is not the disjoint of two nonempty open subsets. (3) A subset A of X is called clopen if it is both open and closed in X. It is clear that X is connected if and only if its only clopen subsets are X and ∅. If A is a nonempty proper subset of X that is clopen, then the pair (U = A, V = X \ A) is a separation of X. Conversely, if X is connected, then for any subset A ⊆ X that is clopen in X, we conclude that (U = A, V = X \ A) is not a separation. (4) A discrete valued map is a continuous map from X to a discrete space D. (5) X is connected if and only if every discrete valued map on X is constant. If X is connected and f : X → D is a discrete valued map and if y ∈ D − is in the range of f , then f 1(y) is clopen and nonempty and so must equal X, and so f is constant with only value y. Conversely, if X is not connected then X = U ∪ V for some disjoint clopen sets U and V. Then the map f : X → {0, 1} which is 0 on U and is 1 on V is a nonconstant discrete value map. (6) A subset A of X is called connected, if A is a connected with respect to the subspace topology. 68 1. BASIC TOPOLOGY

(7) Define a relation ∼ on X by x ∼ y ⇐⇒ x and y belong to a connected subset of X. Then ∼ is an equivalence relation on X, and the equivalence class [x] is called a connected component or component of X. (8) Define a relation ∼ on X by x ∼ y ⇐⇒ f (x) = f (y) for every discrete valued map f on X. Then ∼ is an equivalence relation on X, and the equivalence class [x] is called a quasi-component of X. T → T PROPOSITION 1.5.1. (1) If f : (X, X) (Y, Y) is continuous and X is connected, then f (X) is connected. (2) If (Yi)i∈I is a collection of connected subsets in a topological space X and if no ∪ two of the Yi are disjoint, then i∈IYi is connected. (3) Components of a topological space X are connected and closed. Each connected subset is contained in a component. Thus the components are “maximal connected subsets”. Components are either equal or disjoint, and fill out X. (4) Quasi-components of a topological space X are closed. Each connected subsets is contained in a quasi-component. In particular, each component is contained in a quasi- component. Quasi-components are either equal or disjoint, and fill out X. (5) (Intermediate value theorem) If f : X → R is continuous and x, y ∈ X, then for any M between f (x) and f (y) we have f (z) = M for some z ∈ X. ··· × · · · × (6) If X1, , Xk are connected, then X1 Xk is connected. (7) If ∼ is an equivalence relation on a topological space X, then X/ ∼ is connected. (8) Suppose that (X, T ) is a topological space and U, V are disjoint open subsets of X. If S is a connected subset of X contained in U ∪ V, then either S ⊆ U or S ⊆ V. (9) If (X, T ) is a topological space that contains a dense connected subset S, then X is connected. (10) Let (X, T ) be a topological space. The union of a collection of connected subspaces of X that have a point in common is connected. (11) Suppose that (X, T ) is a topological space and A ⊆ X is connected. If a subset B of X satisfies A ⊆ B ⊆ A, then B is connected. In particular, A is connected.

PROOF. (1) Let d : f (X) → D be a discrete valued map. Then d ◦ f : X → D is a discrete valued map on X and hence must be constant. This implies that d is constant on f (X) and then f (X) is connected. ∪ → (2) Let d : i∈IYi D be a discrete valued map. Let p, q be any two points in ∪ ∈ ∈ ∈ ∩ i∈IYi. Suppose x Yi, y Yj, and z Yi Yj. Because d is constant on each Yi, we conclude that d(x) = d(z) = d(y). Hence d is constant. (3) By deﬁnition, the component of X containing x is the union of all connected subsets containing p, hence it is connected by (2). Since the closure of a connected set is connected5, it follows that components of X is closed. (4) If x ∈ X then the quasi-component containing p is the set ∩ − {y ∈ X : d(y) = d(x) for all discrete valued maps d} = d 1(d(x)) d discrete value maps

5Suppose that A is a connected subset. If U and V ar disjoint open subsets of X that separates A, then A is connected in one of them, say A ⊆ U. Since U ∩ V = ∅, it follows that A ⊆ U, which means that A ∩ V = ∅, a contradiction. 1.5. CONNECTEDNESS AND COMPACTNESS 69 which is an intersection of closed subsets and hence is closed. (5) This follows from the fact that f (X) is an interval (a nonempty subset of R is connected if and only if it is an interval). (6) By induction, we need only to consider a product of two topological spaces. Let X, Y be connected and d : X × Y → D a discrete valued map. Given a point ∈ × (x0, y0) X Y and deﬁne maps ι −→ × 7−→ ι −→ × 7−→ x0 : Y X Y, y (x0, y) y0 : X X Y, x (x, y0). ◦ ι → ◦ ι → Then we get maps d x0 : Y D and d y0 : X D. Since X and Y are ◦ ◦ ι connected, it follows that d x0 = cx0 and d y0 = cy0 for some constants cx0 , cy0

(may depending on (x0, y0)). In particular, cx0 = cy0 = d(x0, y0) =: cx0,y0 . Thus ∈ ∈ d(x0, y) = d(x, y0) = cx0,y0 for any x X, y Y. We now show that the constant cx0,y0 is independent of the choice of (x0, y0). Indeed, for another pair (x1, y1), we ∈ ∈ also have d(x1, y) = d(x, y1) = cx1,y1 for any x X, y Y. Then

cx1,y1 = d(x0, y1) = cx0,y0 .

Thus d(x0, y0) = c for some constant c that is independent of the choice of (x0, y0). × Since (x0, y0) was arbitrary, we get d is constant and then X Y is connected. (7) This follows from the fact that X → X/ ∼ is surjective and (1). (8) Since U, V are open in X, it follows that A ∩ U and A ∩ V are open in A. The connectedness of A implies that A ∩ U = ∅ or A ∩ V = ∅. Thus A ⊆ V or A ⊆ U. (9) Let A ⊆ X be a dense connected subset of X. Assume that (U, V) is a separation of X. By (8), we have A ⊆ U or A ⊆ V. If A ⊆ U, then X = A ⊆ U = U (since U is both open and closed). But this implies V = ∅. Hence X is connected. { } ∈ ∩ (10) Let Aα α∈I be a collection of connected subspaces of X and p α∈I Aα. ∪ ∈ ∈ Set Y := α∈I Aα and suppose (C, D) is a separation of Y. Then p C or p D. Assume that p ∈ C. Since Aα is connected, from (8), we must have Aα ⊆ C for all α ∈ ∪ ⊆ I. Thus Y = α∈I Aα C. This contradiction shows that Y is connected. (11) Consider a separation (C, D) of B. According to (8), A ⊆ C or A ⊆ D. Suppose that A ⊆ C. Then A ⊆ C. Since C ∩ D = C ∩ D = ∅, it follows that B ∩ D = ∅. This contradicts the fact that ∅ ̸= D ⊆ B.

ω EXAMPLE 1.5.2. R is not connected in the box topology, but it is connected in the product topology. Indeed, we have ω R = A ∪ B where A ∩ B = ∅, and

A := {bounded sequences of R} ̸= ∅, B := {unbounded sequences of R} ̸= ∅.

i ∈ ω We claim that both A and B are clopen in the box topology. For all a = (a )i≥1 R consider the open set i i Ua := ∏(a − 1, a + 1). i≥1

Then Ua ⊆ A if a ∈ A and Ua ∈ B if a ∈ B. 70 1. BASIC TOPOLOGY

ω To prove that R is connected in the product topology we shall use the fact that R is connected (see Theorem 1.5.3). Deﬁne { }

e n i ∈ ω i ∼ n R := x = (x )i≥1 R x = 0 for i > n = R , ∪ ∞ R := Re n n≥1 From Proposition 1.5.1 (6), Rn is connected. Since Re n have the common point 0, it ∞ follows from Proposition 1.5.1 (10) that R is also connected. We now claim that ω R∞ = R ω which, by Proposition 1.5.1 (11), implies the connectedness of R in the product topol- i ∈ ω ogy. For a = (a )i≥1 R , let U := ∏i≥1 Ui be a basis element for the product topology ∈ that contains a. Then there exists an integer N N such that Ui = R for i > N. Hence ∞ ∞ x = (x1, ··· , xn, 0, ··· , 0) ∈ R ∩ U. Thus U ∩ R ̸= ∅ and a ∈ R∞. A subset I ⊆ R is said to be an interval if it contains more than one element, and whenever a, b ∈ I, every c such that a < c < b is also in I. An interval must be one of the nine types of sets [a, b], (a, b), [a, b), (a, b], (−∞, b], (−∞, b), [a, +∞), (a, +∞), (−∞, +∞).

THEOREM 1.5.3. R is connected. Moreover a nonempty subset of R is connected if and only if it is a singleton or an interval.

PROOF. We assume that I ⊆ R contains at least two points. Assume first that I is not an interval. By definition, there exists a, b, c ∈ R so that a < c < b, a, b ∈ I but c ∈/ I. Then he sets (−∞, c) ∩ I and (c, +∞) ∩ I form a separation for I. Thus I is not connected. Now we assume that I is an interval. It is not connected, there are open subsets U, V ⊆ R such that (U ∩ I, V ∩ I) forms a separation for I. Choose a ∈ U ∩ I and b ∈ V ∩ I. We may without loss of generality that a < b. Since I is a interval, it follows that [a, b] ⊆ I. Since U and V are open, there exists ϵ > 0 such that [a, a + ϵ) ⊆ U ∩ I and (b − ϵ, b] ⊆ V ∩ I. Let c; = sup(U ∩ [a, b]). Then a + ϵ ≤ ′ ′ c‘b − ϵ and c ∈ (a, b) ⊆ I ⊆ U ∪ V. If c ∈ U, then (c − ϵ , c + ϵ ) ⊆ U ∩ (a, b) ′ ′′ ′′ ′′ for some ϵ > 0; if c ∈ V, then (c − ϵ , c + ϵ ) ⊆ V for some ϵ > 0. Both cases contradict with the definition of c. Therefore I is connected. If a topological space (X, T ) has only finitely many components, then, according to Proposition 1.5.1 (3), each component is also open in X (since its complement is a finite union of closed subsets). But in general the components of X need not to be open or closed in X. For example, consider Q ⊆ R. Then each component of Q consists of a single point so that this component is neither open nor closed. 1.5.2. Path-connectedness. Let (X, T ) be a topological space and x, y ∈ X.A path in X from x to y is a continuous map f : [0, 1] → X such that f (0) = x and f (1) = y. (1) X is path-connected if for any two points x, y in X there is a path in X from x to y. 1.5. CONNECTEDNESS AND COMPACTNESS 71

(2) Deﬁne a relation ∼ on X by x ∼ y ⇐⇒ there exists a path in X from x to y. Then ∼ is an equivalence relation on X, and the equivalence class [x] is called a path-component of X. (3) X is called locally connected if for each x ∈ X and each neighborhood N of x, there is a connected neighborhood V of x with V ⊆ N. Equivalently, X is locally connected if it admits a basis of connected open subsets. (4) X is called locally path-connected if it admits a basis of path-connected open subsets. Equivalently, X is locally path-connected if for each x ∈ X and each neighborhood N of x, there is a path-connected neighborhood V of x with V ⊆ N.

REMARK 1.5.4. (1) The following sets Rn, Rn \{0} (n ≥ 2), Bn (n ≥ 1), Sn (n ≥ 1), Tn, convex sets are all path-connected. (2) The topologist’s sin curve { } { }

1 X = (x, y) ∈ R2 x = 0, −1 ≤ y ≤ 1 ∪ (x, y) ∈ R2 y = sin , 0 < x ≤ 1 x is connected, but not path-connected.

PROOF. Indeed, Write S := {(x, sin(1/x))|x ∈ (0, 1]}. Then S is connected since S = (1 × f )((0, 1]) for f (x) := sin(1/x) (by Proposition 1.5.1 (1)), S is also connected (by Proposition 1.5.1 (11)), and S = X. We now show that S is not path-connected. Suppose there exists a path f : [a, c] → S beginning at (0, 0) and ending at a point of S. Set I := {t ∈ [a, c]| f (t) ∈ {0} × [−1, 1]} . Then I is closed in [a, c] and b := sup I exists. Moreover f : [b, c] → S is a path that maps b into {0} × [−1, 1] and maps (b, c] to S. Write f (t) = (x(t), y(t)), t ∈ [b, c]. Then x(b) = 0, x(t) > 0 for all t ∈ (b, c], and y(t) = sin(1/x(t)) for t ∈ (b, c]. ∈ ∈ 1 1 − n Given n N, there exists u (0, x(b + n )) such that sin u = ( 1) . Then there ∈ 1 − n exists tn (b, b + n ) such that x(tn) = u. Hence y(tn) = sin(1/u) = ( 1) . But tn → b, on the other hand we have y(tn) → y(b) ∈ [−1, 1]. This contradiction indicates X is not path-connected. (3) If U ⊆ R2 is an open and connected subspace, then U is path-connected. ∈ PROOF. Given x0 U, define e ∈ |∃ ⊆ U := {x U path in U joining x0 to x} U. ∈ e e e Since x0 U, we claim that U is both open and closed in U so that U = U. For ∈ e ∈ → x U, there exists a path f [0, 1] U such that f (0) = x0 and f (1) = x. Since U is open, it follows that there exists an open ball B2(x, ϵ) ⊆ U and then 2 ϵ ⊆ e e { } e B (x, ) U. To prove the closedness of U, we take a sequence xn n≥1 in U such 2 that xn → x ∈ U. For any ϵ > 0, there exists N ∈ N such that xn ∈ B (x, ϵ) for ≥ ∈ e all n N. Since xn U, we can find a path connecting x0 to xn. Therefore we can ∈ e find a path joining x0 to x. Thus x U. 72 1. BASIC TOPOLOGY

(4) We have the following relations: +LPC PC ←−−−− C ←−−−− PC x  +C

LC ←−−−− LPC where C = connected, PC = path-connected, LC = locally connected, and LPC = locally path-connected.

PROPOSITION 1.5.5. Let (X, T ) be a topological space. (1) If X is path-connected, then X is connected. (2) Each path-component is contained in a single component, and each components is a disjoint union of path components. (3) If A ⊂ X is path-connected, then A is contained in a single path-component. (4) If X is locally connected, then any open subset of X is locally connected and each component of X is open in X. (5) If X is locally path-connected, then X is locally connected, any open subset of X is locally path-connected, each path-component of X is open, and its path- components are the same as its components. (6) If X is locally path-connected, then X is connected if and only if it is path- connected. (7) If f : X → Y is continuous and X is path-connected, then f (X) is path- connected. (8) Any ﬁnite product of path-connected spaces is path-connected. (9) If ∼ is an equivalence relation on a path-connected space X, then X/ ∼ is path- connected.

PROOF. (1) Suppose that X is path-connected but not connected. Choose a separation (U, V) of X, choose p ∈ U and q ∈ V, and choose a path γ from p to q − − in X. Then f 1(U) and f 1(V) are disjoint open subsets of [0, 1] that cover [0, 1]; − − − − moreover 0 ∈ f 1(U) and 1 ∈ f 1(V), so that ( f 1(U), f 1(V)) is a separation of [0, 1], which is a contradiction. (2) and (3) can be proved as Proposition 1.4.1 (3). (4) The ﬁrst statement is obvious. Let A be a component of X. If x ∈ A, then x has a connected neighborhood U by local connectedness. Then U ⊂ A. Thus A is open. (5) The ﬁrst two statements are obvious. Let B be a path-component and A the single component containing A. By (2), A can be written as a disjoint union of path-components, each of which is open in X and thus in A. If B ̸= A, then (B, A \ B) is a separation of A, which is a contradiction since A is connected. (6) If X is connected, then X has only one component and hence one path- component; thus X is path-connected. Conversely, if X is path-connected, by (5), X has only one component so that X is connected.

EXERCISE 1.5.6. Prove Proposition 1.5.5 (7)-(9). Consider the topologist’s sin curve X = S in Remark 1.5.4. It has one component, two path components S and V := {0} × [−1, 1], S is open in S but not closed, 1.5. CONNECTEDNESS AND COMPACTNESS 73 and V is closed but not open in S. Let b S := S \ ({0} × (Q ∩ [−1, 1])). b Then S has still one component but uncountably many path components.

EXAMPLE 1.5.7. (1) Each interval in R is both connected and locally connected. (2)[−1, 0) ∪ (0, 1] ⊆ R is not connected, but is locally connected. (3) The topologist’s sine curve X is connected but not locally connected. Otherwise, X must be path-connected! (4) Q is neither connected nor locally connected. There are no relations between connectedness and local connecteness.

THEOREM 1.5.8. Any topological manifold is locally path-connected. Consequently, by Proposition 1.5.5, for any topological manifold M, M is connected if and only if it is path-connected.

PROOF. By definition, for any m-dimensional topological manifold M we have M ∪ U φ U → ⊂ m M = i∈N i where i : i Bi R is homeomorphic for each i. Hence is locally path-connected. 1.5.3. Compactness. Let (X, T ) be a topological space. (1) X is compact if any open cover of X has a finite subcover. (2) A collection U of sets has the finite intersection property if the intersection of any finite subcollection is nonempty. A topological space (X, T ) is compact if and only if for every collection of closed subsets of X which has the finite intersection property, the intersection of the entire collection is nonempty.

PROOF. Given a collection U of subsets of X, let C := {X \ U|U ∈ U } be the collection of their complements. (a) U is a collection of open subsets if and only if C is a collection of closed subsets. { } ⊆ U ∩ ∅ (b) Ui 1≤i≤n covers X if and only if 1≤i≤nCi = where Ci := \ X Ui. U ∩ ∅ (c) covers X if and only if C∈C C = . Then { X is compact ⇐⇒ given any collection U of open subsets of X } that covers X, ∃ ﬁnite subcollection of A covers X ⇐⇒ { given any collection U of open subsets of X, if no ﬁnite subcollection of } U covers X, then U does not cover X ⇐⇒ the desired statement

by the above statements (a) − (c). (3) X is locally compact if for any x ∈ X, there is a compact subset of X containing a neighborhood of x. (4) A subset S ⊆ X is precompact or relatively compact if S ⊆ X is compact. 74 1. BASIC TOPOLOGY

(5) A continuous map f : X → Y between topological spaces is proper if − f 1(K) ⊆ X is compact whenever K ⊆ X is compact. Recall that a subset S ⊆ Rn is compact if it is closed and bounded. U { } EXAMPLE 1.5.9. (1) R is not compact, since the open cover = (n, n + 2) n∈Z of R contains no finite subcollection that cover R. { } ∪ { } U (2) X = 0 1/n n≥1 is compact. Given an open cover of X, there exists an open subset U ∈ U containing 0. Take N ∈ N such that 1/n ∈ U for all n > N. For ∈ { ··· } ∈ U ··· each i 1, , N , we choose Ui containing 1/i. Then U, U1, , UN form a finite subcollection of U that contains X. (3) If X is finite, then X is compact. U { } (4)(0, 1] is not compact, because the open cover = (1/n, 1] n≥1 contains no finite subcollection covering (0, 1]. Let Y be a subset of a topological space (X, T ). Then Y is compact (in the subspace topology) if and only if every cover of Y by open subsets of X contains a U { } finite subcollection covering Y. Let = Uα α∈I be a cover of Y by subsets open U { ∩ |α ∈ } in X. Then Y := Uα Y I is a cover of Y by subsets open in Y. Since { α ∩ } Y is compact, there exists a finite subcollection A i Y 1≤i≤n covers Y. Thus { } A U ′ { ′ |α ∈ ′} Uα 1≤i≤n is a subcollection of that covers Y. Conversely, let = Uα I i ′ ′ be a cover of Y by subsets open in Y. For each α ∈ I , Uα = Uα ∩ Y for some open U { } subset Uα in X. Then = Uα α∈I′ is a cover of Y by subsets open in X. By { } { ′ } hypothesis, there is a finite subcollection Uα 1≤i≤n covers Y. Then Uα 1≤i≤n is ′ i i a subcollection of U that covers Y.

PROPOSITION 1.5.10. (1) If f : X → Y is a continuous map between topological spaces and X is compact, then f (X) is compact. (2) (Extreme value theorem) If f : X → R is a continuous map and X is compact, then f is bounded and f attains its minimum and maximum in X. (3) If X is compact and U ⊂ X is closed, then U is compact. (4) If X is Hausdorff and U ⊂ X is compact, then U is closed. (5) If X is Hausdorff, K, L ⊂ X are compact, and K ∩ L = ∅, then K ⊂ U and L ⊂ V for some disjoint open subsets U, V ⊂ X. (6) The ﬁnite product of compact spaces is compact. (7) If X is compact and ∼ is any equivalence relation, then X/ ∼ is compact. (8) (Closed map lemma) Let f : X → Y be a continuous map from a compact space X to a Hausdorff space Y. Then f is closed. If f is moreover surjective (injective, bijective), then f is s quotient map (topological embedding, homeomorphism). (9) If X is Hausdorff, then the following are equivalent: (a) X is locally compact, (b) any x ∈ X has a precompact neighborhood, (c) X has a basis of precompact open sets. (10) Any open or closed subset of a locally compact Hausdorff space are locally compact Hausdorff. (11) If f : X → Y is a proper map between locally compact Hausdorff spaces, then f is a closed map.

− PROOF. (1) Let U be an open cover of f (X). For each U ∈ U , f 1(U) is an − − open subset of X. Since U coves f (X), it follows that f 1(U ) := { f 1(U) : U ∈ 1.5. CONNECTEDNESS AND COMPACTNESS 75

U } is an open cover of X. By compactness of X, there exists N ∈ N such that −1 ( f (Ui))1≤i≤N cover X. Then (Ui)1≤i≤N covers f (X). (2) By (1), f (X) is a compact subset of R so that f (X) is closed and bounded. In particular, it contains its maximum and maximum in X. (3) Let U be a cover of U by open subsets of X. Then U ∪ {X \ U} is an open ··· \ } cover of X, which has a finite subcover (U1, , UN, X A . Therefore A can be covered by (Ui)1≤i≤N and A is compact. (5) If the statement holds for L being a one-point set, then the same result holds for general L. For each y ∈ L there exist disjoint open subsets Uy, Vy ⊂ X such that ⊂ ∈ K Uy and y Vy. By compactness of L, finitely many of these, say (Vy )1≤i≤N, ∩ ∪ i cover L. Then setting U := 1≤i≤NUyi and V := 1≤i≤NVyi proves the result. We now prove the result for the case L = {y}. For each x ∈ K, there exists disjoint open sets Ux containing x and Vx containing y by the Hausdorff property. The collection {Ux : x ∈ K} is an open cover of K, so it has a finite subcover, say { ··· } ∪ ∩ Ux1 , , UxN . Let U := 1≤i≤NUxi and V := 1≤i≤NVxi . Then U and V are disjoint open sets with K ⊂ U and {y} ⊂ V. (4) For any y ∈ X \ U, by (5), we have U ⊂ U and y ∈ V for some disjoint open subsets U and V. Thus X \ U is open. (6) Without loss of generality, we may consider the product X × Y of compact spaces. Let U be an open cover of X × Y. Choose x ∈ X; the slice {x} × Y is U homeomorphic to Y, so finitely many of the sets of cover it, say (Ui)1≤i≤N. Since product open sets are a basis for the product topology, for each y ∈ Y there × ⊂ × ∈ × ⊂ ∪ is a product open set V W X Y such that (x, y) V W 1≤i≤NUi. { } × × Finitely many of these product sets cover x Y, say (Vi Wi)1≤i≤M. If we set ∩ × ∪ Zx := 1≤i≤MVi, then the whole strip Zx Y is contained in 1≤i≤NUi. Because (Zx)x∈X is an open cover of X, by compactness of X, we have a finite subcover, U × say (Zxi )1≤i≤K, of X. Since finitely many sets of cover each strip Zxi Y, and finitely many such strips cover X × Y. (7) This follows from the fact that a quotient of a compact space is the image of a compact space by a continuous map. (8) HW. (9) (c) ⇒ (b) ⇒ (a) are obvious. We now prove (a) ⇒ (c). It suffices to show that every x ∈ X has a neighborhood of precompact open sets. Being locally compact, X has a compact subset K containing a neighborhood U of x. Set

Ux := {neighborhoods of x contained in U} so that Ux is a neighborhood of x. Since X is Hausdorff and K is compact, by (4), K must be closed. If V ∈ Ux, then V ⊂ U ⊂ K implies V ⊂ K = K. Thus V is compact by (3), and V is precompact. (10) We need only to verify the local compactness. If U is open in X, then, Theorem 1.5.12 yields that for any x ∈ Y there exists a precompact neighborhood Vx of x such that Vx ⊂ X. Thus U is locally compact. Suppose that K is closed in X. By (9), any x ∈ K has a precompact neighborhood Ux in X. Since Ux ∩ Z = Ux ∩ Z is closed in Ux that is compact in X, it follows that Ux ∩ Z is compact in X by (3). Hence Ux ∩ Z is a precompact neighborhood of x in Z, and Z is locally compact. (11) For any closed set K ⊂ X, we shall show that f (K) is closed in Y. Let y ∈ ∂( f (K)) ⊂ Y. There exists a precompact neighborhood U of y so that y ∈ ∂( f (K) ∩ 76 1. BASIC TOPOLOGY

− − U). The properness of f implies that f 1(U) is compact in X and then K ∩ f 1(U) − − is closed in f 1(U). Hence K ∩ f 1(U) is compact in X. The continuity of f implies − − that f (K) ∩ U = f (K ∩ f 1(U)) is compact in Y, and then f (K ∩ f 1(U)) is closed in Y. Therefore y ∈ f (K) ∩ U ⊂ f (K). Thus f (K) is closed in Y. Let (X, d) be a metric space and S ⊆ X. Deﬁne (1.5.3.1) diam(S) := sup d(x, y) x,y∈S the diameter of S. If U is an open cover of (X, d), a positive number δ > 0 is called a Lebesgue number for U if any set whose diameter less than δ is contained in one of U ∈ U .

THEOREM 1.5.11. (Lebesgue number lemma) Any open cover of a compact metric space has a Lebesgue number.

PROOF. Let U be an open cover of a compact metric space (X, d). For any x ∈ X, there is an open subset U ∈ U with x ∈ U. The openness of U implies that ⊆ { } B2r(x)(x) U for some r(x) > 0. Since Br(x)(x) x∈X forms an open cover of X, it B (x ) ··· B (x ) X n ∈ N follows that r(x1) 1 , , r(xn) n cover for some . Deﬁne δ := min r(xi) > 0. 1≤i≤n If S ⊆ X with diam(S) < δ, we claim that S ⊆ B (x ) for some i ∈ {1, ··· , n}. 2r(xi) i Let y ∈ S and choose y ∈ B (x ) for some i. Therefore for any z ∈ S, r(xi) i ≤ δ δ d(z, xi) d(z, y) + d(y, xi) < + d(y, xi) < + r(xi) < 2r(xi). Thus δ is a Lebesgue number for U .

THEOREM 1.5.12. (Shrinking lemma) Let X be a locally compact Hausdorff space and x ∈ X. If U is a neighborhood of x, then there exists a precompact neighborhood V of x such that V ⊂ U.

PROOF. By Proposition 1.5.19 (9), there is a precompact neighborhood W of x so that W \ U is closed in W. Then W \ U is compact by Proposition 1.5.19 (3). ′ Part (5) in Proposition 1.5.19 implies that there exist open sets Y, Y ⊆ X such that ′ x ∈ Y and W \ U ⊆ Y . Let V := Y ∩ W be a neighborhood of x. Since V ⊆ W, it ′ follows that V is compact by (3). Because V ⊆ Y, we get V ⊆ W \ Y ⊆ U.

EXAMPLE 1.5.13. (1) Rn is locally compact Hausdorff so that any open or closed subset of Rn is locally compact Hausdorff. (2) Any topological manifold is locally compact Hausdorff. − (3) Bn and Sn 1 are compact. (4) The set A = {(x, 1/x)|0 < x ≤ 1} is closed in R2, but is not compact. (5) The set A = {(x, sin(1/x))|0 < x ≤ 1} is bounded in R2, but is not compact. A point x of a topological space (X, T ) is said to be an isolated point of X if {x} is open in X.

THEOREM 1.5.14. If (X, T ) is a nonempty compact Hausdorff space without any isolated points, then X is uncountable. 1.5. CONNECTEDNESS AND COMPACTNESS 77

PROOF. (1) Given any nonempty open subset U ⊆ X and any x ∈ X, there exists a nonempty open subset V ⊆ U such that x ∈/ V. Choose y ∈ U different from x (when x ∈/ U, we choose any point in U because U ̸= ∅; when x ∈ U, we can also choose such a y because x is not an isolated point of X). Since X is Hausdorff, there exist disjoint open subsets W1 and W2 about x and y respectively. ∩ Then we can set V := W2 U. (2) Given a map f : Z+ → X. We show that f is not surjective (and hence X is uncountable). Let xn := f (n). From (1), we can ﬁnd a nonempty open subset ⊆ ∈ ̸ ∅ V1 X such that x1 / V1. In general, given Vn−1 = , there exists a nonempty ⊆ ∈ { } open subset Vn Vn−1 such that xn / Vn. Now we have a family Vn n≥1 of ⊇ nonempty closed subsets of X with the property that Vn Vn+1. Because X is ∈ ∩ ≥ ̸ compact, we can ﬁnd x n≥1Vn. Consequently, for any n 1, x = xn, so that f is not surjective.

COROLLARY 1.5.15. Any nonempty closed interval in R is uncountable.

EXERCISE 1.5.16. Verify Corollary 1.5.15.

1.5.4. Paracompact spaces. Let (X, T ) be a topological space. (1) A collection U ⊆ 2X is locally finite if any point x ∈ X has a neighborhood that intersects at most finitely many of the sets in U . (2) Given a cover U of X, a cover V is called a refinement of U (written as V ≺ U ) if any V ∈ V is contained in some U ∈ U . It is an open refinement if each V ∈ V is open in X. (3) X is paracompact if any open cover of X has a locally finite open refinement. U { } EXAMPLE 1.5.17. (1) = (n, n + 2) n∈Z is locally finite in R. U { } { } (2) = (0, 1/n) n≥1 or (1/(n + 1), 1/n) n≥1 is locally finite in (0, 1) but not in R.

PROPOSITION 1.5.18. (1) Let X be a topological space and U ⊂ 2X. (a) U is locally finite if and only if U := {U : U ∈ U } is locally finite. (b) If U is locally finite, then ∪ ∪ U = U. U∈U U∈U (2) Any second countable, locally compact Hausdorff space X is paracompact. In fact, each open cover of X has a countable, locally finite refinement consisting of open subsets with compact closure (i.e., there exists a countable, locally finite, precompact refinement). (3) Any paracompact Hausdorff space is normal. (4) Any closed subspace of a paracompact space is paracompact.

PROOF. (1) (a) Suppose that U is locally ﬁnite. For any x ∈ X, there exists a ··· ∈ U ∈ neighborhood W of x such that W intersects U1, , Um for some m N. If ∩ ̸ ∅ ∩ ̸ ∅ ∈ { ··· } W U = , then W U = so that U U1, , Um . Thus W intersects U for ﬁnitely many U ∈ U . The converse is obvious. 78 1. BASIC TOPOLOGY

∪ ⊆ ∪ ∈ ∪ (b) Observe that U∈U U U∈U U. Let x / U∈U U. There exists some ∈ U ∈ ··· ∈ U \ ∪ U such that x U and intersects U1, , Uk by (1). Then U 1≤i≤kUk U ∈ ∪ contains x and intersects none of the sets in ; hence x / U∈U U. (2) Let X be a second countable locally compact Hausdorff space. We claim ⊆ that there exists a sequence (Gi)i∈N of open sets such that Gi is compact, Gi Gi+1, ∪ and X = i∈NGi. In fact, local compactness implies that X has a basis of precompact open sets by Proposition 1.5.19. The second countable assumption implies that there exists a countable basis (Ui)i∈N of precompact open sets by Theorem 1.2.34. Let G1 := U1. = ∪ · · · ∪ Assume Gk : U1 Ujk . Let jk+1 be the smallest positive integer greater ⊆ ∪ ≤ ≤ = ∪ ≤ ≤ ( ) ∈ than jk such that Gk 1 i jk+1 Ui. Define Gk+1 : 1 i jk+1 Ui. Then Gi i N is ⊆ countable, Gi is compact, Gi Gi+1, and ∪ ∪ ∪ ⊆ ⊆ ⊆ X = Ui Gi Gi X. i∈N i∈N i∈N U \ ⊆ \ Let = (Uα)α∈I be an arbitrary open cover of X. Since Gi Gi−1 Gi+1 ≥ Gi−2 (i 3) is compact, it follows that we can choose a finite subcover of the open ∩ \ \ cover (Uα (Gi+1 Gi−2))α∈I of Gi Gi−1 and choose a finite subcover of the open ∩ \ cover (Uα G3)α∈I of G2. Indeed, let Wi := Gi Gi−1; the local compactness of ( ) ≤ ≤ ( α ∩ X implies that there exists a finite subcover Uij 1 j mi of the open cover U \ ∩ \ ≤ ≤ (Gi+1 Gi−2)α∈I of Wi. Set Vij := Uij (Gi+1 Gi−2) for 1 j mi. Similarly, we can define Vij for i = 1, 2. Hence (Vij)1≤j≤m is an open subcover of Wi and ∪ ∪ i∪ Vij = Wi = X. ∈ ≤ ≤ ∈ i N 1 j mi i N V = ( ) ∈ ≤ ≤ V ≺ U ∩ ∪ ≤ ≤ = ∅ ̸= − − If : Vij i N,1 j mi , then and Wi 1 j mi Vkj for k i 2, i 1, i, i + 1, i + 1. Thus V is locally finite. (3) We first prove that X is regular. Let a ∈ X and B be closed subset of X disjoint from a. The Husdorff property implies that for any b ∈ B there is an open ∋ ∩ { } ∅ { } \ subset Ub b such that Ub a = . Cover X by Ub b∈B and X B, and take a locally finite open refinement C that covers X. Set D := {C ∈ C |C ∩ B ̸= ∅}. Then D covers B. For any D ∈ D, D ∩ {a} = ∅. Let ∪ V := D D∈D that is open in X and contains B. Since D is locally finite, it follows that V = ∪ ∩ { } ∅ \ D∈D D. Hence V a = so that X is regular (because V and X V are disjoint open subsets of B and a respectively. Let A and B be two disjoint closed subsets of X. For any a ∈ A, there exist open subsets Ua of a and Va containing B, respectively, such that Ua ∩ Va = ∅. As above, we can find two disjoint open subsets V and X \ V of B and A respectively. (4) Let Y be a closed subspace of the paracompact space X and let U be an ′ open cover of Y. For any U ∈ U , there exists an open subset U of X such that ′ ′ U ∩ Y = U. Cover X by the open subsets U , together with X \ Y. We then can take a locally finite open refinement V of U that still covers X. The collection C := {V ∩ Y|V ∈ V } is a locally finite open refinement of U . 1.5. CONNECTEDNESS AND COMPACTNESS 79

REMARK 1.5.19. (1) A paracompact subspace of a Hausdorff space X need not be closed in X. For example, (0, 1) is paracompact, but it is not closed in R. (2) (A. H. Stone) Every metric space is paracompact. (3) Every regular Lindelöfspace is paracompact. (4) The product of two paracompact spaces need not be paracompact. For example, 2 Rℓ is regular and Lindelöf,it follows from (3) that Rℓ is paracompact, but Rℓ, that is Hausdorff and not normal, is not paracompact. ω (5) R is paracompact in te product topology. (6) RJ is not paracompact of J is unvountable. U { } Let = Uα α∈J be an index open cover of a topological space X. An indexed family of continuous functions ϕα : X → [0, 1] is a partition of unity on X dominated by U , if (1) supp(ϕα) ⊆ Uα for all α ∈ J, { ϕ } (2) supp( α) α∈J is locally finite, and ϕ ≡ ∈ (3) ∑α∈J α(x) 1 for all x X.

LEMMA 1.5.20. (Shrinking lemma) Suppose that(X, T ) is a paracompact Haus- U { } U dorff space. If = Uα α∈J is an indexed family of open cover of X, then admits a V { } locally ﬁnite open reﬁnement = Vα α∈J indexed by the same index set J, such that Vα ⊆ Uα for each α ∈ J.

PROOF. Let A := {A ∈ T |A ⊆ Uα for some α ∈ J}. According to Proposi- tion 1.5.18, X is normal so that for any x ∈ X we can find a open subset Ax such that Ax ⊆ Uα for some α ∈ J. Thus A covers X. The paracompactness implies A B { } that has a locally finite open refinement = Bβ β∈K. We then find a map → β ∈ ⊆ ⊆ α ∈ f : K J such that for any K one has Bβ A Uf (β). Define, for each J, { ∪ B { ∈ B| β α} ̸ ∅ Bβ∈Bα Bβ, α := Bβ f ( ) = = , Vα := ∅, Bα = ∅. ∈ B ⊆ B For Bβ α, we have Bβ Uf (β) = Uα. Since α is locally finite, by Proposition 1.5.18, ∪ Vα = Bβ ⊆ Uα. Bβ∈Bα V { } ∈ Finally we check the local finiteness of = Vα α∈J. Given x X. There is a neighborhood W of x that intersects Bβ for only finitely many values of β = β ··· β α ∈ { β ··· β } 1, , N. Then W can intersects Vα only if f ( 1), , f ( N) .

T U { } THEOREM 1.5.21. Let (X, ) be a paracompact Hausdorff space. If = Uα α∈J is an indexed open cover of X, then there is a partition of unity on X dominated by U .

PROOF. By Lemma 1.5.20, we can find two locally finite indexed families of W { } V { } open sets = Wα α∈J and = Vα α∈J covering X such that Wα ⊆ Vα, Vα ⊆ Uα. The normality, coming from Proposition 1.5.18, implies that there is a continuous function ψα : X → [0, 1] such that ψα(Wα) = 1, ψα(X \ Vα) = 0, 80 1. BASIC TOPOLOGY by Theorem 1.2.30. Hence supp(ψα) ⊆ Vα ⊆ Uα. Because {W covers X, for any x ∈ X we can find some ψα such that ψα(x) > 0. Define Ψ(x) := ∑ ψα(x) ≥ 0 α∈J which is strictly positive for each x ∈ X. Since for each x ∈ X there is a neighborhood Wx of x that intersects supp(ψα) for only finitely many values of α, it follows that the above infinite sum at xis actually finite. Hence Ψ is well-defined and continuous on Wx and then on X. Finally set ϕα(x) := ψα(x)/Ψ(x). CHAPTER 2

Analysis on topological groups

2.1. Haar measures From now on let G be a local compact group, which in this note means a locally compact and Hausdorff topological group.

A collection S of subsets of G is called a σ-algebra if (1) ∅, X ∈ S , (2) A ∈ S implies X \ A ∈ S , and { } ⊆ S ∪ ∈ S (3) An n≥1 implies n≥1 An . Clearly that the topology T for X is a σ-algebra. Deﬁne ∩ BG := {S |S is a σ-algebra of X and contains T }.

EXERCISE 2.1.1. Show that BG is also a σ-algebra and is the smallest σ-algebra containing the open subsets of G.

Elements in BG are called Borel subsets in G.

For any locally compact group G, Haar’s theorem gives, up to a positive multiplicative constant, a unique countably additive, nontrivial measure µ on the Borel sets of G satisfying the following properties: (1) the measure µ is left-invariant: µ(gS) = µ(S) for every g ∈ G and all Borel sets S ⊆ G; (2) the measure µ is ﬁnite on every compact set: µ(K) < ∞ for all compact K ⊆ G; (3) the measure µ is outer regular on Borel sets S ⊆ G: µ(S) = inf{µ(U) : S ⊆ U, U open}; (4) the measure µ is inner regular on open sets U ⊆ G: µ(U) = sup{µ(K) : K ⊆ U, K compact}. Such a measure on G is called a (left) Haar measure. If µ is a Haar measure and ∈ G B ∋ S 7→ µ S g1 , then G ( g1) is left-invariant. Thus, by uniqueness of the Haar measure, there exists a function ∆ from the group G to R+, called the modular function, such that µ S ∆ µ S ( g1) = (g1) ( ) for every Borel set S ⊆ G.

We say G is unimodular if the modular function ∆ is identically 1, or equivalently, if the Haar measure is both left and right-invariant. Examples of unimodular groups are Abelian groups, compact groups, discrete groups, semisimple Lie groups and connected nilpotent Lie groups.

81 82 2. ANALYSIS ON TOPOLOGICAL GROUPS

An example of a non-unimodular group is the group of afﬁne transformations {[ ] }

∗ a b ∗ {x 7→ ax + b : a ∈ R , b ∈ R} = a ∈ R , b ∈ R 0 1 on the real line. This example shows that a solvable Lie group need not be unimodular. In this group a left Hasr measure is given by da ∧ db/a2, and a right Haar measure by da ∧ db/|a|.

Let G be a locally compact group with Haar measure µ. For 0 < p < +∞, we b deﬁne Lp(G, µ) the set of all complex-valued µ-measurable functions on G whose b∞ modulus to the p-th power is integrable. L (G, µ) is the set of all complex-valued µ-measurable functions f on G such that for some M > 0 the set {x ∈ G|| f (x)| > M} has µ-measure zero. We say two functions in Lp(G, µ) are considered equal (or equivalent) if they are equal µ-almost everywhere. The equivalence class of b ∞ Lp(G, µ) is denoted Lp(G, µ). Similarly we can deﬁne L (G, µ).

We simply write Lp(G) for Lp(G, µ), when p > 0 or p = ∞. For 0 < p < +∞ deﬁne ( ) ∫ 1/p p || f || p G := | f (x)| dµ(x) L ( ) G and for p = ∞ deﬁne || || { |µ { ∈ G|| | } } f L∞(G) := inf M > 0 ( x f (x) > M ) = 0 . The left invariance of µ is equivalent to the fact that for all t ∈ G and all nonnegative measurable functions f on G we have ∫ ∫ f (tx)dµ(x) = f (x)dµ(x). G G ∗ 2.1.1. Examples of Haar measures. Let G := R = R \{0} with group law the usual multiplication. Then G is locally compact and the measure dx µ := |x| is invariant under multiplicative translation, because of the following identity ∫ ∫ +∞ dx +∞ dx f (tx) = f (x) −∞ |x| −∞ |x| ∗ ∗ for all f ∈ L1(R , µ) and all t ∈ R . Then ∫ ∫ +∞ µ dx χ dx (A) = = A(x) . A x −∞ x

+ EXAMPLE 2.1.2. (1) A Haar measure on the multiplicative group R is dx/x. That is ∫ +∞ µ χ dx (A) = A(x) . −∞ x (2) The Heisenberg group Hn is the set Cn × R with the group operation ( ) (z, t)(w, s) := z + w, t + s + 2Im ∑ ziwi . 1≤i≤n 2.1. HAAR MEASURES 83

− The identity in Hn is (0, 0) ∈ Cn × R and (z, t) 1 = (−z, −t). Hn is topologically identiﬁed with Cn × R and hence is a locally compact group. Both the left and right Haar measure on Hn is Lebesgue measure dz ∧ dt. (3) Let G := R2 \{(0, y)|y ∈ R} with group operation [ ][ ] [ ] x y z w xz xw + y (x, y)(z, w) := (xz, xw + y) ⇐⇒ = 0 1 0 1 0 1 ⊆ B For A G deﬁne ∫ ∫ +∞ +∞ µ χ dxdy (A) := A(x, y) . −∞ −∞ x2 ∈ G ∈ When t = (t1, t2) and (x, y) A we have

t(x, y) = (t1x, t1y + t2) and then ∫ ∞ ∫ ∞ ∫ ∞ ∫ ∞ ′ ′ + + dxdy + + ′ ′ d(x /t)d((y − t )/t ) µ( ) = χ ( ) = χ ( ) 2 1 tA tA x, y 2 A x , y ′ 2 −∞ −∞ x −∞ −∞ (x /t1) ∫ ∫ +∞ +∞ ′ ′ χ ′ ′ dx dy µ = A(x , y ) ′ = (A). −∞ −∞ x 2 Thus µ is a left Haar measure on G. Similarly we can prove that ∫ ∫ +∞ +∞ ν χ dxdy (A) := A(x, y) −∞ −∞ |x| is a right Haar measure on G.

2.1.2. Convolution. Let f , g ∈ L1(G). Deﬁne the convolution f ∗ g by ∫ − (2.1.2.1) ( f ∗ g)(x) := f (y)g(y 1x)dµ(y). G Since ∫ ∫ ∫ ∫ − − f (y)g(y 1x)dµ(y)dµ(x) ≤ | f (y)||g(y 1x)|dµ(x)dµ(y) G G G G ∫ ∫ ∫ ∫ − = | f (y)| |g(y 1x)|dµ(x)dµ(y) = | f (y)| |g(x)|dµ(x)dµ(y) G G G G || || || || ∞ = f L1(G) g L1(G) < + , it follows that f ∗ g is well-deﬁned and || ∗ || ≤ || || || || (2.1.2.2) f g L1(G) f L1(G) g L1(G). − Letting z = x 1y in (2.1.2.1) yields ∫ − (2.1.2.3) ( f ∗ g)(x) = f (xz)g(z 1)dµ(z). G

EXERCISE 2.1.3. (1) For all f , g, h ∈ L1(G), we have f ∗ (g ∗ h) = ( f ∗ g) ∗ h and f ∗ (g + h) = f ∗ g + f ∗ h, ( f + g) ∗ h = f ∗ h + g ∗ h. (2) Verify (2.1.2.4). 84 2. ANALYSIS ON TOPOLOGICAL GROUPS

For f , g ∈ L1(G) and t ∈ G, deﬁne t f (x) := f (tx), f t(x) := f (xt). Then we have (2.1.2.4) t f ∗ g = t( f ∗ g), f ∗ gt = ( f ∗ g)t.

THEOREM 2.1.4. (Minkowski’s inequality) Let 1 ≤ p ≤ +∞. For f ∈ Lp(G) and g ∈ L1(G) we have g ∗ f exists µ-a.e. and satisﬁes || ∗ || ≤ || || || || (2.1.2.5) g f Lp(G) g L1(G) f Lp(G).

PROOF. The case p = 1 was proved in (2.1.2.2), and the case p = ∞ is obvious. ∞ ′ p Now we assume 1 < p < + . Write p = p−1 , and compute ∫ ∫ 1 − − 1 ′ (|g| ∗ | f |)(x) = | f (y 1x)||g(y)|dµ(x) = | f (y 1x)||g(y)| p |g(y)| p dµ(x) G G ( ) ( ) ′ ∫ 1/p ∫ 1/p − ≤ | f (y 1x)|p|g(y)|dµ(y) |g(y)|dµ(y) G G so that ( ) ∫ ∫ 1/p p−1 −1 p |||g| ∗ | f ||| p G ≤ ||g|| | f (y x)| |g(y)|dµ(y)dµ(x) L ( ) L1(G) G G ( ) ∫ ∫ 1/p p−1 p = ||g|| | f (x)| dµ(x)|g(y)|dµ(y) = || f || p G ||g|| 1 G . L1(G) G G L ( ) L ( ) Now the inequality (2.1.2.5) follows from |g ∗ f | ≤ |g| ∗ | f |. If f is a function on G, we deﬁne e − (2.1.2.6) f (x) := f (x 1), x ∈ G. Then (2.1.2.3) can be written as ∫ ( f ∗ g)(x) = f (xz)ge(z)dµ(z). G

THEOREM 2.1.5. (Young’s inequality) Let 1 ≤ p, q, r ≤ +∞ satisfy 1 1 1 + 1 = + . q p r ∈ p G ∈ r G || || ||e|| ∗ Then for all f L ( ) and all g L ( ) satisfying g Lr(G) = g Lr(G) we have f g exists µ-a.e. and satisﬁes || ∗ || ≤ || || || || (2.1.2.7) f g Lq(G) g Lr(G) f Lp(G).

PROOF. If r = +∞, then p = 1 and q = +∞ in which case the inequality ∞ 1 1 1 (2.1.2.7) is trivial. Now we assume that r < + . From q + 1 = p + r we have 1 1 1 1 1 1 1 + + = 1, + = 1 = + . r′ p′ q r r′ p p′ By Holder’s¨ inequality we get ∫ − |(| f | ∗ |g|)(x)| ≤ | f (y)||g(y 1x)|dµ(x) G 2.1. HAAR MEASURES 85 ∫ ( ) ′ − − ′ = | f (y)|p/r | f (y)|p/q|g(y 1x)|r/q |g(y 1x)|r/p dµ(y) G (∫ ) (∫ ) ′ ′ 1/q 1/p p r − − ≤ || f || / | f (y)|p|g(y 1x)|rdµ(y) |g(y 1x)|rdµ(y) Lp(G) G G (∫ ) (∫ ) ′ ′ 1/q 1/p p r − − = || f || / | f (y)|p|g(y 1x)|rdµ(y) |ge(x 1y)|rdµ(y) Lp(G) G G (∫ ) 1/q ′ ′ − p r r p = | f (y)|p|g(y 1x)|rdµ(y) |||| / ||ge|| / G Lp(G) Lr(G) so that (∫ ∫ ) ′ ′ 1/q p/r r/p p −1 r ||| f | ∗ |g||| q G ≤ || f || ||ge|| | f (y)| |g(y x)| dµ(x)dµ(y) L ( ) Lp(G) Lr(G) G G ′ ′ p/r r/p p/q r/q = || || ||e|| || || || || = || || r || || p f Lp(G) g Lp(G) f Lp(G) g Lr(G) g L (G) f L (G) || || ||e|| using again the assumption g Lr(G) = g Lr(G).

EXERCISE 2.1.6. (Hardy) Use Theorem 2.1.4 to prove the following inequalities. (1) For 1 < p < +∞, (∫ ( ∫ ) ) +∞ x p 1/p 1 | | ≤ p || || f (t) dt dx f Lp((0,+∞)) 0 x 0 p − 1 and (∫ (∫ ) ) (∫ ) +∞ +∞ p 1/p +∞ 1/p | f (t)|dt dx ≤ p | f (t)|ptpdt . 0 x 0 (2) For 0 < b < +∞ and 1 ≤ p < +∞, ( ( ) ) ( ) ∫ ∞ ∫ p 1/p ∫ ∞ 1/p + x − − p + − − | f (t)|dt x b 1dx ≤ | f (t)|ptp b 1dt 0 0 b 0 and ( ( ) ) ( ) ∫ ∞ ∫ ∞ p 1/p ∫ ∞ 1/p + + − p + − | f (t)|dt xb 1dx ≤ | f (t)|ptp+b 1dt . 0 x b 0

2.1.3. Modular functions on G. Let G be a locally compact group with Haar measure µ := dg. Recall that the modular function ∆ ≡ ∆G on G is the function (actually continuous homomorphism into R+) such that for all f ∈ C (G), ∫ ∫ c (2.1.3.1) f (gg )dg = ∆(g ) f (g)dg. G 1 1 G

PROPOSITION 2.1.7. Let P be a locally compact group with two closed subgroups A, U such that A normalizes U, and such that the product (2.1.3.2) A × U −→ AU = P is a topological isomorphism. Then for f ∈ C (P) the functional ∫ ∫ c ∫ ∫ (2.1.3.3) f 7−→ f (au)dadu = f (au)duda U A A U is a Haar (left-invariant) functional on P. 86 2. ANALYSIS ON TOPOLOGICAL GROUPS

A ∈ U PROOF. Left invariant by is obvious. Let u1 . Then ∫ ∫ ∫ ∫ −1 f (u1au)duda = f (aa u1au)duda A U ∫A ∫U ◦ a a −1 ∈ U = ( f a)(u1u)duda where u1 := a u1a ∫A ∫U = f (au)duda A U by the Haar measure property.

PROPOSITION 2.1.8. Notation is as in Proposition 2.1.7. There exists a unique + continuous homomorphism δ : A → R such that for f ∈ Cc(U), ∫ ∫ − (2.1.3.4) f (a 1ua)du = δ(a) f (u)du, U U or in other words, for f ∈ Cc(P), ∫ ∫ (2.1.3.5) f (ua)du = δ(a) f (au)du. U U If U is unimodular, then δ is the modular function on P, that is,

(2.1.3.6) ∆P (p) = ∆P (au) = δ(a).

− PROOF. For the ﬁrst statement, we note that the map u 7→ ua = a 1ua is a topological group automorphism of U, which preserves Haar measure up to a constant factor. For the second statement, we ﬁrst note that the functional ∫ ∫ f 7−→ f (au)dadu U A U ∈ A is right-invariant under . Let a1 . Then ∫ ∫ ∫ ∫ ∫ ∫ −1 f (aua1)dadu = f (aa1a ua1)dadu = f (aa1u)dadu U A ∫U ∫A U A∫ ∫ − = f (aa a 1au)dadu = δ(a ) f (au)dadu, U A 1 1 U A which proves the formula (2.1.3.6).

PROPOSITION 2.1.9. Let G be a locally compact group with two closed subgroups P, K such that (2.1.3.7) P × K −→ PK = G is a topological isomorphism (not group isomorphism). Assume that G, K are unimodular. Let dg, dp, dk be given Haar measures on G, P, K respectively. Then there is a constant c such that for all f ∈ Cc(G), ∫ ∫ ∫ (2.1.3.8) f (g)dg = c f (pk)dpdk. G P K If in addition P = AU as in Proposition 2.1.7, with U unimodular, so we have the product decomposition (2.1.3.9) U × A × K −→ G, 2.2. IWASAWA’S DECOMPOSITION 87 then ∫ ∫ ∫ ∫ − (2.1.3.10) f (g)dg = f (uak)δ(a) 1dudadk. G U A K

PROOF. The ﬁrst assertion comes from a standard fact of homogeneous spaces, that is, there exists a left-invariant Haar measure on G/K = P. The second integral formula comes from Proposition 2.1.8.

2.2. Iwasawa’s decomposition G C G UAK The Iwasawa decomposition for := SL2( ) = SL(2, C) is = , where • U is the subgroup of upper triangular unipotent matrices; • A is the subgroup of diagonal matrices with positive diagonal components; • K is the unitary subgroup.

2.2.1. Iwasawa’s decomposition. Thus U is the group of 2 × 2 matrices of the form [ ] 1 x (2.2.1.1) u(x) := with x ∈ C. 0 1 A The elements of are of[ the form] a1 0 (2.2.1.2) a = with a1, a2 > 0 and a1a2 = 1. 0 a2 We also deﬁne the diagonal group D consisting of all elements [ ] z1 0 ̸ ∈ C (2.2.1.3) z = −1 with 0 = z1 . 0 z1 The elements k of K are the matrices satisfying (besides determinant 1) − ∗ ∗ (2.2.1.4) k 1 = k , where M := t M for all matrix M.

THEOREM 2.2.1. (Iwasawa’s decomposition) Every element g ∈ G has a unique expression as a product (2.2.1.5) g = uak with u ∈ U, a ∈ A, k ∈ K. C × Furthermore, let Pos2 := Pos2( ) be the space of positive deﬁnite 2 2 Hermitian ma- 7→ ∗ G K ≈ UA ≈ trices. Then the map g gg is a bijection / SPos2, so of SPos2.

∈ G C2 PROOF. Let g and e1, e2 be the standard (vertical) unit vectors of . Let (i) (1) (2) g := gei. We orthonormalize g , g by the standard Gram-Schmidt process, which means we can ﬁnd an upper triangular matrix B = (bij) (b21 = 0) such that if we let ′ (1) ′ (1) (2) e1 := b11g , e2 := b12g + b22g , ′ ′ then e1, e2 are orthonormal unit vectors, and we can choose both b11 and b22 greater than 0 (This corresponds to dividing by the norm). Thus B ∈ AU = UA. ′ Let k be the matrix such that kei = ei for i = 1, 2. Then k is unitary and gB = k. Since det g = 1 and the diagonal components of B are positive, it follows that 88 2. ANALYSIS ON TOPOLOGICAL GROUPS

∗ det k = 1, so G = KAU = UAK. To show uniqueness, we use the map g 7→ gg . ∗ Note that gg = 1 (identity) if and only if g ∈ K. Suppose that ∗ 2 ∗ 2 ∗ ∗ u1ak1(u1ak1) = u1a u1 = u2b u2 = u2bk2(u2bk2) ∈ U ∈ A ∈ K −1 2 with u1, u2 , a, b , and k1, k2 . Putting u := u2 u1, we obtain ua = ∗ − ∗ b2(u ) 1. Since u, u are triangular in opposite directions, they must be diagonal, and ﬁnally a2 = b2, so a = b because the diagonal elements are positive.

COROLLARY 2.2.2. Every element g ∈ G has a decomposition, called polar, ∈ K ∈ A (2.2.1.6) g = k1bk2 with k1, k2 and b . The element b is uniquely determined up to permutation of diagonal components. Let A+ ∈ ≥ G KA+K be the set of a = diag(a1, a2) A such that a1 1. Then = , and the decomposition (2.2.1.6) determines b uniquely in A+.

∗ ∗ 2 −1 ∈ A ∈ K PROOF. Since gg can be written as gg = k1b k with b and k1 , ∗ 1 where b2 is the matrix of eigenvalues of gg . By the bijection in Theorem 2.2.1, ∈ K there exists k2 such that g = k1bk2. Given g = uak ∈ G with its Iwasawa decomposition, we let

(2.2.1.7) gA := a or also ag := a denotes the A-projection, and similarly for gU and gK. 2.2.2. Characters. A character of A is a continuous homomorphism χ : A → × C into the multiplication group of nonzero complex numbers. Additively, let A R a := Lie( ) be the -vector space of diagonal matrices H = diag(h1, h2) with ∨ trace zero, so A = exp a. Let a be the dual space. Let α be the (additive) character α − ∈ A on a deﬁned by (H) := h1 h2. If a = exp H = diag(a1, a2) = exp a, then we deﬁne the Iwasawa coordinate y : A → R+ by ≡ ≡ χ ≡ α a1 2 (2.2.2.1) y(a) ya α(a) a := = a1. a2 ∨ We note that α is a basis of a . Furthermore, y gives a group isomorphism of A with the positive multiplicative group. For complex s, we get s sα χ (2.2.2.2) ya = a = sα(a), with a complex (multiplicative) character χ α. We often write y instead of y and [s ] a y1/2 0 (2.2.2.3) a = a or a(y) = − . y 0 y 1/2 Let D = diagonal subgroup of G with elements [ ] z1 0 (2.2.2.4) z = −1 0 z1 and T = torus subgroup = subgroup of D consisting of elements [ ] ϵ ϵ 1 0 |ϵ | (2.2.2.5) = ϵ−1 with 1 = 1. 0 1 Then we have a direct product decomposition

(2.2.2.6) D = AT with z = azϵ and T = D ∩ K. 2.2. IWASAWA’S DECOMPOSITION 89

| |ϵ Indeed, writing z1 = z1 1, [ ] [ ][ ] | | ϵ z1 0 z1 0 1 0 ϵ z = −1 = | |−1 ϵ−1 =: az . 0 z1 0 z1 0 1 The character α is actually the restriction of a character on D, namely

α z1 2 (2.2.2.7) z = = z1. z2 The kernel of α in D is the group {±1}. We deﬁne z ∈ D to be regular if its diago- α nal components are distinct, in other words if z ̸= 1. More generally, an arbitrary square matrix is called regular if it can be conjugated to a regular diagonal matrix.

2.2.3. K-bi-invariant functions. Let f be a function on G, which we write f ∈ Fu(G). We say that f is K-bi-invariant if ∈ K ∈ G (2.2.3.1) f (k1gk2) = f (g), for all k1, k2 and g . By the polar decomposition, if b is the A-polar component of g, then f (g) = f (b) and f is determined by its values on A. Let [ ] 0 1 (2.2.3.2) w := −1 0 be called the Weyl element, and let the group of order 2 that it generates mod ± 1 be called the Weyl group W ⊆ K. Then w acts on D and A by conjugation, and for z ∈ D, [ ][ ][ ] [ ] − − z − 1 − w 1 0 1 1 0 0 1 z1 0 1 (2.2.3.3) z := wzw = − −1 = = z . 1 0 0 z1 1 0 0 z1

PROPOSITION 2.2.3. A K-bi-invariant function is necessarily even.

∈ A φ PROOF. By the polar decomposition (2.2.1.6), g = k1bk2, where b , so if is K-bi-invariant, then − − φ(g) = φ(b) = φ(wbw 1) = φ(bw) = φ(b 1) − because of b = wb 1 and (2.2.3.3). Thus the restriction map from G to A induces bijections

(2.2.3.4) Fu(K\G/K) Fueven(A) and C(K\G/K) Ceven(A). Since the Iwasawa coordinate y gives an isomorphism of A with R+, the function ν := ln y gives an isomorphism of A with R. We obtain a linear isomorphism

(2.2.3.5) Fu(K\G/K) Fueven(R). Given a K-bi-invariant function f ∈ Fu(K\G/K), the corresponding even function on R will be denoted by f+, so that we have f (g) = f (b) = f+(ν). We denote the conjugation action of G on itself by c, so ′ ′ − (2.2.3.6) c(g)g := gg g 1. This action induces what we also call the conjugation action on various functors. 90 2. ANALYSIS ON TOPOLOGICAL GROUPS

Let u be the Lie algebra of U, by deﬁnition the algebra of strictly upper triangular matrices, so of the form [ ] 0 x (2.2.3.7) , x ∈ C. 0 0 Then u is 1-dimensional as a complex vector space, with C-basis [ ] 0 1 (2.2.3.8) E = 12 0 0 R For u as a vector space over , a natural basis consists of E12 and iE12. We have (2.2.3.9) U = 1 + n = exp n, where exp is the exponential given by the usual power series.

−1 ∈ A Let a = diag(a1, a1 ) as above. Then E12 and iE12 are eigenvectors for the conjugation action by elements of A. In fact1, −1 α −1 α (2.2.3.10) a(E12)a = a (E12), a(iE12)a = a (iE12). Thus u = uα is an eigenspace, with eigencharacter α, which has multiplicity 2 viewing u as a vector space of dimension 2 over R. Note that tu is a (−α)-eigenspace. We call α, −α the regular characters. The trace of the representation on u is 2α, and the half-trace is ρ := α = ρG (this is notation that becomes significant in the higher-dimensional case). More generally, for z ∈ D, [ ] − 1 z2x α (2.2.3.11) zu(x)z 1 = 1 = u(z x). 0 1 We let δ = δG be the Iwasawa character on A, namely δ 2α 2 4 (2.2.3.12) G (a) := a = ya = a1. The exponent is the trace of the representation, viewed as being on a 2-dimensional R-space. The absolute Jacobian of conjugation on u viewed as an R-space is |δ | 2α (2.2.3.13) (z) = az . G C 2.2.4. Group Fubini theorem. In the concrete application to = SL2( ) and its Iwasawa decomposition G = UAK = AUK, we have (2.2.4.1) P = G/K = UA = AU, and U is normal in P. All four of G, K, U, A are unimodular, so all the hypotheses in the previous propositions are satisfied. The character a 7→ δ(a) in Proposition α 2.1.8 is the Iwasawa character, δ(a) = a2 . We normalize the Haar measure by taking c = 1. Thus we define the Iwasawa measure to be the Haar measure such that for f ∈ C (G) we have ∫ c ∫ ∫ ∫ ∫ ∫ ∫ − α (2.2.4.2) f (g)dg := f (auk)dadudk = f (uak)a 2 dudadk, G K U A K A U

1For example, [ ][ ][ ] [ ] −1 2 −1 a1 0 0 1 a1 0 0 a1 2 α a(E12)a = −1 = = a1E12 = a E12. 0 a1 0 0 0 a1 0 0 2.2. IWASAWA’S DECOMPOSITION 91 where dy α da := with y = a , y du := Euclidean measure dx = dx1dx2 if x = x1 + ix2 with x1, x2 ∈ R, dk := Haar measure on K giving K total measure 1. Unless otherwise speciﬁed, all integral formulas are with respect to the Iwa- sawa measure.

The Iwasawa measure is canonically determined by the Iwasawa decomposition. We may use the notation

(2.2.4.3) dg ≡ dIWg ≡ dµIW(g) for the Iwasawa measure, when comparing it with other measures. Thus the basic Iwasawa measure integration formula can be written ∫ ∫ ∫ ∫ −2α (2.2.4.4) f (g)dµIW(g) = f (uak)a dudadk, INT1. G K A U The above normalization of the Haar measure ﬁts perfectly the normalization that ∼ comes from the quaternionic model for G/K = H3, i.e., the set of quaternions z = x1 + ix2 + jy, and the standard normalization of the measure on this homo- C geneous space for the action of SL2( ) given by (az + b)/(cz + d) with complex a, b, c, d. This measure is dx1dx2dy/y3.

− For z diagonal regular in G, and u ∈ U, the commutator is [z, u] = zu(uz) 1 = − − zuz 1u 1. Deﬁne ≡ | − α|m(α) α (2.2.4.5) JG,com(z) Jcom(z) := 1 z with m( ) = 2. The exponent is the multiplicity m(α) of α. Then ∫ ∫ − − 1 ( 1 1) = ( ) (2.2.4.6) f z uzu du −1 f u du, INT2. U Jcom(z ) U Indeed, matrix multiplication shows that [ ][ ][ ][ ] [ ] − − − − − 1 z − 2 − 1 1 1 z1 0 1 x 1 0 1 x 1 (z1 1)x [z , u] = z uzu = −1 = 0 z1 0 1 0 z1 0 1 0 1 whence ∫ ∫ − − −α i f (z 1uzu 1)du = f [u((z − 1)x)] dx ∧ dx¯ U C 2 [ ]− ∫ ∫ −α 1 i 1 = ( − )( −α − ) ( ( )) ∧ = ( ) z 1 z 1 f u x dx dx¯ −1 f u du. C 2 Jcom(z ) U From the product decomposition D = AT, we give the compact group T the total Haar measure 1. Then we give D the product measure of da = dy/y and this normalized measure on T, which we write dw. Then D \G has the homogeneous µ space measure, which makes the group Fubini theorem valid, that is, D\G such that ∫ ∫ ∫ µ µ (2.2.4.7) f (g)d IW(g) = f (wg˙)dwd D\G (g˙). G D\G D 92 2. ANALYSIS ON TOPOLOGICAL GROUPS

Under the correspondence D \ G ↔ U(T \K), for right K-invariant functions, we have with respect to a decomposition g = wuk (w ∈ D, u ∈ U, k ∈ K), µ (2.2.4.8) d D\G (g) = du.

THEOREM 2.2.4. Let φ ∈ Cc(K\G/K) and let z be a regular diagonal element. Then − (2.2.4.9) g 7→ φ(g 1zg) has compact support on D \G, and we have ∫ ∫ −1 1 φ( ) µ \G ( ) = φ( ) (2.2.4.10) g zg d D g −1 azu du. D\G Jcom(z ) U

− − − − PROOF. For (2.2.4.10), using g 1zg = k 1(u 1w 1zwu)k, one has ∫ ∫ ∫ − − − − φ(g 1zg)dg = φ(u 1zu)du = φ(zz 1u 1zu)du D\G U U ∫ ∫ ∫ − − − − 1 = (φ ◦ )( 1 1 ) = (φ ◦ )( 1 1) = φ( ) z z u zu du z z uzu du −1 zu du U U Jcom(z ) U by (2.2.4.6). Since, writing z = |z |ϵ, [ ][ 1 1][ ] [ ] −1 | | | | ϵ 0 z1 z1x 1 0 z1 z1 x ϵ −1 = | |−1 = azu, 0 0 z1 0 1 0 z1 we see that φ(zu) = φ(azu). 2.3. Harish, Mellin and spherical transforms The Iwasawa decomposition leads into the Harish transform and Mellin transform, whose composite is the spherical transform, of Harish-Chandra, relating analytic (Fourier-type) expansion on G/K to those on A.

2.3.1. The Harish transform and the orbital integral. We factorize Jcom(z) for α/2 −α/2 diagonal z as follows. For z as given by (2.2.2.4), we have z = z1 and z = −1 | α/2 − −α/2| | − −1| | −1 − | | −1 α/2 − −1 −α/2| z1 , so that z z = z1 z1 = z1 z1 = (z ) (z ) . Put α − α − α m( ) α − α 2 − 2 | | 1 | | 2 − 2 2 − 2 − 1 (2.3.1.1) D (z ) = D (z) := z z = z z = z1 z1 . Then −1 −1/2 −α (2.3.1.2) Jcom(z ) = |D|(z)δ(az) = |D|(z)z . α/2 −α/2 −1 Note that we can take z = z1 and z = z1 . We define the orbital integral by ∫ φ φ −1 µ (2.3.1.3) OrbD\G ( , z) := (g zg)d D\G (g) D\G µ φ ∈ K\G K with D\G as in (2.2.4.8). We define the Harish transform of Cc( / ) by ∫ 1/2 (2.3.1.4) (Hφ)(z) := δ(az) φ(azu)du = |D|(z)Orb \G (φ, z). U D The equality on the right comes from Theorem 2.2.4 and (2.3.1.2). The first formula is applicable to all diagonal z, but the second, with the orbital integral, is applicable only to regular z. As usual in measure-integration theory, the equality extends to 2.3. HARISH, MELLIN AND SPHERICAL TRANSFORMS 93 functions for which the integrals are absolutely convergent, by continuity. This will be important for us, and certain convergence propertied will be dealt with at the appropriate time and place.

THEOREM 2.3.1. For φ ∈ Cc(K\G/K), the Harish transform Hφ is invariant under the Weyl group, i.e., − (2.3.1.5) (Hφ)(a) = (Hφ)(a 1).

− PROOF. Hφ is invariant if and only if (Hφ)(wzw 1) = (Hφ)(z), or equiva- − lently, if and only if (Hφ)(z 1) = (Hφ)(z) by (2.2.3.3). Using (2.2.2.6) and T ⊆ K − and the fact that φ is K-bi-invariant, it suffices to verify that (Hφ)(a) = (Hφ)(a 1). By continuity, it suffices to prove the assertion when a is regular, so |D|(a) = − |D|(a 1). From the fact that K-bi-invariant implies that φ is even, we get ∫ −1 −1 −1 −1 −1 (Hφ)(a ) = |D|(a )Orb \G (φ, a ) = |D|(a) φ(g a g)dµ \G (g) D \G D ∫ D | | φ −1 −1 −1 µ φ = D (a) ((g a g) )d D\G (g) = (H )(a). D\G One can also argue directly on the other integral defining H, using the Jacobian − for u 7→ a 1ua.

The Harish transform is therefore a linear map W (2.3.1.6) H : Cc(K\G/K) −→ Cc(A) , where the upper index W means the space of functions invariant under W.

We consider next the behavior under products, notably the convolution product on G, defined for a specified Haar measure by ∫ −1 (2.3.1.7) (h ∗ f )(z) := h(zg ) f (g)dµIWg G on a class of pairs of functions for which the integral is absolutely convergent. For non-commutative G, the K-bi-invariance (and the evenness) allows a choice of writing variables on one side or another, with inverses or not, namely for ψ, f , K- bi-invariant such that for all z the function z 7→ ψ(zg) f (g) is in L1, the convolution (ψ ∗ f )(z) is defined by ∫ ∫ −1 −1 (ψ ∗ f )(z) := ψ(zg) f (g)dµIW(g) = ψ(zg ) f (g )dµIW(g) ∫G G∫ −1 −1 (2.3.1.8) = ψ(zg ) f (g)dµIW(g) = ψ(g ) f (gz)dµIW(g) ∫G G

= ψ(g) f (gz)dµIW(g). G

THEOREM 2.3.2. (Gelfand) Let G be a locally compact unimodular group, and let K be a compact subgroup. Let τ be an anti-automorphism of G of order 2 such that given ∈ G ∈ K τ g , there exist k1, k2 satisfying g = k1gk2. Then the convolution is commutative (whenever the convolution integral is absolutely convergent). 94 2. ANALYSIS ON TOPOLOGICAL GROUPS

PROOF. Since ψ, f are K-bi-invariant and G is unimodular, it follows that ψ, f are τ-invariant and so is the Haar measure. Therefore ∫ ∫ − τ τ − τ (ψ ∗ f )(z) = ψ(zg 1) f (g)dg = ψ( z g 1) f ( g)dg ∫G G ∫ ψ τ −1 τ τ 7→ ψ −1 τ = ( g ) f ( z g)dg (by g gz) = (g ) f (k1zk2 g)dg ∫G ∫ G ψ τ −1 ψ −1 = ( g ) f (k1zk2g)dg = (g k2) f (k1zg)dg ∫G ∫G − − − − − = f (zgk 1)ψ(k g 1)dg = f (z(k g 1) 1)ψ(k g 1)dg. G 2 2 G 2 2 Thus ψ ∗ f = f ∗ ψ.

THEOREM 2.3.3. For φ, ψ ∈ Cc(K\G/K), we have (2.3.1.9) H(φ ∗ ψ) = Hφ ∗ Hψ, i.e., on Cc(K\G/K) the Harish transform is an algebra homomorphism for the convolution product.

PROOF. By continuity, we may assume that z = a ∈ A. Then, in this case, α 1/2 a = δ(a) and az = aa = a. Hence ∫ ∫ ∫ α α − (H(φ ∗ ψ))(a) = a (φ ∗ ψ)(au)du = a φ(aug)ψ(g 1)dgdu ∫U ∫ U G α − = a φ(ag)ψ(g 1u)dgdu. U G Let g = bvk be the Iwasawa decomposition with b ∈ A, ν ∈ U, and k ∈ K. Then, using (2.2.4.2), we get ∫ ∫ ∫ ∫ α − − − (H(φ ∗ ψ))(a) = a φ(abνk)ψ(k 1ν 1b 1u)dbdνdkdu U K U A ∫ ∫ ∫ ∫ α − − = a φ(abν)ψ(ν 1b 1u)dbdνdkdu U K U A ∫ ∫ ∫ α − − = a φ(abν)ψ(ν 1b 1u)dbdνdu U U A ∫ ∫ ∫ α − − = a φ(ab 1ν)ψ(ν 1bu)dbdudv U U A ∫ ∫ ∫ α − − − = a φ(ab 1ν)ψ(ν 1ub)δ(b) 1dbdνdu U U A ∫ ∫ ∫ α − − = a φ(ab 1ν)ψ(ub)δ(b) 1dbdνdu U U A ∫ ∫ ∫ α − = a φ(ab 1ν)ψ(bu)dbdνdu U U A ∫ ∫ ∫ − α α − = (ab 1) b φ(ab 1ν)ψ(bu)dbdνdu U U A which is exactly the convolution (Hφ ∗ Hψ)(a). 2.3. HARISH, MELLIN AND SPHERICAL TRANSFORMS 95

2.3.2. The Mellin and spherical transforms. Let χ be a character on A (con- × tinuous homomorphism into C ). Let F ∈ Cc(A). We define the Mellin transform of F to be the function on the group ch(A) of characters given by ∫ (2.3.2.1) (MF)(χ) := F(a)χ(a)da. A We use the same Haar measure da on A as before, with the coordinate character y. Viewing y as a basis for the characters, we can write a character in the form χ s sα (2.3.2.2) (a) := ya = a with a complex variable s. Then the Mellin transform is written with Fα(y) = F(a): ∫ ∞ dy (2.3.2.3) (MαF)(s) ≡ (MFα)(s) = Fα(y)ys . 0 y ∈ ∞ Here y := ya [0, ). This is the orbidary Mellin transform of elementary analysis. If F does not have compact support, the integral may converge only in a restricted domain of s, usually some half-plane where the transform is analytic. − If F is even (F(a) = F(a 1)), then MF is also even. In other words, − (2.3.2.4) (MF)(χ) = (MF)(χ 1). For g ∈ G let gA be its Iwasawa projection on A. Then the function g 7→ χ(gA), also written χ(g), is well-defined, and is called the Iwasawa lifting of χ ∈ ch(A) α to G. As before, let δ = δG be the Iwasawa character, δ(a) = a2 . We define the spherical kernel on the product of G with the character group ch(A) to be the function given by ∫ ∫ (2.3.2.5) Φχ(g) ≡ Φ(χ, g) := (χδ1/2)(kg)dk = (χδ1/2)((kg)A)dk. K K Let a := Lie(A) be the real vector space of diagonal 2 × 2 matrices of trace 0, [ ] h1 0 − (2.3.2.6) H = with h2 = h1. 0 h2 Let ζ denote a complex (additive) character of A. Then χ may be written as χζ, with ζ such that if a = exp H, then ζ(H) ζ(ln a) (2.3.2.7) χζ(a) := e = e . Thus the Mellin transform can also be viewed as defined on the complex dual ∨ ξ (aC) , and the spherical kernel may be written in terms of the additive variable , in the form ∨ (2.3.2.8) Φζ (g) ≡ Φ(ζ, g) = Φ(χζ, g), ζ ∈ (aC) . We let S ≡ S be the integral (spherical) transform defined by this kernel, IW ∫ (2.3.2.9) (S f )(χ) := Φ(χ, g) f (g)dg. G

THEOREM 2.3.4. Let f be measurable and K-bi-invariant on G. Also suppose that the repeated integral ∫ ∫ (|χ|δ1/2)((kg)A)| f (g)|dkdg G K is absolutely convergent. Then (2.3.2.10) S f = M(H f ), 96 2. ANALYSIS ON TOPOLOGICAL GROUPS that is, S = M ◦ H on Cc(K\G/K).

PROOF. From (2.3.1.6), consider the following diagram

(K\G K) H / (A)W ⊆ (A) Cc / RR Cc Cc RRR RRR RRR M S RR( Fu(ch(A)) By Fubini’s theorem and K-invariance of f and dg, we have ∫ ∫ ∫ (S f )(χ) = Φ(χ, g) f (g)dg = (χδ1/2)((kg)A) f (g)dgdk G K G ∫ ∫ ∫ ∫ = (χδ1/2)(gA) f (g)dgdk = χ(a)δ1/2(a) f (au)dadu K G A U ∫ = χ(a)(H f )(a)da = M(H f )(χ). A Thus S f = M(H f ).

COROLLARY 2.3.5. The spherical kernel is invariant under W in the variable χ, that is,

(2.3.2.11) Φχ = Φχ−1 .

PROOF. From (2.3.2.10), we have, for any f ∈ C (K\G/K), ∫ ∫c − − − (S f )(χ 1) = χ 1(a)(H f )(a)da = χ(a 1)(H f )(a)da A A ∫ ∫ − = χ(a)(H f )(a 1)da = χ(a)(H f )(a)da = (S f )(χ). A A

Therefore Φχ = Φχ−1 .

THEOREM 2.3.6. The Mellin transform and the spherical transform are multiplicative homomorphisms, that is, for φ, ψ ∈ Cc(K\G/K), and f , h ∈ Cc(A), we have (2.3.2.12) S(φ ∗ ψ) = S(φ)S(ψ), M( f ∗ g) = M( f )M(h).

PROOF. By deﬁnitions, ∫ ∫ ∫ − M( f ∗ h)(χ) = (g ∗ h)(a)χ(a)da = f (ab 1)h(b)χ(a)dbda. A A A ∫ ∫ = f (a)h(b)χ(a)χ(b)dbda = (M f )(χ)(Mh)(χ). A A Combining this with (2.3.1.9) and (2.3.2.10) gets the result for S. χ χ χ χ 2s If = sα (that is, (a) = sα(a) = a1 ), we may write χ (2.3.2.13) (Sα f )(s) := (S f )( sα). ∨ If χ = χζ with ζ ∈ (aC) , we may write

(2.3.2.14) (S f )(ζ) := (S f )(χζ). 2.3. HARISH, MELLIN AND SPHERICAL TRANSFORMS 97

2.3.3. Computation of the orbital integral. Let z ∈ G be regular, that is, z is conjugate to a diagonal matrix with distinct diagonal elements (the eigenvalues) [ ] η 0 (2.3.3.1) z ∼ η˜ = − with 0 ̸= η ∈ C. 0 η 1 Without loss of generality, we may suppose that |η| ≥ 1. If |η| > 1, then η is − uniquely determined among the two eigenvalues η, η 1 of z. From the conjugation action on matrices, one sees that the centralizer in G of a regular diagonal element is the set of diagonal matrices. In particular, the isotropy group of η˜ under the conjugation action is the diagonal group D. Hence the isotropy group Gz is conjugate to D by the same conjugation that diagonalizes z to η˜. We write η ≡ η(z) and η˜ ≡ η˜(z). Since conjugation preserves Haar measure, we have ∫ ∫ − − (2.3.3.2) φ(g 1zg)dg = φ(g 1η˜g)dg. Gz\G D\G Thus the Harish transform can be computed for z in diagonal form, and we use indiscriminately identity ( φ)( ) = | |(η( )) (φ ) = | |(η) (φ η) (2.3.3.3) H z D ˜ z OrbGz\G , z D OrbD\G , ˜ . We now go on with a theorem that allows for a more explicit determination of the orbital integral, and therefore of the Harish transform. As in Section 2.3.2, we normalize Haar measure so that T as well as K has measure 1. On D = AT we put the product measure of da = dy/y on A and this normalized measure on T, which we write dw (w variable in D). Then for f ∈ L1(G/K), we have ∫ ∫ ∫ (2.3.3.4) f (g)dg = f (wu)dwdu. G U D The dg on the left is of course the Iwasawa measure. From the modiﬁed Iwasawa decomposition G = DUK, we get ∫ ∫ −1 −1 −1 −1 Orb \G (φ, η˜) = φ(g η˜g)dµ \G (g) = g(k u z η˜zuk)dudk D \G D UK ∫D ∫ − − − (2.3.3.5) = φ(u 1z 1η˜zu)du = φ(u 1η˜u)du ∫U U∫ = φ(u(−x)η˜u(x))dx = φ(Mη(x))dx, C C where [ ] η η − η−1 −1 x( ) (2.3.3.6) Mη(x) := u(x) η˜u(x) = − 0 η 1 The function φ is given in terms of polar coordinates, so we have to express the polar A-coordinate of Mη(x) in terms of x. We shall deal with the (x, y) coordinates, polar A-coordinates, and ordinary polar coordinates in C.

THEOREM 2.3.7. Let the Haar measure on G be the Iwasawa measure as in Subsec- tion 2.2.4. Let φ be measurable on G, K-bi-invariant (so even), and such that the orbital φ η η ∈ A α ≥ integral OrbD\G ( , ˜) is absolutely convergent for regular ˜. For b , y = b 1, write (2.3.3.7) φα(y) ≡ φα(y(b)) := φ(b). 98 2. ANALYSIS ON TOPOLOGICAL GROUPS

Then for |η| ≥ 1 and η˜ regular, we have ∫ π ∞ − −1 φ η 2 φ y y dy (2.3.3.8) OrdD\G ( , ˜) = − α(y) . |η − η 1|2 |η|2 2 y Or in terms of the additive variable ν := ln y, using the notation

(2.3.3.9) φα,+(ν) ≡ φα,+(ln y) := φ(b), the formula becomes ∫ π ∞ − −1 φ η 2 φ y y dy OrdD\G ( , ˜) = − α,+(ln y) |η − η 1|2 |η|2 2 y ∫ π ∞ 2 φ ν ν ν (2.3.3.10) = − α,+( ) sinh( )d . |η − η 1|2 2 ln |η|

PROOF. To go back and forth between the variable x in Mη(x) and the A- variable, we use the quadratic map, namely, ∗ − η ∈ K ∈ A η η 2 1 (2.3.3.11) If M (x) = k1bk2 (k1, k2 , b ), then M (x)M (x) = k1b k1 . For (2.3.3.6), we have [ ] − − − ∗ |η|2 + |x|2(η − η 1)2 xη¯ 1(η − η 1) (2.3.3.12) Mη(x)Mη(x) = − − − x¯η 1(η¯ − η¯ 1) |η| 2 ∗ So b2 can be computed from the eigenvalues of Mη(x)Mη(x) , namely, the roots of the characteristic polynomial T2 − τ(r)T + 1 = 0, ∗ where r = |x|, and τ(r) is the trace of the matrix Mη(x)Mη(x) : − − (2.3.3.13) τ(r) := |η|2 + |η| 2 + r2|η − η 1|2. 2 1 2 θ We have r = xx¯ and dx = dx dx = rdrd . With b = diag(b1, b2), b1 > 1, we can solve for b as a function of r (and so as a function of x). Introduce the discriminant of the characteristic polynomial ∆ := τ(r)2 − 4. Then the roots of the characteristic polynomial are τ(r) + ∆1/2 τ(r) − ∆1/2 (2.3.3.14) y ≡ y = y(b) = b2 = and b2 = , b 1 2 2 2 τ ∆1/2 2 −2 ∆ or also + = 2y. Since b2 = b1 we get from (2.3.3.14) the value of in terms of y, − (2.3.3.15) ∆1/2 = y − y 1. Note that y is a variable on the positive multiplicative group. In terms of the α multiplicative coordinate y = y(b) = b , we write φ(b) = φα(y). We change variables in the integral form (2.3.3.5), dx1dx2 = rdrdθ, to get ∫ ∫ ∞ φ η φ π φ (2.3.3.16) OrbD\G ( , ˜) = (Mη(x))dx = 2 (b(r))rdr. C 0 2.3. HARISH, MELLIN AND SPHERICAL TRANSFORMS 99

We have to write down the Jacobian between the variables y and r. By (2.3.3.13), we have − dτ = |η − η 1|22rdr. By (2.3.3.14), ( ) 1 2τdτ ∆1/2 + τ y dy = dτ + = dτ = dτ. 2 2∆1/2 2∆1/2 ∆1/2 Hence, using (2.3.3.15), − − dy |η − η 1|2 y − y 1 1 dy (2.3.3.17) = 2rdr or rdr = . y ∆1/2 2 |η − η−1|2 y Putting (2.3.3.16) and (2.3.3.17) together yields (2.3.3.8), except determining the limits of integration. The polar r ranges over (0, +∞). The expression (2.3.3.13) for τ(r) shows that − if we put ℓ := |η|2 + |η| 2, then (2.3.3.18) √ − τ ∈ (ℓ, +∞), ∆ ∈ (ℓ2 − 4, +∞), ℓ2 − 4 = (|η|2 − |η| 2)2, ℓ + ℓ2 − 4 = 2|η|2. From (2.3.3.14) we get ( √ ) τ + ∆1/2 ℓ + ℓ2 − 4 (2.3.3.19) y = ∈ , ∞ = (|η|2, ∞), 2 2 determining the limits of integration. The last formula (2.3.3.10) is obvious.

+ α COROLLARY 2.3.8. For a ∈ A (i.e., a > 1, so a regular), we have ∫ − ∞ y − y 1 dy (2.3.3.20) (Hφ)(a) = 2π φα(y) . aα 2 y

PROOF. It follows from (2.3.1.1), (2.3.3.4), and (2.3.3.9).

2.3.4. Gaussians on G and their spherical transform. By a basic G-Gaussian (basic Gaussian for short) we mean a function on G that is K-bi-invariant, and with some constant c > 0, for all b ∈ A has the formula

− α 2 α(ln b) φ (b) = e ( (ln b)) /2c c sinh(α(ln b)) − 2 ln y (2.3.4.1) = e (ln y) /2c with y = y(b), (y − y−1)/2 −ν2 ν = e /2c with ν = ln y. sin hν We can then deal with the vector space Gauss(G) generated by the basic G-Gaussian functions as a natural space of test functions.

THEOREM 2.3.9. For the orbital integral, with the basic G-Gaussian φc ∈ Gauss(G), and regular diagonal z, we have the value

π α 2 c −(ln |z |)2/2c (2.3.4.2) Orb \G (φc, z) = e . D |zα/2 − z−α/2|2 100 2. ANALYSIS ON TOPOLOGICAL GROUPS

For a ∈ A regular, y = y(a), the Harish transform is given by

−(α(ln a))2/2c −(ln y)2/2c (2.3.4.3) (Hφc)(a) = 2πce = 2πce . Here we use the y-variable, α(ln a) = ln y.

α/2 PROOF. Letting η˜ = z (so that η = z ) and φ = φc in (2.3.3.10), we have ∫ π ∞ ν2 φ 2 c −ν2/2c OrbD\G ( c, z) = − e d |η − η 1|2 2 ln η 2c 2πc − η 2 2πc − | α| 2 = e (2 ln ) /2c = e (ln z ) /2c. |η − η−1|2 |zα/2 − z−α/2|2 The last formula follows from (2.2.2.2), (2.3.1.4), and (2.3.4.2).

The factor 2π came originally from having taken the ordinary Euclidean mea- ∼ sure on C = U, so polar coordinates for which the circle has measure 2π, which is standard, but is given priority over the counterconvention to given compact groups measure 1.

THEOREM 2.3.10. Write a character on a = Lie(A) in the form χ = sα with complex s. Then the spherical transform is

3/2 s2c/2 (2.3.4.4) (Sφc)(sα) = (M(Hφc))(sα) = (2πc) e . On the imaginary axis s = ir, this gives

3/2 −r2c/2 (2.3.4.5) (Sφc)(irα) = (M(Hφc))(irα) = (2πc) e .

PROOF. By Theorem 2.3.9 and S = M ◦ H (cf., Theorem 2.3.4) we have (cf., equations (2.3.2.3) and (2.3.2.13)) ∫ ∞ ∫ ∞ 1 1 −(ln y)2/2c s dy −ν2/2c νs (Sα φc)(s) = (Sφc)(sα) = e y = e e dν. 2πc 2πc 0 y −∞ Because of the rapid decay of Gaussians, the Mellin transform is entire in s, so it sufﬁces to determine the integral when s = ir (r real) is pure imaginary.√ One then −ν2 uses the self-duality of the function e /2 with respect to dν/ 2π, and a change of variables to get rid of c.

We let the Gauss space on R, Gauss(R), be the vector space generated by − 2 the Gaussian functions r 7→ e r c/2, with all constants c > 0. Sometimes, we deal with the imaginary axis iR, and Gauss(iR) is the space generated by the functions having the above value at ir.

COROLLARY 2.3.11. The spherical transform gives a linear isomorphism

(2.3.4.6) Gauss(G) −→ Gauss(R), φc 7−→ (Sφc)(·α).

COROLLARY 2.3.12. With the Iwasawa measure dµIW, the total integral of φc is ∫ 3/2 c/2 (2.3.4.7) φc(g)dµIW(g) = (2πc) e . G 2.3. HARISH, MELLIN AND SPHERICAL TRANSFORMS 101

PROOF. Taking s = 1 in (2.3.4.4) yields ∫ 3/2 c/2 (2πc) e = (Sφc)(−α) = Φ(−α, g)φc(g)dg G by (2.3.2.9). The spherical kernel is given by the integral ∫ ∫ ∫ −α+α Φ(−α, g) = (−αδ1/2)((kg)A)dk = ((kg)A) dk = dk = 1. K K K Thus, we get (2.3.4.7). Let g ∈ G with polar decomposition ∈ K ∈ A (2.3.4.8) g = k1bk2, k1, k2 and b . The element b ∈ A is determined only up to a permutation of its diagonal compo- α − nents. We call y = y(b) = b or y 1 a polar coordinate (or polar A-coordinate) of g. Note that (ln y)2 is uniquely determine by g, and is real analytic on G. Deﬁne the polar height σ to be the K-bi-invariant function such that σ ≥ 0 and (2.3.4.9) σ2(g) = σ2(b) = (ln y)2 = ν2.

Then the basic G-Gaussian φc, given in (2.3.4.1), can be written in the form

−σ2 σ − 2 ln y (2.3.4.10) φ = e /2c = e (ln y) /2c . c sinh σ sinh(ln y) Note that ν/ sinh ν is a power series in ν2 = (ln y)2, so real analytic on G.

For H ∈ a given in (2.3.2.6), we define the symmetric bilinear form B : a × a → R by 2 (2.3.4.11) B(H, H) := tr(HH) = |H|B. ∈ A ∈ For b = diag(b1, b2) , we have ln b = diag(ln b1, ln b2) a, and 2 2 2 2 2 (ln y) | ln b|B = (ln b ) + (ln b ) = 2(ln b ) = , 1 2 1 2 so 2 2 2 (2.3.4.12) σ (b) = 2| ln b|B = (ln y) . The trace form on a can be extended in two natural ways to the complexification, namely, a symmetric way and a Hermitian way. The symmetric way will be relevant in the next subsection. The trace form induces a corresponding positive ∨ definite scalar product on the dual space a , because it induces a natural isomor- ∨ ∨ phism of a and a . Specifically, if λ ∈ a , then there is an element Hλ ∈ a such that for all H ∈ a, (2.3.4.13) B(Hλ, H) = λ(H). ∨ Then by definition, for λ, ξ ∈ a , we have ∨ (2.3.4.14) B (ξ, λ) := B(Hξ, Hλ), ∨ so for example2, B (α, α) = 2.

∨ 2Let Hα ∈ a be the correspondence of α ∈ a . According to tr(Hα H) = α(H), we have Hα = ∨ diag(1, −1) and then B (α, α) = tr(diag(1, 1)) = 2. 102 2. ANALYSIS ON TOPOLOGICAL GROUPS

∨ ∨ ∨ We can then extend the form from a to a C-bilinear form BC := (B )C = ∨ ∨ ∨ ∨ B ⊗R C on its complexification, i.e., to characters ζ ∈ aC = (a )C = a ⊗R C ∨ that can be written in the form ζ = ξ + iλ, with ξ, λ ∈ a . We call ξ and λ the real and imaginary parts of ζ respectively. Then by definition ∨ ∨ ∨ ∨ (2.3.4.15) BC(ζ, ζ) := B (ξ, ξ) + 2iB (ξ, λ) − B (λ, λ). In particular, for ζ = sα with s ∈ C, we have ∨ ∨ ∨ 2 (2.3.4.16) BC(iλ, iλ) = −B (λ, λ), BC(sα, sα) = 2s . ∨ Thus the form BC is C-bilinear, not Hermitian. Its restriction to the imaginary axis ∨ Ia is negative definite, corresponding to minus the original form on a. 2.3.5. The polar Haar measure and inversion. A K-bi-invariant function is determined by its values on the polar A-component b in the polar decomposition G KAK = , i.e., g = k1bk2. We define the polar Jacobian ( α −α ) (α) ( − )2 b − b m y(b) − y(b) 1 (2.3.5.1) J (b) := = P 2 2 α 2 α µ with y = y(b) = b = b1 and m( ) = 2. The associated polar measure d P is given by ( ν −ν ) e − e 2 (2.3.5.2) dµ (b) := J (b)db = = (sinh ν)2dν P P 2 in terms of the ν-variable, ν = ln y. It is a Jacobian computation due to Harish- Chandra in general that the functional ∫ ∫ ∫

(2.3.5.3) φ 7−→ φ(k bk )JP(b)dbdk dk K K A 1 2 1 2 is a Haar functional on G. For K-bi-invariant functions, one does not need the two K integrations in the above formula. The total dk-measure of K is assumed to be normalized to 1. The question arises how the polar Haar measure is related to the Iwasawa measure.

THEOREM 2.3.13. For φ ∈ Gauss(G), we have ∫ ∫ ∫ 1 (2.3.5.4) φ(g)dµIW(g) = φ(b)JP(b)db = φdµP. 2π G A A

PROOF. For φc ∈ Gauss(G), one has ∫ 1 1/2 3/2 c/2 φc(g)dµIW(g) = (2π) c e . 2π G On the other hand, ( ) ∫ ∫ ∞ ν − −ν ∫ ∞ −ν2/2c e e −ν2/2c ν φc(b)dµP(b) = e ν dν = e e νdν. A −∞ 2 −∞ Using the identity ∫ ∞ 1 − 2 2 (2.3.5.5) esxe x /2xxdx = (2π)1/2c3/2es c/2, s ∈ C, s −∞ we get (2.3.5.4). 2.3. HARISH, MELLIN AND SPHERICAL TRANSFORMS 103

COROLLARY 2.3.14. For all f ∈ Cc(K\G/K) one has ∫ ∫ 1 (2.3.5.6) f (g)dµIW(g) = f (b)dµP(b). 2π G A

PROOF. It follows from Corollary 2.3.11 and the fact the Gauss(R) is dense in Cc(R). Having two natural measures, the Iwasawa measure and the polar measure, we get two normalizations of the spherical transform, which we denote by 1 (2.3.5.7) S and S so that S = S . IW P P 2π IW The S of proceeding sections is SIW. We may call SP the polar normalized spherical transform, or simply the polar spherical transform.

We let B∨ α α ζ − −ζ ζ − −ζ Φ ζ ( , ) a a 2 a a (2.3.5.8) P( , a) := ∨ α −α = ∨ α −α . BC(α, ζ) a − a BC(α, ζ) a − a Writing ζ = sα with s ∈ C, we can write this formula in the form − ν − ν 1 ys − y s 1 es − e s 1 sinh(sν) (2.3.5.9) Φ α(s, a) := Φ (sα, a) = = = . P, P s y − y−1 s eν − e−ν s sinh ν

THEOREM 2.3.15. For all basic G-Gaussians, taking the polar measure convolution ∗ A A,P on , we have Φ ∗ φ α φ α (2.3.5.10) ( P A,P c) (s ) = (SP c)(s ) so ΦP is an integral kernel for SP ( with polar measure).

PROOF. By deﬁnition, (2.3.4.1) and, (2.3.5.2), ∫

(ΦP ∗A φc) (sα) = ΦP(sα, a)φc(a)dµP(a) ,P A ∫ ( ) ∞ sν − −sν ν ν − −ν 2 1 e s −ν2/2c 2 e e ν = ν −ν e ν −ν d s −∞ e − e e − e 2 ∫ ( ) ∫ ∞ sν − −sν ∞ 1 e e −ν2/2c 1 sν −ν2/2c = e νdν = e e νdν = (SP φc)(sα) s −∞ 2 s −∞ where we used (2.3.5.5), (2.3.4.4), and (2.3.5.7). C The spherical kernel (2.3.5.8) in polar coordinates on SL2( ) is originally due to Gelfand and Naimark (1950).

We observe that the polar spherical kernel has a product structure. The ﬁrst factor in (2.3.5.8) has a name, the Harish-Chandra c-function, deﬁned for our pur- pose by ∨ B (α, α) 1 (2.3.5.11) c(ζ) := ∨ = if ζ = sα. BC(α, ζ) s 104 2. ANALYSIS ON TOPOLOGICAL GROUPS

The second factor in (2.3.5.8) comes from a skew-symmetrization procedure, namely, ζ −ζ a − a . aα − a−α We know from Corollary 2.3.11 that the spherical transform is a linear iso- ∨ morphism between the Gauss spaces. We want to exhibit the measure on a with respect to which the transpose of the spherical kernel induces the inverse ∨ integral transform. As to the Haar measure on a , we have the Haar measure ν ∨ ∼ R µ λ ∨ da = dy/y = d on a = . We let d FOU( ) be the Haar measure on a that makes Fourier inversion come out without any extra constant factor. The Fourier transform F is defined by ∫ − λ ∨ (2.3.5.12) (F f )(λ) := f (H)e i (H)dH, λ ∈ a . a The isomorphism of a with R is taken via the coordinate ν. That is, [ ] [ ] − exp 1 / a 0 / h 0 A a B a = 1 H = 1 BB 0 a 0 h BBν 2 2LL y B LLν BB LL LLL R+ ln / R L% / − a1/a2 h1 h2 Thus if λ = rα on a with λ ∈ R, then ∫ ∫ −irα(H) −irν (2.3.5.13) (Fα,+ f )(r) ≡ (F f )(λ) = f (H)e dH = fα,+(ν)e dν. a R The Fourier normalized measure is then defined to be dr (2.3.5.14) dµ (λ) := . FOU 2π ∨ We define the Harish-Chandra measure on a by 1 r2dr (2.3.5.15) dµ (λ) := dµ (λ) = , λ = sα. HC |c(iλ)|2 FOU 2π Φ Φ λ We let P,HC be the integral operator having the kernel function P(i , a) with µ λ respect to the Harish-Chandra measure d HC( ), so for an even function h such that the integral is absolutely convergent, by definition ∫ Φ Φ λ λ µ λ ( P,HCh)(a) := P(i , a)h(i )d HC( ) a∨ ∫ irν − −irν − e e dr ν α (2.3.5.16) = hα(ir)( ir) ν −ν with = (ln a). R e − e 2π

Φ THEOREM 2.3.16. The integral operator P,HC is the inverse of SP on the Gauss spaces

SP : Gauss(G) −→ Gauss(iR). 2 The map SP is an L -isometry with respect to the polar measure on G and the Harish- ∨ ∨ Chandra measure on a , the integral scalar product taken on a . 2.3. HARISH, MELLIN AND SPHERICAL TRANSFORMS 105

PROOF. For given φc ∈ Gauss(G), let h := SP φc. By (2.3.4.5), − 2 hα(ir) = (2π)1/2c3/2e r c/2. Then ∫ irν − −irν Φ π 1/2 3/2 −r2c/2 − e e dr ( P,HCh)(a) = (2 ) c e ( ir) ν −ν R e − e 2π ∫ −1/2 3/2 (2π) c − 2 ν = (−ir)e r c/2+i rdr R sinh ν −ν2 ν = e /2c = φ (a). sinh ν c Φ ◦ φ φ Thus ( P,HC SP) c = c. To prove the second statement, we ﬁrst make explicit the L2-scalar product on G, for the polar measure. For two basic Gaussians φ , φ ′ , we get, integrating over ∼ c c a = R: ∫ ( ν −ν ) ν ′ ν − 2 ⟨φ φ ⟩ −ν2/2c 2 −ν2/2c 2 e e ν c, ′ P = e ν −ν e ν −ν d c R − − ∫ e e e e ∫ 2 −ν2 ′ ′ ′ −ν2 ′ = e (c+c )/2cc ν2dν = (cc )3/2 e (c+c )/2ν2dν. R R On the other hand, ∫ 2 ⟨ φ φ ⟩ φ α φ α r dr SP c, SP ′ HC = (SP c)(ir )(SP ′ )(ir ) c R c 2π ∫ 2 − 2 ′ − 2 ′ r dr = (2π)1/2c3/2e r c/2(2π)1/2c 3/2e r c /2 R 2π ∫ ( ′ ) ′ 3/2 ′ 3/2 −r2(c+c )/2 2 2cc Γ = (cc ) e r dr = ′ (3/2). R c + c ⟨φ φ ⟩ ⟨ φ φ ⟩ Thus c, c′ P = SP c, SP c′ HC. Jorgenson and Lang (2005) showed that the above isometry can be extended by continuity to L2(K\G/K) because of the denseness of the Gauss space.

2.3.6. Point-pair invariants, the polar height, and the polar distance. Let φ ∈ Fu(K\G/K) be a K-bi-invariant function on G. Following Selberg, we define its associated point-pair invariant Kφ to be the function defined by − (2.3.6.1) Kφ(z, w) := φ(z 1w), z, w ∈ G. The K-bi-invariance of φ implies that Kφ is defined on G/K × G/K. (1) Kφ is symmetric, i.e., Kφ(z, w) = Kφ(w, z), because φ is even. In higher dimensions, one has to assume the evenness condition in addition to the K-bi-invariance. (2) Kφ is G-invariant, i.e., Kφ(gz, gw) = Kφ(z, w). A K-bi-invariant function on G is determined by its values on A by the polar G KAK decomposition = . If g = k1ak2, then b is uniquely determined up to a permutation of its diagonal elements. Hence a function on A that ia invariant under permutations extends to a K-bi-invariant function on G. 106 2. ANALYSIS ON TOPOLOGICAL GROUPS

Recall that the polar height σ is the K-bi-invariant function such that σ ≥ 0 and α σ2(g) = σ2(b) = (ln y(b))2 = (ln b )2, ∈ G where g = k1bk2 is the polar decomposition of g .

PROPOSITION 2.3.17. Let g = uak = k1bk2 be the Iwasawa, respectively polar, decomposition. Then (2.3.6.2) σ(a) ≤ σ(b) or | ln a| ≤ | ln b|. In general reductive Lie group theory, this result is due to Harish-Chandra. The inequality (2.3.6.2) can be viewed from a differential geometric point of view. The space G/K is a Cartan-Hadamard manifold, whose exponential map is metric semi-increasing. The U-components are perpendicular to A. The metric increasing property shows that the law of cosines with the inequality in the right direction is valid on G/K, and the desired inequality (2.3.6.2) then follows at once.

C2 PROOF. Let e1, e2 be the unit column vectors of . Let a = diag(a1, a2) and ≥ b = diag(b1, b2). We may without loss of generality that a1, b1 1. It sufﬁces to ≤ UA AU ∈ U prove that a1 b1. From = , we have ua = av for some v . Writing −1 ′ ′ −1 ∈ K av = ua = k1bk2k = k1bk with k := k2k , we have 2 | |2 | ′ |2 a1 = ave1 = k1bk e1 , ′ where | · | stands for he Euclidean metric on C2. Since k is unitary, there are num- ∈ C ′ | |2 | |2 bers c1, c2 such that k e1 = c1e1 + c2e2 and c1 + c2 = 1. Then | ′ |2 | |2 | |2 2| |2 2| |2 bk e1 = b1c1 + b2c2 = b1 c1 + b2 c2 . ≥ 2 It sufﬁces to show that for y 1 (y := b1) we have | |2 −1| |2 ≤ y c1 + y c2 y. | |2 − | |2 Using c2 = 1 c1 yields 2 2 2 − |c | (y − 1) + 1 (y − 1) + 1 y|c |2 + y 1|c |2 = 1 ≤ = y 1 2 y y which proves the inequality (2.3.6.2).

The following deep property is due to Cartan and Harish-Chandra. ′ THEOREM 2.3.18. (Triangle inequality) For g, g ∈ G, we have ′ ′ (2.3.6.3) σ(gg ) ≤ σ(g) + σ(g ). Lang (1999) provided an elementary self-contained treatment of Theorem 2.3.18 on Bruhat-Tits spaces.

DEFINITION 2.3.19. We then deﬁne the polar distance −1 (2.3.6.4) dP(z, w) := σ(z w), z, w ∈ G, which is the associated point-pair invariant. According to Theorem 2.3.18, dP is a distance deﬁned on G/K. 2.3. HARISH, MELLIN AND SPHERICAL TRANSFORMS 107

G K ≈ 7→ ∗ ∗ Recall from Theorem 2.2.1 that / Pos2 by g gg (with g = g¯). Let

α b1 2 yP(g) = b = = b1 b2 ∈ G be the polar y-coordinate of g = k1bk2 . Then σ | | K (2.3.6.5) (g) = ln yP(g) = dP(g , eG/K), G K where eG/K is the unit coset of / .

THEOREM 2.3.20. The function dP is a distance function on G/K.

R PROOF. Lang (1999) deﬁned a distance on Pos2( ). The analogous treatment for Posn(C) and SPosn(C) runs exactly the same way. Thus we get a natural G- invariant distance on G/K, which is a Riemannian manifold, with G acting by translation as a group of isometries.

CHAPTER 3

Fundamental groups

109