Topology I: Elementary Topology Yi Li

Home , Excluded point topology, Sorgenfrey plane, Topological pair, Trivial topology

Topology I: Elementary Topology

Yi Li

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

Contents

Main References 1 Chapter 1. Basic topology 3 1.1. Metric spaces 3 1.2. Topological spaces 13 1.3. Constructions 38 1.4. Topological groups and matrix Lie groups 52 1.5. Connectedness and compactness 69 Chapter 2. Analysis on topological groups 83 2.1. Haar measures 83 2.2. Iwasawa’s decomposition 89 2.3. Harish, Mellin and spherical transforms 94 Chapter 3. Fundamental groups 111 3.1. Homotopy 111 3.2. H-spaces 122 1 3.3. π1(S ) 137

v Main References

• Kelley, John L. General topology, Reprint of the 1955 edition [Van Nos- trand, Toronto, Ont.], Graduate Texts in Mathematics, 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. • 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 • Munkres, James R. Topology: a ﬁrst course, Prentice-Hall, Inc., Englewood Cliffs, N. J., 1975. xvi+413 pp. • Burago, Dmitri; Burago, Yuri; Ivanov, Sergei. A course in metric geometry, Graduate Studies in Mathematics, 33, American Mathematical Society, Providence, RI, 2001. xiv+415 pp. ISBN: 0-8218-2129-6 • Spanier, Edwin H. Algebraic topology, Corrected reprint of the 1966 original, Springer-Verlag, New York. xvi+528 pp. ISBN: 0-387-94426-5

CHAPTER 1

Basic topology

1.1. Metric spaces Over the n-dimensional Euclidean space Rn, we have a natural distance function r d(x, y) := ∑ (xi − yi)2, x = (x1, ··· , xn), y = (y1, ··· , yn), 1≤i≤n which satisﬁes the following properties: d is always nonnegative (with zero if and only if x = y), symmetric, and d(x, z) ≤ d(x, y) + d(y, z).

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) 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.

3 4 1. BASIC TOPOLOGY

Then ∼ is an equivalence relation on X. Deﬁne Xb := X/ ∼= {[x] : x ∈ X}, db([x], [y]) := d(x, y). Then (X/d, d) := (Xb, db) is a metric space. Exercise 1.1.2. Show that dbis well-deﬁned and (X/d, d) is a metric space.

Exercise 1.1.3. Let X = R2 and deﬁne d((x, y), (x0, y0)) := |(x − x0) + (y − y0)|. 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 Z 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 q 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 Eu- 2 clidean metric from R . We can deﬁne another (intrinsic) metric dint by

dint(x, y) := the length of the shorter arc between them. 1 Note that dint(x, y) ∈ [0, π] 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. 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 e > 0 such that B(x, e) ⊆ 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 ; (2) ∪i∈IUi ∈ T for any collection {Ui}i∈I ⊂ T ; (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 ; 1.1. METRIC SPACES 5

(2) ∪i∈IUi ∈ T for any collection {Ui}i∈I ⊂ T ; (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. • S ⊂ X is an e-net if Se := {x ∈ X : d(x, S) = infy∈X d(x, y) ≤ e} = X. • X is totally bounded if for any e > 0 there exists a ﬁnite e-net in X. The following theorem is a basic result. Theorem 1.1.7. Let (X, d) be 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

(a)( A is closed under restrictions) If γ : [a, b] → X in A , then γ|[c,d] ∈ A for any subinterval [c, d] ⊆ [a, b]; (b)( A is closed under concatenations of paths) If γ1 : [a, c] → X and γ2 : [c, d] → X are admissible with γ1(c) = γ2(c), then the obvious product γ := γ1 · γ2 : [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; (d) (Length of paths is additive) L(γ|[a,b]) = L(γ|[a,c]) + L(γ|[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) := L(γ[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);

1 A sequence {xn}n∈N in X is said to be Cauchy if for any e > 0 there exists an integer n0 ∈ N such that d(xn, xm) < e whenever n, m ≥ n0. 2That is, γ−1(U) is open in I for any open set U in X. 6 1. BASIC TOPOLOGY

(g) (Length structures agree with the topology of X) For any x ∈ X and open set Ux 3 x, one has

inf {L(γ) : γ(a) = x, γ(b) ∈ X \ Ux} > 0. γ∈A Let (X, T ) is a Hausdorff space with a length structure (A , L). Deﬁne

(1.1.2.1) dL(x, y) := inf {L(γ)|γ : [a, b] → X, γ(a) = x, γ(b) = y} . γ∈A

If there is no path between x and y, we set dL(x, y) = +∞. We say a length structure (A , L) is complete if for any x, y ∈ X with dL(x, y) < +∞, there exists a path γ ∈ A joining x and y such that dL(x, y) = L(γ).

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 (X, d). If γ : [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, {yi}0≤i≤N) := ∑ d (γ(yi−1), γ(yi)) . 1≤i≤N The length of γ is given by

(1.1.2.2) Ld(γ) := sup Σ (γ, {yi}0≤i≤N) . {yi}0≤i≤N

The path γ is said to be rectiﬁable if Ld(γ) < +∞. The length structure (A , L) now is deﬁned by

(1.1.2.3) A := {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) defined in (1.1.2.3), we have the following properties: (a) L(γ) ≥ d(γ(a), γ(b)) for any path γ : [a, b] → X; (b) L(γ|[a,c]) + L(γ|[c,b]) = L(γ) for any path γ : [a, b] → X and a < c < b; (c) If γ is rectifiable, then L(γ, c, d) := L(γ|[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- wise convergence; that is, if {γi}i∈N is a sequence of rectifiable paths in X with the same domain and γi(t) converges to γ(t) for every given t ∈ [a, b], then L(γ) ≤ lim infi→∞ L(γi).

PROOF. (a) For any partition {yi}0≤i≤N of [a, b] we have

Σ (bγ, {yi}0≤i≤N) = ∑ d (γ(yi−1), γ(yi)) ≥ d(γ(y0), γ(yN)) = d(γ(a), γ(b)) 1≤i≤M by the triangle inequality, so that L(γ) ≥ d(γ(a), γ(b)). 1.1. METRIC SPACES 7

0 00 (b) Given partitions {yi}0≤i≤N0 and {yj }0≤j≤N00 of [a, c] and [c, b], respectively. 0 00 Then {yi}0≤i≤N0 ∪ {yj }0≤j≤N00 forms a partition of [a, b] and

0 00 L(γ) ≥ Σ γ, {yi}0≤i≤N0 ∪ {yj }0≤j≤N00 0 0 00 00 = ∑ d γ(yi−1), γ(yi) + ∑ d γ(yj−1), γ(yj ) 1≤i≤N 1≤j≤N00 0 00 = Σ γ|[a,c], {yi}0≤i≤N0 + Σ γ|[c,b], {yj }0≤j≤N00 .

Hence L(γ) ≥ L(γ|[a,c]) + L(γ|[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

L(γ|[a,c]) + L(γ|[c,b]) ≥ Σ (γ, {yi}0≤i≤N ∪ {c}) ≥ Σ (γ, {yi}0≤i≤N) and then L(γ|[a,c]) + L(γ|[c,b]) ≥ L(γ). (c) We only prove that L(γ, c, d) is continuous in d ∈ (a, b]. For any e > 0, since L(γ) < +∞), there exists a partition {yi}0≤i≤N of [a, b] such that 0 ≤ L(γ) − Σ(γ, {yi}0≤i≤N) < e. 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 ) ) < e [yj−1,d] j−1 i 0≤i≤N 0 by (a) and (b). For any d ∈ [yj−1, d], one has L γ| − L γ| 0 ≤ e + d γ(y ), γ(d) − L γ| 0 [yj−1,d] [yj−1,d ] j−1 [yj−1,d ] 0 ≤ e + d γ(yj−1), γ(d) − d γ(yj−1), γ(d ) ≤ e + d(γ(d0), γ(d)).

Since γ is continuous, it follows that L(γ, c, d0) − L(γ, c, d) = L(γ, d0, d) ≤ 2e when d0 is very close to d. Thus L(γ, c, d) is continuous in d. (d) We ﬁrst assume that L(γ) < +∞. For any e > 0 there exists a partition {yi}0≤i≤N of [a, b] such that L(γ) − Σ(γ, {yi}0≤i≤N) < e. Choose j 1 so that d(γj(yi), γ(yi)) < e/N for each 0 ≤ i ≤ N. Compute

L(γ) ≤ Σ(γ, {yi}0≤i≤N) + e e ≤ Σ(γ , {y } ) + e + · 2 · N j i 0≤i≤N N ≤ L(γj) + 3e where we used d (γ(yi−1), γ(yi)) ≤ d γ(yi−1), γj(yi−1) + d γj(yi−1), γj(yi) + d γ(yi), γj(yi) e ≤ · 2 + d γ (y ), γ (y ) . N j i−1 j i

Therefore L(γ) ≤ lim infj→∞ L(γj). 8 1. BASIC TOPOLOGY

When L(γ) = +∞, we choose a partition {yi}0≤i≤N such that Σ(γ, {yi}0≤i≤N) > 1/e. Then 1 e < Σ(γ, {y } ≤ ≤ ) ≤ Σ(γ , {y } ≤ ≤ ) + · 2 · N ≤ L(γ ) + 2e e i 0 i N j i 0 i N N j for j 1. Letting j → ∞ yields L(γj) → ∞.

For a metric space (X, d), we have a length structure (A , 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 := dR2 |X we see that d((1/n, 0), (0, 0)) = 1/n → 0 as ∧ 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 r r 1 1 = 1 + + 1 + > 2. n2 m2 Given a metric space (X, d), we have the following diagram:

(X, d) −−−−→ (A , L = Ld)   y ∧ (X, d = dL)   y ∧ ∧ (X (d ) = d ) ←−−−− ( L = L ∧ ) , bL A , b d by Theorem 1.1.10. A natural question is whether d = (d∧)∧.

Theorem 1.1.12. Let (X, d) be a metric space and db the intrinsic metric induced by d. (a) If Ld(γ) < +∞, then Ld(γ) = Ld∧ (γ). (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 Ld(γ) ≥ d(γ(a), γ(b)) = d(x, y) ∧ which implies d(x, y) ≤ d (x, y) and hence Ld(γ) ≤ Ld∧ (γ). Conversely, for any path γ : [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 db 1≤i≤N 1≤i≤N 1.1. METRIC SPACES 9 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 Z b (1.1.2.5) L(x, y) = Ld(γ) := |γ˙ (t)|gdt, a where n 1 o (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”.

Let (Z, dZ) be a metric space. Given an e > 0, the e-neighborhood of a subset S ⊂ Z is given by (1.1.3.1) Se := z ∈ Z : dZ(z, S) = inf dZ(z, x) ≤ e . x∈S

We say a subset S ⊂ X is an e-net if Se = X. The Hausdorff distance of A, B ⊂ Z is deﬁned to be the quantity Z (1.1.3.2) dH(A, B) := inf {e > 0 : A ⊂ Be and B ⊂ Ae} . Z If no such e, 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, d ) → (Z, d )  X Z   Z g : (Y, dY) → (Z, dZ)  := inf dH ( f (X), g(Y)) : . ( ) are isometric embeddings Z,dZ , f,g    into a metric space (Z, dZ) 

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 e-isometry if dis( f ) ≤ e and f (X) is an e-net in Y. The following result characterizes the Gromov-Hausdorff distances with e- isometries. 10 1. BASIC TOPOLOGY

Lemma 1.1.14. Let (X, dX) and (Y, dY) be metric spaces and e > 0. (a) If dGH((X, dX), (Y, dY)) < e, then there exist e-isometries f : (X, dX) → (Y, dY) and g : (Y, dY) → (X, dX). (b) If f : (X, dX) → (Y, dY) is an e-isometry, then dGH((X, dX), (Y, dY)) < 2e.

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 e-pointed-isometry f : (X, dX, x0) → (Y, dY, y0) is a pointed map f satisfying

|dX(x1, x2) − dY( f (x1), f (x2))| < e −1 −1 −1 for any x1, x2 ∈ BX(x0, e ), and BY(y0, e ) ⊂ ( f (BX(x0, e )))e. 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 e-pointed-isometries  := inf e > 0 : f : (X, dX, x0) → (Y, dY, y0) and .(1.1.3.5)  g : (Y, dY, y0) → (X, dX, x0) 

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

PROOF. By deﬁnition, there exist 2e-pointed-isometries f : (X, dX, x0) → (Y, dY, y0) −1 and g : (Y, dY, y0) → (X, dX, x0). For any x ∈ BX(x0, e ) we have |dX(x, x0) − −1 −1 dY( f (x), y0)| < e and then f (BX(x0, e )) ⊆ BY(y0, e + e). ( ) 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. 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 e > 0 there exists N(e) ∈ N such that any X ∈ M contains an e-net S ⊆ X consisting of at most N(e) 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 pt for any e, ρ > 0, there exists N(e, ρ) ∈ N such that any (X, x0) ∈ M contains an e-net S ⊆ B(x0, ρ) ⊆ X consisting of at most N(e, ρ) points, 1.1. METRIC SPACES 11 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

(1.1.3.6) (TxX, dx, 0x) := lim (X, αkd, x) αk→∞ provided this limit (in the sense of pointed Gromov-Hausdorff distance) exists for any sequence αk → ∞ and is independent of {αk}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 (TxM, g(x), 0x). ∼ (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, dAX, 0) := lim (X, ωid, x) ωi→0 provided this limit exists and is independent (up to isometry) of the sequence {ωi}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.)

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 q 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)]) . 12 1. BASIC TOPOLOGY

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

(1.1.3.12) dgCone ((x1, r1), (x2, r2)) = dCone(M) ([(x1, r1)], [(x2, r2)]) .

In fact, given two points x1, x2 ∈ M, let γ : [0, d(x1, x2)] → M be a unit speed minimal geodesic joining x1 to x2, where d := dg. Given r1, r2 ∈ (0, ∞), join the two points (x1, r1), (x2, r2) ∈ M × (0, ∞) by paths

γ : [0, d(x1, x2)] −→ M × (0, ∞), u 7−→ (γ(u), r(u)) where r : [0, d(x1, x2)] → (0, ∞) satisﬁes r(0) = r1 and r(d(x1, x2)) = r2. The length of γ with respect to gCone is given by

Z d(x1,x2) q 2 2 Lg (γ) = r(u) + r˙(u) du. Cone 0 For any variation ar = s with s(0) = s(d(x1, x2)) = 0, we have Z d(x ,x ) −1/2 1 2 h 2 2i aLgCone (γ) = r(u) + r˙(u) [r(u)s(u) + r˙(u)s˙(u)] du 0 Z d(x1,x2) h i−1/2 = r(u)2 + r˙(u)2 r(u)s(u) 0 −3/2 Z 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

Z d(x1,x2) h i−3/2 h i = r(u)2 + r˙(u)2 −r(u)r¨(u) + 2r˙(u)2 + r(u)2 r(u)s(u)du. 0 The Euler-Lagrange equation is now of the form 1 r(u) = a cos u + b sin u for some a, b ∈ R. Let α := d(x1, x2) ∈ [0, π] and consider the Euclidean planar triangle with side-angle-side equal to r1-α-r2 and vertices with polar coordinates (r1, 0), (0, 0), and (r2, α). By the law of sines, we obtain −1 −1 r1 1 r − r cos α r − cos α a = , b = 2 1 = 2 r1 sin α r1 sin α and Z α q 2 2 dgCone ((x1, r1), (x2, r2)) = r(u) + r˙(u) du 0 p Z α du p − cos u α = 2 + 2 = 2 + 2 a b 2 a b 0 (a cos u + b sin u) b(b sin u + a cos u) 0 1.2. TOPOLOGICAL SPACES 13

p ab2 sin α q = 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 ; (2) ∪i∈IUi ∈ T for any collection {Ui}i∈I ⊆ T ; (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

TA := {A ∩ U|I ∈ T }.

Then clearly (A, TA) 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   Bm := x = (x1, ··· , xm) ∈ Rm |x| = ∑ (xi)2 < 1 .  1≤i≤m 

When m = 2, we call D := B2 the (open) unit disk. • The (closed) unit ball in Rm: Dm ≡ Bm ≡ Bm := {x ∈ Rm||x| ≤ 1}. When m = 2, we call D := D2 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. 14 1. BASIC TOPOLOGY

• 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) ä Ext(A) ä Bdy(A). 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. 1.2. TOPOLOGICAL SPACES 15

– 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 • 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 .

A continuous map f : (X, TX) → (Y, TY) between topological spaces is a −1 map f : X → Y such that f (U) ∈ TX whenever U ∈ TY. We say that a map f : (X, TX) → (Y, TY) 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.

Theorem 1.2.4. A map f : (X, TX) → (Y, TY) is continuous if and only if f is continuous at each point in X. 16 1. BASIC TOPOLOGY

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 [ f (U) = Vx ⊂ TX x∈ f −1(U) and f is continuous.

Example 1.2.5. (1) Every constant map f : (X, TX) → (Y, TY) is continuous. (2) The identity map 1X : (X, TX) → (X, TX) is continuous. (3) If f : (X, TX) → (Y, TY) is continuous, so is the restriction of f to any open subset U ⊆ X. (4) For continuous maps f : (X, TX) → (Y, TY) and g : (Y, TY) → (Z, TZ), the composite map g ◦ f : (X, TX) → (Z, TZ) is also continuous. We also observe that for any continuous map f : (X, TX) → (Y, TY) we have f ◦ 1X = f and 1Y ◦ f = f.

We say a map f : (X, TX) → (Y, TY) is a homeomorphism if f is bijective and −1 f, f are continuous. In this case, we say that (X, TX) and (Y, TY) are homeomorphic or topologically equivalent, and denote by X ≈ Y.

Exercise 1.2.6. (1) Let f : (X, TX) → (Y, TY) be a bijection between topological spaces. Show that f is homeomorphic if and only if f (TX) = TY, that is, U ∈ TX if and only if f (U) ∈ TY. (2) Let f : (X, TX) → (Y, TY) be a homeomorphism and U ∈ TX. Show that f (U) ∈ TY and f |U is a homeomorphism from (U, TU) → ( f (U), T f (U)). (3) Any translation m m R −→ R , x 7−→ 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

2 (x, y, z) F : C −→ S , (x, y, z) 7−→ p . x2 + y2 + z2 Show that F is homeomorphic. (6) Let X := [0, 1) ⊆ R and Y := S1 ⊆ C. Deﬁne √ √ e : X −→ Y, e(s) := e2π −1s = cos(2πs) + −1 sin(2πs). Show that e is continuous and bijective, but is not homeomorphic.

A map f : (X, TX) → (Y, TY) is open if f (U) ⊆ TY for all U ∈ TX; f is closed if f (C) is closed in Y for all closed set C ⊆ X. 1.2. TOPOLOGICAL SPACES 17

Exercise 1.2.7. Let f : (X, TX) → (Y, TY) be a bijective continuous map. Show that the following are equivalent: • f is a homeomorphism. • f is open. • f is closed.

Proposition 1.2.8. Let f : (X, TX) → (Y, TY) 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 f (V) ∈ TX and x ∈ U. Hence by Exercise 1.2.1 (4) again, we can find a point x0 ∈ A ∩ U, so that y0 := f (x0) ∈ f (A) ∩ V and f (x) ∈ f (A). 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 so that x ∈ A. Hence A = A. Next we show the continuity of f. Let V ∈ TY 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 a map f : (X, TX) → (Y, TY) 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). 18 1. BASIC TOPOLOGY

A map f : (X, TX) → (Y, TY) between topological spaces is said to be locally homeomorphic if every point x ∈ X has a neighborhood U ⊆ X such that f (U) ∈ TY 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 – if B1, B2 ∈ B and x ∈ B1 ∩ B2, there exists an element B3 ∈ B such that x ∈ B3 ⊆ B1 ∩ B2. (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. (2) If B1, B2 ∈ B and x ∈ B1 ∩ B2, there exists an element B3 ∈ B 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. • TB is closed under arbitrary unions. For any collection {Ui}i∈I ⊆ TB, write each Ui as [ Ui = Bij. j∈Ji Then [ [ [ Ui = Bij ∈ TB. i∈I i∈I j∈Ji 1.2. TOPOLOGICAL SPACES 19

• 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 B1, B2 ∈ B such that x ∈ B1 ⊆ U and x ∈ B2 ⊆ V. By condition (2), there exists Bx ∈ B 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 0 is another topology on X for which B is a basis, then B ⊆ T and, 0 for every U ∈ T , we have U = ∪i∈I Bi, for a collection {Bi}i∈I of B. Hence 0 0 0 U ∈ TB and then T ⊆ TB. Since B ⊆ T , it follows that TB ⊆ T . Therefore 0 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).

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 induces the metric topology T . 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.

PROOF. If X is second countable, then X has a countable basis B = {Bi}i∈N. Choose xi ∈ Bi for each i ∈ N and let A := {xi}i∈N. Then A is a countable dense set. 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 B is a countable basis. 20 1. BASIC TOPOLOGY

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 sequence of maps { f n}n∈N between X and Y converges uniformly to a map f : X → Y if ! lim sup d( f (x), f (x)) = 0. → i n ∞ x∈X

In this case, we write it as f n ⇒ f.

Theorem 1.2.12. Let { f n}n∈N, f be maps between a topological space (X, T ) and a metric space (Y, d). If f n are continuous and f n ⇒ f, then f is continuous.

e PROOF. Given e > 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 e d( f (x), f (x0)) < , x ∈ N. m0 n0 3 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 0 e e e < + + = e. 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 0 on X whose open sets satisfying P, then T -open sets are T 0-open. In this case we use the notion T ≺ T 0. (2) T is the largest topology or ﬁnest topology satisfying P, if for any topology T 0 on X whose open sets satisfying P, then T 0-open sets are T - open. In this case, we use the notion T 0 ≺ 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. (2) (Discrete topology) Any set X can also be equipped with another topology Td = 2X. This is the largest topology on X. (3) (Finite complement topology) Given an infinite set X. The collection of subsets whose complement are finite, together with the empty set, forms a topology Tfc. (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. 1.2. TOPOLOGICAL SPACES 21

(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. (9) Show that (R, Tfc) 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

(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 g0 : Y → X is a left inverse and 00 0 00 g : Y → X is a right inverse, then g ◦ f = 1X and f ◦ g = 1Y so that 0 0 0 00 00 00 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. An equivalence f : X → Y is also called an isomorphism. 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 22 1. BASIC TOPOLOGY

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 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.

The fundamental groups π1 roughly speaking is a map from Top∗ to Group; namely associated to a pointed topological space (X, x0) a group π1(X, x0). We also can deﬁne higher homotopy groups πi(X, x0) which turns out to be Abelian for any i ≥ 2.

Let C be a category. A subcategory C0 of C is itself a category and satisﬁes • any object in C0 is also object in C, 0 • For any X, Y ∈ Ob(C ), we have HomC0 (X, Y) ⊆ HomC(X, Y), and 0 • 1X ∈ HomC0 (X, X) for any X ∈ Ob(C ). A subcategory C0 is called a full subcategory if moreover

HomC0 (X, Y) = HomC(X, Y) for any X, Y ∈ Ob(C0). 1.2. TOPOLOGICAL SPACES 23

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 → C0 between two categories consists of (1) a map F : Ob(C) → Ob(C0), and (2) for any two objects X, Y ∈ Ob(C), a map 0 F : HomC(X, Y) → HomC0 (F(X), F(X )). These data satisfy

(1.2.2.1) F(1X) = 1F(X), F( f ◦ g) = F( f ) ◦ F(g). A contravariant functor G : C → C0 is a covariant functor from C◦ to C0. Example 1.2.16. (1) Let C be a category and take an object X ∈ Ob(C). Deﬁne

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

Then HomC(X, ·) is a functor, while HomC(·, X) is a contravariant functor. (2) The forgetful functor F : Top → Set. (3) The fundamental group functor π1 : Top∗ → Group, given by (X, x) 7→ π1(X, x). 0 Let F1, F2 : C → C be two functors. A morphism or natural transformation Θ from F1 to F2 consists of

Θ(X) ∈ HomC0 (F1(X), F2(X)) whenever X ∈ Ob(C). These data satisfy the following commutative diagram: Θ(X) F1(X) −−−−→ F2(X)   ( )  ( ) (1.2.2.4) F1 f y yF2 f F2( f ) ◦ Θ(X) = Θ(Y) ◦ F1( f )

F1(Y) −−−−→ F2(Y) Θ(Y) for any X, Y ∈ Ob(C) and any f ∈ HomC(X, Y). Deﬁnition 1.2.17. For two categories C and C0, deﬁne a new category Func(C, C0) as follows: Ob(Func(C, C0)) := {functors from C to C0},

HomFunc(C,C0)(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). 24 1. BASIC TOPOLOGY

Exercise 1.2.18. (1) Show that Func(C, C0) 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 → C0 be a functor. (1) F is fully faithful if for any objects X, Y ∈ Ob(C), the map

HomC(X, Y) −→ HomC0 (F(X), F(Y)) is bijective. (2) F is an equivalence of categories if F is fully faithful and for any object X0 ∈ Ob(C0) there exists X ∈ Ob(C) and an isomorphism X0 → F(X).

Exercise 1.2.19. F ∈ Ob(Func(C, C0)) is an equivalence of categories if and only if there exists F0 ∈ Ob(Func(C0, C)) and isomorphisms 0 0 0 0 F ◦ F → 1C0 in Func(C , C ) and F ◦ F → 1C in Func(C, C). 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.

PROOF. (1) To f ∈ HomC∨ (HomC(·, X), F), we associate φ( f ) ∈ F(X) as follows:

f (X) : HomC(X, X) −→ F(X), 1X 7−→ φ( f ) := f (X)(1X).

Conversely, to s ∈ F(X), we can associate ψ(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. TOPOLOGICAL SPACES 25

1.2.3. Subspaces. Let (X, T ) be a topological space and S ⊆ X is a subset. S ⊆ X is a subspace if (S, TS) is a topological space, where TS := {U ∩ S|U ∈ T } is the relative topology or subspace topology. The inclusion map

(1.2.3.1) ιS : (S, TS) −→ (X, T ) 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 relatively closed in S if U is open or closed in the subspace topology TS. 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. (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 N0 := N ∩ S is a neighborhood of x in S, so N0 ∩ U 6= ∅. Because U ⊂ S, we get N ∩ U 6= ∅ and x ∈ U ∩ S. Conversely, let x ∈ U ∩ S and N0 is any neighborhood of x in S. By deﬁnition, N0 0 0 0 contains an open subset V ∈ TS satisfying V 3 x. Write V = N ∩ S for some N ∈ T . Then N itself is a neighborhood of x in X, so N ∩ U 6= ∅. Consequently, V0 ∩ U 6= ∅ and N0 ∩ U 6= ∅, 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 N0 := 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) Suppose that (X, T ) is a topological space and S ⊆ X is a subspace. For any topological space (Y, TY), a map f : (Y, TY) → (S, TS) is continuous if and only if the composite map ιS ◦ f : (Y, TY) → (X, T ) is continuous: X ? O ιS◦ f ι S Y / S f 26 1. BASIC TOPOLOGY

(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.

PROOF. (1) Suppose that ιS ◦ f : (Y, TY) → (X, T ) is continuous. If U ∈ TS, −1 then U = V ∩ S = ιS (V) for some V ∈ T . Thus −1 −1 −1 −1 f (U) = f (ιS (V)) = (ιS ◦ f ) (V) ∈ TY. Conversely, suppose that f is continuous. For any V ∈ T , we have −1 −1 −1 −1 (ιS ◦ f ) (V) = f (ιS (V)) = f (S ∩ V) ∈ TY. Hence ιS ◦ f is continuous. 0 (2) Suppose S is equipped with some topology TS that satisﬁes the characteristic property (1). For convenience, let Ss denote the topological space (S, TS) and 0 0 Ss0 denote the topological space (S, TA). To prove TS = TS we should verify that (by Exercise 1.2.6) the identity map 1S : S → S is a homeomorphism between Ss and Ss0 . Clearly that the inclusion map ιs : Ss → (X, T ) is continuous. Since Ss0 satis- ﬁes the characteristic property, it follows that the inclusion map ιs0 : Ss0 → (X, T ) is also continuous. Consider the following two diagrams: X X ◦ > O ◦ > O ιs0 1ss0 }} ιs 1s0s }} }} ιs0 } ιs }} }} }} }} Ss / Ss0 Ss0 / Ss 1ss0 1s0s Here 1ss0 : Ss −→ Ss0 , 1s0s : Ss0 −→ Ss. Because ιs0 ◦ 1ss0 = ιs and ιs ◦ 1s0s = ιs0 , we, using again the characteristic property, obtain the continuities of both 1ss0 and 1s0s. Thus 1S : S → S is a homeomorphism between Ss and Ss0 .

Corollary 1.2.24. Let f : (X, TX) → (Y, TY) 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.

PROOF. (1) Because f |S = f ◦ ιS. (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. (2) If B is a basis for T , then BS := {B ∩ S|B ∈ B} is a basis for TS. 1.2. TOPOLOGICAL SPACES 27

(3) Every subspace of a ﬁrst countable space is ﬁrst countable. (4) Every subspace of a second countable space if second countable.

PROOF. (1) It sufﬁces to prove (TS)R = TR. Let U ∈ (TS)R. Then U = R ∩ V for some V ∈ TS. But V can be written as V = S ∩ W for some W ∈ T , we obtain U = R ∩ (S ∩ W) = R ∩ W ∈ TR. Conversely, Let U ∈ TR. Then U = R ∩ W for some W ∈ T and

U = R ∩ V, V := S ∩ W ∈ TS.

Hence U ∈ (TS)R. (2) By deﬁnition of subspace topology, BS ⊆ TS. For any U ∈ TS, we have U = S ∩ W for some W ∈ T . Since B is a basis for T , it follows that [ W = Bi, Bi ∈ B. i∈I

Therefore U = ∪i∈I Bi ∩ S with Bi ∩ S ∈ BS. Thus BS is a basis for TS. (3) Assume that S is a subspace of a first countable topological space (X, T ). For any x ∈ S, we can find a countable neighborhood basis Bx = {Bi}i∈N at x in X. Then BS,x := {Bi ∩ S}i∈N is a countable neighborhood basis at x in S, so (S, TS) 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 of continuity below Proposition 1.2.8, that is, f : (X, TX) → (Y, TY) 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 −1 B respectively. Hence h (C) is closed in X. 1.2.4. Separation axioms. Let (X, T ) be a topological space. (1) T0-space: for any x 6= 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 6= 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 6= y ∈ X, there exist U, V ∈ T such that x ∈ U, y ∈ V, and U ∩ V = ∅. 28 1. BASIC TOPOLOGY

(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 6= 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) 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) R` is regular. Recall that R` 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 R` so that, according to Theorem 1.2.27 below, R` is T1. To prove normality, we suppose that A and B are disjoint closed subsets in R`. For a ∈ A choose a basis element [a, xa) not intersecting B and similarly for any b ∈ B choose a basis element [b, xb) not intersecting A. Then open subsets [ [ U := [a, xa), V := [b, xb) a∈A b∈B are disjoint open subsets about A and B respectively. 2 (3) The Sorgenfrey plane R` := R` × R` 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.

PROOF. (1) Let (X, T ) 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. 1.2. TOPOLOGICAL SPACES 29

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 that y ∈ Yy but x ∈/ Uy. The similar result holds for Vx. Hence (X, T ) is T1. (2) Assume first that (X, T ) is T2 and S is a subspace of (X, T ). For any x 6= y in S, we can find two disjoint open subsets U, V ∈ T of x, y respectively, satisfying U ∩ V = ∅. Then (U ∩ S) ∩ (V ∩ S) = ∅ so (S, TS) is Hausdorff. Next assume that (X, T ) is T3 and S is subspace of (X, T ). Clearly (S, TS) 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 3 x and V ⊇ B. Then U ∩ S and V ∩ S are disjoint open subsets in Y containing x and B respectively. Therefore (S, TS) is regular. (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 3 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 ea > 0 so that B(a, ea) ∩ B = ∅. Similarly, for any b ∈ B there exists eb > 0 so that B(b, eb) ∩ A = ∅. Define [ [ U := B(a, ea/2), V := B(b, eb/2). a∈A b∈B Then U and V both are open and contain A and B respectively. We claim that U ∩ V = ∅. Otherwise, for z ∈ U ∩ V, we have z ∈ B(a, ea/2) ∩ B(b, eb/2) for some a ∈ A and b ∈ B. Then e + e d(a, b) ≤ d(a, z) + d(z, b) < a b . 2

If ea ≤ eb, then d(a, b) < eb so that a ∈ B(b, eb). If eb ≤ ea, then b ∈ B(a, ea. 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. 30 1. BASIC TOPOLOGY

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 obtain a countable open subsets {Un}n∈N ⊆ B 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 0 [ 0 [ Un := Un \ Vi, Vn := Vn \ Ui. 1≤i≤n 1≤i≤n 0 0 Then Un and Vn are open, and [ 0 [ 0 A ⊆ Un, B ⊆ Vn. n∈N n∈N Finally, deﬁne 0 [ 0 0 [ 0 U := Un, V := Vn. n∈N n∈N 0 0 0 0 0 0 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 0 0 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)] 1.2. TOPOLOGICAL SPACES 31

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.

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 3 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. 32 1. BASIC TOPOLOGY

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 3 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. 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 1.2. TOPOLOGICAL SPACES 33

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.

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) 6= 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 34 1. BASIC TOPOLOGY

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

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|D = 0, φ|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. TOPOLOGICAL SPACES 35

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 0 ⊆ 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.

PROOF. Let B be a countable basis for X and let U be any open cover of X. Deﬁne B0 := {B ∈ B : B ⊂ U for some U ∈ U } 6= ∅. 0 0 It is clear that B is countable. For any B ∈ B let UB ∈ U be such that UB ⊃ B as above. Let 0 0 U := {UB ∈ U : B ∈ 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,¨ (R, Tep) is Lindelof,¨ and (R, Tfc) 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. 36 1. BASIC TOPOLOGY

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}. 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 SL2(Z) generated by 1 1 0 −1 , . 0 1 1 0

Prove Γ = SL2(Z). 0 (2) For any γ, γ ∈ SL2(Z) and any τ ∈ H, one has (γ · γ0)(z) = γ(γ0(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 1.2. TOPOLOGICAL SPACES 37 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 6= ∅.

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.

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) ä ∂Hm, a disjoint union. Actually this fact holds for any m- dimensional topological manifold with boundary M, i.e., M = Int(M) ä ∂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.

Let U := {Ui}1≤i≤n be an open cover of a topological space (X, T ).A partition of unity subordinate to U is a ﬁnite family of continuous functions φi : X → R, with the following properties: • 0 ≤ φi(x) ≤ 1 for all i and all x ∈ X; • Supp(φi) ⊆ Ui; • ∑1≤i≤n φi(x) = x for all x ∈ X.

Theorem 1.2.41. (Existence of ﬁnite partitions of unity) If U = {Ui}1≤i≤n is a ﬁ- nite open cover of a normal space (X, T ), then there exists a partition of unity subordinate to U .

PROOF. Basic idea is to construct two ﬁnite open covers W = {Wi}1≤i≤n and V = {Vi}1≤i≤n of X so that

Wi ⊆ Vi ⊆ Vi ⊆ Ui.

Then, by Theorem 1.2.30 we can ﬁnd {φi}1≤i≤n.

Step 1: We can shrink U = {Ui}1≤i≤n to an open cover V = {Vi}1≤i≤n of X such that Vi ⊆ Ui for each i. 38 1. BASIC TOPOLOGY

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 k+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.

By Theorem 1.2.30, for each i, there exists a continuous function ψi : X → [0, 1] −1 such that ψi(Wi) = 1 and ψi(X \ Vi) = 0. Because ψi (R \{0}) ⊆ Vi, we have Supp(ψi) ⊆ Vi ⊆ Ui. Since W covers X, the ﬁnite sum ∑1≤i≤n ψi(x) > 0 for all x ∈ X. Deﬁne ψi(x) φi(x) := , x ∈ X. ∑1≤j≤n ψj(x)

Then {φi}1≤i≤n is a partition of unity subordinate to U .

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.

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

If S is a subspace of a topological space (X, T ), then the inclusion map ιS : (S, TS) → (X, T ) is a topological embedding. 1.3. CONSTRUCTIONS 39

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 2 map. Denote π : R → R by π(x, y) := x. Then π|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 n+k of f is the subset Γ f ⊆ R defined by k Γ f := {(x, y) ∈ U × R |y = f (x)}, with the subspace topology inherited from Rn+k. Define two maps n+k Φ f : U −→ R , x 7−→ (x, f (x)) and π : Rn+k −→ Rn, (x1, ··· , xn, xn+1, ··· , xn+k) 7−→ (x1, ··· , xn). | Then Φ f is a continuous bijection from U onto Γ f and π Γ f is a continuous inverse for Φ f . Hence Φ f is a topological embedding and Γ f is homeomorphic to U, so that Γ 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 ± n ± On each Ui , we see that x ∈ S ∩ Ui if and only if s xi = ± 1 − ∑ (xj)2. 1≤j6=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, 2u1 2un |u|2 − 1 ϕ−1(u1, ··· , un) = , ··· , , , u = (u1, ··· , un). 1 + |u|2 1 + |u|2 1 + |u|2 40 1. BASIC TOPOLOGY

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,

2u1 2un 1 − |u|2 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).

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)) .

1.3.1. Product spaces. Let (Xi, Ti), 1 ≤ i ≤ n be topological spaces. Deﬁne

B := {U1 × · · · × Un|Ui ∈ Ti, 1 ≤ i ≤ n}. According to Proposition 1.2.10, B is a basis for some topology. We call such a ⊗ · · · ⊗ × · · · × topology the product topology TX1×···×Xn or T1 Tn on X1 Xn. The topological space (X1 × · · · × Xn, T1 ⊗ · · · ⊗ Tn) 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.

Consider the above product space (X1 × · · · × Xn, T1 ⊗ · · · ⊗ Tn). The canon- ical projection

(1.3.1.1) πi : X1 × · · · × Xn −→ Xi is actually continuous. Indeed, for any Ui ∈ Ti we have −1( ) = × · · · × × × × · · · × ∈ π Ui X1 Xi−1 Ui Xi+1 Xn TX1×···×Xn so that πi is continuous. 1.3. CONSTRUCTIONS 41

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 is continuous if and only if each of its component functions f i := πi ◦ f is continuous:

X1 × · · · × Xn r9 f rr rr πi rr rr rr 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. (a) The projection maps πi : X1 × · · · × Xn → Xi are all continuous and open. (b) The product topology is “associative” in the sense that the three product topologies 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 6= 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. (d) If for each i, Bi is a basis for the topology of Xi, then the set {B1 × · · · × Bn : Bi ∈ Bi} is a basis for the product topology on X1 × · · · × Xn. (e) If Ai is the subspace of Xi for i = 1, ··· , n, the product topology and the subspace 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. (5) If M1, ··· , Mn are topological manifolds of dimensions m1, ··· , mn, respectively, the product spade M1 × · · · × Mn 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. Conversely, if f is continuous, then for any Ui ∈ Ti 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 (1.3.1.2) Tm := S1 × · · · × S1 | {z } m 42 1. BASIC TOPOLOGY which is an m-dimensional topological manifold. When m = 2, we call T2 the torus.

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

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

∏ Uα Uα ∈ Tα and Uα = Xα for all but a ﬁnite number of α . α∈J

We call T 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.

Theorem 1.3.5. (1) Suppose that the topology Tα on each Xα is generated by a basis Bα, 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 B . ∏ α B = X for all the remaining indices α∈J α α

(2) Let πβ : ∏α∈J Xα → Xβ be the natural projection, that is, πβ(x) = xβ. Deﬁne [ n −1 o 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

(4) If each Xα, α ∈ J, is Hausdorff, then ∏α∈J Xα is Hausdorff in both the box and product topologies. (5) (Characteristic property of inﬁnite product spaces) Let (Xα, Tα)α∈J be an indexed family of topological spaces. For any topological space (B, TB), a map f : B → 1.3. CONSTRUCTIONS 43

∏α∈J Xα (with the product topology) ia continuous if and only if its component functions f α = πα ◦ f is continuous. The product topology is the unique topology on ∏α∈J Xα that satisﬁes this property.

PROOF. (1) is clear. (2) Let BS be the basis on ∏α∈A Xα consisting of ﬁnite intersections of elements of S . Then for any B ∈ BS we have \ B = π−1(U ) βi βi 1≤i≤n ··· ∈ = 6= ··· where β1, , βn are distinct and Uβi Tβi . Deﬁning Uα Xα if α β1, , βn, 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 ∏ Xα B F / α∈J FF FF FF πβ f FF β F" Xβ

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

−1 −1 −1 f (πβ (Uβ)) = f β (Uβ) ∈ TB. 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 product ω topology is replaced by the box topology. Consider R := ∏n≥1 Xn with Xn ≡ R, and deﬁne f R / Rω GG GG GG πn f GG n G# 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 44 1. BASIC TOPOLOGY be an open subset of Rω 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!

Theorem 1.3.8. A topological space (X, T ) is Hausdorff if and only if ∆X := {(x, y) ∈ X × X : x = y} ⊂ X × X is closed in (X × X, TX×X).

PROOF. Suppose ﬁrst that ∆X is closed in X × X with respect to the product topology. Then X × X \ ∆X is open. If x 6= y in X, then (x, y) ∈ U × V ⊂ TX×X ∩ (X × X \ ∆X). Hence X is Hausdorff. For any (x, y) ∈ X × X \ ∆X, we have x ∈ U ∈ TX and y ∈ V ∈ TY 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.

1.3.2. Quotient spaces. Let (X, TX) be a topological space, Y be a set, and q : X → Y a surjective map.

(1) Deﬁne a topology TY on Y by −1 (1.3.2.1) TY := {V ⊆ Y : q (V) ∈ TX}. This is called the quotient topology induced by q. In this topology, q : (X, TX) → (Y, TY) is continuous (called a quotient map). (2) (Y, TY) is a topological space and TY is the largest topology on Y which makes q continuous.

Let (X, TX) 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.

More generally, a surjective map π : (X, TX) → (Y, TY) between topological spaces is called a quotient map if the following condition holds: −1 V ∈ TY ⇐⇒ π (V) ∈ TX. It is clear that a quotient map is continuous. Example 1.3.9. (1) (Quotient spaces may not be Hausdorff) Let X = R and x ∼ y if and only x − y ∈ Q. Then X/ ∼ is noncountable. If TX := TR, then TX/∼ 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 6= 0. 1.3. CONSTRUCTIONS 45

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].

If Ui is given by n n i o Ui := [x] ∈ P : x 6= 0 , 1 ≤ i ≤ n + 1, n then (Ui)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 n then ϕi 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. (5) Let (Xi, Ti)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 ä1≤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

π1 : R × R −→ R, (x, y) 7−→ x and A := {(x, y) ∈ R × R|x ≥ 0 or y = 0}. Let q : A → R be obtained by restricting π1 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. (8) Consider the projection π1 : R × R → R, (x, y) 7→ x. Then π1 is continuous, surjective, open, but is not closed (consider C := {(x, y) ∈ R × R|xy = 1}). Let A := C ∪ {(0, 0)} and q := π1|A; then q is continuous, surjective, but is not a quotient map. 46 1. BASIC TOPOLOGY

(9) Deﬁne   a, x > 0, q : R −→ A := {a, b, c}, x 7−→ b, x < 0,  c, x = 0.

Then the quotient topology on A induced by q is TA = {{a}, {b}, {a, b}, {a, b, c}, ∅}.

If q : (X, TX) → (Y, TY) 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 : (X, TX) → (Y, TY) be a quotient map. For any topological space (B, TB), 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

(2) (Characterization of quotient maps) Let (X, TX) and (Y, TY) 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. (3) (Passing to the quotient) Suppose that q : (X, TX) → (Y, TY) is a quotient map. (B, TB) is a topological space, and f : (X, TX) → (B, TB) is any continuous map that is constant on the ﬁbers of q (i.e., if q(x) = q(x0), then f (x) = f (x0)). Then there exists a unique continuous map ef : (Y, TY) → (B, TB) such that f = ef ◦ q: X ?? f q ?? ?? ? Y / B ef In this situation, we say that the map f passes to the quotient or descends to the quotients. ( ) ( ) → ( ) 4 (Uniqueness of quotient spaces) Suppose q1 : X, TX Y1, TY1 and ( ) → ( ) q2 : X, TX Y2, TY2 are quotient maps that make the some identiﬁcations (i.e., 0 0 q1(x) = q1(x ) if and only if q2(x) = q2(x )). Then there is a unique homeomorphism ( ) → ( ) ◦ = ϕ : Y1, TY1 Y2, TY2 such that ϕ q1 q2. ( ) ( ) → ( ) 5 (Composition property of quotient maps) Suppose q1 : X, TX Y, TY1 and q2 : (Y, TY) → (Z, TZ) are quotient maps. Then their composition q2 ◦ q1 : (X, TX) → (Z, TZ) is also a quotient map. (6) A continuous surjective map q : (X, TX) → (Y, TY) is a quotient map if and only if it takes saturated open sets to open sets, or saturated closed sets to closed set. (7) Suppose that q : (X, TX) → (Y, TY) is a quotient map. The restriction of q to any saturated open or closed set is a quotient map. (8) If q : (X, TX) → (Y, TY) is a surjective continuous map that is also an open or closed map, then it is a quotient map. 1.3. CONSTRUCTIONS 47

−1 −1 −1 PROOF. (1) For any U ∈ TB, 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 ef (y) := f (x). Since f is constant on the fibers of q, it follows that ef is well-defined. The continuity of ef follows from (1). 0 If there exists another continuous map ef : (Y, TY) → (B, TB) such that f = 0 ef ◦ q. Then 0 0 ef (y) = ef (q(x)) = f (x) = (ef ◦ q)(x) = ef (y). (4) Consider three diagrams

X X X A A A A q A q A q q AA 2 q AA 1 q AA 1 1 AA 2 AA 1 AA A A 1 A Y1 Y1 / Y2 Y2 / Y1 Y1 / Y1 ∃eq2 ∃eq1 eq1◦eq2

By (3), there exist eq2 and qe1. Since

eq1 ◦ eq2 ◦ q1 = eq1 ◦ q2 = q1, ( ) q ◦ q = q ◦ q = we obtain from 2 that e1 e2 1Y1 . Similarly we have e2 e1 1Y2 . (5) − (6): as an exercise. (7) Let A be a saturated open set and let π = 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 q : (X, TX) → (Y, TY) is a quotient map and TX is Hausdorff, then TY need not be Hausdorff.

Lemma 1.3.12. Suppose π : (X, TX) → (Y, TY) is a quotient map and X is second countable. If Y is locally Euclidean, it is second countable. 48 1. BASIC TOPOLOGY

PROOF. Since Y is locally Euclidean, it follows that Y has an open cover U = −1 −1 (Uα)α∈A with Uα is homeomorphic to some open ball. Then π (U ) := {π (Uα) : 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 0 ⊆ U denote a subcover such that π−1(U 0) is a countable subcover of X and U 0 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 I ω = ∑ ωI dx . |I|=p Then M (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: (i) if f ∈ Ω0(U), then ∂ f = i = d f ∑ i dx d f ; 1≤i≤n ∂x I p (ii) if ω = ∑|I|=p ωI dx ∈ Ω (U), then I I dω := ∑ dωI ∧ dx = ∑ dωI ∧ dx . |I|=p |I|=p It is easy to show that d2 = d ◦ d = 0 and (1.3.3.4) d(ω ∧ η) = dω ∧ η + (−1)deg(ω)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 • M p (1.3.3.6) HDR(U) := HDR(U). 0≤p≤n

3Consider U = R2 \{(0, 0)} and ω = (xdy − ydx)/(x2 + y2) 1.3. CONSTRUCTIONS 49

• For any [ω], [η] ∈ 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

∗ i1 ip (1.3.3.8) (F ω)(x) = ∑ ωI (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 ∗ (1.3.3.11) F : HDR(V) −→ HDR(U), [ω] 7−→ [F ω]. Let ω be a smooth form on [0, 1] × U. For (t, x) ∈ [0, 1] × U we write I J (1.3.3.12) ω(t, x) = ∑ ωI (t, x)dx + ∑ ωeJ(t, x)dt ∧ dx . |I|=p |J|=p−1 Deﬁne Z 1 p p−1 J (1.3.3.13) K : Ω ([0, 1] × U) −→ Ω (U), ω 7−→ ∑ ωeJ(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) Z 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 ∗ ∗ (1.3.3.15) (Kd + dK) ω = ι1ω − ι0ω 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 ∈ Zp(U) p−1 p and p ≥ 1, then ω = dη for some η ∈ Ω (U). Consequently, HDR(U) = {0} for 0 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. 50 1. BASIC TOPOLOGY

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 F = H ◦ ι1, G = H ◦ ι0, 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 M C• := Cp p∈Z is called a differential complex if there are homomorphisms

dp−1 dp dp+1 · · · −−−−→ Cp−1 −−−−→ Cp −−−−→ Cp+1 −−−−→ · · · 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 M (1.3.3.16) H•(C•) := Hp(C•) p∈Z where Ker(dp) ∩ Cp Hp(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 · · ·     p−1  p  p+1 y f y f y f 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•) −−−−→ 1.3. CONSTRUCTIONS 51

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•). Next we define δp : Hp(C•) → Hp+1(A•) as follows......

x p+ x p+ x p+ d 1 d 1 d 1  A•  B•  C• p+1 p+1 p+1 f p+1 p g p+1 0 −−−−→ A 3 a −−−−→ B 3 dB• (b) −−−−→ C −−−−→ 0 x p x p x p d d d  A•  B•  C• f p gp 0 −−−−→ Ap −−−−→ Bp 3 b −−−−→ Cp 3 c −−−−→ 0

x p− x p− x p− d 1 d 1 d 1  A•  B•  C• ...... 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 f , 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. Proposition 1.3.14. The map δp is independent of the choice of a, b, c.

0 0 p−1 p−1 0 PROOF. Assume [c] = [c ]. Then c = c + dC• (ce) for some ce ∈ C . For c , we can also ﬁnd b0 ∈ Bp and a0 ∈ Ap+1 such that p 0 0 p+1 0 p+1 0 p 0 g (b ) = c , dA• (a ) = 0, f (a ) = dB• (b ). p 0 0 p−1 Since g (b − b) = c − c = dC• (ce), it follows that p p 0 p+1 p 0 0 = dC• ◦ g (b − b) = g ◦ dB• (b − b) 52 1. BASIC TOPOLOGY

p 0 p+1 p+1 p+1 p 0 and then dB• (b − b) ∈ Ker(g ) = Im( f ). Hence f (ea) = dB• (b − b) for p+1 some ea ∈ A . Moreover p+1 p+1 p+2 p+1 p+1 0 = dB• ◦ f (ea) = f ◦ dA• (ea) =⇒ dA• (ea) = 0. p+1 p+1 0 p+1 0 From f (ea) = f (a − a) and the injectivity of f , we obtain a = a + ea for 0 some closed vector ea. Therefore [a ] = [a].

1.4. Topological groups and matrix Lie groups A topological group is a group G endowed with a topology such that

µ : G × G −→ G, (g1, g2) 7−→ 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 topology, 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. 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) to be added! By Proposition 1.4.1, Rn, S1, Tn are topological groups. Since the orthogonal group O(n) = {A ∈ GL(m, R) : AAT = I} 1.4. TOPOLOGICAL GROUPS AND MATRIX LIE GROUPS 53 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 0 0 (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 0 0 −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 0 0 −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

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 6= ∅, 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. 54 1. BASIC TOPOLOGY

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 −1 (π × π) ∆G\X = {(x, y) ∈ X × X : g · x = y for some g ∈ G}.

Since π × π is open, surjective, and continuous, it follows that ∆G\X is closed if −1 and only if (π × π) (∆G\X) is closed. 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). 1.4. TOPOLOGICAL GROUPS AND MATRIX LIE GROUPS 55

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 (" √ # ) e −1t 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 (" √ # ) e −1t 0 G = √ s, t ∈ R =∼ S1 × S1 = T2. 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

∑ Aki Akj = δij, 1 ≤ i, j ≤ n. 1≤k≤n The usual inner product on Rn is hx, yi := ∑ xkyk, x = (x1, ··· , xn), y = (y1, ··· , yn) ∈ Rn. 1≤k≤n Then A is orthogonal is equivalent to hx, yi = hAx, Ayi 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 ∑ Aki Akj = δij, 1 ≤ i, j ≤ n. 1≤k≤n The usual inner product on Cn is hz, wi := ∑ zkwk, z = (z1, ··· , zn), w = (w1, ··· , wn) ∈ Cn. 1≤k≤n 56 1. BASIC TOPOLOGY

Then A is unitary is equivalent to hz, wi = hAz, Awi 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 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   A1,i (i)  .  A :=  .  , 1 ≤ i ≤ n + k. An+k,i Then  [A(i), A(j)] = 0, i 6= j,  n,k (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 I 0 g := n 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). 1.4. TOPOLOGICAL GROUPS AND MATRIX LIE GROUPS 57

(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] = hx, Jyi 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). 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 6= 0} =∼ GL(1, R), C∗ = {a ∈ C|a 6= 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.

4Some books used Sp(n) for our Sp(n, C) and USp(n) for our Sp(n). 58 1. BASIC TOPOLOGY

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)}. 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. 1.4. TOPOLOGICAL GROUPS AND MATRIX LIE GROUPS 59

(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 A = , α, β ∈ C, |α|2 + |β|2 = 1. β α 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.

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. 60 1. BASIC TOPOLOGY

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   λ1 ∗ · · · ∗  .. .  −  0 . ∗ .  A = CBC 1, B =    . . .   . .. .. ∗  0 ··· 0 λn

If A ∈ GL(n, C), then all λi are nonzero. Set   λ1 (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 A(t) ∈ GL(n, C). Next, for each i ∈ {1, ··· , n}, choose a continuous path λi(t) ∈ ∗ C , 1 ≤ t ≤ 2, such that λi(1) = λi and λi(2) = 1. Then Ae(t) is a continuous path −1 in GL(n, C) from CDC to In, where

−1 Ae(t) = C · diag(λ1(t), ··· , λn(t)) · C , 1 ≤ t ≤ 2.

Hence we ﬁnd a continuous path in GL(n, C) from A to In. (3) In the proof of (2), deﬁne λi(t), 1 ≤ i ≤ n − 1 as before and

1 λn(t) = . λ1(t) ··· λn−1(t)

(4) From linear algebra, any unitary matrix U ∈ U(n) can be written as √ √ −1θ1 −1θn −1 U = U1 · diag(e , ··· , e ) · U1 , U1 ∈ U(n), θ1, ··· , θn ∈ R.

If we deﬁne √ √ −1(1−t)θ1 −1(1−t)θn −1 U(t) = U1 · diag(e , ··· , e ) · U1 , 0 ≤ t ≤ 1, then U(t) ∈ U(n) and connects U to In. 1.4. TOPOLOGICAL GROUPS AND MATRIX LIE GROUPS 61

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 A = , α, β ∈ C with |α|2 + |β|2 = 1. β α

Deﬁne α −β SU(2) −→ S3 ⊆ C2 =∼ R4, 7−→ (α, β). β α

The above map is homeomorphic so that SU(2) is simply-connected. 62 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 − X`|| → 0 as m, ` → ∞. 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 63

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 6= YX, eY = , eXeY 6= 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. (4) e(α+β)X = eαXeβX 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) in Proposition 1.4.15. 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 64 1. BASIC TOPOLOGY

From Proposition 1.4.15 (6), we have √ " √ #" √ # 1 −1 e− −1a 0 1 − −1 eX = CeDC−1 = √ √ √2 2 −1 1 −1a −1 1 0 e − 2 2 a − a = cos sin 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 tX −tX d 2 X 2 X e Ye = In + tX + t + ··· Y In − tX + t + ··· dt t=0 dt t=0 2 2

d = Y + t(XY − YX) + t2(··· ) = XY − YX. dt t=0 Deﬁnition 1.4.17. For any X ∈ Max(n, K) deﬁne

(1.4.3.26) adX : Max(n, K) −→ Max(n, K), Y 7−→ [X, Y] := XY − YX. We also call [X, Y] the Lie bracket of X and Y. m If g(z) = ∑m≥0 am(z − a) is a formal series, then define 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, ! (ad )m XkY (−X)m−k 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 0≤k≤m k m m = XkY(−X)m+1−k + XkY(−X)m+1−k ∑ − ∑ 1≤k≤m+1 k 1 0≤k≤m k 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 65

(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] 6= 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 1 1 eXeY = I + (X + Y) + X2 + XY + Y2 n 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 66 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 Z 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 67

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   (1.4.3.34) h := 0 0 γ α, β, γ ∈ 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). −1 Write S = Cdiag(λ1, ··· , λn)C for some C ∈ GL(n, C). Then eS = Cdiag(eλ1 , ··· , eλn )C−1 and det(eS) = ∏ eλi = e∑1≤i≤n λi = etr(S). 1≤i≤n 68 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 69

(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 ∅ 6= 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. 70 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.

Proposition 1.5.1. (1) If f : (X, TX) → (Y, TY) 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 71 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

ιx0 : Y −→ X × Y, y 7−→ (x0, y) ιy0 : X −→ X × Y, x 7−→ (x, y0).

Then we get maps d ◦ ιx0 : Y → D and d ◦ ιy0 : X → D. Since X and Y are connected, it follows that d◦x0 = cx0 and d ◦ ιy0 = 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 ∅ 6= D ⊆ B.

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

A := {bounded sequences of R} 6= ∅, B := {unbounded sequences of R} 6= ∅.

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. 72 1. BASIC TOPOLOGY

To prove that Rω is connected in the product topology we shall use the fact that R is connected (see Theorem 1.5.3). Deﬁne n i ω i ∼ n Re := x = (x ) ≥ ∈ R x = 0 for i > n = R , i 1 [ 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 R∞ = Rω which, by Proposition 1.5.1 (11), implies the connectedness of Rω 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∞ 6= ∅ 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 e > 0 such that [a, a + e) ⊆ U ∩ I and (b − e, b] ⊆ V ∩ I. Let c; = sup(U ∩ [a, b]). Then a + e ≤ c‘b − e and c ∈ (a, b) ⊆ I ⊆ U ∪ V. If c ∈ U, then (c − e0, c + e0) ⊆ U ∩ (a, b) for some e0 > 0; if c ∈ V, then (c − e00, c + e00) ⊆ V for some e00 > 0. Both cases contradict with the definition of c. Therefore I is connected. But in general the components of X need not to be open in X. For example, consider Q ⊆ R. Then each component of Q consists of a single point so that this component is not open.

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. (2) Deﬁne a relation ∼ on X by x ∼ y ⇐⇒ there exists a path in X from x to y. 1.5. CONNECTEDNESS AND COMPACTNESS 73

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

Ue := {x ∈ U|∃ path in U joining x0 to x} ⊆ U.

Since x0 ∈ Ue, we claim that Ue is both open and closed in U so that Ue = U. For x ∈ Ue, 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, e) ⊆ U and then 2 B (x, e) ⊆ Ue. To prove the closedness of Ue, we take a sequence {xn}n≥1 in Ue such 2 that xn → x ∈ U. For any e > 0, there exists N ∈ N such that xn ∈ B (x, e) for all n ≥ N. Since xn ∈ Ue, we can ﬁnd a path connecting x0 to xn. Therefore we can ﬁnd a path joining x0 to x. Thus x ∈ Ue. 74 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 6= 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 75 and V is closed but not open in S. Let Sb := S \ ({0} × (Q ∩ [−1, 1])). Then Sb 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 M = ∪i∈NUi where ϕi : Ui → Bi ⊂ R is homeomorphic for each i. Hence M 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. (b) {Ui}1≤i≤n ⊆ U covers X if and only if ∩1≤i≤nCi = ∅ where Ci := X \ Ui. (c) U 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. 76 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.

Example 1.5.9. (1) R is not compact, since the open cover U = {(n, n + 2)}n∈Z of R contains no finite subcollection that cover R. (2) X = {0} ∪ {1/n}n≥1 is compact. Given an open cover U of X, there exists an open subset U ∈ U containing 0. Take N ∈ N such that 1/n ∈ U for all n > N. For each i ∈ {1, ··· , N}, we choose Ui ∈ U 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. (4)(0, 1] is not compact, because the open cover U = {(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 finite subcollection covering Y. Let U = {Uα}α∈I be a cover of Y by subsets open in X. Then UY := {Uα ∩ Y|α ∈ I} is a cover of Y by subsets open in Y. Since

Y is compact, there exists a finite subcollection {Aαi ∩ Y}1≤i≤n covers Y. Thus 0 0 0 {Uαi }1≤i≤n is a subcollection of A that covers Y. Conversely, let U = {Uα|α ∈ I } 0 0 be a cover of Y by subsets open in Y. For each α ∈ I , Uα = Uα ∩ Y for some open subset Uα in X. Then U = {Uα}α∈I0 is a cover of Y by subsets open in X. By {U } Y {U0 } hypothesis, there is a finite subcollection αi 1≤i≤n covers . Then αi 1≤i≤n is a subcollection of U 0 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 77

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 (Vyi )1≤i≤N, 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 homeomorphic to Y, so finitely many of the sets of U 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, say (Zxi )1≤i≤K, of X. Since finitely many sets of U 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) ∩ 78 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 S ⊆ X (S) < S ⊆ B (x ) i ∈ { ··· n} If with diam δ, we claim that 2r(xi) i for some 1, , . y ∈ S y ∈ B (x ) i z ∈ S Let and choose r(xi) i for some . Therefore for any ,

d(z, xi) ≤ d(z, y) + d(y, xi) < δ + d(y, xi) < δ + 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, Y0 ⊆ X such that x ∈ Y and W \ U ⊆ Y0. Let V := Y ∩ W be a neighborhood of x. Since V ⊆ W, it 0 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 79

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 6= ∅; 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 6= ∅, 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 6= 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.

Example 1.5.17. (1) U = {(n, n + 2)}n∈Z is locally ﬁnite in R. (2) U = {(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 neighborhood W of x such that W intersects U1, ··· , Um ∈ U for some m ∈ N. If W ∩ U 6= ∅, then W ∩ U 6= ∅ so that U ∈ {U1, ··· , Um}. Thus W intersects U for ﬁnitely many U ∈ U . The converse is obvious. 80 1. BASIC TOPOLOGY

(b) Observe that ∪U∈U U ⊆ ∪U∈U U. Let x ∈/ ∪U∈U U. There exists some U ∈ U such that x ∈ U and intersects U1, ··· , Uk ∈ U by (1). Then U \ ∪1≤i≤kUk contains x and intersects none of the sets in U ; 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

Let U = (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 ( ) ( ∩ X implies that there exists a finite subcover Uij 1≤j≤mi of the open cover Uα (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≤mi is an open subcover of Wi and [ [ [ Vij = Wi = X. i∈N 1≤j≤mi i∈N = ( ) ≺ ∩ ∪ = 6= − − If V : Vij i∈N,1≤j≤mi , then V U 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 3 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 6= ∅}. 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 U0 of X such that U0 ∩ Y = U. Cover X by the open subsets U0, 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 81

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 Lindel¨ofspace is paracompact. (4) The product of two paracompact spaces need not be paracompact. For example, 2 R` is regular and Lindel¨of,it follows from (3) that R` is paracompact, but R`, that is Hausdorff and not normal, is not paracompact. (5) Rω is paracompact in te product topology. (6) RJ is not paracompact of J is unvountable.

Let U = {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, (2) {supp(φα)}α∈J is locally finite, and (3) ∑α∈J φα(x) ≡ 1 for all x ∈ X. Lemma 1.5.20. (Shrinking lemma) Suppose that(X, T ) is a paracompact Hausdorff space. If U = {Uα}α∈J is an indexed family of open cover of X, then U admits a locally finite open refinement V = {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 that A has a locally finite open refinement B = {Bβ}β∈K. We then find a map f : K → J such that for any β ∈ K one has Bβ ⊆ A ⊆ Uf (β). Define, for each α ∈ J, ∪ = { ∈ | ( ) = } 6= Bβ∈Bα Bβ, Bα : Bβ B f β α ∅, Vα := ∅, Bα = ∅.

For Bβ ∈ Bα, we have Bβ ⊆ Uf (β) = Uα. Since Bα is locally ﬁnite, by Proposition 1.5.18, [ Vα = Bβ ⊆ Uα. Bβ∈Bα

Finally we check the local ﬁniteness of V = {Vα}α∈J. Given x ∈ X. There is a neighborhood W of x that intersects Bβ for only ﬁnitely many values of β = β1, ··· , βN. Then W can intersects Vα only if α ∈ { f (β1), ··· , f (βN)}.

Theorem 1.5.21. Let (X, T ) be a paracompact Hausdorff space. If U = {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 open sets W = {Wα}α∈J and V = {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, 82 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 (3) {An}n≥1 ⊆ S implies ∪n≥1 An ∈ S . 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 g1 ∈ G, then BG 3 S 7→ µ(Sg1) 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(Sg1) = ∆(g1)µ(S) 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.

83 84 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 deﬁne bLp(G, µ) the set of all complex-valued µ-measurable functions on G whose modulus to the p-th power is integrable. bL∞(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 bLp(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 Z 1/p p || f ||Lp(G) := | f (x)| dµ(x) G and for p = ∞ deﬁne

|| f ||L∞(G) := inf {M > 0|µ({x ∈ G|| 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 Z Z 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 Z +∞ dx Z +∞ dx f (tx) = f (x) −∞ |x| −∞ |x| for all f ∈ L1(R∗, µ) and all t ∈ R∗. Then Z dx Z +∞ dx µ(A) = = χA(x) . A x −∞ x

Example 2.1.2. (1) A Haar measure on the multiplicative group R+ is dx/x. That is Z +∞ dx µ(A) = χA(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 85

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

For A ⊆ BG deﬁne Z +∞ Z +∞ dxdy µ(A) := χA(x, y) . −∞ −∞ x2 When t = (t1, t2) ∈ G and (x, y) ∈ A we have

t(x, y) = (t1x, t1y + t2) and then Z +∞ Z +∞ dxdy Z +∞ Z +∞ d(x0/t)d((y0 − t )/t ) ( ) = ( ) = ( 0 0) 2 1 µ tA χtA x, y 2 χA x , y 0 2 −∞ −∞ x −∞ −∞ (x /t1) Z +∞ Z +∞ 0 0 0 0 dx dy = χA(x , y ) 0 = µ(A). −∞ −∞ x 2 Thus µ is a left Haar measure on G. Similarly we can prove that Z +∞ Z +∞ dxdy ν(A) := χA(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 Z (2.1.2.1) ( f ∗ g)(x) := f (y)g(y−1x)dµ(y). G Since Z Z Z Z − − f (y)g(y 1x)dµ(y)dµ(x) ≤ | f (y)||g(y 1x)|dµ(x)dµ(y) G G G G Z Z Z Z = | 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 Z (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). 86 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. 0 p Now we assume 1 < p < +∞. Write p = p−1 , and compute

Z Z 1 1 (|g| ∗ | f |)(x) = | f (y−1x)||g(y)|dµ(x) = | f (y−1x)||g(y)| p |g(y)| p0 dµ(x) G G 0 Z 1/p Z 1/p ≤ | f (y−1x)|p|g(y)|dµ(y) |g(y)|dµ(y) G G so that Z Z 1/p p−1 −1 p |||g| ∗ | f |||Lp(G) ≤ ||g|| 1 | f (y x)| |g(y)|dµ(y)dµ(x) L (G) G G Z Z 1/p p−1 p = ||g|| 1 | f (x)| dµ(x)|g(y)|dµ(y) = || f ||Lp(G)||g||L1(G). L (G) G G Now the inequality (2.1.2.5) follows from |g ∗ f | ≤ |g| ∗ | f |. If f is a function on G, we deﬁne (2.1.2.6) fe(x) := f (x−1), x ∈ G. Then (2.1.2.3) can be written as Z ( f ∗ g)(x) = f (xz)g(z)dµ(z). G e

Theorem 2.1.5. (Young’s inequality) Let 1 ≤ p, q, r ≤ +∞ satisfy 1 1 1 + 1 = + . q p r p r Then for all f ∈ L (G) and all g ∈ L (G) satisfying ||g||Lr(G) = ||ge||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 = + . r0 p0 q r r0 p p0 By Holder’s¨ inequality we get Z |(| f | ∗ |g|)(x)| ≤ | f (y)||g(y−1x)|dµ(x) G 2.1. HAAR MEASURES 87

Z 0 0 = | f (y)|p/r | f (y)|p/q|g(y−1x)|r/q |g(y−1x)|r/p dµ(y) G 0 0 Z 1/q Z 1/p p/r p −1 r −1 r ≤ || f || p | f (y)| |g(y x)| dµ(y) |g(y x)| dµ(y) L (G) G G 0 0 Z 1/q Z 1/p p/r p −1 r −1 r = || f || p | f (y)| |g(y x)| dµ(y) |g(x y)| dµ(y) L (G) G G e

Z 1/q 0 0 p −1 r p/r r/p = | f (y)| |g(y x)| dµ(y) |||| p ||g|| r G L (G) e L (G) so that

0 0 Z Z 1/q p/r r/p p −1 r ||| f | ∗ |g|||Lq(G) ≤ || f || p ||g|| r | f (y)| |g(y x)| dµ(x)dµ(y) L (G) e L (G) G G p/r0 r/p0 p/q r/q = || || || || || || || || = || || r || || p f Lp(G) ge Lp(G) f Lp(G) g Lr(G) g L (G) f L (G) using again the assumption ||g||Lr(G) = ||ge||Lr(G).

Exercise 2.1.6. (Hardy) Use Theorem 2.1.4 to prove the following inequalities. (1) For 1 < p < +∞, 1/p Z +∞ 1 Z x p p | f (t)|dt dx ≤ || f ||Lp((0,+∞)) 0 x 0 p − 1 and Z +∞ Z +∞ p 1/p Z +∞ 1/p | f (t)|dt dx ≤ p | f (t)|ptpdt . 0 x 0 (2) For 0 < b < +∞ and 1 ≤ p < +∞, Z +∞ Z x p 1/p p Z +∞ 1/p | f (t)|dt x−b−1dx ≤ | f (t)|ptp−b−1dt 0 0 b 0 and Z +∞ Z +∞ p 1/p p Z +∞ 1/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 ∈ Cc(G), Z Z (2.1.3.1) f (gg1)dg = ∆(g1) f (g)dg. G 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 ∈ Cc(P) the functional Z Z Z Z (2.1.3.3) f 7−→ f (au)dadu = f (au)duda U A A U is a Haar (left-invariant) functional on P. 88 2. ANALYSIS ON TOPOLOGICAL GROUPS

PROOF. Left invariant by A is obvious. Let u1 ∈ U. Then Z Z Z Z −1 f (u1au)duda = f (aa u1au)duda A U A U Z Z a a −1 = ( f ◦ a)(u1u)duda where u1 := a u1a ∈ U A U Z Z = 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), Z Z (2.1.3.4) f (a−1ua)du = δ(a) f (u)du, U U or in other words, for f ∈ Cc(P), Z Z (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 Z Z f 7−→ f (au)dadu U A is right-invariant under U. Let a1 ∈ A. Then Z Z Z Z Z Z −1 f (aua1)dadu = f (aa1a ua1)dadu = f (aa1u)dadu U A U A U A Z Z Z Z −1 = f (aa1a au)dadu = δ(a1) f (au)dadu, U A 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), Z Z Z (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 89 then Z Z Z Z (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

The Iwasawa decomposition for G := SL2(C) = SL(2, C) is G = UAK, 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 The elements of A 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 (2.2.1.3) z = −1 with 0 6= z1 ∈ C. 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.

Furthermore, let Pos2 := Pos2(C) be the space of positive deﬁnite 2 × 2 Hermitian ma- ∗ trices. Then the map g 7→ gg is a bijection G/K ≈ SPos2, so of UA ≈ SPos2.

2 PROOF. Let g ∈ G and e1, e2 be the standard (vertical) unit vectors of C . 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 0 (1) 0 (1) (2) e1 := b11g , e2 := b12g + b22g , 0 0 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. 0 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 90 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) −1 2 with u1, u2 ∈ U, a, b ∈ A, and k1, k2 ∈ K. 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,

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

∗ ∗ 2 −1 PROOF. Since gg can be written as gg = k1b k1 with b ∈ A and k1 ∈ K, where b2 is the matrix of eigenvalues of gg∗. By the bijection in Theorem 2.2.1, there exists k2 ∈ K 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 := Lie(A) be the R-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 on a deﬁned by α(H) := h1 − h2. If a = exp H = diag(a1, a2) ∈ A = exp a, then we deﬁne the Iwasawa coordinate y : A → R+ by α a1 2 (2.2.2.1) y(a) ≡ ya ≡ χα(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 (2.2.2.2) ya = a = χsα(a), with a complex (multiplicative) character χsα. We often write y instead of ya and 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 e1 0 (2.2.2.5) e = −1 with |e1| = 1. 0 e1 Then we have a direct product decomposition

(2.2.2.6) D = AT with z = aze and T = D ∩ K. 2.2. IWASAWA’S DECOMPOSITION 91

Indeed, writing z1 = |z1|e1, z 0 |z | 0 e 0 z = 1 = 1 1 = a e −1 | |−1 −1 : z . 0 z1 0 z1 0 e1 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 diagonal components are distinct, in other words if zα 6= 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

(2.2.3.1) f (k1gk2) = f (g), for all k1, k2 ∈ K and g ∈ 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, − 0 1 z 0 0 −1 z 1 w = −1 = 1 = 1 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.

PROOF. By the polar decomposition (2.2.1.6), g = k1bk2, where b ∈ A, so if ϕ 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)g0 := gg0g−1. This action induces what we also call the conjugation action on various functors. 92 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

For u as a vector space over R, 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 Let a = diag(a1, a1 ) ∈ A 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 (2.2.3.11) zu(x)z−1 = 1 = u(zαx). 0 1

We let δ = δG be the Iwasawa character on A, namely 2α 2 4 (2.2.3.12) δG (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 2α (2.2.3.13) |δ(z)| = az .

2.2.4. Group Fubini theorem. In the concrete application to G = SL2(C) 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 2.1.8 is the Iwasawa character, δ(a) = a2α. We normalize the Haar measure by taking c = 1. Thus we deﬁne the Iwasawa measure to be the Haar measure such that for f ∈ Cc(G) we have Z Z Z Z Z Z Z (2.2.4.2) f (g)dg := f (auk)dadudk = f (uak)a−2αdudadk, 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 93 where dy da := with y = aα, 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 Z Z Z Z −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 homogeneous space for the action of SL2(C) 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) (2.2.4.5) JG,com(z) ≡ Jcom(z) := |1 − z | with m(α) = 2. The exponent is the multiplicity m(α) of α. Then Z 1 Z ( −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 Z Z i f (z−1uzu−1)du = f [u((z−α − 1)x)] dx ∧ dx¯ U C 2 h i−1 Z i 1 Z = ( −α − )( −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 space measure, which makes the group Fubini theorem valid, that is, µD\G such that Z Z Z (2.2.4.7) f (g)dµIW(g) = f (wg˙)dwdµD\G (g˙). G D\G D 94 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),

(2.2.4.8) dµD\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 Z 1 Z ( −1 ) ( ) = ( ) (2.2.4.10) ϕ g zg dµD\G g −1 ϕ azu du. D\G Jcom(z ) U

− − − − PROOF. For (2.2.4.10), using g 1zg = k 1(u 1w 1zwu)k, one has Z Z Z ϕ(g−1zg)dg = ϕ(u−1zu)du = ϕ(zz−1u−1zu)du D\G U U Z Z 1 Z = ( ◦ )( −1 −1 ) = ( ◦ )( −1 −1) = ( ) ϕ z z u zu du ϕ z z uzu du −1 ϕ zu du U U Jcom(z ) U by (2.2.4.6). Since, writing z1 = |z1|e, e−1 0 z z x 1 0 |z | |z |x 1 1 = 1 1 = a u −1 | |−1 z , 0 e 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 m(α) 2 2 −1 α − α α − α −1 | |( ) = | |( ) = 2 − 2 = 2 − 2 = − (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 Z −1 (2.3.1.3) OrbD\G (ϕ, z) := ϕ(g zg)dµD\G (g) D\G with µD\G as in (2.2.4.8). We define the Harish transform of ϕ ∈ Cc(K\G/K) by Z 1/2 (2.3.1.4) (Hϕ)(z) := δ(az) ϕ(azu)du = |D|(z)OrbD\G (ϕ, z). U 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 functions for which the integrals are absolutely convergent, by continuity. This 2.3. HARISH, MELLIN AND SPHERICAL TRANSFORMS 95 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 equivalently, 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 Z −1 −1 −1 −1 −1 (Hϕ)(a ) = |D|(a )OrbD\G (ϕ, a ) = |D|(a) ϕ(g a g)dµD\G (g) D\G Z −1 −1 −1 = |D|(a) ϕ((g a g) )dµD\G (g) = (Hϕ)(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 Z −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 Z Z −1 −1 (ψ ∗ f )(z) := ψ(zg) f (g)dµIW(g) = ψ(zg ) f (g )dµIW(g) G G Z Z −1 −1 (2.3.1.8) = ψ(zg ) f (g)dµIW(g) = ψ(g ) f (gz)dµIW(g) G G Z = 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 ∈ G, τ there exist k1, k2 ∈ K satisfying g = k1gk2. Then the convolution is commutative (whenever the convolution integral is absolutely convergent). 96 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 Z Z (ψ ∗ f )(z) = ψ(zg−1) f (g)dg = ψ(τzτ g−1) f (τ g)dg G G Z Z τ −1 τ τ −1 τ = ψ( g ) f ( z g)dg (by g 7→ gz) = ψ(g ) f (k1zk2 g)dg G G Z Z τ −1 −1 = ψ( g ) f (k1zk2g)dg = ψ(g k2) f (k1zg)dg G G Z Z −1 −1 −1 −1 −1 = f (zgk )ψ(k2g )dg = f (z(k2g ) )ψ(k2g )dg. G 2 G 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 Z Z Z (H(ϕ ∗ ψ))(a) = aα (ϕ ∗ ψ)(au)du = aα ϕ(aug)ψ(g−1)dgdu U U G Z Z = aα 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 Z Z Z Z (H(ϕ ∗ ψ))(a) = aα ϕ(abνk)ψ(k−1ν−1b−1u)dbdνdkdu U K U A Z Z Z Z = aα ϕ(abν)ψ(ν−1b−1u)dbdνdkdu U K U A Z Z Z = aα ϕ(abν)ψ(ν−1b−1u)dbdνdu U U A Z Z Z = aα ϕ(ab−1ν)ψ(ν−1bu)dbdudv U U A Z Z Z = aα ϕ(ab−1ν)ψ(ν−1ub)δ(b)−1dbdνdu U U A Z Z Z = aα ϕ(ab−1ν)ψ(ub)δ(b)−1dbdνdu U U A Z Z Z = aα ϕ(ab−1ν)ψ(bu)dbdνdu U U A Z Z Z = (ab−1)αbα ϕ(ab−1ν)ψ(bu)dbdνdu U U A which is exactly the convolution (Hϕ ∗ Hψ)(a). 2.3. HARISH, MELLIN AND SPHERICAL TRANSFORMS 97

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 Z (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 s sα (2.3.2.2) χ(a) := ya = a with a complex variable s. Then the Mellin transform is written with Fα(y) = F(a): Z ∞ s dy (2.3.2.3) (MαF)(s) ≡ (MFα)(s) = Fα(y)y . 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) 2α to G. As before, let δ = δG be the Iwasawa character, δ(a) = a . We deﬁne the spherical kernel on the product of G with the character group ch(A) to be the function given by Z Z 1/2 1/2 (2.3.2.5) Φχ(g) ≡ Φ(χ, g) := (χδ )(kg)dk = (χδ )((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 ∨ (aC) , and the spherical kernel may be written in terms of the additive variable ξ, in the form ∨ (2.3.2.8) Φζ (g) ≡ Φ(ζ, g) = Φ(χζ, g), ζ ∈ (aC) .

We let S ≡ SIW be the integral (spherical) transform deﬁned by this kernel, Z (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 Z Z 1/2 (|χ|δ )((kg)A)| f (g)|dkdg G K is absolutely convergent. Then (2.3.2.10) S f = M(H f ), 98 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

H W Cc(K\G/K) / Cc(A) ⊆ Cc(A) RR RRR RRR M S RRR RR( Fu(ch(A)) By Fubini’s theorem and K-invariance of f and dg, we have Z Z Z 1/2 (S f )(χ) = Φ(χ, g) f (g)dg = (χδ )((kg)A) f (g)dgdk G K G Z Z Z Z 1/2 1/2 = (χδ )(gA) f (g)dgdk = χ(a)δ (a) f (au)dadu K G A U Z = 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 ∈ Cc(K\G/K), Z Z (S f )(χ−1) = χ−1(a)(H f )(a)da = χ(a−1)(H f )(a)da A A Z Z = 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, Z Z Z M( f ∗ h)(χ) = (g ∗ h)(a)χ(a)da = f (ab−1)h(b)χ(a)dbda. A A A Z Z = 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. 2s If χ = χsα (that is, χ(a) = χsα(a) = a1 ), we may write

(2.3.2.13) (Sα f )(s) := (S f )(χsα). ∨ 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 99

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 6= 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 Z Z (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 ( )( ) = | |( ( )) ( ) = | |( ) ( ) (2.3.3.3) Hϕ z D η˜ z OrbGz\G ϕ, z D η OrbD\G ϕ, η˜ . 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 Z Z Z (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 Z Z −1 −1 −1 −1 OrbD\G (ϕ, η˜) = ϕ(g η˜g)dµD\G (g) = g(k u z η˜zuk)dudk D\G UK Z Z (2.3.3.5) = ϕ(u−1z−1η˜zu)du = ϕ(u−1η˜u)du U U Z Z = ϕ(u(−x)η˜u(x))dx = ϕ(Mη(x))dx, C C where η x(η − η−1) (2.3.3.6) M (x) := u(x)−1η˜u(x) = η 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 Subsection 2.2.4. Let ϕ be measurable on G, K-bi-invariant (so even), and such that the orbital in- α tegral OrbD\G (ϕ, η˜) is absolutely convergent for regular η˜. For b ∈ A, y = b ≥ 1, write

(2.3.3.7) ϕα(y) ≡ ϕα(y(b)) := ϕ(b). 100 2. ANALYSIS ON TOPOLOGICAL GROUPS

Then for |η| ≥ 1 and η˜ regular, we have 2π Z ∞ y − y−1 dy (2.3.3.8) OrdD\G (ϕ, η˜) = − ϕα(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 2π Z ∞ y − y−1 dy OrdD\G (ϕ, η˜) = − ϕα,+(ln y) |η − η 1|2 |η|2 2 y 2π Z ∞ = − ϕα,+(ν) sinh(ν)dν.(2.3.3.10) |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, ∗ 2 −1 (2.3.3.11) If Mη(x) = k1bk2 (k1, k2 ∈ K, b ∈ A), then Mη(x)Mη(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) M (x)M (x)∗ = η η x¯η−1(η¯ − η¯−1) |η|−2 2 ∗ So b 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 We have r = xx¯ and dx = dx dx = rdrdθ. 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 1/2 2 −2 or also τ + ∆ = 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 multiplicative coordinate y = y(b) = bα, we write

ϕ(b) = ϕα(y). We change variables in the integral form (2.3.3.5), dx1dx2 = rdrdθ, to get Z Z ∞ (2.3.3.16) OrbD\G (ϕ, η˜) = ϕ(Mη(x))dx = 2π ϕ(b(r))rdr. C 0 2.3. HARISH, MELLIN AND SPHERICAL TRANSFORMS 101

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 ` := |η|2 + |η|−2, then (2.3.3.18) p τ ∈ (`, +∞), ∆ ∈ (`2 − 4, +∞), `2 − 4 = (|η|2 − |η|−2)2, ` + `2 − 4 = 2|η|2. From (2.3.3.14) we get √ ! τ + ∆1/2 ` + `2 − 4 (2.3.3.19) y = ∈ , ∞ = (|η|2, ∞), 2 2 determining the limits of integration. The last formula (2.3.3.10) is obvious.

Corollary 2.3.8. For a ∈ A+ (i.e., aα > 1, so a regular), we have Z ∞ 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

2 α(ln b) ϕ (b) = e−(α(ln b)) /2c c sinh(α(ln b)) 2 ln y = e−(ln y) /2c with y = y(b),(2.3.4.1) (y − y−1)/2 2 ν = e−ν /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

2πc −(ln |zα|)2/2c (2.3.4.2) Orb (ϕc, z) = e . D\G |zα/2 − z−α/2|2 102 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 Z ∞ 2 2πc −ν2/2c ν OrbD\G (ϕc, z) = − e d |η − η 1|2 2 ln η 2c 2πc 2 2πc α 2 = e−(2 ln η) /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 measure 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)) Z ∞ Z ∞ 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 2 √ uses the self-duality of the function e−ν /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 Z 3/2 c/2 (2.3.4.7) ϕc(g)dµIW(g) = (2πc) e . G 2.3. HARISH, MELLIN AND SPHERICAL TRANSFORMS 103

PROOF. Taking s = 1 in (2.3.4.4) yields Z 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 Z Z Z 1/2 −α+α Φ(−α, g) = (−αδ )((kg)A)dk = ((kg)A) dk = dk = 1. K K K Thus, we get (2.3.4.7). Let g ∈ G with polar decomposition

(2.3.4.8) g = k1bk2, k1, k2 ∈ K and b ∈ A. The element b ∈ A is determined only up to a permutation of its diagonal components. We call y = y(b) = bα 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

2 σ 2 ln y (2.3.4.10) ϕ = e−σ /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.

For b = diag(b1, b2) ∈ A, we have ln b = diag(ln b1, ln b2) ∈ a, and (ln y)2 | ln b|2 = (ln b )2 + (ln b )2 = 2(ln b )2 = , 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.

2 ∨ Let 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. 104 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 ( ) 2 bα − b−α m α y(b) − y(b)−1 (2.3.5.1) J (b) := = P 2 2 α 2 with y = y(b) = b = b1 and m(α) = 2. The associated polar measure dµP is given by eν − e−ν 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 Z Z Z (2.3.5.3) ϕ 7−→ ϕ(k1bk2)JP(b)dbdk1dk2 K K A 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 Z Z Z (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 Z 1 1/2 3/2 c/2 ϕc(g)dµIW(g) = (2π) c e . 2π G On the other hand, Z Z ∞ ν −ν Z ∞ −ν2/2c e − e −ν2/2c ν ϕc(b)dµP(b) = e ν dν = e e νdν. A −∞ 2 −∞ Using the identity

1 Z ∞ 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 105

Corollary 2.3.14. For all f ∈ Cc(K\G/K) one has 1 Z Z (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) aζ − a−ζ 2 aζ − a−ζ ( ) = = (2.3.5.8) ΦP ζ, 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 1 ys − y−s 1 esν − e−sν 1 sinh(sν) (2.3.5.9) Φ (s, a) := Φ (sα, a) = = = . P,α 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,P on A, we have

(2.3.5.10) (ΦP ∗A,P ϕc) (sα) = (SP ϕc)(sα) so ΦP is an integral kernel for SP ( with polar measure).

PROOF. By deﬁnition, (2.3.4.1) and, (2.3.5.2), Z (ΦP ∗A,P ϕc) (sα) = ΦP(sα, a)ϕc(a)dµP(a) A

sν −sν ν −ν 2 1 Z ∞ e − s 2 2ν e − e = −ν /2c ν −ν e ν −ν dν s −∞ e − e e − e 2 Z ∞ sν −sν Z ∞ 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).

The spherical kernel (2.3.5.8) in polar coordinates on SL2(C) 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 106 2. ANALYSIS ON TOPOLOGICAL GROUPS

The second factor in (2.3.5.8) comes from a skew-symmetrization procedure, namely, aζ − a−ζ . aα − a−α We know from Corollary 2.3.11 that the spherical transform is a linear isomorphism 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 ∨ ∼ ∨ da = dy/y = dν on a = R. We let dµFOU(λ) be the Haar measure on a that makes Fourier inversion come out without any extra constant factor. The Fourier transform F is deﬁned by Z (2.3.5.12) (F f )(λ) := f (H)e−iλ(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 a = 1 / H = 1 BB 0 a2 0 h2 BB ν L y BB LLLν BB LL ln LLL R+ / R L% a1/a2 / h1 − h2 Thus if λ = rα on a with λ ∈ R, then Z Z −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

We let ΦP,HC be the integral operator having the kernel function ΦP(iλ, a) with respect to the Harish-Chandra measure dµHC(λ), so for an even function h such that the integral is absolutely convergent, by deﬁnition Z (ΦP,HCh)(a) := ΦP(iλ, a)h(iλ)dµHC(λ) a∨ Z eirν − e−irν dr (2.3.5.16) = hα(ir)(−ir) − with ν = α(ln a). R eν − e ν 2π

Theorem 2.3.16. The integral operator ΦP,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 107

PROOF. For given ϕc ∈ Gauss(G), let h := SP ϕc. By (2.3.4.5), 1/2 3/2 −r2c/2 hα(ir) = (2π) c e . Then Z irν −irν 1/2 3/2 −r2c/2 e − e dr (ΦP,HCh)(a) = (2π) c e (−ir) − R eν − e ν 2π −1/2 3/2 Z (2π) c 2 = (−ir)e−r c/2+iνrdr R sinh ν 2 ν = e−ν /2c = ϕ (a). sinh ν c

Thus (ΦP,HC ◦ SP)ϕc = ϕc. To prove the second statement, we ﬁrst make explicit the L2-scalar product on G, for the polar measure. For two basic Gaussians ϕc, ϕc0 , we get, integrating over a =∼ R: Z ν −ν 2 −ν2/2c 2ν −ν2/2c0 2ν e − e hϕc, ϕc0 iP = e − e − dν R eν − e ν eν − e ν 2 Z 2 0 0 Z 2 0 = e−ν (c+c )/2cc ν2dν = (cc0)3/2 e−ν (c+c )/2ν2dν. R R On the other hand, Z r2dr hSP ϕc, SP ϕc0 iHC = (SP ϕc)(irα)(SP ϕc0 )(irα) R 2π 2 Z 2 2 0 r dr = (2π)1/2c3/2e−r c/2(2π)1/2c03/2e−r c /2 R 2π Z 0 3/2 0 3/2 −r2(c+c0)/2 2 2cc = (cc ) e r dr = 0 Γ(3/2). R c + c

Thus hϕc, ϕc0 iP = hSP ϕc, SP ϕc0 iHC. 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 −1 (2.3.6.1) Kϕ(z, w) := ϕ(z w), 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 decomposition G = KAK. 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. 108 2. ANALYSIS ON TOPOLOGICAL GROUPS

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

2 PROOF. Let e1, e2 be the unit column vectors of C . Let a = diag(a1, a2) and b = diag(b1, b2). We may without loss of generality that a1, b1 ≥ 1. It suffices to prove that a1 ≤ b1. From UA = AU, we have ua = av for some v ∈ U. Writing −1 0 0 −1 av = ua = k1bk2k = k1bk with k := k2k ∈ K, we have 2 2 0 2 a1 = |ave1| = |k1bk e1| , where | · | stands for he Euclidean metric on C2. Since k0 is unitary, there are num- 0 2 2 bers c1, c2 ∈ C such that k e1 = c1e1 + c2e2 and |c1| + |c2| = 1. Then 0 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 |c |2(y2 − 1) + 1 (y2 − 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, g0 ∈ G, we have (2.3.6.3) σ(gg0) ≤ σ(g) + σ(g0). 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 109

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

α b1 2 yP(g) = b = = b1 b2 be the polar y-coordinate of g = k1bk2 ∈ G. Then

(2.3.6.5) σ(g) = | ln yP(g)| = dP(gK, eG/K), where eG/K is the unit coset of G/K.

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

PROOF. Lang (1999) deﬁned a distance on Pos2(R). 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

3.1. Homotopy Recall the category Top of topological spaces and continuous maps. For any f , g ∈ HomTop(X, Y), a homotopy from f to g is a continuous map H : X × [0, 1] → Y, (x, t) 7→ H(x, t) ≡ Ht(x), such that

H0 = f , H1 = g.

In this case, we also say that f and g are homotopic and write f ' g or H : f ' g. If f is homotopic to a constant map, we say f is null homotopic.

3.1.1. Homotopy relation. It is clear that f ' f . Actually, we can prove that homotopy is an equivalence relation.

Proposition 3.1.1. (1) Homotopy is an equivalence relation on HomTop(X, Y). (2) Homotopy relation is preserved by composition. That is, if

f0, f1 ∈ HomTop(X, Y), g0, g1 ∈ HomTop(Y, Z), and f0 ' f1 and g0 ' g1, then g0 ◦ f0 ' g1 ◦ f1.

− PROOF. (1) If H : f ' g, then H 1 : g ' f , where

H(x, t) := H(x, 1 − t), (x, t) ∈ X × [0, 1].

If F : f ' g and G : g ' h, then H : f ' h, where

1 F(x, 2t), 0 ≤ t ≤ 2 , H(x, t) := 1 G(x, 2t − 1), 2 ≤ t ≤ 1. Since F(x, 1) = g(x) = G(x, 0), the map H is well-deﬁned and hence, by Theorem 1.2.26, H is continuous. (2) Let F : f0 ' f1 and G : g0 ' g1. Then we deﬁne Ht := Gt ◦ Ft which implies that H : g0 ◦ f0 ' g1 ◦ f1.

Because ' is an equivalence relation by Proposition 3.1.1, we denote by [X, Y] the set of equivalence classes of continuous maps from X to Y, that is,

(3.1.1.1) [X, Y] := HomTop(X, Y)/ ' .

111 112 3. FUNDAMENTAL GROUPS

3.1.2. The fundamental group. Recall the category Top∗ consisting of pointed topological space (X, x) and point-preserving continuous maps. For any (X, x) ∈ Ob(Top∗), denote by Ω(X, x) the set of all loops in X based at x, that is, (3.1.2.1) Ω(X, x) := { f ∈ [0, 1] → X| f is continuous and f (0) = f (1) = x}.

The set Ω(X, x) consists at least one element cx given by

(3.1.2.2) cx(s) := x, s ∈ [0, 1]. For f ∈ Ω(X, x), deﬁne its reverse path f by f (s) := f (1 − s), s ∈ [0, 1]. Then f ∈ Ω(X, x). We say f , g ∈ Ω(X, x) is path homotopic and write f ∼ g, if there exists a continuous map G : [0, 1] × [0, 1] → X, (s, t) 7→ H(s, t) = Ht(s), such that H0 = f , H1 = g, Ht(0) = Ht(1) = x (t ∈ [0, 1]). We also say that H is a path homotopy from f to g and write H : f ∼ g. Proposition 3.1.2. ∼ is an equivalence relation on Ω(X, x).

PROOF. The proof is almost the same as Proposition 3.1.1. We use the above proposition to deﬁne the fundamental group of X based at x as

(3.1.2.3) π1(X, x) := Ω(X, x)/ ∼= {[ f ]| f ∈ Ω(X, x)}, the set of path classes of loops based at x.

For f , g ∈ Ω(X, x), we can define their product f ∗ g : [0, 1] → X by 1 f (2s), 0 ≤ s ≤ 2 , (3.1.2.4) f ∗ g(s) := 1 g(2s − 1), 2 ≤ s ≤ 1. For [ f ], [g] ∈ π1(X, x), we can also define (3.1.2.5) [ f ] ∗ [g] := [ f ∗ g]. Firstly, we shall verify that the above definition (3.1.2.5) is well-defined, that is, if F : f ∼ f 0 and G : g ∼ g0, then H : f ∗ g ∼ f 0 ∗ g0. Indeed, define 1 F(2s, t), 0 ≤ s ≤ 2 , 0 ≤ t ≤ 1, H(s, t) = 1 G(2s − 1, t), 2 ≤ s ≤ 1, 0 ≤ t ≤ 1.

Theorem 3.1.3. For any (X, x) ∈ Ob(Top∗), (π1(X, x), ∗) is a group.

PROOF. We shall verify

(1) [cx] is the identity element in π1(X, x):

[cx] ∗ [ f ] = [ f ] ∗ [cx] = [ f ]. (2) The inverse of [ f ] is [ f ]:

[ f ] ∗ [ f ] = [ f ] ∗ [ f ] = [cx]. (3) ∗ is associative: [ f ] ∗ ([g] ∗ [h]) = ([ f ] ∗ [g]) ∗ [h]. 3.1. HOMOTOPY 113

For (1): we have H : f ∼ cx ∗ f , where x, t ≥ 2s, H(s, t) = 2s−t f 2−t , t ≤ 2s.

Also, we have G : f ∼ f ∗ cx, where f 2s , t ≤ 2 − 2s, G(s, t) = 2−t x, t ≥ 2 − 2s.

For (2): we have H : cx ∼ f ∗ f , where  t f (2s), 0 ≤ s ≤ 2 ,  t t H(s, t) = f (t), 2 ≤ s ≤ 1 − 2 ,  t f (2 − 2s), 1 − 2 ≤ s ≤ 1.

Also, we have G : cx ∼ f ∗ f , where  t f (1 − 2s), 0 ≤ s ≤ 2 ,  t t H(s, t) = f (1 − t), 2 ≤ s ≤ 1 − 2 ,  t f (2s − 1), 1 − 2 ≤ s ≤ 1. For (3): we have H : ( f ∗ g) ∗ h ∼ f ∗ (g ∗ h).

Exercise 3.1.4. Prove ( f ∗ g) ∗ h ∼ f ∗ (g ∗ h) in Theorem 3.1.3. Suppose now that X is path-connected. For any x, y ∈ X we can deﬁne a natural map

(3.1.2.6) Φg : π1(X, x) −→ π1(X, y), [ f ] 7−→ [g ∗ f ∗ g] = [g] ∗ [ f ] ∗ [g], where g is a path from x to y. Proposition 3.1.5. If X is path-connected, then the map (3.1.2.6) is a group isomorphism with inverse Φg.

PROOF. For any [ f1], [ f2] ∈ π1(X, x) we have

Φg([ f1]) ∗ Φg([ f2]) = ([g] ∗ [ f1] ∗ [g]) ∗ ([g] ∗ [ f2] ∗ [g])

= [g] ∗ [ f1] ∗ ([g] ∗ [g]) ∗ [ f2] ∗ [g] = [g] ∗ [ f1] ∗ [ f2] ∗ [g] = Φg([ f1] ∗ [ f2]).

Hence Φg is a group homomorphism. To prove that Φg is an isomorphism we notice that

Φg ◦ Φg([ f ]) = Φg([g] ∗ [ f ] ∗ [g]) = [g] ∗ ([g] ∗ [ f ] ∗ [g]) ∗ [g] = [ f ].

Similarly one has Φg ◦ Φg([ f ]) = [ f ]. As a consequence of Theorem 3.1.5, for any path-connected topological space X and any two points x, y ∈ X, we have that π1(X, x) is isomorphic to π1(X, y). Hence we can use the notation “π1(X)” to denote “the” group π1(X, x), relative to any point x ∈ X. We say a path-connected topological space is simply-connected if π1(X, x) = {[cx]} (that is, π1(X) is trivial). 114 3. FUNDAMENTAL GROUPS

n n n Example 3.1.6. (1) π1(R ) is trivial. For any ﬁxed point x ∈ R and any f ∈ Ω(R , x), we have

f ∼ cx, since

H(s, t) = (1 − t) f (s) + tcx(s), (s, t) × [0, 1] × [0, 1], n n is a homotopic from f to cx. Hence [ f ] = [cx] for any f ∈ Ω(R , x) and then c1(R ) = {[cx]}. n (2) For any convex domain X ⊆ R , the fundamental group c1(X, x) is trivial. Indeed, for any f ∈ Ω(X, x) deﬁne H as in (1), so that f ∼ cx. n (3) π1(S ) is trivial, for any n ≥ 2. ∈ (( ) ( )) → Consider a morphism ϕ HomTop∗ X, x , Y, y . Then ϕ : X Y is a continuous map with ϕ(x) = y. Deﬁne

(3.1.2.7) ϕ∗ : π1(X, x) −→ π1(Y, y), [ f ] 7−→ [ϕ ◦ f ].

The same argument as for Proposition 3.1.5 shows that ϕ∗ is a group homomorphism. We call ϕ∗ the homomorphism induced by ϕ. The above discussion lets us to deﬁne a functor

(3.1.2.8) π1 : Top∗ −→ Group, (X, x) 7−→ π1(X, x), with π1(ϕ) = ϕ∗.

Theorem 3.1.7. π1 deﬁned in (3.1.2.8) is a functor from Top∗ to Group.

PROOF. We shall verify the following two things: ( ) ∈ (( ) ( )) ∈ (( ) ( )) 1 If ϕ HomTop∗ X, x , Y, y and ψ HomTop∗ Y, y , Z, z , then

(ψ ◦ ϕ)∗ = ψ∗ ◦ ϕ∗.

(2) For any x ∈ X, we have

(1 ) = 1 X ∗ π1(X,x).

For (1),

(ψ ◦ ϕ)∗([ f ]) = [ψ ◦ ϕ ◦ f ] = ψ∗([ϕ ◦ f ]) = ψ∗(ϕ∗([ f ])). For (2), (1 ) ([ f ]) = [1 ◦ f ] = [ f ] = 1 ([ f ]) X ∗ X π1(X,x) .

Consequently, π1 is a functor.

Corollary 3.1.8. If ϕ : (X, x) → (Y, y) is a homeomorphism in Top∗, then the induced homomorphism ϕ∗ : π1(X, x) → π1(Y, y) is a group isomorphism. 3.1. HOMOTOPY 115

3.1.3. Fundamental groups of manifolds. Recall the Lebesgue number lemma, Theorem 1.5.11. Theorem 3.1.9. The fundamental group of a topological manifold is countable.

PROOF. Let M be a topological m-manifold and take any ﬁxed point x ∈ M. To prove π1(M, x) is countable, we shall analyze any loop f : [0, 1] → M at x. (1) According to Proposition 1.5.18, we can choose a countable open cover U of M in such a way that each element U ∈ U is isomorphic to some open ball in Rm (so that U is path-connected). For any given U, U 0 ∈ U , the intersection U ∩ U 0 0 0 has at most countably many connected components (U ∩ U )[i], i ∈ I(U, U ) ⊆ N; 0 0 otherwise U is uncountable. For each i ∈ I(U, U ), choose a point xi ∈ (U ∩ U )[i]. Consider the set n 0 0 0 o ∆ := xi ∈ (U ∩ U )[i]|i ∈ I(U, U ), U, U ∈ U . Because U and I(U, U 0) are countable, we see that the set ∆ is also countable. (2) For each U ∈ U and each pair (x, x0) in ∆ with x, x0 ∈ U, we can take a U 0 path hx,x0 from x to x in U, since U is path-connected. Applying Theorem 1.5.11 to the open cover

−1 n −1 o U f := f ( f ([0, 1]) ∩ U ) = f (U ∩ f ([0, 1])) of the compact interval [0, 1], we obtain a number δ > 0 such that any subset A ⊆ [0, 1] with diameter diam(A) < δ is contained in some element of U f . In particular, taking n ∈ N with n > 1/δ yields that f ([(k − 1)/n, k/n]) ⊆ Uk ∈ U. Write s + k − 1 f (s) := f | , s ∈ [0, 1]. k [(k−1).n,k/n] n

Then [ f ] = [ f1] ··· [ fn]. (3) For each k = 1, ··· , n − 1, take a point xk ∈ Uk ∩ Uk+1. Since f (k/n) ∈ Uk ∩ Uk+1 and Uk ∩ Uk+1 is path-connected, it follows that we can choose a path gk in Uk ∩ Uk+1 from xk to f (k/n). Let 0 fk := gk−1 ∗ fk ∗ gk.

For k = 0 or k = n, take xk = p and gk = cx the constant path. Therefore 0 0 [ f ] = [ f1] ∗ · · · ∗ [ fn]. 0 Uk U Now fk and hxk−1,xk both connect xk−1 to xk in k, using the simple-connectedness U 0 Uk of k, we see that fk is homotopic to hxk−1,xk . Consequently h U i h i [ f ] = h 1 ∗ · · · ∗ hUn x0,x1 xn−1,xn .

Hence π1(M, x) is countable. The above argument also shows that the fundamental group of a compact manifold is ﬁnite.

To improve the result in Theorem 3.1.9 we shall assume some curvature conditions. 116 3. FUNDAMENTAL GROUPS

Theorem 3.1.10. (Bonnet-Myers) If (M, g) is a complete Riemannian√ m-manifold with Ricg ≥ (m − 1)Kg, where K > 0, then diam(M, g) ≤ π/ K. In particular, M is compact and π1(M) < +∞.

PROOF. We will give a proof in the course of Differential Manifold. If the Ricci curvature bound is replaced by the sectional sectional curvature bound, we have the following outstanding result. Theorem 3.1.11. (Cheeger-Gromoll soul theorem) If (M, g) is a complete Riemann- m ian m-manifold with Secg > 0 everywhere, then M is diffeomorphic to R . In 1994, Perelman proved the so-called soul conjecture. Theorem 3.1.12. (Soul conjecture, solved by Perelman)If (M, g) is a complete Rie- mannian m-manifold with Secg ≥ 0 everywhere and > 0 at some point, then M is diffeomorphic to Rm. For nonnegative Ricci curvature we have the following well-known conjecture. Conjecture 3.1.13. (Milnor’s conjecture) If (M, g) is a complete Riemannian m- manifold with Ricg ≥ 0, then π1(M) is ﬁnite generated. When m = 3, the above conjecture was recently solved by Gang Liu under the previous work of Schoen-Yau. Under an extra condition that g has maximum volume growth, Peter Li proved Conjecture 3.1.13 in 1986 (actually, he proved that π1(M) is ﬁnite). Recall that if Ricg ≥ 0, then the volume radio

Volg(Bg(x, r)) m ωmr is nonincreasing. Then Vol (B (x, r)) g g = ∈ [ ] lim m θ 0, 1 . r→∞ ωmr We say the Riemannian metric g has maximum volume growth, if θ ∈ (0, 1]. When θ = 1, we can prove that in this case (M, g) is isometric to Rm.

On the other hand, a simply-connected manifold with suitable curvature conditions has a rigidity result. Theorem 3.1.14. (Rauch-Klingenberg-Berger topological sphere theorem) If (M, g) is a complete, simply-connected Riemannian n-manifold with 1 ≤ Sec (Π) ≤ 1 4 g for all 2-planes Π, then M is homeomorphic to Sm. In particular, M3 is diffeomorphic to S3. How about the diffeomorphisms between Mm and Sm when the dimension m ≥ 4? The answer is YES! It was proved by Brendle and Schoen by using the Ricci ﬂow. 3.1. HOMOTOPY 117

Theorem 3.1.15. (Diffeomorphic spheral theorem) If (M, g) is a complete, simply- connected Riemannian n-manifold with 1 ≤ Sec (Π) ≤ 1 4 g for all 2-planes Π, then M is diffeomorphic to Sm.

3.1.4. Fundamental groups of product spaces. Let {(Xi, Ti)}1≤i‘n be a ﬁnite family of topological spaces, and let

(3.1.4.1) pi : X1 × · · · × Xn −→ Xi, (x1, ··· , xn) 7−→ xi denote the i-th projection. Then we get group homomorphisms

(3.1.4.2) pi,∗ : π1(X1 × · · · × Xn, (x1, ··· , xn)) −→ π1(Xi, xi) by (3.1.2.7). Consequently, we can deﬁne a group homomorphism

(3.1.4.3) p∗ : π1(X1 × · · · × Xn, (x1, ··· , xn)) −→ π1(X1, x1) × · · · × π1(Xn, xn) given by

(3.1.4.4) p∗([ f ]) := (p1,∗[ f ], ··· , pn,∗[ f ]).

Theorem 3.1.16. The map (3.1.4.4) is an isomorphism.

PROOF. We shall prove the injectivity and surjectivity of p∗.

(1) p∗ is injective. Suppose p∗([ f ]) = ([cx1 ], ··· , [cxn ]). Write

f = ( f1, ··· , fn).

Then pi ◦ f = fi and

([cx1 ], ··· , [cxn ]) = ([ f1], ··· , [ fn]). f ∼ c ≤ i ≤ n f ∼ c Thus i xi , 1 and hence (x1,··· ,xn). (2) p∗ is surjective. For ([ f1], ··· , [ fn] ∈ π1(X1, x1) × · · · × π1(Xn, xn), deﬁne

f (s) := ( f1(s), ··· , fn(s)), s ∈ [0, 1].

Then p∗([ f ]) = ([ f1], ··· , [ fn]). 3.1.5. Retractions and deformation retractions. Let (X, T ) be a topological space and A ⊆ X a subset. Under the subspace topology TA := T ∩ A, we see that A is itself a topological space.

(1) A continuous map r : X → A is called a retraction if r ◦ ιA = 1A, where ιA : A → X is the inclusion. If there exists a retraction from X to A, we say that A is a retract of X. It is clear that a retract of a connected space (resp., compact space) is connected (resp., compact). Moreover, if A ⊆ B ⊆ X, A is a retract of B, and B is a retract of X, then A is a retract of X. (2) A retraction r : X → A is said to be a deformation retraction if ιA ◦ r is homotopic to 1X. If there exists a deformation retraction from X to A, then A is said to be a deformation retract of X. That is, A ⊆ X is 118 3. FUNDAMENTAL GROUPS

a deformation retract of X if and only if there exists a homotopy H : X × [0, 1] → X satisfying

H(x, 0) ≡ H0(x) = x, x ∈ X, H(x, 1) ≡ H1(x) ∈ A, x ∈ X, H(a, 1) ≡ H1(a) = a, a ∈ A.

Let f ∈ HomTop(X, Y). A morphism g ∈ HomTop(Y, X) is a homotopy inverse for f if g ◦ f ' 1X, f ◦ g ' 1Y. If there exists a homotopy inverse for , then is called a homotopy equivalence and we say that X is homotopic to Y (write X ' Y). Exercise 3.1.17. Homotopy equivalence is an equivalence relation on the class of all topological spaces Ob(Top).

If r : X → A is a deformation retraction, then, according to r ◦ ιA = 1A, r and ιA are homotopy equivalences.

Proposition 3.1.18. (1) If r : X → A is a retraction, then for any x ∈ A, ιA,∗ : π1(X, x) → π1(X, x) is injective and r∗ : π1(X, x) → π1(A, x) is surjective. (2) A retract of a simply-connected space is also simply-connected.

ROOF ( ) r ◦ ι = 1 1 = r ◦ ι P . 1 From A A we obtain π1(A,x) ∗ A,∗ which implies the desired results. (2) Since π1(X, x) is trivial and ιA,∗ : π1(A, x) → π1(X, x) is injective, it follows that π1(A, x) is also trivial.

Example 3.1.19. (1) The map r : Rm \{0} → Sm−1 given by x r(x) := |x| 1 ∼ is a retraction. Since π1(S ) = Z (we shall prove soon), it follows from Proposition 3.1.18 that R2 \{0} is not simply-connected. Consequently, R2 \{0} is not homeomorphic to R2. (2) Since the map r : T2 = S1 × S1 → S1 × {1} given by r(x, y) := (x, 1) is a retraction, it follows that T2 is not simply-connected and then T2 is not homeomorphic to S2.

Let X, Y ∈ Ob(Top) and A ⊆ X. A homotopy H between f , g ∈ HomTop(X, Y) is said to be stationary on A if H(x, t) = f (x), ∀ (x, t) ∈ A × [0, 1]. If there exists such a homotopy H, we say that f and g are homotopic relative to A, written as H : f ' g (rel A), and H is called a homotopy relative to A. In this case, f = g on A.

It is clear that homotopy relative to A is an equivalence relation in HomTop(X, Y). 3.1. HOMOTOPY 119

A retraction r : X → A is called a strong deformation retraction if 1X is homotopic to ιA ◦ r relative to A. That is, A ⊆ X is a strong deformation retract of X if and only if there exists a homotopy H : X × [0, 1] → X satisfying

H(x, 0) ≡ H0(x) = x, x ∈ X, H(x, 1) ≡ H1(x) ∈ A, x ∈ X, H(a, 1) ≡ H1(a) = a, a ∈ A, H(a, t) ≡ Ht(a) = a, a ∈ A, t ∈ [0, 1].

Example 3.1.20. (1) For any m ≥ 1, Sm−1 is a strong deformation retract of Rm \{0} and of Bm \{0}. Indeed, consider the homotopy x H : (Rn \{0}) × [0, 1] −→ Rm \{0}, (x, t) 7−→ (1 − t)x + t . |x| (2) For m ≥ 3, Rm \{0} and Bm \{0} are both simply-connected. Next result generalizes Corollary 3.1.8. Theorem 3.1.21. (Homotopy invariance) If ϕ : X → Y is a homotopy equivalence, then for any point x ∈ X, ϕ∗ : π1(X, x) → π1(Y, ϕ(x)) is an isomorphism.

PROOF. Let ψ : Y → X be a homotopy inverse for ϕ. Then, by deﬁnition, we have ψ ◦ ϕ ' 1X and ϕ ◦ ψ ' 1Y. Moreover ϕ, ψ induce ϕ∗ ψ∗ ϕ∗ π1(X, x) −−−−→ π1(Y, ϕ(x)) −−−−→ π1(X, ψ(ϕ(x))) −−−−→ π1(Y, ϕ(ψ(ϕ(x)))).

Lemma 3.1.22. If ϕ, ψ ∈ HomTop(X, Y) and H : ϕ ' ψ is a homotopy. For any x ∈ X, let h be the path in Y from ϕ(x) to ψ(x) deﬁned by h(t) := H(x, t), t ∈ [0, 1], and let Φh : π1(Y, ϕ(x)) → π1(Y, ψ(x)) be the isomorphism deﬁned in (3.1.2.6). Then the following diagram ϕ∗ π1(X, x) −−−−→ π1(Y, ϕ(x))   yΦh ψ∗ π1(X, x) −−−−→ π1(Y, ψ(x)) is commutative.

PROOF. For [ f ] ∈ π1(X, x) we need to verify

ψ∗([ f ]) = Φh(ϕ∗([ f ])) which is equivalent to h ∗ (ψ ◦ f ) ∼ (ϕ ◦ f ) ∗ h or (ϕ ◦ f ) ∗ h ∼ h ∗ (ψ ◦ f ). Consider the map F : [0, 1] × [0, 1] −→ Y, (s, t) 7−→ H( f (s), t). Then F(s, 0) = H( f (s), 0) = ϕ( f (s)), F(1, s) = H( f (1), s) = H(x, s) = h(s) and F(0, s) = H( f (0), s) = H(x, s) = h(s), F(s, 1) = H( f (s), 1) = ψ( f (s)). Lemma 3.1.23 shows that (ϕ ◦ f ) ∗ h ∼ h ∗ (ψ ◦ f ). 120 3. FUNDAMENTAL GROUPS

Lemma 3.1.23. (Square lemma) Let F : [0, 1] × [0, 1] → X be a continuous map, and let f , g, h, k be the paths in X deﬁned by f (s) := F(s, 0), g(s) := F(1, s), h(s) := F(0, s), k(s) := F(s, 1). Then f ∗ g ∼ h ∗ k.

PROOF. Consider the diagram map d(s) := F(s, s). It is clear that f ∗ g ∼ d (Exercise 3.1.24). Similarly, h ∗ k ∼ d. Hence f ∗ g ∼ h ∗ k.

Since H : 1X ' ψ ◦ ϕ, it follows that h deﬁned by h(t) := H(x, t), t ∈ [0, 1], is a path in X from x to ψ(ϕ(x)). By Lemma 3.1.22, we have the following commutative diagram: 1X,∗ π1(X, x) −−−−→ π1(X, x)   yΦh

(ψ◦ϕ)∗ π1(X, x) −−−−→ π1(Y, ψ(ϕ(x))) Thus Φh = ψ∗ ◦ ϕ∗. Consequently, ψ∗ is surjective and ϕ∗ is injective. Similarly, since G : 1Y ' ϕ ◦ ψ, it follows that g deﬁned by g(t) := G(ϕ(x), t), t ∈ [0, 1], is a path in Y from ϕ(x) to ϕ(ψ(ϕ(x))). By Lemma 3.1.22, we have the following commutative diagram:

1Y,∗ π1(Y, ϕ(x)) −−−−→ π1(Y, ϕ(x))   yΦg

(ϕ◦ψ)∗ π1(Y, ϕ(x)) −−−−→ π1(Y, ϕ(ψ(ϕ(x)))) Thus Φg = ϕ∗ ◦ ψ∗. Consequently, ϕ∗ is surjective and ψ∗ is injective. Therefore ϕ∗ is isomorphic.

Exercise 3.1.24. Verify f ∗ g ∼ d in Lemma 3.1.23.

3.1.6. Contractible spaces. A topological space X is said to be contractible if 1X is homotopic to some constant map. Equivalently, X is contractible, if there exists a point x0 ∈ X and a continuous map H : X × [0, 1] → X such that

H(x, 0) = x (x ∈ X), H(x, 1) = x0 (x ∈ X). Thus, X is contractible if, intuitively, X can be continuously shrunk to a point. (1) Rn is contractible, because we can take H(x, t) = (1 − t)x. (2) [0, 1] is contractible, because we can take H(x, t) := (1 − t)x. (3) Any convex subset X of Rn is contractible, because we can take H(x, t) = (1 − t)x + tx0. (4) Any star-shaped subset X of Rn is contractible. 3.1. HOMOTOPY 121

Proposition 3.1.25. (1) If f , g ∈ HomTop(X, Y) and Y is contractible, then f ' g. (2) If X is contractible, then any two constant maps on X are homotopic and the identity map 1X is homotopic to any constant map on X. (3) A topological space X is contractible if and only if X is homotopic to a one-point space. (4) Two contractible spaces are homotopic and any continuous map between contractible spaces is a homotopy equivalence.

PROOF. (1) Assume 1Y ' cy0 , where cy0 is a constant map on Y. Then f =

1Y f ' cy0 f = cy0 g ' 1Y g = g. (2) Trivial. (3) If X is homotopic to a one-point space Y = {y0}, then we have a homotopy equivalence f : X → Y with homotopy inverse g : Y → X. Then 1X ' g f = cx0 , where x0 := g(y0). Thus X is contractible. Conversely, assume that X is contractible. We can ﬁnd a point x0 ∈ X and continuous map H : X × [0, 1] → X such that H(x, 0) = x and F(x, 1) = x0 for all x ∈ X. Set Y := {x0} and consider two continuous maps f : X → Y given by f (x) := x0 and g : Y → X given by g(x0) = x0. Then

f ◦ g = 1Y, g ◦ f = cx0 ' 1X. Therefore f is a homotopy equivalence from X to Y with homotopy inverse g. (4) Let X and Y be two contractible spaces. By (3), X is homotopic to some one-point space {x0}; thus X '{x0}. Similarly, Y '{y0} for some y0 ∈ Y. Since {x0}'{y0}, it follows that X ' Y. Next, take a map f ∈ HomTop(X, Y) with X, Y being contractible. The ﬁrst part gives two maps f1 ∈ HomTop(X, Y) and g1 ∈ HomTop(Y, X) such that f1 ◦ g1 ' 1Y and g1 ◦ f1 ' 1X. Because f ◦ g1 ∈ HomTop(Y, X) and g1 ◦ f ∈ HomTop(X, Y). According to (1), we get f ◦ g1 ' 1Y and g1 ◦ f ' 1X. Thus f is a homotopy equivalence.

Example 3.1.26. Proposition 3.1.25 (1) is not true in the case of relative homotopy. That is, if f , g ∈ HomTop(X, Y), f |A = g|A for some open subset A ⊆ X, and Y is contractible, then we may not have f ' g (rel A). Consider the comb space Y deﬁned by 1 Y = ({0} × [0, 1]) ∪ ([0, 1] × {0}) ∪ (x, y) ∈ R2 y ∈ [0, 1], x = , n = 1, 2, ··· . n Let H : Y × [0, 1] → Y be deﬁned by H((x, y), t) := (x, (1 − t)y).

Then H is a homotopy from 1Y to the projection of Y to the x-axis, and so 1Y is homotopic to some constant map on Y. Thus Y is contractible. Deﬁne c : Y → Y, (x, y) 7→ (0, 1). Using Proposition 3.1.25 yields that 1Y ' c and 1Y|A = c|A where A := {(0, 1)}. However we can not ﬁnd a homotopy from 1Y to c relative to A.

We say a map f ∈ HomTop(X, Y) is null homotopic if it is homotopic to some constant map. n n Theorem 3.1.27. Let x0 ∈ S and let f ∈ HomTop(S , Y). Then TFAE: (1) f is null homotopic. 122 3. FUNDAMENTAL GROUPS

(2) f can be continuously extend over Dn+1 := Bn+1. (3) f is null homotopic relative to x0, that is, f is homotopic to some constant map relative to x0.

ROOF ( ) ⇒ ( ) ' ∈ P . 1 2 : Let H : f cy0 for some y0 Y. Deﬁne an extension feof f over Dn+1 by 1 y0, 0 ≤ |x| ≤ 2 , fe(x) := 1 H (x/|x|, 2(1 − |x|)) , 2 ≤ |x| ≤ 1. Then feis continuous. (2) ⇒ (3): Assume that f has a continuous extension fe : Dn+1 → Y. Deﬁne H : Sn × [0, 1] → Y by

H(x, t) := fe((1 − t)x + tx0) . n Then H(x, 0) = fe(x) = f (x) for all x ∈ S and H(x, 1) = fe(x0) = cx0 . (3) ⇒ (1): Trivial.

Corollary 3.1.28. Any continuous map from Sn to a contractible space has a continuous extension over Dn+1.

PROOF. By Proposition 3.1.25 (1) and Theorem 3.1.26.

3.2. H-spaces The homotopy category HTop is the category whose objects are topological spaces and whose morphisms are homotopy classes of continuous maps. That is

Ob(HTop) = Ob(Top), HomHTop(X, Y) := [X, Y] where the last one was deﬁned in (3.1.1.1). By Proposition 3.1.1, we see that HTop is indeed a category and the composition is deﬁned as

HomHtop(X, Y) × HomHTop(Y, Z) −→ HomHtop(X, Z), ([ f ], [g]) 7−→ [g ◦ f ]. Deﬁne a functor [·] : Top → HTop to be [·] : Ob(Top) −→ Ob(HTop), X 7−→ X and [·] : HomTop(X, Y) −→ HomHTop(X, Y), f 7−→ [ f ]. Thus the above functor [·] is covariant.

We extend the homotopy category of topological spaces to the homotopy category of topological pairs. Deﬁnition 3.2.1. A topological pair (X, A) consists of a topological space X and a subspace A ⊆ X. If A is empty, we write simply X for (X, ∅).A topological subpair (X0, A0) ⊆ (X, A) consists of a topological pair (X0, A0) with X0 ⊆ X and A0 ⊆ A.A topological pair map f : (X, A) → (Y, B) between topological pairs is a continuous map f ∈ HomTop(X, Y) such that f (A) ⊆ B. 3.2. H-SPACES 123

The above concept generalizes the pointed topological spaces and the pointed- preserving maps. Let Top∗∗ denote the category of topological pairs and topological pair maps. Hence Top and Top∗ are both (fully) subcategories of Top∗∗.

Given a topological pair (X, A) ∈ Ob(Top∗∗), we let (X, A) × [0, 1] denote the topological pair (X × [0, 1], A × [0, 1]) ∈ Ob(Top∗∗). ( ) 0 ⊆ ∈ (( ) ( )) 0 1 Let X X and suppose f0, f1 HomTop∗∗ X, A , Y, B agree on X , 0 that is, f0|X0 = f1|X0 . We say f0 is homotopic to f1 relative to X , de- ' ( 0) ∈ (( ) × noted by f0 f1 rel X , if there exists a map F HomTop∗∗ X, A [0, 1], (Y, B)) such that

F(x, 0) = f0(x), F(x, 1) = f1(x), x ∈ X and 0 F(x, t) = f0(x), x ∈ X , t ∈ [0, 1]. 0 We also say F is a homotopy relative to X from f0 to f1 and denote by 0 F : f0 ' f1 (rel X ). 0 (2) When X = ∅, we return back to the previous definition of f0 ' f1. 0 00 00 0 (3) Note that f0 ' f1 (rel X ) implies f0 ' f1 (rel X ) whenever X ⊆ X . ( ) ( ) = ( ) ∈ (( ) ( )) 4 We also write Ft x : F x, t so that Ft HomTop∗∗ X, A , Y, B is a continuous one-parameter family of topological pair maps from (X, A) to (Y, B). Proposition 3.1.1 can be generalized to the following Proposition 3.2.2. (1) Homotopy relative to X0 is an equivalence relation in (( ) ( )) HomTop∗∗ X, A , Y, B . (2) Homotopy relative to X0 is preserved by composition. That is, if ∈ (( ) ( )) ∈ (( ) ( )) f0, f1 HomTop∗∗ X, A , Y, B , g0, g1 HomTop∗∗ Y, B , Z, C , 0 0 0 0 0 and f0 ' f1 (rel X ), g0 ' g1 (rel Y ), and f0(X ) = f1(X ) ⊆ Y , then g0 ◦ f0 ' 0 g1 ◦ f1 (rel X ). Exercise 3.2.3. Prove Proposition 3.2.2. The above proposition allows us to define the homotopy class relative to X0 0 [( ) ( )] 0 = [ ] 0 ∈ (( ) ( )) X, A , Y, B X : f X : f HomTop∗∗ X, A , Y, B . We also can define the homotopy category of topological pairs HTop∗∗ whose objects are topological pairs and whose morphisms are homotopy classes (relative to ∅). We also have a covariant functor from Top∗∗ to HTop∗∗. (1) For any (P, Q) ∈ Ob(HTop∗∗), there is a covariant functor [( ) (· ·)] = (( ) (· ·)) P, Q , , : HomHTop∗∗ P, Q , , from HTop∗∗ to Set, satisfying [( ) (· ·)](( )) = (( ) ( )) P, Q , , X, A : HomHTop∗∗ P, Q , X, A [ ] ∈ (( ) ( )) and for f HomHTop∗∗ X, A , Y, B ,

f# ≡ [(P, Q), (·, ·)]([ f ]) ∈ HomSet ([(P, Q), (X, A)], [(P, Q), (Y, B)]) where f#([g]) := [ f ◦ g], [g] ∈ [(P, Q), (X, A)]. 124 3. FUNDAMENTAL GROUPS

(2) A topological pair map f : (X, A) → (Y, B) is called a homotopy equivalence if [ f ] is an equivalence in HTop∗∗. Equivalently, there is a topological pair map g : (Y, B) → (X, A), called a homotopy inverse of f , such that f ◦ g ' 1Y, g ◦ f ' 1X. Pairs (X, A) and (Y, B) are said to have the same homotopy type if there are equivalent in HTop.

3.2.1. Two notions on functors. Let C be a category and take an object Y ∈ C.

(1) Deﬁne a covariant functor πY : C → Set by

(3.2.1.1) πY := Hom(Y, ·) on Ob(C) and 0 (3.2.1.2) πY(h) := h# : HomC(Y, Z) −→ HomC(Y , Z), g 7−→ h ◦ g, 0 for any h ∈ HomC(Z, Z ). (2) Deﬁne a covariant functor πY : C → Set by (3.2.1.3) πY := Hom(·, Y) on Ob(C) and Y 0 0 0 (3.2.1.4) π ( f ) := h# : HomC(X , Y) −→ HomC(X, Y), g 7−→ g ◦ f , 0 for any f ∈ HomC(X, X ).

3.2.2. H-spaces. If (X, x0), (Y, y0) ∈ Ob(Top∗), we denote by

(3.2.2.1) [X, Y] := [(X, x0), (Y, y0)] the set of morphisms from X to Y in HTop∗. Theorem 3.2.4. If P is a topological group with the identity element, πP is a contravariant functor from HTop∗ to Group.

PROOF. (1) [X, P] is a group for any given pointed topological space X. If [ ] [ ] ∈ ( ) g1 , g2 HomHTop∗ X, P , we deﬁne

[g1][g2] := [g1g2] with g1g2 : X → P given by (g1g2)(x) := g1(x)g2(x). Clearly that under this multiplication, [X, P] is a group. (2) πP is a contravariant functor.

Exercise 3.2.5. Complete the proof of Theorem 3.2.4. It is clear that [X, P] is Abelian if P is Abelian. For example, [X, S1] is an Abelian group.

An H-space (P, µ) consists of a pointed topological space P = (P, p0) together with a continuous multiplication (3.2.2.2) µ : P × P −→ P 3.2. H-SPACES 125

for which the constant map c : P → P, given by c = cp0 , is a homotopy identity, that is, each composite, 1 := 1P

(c,1) µ (1,c) µ P −−−−→ P × P −−−−→ P, P −−−−→ P × P −−−−→ P is homotopic to 1. (1) The multiplication µ is said to be homotopy associative if the square

µ×1 P × P × P −−−−→ P × P    µ 1×µy y P × P −−−−→ P µ is homotopy commutative, that is, µ ◦ (µ × 1) ' µ ◦ (1 × µ). (2) A continuous map ϕ : P → P is called a homotopy inverse for P and µ if each composite

(ϕ,1) µ (1,ϕ) µ P −−−−→ P × P −−−−→ P, P −−−−→ P × P −−−−→ P is homotopic to c. (3) An H-group (P, µ, ϕ) is a homotopy associative H-space with a homotopy inverse. (4) Any topological group is an H-group. (5) A multiplication µ in an H-space P is said to be homotopy Abelian if the triangle

T P × P −−−−→ P × P   µ µ T(p1, p2) := (p2, p1), y y P P is homotopy commutative, that is, µ ◦ T ' µ. (6) An H-group with homotopy Abelian multiplication is called an Abelian H-group. (7) An H-space homomorphism between two H-spaces (P, µ) and (P0, µ0) is a continuous map α : P → P0 such that the square µ P × P −−−−→ P     α×αy yα P0 × P0 −−−−→ P0 µ0 is homotopy commutative, that is, α ◦ µ ' µ0 ◦ (α × α). 126 3. FUNDAMENTAL GROUPS

Theorem 3.2.6. A pointed topological space having the same homotopy type as an H- space (or an H-group, or an Abelian H-group), is itself an H-space (or H-group, or an Abelian H-group) in such a way that the homotopy equivalence is an H-space homomorphism.

0 0 PROOF. (1) Let (P, µ) be an H-space, and (P , p0) is a pointed topological space that has the same homotopy type as P. Let f : P → P0 and g : P0 → P be homotopy inverses. Deﬁne g×g µ f P0 × P0 −−−−→ P × P −−−−→ P −−−−→ P0 µ as µ0 := f ◦ µ ◦ (g × g). Then µ0 is a continuous map. Next we shall verify that each composite (c0,10) µ0 (10,c0) µ0 P0 −−−−→ P0 × P0 −−−−→ P0, P0 −−−−→ P0 × P0 −−−−→ P0 is homotopic to 1P0 . Since (10,c0) µ0 P0 −−−−→ P0 × P0 −−−−→ P0  x g  y  f (1,c) µ P −−−−→ P × P −−−−→ P 0 0 the above diagram shows f ◦ µ ◦ (1, c) ◦ g = µ ◦ (1, c ) and µ ◦ (1, c) ' 1P, it follows that 0 0 0 µ ◦ (1 , c ) ' 1P0 . Similarly, from (c0,10) µ0 P0 −−−−→ P0 × P0 −−−−→ P0  x g  y  f (c,1) µ P −−−−→ P × P −−−−→ P 0 0 0 the above diagram shows f ◦ µ ◦ (c, 1) ◦ g = µ ◦ (c , 1 ) and µ ◦ (c, 1) ' 1P, it follows that 0 0 0 µ ◦ (c , 1 ) ' 1P0 . Consequently, (P0, µ0) is an H-space. (2) Consider the diagrams µ µ0 P × P −−−−→ P P0 × P0 −−−−→ P0     f × f   f ×  g y y g gy y 0 0 0 P × P −−−−→ P P × P −−−−→ P µ0 µ Because µ0 ◦ ( f × f ) = f ◦ µ ◦ (g × g) ◦ ( f × f )

= f ◦ µ ◦ (g ◦ f ) × (g ◦ f ) ' f ◦ µ ◦ (1P × 1P) ' f ◦ µ and g ◦ µ0 = g ◦ f ◦ µ ◦ (g × g) ' µ ◦ (g × g), we see that g and f are both H-space homomorphisms. 3.2. H-SPACES 127

(3) If µ is homotopy associative, so is µ0. (4) If µ is homotopy Abelian, so is µ0. (5) If ϕ : P → P is a homotopy inverse for P, then ϕ0 := f ◦ ϕ ◦ g : P0 → P0 is 0 a homotopy inverse for P .

Exercise 3.2.7. Verify (3), (4), and (5) in Theorem 3.2.6.

Given an H-space (P, µ) (with base point p0) and a pointed topological space (X, x0), for any [g1], [g2] ∈ [X, P] deﬁne

(3.2.2.3) [g1] ∗ [g2] := [µ ◦ (g1, g2)]. Clearly the above multiplication is well-deﬁned. If (P, µ, ϕ) is an H-group, then [X, P] is a group under (3.2.2.3). Note that in (3.2.2.3), the multiplication ∗ is determined only by µ and then by P. If we consider [g1], [g2] ∈ [Y, P], we still use the same notion ∗ for the multiplication of [Y, P]. Exercise 3.2.8. Show that if (P, µ, ϕ) is an H-group, then [X, P] is a group under (3.2.2.3). We then have a contravariant functor P π : HTop∗ −→ Group, X 7−→ [X, P] [ ] ∈ ( ) and for any morphism f HomHTop∗ X, Y , # P 0 0 f := π ([ f ]) ∈ HomGroup([Y, P], [X, P]), [Y, P] 3 g 7−→ g ◦ f . Consequently, we obtain the following P Theorem 3.2.9. If P ≡ (P, µ, ϕ) is an H-group, then π : HTop∗ → Group is a contradvariant functor. If P is an Abelian H-group, then πP is a contravariant functor from HTop∗ into AG.

P Theorem 3.2.10. If P ≡ (P, p0) is a pointed topological space such that π takes values in Group, then P is an H-group (or Abelian H-group if πP takes values in AG).

PROOF. Consider the pointed topological space P × P ≡ (P × P, (p0, p0)) and πP(P × P) = [P × P, P] ∈ Ob(Group). Denote by ∗ the group multiplication of [P × P, P]. The projections pr P × P −−−−→1 P  pr  2y P gives rise to a continuous map µ : P × P → P such that

[pr1] ∗ [pr2] = [µ].

Write c : P → P the constant map cp0 . (1)(P, µ) is an H-space. Consider the following diagrams

(c,1) µ P −−−−→ P × P −−−−→ P (1,c) 128 3. FUNDAMENTAL GROUPS and try to prove µ ◦ (c, 1) ' 1 ' µ ◦ (1, c), where 1 := 1P. We ﬁrst establish a [ ◦ ( )] ∈ ( ) general formula for µ f , g , where f , g HomTop∗ X, P . In fact, # # [µ ◦ ( f , g)] = ( f , g) [µ] = ( f , g) ([pr1] ∗ [pr2])

# # = ( f , g) [pr1] ∗ ( f , g) [pr2] = [ f ] ∗ [g]. Consequently, [µ ◦ (1, c)] = [1] ∗ [c]. Consider the following two continuous maps

f : P −→ {p0}, g : {p0} −→ P deﬁned, respectively, by

f (p) := p0, g(p0) := p0. Since πP takes values in Group, it follows that P π ([ f ]) ∈ HomGroup([{p0}, P], [P, P]). In particular, P([ f ])(e ) = e π [{p0},P] [P,P], where eG stands for the identity element of the given group G. Because [{p0}, P] [g] [g] = e consists only one element , we must have [{p0},P]. Hence P e[P,P] = π ([ f ])([g]) = [g ◦ f ] = [c]. Thus [c] is the identity element in the group [P, P]. Consequently, [µ ◦ (1, c)] = [1] ∗ [c] = [1] ⇐⇒ µ ◦ (1, c) ' 1. Similarly, one has µ ◦ (c, 1) ' 1. (2)(P, µ, ϕ) is an H-group. We next show that µ is homotopy associative, i.e.,

µ×1 P × P × P −−−−→ P × P    µ 1×µy ‘y , µ ◦ (µ × 1) ' µ ◦ (1 × µ). P × P −−−−→ P µ

Let q1, q2, q3 : P × P × P → P be three projections. Then # # [µ ◦ (1 × µ)] = (1 × µ) [µ] = (1 × µ) ([pr1] ∗ [pr2])

# # = (1 × µ) [pr1] ∗P×P×P (1 × µ) [pr2] = [q1] ∗ [µ ◦ (q2, q3)] = [q1] ∗ ([q2] ∗ [q3]). Similarly,

[µ ◦ (µ × 1)] = ([q1] ∗ [q2]) ∗ [q3]. Since ∗ is associative, it follows that [µ ◦ (1 × µ)] = [µ ◦ (µ × 1)]. Thirdly, we show that P has a homotopy inverse. Let [ϕ] ∈ [P, P] be the inverse of [1] in [P, P], i.e., [1] ∗ [ϕ] = [c] = [ϕ] ∗ [1]. Then c ' µ(1, ϕ) and c ' µ(ϕ, 1). 3.2. H-SPACES 129

(3)(P, µ, ϕ) is an Abelian H-group if πP takes values in AG. In this case, ∗ is Abelian. Consider the following diagram T P × P −−−−→ P × P   µ µ , T(p1, p2) = (p2, p1). y y P P

Because T = (pr2, pr1), we obtain

[µ ◦ T] = [µ ◦ (pr2, pr1)] = [pr2] ∗ [pr1] = [pr1] ∗ [pr2] = [µ]. Thus P is Abelian.

0 Corollary 3.2.11. Let α : P → P be a map between H-groups. Then α# is a natural 0 transformation from πP to πP in Group if and only if α is an H-space homomorphism.

[ ] ∈ ( ) PROOF. For f HomHTop∗ X, Y , consider the following diagram

α (X) 0 πP(X) −−−−→# πP (X) x x P   P0 π ( f ) π ( f ) 0 πP(Y) −−−−→ πP (Y) α#(Y) For any [g] ∈ [Y, P], we have

P0 P0 π ( f ) ◦ α#(Y) [g] = π ( f )[α ◦ g] = [(α ◦ g) ◦ f ] and P α#(X) ◦ π ( f ) [g] = α#(X)[g ◦ f ] = [α ◦ (g ◦ f )]. P0 P Hence π ( f ) ◦ α#(Y) = α#(X) ◦ π ( f ). We need only to verify that α#(X) : 0 [X, P] → [X, P ] is a group homomorphism. For [g1], [g2] ∈ [X, P] we have

α#(X)([g1] ∗ [g2]) = α#(X)[µ ◦ (g1, g2)] = [α ◦ µ ◦ (g1, g2)] and 0 α#(X)[g1] ∗ α#(X)[g2] = [α ◦ g1] ∗ [α ◦ g2] = [µ ◦ (α ◦ g1, α ◦ g2)]. 0 Consequently, α#(X) is a group homomorphism if and only if α ◦ µ ' µ ◦ (α × α), i.e., α#(X) is a group homomorphism for any X ∈ Ob(HTop∗) if and only if α is a H-space homomorphism. 3.2.3. A classical example of H-groups. A classical example of H-groups is as follows. Let Y ≡ (Y, y0) be a pointed topological space. The loop space of Y (based at y0), denoted by Ω(Y, y0), is deﬁned to be (3.2.3.1)   continuous functions   Ω(Y, y ) := ω : ([0, 1], {0, 1}) −→ (Y, y ) topologized by the . 0 0  compact-open topology 

We recall the compact-open topology. If (X, A), (Y, B) ∈ Ob(Top∗∗), then (( ) ( )) → HomTop∗∗ X, A , Y, B denotes the set of continuous functions f : X Y such 130 3. FUNDAMENTAL GROUPS that f (A) ⊆ B. Let YX denote the space of continuous functions from X to Y, given the compact-open topology (which is the topology generated by the subbasis (( ) ( )) (3.2.3.2) HomTop∗∗ X, K , Y, U : K compact in X and U open in Y . Thus the compact-open topology is the collection of subsets consisting of YX, ∅, and all unions of ﬁnite intersections of elements of (3.2.3.2). (1) Let (3.2.3.3) E : YX × X −→ Y, ( f , x) 7−→ f (x), denote the evaluation map. Given a function g : Z → YX, the composite

g×1 E (3.2.3.4) Z × X −−−−→ YX × X −−−−→ Y is a function from Z × X to Y. (2) (Theorem of exponential correspondence) If X is locally compact Haus- dorff space and Y, Z ∈ Ob(Top), a map g : Z → YX is continuous if and only if E ◦ (g × 1) : Z × X → Y is continuous.

Theorem 3.2.12. Ω(Y, y0) is an H-group.

PROOF. Deﬁne ω0 : [0, 1] → Y to be ω0(t) ≡ y0. Then ω0 ∈ Ω(Y, y0) and then Ω(Y, y0) ∈ Ob(Top∗). (1) Deﬁne a map

µ : Ω(Y, y0) × Ω(Y, y0) −→ Ω(Y, y0) by 1 0 0 ω(2t), 0 ≤ t ≤ 2 , µ(ω, ω )(t) := (ω ∗ ω )(t) = 0 1 ω (2t − 1), 2 ≤ t ≤ 1. To prove µ is continuous, we use the above Theorem of exponential correspondence. Let E : Ω(Y, y0) × [0, 1] → Y be the evaluation map. It sufﬁces to show that the composite

µ×1 E Ω(Y, y0) × Ω(Y, y0) × [0, 1] −−−−→ Ω(Y, y0) × [0, 1] −−−−→ Y is continuous. By deﬁnition of µ, we see that E ◦ (µ × 1) is continuous so that µ is continuous. (2) To prove

(c,1) µ Ω(Y, y0) −−−−→ Ω(Y, y0) × Ω(Y, y0) −−−−→ Ω(Y, y0), µ ◦ (c, 1) ' 1 ' µ ◦ (1, c) (1,c) = 1 where 1 : Ω(Y,y0), we deﬁne a map 2s t+1 ω( t+1 ), 0 ≤ s ≤ 2 , F : Ω(Y, y0) × [0, 1] −→ Ω(Y, y0), F(ω, t)(s) = t+1 y0, 2 ≤ s ≤ 1.

Since E ◦ (F × 1) : Ω(Y, y0) × [0, 1] × [0, 1] → Y is continuous, it follows that F is continuous and then F : µ ◦ (1, c) ' 1. Similarly, we have µ ◦ (c, 1) ∼ 1. Hence (Ω(Y, y0), µ) is an H-space. 3.2. H-SPACES 131

(3) To show that µ is homotopy associative, consider the following diagram µ×1 Ω(Y, y0) × Ω(Y, y0) × Ω(Y, y0) −−−−→ Ω(Y, y0) × Ω(Y, y0)    µ 1×µy y

Ω(Y, y0) × Ω(Y, y0) −−−−→ Ω(Y, y0) µ Deﬁne a homotopy by

G : Ω(Y, y0) × Ω(Y, y0) × Ω(Y, y0) × [0, 1] −→ Ω(Y, y0) by  ( 4s ) ≤ s ≤ t+1  ω t+1 , 0 4 , ( 0 00 )( ) = 0 t+1 t+2 G ω, ω , ω , t s : ω (4s − t − 1), 4 ≤ s ≤ 4 ,  00 4s−t−2 t+2 ω ( 2−t ), 4 ≤ s ≤ 1. Then G is continuous and G : µ ◦ (µ × 1) ' µ ◦ (1 × µ). (4) Deﬁne a homotopy inverse ϕ : Ω(Y, y0) → Ω(Y, y0) by ϕ(ω)(s) := ω(1 − s), s ∈ [0, 1]. Deﬁne a homotopy by

H : Ω(Y, y0) × [0, 1] −→ Ω(Y, y0) by  y ≤ s ≤ t  0, 0 2 ,  t 1 ω(2s − t), 2 ≤ s ≤ 2 , H(ω, t)(s) = 1 t  ω(2 − 2s − t), 2 ≤ s ≤ 1 − 2 ,  t y0, 1 − 2 ≤ s ≤ 1. Then H is continuous and H : µ ◦ (1, ϕ) ' c. Similarly, we have µ ◦ (ϕ, 1) ' c. In summary, (Ω(Y, y0), µ, ϕ) is an H-group. Let HGroup denote the category of H-groups and continuous H-space homomorphisms.

Proposition 3.2.13. The loop functor Ω is a covariant functor from Top∗ to HGroup.

PROOF. For (Y, y0) ∈ Ob(Top∗), deﬁne Ω((Y, y0)) := (Ω(Y, y0), µ, ϕ) ∈ ( ) ∈ ( 0) ( ) = 0 Ob HGroup . If h HomTop∗ Y, Y (h y0 y0), we deﬁne 0 0 Ω(h) : Ω(Y, y0) −→ Ω(Y , y0) by Ω(h)(ω)(s) := h(ω(s)), s ∈ [0, 1]. 0 0 To verify Ω(h) ∈ HomHGroup(Ω(Y, y0), Ω(Y , y0)) we consider the following diagram µ Ω(Y, y0) × Ω(Y, y0) −−−−→ Ω(Y, y0)     Ω(h)×Ω(h)y yΩ(h) 0 0 0 0 0 0 Ω(Y , y0) × Ω(Y , y0) −−−−→ > Ω(Y , y0) µ0 0 For any ω, ω ∈ Ω(Y, y0) we have (Ω(h) ◦ µ)(ω, ω0) = Ω(h)(µ(ω, ω0)) = h ◦ µ(ω, ω0) = h ◦ (ω ∗ ω0) 132 3. FUNDAMENTAL GROUPS

' (h ◦ ω) ∗ (h ◦ ω0) = µ0(h ◦ ω, h ◦ ω0) = µ0(Ω(h)(ω), Ω(h)(ω0)) = µ0 ◦ (Ω(h) × Ω(h)) (ω, ω0). 0 Thus Ω(h) ◦ µ ' µ ◦ (Ω(h) × Ω(h)). In summary, Ω is a covariant functor.

3.2.4. Suspension. If (X, x0) and (Y, y0) are pointed topological spaces, we deﬁne

(3.2.4.1) X ∨ Y := (X × {y0}) ∪ {x0} × Y) × ∈ ( ) ∈ ( ) the subspace of X Y. If f HomTop∗ X, Z and g HomTop∗ Y, Z , then (3.2.4.2) ( f , g) : X ∨ Y −→ Z is given by ( f , g)|X = f and ( f , g)|Y = g. Note that X ∨ Y can be regarded as a pointed topological space (X ∨ Y, (x0, y0)).

Deﬁnition 3.2.14. Let C be a category and Xi ∈ Ob(C), i ∈ I. The direct sum of L Xi(i ∈ I) is an object i∈I Xi together with a family of morphisms M ki : Xi −→ Xi i∈I with the universal property: For any object Y and any family of morphisms fi : L Xi → Y, there exists a unique morphism f : i∈I Xi → Y such that f ◦ ki = fi for any i ∈ I. L ki i∈I Xi ←−−−− Xi     y∃! f y fi C C If the direct sum of Xi exists, it is unique up to unique morphism.

We now show that X ∨ Y, X ≡ (X, x0), Y ≡ (Y, y0) are pointed topological spaces, is the direct sum X ⊕ Y of X and Y in Top∗. Consider

k1 : X → X ∨ Y, k2 : Y → X ∨ Y given, respectively, by k1(x) := (x, y0) and k1(y) := (x0, y). For any Z ∈ Ob(Top∗) ∈ ( ) ∈ ( ) and f1 HomTop∗ X, Z and f2 HomTop∗ Y, Z , deﬁne f (x), (x, y) ∈ X × {y }, f : X ∨ Y −→ Z, f (x, y) := 1 0 f2(y), (x, y) ∈ {x0} × Y. ∼ Hence f ◦ ki = fi, i = 1, 2 and therefore X ∨ Y = X ⊕ Y in C.

An H-cogroup consists of a pointed topological space Q ≡ (Q, q0) together with a continuous multiplication (3.2.4.3) ν : Q −→ Q ∨ Q for which the constant c : P → P, given by c = cq0 , is a homotopy identity, that is, each composite, 1 = 1Q,

ν (c,1) ν (1,c) Q −−−−→ Q ∧ Q −−−−→ Q, Q −−−−→ Q ∨ Q −−−−→ Q is homotopic to 1. 3.2. H-SPACES 133

(1) The multiplication ν is said to be homotopy associative if the square ν Q −−−−→ Q ∨ Q     νy y1∨ν Q ∨ Q −−−−→ Q ∨ Q ∨ Q ν∨1 is homotopy commutative, i.e., (1 ∨ ν) ◦ ν ' (ν ∨ 1) ◦ ν. (2) A continuous map ψ : Q → Q is called a homotopy inverse for Q and ν if each composite

ν (1,ψ) ν (ψ,1) Q −−−−→ Q ∨ Q −−−−→ Q,, Q −−−−→ Q ∨ Q −−−−→ Q is homotopic to c. (3) An H-cogroup (Q, ν, ψ) is a homotopy associative H-cospace with a homotopy inverse. (4) A multiplication ν in an H-cospace Q is said to be homotopy Abelian if the triangle ν Q −−−−→ Q ∨ Q   yT T(q1, q2) := (q2, q1), Q −−−−→ Q ∨ Q ν is homotopy commutative, that is, T ◦ ν ' ν. (5) An H-cogroup with homotopy Abelian multiplication is called an Abelian H-cogroup. (6) An H-cospace homomorphism between two H-cospaces (Q, ν) and (Q0, ν0) is a continuous map β : Q → Q0 such that he square ν Q0 −−−−→ Q ∨ Q     βy yβ∨β Q0 −−−−→ Q0 ∨ Q0 ν0 is homotopy commutative, that is, (β ∨ β) ◦ ν ' ν0 ◦ β.

Theorem 3.2.15. A pointed topological space having the same homotopy type as an H- cospace (or an H-cogroup, or an Abelian H-cogroup), is itself an H-cospace (or an H- cogroup, or an Abelian H-cogroup) in such a way that the homotopy equivalence is an H-cospace homomorphism.

PROOF. Similar to Theorem 3.2.6. 134 3. FUNDAMENTAL GROUPS

Given an H-cospace (Q, ν) (with base point q0) and a pointed topological space (X, x0), for any [ f1], [ f2] ∈ [Q, X] deﬁne

(3.2.4.4) [ f1] ∗ [ f2] := [( f1, f2) ◦ ν]. Clearly the above multiplication is well-deﬁned. If (P, µ, ϕ) is an H-group, then [X, P] is a group under (3.2.4.4). We then have a covariant functor

πQ : HTop∗ −→ Group, X 7−→ [Q, X] [ ] ∈ ( ) and for any morphism f HomHTop∗ X, Y , 0 0 f# := πQ([ f ]) ∈ HomGroup([Q, X], [Q, Y]), [Q, X] 3 g 7−→ f ◦ g . Consequently, we obtain the following

Theorem 3.2.16. If Q ≡ (Q, ν, ψ) is an H-cogroup, then πQ : HTop∗ → Group is a covariant functor. If P is an Abelian H-cogroup, then πQ is a covariant functor from HTop∗ into AG.

PROOF. Similar to Theorem 3.2.9.

Theorem 3.2.17. If Q ≡ (Q, q0) is a pointed topological space such that πQ takes values in Group, then Q is an H-cogroup (or Abelian H-cogroup if πQ takes values in AG).

PROOF. Similar to Theorem 3.2.10.

Corollary 3.2.18. Let β : Q → Q0 be a map between H-cogroups. Then β# is a natural transformation from πQ0 to πQ in Group if and only if β is an H-cospace homomorphism.

PROOF. Similar to Corollary 3.2.11.

Exercise 3.2.19. Verify Theorem 3.2.15, Theorem 3.2.16, Theorem 3.2.17, and Corol- lary 3.2.18.

3.2.5. A classical example of H-cogroups. A classical example of H-cogroups is as follows. Let Z ≡ (Z, z0) be a pointed topological space. The suspension of Z (based at z0), denoted by Ξ(Z, z0), is deﬁned to be

(3.2.5.1) Ξ(Z, z0) := Z × [0, 1]/((Z × {0}) ∪ ({z0} × [0, 1]) ∪ (Z × {1})). The equivalence class of (z, t) ∈ Z × [0, 1] is denoted by [z, t] so that 0 0 [z, 0] = [z , 1] = [z0, t], for all z, z ∈ Z and t ∈ [0, 1].

Moreover, Ξ(Z, z0) is a pointed topological space with base point [z0, 0]. Deﬁne

ν : Ξ(Z, z0) −→ Ξ(Z, z0) ∨ Ξ(Z, x0) by 1 ([z, 2t], [z0, 0]), 0 ≤ t ≤ 2 , ν([z, t]) := 1 ([z0, 0], [z, 2t − 1]), 2 ≤ t ≤ 1.

Theorem 3.2.20. Ξ(Z, z0) is an H-cogroup. Let HCGroup denote the category of H-cogroups and continuous H-cospace homomorphisms. 3.2. H-SPACES 135

Proposition 3.2.21. The loop functor Ξ is a covariant functor from Top∗ to HCGroup.

Lemma 3.2.22. For n ≥ 0, Ξ(Sn) is homeomorphic to Sn+1.

n n PROOF. Let p0 := (1, 0, ··· , 0) ∈ S be a base point of S . Then Sn = {z ∈ Rn+2|x = (x1, ··· , xn+2), |x| = 1, xn+2 = 0} and Dn+1 = {x ∈ Rn+2|x = (x1, ··· , xn+2), |x| ≤ 1, xn+2 = 0}. It is clear that n+1 n S = H+ ∪ H−, S = H+ ∩ H−, where n+1 n+2 H± := {x ∈ S | ± x ≥ 0}. The natural projection Rn+2 → Rn+1 gives rise two projections n+1 1 n+2 1 n+1 p± : H± −→ D , x = (x , ··· , x ) 7−→ (x , ··· , x ). Deﬁne F : Ξ(Sn) → Sn+1 to be, (x, t) ∈ Sn × [0, 1], −1 1 p− (2tx + (1 − 2t)p0), 0 ≤ t ≤ 2 , F([x, t]) := −1 1 p+ (2(1 − t)x + (2t − 1)p0), 2 ≤ t ≤ 1. Since −1 −1 n+1 F([x, 0]) = p− (p0) = (1, 0, ··· , 0, 0) = p+ (p0) = F([x, 1]) = F([p0, t]) ∈ S , ∈ ( ( n) n+1) ◦ = ◦ = it follows that F HomTop∗ Ξ S , S . Solving F G 1 and G F 1, we can construct the inverse G : Sn+1 → Ξ(Sn). Therefore F : Ξ(Sn) → Sn+1 is homeomorphic.

Deﬁnition 3.2.23. For n ≥ 1 the n-th homotopy group functor πn is the covariant functor on HTop∗ deﬁned by πn := πSn . For any X ≡ (X, x0) ∈ Ob(HTop∗), we call πn(X) ≡ πn(X, x0) the n-th homotopy group.

Note π1 is just the fundamental group functor proved in Subsection 3.3.2. n Proposition 3.2.24. A map α : S → X represents the trivial element of πn(X) for n+1 n ≥ 1, where X ∈ Ob(Top∗), if and only if α can be continuously extended over D .

PROOF. By Theorem 3.1.27. 3.2.6. Higher homotopy groups. If P ≡ (P, µ, ϕ) ∈ Ob(HGroup) and Q ≡ (Q, ν, ψ) ∈ OB(HCGroup), then we have two different group multiplication on [P, Q] given, respectively, by (3.2.2.3) and (3.2.4.4). Proposition 3.2.25. Let X and Y be two objects in a category C and let ∗ and ∗0 be two laws of composition in HomC(X, Y) such that (a) ∗ and ∗0 have a common two-sided identity element, that is, there is a morphism f0 ∈ HomC(X, Y) such that 0 0 f ∗ f0 = f0 ∗ f = f = f ∗ f0 = f0 ∗ f ,

for any f ∈ HomC(X, Y); 136 3. FUNDAMENTAL GROUPS

0 (b) ∗ and ∗ are mutually distributive, i.e., for any f1, f2, g1, g2 ∈ HomC(X, Y) 0 0 0 ( f1 ∗ f2) ∗ (g1 ∗ g2) = ( f1 ∗ g1) ∗ ( f2 ∗ g2). Then ∗ and ∗0 are equal, and each is commutative and associative.

PROOF. For any f , g ∈ HomC(X, Y) one has 0 0 0 0 f ∗ g = ( f ∗ f0) ∗ ( f0 ∗ g) = ( f ∗ f0) ∗ ( f0 ∗ g) = f ∗ g and 0 0 0 0 g ∗ f = ( f0 ∗ g) ∗ ( f ∗ f0) = ( f0 ∗ f ) ∗ (g ∗ f0) = f ∗ g. 0 Therefore f ∗ g = f ∗ g = g ∗ f . For any f , g, h ∈ HomC(X, Y) we have 0 0 0 0 ( f ∗ g) ∗ h = ( f ∗ g) ∗ h = ( f ∗ g) ∗ ( f0 ∗ h) = ( f ∗ f0) ∗ (g ∗ h) = f ∗ (g ∗ h). 0 This proves the associativity of ∗ and then of ∗ .

Corollary 3.2.26. If P = (P, µ) is an H-space and Q ≡ (Q, ν, ψ) is an H-cogroup, then [Q, P] is an Abelian group and the group structure is deﬁned by the multiplication map in P.

∗ ∗0 [ ] = ( ) PROOF. Deﬁne two multiplications and in Q, P HomHTop∗ Q, P by [ f ] ∗ [g] := [µ ◦ ( f , g)], [ f ] ∗0 [g] := [( f , g) ◦ ν], [ f ], [g] ∈ [Q, P].

Because µ ◦ (c, 1) ' 1 ' µ ◦ (1, c), with c := cp0 , we get µ ◦ ( f , c) ' f ' µ ◦ (c, f ) =⇒ [ f ] ∗ [c] = [ f ] = [c] ∗ [ f ].

Similarly, using the deﬁnition of ν and noting that f (q0) = p0, we have [ f ] ∗0 [c] = [ f ] = [c] ∗0 [ f ]. The mutual distribution of ∗ and ∗0 can be proved, see Exercise 3.2.29, in the same way. According to Proposition 3.2.25, we can conclude that ∗ and ∗0 are equal, commutative and associative.

Corollary 3.2.27. If P = (P, µ) is an H-space, then πn(P) is Abelian for all n ≥ 1.

PROOF. Recall from Lemma 3.2.22 that n n−1 πn(P) = πSn (P) = [S , P] = [Ξ(S ), P], n ≥ 1.

Using Theorem 3.2.20 and Corollary 3.2.26 yields that πn(P) is Abelian for any H-space and any n ≥ 1. To extend Corollary 3.2.27 to general pointed topological spaces, we consider double suspensions n−1 h n−2 i πn(X) = [Ξ(S ), X] = Ξ Ξ(S ) , X , n ≥ 2. Deﬁne two multiplications ∗ and ∗0 on [Ξ(Ξ(Sn−2)), P] as follows: [ f ] ∗ [g] := [ f ∗ g], [ f ] ∗0 [g] := [ f ∗0 g], [ f ], [g] ∈ [Ξ(Ξ(Sn−2)), P], where 1 f ([[z, 2t], s]), 0 ≤ t ≤ 2 , ( f ∗ g)([[z, t], s]) := 1 g([[z, 2t − 1], s]), 2 ≤ t ≤ 1, 1 3.3. π1(S ) 137 and 1 0 f ([[z, t], 2s]), 0 ≤ s ≤ 2 , ( f ∗ g)([[z, t], s]) := 1 g([[z, t], 2s − 1]), 2 ≤ s ≤ 1. n−2 Let c : Ξ(Ξ(S )) → P be the constant map, i.e., c([[z, t], s]) = p0.

Corollary 3.2.28. For n ≥ 2, πn : HTop∗ → AG is a covariant functor.

0 PROOF. (1)[ f ] ∗ [c] = [ f ] = [ f ] ∗ [c]. We ﬁrst prove that f ∗ c ' f . Consider the map 2t u+1 n−2 f ([[z, u+1 ], s]), 0 ≤ t ≤ 2 , F : Ξ(Ξ(S )) × [0, 1] −→ P, ([[z, t], s], u) 7−→ u+1 p0, 2 ≤ t ≤ 1. u+1 2t When t = 2 , we see that f ([[z, u+1 ], s]) f ([[z, 1], s]) = p0, since [z, 1] is the base point of Ξ(Sn−2). Hence F is continuous. Observe that F([[z, t], s], 0) = f ∗ c, F([[z, t], s], 1) = f . Hence f ∗ c ' f . Similarly, one can prove c ∗ f ' f and f ∗0 c ' f ' c ∗0 f . The mutual distribution of ∗ and ∗0 can be proved, see Exercise 3.2.29, in the same way. According to Proposition 3.2.25, we can conclude that ∗ and ∗0 are equal, commutative and associative. Thus πn(X) is Abelian for all n ≥ 2.

Exercise 3.2.29. Prove the mutual distribution in Corollary 3.2.26 and Corollary 3.2.28.

1 3.3. π1(S ) Recall some facts from categories. A category C is said to be small if Ob(C) is a set. A morphism f ∈ HomC(X, Y) is called an equivalence if there is a morphism g ∈ HomC(Y, X) satisfying

g ◦ f = 1X, f ◦ g = 1Y. 0 If g ∈ HomC(Y, X) is another morphism satisfying 0 0 g ◦ f = 1X, f ◦ g = 1Y, then 0 0 0 0 g = g ◦ 1Y = g ◦ ( f ◦ g) = (g ◦ f ) ◦ g = 1X ◦ g = g. Thus, g is unique and denoted by f −1.A groupoid P is a small category in which each morphism is an equivalence. Proposition 3.3.1. Let P be a groupoid. (1) For any A, B ∈ Ob(P) deﬁne a relation

A ∼ B ⇐⇒ HomP(A, B) 6= ∅. Then ∼ is an equivalence relation on the set Ob(P). The equivalence classes are called the components of P. The groupoid P is said to be connected if it has only one component. (2) For any A ∈ Ob(P), the law of composition

HomP(A, A) × HomP(A, A) −→ HomP(A, A), (g, f ) 7−→ f ◦ g is a group operation in HomP(A, A). 138 3. FUNDAMENTAL GROUPS

(3) There is a covariant functor ad : P → Group which assigns to an object A ∈ Ob(P) the group ad(A) = HomP(A, A) and to a morphism f ∈ HomP(A, B) the group homomorphism ad f ∈ HomGroup(ad(A), ad(B)),

ad f : HomP(A, A) −→ HomP(B, B), g 7−→ ad f (g), −1 deﬁned by ad f (g) := f ◦ g ◦ f . Actually, ad f is a group isomorphism. (4) If A ∼ B in P, the collection of isomorphisms {ad f : f ∈ HomP(A, B)} is a conjugacy class of isomorphisms HomP(A, A) → HomP(B, B). (5) Choose another groupoid P0 and let F be a covariant functor from P to P0. Then F maps each component of P into some component of P0.

PROOF. (1) Obvious. (2) The identity element of HomP(A, A) is the identity morphism

1A ∈ HomP(A, A). It is clearly that the group axioms are satisﬁed. (3) For any g1, g2 ∈ HomP(A, A), one has −1 −1 ad f (1A) = f ◦ 1A ◦ f = f ◦ f = 1A and −1 ad f (g1 ◦ g2) = f ◦ (g1 ◦ g2) ◦ f −1 −1 = ( f ◦ g1 ◦ f ) ◦ ( f ◦ g2 ◦ f ) = ad f (g1) ◦ ad f (g2).

Thus ad f is a group homomorphism. Hence ad is a covariant functor from P to −1 Group. Since ad f = ad f −1 , it follows that ad f is a group isomorphism. (4) It follows from (3). (5) Obvious.

Exercise 3.3.2. Complete the proof of Proposition 3.3.1.

3.3.1. The fundamental groupoid. A path ω in a topological space X is a continuous map ω : [0, 1] → X; the point ω(0) is called the origin of ω, while the point ω(1) the end of ω. Two paths ω, ω0 in X are said to be homotopic if they are homotopic relative to ∂[0, 1] = {0, 1}; in this case we denote by ω ' ω0 instead of ω ' ω0(rel {0, 1}). It is clear that “'” is an equivalence relation so the resulting equivalence classes, denoted as [ω], are called path classes.

Recall the first fundamental group (π1(X, x), ∗) defined in (3.1.2.3). We now relax one point x to two points x1, x2, and then define a “fundamental group”, called fundamental groupoid, which is a grouopoid.

Given a topological space X ∈ Ob(Top), we deﬁne a category P(X) whose objects are the points of X, whose morphisms from x1 to x0, x0, x1 ∈ X, are the path class [ω] with x0 as an origin and x1 as end, and whose composite is the product of path classes: 0 0 0 [ω] ∗ [ω ] = [ω ∗ ω ], [ω] ∈ HomP(X)(x0, x1), [ω ] ∈ HomP(X)(x1, x2), 1 3.3. π1(S ) 139 given by 1 0 ω(2s), 0 ≤ s ≤ 2 , (ω ∗ ω )(s) := 1 ω(2s − 1), 2 ≤ s ≤ 1. Theorem 3.3.3. P(X) is a groupoid for any topological space X ∈ Ob(Top).

PROOF. (1) P(X) is s category. To prove the associativity of the composite, consider 0 00 [ω] ∈ HomP(X)(x0, x1), [ω ] ∈ HomP(X)(x1, x2), [ω ] ∈ HomP(X)(x2, x3). We should verify (ω ∗ ω0) ∗ ω00 ' ω ∗ (ω0 ∗ ω00). Deﬁne the corresponding homotopy F : [0, 1] × [0, 1] → X, (s, t) 7→ F(s, t), by  ( 4s ) ≤ s ≤ t+1  ω t+1 , 0 4 , ( ) = 0 t+1 t+2 F s, t ω (4s − t − 1), 4 ≤ s ≤ 4 ,  00 4s−t−2 t+2 ω ( 2−t ), 4 ≤ s ≤ 1.

For any x ∈ X, we should construct the identity morphism 1x ∈ HomP(X)(x, x). Let cx : [0, 1] → X be the constant map given by cx(s) := x for any s ∈ [0, 1]. We claim that [cx] = 1x ∈ HomP(X)(x, x), x ∈ Ob(P(X)). 0 For any [ω] ∈ HomP(X)(x , x), define a homotopy G : [0, 1] × [0, 1] → X by 2s t+1 ω( t+1 ), 0 ≤ s ≤ 2 , G(s, t) = t+1 x, 2 ≤ s ≤ 1. 0 Then G : ω ∗ cx ' ω. Thus [ω] ∗ [cx] = [ω] for any [ω] ∈ HomP(X)(x , x). Simi- larly, we have 0 [cx] ∗ [ω] = [ω], for all [ω] ∈ HomP(X)(x, x ). 0 (2) P(X) is a groupoid. For any [ω] ∈ HomP(X)(x, x ) we should prove that [ω]−1 = [ω−1], where ω−1(s) := ω(1 − s). Equivalently, we should prove that ω ∗ −1 −1 ω ' cx and ω ∗ ω ' cx0 . We give a proof of the first claim. The corresponding homotopy H : [0, 1] → [0, 1] → X is  x ≤ s ≤ t  , 0 2 ,  t 1 ω(2s − t), 2 ≤ s ≤ 2 , H(s, t) = 1 t  ω(2 − 2s − t), 2 ≤ s ≤ 1 − 2 ,  t x, 1 − 2 ≤ s ≤ 1. In summary, P(X) is a groupoid. The category P(X) is called the category of path classes of X or the fundamental groupoid of X. From the definition, we see that the components of the fundamental groupoid are path components of X.

Given a continuous map f ∈ HomTop(X, Y), we can construct a covariant functor f# : P(X) → P(Y) as follows. For x ∈ Ob(P(X)), let f#(x) = f (x) ∈ 0 Ob(P(Y)), and for [ω] ∈ HomP(X)(x, x ), let 0 f#([ω]) := [ f ◦ ω] ∈ HomP(Y)( f (x), f (x )). 140 3. FUNDAMENTAL GROUPS

Then, f# maps each path component of X into some path component of Y. Hence, we obtain a covariant functor

(3.3.1.1) π0 : Top −→ Set with

π0(X) = {path components of X} and for f ∈ HomTop(X, Y),

π0( f ) := f# : π0(X) −→ π0(Y) maps the path component of x in X to the path component of f (x) in Y. The above covariant functor (3.3.1.1) can be regarded as a covariant functor

(3.3.1.2) π0 : HTop −→ Set with

π0(X) = {path components of X} and for [ f ] ∈ HomHTop(X, Y),

π0([ f ]) := f# : π0(X) −→ π0(Y). 0 Indeed, if F : f ' f , then, for any x ∈ X, the path ωx(t) := F(x, t) connects f (x) to f 0(x). Therefore f (x) and f 0(x) belong to the same path component of Y and 0 then f# = f#.

Proposition 3.3.4. If X is contractible, then πX and π0 are naturally equivalent functors on HTop.

PROOF. Since X and one-point space P have the same homotopy type, it suf- ﬁces to prove that πP and π0 are naturally equivalent functors on HTop. For any X ∈ Ob(HTop), deﬁne

ψ(X)([g]) := g#(P), for all [g] ∈ πP(X) = [P, X].

Here we observe that π0(P) consists of the only path component P. Consider the digram ψ(X) πP(X) −−−−→ π0(X)     πP([ f ])y yπ0([ f ])

πP(Y) −−−−→ π0(Y) ψ(Y)

Because, for any [g] ∈ πP(X),

(π0([ f ]) ◦ ψ(X)) ([g]) = π0([ f ])(g#(P)) = f#(g#(P)) = ( f ◦ g)#(P) and

(ψ(Y) ◦ πP([ f ])) ([g]) = ψ(Y)( f#([g])) = ψ(Y)([ f ◦ g]) = ( f ◦ g)#(P). Thus ψ is a natural transformation. 1 3.3. π1(S ) 141

3.3.2. The equivalent of π1. From Lemma 3.1.22 we obtain a covariant functor

(3.3.2.1) π : HTop∗ −→ Group given by, for X ≡ (X, x0) ∈ Ob(HTop∗),

π(X) := π1(X, x0) [ ] ∈ ( ) and, for f HomHTop∗ X, Y ,

π([ f ]) := f∗ : π1(X, x0) −→ π1(Y, y0).

Recall from Deﬁnition 3.2.23 that π1 := πS1 is a covariant functor from HTop∗ to Group.

Theorem 3.3.5. Two covariant functors π, π1 : HTop∗ → Group are naturally equivalent.

PROOF. Deﬁne a continuous map λ : [0, 1] → Ξ(S0) by λ(s) := [−1, s], where S0 = {−1, 1} and 1 is the base point of S0. Then λ induces, by the deﬁnition of Ξ(S0) =∼ S1, a bijection # 0 λ : [Ξ(S ), X] = π1(X) −→ π(X) = π1(X, x0), [g] 7−→ [g ◦ λ]. # Moreover we see that λ is a natural transformation between π1 and π. A topological space X ∈ Ob(Top) is said to be n-connected for n ≥ 0 if every continuous map f : Sk → X, for k ≤ n, has a continuous extension over Dk+1. (1) A 1-connected space is also said to be simply-connected. (2) If 0 ≤ m ≤ n, an n-connected space is m-connected. (3) According to Proposition 3.2.24, a topological space X is n-connected for n ≥ 1 if and only if it is path connected and πk(X, x) is trivial for every base point x ∈ X and 1 ≤ k ≤ n. (4) From Corollary 3.1.28, any contractible space is n-connected for every n ≥ 0.

1 1 1 3.3.3.√ π1(S ) = Z. We compute the fundamental group π(S ) = π1(S , 1) of 1 −1θ S = {e }θ∈[0,2π] following Tucker (1945). Consider the exponential map √ (3.3.3.1) exp : R −→ S1, t 7−→ exp(t) = e2π −1t ≡ e(t). Observe that

e(t1 + t2) = e(t1)e(t2), e(t1) = e(t2) ⇐⇒ t1 − t2 ∈ Z. √ 1 1 1 π −1 Moreover, e|(−1/2,1/2) is a homeomorphism of (− 2 , 2 ) onto S \{e }. We then can deﬁne the inverse of e|(−1/2,1/2) to be √ 1 1 (3.3.3.2) e−1 : S1 \{eπ −1} −→ − , . 2 2 n Recall that a subset X ⊆ R is star-shaped with respect to x0 ∈ X if x0x ⊆ X whenever x ∈ X. 142 3. FUNDAMENTAL GROUPS

n Lemma 3.3.6. Let X ⊂ R be a star-shaped (with respect to x0 ∈ X), compact, and 1 continuous set. For any given continuous map f : X → S with e(t0) = f (x0) for some t0 ∈ R, we can ﬁnd a unique continuous map fe : X → R such that

(3.3.3.3) fe(x0) = t0, e ◦ fe = f . Thus, f can be uniquely lifted to R: R ? ∃! fe e X / 1 f S

PROOF. (1) Uniqueness. Let ge : X → R be another continuous map satisfying ge(x0) = t0 and e ◦ ge = f . Then e ◦ ( fe− ge) = 0 and ( fe− ge)(x0) = 0. Thus fe− ge is a continuous map from X onto Z. The continuity of X implies that fe− ge is constant so that fe = ge on X. (2) Existence. Because the parallel translation is continuous, we may without loss of generality assume that x0 = 0. Since X is compact, it follows that f is uniformly continuous. There exists δ > 0 such that | f (x) − f (y)| < 2 whenever |x − y| < δ (in particular, f (x) and f (y) are not antipodes in S1). The boundedness of X implies that there exists n ∈ N such that |x|/n < δ for all x ∈ X. For each j ∈ {0, 1, ··· , n − 1} and all x ∈ X, we have j + 1 j f x − f x < 2 n n because |(j + 1)x/n − jx/n| = |x|/n < δ. Particularly, f ((j + 1)x/n) 6= − f (jx/n). Then we, for each 0 ≤ j ≤ n − 1, can deﬁne a continuous map √ f ((j + 1)x/n) g : X −→ S1 \{eπ −1}, x 7−→ . j f (jx/n) Consequently, f (x) g (x) = or f (x) = f (0) g (x). ∏ j ( ) ∏ j 0≤j≤n−1 f 0 0≤j≤n−1

Deﬁne fe : X → R to be fe(x) := e−1( f (x)); precisely, −1 fe(x) := t0 + ∑ e (gj(x)). 0≤j≤n−1

Then fe(0) = t0 and e ◦ fe = f . Taking X = [0, 1] in Lemma 3.3.6 yields that for any loop α : [0, 1] → S1 based at p0 = 1, there exists a unique continuous map eα : [0, 1] → R such that eα(0) = 0, e ◦ eα = α. In particular, e(eα(1)) = α(1) = p0 = 1. 1 3.3. π1(S ) 143

Thus eα(1) ∈ Z. The degree of α is now given by (3.3.3.4) deg(α) := eα(1).

1 Lemma 3.3.7. For [α] = [β] ∈ π1(S , p0), we have (3.3.3.5) det(α) = deg(β). 1 Hence we have a well-deﬁned function deg : π1(S , p0) → Z given by (3.3.3.6) deg([α]) := det(α).

PROOF. Let F : α ' β be a homotopy relative to ∂[0, 1] = {0, 1} from α to β. Then F is a continuous map from [0, 1] × [0, 1] → S1. By Lemma 3.3.6, there is a unique continuous map Fe : [0, 1] × [0, 1] → R such that Fe(0, 0) = 0, r ◦ Fe = F.

Because Fe(0, t) : [0, 1] → R is continuous and e(Fe(0, t)) = F(0, t) = p0 = 1, we obtain Fe(0, t) is constant for all t ∈ [0, 1] and then Fe(0, t) = Fe(0, 0) = 0. Similarly one has Fe(1, t) = Fe(1, 1) for all t ∈ [0, 1]. Since e(Fe(s, 0)) = F(s, 0) = α(s), e(Fe(s, 1)) = F(s, 1) = β(s) the uniqueness in Lemma 3.3.6 implies that eα(s) = Fe(s, 0) and βe = Fe(s, 1). In particular, eα(1) = Fe(1, 0) = Fe(1, 1) = βe(1). Thus deg(α) = deg(β).

Theorem 3.3.8. The function deg is a group isomorphism 1 (3.3.3.7) deg : π1(S , p0) −→ Z, [α] 7−→ deg(α).

1 PROOF. For any [α], [β] ∈ π1(S , p0), we have continuous maps eα, βe : [0, 1] → R such thst eα(0) = βe(0) = 0, e ◦ eα = α, e ◦ βe = β. Since (eα + βe)(0) = 0 and e ◦ (eα + βe) = αβ, it follows from Lemma 3.3.6 that αβf = eα + βe. Therefore deg([α] ∗ [β]) = deg([αβ]) = αβf(1) = eα(1) + βe(1) = deg([α]) + deg([β]) where the multiplication is determined by (3.2.2.3). Hence deg is a group homomorphism. If deg([α]) = 0, then the lifted map eα : [0, 1] → R is also a loop based at 0 and 1 satisﬁes e ◦ eα = α. Since eα can be viewed as a continuous map from S to R and R ' ◦ ' is contractible, it follows from Corollary 3.1.28 that eα c0 and hence e eα cp0 . Hence α ' cp0 . Thus deg is injective. For any n ∈ Z, deﬁne eα(s) := ns and α(s) := e(eα(s)). Then α(0) = e(eα(0)) = e(0) = p0, α(1) = e(eα(1)) = e(n) = p0. Hence deg(α) = eα(1) = n.