Some Basic Facts 1 2. Topological Groups - Basics 7 3

TOPOLOGICAL GROUPS - PART 2/3

T.K.SUBRAHMONIAN MOOTHATHU

Contents

1. Linear groups - some basic facts 1 2. Topological groups - basics 7 3. Characters 17

1. Linear groups - some basic facts

We discuss a few things about subgroups of GL(n, E) where E = R or C. These linear groups are examples of non-abelian topological groups. We discuss them separately here since most of our later discussions will be about abelian topological groups. Think of GL(n, E) as a subspace of En2 .

[130] The space GL(n, E) is open and dense in En2 . The space GL(n, C) is path connected and hence connected; but GL(n, R) is not connected.

Proof. GL(n, E) is open since GL(n, E) = det−1(E \{0}) and determinant is a continuous function (being a ﬁnite sum of products of projections) [as an aside it may be noted that det : GL(n, E) → E \{0} is also a group homomorphism]. Denseness is clear since given any n × n matrix over E and ² > 0, we can modify the entries of the matrix upto ±² to obtain a matrix with non-zero determinant. Let G = GL(n, C). If A ∈ G is an upper triangular matrix, it is easy to see that there is a path in G connecting A and the identity matrix I (work with each entry separately). Now if B ∈ G, then there is P ∈ G such that PBP −1 is upper triangular [p.130, Artin’s Algebra]. Conjugation by P is a homeomorphism of G so that if α is a path in G connecting I to PBP −1, then P −1αP is a path in G connecting I and B. Thus G = GL(n, C) is path connected. But GL(n, R) is not connected since det : GL(n, R) → R \{0} is onto (check) and R \{0} is not connected. ¤

Remark: It can be shown (using the rational canonical form) that GL(n, R) has exactly two (path) connected components, {A : det[A] > 0} and {A : det[A] < 0}. Try to ﬁnd a path in GL(2, R) from I to −I to get a feeling of the problem.

Some standard subgroups of GL(n, E) are: SL(n, E) = ker(det) = {A ∈ GL(n, E): det[A] = 1} [linear maps preserving volume and orientation], 1 2 T.K.SUBRAHMONIAN MOOTHATHU

O(n) = {A ∈ GL(n, R): AAt = I} [linear isometries], U(n) = {A ∈ GL(n, C): AA∗ = I} [complex analogue of O(n); A∗ is conjugate transpose], SO(n) = SL(n, R) ∩ O(n), SU(n) = SL(n, C) ∩ U(n).

[131] The space SL(n, E) is closed in En2 for every n, but unbounded (hence not compact) for n ≥ 2. The spaces O(n),U(n),SO(n),SU(n) are all compact.

Proof. Since SL(n, E) = det−1({1}), it is closed. It is unbounded for n ≥ 2 since there are diagonal matrices in SL(n, E) with arbitrarily large entries. We prove now that U(n) is compact. If A = [aij], ∗ 2 Pn the condition AA = I is equivalent to the collection of n conditions k=1 aikajk = δij. Since these are closed conditions, U(n) is closed in Cn2 . Taking i = j in the above condition, we have Pn 2 k=1 |aik| = 1 which implies |aik| ≤ 1 for every i, k so that U(n) is bounded. By Heine-Borel Theorem U(n) is compact. Compactness of O(n) is argued similarly. Now, SO(n),SU(n) are compact since they are the intersections of O(n) and U(n) with the closed set SL(n, E). ¤

Orthogonal matrics, O(n): Let A be an n × n real matrix. Then, A ∈ O(n) iff the columns of A form an orthonormal basis for Rn iff kAxk = kxk for every x ∈ Rn iff hAx, Ayi = hx, yi for every x, y ∈ Rn [Chapter 4 of Artin’s Algebra]. Moreover, any isometry T : Rn → Rn is of the form Ax + v for some A ∈ O(n) and v ∈ Rn. n Now,Ã suppose! T x = Ax + v is an isometry of R , and consider the (n + 1) × (n + 1) matrix A v B = . Now if x ∈ Rn, then (x, 1) ∈ Rn+1 and B(x, 1) = (Ax + v, 1) = (T x, 1). So T can 0 1 be expressed using the matrix B - this is useful in applications.

[132] The (topological) groups SO(2) and U(1) are both isomorphically homeomorphic to S1. Ã ! a b Proof. The case of U(1) is clear. Now, suppose A = ∈ SO(2). Then a2 +c2 = 1 = b2 +d2, c d which implies a = cos θ, c = sin θ, b = cos α, d = sin α for some θ, α ∈ [0, 2π). From ab + cd = 0 and ad − bc = 1 we get cos (α − θ) = 0 and sin (α − θ) = 1 and hence α = θ + π . Therefore, A = Ã ! 2 cos θ sin θ , which corresponds to eiθ ∈ S1. [Aside: note that A induces an anticlockwise − sin θ cos θ rotation by angle θ; and for B ∈ O(2) \ SO(2), det[B] = −1, and B corresponds to the composition of a rotation and a reﬂection.] ¤

Exercise-18 : If A ∈ SO(3), then 1 is an eigenvalue of A.[Hint: det[I − A] = det[A(At − I)] = det[A]det[(A − I)t] = det[A − I]. But det[I − A] = −det[A − I] since A is 3 × 3. So det[I − A] = 0.]

[133] SO(3) is the collection of all rotations of R3. Given A ∈ SO(3) and, there exist a unit vector v ∈ R3 and an angle θ such that A is the anticlockwise rotation by θ around the axis given by v, ﬁxing v. And SO(3) is homeomorphic to RP 3. TOPOLOGICAL GROUPS - PART 2/3 3

3 Proof. Let A ∈ SO(3). Then there is a unit vector v1 ∈ R such that Av1 = v1 by Exercise-18. Ex- 3 tend to an orthonormal basis {v1, v2, v3} of R . LetÃP ∈ O(3)! be the matrix with columns v1, v2, v3. 1 0 Then P −1AP ∈ SO(3) and P −1AP has the form , where B ∈ SO(2) necessarily. Since 0 B B is a rotaion, we have that A is a rotation in the plane spanned by {v2, v3} around the axis given 3 by v1. Conversely, given a unit vector v ∈ R and angle θ, the rotation by angle θ around the axis given by v and ﬁxing v, is easily seen to be a member of SO(3) (∵ after conjugating by an element of O(3), assume v = e3). An argument for showing that SO(3) is homeomorphic to RP 3 is possibly the following [see Artin’s Algebra for the details]. Consider {(v, θ): v ∈ S2, 0 ≤ θ < 2π}, which may be identiﬁed with S3. Now, the antipodal points (v, θ) and (−v, −θ) = (−v, 2π − θ) represent the same rotation. If we identify (v, θ) and (−v, −θ), we get the real projective space RP 3. ¤

[134] SU(2) is homeomorphic to S3 (in particular, it follows that S3 admits a group structure compatible with the topology). Ã ! a b Proof. Let A = ∈ SU(2), a, b, c, d ∈ C. Since A−1 = A∗, we have c = −b and d = a so d c Ã ! a b 2 2 that A = . Now using det[A] = 1, we have |a| + |b| = 1. If a = x1 + ix2, b = x3 + ix4, −b a 3 then A corresponds to (x1, x2, x3, x4) ∈ S . ¤

Remark: It is a fact that Sn admits a group structure compatible with the topology iﬀ n = 1 or 3.

Latitudes and longitudes on S3 have algebraic interpretations in terms of the conjugacy classes in

SU(2) [p.274, Artin’s Algebra]. Fix c ∈ (−1, 1). Then the corresponding latitute is {(x1, x2, x3, x4) ∈ √ 3 2 S : x1 = c}, which is aÃ 2-sphere! of radius 1 − c . This latitude is precisely {A ∈ SU(2) : a b trace[A] = 2c} (∵ if A = with a = x1 + ix2 and b = x3 + ix4, then trace[A] = 2x1), −b a and the conjugacy classes in SU(2) are these latitudes together with {I}, {−I} (corresponding to c = ±1). Longitudes of S3 are the circles obtained by intersecting S3 with planes containing the north 3 3 and south poles. If P = {x ∈ S : x3 = x4 = 0}, then the longitude S ∩ P is the collection of all diagonal matrices in SU(2), which is a subgroup. All other longitudes are conjugates of this subgroup. Let G = {A ∈ SU(2) : trace[A] = 0}, which corresponds to the equator on S3. Every B ∈ SU(2) −1 acts on G by conjugation: fB : G → G, fB(A) = BAB . On the 2-sphere G, fB acts as a rotation 3 and then fB can be thought of as a rotation of R and thus fB ∈ SO(3). [If A ∈ G, then it can be seen that A is skew-Hermitian, i.e, A∗ = −A; the collection of all 2 × 2 skew-Hermitian matrices 3 has real dimension 3, and this collection plays the role of R for fB to act, see p.278, Artin.] The 4 T.K.SUBRAHMONIAN MOOTHATHU map B 7→ fB from SU(2) to SO(3) is a surjective group homomorphism with kernel {±I}. Hence people say that:

[135] SU(2) is a double cover of SO(3).

2 SL(2, R and SL(2, C): If v ∈ R \{0}, let Γv = {rv : r ≥ 0} be the ray determined by v. If A ∈ SL(2, R), then A(Γ ) = Γ . Let G = {A ∈ SL(2, R): A(Γ ) = Γ }. Then G consists Ãv A(v)! e1 e1 a b of matrices of the form , where a > 0 and b ∈ R. If A ∈ SL(2, R), then there is a 0 1/a −1 unique rotation B ∈ SO(2) such that B(Γe1 ) = ΓA(e1) = A(Γe1 ). Then C := B A ∈ G, and this C is unique since B was unique. This says that the map (B,C) 7→ BC from SO(2) × G to SL(2, R) is a bijection. It may be veriﬁed that this map is continuous both ways. Also note that G is homeomorphic to (0, ∞) × R =∼ R2. Hence we have:

[136] The map f : SO(2) × G → SL(2, R), f((B,C)) = BC is a homoemorphism (not a group homomorphism). Consequently, SL(2, R) is homeomorphic to S1 × R2.

Exercise-19 : Let f : C → C be an injective entire function. Then f(z) = az +b for some a ∈ C\{0} and b ∈ C.[Hint: Enough to show f is a polynomial, for then |f −1(w)| = deg(f) except for finitely many w ∈ C (∵ f 0 vanishes if there is a repeated solution). Since z 7→ f(1/z) is one-one, 0 cannot be an essential singularity of f(1/z) by Casorati-Weierstrass Theorem and Open Mapping Theorem. By considering the power series expansion, f is a polynomial.] az + b Exercise-20 : Let f(z) = be a Mobius map such that f(R ∪ {∞}) ⊂ R ∪ {∞}. Then cz + d a, b, c, d ∈ R.[Hint: There is a Mobius map g with real coefficients such that g takes the three points f(0), f(1), f(∞) ∈ R ∪ {∞} to 0, 1, ∞ respectively. Then g ◦ f fixes 0, 1, ∞ and hence g ◦ f = Id or f = g−1.]

Exercise-21 : Let D be the open unit disc in C. If f : D → D is a biholomorphism ﬁxing 0, then f is a rotation. [Hint: By Schwarz Lemma, |f(z)| ≤ |z| and |f −1(w)| ≤ |w| for every z, w ∈ D. Taking w = f(z) we have |z| = |f −1(f(z))| ≤ |f(z)| so that |f(z)| = |z|. Apply Schwarz once more.]

SL(2, R) and SL(2, C) are naturally related to the theory of Riemann surfaces. Let Aut(C∞) and

Aut(H) be the groups of all biholomorphisms of the Riemann sphere C∞ and the upper half-plane H = {z ∈ C : Im(z) > 0} respectively.

[137] We have the following group isomorphisms: Aut(C∞) = SL(2, C)/{±I} and Aut(H) = SL(2, R)/{±I}.

Ã ! a b Proof. Let M be the group of all Mobius maps. If A = ∈ SL(2, C), then we have c d az + b the corresponding Mobius map f (z) = , which belongs to Aut(C ). Moreover, A and A cz + d ∞ −A induce the same Mobius map, and so A 7→ fA from SL(2, C) to M is a surjective group TOPOLOGICAL GROUPS - PART 2/3 5 homomorphism with kernel {±I}. Thus it suﬃces to show that every f ∈ Aut(C∞) is a Mobius map.

After composing with a Mobius map, assume f(0) = 0 and f(∞) = ∞. Then f|C is an entire function. There are two arguments to say f is a Mobius map. 0 Argument-1 : Since limz→0 f(z)/z = f (0) ∈ C, f(z)/z is bounded in a deleted neighborhood of 0. Let φ(z) = 1/z. Then φ ∈ M and φ−1 = φ. If g(z) = (φ ◦ f ◦ φ)(z) = 1/f(1/z), then g ﬁxes 0 and −1 0 ∞, so that g|C is entire. Since g has an inverse φ ◦ f ◦ φ, g never vanishes on C. In particular, 0 limz→∞ f(z)/z = limw→0 w/g(w) = 1/g (0) ∈ C. So f(z)/z is bouded in C \ K for some compact K ⊂ C. Thus f(z)/z is bounded on C \{0}. Hence the entire function z 7→ f(ez)/ez is constant, say c 6= 0, by Liouville’s Theorem. So f(z) = cz, which is a Mobius map.

Argument-2 : Since f|C is an injective entire function, f(z) = az + b by Exercise-19 (in fact, b = 0 since f(0) = 0).

Let MR be the subgroup of all Mobius maps with real coeﬃcients. It may be seen that

SL(2, R)/{±I} is isomorphic to MR. So it suﬃces to show MR = Aut(H). Consider f ∈ MR, az + b f(z) = , where we may assume ad − bc = 1. Then f(R ∪ {∞}) ⊂ R ∪ {∞} and Im(f(i)) = cz + d 1/(c2 + d2) > 0 so that f(i) ∈ H. We conclude that f ∈ Aut(H). Conversely, consider f ∈ Aut(H).

Since MR acts transitively on H (∵ z 7→ az + b takes i to b + ia), we may assume f(i) = i. Now i − z there is φ ∈ M (in fact, φ(z) = ) taking H biholomorphically onto the open unit disc D in i + z such a way that φ(i) = 0. Then φ ◦ f ◦ φ−1 is a biholomorphism of D ﬁxing 0 and therefore it is a rotation, say g, by Exercise-21. Then, f = φ−1 ◦ g ◦ φ ∈ M and also f(R ∪ {∞}) ⊂ R ∪ {∞}.

Hence by Exercise-20, f ∈ MR. ¤

Remarks: (i) Any Mobius map can be written as a ﬁnite composition of: translations (z 7→ z + b, b ∈ C), dilationsÃ (z 7→ az!, a ∈ C \{Ã0}), and -(inversion)! (z 7→ −1/z). TheÃ corresponding! matrices 1 b b 1 0 1 in SL(2, C) are (b ∈ C), (b ∈ C \{0}, b2 = a), . They generate 0 1 0 1/b −1 0 Ã ! b 1 SL(2, C)/{±I} and hence SL(2, C) since −I is of the form . 0 1/b (ii) Suppose f ∈ Aut(H). If f(∞) 6= ∞, then compose with z 7→ z − f(∞) and z 7→ −1/z az + b and assume f(∞) = ∞. If f(z) = , then c = 0 so that f(z) = (az/d) + b/d. Since Ã ! cz + d a b 1 = det = ad, we have a/d > 0. Also b/d ∈ R since d 6= 0. This argument shows that 0 d Ã ! Ã ! 1 y y 1 SL(2, R)/{±I} is generated by the following types of matrices: (y ∈ R), 0 1 0 1/y 6 T.K.SUBRAHMONIAN MOOTHATHU Ã ! Ã ! 0 1 0 1 (y > 0), . We may write y = et so that 1/y = e−t. Also note that −I and −1 0 −1 0 Ã ! cos t − sin t can be written in the form . Summarizing: sin t cost Ã ! Ã ! Ã ! 1 b b 1 0 1 [138] (i) SL(2, C) is generated by: (b ∈ C), (b ∈ C \{0}), and . 0 1 0 1/b −1 0 Ã ! Ã ! Ã ! 1 t et 1 cos t − sin t (ii) SL(2, R) is generated by: (t ∈ R), (t > 0), and 0 1 0 e−t sin t cos t (t ∈ R).

An informal introduction to the Lie algebra of some linear groups: Consider GL(n, R),SL(n, R), O(n) for our discussion (complex analogues can be treated similarly). They are examples of Lie groups (i.e, smooth manifolds having compatible group structure). If G is a Lie group with identity e, then the tangent space at e to G is called the Lie algebra of G. So the Lie algebra is a vector space (it has additional structures). Some advantages in working with the Lie algebra are: (i) G may not be commutative but the addition in the Lie algebra is commutative, (ii) G may have nontrivial ‘curvature’, but the Lie algebra is ‘ﬂat’.

We loosely deﬁne a tangent vector: v ∈ TeG iﬀ there is a smooth path α on G such that α(0) = e and α0(0) = v. Now by Taylor expansion, such α can be written as α(t) = e+tv+E(t2), where E(t2) denotes a negligible error term containing powers tk with t ≥ 2. Hence if G = GL(n, R),SL(n, R) 2 or O(n), then TI (G) = {M ∈ M(n, R): I + tM ∈ G up to an error of t for small t}. We have:

[139] [Lie Algebras of some linear groups] (i) If G = GL(n, R), then TI G = M(n, R).

(ii) If G = SL(n, R), then TI G = {A ∈ M(n, R): trace[A] = 0}. t (iii) If G = O(n), then TI G = {A ∈ M(n, R): A = −A} (skew-symmetric matrices).

Proof. (i) It is intuitively clear since GL(n, R) is open in M(n, R). Or, note that for any M ∈ M(n, R), det[I + tM] ' det[I] = 1 so that det[I + tM] 6= 0 for small t. 2 (ii) TI G = {A ∈ M(n, R): det[I + tM] = 1 up to an error of t for small t}. If M = [mij], then Qn 2 Pn 2 2 2 det[I +tM] = i=1(1+tmii)+E1(t ) = 1+t( i=1 mii)+E2(t )+E1(t ) = 1+t(trace[M])+E(t ). Hence det[I + tM] = 1 up to an error of t2 for small t iﬀ trace[M] = 0. t −1 2 (iii) TI G = {A ∈ M(n, R):[I + xM] = [I + xM] up to an error of x for small x}. But [I + xM][I + xM]t = [I + xM][I + xM t] = I + x(M + M t) + x2MM T . This is equal to I up to an error of x2 iﬀ M + mt = 0. ¤

∗ If G = U(n), then we can show TI G = {A ∈ M(n, C : A = −A}.

If we think of the ‘dimension’ of G as that of TI G, then we may deduce: TOPOLOGICAL GROUPS - PART 2/3 7

[140] Real dimension of some linear groups are as follows: dim[SL(n, R)] = n2 −1, dim[SL(n, C)] = 2n2 − 2, dim[O(n)] = n(n − 1)/2, and dim[U(n)] = n(n − 1).

2. Topological groups - basics

We say (X, T , ∗) is a topological group if (i) (X, T ) is a T1 topological space, (ii) (X, ∗) is a group, and (iii) the maps x 7→ x−1 (from X to X) and (x, y) 7→ x ∗ y (from X2 to X) are continuous.

Examples:(Rn, +), (Cn, +), ((0, ∞), ·), (C \{0}, ·), (S1, ·), GL(n, E), SL(n, E)(E = R or C), O(n),U(n), SO(n),SU(n), (Q, +), all normed spaces, any group with discrete topology, the group of isometries of Rn, and ...

Exercise-22: Let X be a compact metric space and H(X,X) = {f : X → X : f is a homeomorphism}. Then H(X,X) is a topological group with respect to compact-open topology and composition. [Hint: The inverse map f 7→ f −1 is continuous since it maps S(K,V ) bijectively onto S(X\V,X\K) (compactness of X is used to say X \ V is compact). Now, given f, g ∈ H(X,X) and S(K,V ) with f ◦ g ∈ S(K,V ), choose open U such that g(K) ⊂ U ⊂ U ⊂ f −1(V ). Then the neighborhood S(U,V ) × S(K,U) of (f, g) is taken into S(K,V ) by the product map.]

Remark: With respect to the compact-open topology, H(X,X) need not be a topological group if X is only LCSC [J.J. Dijkstra, On homeomorphism groups and the compact-open topology, Amer. Math. Monthly, 112, (2005), 910-912]; is a topological group if X is LCSC and locally connected [R.Arens, Topologies for homeomorphism groups, Amer. J. of Math., 68, (1946), 593-610].

Exercise-23 : Let (X, T ) be a T1 topological space and suppose (X, ∗) is a group. Then (X, T , ∗) is a topological group iﬀ (x, y) 7→ x ∗ y−1 is continuous.

Exercise-24 : Show that the Cantor space {0, 1}N with coordinatewise addition modulo 2 is a Q topological group. More generally, if Gα’s are topological groups, then the direct product α Gα with product topology and coordinatewise operation is a topological group, and the direct sum L Q Q α Gα = {(xα) ∈ α Gα : xα = eα except for ﬁnitely many α} is a dense subgroup of α Gα.

[141] Let X be a topological group. Then for each a ∈ X, the left translation La : X → X, −1 La(x) = ax, is a homeomorphism with (La) = La−1 . Similarly, the right translation Ra is also a homeomorphism. Consequently, every topological group is homogeneous. Then inverse map x 7→ x−1 is also a homeomorphism.

Non-examples: (i) (R, +) with cofinite topology is T1 and homogeneous - translations are homeomorphisms. Also x 7→ −x is continuous. Now note that R \{1} is an open neighborhood of 0, and that for any two nonempty open U, V ⊂ R, there exist a ∈ U, b ∈ V with a + b = 1. So the map (x, y) 7→ x + y is not continuous at (0, 0). Another argument to say (R, +) with cofinite topology is not a topological group is to note that the space is not Hausdorff and then to use [142] below. 8 T.K.SUBRAHMONIAN MOOTHATHU

(ii) Consider (R, +) with the usual topology. Then R/Q is a quotient group. But the quotient topolgy on R/Q is indiscrete (∵ if A ⊂ R/Q is nonempty open, then q−1(A) is nonempty open in −1 −1 −1 R and q (A) + Q = q (A) so that q (A) = R) and hence not T1. (iii) Let T = {U ⊂ R : U open in R and there is M > 0 such that [M, ∞) ⊂ R \ U} ∪ {R}. Then, in (R, T , +), (x, y) 7→ x + y is continuous but x 7→ −x is not continuous. (iv) The Hilbert cube [0, 1]N is homogeneous (non-trivial to prove; see the book by van Mill - Infinite dimensional topology of function spaces). We note that any continuous self-map f on [0, 1]N has n n a fixed point (∵ define gn : [0, 1] → [0, 1] as gn(x1, . . . , xn) = Πn(f(x1, . . . , xn, 0, 0, 0,...)), where N n Πn : [0, 1] → [0, 1] is the projection to the first n coordinates. Let (a1(n), . . . , an(n)) be a N fixed point of gn, and let b(n) = (a1(n), . . . , an(n), 0, 0, 0,...) ∈ [0, 1] . If (b(nk)) → b, then f(b) = b). Now observe that a topological space X having fixed point property cannot be made into a topological group if |X| ≥ 2 (∵ if a ∈ X \{e}, then La does not have fixed points). Thus the Hilbert cube does not admit the structure of a topological group.

[142] Let X be a topological group with identity e. (i) If V ⊂ X is a symmetric (i.e, V −1 = {v−1 : v ∈ V } = V ) open set containing e, then V ⊂ V 2 = {vw : v, w ∈ V }. (ii) If U ⊂ X is a neighborhood of e, then there is a symmetric open set V ⊂ X containing e such that V 2 ⊂ U. (iii) Every topological group is regular.

Proof. (i) Let x ∈ V \ V . Since xV is a neighborhood of x, we have xV ∩ V 6= ∅. So there are −1 2 v1, v2 ∈ V such that xv1 = v2 or x = v2v1 ∈ V .

(ii) Since (x, y) → xy is continuous at e, there are neighborhoods V1,V2 of e such that V1V2 ⊂ U. −1 Let W = V1 ∩ V2 and V = W ∩ W . (iii) Because of homogeneity, it suﬃces to show the following: if U is a neighborhood of e, then there is an open set V with e ∈ V ⊂ V ⊂ U. But this follows from (i) and (ii). ¤

Remark: From Urysohn’s metrization theorem, it follows that every second countable topological group is metrizable. In fact the following are true: (i) every topological group having a countable base at the identity is metrizable, (ii) every topological group is completely regular [p.70 of Hewitt and Ross].

[143] Let X be a topological group and A, B ⊂ X. (i) If A or B is open, then AB is open. (ii) If A and B are compact, then AB is compact. (iii) If A is compact and B is closed, then AB is closed. (iv) Even if A, B are closed, AB need not be closed. S S Proof. (i) AB = a∈A aB = b ∈ BAb. TOPOLOGICAL GROUPS - PART 2/3 9

(ii) AB is the continuous image of the compact set A × B under the product map.

(iii) First assume X is metrizable and let x ∈ AB. Then there are (an) in A and (bn) in B such that (a b ) → ab. Since A is compact, assume (a ) → a ∈ A. Then, (b ) = (a−1a b ) → a−1x n n nk nk nk nk nk and therefore a−1x ∈ B since B is closed. So x = aa−1x ∈ AB. If X is not metrizable, argue similarly using nets instead of sequences. (iv) Let X = (R, +), A = Z and B = αZ, where α is irrational. Then A, B are closed subgroups of R, and A + B 6= R since A + B is countable. To show A + B is not closed, it now suﬃces to show that A + B is dense in R. Let ² = inf{x ∈ A + B : x > 0}. Enough to show ² = 0 since A + B is a subgroup. Let k ∈ N and let if possible (A + B) ∩ (0, 1/k) = ∅; we will derive a contradiction. Since (A + B) ∩ (0, 1) 6= ∅, there is a maximal i ∈ {1, 2, . . . , k − 1} such that (A + B) ∩ (0, i/k) = ∅. i i+1 Then there is x ∈ (A + B) ∩ ( k , k ). Choose n ∈ N such that nx < 1 < (n + 1)x. Since 1 − nx > 0 and 1 − nx ∈ A + B, we have 1 − nx > 1/k or nx < 1 − 1/k. Then, 1 < (n + 1)x < 1 + i/k and this means (n + 1)x − 1 ∈ (A + B) ∩ (0, i/k), a contradiction. ¤

[144] Let X be a topological group. (i) If U is a neighborhood of e and x ∈ X, then there is a neighborhood V of e with xV x−1 ⊂ U. (ii) If U is a neighborhood of e and K ⊂ X is compact, then there is a neighborhood V of e with KVK−1 ⊂ U.

Proof. (i) y 7→ xyx−1 is continuous at e. (ii) Let h : X2 → X be h(x, y) = xyx−1, which is continuous. For each x ∈ K, there exist neighborhoods W (x) of x and V (x) of e with h(W (x) × V (x)) = W (x)V (x)W (x)−1 ⊂ U. Let Sn Tn x1, . . . , xn ∈ K be such that K ⊂ i=1 W (xi) and let V = i=1 V (xi). ¤

Exercise-25 : Let X be a topological group, A ⊂ X be compact, B ⊂ X be closed, and A ∩ B = ∅. Then there is a neighborhood U of e such that AU ∩ BU = ∅ = UA ∩ UB.[Hint: Let V be a neighborhood of e disjoint with the closed sets A−1B and AB−1, and let U be a symmetric neighborhood of e with U 2 ⊂ V .]

Exercise-26 : Let X be a topological group, K ⊂ U ⊂ X where K is compact and U is open. Then there is a neighborhood V of e such that KV ∪ VK ⊂ U. If X is locally compact, V can be chosen so that KV ∪ VK ⊂ U is compact. [Hint: For each x ∈ K, there is a neighborhood V (x) of e such Sn Sn Tn that xV (x) ⊂ U and V (x)x ⊂ U. If K ⊂ [ i=1 xiV (xi)] ∩ [ i=1 V (xi)xi], take V = i=1 V (xi).] Exercise-27 : Let X be a topological group and A, B ⊂ X. Then, (i) A B ⊂ AB, (ii) (A)−1 = (A−1), (iii) xAy = xAy for every x, y ∈ X.[Hint:(x, y) 7→ xy is continuous, and z 7→ z−1, z 7→ xzy are homeomorphisms.]

Exercise-28 : Suppose X is a group and {Uα : α ∈ I} is a collection of subsets of X such that (i) e ∈ Uα for every α, (ii) for every α, β, there is γ such that Uγ ⊂ Uα ∩ Uβ, (iii) for every α, there 2 −1 is β with Uβ ⊂ Uα, (iv) for every α, there is β with Uβ ⊂ Uα, (v) for every α ∈ I and for every 10 T.K.SUBRAHMONIAN MOOTHATHU

−1 x ∈ X, there is β such that xUβx ⊂ Uα. Then the topology generated by {Uα : α ∈ I} makes X into a topological group; and {Uα : α ∈ I} is a base at e.

Exercise-29 : Let X be a topological group, let Z = {z ∈ Z : zx = xz for every x ∈ X} be the center of X, and let Y ⊂ X be the commutator subgroup, i.e., Y is generated by {aba−1b−1 : a, b ∈ X}. We know from Algebra that Z,Y are normal subgroups. Show that: (i) Z is closed. (ii) Y is connected if X is connected. (iii) If X is connected and H ⊂ X is a totally disconnected normal subgroup, then T −1 −1 −1 −1 −1 H ⊂ Z.[Hint: Z = x∈X hx (e), where hx(z) = zxz x . (ii) Let A = {aba b : a, b ∈ X}, −1 n n+1 S∞ n which is seen to be connected. And we have A = A, A ⊂ A and Y = n=1 A . (iii) If a ∈ H, then h : X → H, h(x) = xax−1, is continuous. Hence the connected set h(X) is equal to {h(e)} = {a}. That is, xax−1 = a or xax−1a−1 = e for every x ∈ X, and thus a ∈ Z. Also Z is closed.]

[145] Let X be a topological group and Y be a subgroup. Then: (i) Y is a subgroup. (ii) If Y is a normal or abelian subgroup, so is Y . (iii) If int[Y ] 6= ∅, then Y is open. (iv) If Y is open, then Y is closed. (v) If Y is closed and has ﬁnite index in X, then Y is open.

Proof. (i) Y Y ⊂ YY = Y and (Y )−1 = (Y −1) be Exercise-27. (ii) If Y is normal, xY x−1 = xY x−1 = Y . If Y is abelian, the continuous map (a, b) 7→ aba−1b−1 takes Y × Y to {e} and hence Y × Y also to {e} since {e} is closed. (iii) If x ∈ U = int[Y ], then for any y ∈ Y , yx−1U is a neighborhood of y contained in Y . S (iv) X \ Y = x/∈Y xY , union of open sets. Sk (v) If X = Y ∪ x1Y ∪ · · · ∪ xkY , a disjoint union of cosets, then X \ Y = i=1 xiY is closed, being a ﬁnite union of closed sets. ¤

Exercise-30 : If X is a compact group and Y ⊂ X is an open subgroup, then Y has ﬁnite index in X.[Hint: Each coset xY is open and they cover X.]

[146] Let X be a topological group and Y ⊂ X be a closed, normal subgroup. Then: (i) X/Y is a topological group. (ii) The quotient map q : X → X/Y is (continuous and) open. (iii) If X is compact, connected, path connected, locally compact or second countable, so is X/Y .

Proof. (i) Note that the quotient space X/Y is obtained by collapsing each coset xY to a singleton. Since Y is closed, xY is closed in X and therefore {xY } is closed in X/Y . Therefore X/Y is −1 T1. Let a, b ∈ X and let U ⊂ X/Y be a neighborhood of q(ab ). We need to ﬁnd open sets V,W ⊂ X/Y containing q(a), q(b) respectively such that VW −1 ⊂ U. Choose open sets A, B ⊂ X containing a, b respectively such that AB−1 ⊂ q−1(U), and put V = q(A),W = q(B). Since TOPOLOGICAL GROUPS - PART 2/3 11 q−1(V ) = AY, q−1(W ) = BY are open in X, we have that V,W are open in X/Y . Check that VW −1 ⊂ U using the fact that q is a homomorphism. (ii) If Z ⊂ X is open, then q−1(q(Z)) = ZY is open in X so that q(Z) is open in X/Y . (iii) Compactness, connectedness and path connectedness are preserved by continuous maps. X/Y is locally compact: if U ⊂ X is an open neighborhood of e with compact closure, then q(U) is compact and hence closed, and therefore q(U) is an open neighborhood of Y with compact closure q(U) = q(U). X/Y is second countable (totally disconnected): if B is a countable base for X, check that {q(B): B ∈ B} is a base for X/Y . ¤

Remark: In general, quotient maps of topological spaces are not open maps (∵ let X = [0, 2], collapse [1, 2] to the singleton {1} and note that the image of (1, 2) is {1}), and quotient maps of topological groups are not closed maps (∵ under the map (x, y) 7→ x from R2 to R, consider the image of {(x, y) ∈ R2 : x > 0, y > 0, xy = 1}); an exception is:

Exercise-31 : Let X be a topological group and Y ⊂ X be a compact normal subgroup. Then the quotient map q : X → X/Y is a closed map. [Hint: If A ⊂ X is closed, then q−1(q(A)) = AY is closed in X by [143].]

Exercise-32 : Let X be a topological group and Y ⊂ X be a compact normal subgroup such that

X/Y is compact. Then X is also compact. [Hint: Let {Uα : α ∈ I} be an open cover for X. If S x ∈ X, then there is ﬁnite I(x) ⊂ I such that xY ⊂ Uα since xY is compact. Choose S α∈I(x) open neighborhood V (x) of x such that V (x)Y ⊂ α∈I(x) Uα. Since q(V (x)) is open, there exist Sn x1, . . . , xn ∈ X such that X/Y = i=1 q(V (xi)). Then {Uα : 1 ≤ i ≤ n, α ∈ I(xi)} is a ﬁnite cover for X.] (It is also true that if Y and X/Y are locally compact, so is X - p.39 of Hewitt and Ross.)

[147] Let X be a connected topological group. Then any neighborhood of e generates X. In fact, if S∞ n V is a symmetric neighborhood of e, then X = n=1 V . In particular, if X is also locally compact, then X is compactly generated.

Proof. Given a neighborhood U of e, chose a symmetric open neighborhood V of e with V ⊂ U, S∞ n and let Y = n=1 V . Clearly Y is a subgroup of X. Since Y is open, it is also closed by [145]. Thus Y = X since X is connected. If X is locally compact, we may choose V with compact closure. Sn n Then, X = n=1 V . ¤

As topological groups R and {0, 1}N are two extreme cases; one is connected and the other is totally disconnected. The following result helps to reduce some arguments about topological groups to arguments about the two extreme cases: connected groups and totally disconnected groups.

[148] Let X be a topological group and let Y ⊂ X be the connected component of e. Then Y is a closed normal subgroup of X and X/Y is totally disconnected. 12 T.K.SUBRAHMONIAN MOOTHATHU

Proof. Being a connected component in a topological space, Y is closed. Since YY −1 is the continuous image of Y under the map (a, b) 7→ ab−1, YY −1 is connected. And e ∈ Y ∩ YY −1. Hence YY −1 ⊂ Y by the maximality of Y . Similarly, being the continuous image of Y under the map a 7→ xax−1, xY x−1 is connected and hence is contained in Y for each x ∈ X. Thus Y is a normal subgroup. By homogeneity, the connected components of X are precisely the cosets xY . If A ⊂ X/Y is a set with |A| ≥ 2, there is F ⊂ X with |F | ≥ 2 such that q−1(A) is the disjoint union −1 S −1 q (A) = x∈F xY . Let U, V ⊂ X be open sets separating q (A). Then for each x ∈ F , we have either xY ⊂ U or xY ⊂ V since xY is connected. So F is the disjoint union of two nonempty sets F1 = {x ∈ F : xY ⊂ U}, F2 = {x ∈ F : xY ⊂ V }. Then q(U), q(V ) are open sets in X/Y intersecting A with A ⊂ q(U) ∪ q(V ) and we have A ∩ q(U) = {q(x): x ∈ F1},

A∩q(V ) = {q(x): x ∈ F2} so that A∩q(U)∩q(V ) = ∅. Thus the sets q(U), q(V ) form a separation for A and so A cannot be connected. ¤

Exercise-33 : Let X be a topological group and Y ⊂ X be a closed subgroup. If the topological spaces Y and X/Y are connected, then X is connected. [Hint: Similar to the proof of [148].]

[149] Let X be a topological group and Y ⊂ X be a subgroup. (i) If {e} is open in Y , then Y is discrete. (ii) If Y is locally compact (in particular, if Y is discrete), then Y is closed in X [note that this is not true for topological spaces]. (iii) If Y is closed in X, the topological space X/Y is discrete iﬀ Y is open.

Proof. (i) By homogeneity. (ii) First suppose Y is discrete. Then there is a neighborhood U os e such that Y ∩U = {e}. Choose a symmetric neighborhood V of e with V 2 ⊂ U. Now, if Y is not closed, let x ∈ X \ Y be a limit −1 point of Y . Then there are distinct y1, y2 ∈ Y in the neighborhood xV of x. Now, e 6= y1 y2 ∈ Y −1 −1 −1 −1 −1 −1 −1 2 and y1 y2 = y1 xx y2 = (x y1) x y2 ∈ V V = V ⊂ U, a contradiction.

Now suppose Y is locally compact. There is a neighborhood U of e in X such that clY [Y ∩ U] is compact. Let W ⊂ X be open such that e ∈ W ⊂ W ⊂ U. Then Y ∩ W is compact, being closed in clY [Y ∩ U]. Therefore, Y ∩ W is closed in X. Let V be a symmetric neighborhood of e with V 2 ⊂ W . If x ∈ Y is a limit point of Y , then there is y ∈ Y ∩ xV . We claim that y−1x is a limit point of Y ∩W . If the claim is true, then y−1x ∈ Y ∩W or x ∈ yY = Y . To prove the claim consider a neighborhood O of y−1x. Then yO ∩ xV is a neighborhood of x and so there is z ∈ Y ∩ yO ∩ xV . Then y−1z ∈ Y ∩ O. Since y, z ∈ xV , we also have y−1z = (x−1y)−1x−1z ∈ V −1V = V 2 ⊂ W . Hence y−1z ∈ O ∩ (Y ∩ W ). (iii) X/Y is discrete iﬀ {Y } is open in X/Y iﬀ Y is open in X. ¤ TOPOLOGICAL GROUPS - PART 2/3 13

Corollary: Let X be a connected topological group and let Y be a locally compact subgroup having ﬁnite index in X. Then Y = X.

Proof. Y is closed by [149] and then open by [145]. ¤

[150] Let X be a topological group and let U be a compact open neighborhood of e. Then there is a compact open subgroup Y of X contained in U. Consequently, we have the following: (i) If X is locally compact and totally disconnected, then there is a base at e consisting of compact open subgroups. (ii) If X is compact and totally disconnected, then there is a base at e consisting of compact open normal subgroups.

Proof. Since the compact U is contained in the open U, by Exercise-26 there is an open neighborhood W of e such that UW ⊂ U. We may assume that W is symmetric and W ⊂ U. Then, S n 2 Y = n=1 W is an open and hence closed subgroup of X. Since W ⊂ UW ⊂ U, by induction W n+1 = W nW ⊂ UW ⊂ U and therefore Y ⊂ U. (i) Fix a base B at e consisting of compact open subsets, and for each U ∈ B choose a compact open subgroup Y ⊂ U. T −1 (ii) Let Y be a base at e consisting of compact open subgroups. If Y ∈ Y, let Z = x∈X xY x . Then Z is a compact normal subgroup. Since X is compact and Y is a neighborhood of e, by [144] −1 T −1 there is a neighborhood W of e such that x W x ⊂ Y for every x ∈ X, or W ⊂ x∈X xY x = Z. So int[Z] 6= ∅ and therefore Z is open. Such Z’s form a base at e. ¤

Remarks: (i) A deep theorem (answer to Hilbert’s ﬁfth problem) says that a topological group admits the structure of a Lie group iﬀ it is locally compact and does not have ‘small subgroups’ (i.e., there is no base at e consisting of subgroups). In this sense, locally compact totally disconnected groups are and Lie groups are at two extremes. (ii) Now we have yet another proof for the fact that Q is not locally compact: (Q, +) is totally disconnected, but does not have any compact open subgroup in (−², ²) if ² > 0.

Example: X = {0, 1}N is a compact totally disconnected abelian group. A base consisting of compact open (normal) subgroups at e = (0, 0,...) is {Yk : k ∈ N}, where Yk = {x ∈ X : xn = Q 0 for 1 ≤ n ≤ k}. More generally, if X = α∈I Xα, where Xα’s are ﬁnite discrete groups, then a base consisting of compact open normal subgroups at e ∈ X is {YF : F ⊂ I is ﬁnite}, where

YF = {x ∈ X : xα = eα for every α ∈ F }.

Exercise-34 : Let X be a locally compact group and Y ⊂ X be the connected component of e. Then Y is the intersection of all open subgroups of X.[Hint: If Z ⊂ X is an open subgroup, then Z is clopen and hence Y ⊂ Z. Now, X/Y is locally compact and totally disconnected so that if x ∈ X \ Y , then there is a compact open subgroup H of X/Y such that q(x) ∈/ H by [150]. Then, q−1(H) is an open subgroup of X not containing x.] 14 T.K.SUBRAHMONIAN MOOTHATHU

Exercise-35 : Let X be a locally compact group. Then X is a normal topological space. [Hint: Sn n Choose a symmetric open neighborhood W of e with compact closure. Then Y = n=1 W is a n+1 n S∞ clopen subgroup of X. If K1 = W and Kn+1 = W \ W , then Kn’s are compact, Y = n=1 Kn and Kn+j ∩Kn = ∅ if j ≥ 2. If A, B ⊂ Y are relatively closed, disjoint sets, choose open Un,Vn ⊂ X S Sn separating Kn ∩ A and Kn ∩ B. Then the open sets U = Un, V = Vn separate A and B, Sn=1 n=1 and thus Y is a normal topological space. Now write X = α xαY , a disjoint union of cosets, and let

A, B ⊂ X be disjoint closed sets. Separate A∩xαY and B ∩xαY by open sets Uα,Vα ⊂ xαY . Then S S U = α Uα and V = α Vα are open sets separating A, B.] (Related fact: σ-compact Hausdorﬀ topological space is normal.)

Continuous homomorphisms between topological groups: If X,Y are topological groups and f : X → Y is a bijection such that f, f −1 are continuous group homomorphisms, then we say f is a topological isomorphism. An example is f :(R, +) → ((0, ∞), ·), f(x) = ex. A non-example: det : GL(n, R) → (R \{0}, ·) is a continuous surjective homomorphism, but it is not 1-1 if n ≥ 2. Considering the identity map from (R, discrete topology, +) to (R, usual topology, +), we see that the analogue of the First Isomorphism Theorem of groups fails in the category of topological groups in the absence of extra assumptions.

[151] [First Isomorphism Theorem] Let X,Y be topological groups and let f : X → Y be a continuous open surjective homomorphism. Then Z := ker(f) is a closed normal subgroup of X and Y is topologically isomorphic to X/Z via the map y 7→ f −1(y).

Proof. From Algebra we know that Z is a normal subgroup, and Z is closed since f is continuous. Note that if f(x) = y, then f −1(y) = xZ. If g : Y → X/Z is g(y) = f −1(y), then g is a group isomorphism. For V ⊂ Y , we have g(V ) = q(f −1(V )), where q : X → X/Z is the quotient map. Since both q and f are open and continuous, so is g. ¤

[152] [Open Mapping Theorem] Let X be a locally compact σ-compact group, Y be a locally compact group and f : X → Y be a continuous surjective group homomorphism. Then f is an open map.

Proof. By homogeneity, it suﬃces to show the following: if U ⊂ X is a neighborhood of eX , then there is a neighborhood W ⊂ Y of eY such that W ⊂ f(U). 2 Let V ⊂ X be a symmetric neighborhood of eX such that V is compact and V ⊂ U. Write S X = j=1 Kj, a countable union of compact sets. Since each Kj can be covered by ﬁnitely many S∞ S∞ xV ’s, there exist x1, x2,... ∈ X such that X = n=1 xnV = n=1 xnV . Then, Y = f(X) = S∞ n=1 f(xn)f(V ) since f is a surjective homomorphism. Now, f(V ) is compact and hence closed. Since Y is locally compact, Y has Baire property by [104]. We conclude that Z := int[f(V )] 6= ∅. −1 −1 −1 Let z ∈ Z and W = z Z. Then W is a neighborhood of eY and W ⊂ Z Z ⊂ f(V ) f(V ) = 2 f((V )−1V ) = f(V ) ⊂ f(U). ¤ TOPOLOGICAL GROUPS - PART 2/3 15

[1520] [Corollary] Let X be a locally compact σ-compact group, Y be a locally compact group and let f : X → Y be a continuous surjective group homomorphism. Then f is an open map and X/ker(f) is topologically isomorphic to Y .

Examples: (i) Let f : R → S1 be f(x) = e2πix. Then f is a continuous surjective group homomorphism with ker(f) = Z. Hence R/Z is topologically isomorphic to S1 by [1520]. (ii) If n ≥ m, any surjective linear map f : Cn → Cm is an open map. (iii) For E = R or C, det : GL(n, E) → (E \{0}, ·) is an open map and GL(n, E)/SL(n, E) is topologically isomorphic to (E \{0}, ·) by [1520].

Exercise-36 : Let X,Y be compact groups with at least two elements. If X is totally disconnected and Y is connected, then there does not exist any continuous surjective group homomorphism f : X → Y (contrast with the topological result [128]). [Hint: Let y ∈ Y \{eY } and let G ⊂ X be a compact open subgroup. Then f(G) is a compact open subgroup of Y and hence f(G) = Y . So y ∈ f(G). Since X has a base at e consisting of such G’s, we get by continuity that f(eX ) = y 6= eY , a contradiction.]

Exercise-37 : [Second Isomorphism Theorem] Let X,Y be topological groups, f : X → Y be an open continuous surjective group homomorphism, N = ker(f), Z ⊂ Y be a closed normal subgroup and let T = f −1(Z). Then X/T , Y/Z and (X/N)/(T/N) are topologically isomorphic. [Hint: Verify ﬁrst that T in X and (T/N) in (X/N) are closed normal subgroups. If q : Y → Y/Z is the quotient map, then q ◦ f : X → Y/Z is an open continuous surjective homomorphism with kernel T and hence X/T =∼ Y/Z. Now, y 7→ f −1(y) is a topological isomorphism from Y to X/N and the image of Z under this map is T/N. So Y/Z =∼ (X/N)/(T/N).]

[153] Let X,Y be topological groups and f : X → Y be a group homomorphism that is continuous at some point of X. Then, (i) f is continuous. (ii) If X,Y admit left-invariant (or, right-invariant) metrics, then f is uniformly continuous.

Proof. (i) By homogeneity. (ii) Let d, d0 be left-invariant metrics on X,Y respectively. Given ² > 0 −1 choose δ > 0 for the point eX . Then for a, b ∈ X with d(a, b) < δ, we have d(eX , a b) < δ so that 0 −1 0 −1 0 ² > d (eY , f(a b)) = d (eY , f(a) f(b)) = d (f(a), f(b)). ¤

Exercise-38 : Let X be a locally compact group and f : X → R be a continuous function with compact support. Then for every ² > 0, there is a neighborhood U of e such that |f(x) − f(y)| < ² for every x, y ∈ X with x−1y ∈ U.[Hint: For each a ∈ K := supp(f), there is a neighborhood W (a) of e such that |f(a) − f(b)| < ²/2 for every b ∈ aW (a). Let U(a) be a symmetric neighborhood of e 2 Sn Tn with U(a) ⊂ W (a). Choose a1, . . . , an ∈ K such that K ⊂ i=1 aiU(ai) and let U = i=1 U(ai). −1 −1 Suppose x y ∈ U. If x ∈ K, then x ∈ aiU(ai) for some i, and then y = x(x y) ∈ aiU(ai)U ⊂ 16 T.K.SUBRAHMONIAN MOOTHATHU

Example: d(x, y) = | log x − log y| is an invariant metric on ((0, ∞), ·).

Example: Let (X, d) be a compact metric space and consider H(X,X) with the supremum metric ρ. Easily seen that ρ is right-invariant. But it is not left-invariant in general. Let X = [0, 1] and f, g ∈ H(X,X) be f(x) = x, g(x) = x2. Then ρ(f, g) = |f(1/2) − g(1/2)| = 1/4. Let h ∈ H(X,X) be such that h(f(1/2)) − h(g(1/2)) = h(1/2) − h(1/4) > 1/4. Then ρ(h ◦ f, h ◦ g) > 1/4.

Example: Let n ≥ 2 and consider GL(n, E), E = R or C. If A ∈ GL(n, E), let kAk be either the Euclidean norm or the operator norm. The metric induced by this norm is neither left-invariant nor right-invariant: if A = [aij] ∈ GL(n, E) is the diagonal matrix with a11 = 2 and aii = 1 for i ≥ 2, then kI − Ak = 1 6= 1/2 = kA−1 − Ik. However, d(A, B) = log (1 + kI − A−1Bk + kI − B−1Ak) is an admissible left-invariant metric on GL(n, E).

Observation: If d is an invariant metric (i.e., both left-invariant and right-invariant) on a topological −1 group X and if (xn), (yn) are sequences in X with (xnyn) → e, then (ynxn) → e [∵ ((xnyn) ) → e −1 −1 −1 −1 and d(e, ynxn) = d(yn , xn) = d(yn xn , e) = d((xnyn) , e)]. Example: For n ≥ 2, GL(n, E) and SL(n, E) do not admit invariant metrics. We argue as follows.

For reals x, y with x 6= 0, let A(x, y) ∈ SL(n, E) ⊂ GL(n, E) be the matrix with a11 = x, a12 = y, a22 = 1/x, aii = 1 for i > 2 and aij = 0 for all other ij. Note that A(x, y)A(s, t) = A(xs, xt + y/s). Hence A(1/m, 1/m)A(m, 1/m) = A(1, 2/m2) → I as m → ∞, but A(m, 1/m)A(1/m, 1/m) = A(1, 2) 6= I. The argument is completed by the Observation given above.

Example: Every discrete topological group trivially admits an invariant metric, namely the discrete metric. Therefore, there are non-abelian topological groups admitting an invariant metric. More examples of such groups follow from [154] below.

Remarks: (i) It is a fact that if a topological group is metrizable, the it admits a left-invariant (right-invariant) metric [p.70 of Hewitt and Ross]. (ii) As mentioned earlier, a topological group is metrizable iﬀ it has a countable base at the identity. For compact groups, this can be improved to:

[154] For a compact topological group X, the following are equivalent: (i) X has a countable base at e. (ii) There is a continuous map f : X → R such that f(x) = 0 iﬀ x = e. (iii) X admits an invariant metric. (iv) X is metrizable. In particular, compact ﬁrst countable groups admit an invariant metric.

Proof. (i) ⇒ (ii): Let {Un : n ∈ N} be a countable base at e. Since X is compact Hausdorﬀ,

X is normal and hence completely regular. So for each n, there is continuous fn : X → [0, 1] TOPOLOGICAL GROUPS - PART 2/3 17

P∞ −n such that fn(e) = 0 and fn ≡ 1 on X \ Un. If f : X → R is f(x) = n=1 2 fn(x), then f is continuous since the series is uniformly convergent by Weierstrass M-test. Also f(x) = 0 iff x = e T∞ since n=1 Un = {e}. (ii) ⇒ (iii): Define d(x, y) = sup {|f(axb) − f(ayb)| : a, b ∈ X}. Then d is an invariant metric on the group X. Let T be the original topology on X and let T 0 be the topology induced by d. First we show T 0 ⊂ T . Given ² > 0, by Exercise-38 choose a neighborhood U of e such that |f(x) − f(y)| < ²/2 for every x, y ∈ X with x−1y ∈ U. Then by [144], there is a neighborhood W of e such that xW x−1 ⊂ U for every x ∈ X. If z ∈ W and a, b ∈ X, then (aeb)−1(azb) = b−1zb ∈ b−1W b ⊂ U so that |f(aeb) − f(azb)| < ²/2 and therefore d(e, z) ≤ ²/2 < ². That is, e ∈ W ⊂ Bd(e, ²) and thus T 0 ⊂ T. Hence Id :(X, T ) → (X, T 0) is a continuous bijection from a compact space to a Hausdorff space, and therefore is a homeomorphism. That is, T 0 = T . The implications (iii) ⇒ (iv) and (iv) ⇒ (i) are clear. ¤

Example: By the above result, we conclude that if G = O(n),U(n),SO(n) or SU(n), then G admits an invariant metric. In fact, an invariant metric is d(A, B) = sup {|f(P AQ) − f(PBQ)| : P,Q ∈ G}, where f(A) = kI − Ak.

Remarks: (i) A corollary of [154] is that if X is a compact ﬁrst countable group, then card[X] ≤ card[R]. (ii) f : GL(n, E) → R given by f(A) = kI −Ak is a continuous map satisfying f(A) = 0 iﬀ A = I, but we know that there is no invariant metric on GL(n, E) if n ≥ 2. Thus we cannot replace compactness with local compactness in [154]. However, all locally compact groups are normal by Exercise-35.

3. Characters

While studying spaces it is important to know whether distinct points can be separated using a relevant class of functions. Examples: (i) If X is a normal topological space and x, y ∈ X are distinct points, then there is continuous f : X → [0, 1] such that f(x) = 0 6= 1 = f(y). (ii) [Corollary of Hahn-Banach Theorem] If X is a normed space and x, y ∈ X are distinct points, then there is a bounded linear functional f on X such that kfk = 1 and f(x) − f(y) = kx − yk. The ‘relevant’ class of functions for a topological group is the class of ‘characters’ deﬁned as follows. If X is a topological group, a character of X is a continuous group homomorphism f : X → (S1, ·).

Question: If X is a topological group and x, y ∈ X are distinct points, does there exist a character f of X such that f(x) 6= f(y)? Equivalently, if x ∈ X \{e}, does there exist a character f of X such that f(x) 6= 1?

Remark: Suppose a topological group X is not abelian. Let x, y ∈ X be such that xy 6= yx, i.e, xyx−1y−1 6= e. Since S1 is abelian, for any character f of X we have f(xyx−1y−1) = f(x)f(y)f(x)−1f(y)−1 = 1. Therefore, being abelian is a necessary condition for a topological 18 T.K.SUBRAHMONIAN MOOTHATHU group to have suﬃciently many characters to separate points. Later we will see that locally compact abelian groups have enough characters (to separate points). Because of this, locally compact abelian groups have a nice theory.

Exercise-39 : If H is a closed subgroup of (R, +), then H = {0}, R or αZ for some α > 0. [Hint: Suppose H 6= {0}. If H ∩ (0, ²) 6= ∅ for every ² > 0, then H will be dense in R. So if H 6= R, then α := inf{² > 0 : H ∩ (0, ²) = ∅} > 0. Show that H = αZ.]

Exercise-40 : If H is a closed subgroup of (S1, ·), then H = S1 or {nth roots of unity} for some n ∈ N.[Hint: If g : R → S1 is g(x) = e2πix, then g−1(H) is a closed subgroup of (R, +). Also g(αZ) = g(Z+αZ) for every α ∈ R and Z+αZ is dense in R if α ∈ R\Q. Hence if g−1(H) 6= {0}, R, then g−1(H) = αZ for some α ∈ Q.]

[155] (i) f is a character of (Z, +) iff there is a ∈ S1 such that f(n) = an for every n ∈ Z. (ii) f is a character of (R, +) iff there is y ∈ R such that f(x) = e2πixy for every x ∈ R. (iii) f is a character of (S1, ·) iff there is n ∈ Z such that f(a) = an for every a ∈ S1.

Proof. The implication “⇐” is easy to verify in all the cases. We prove “⇒”. (i) Let a = f(1). (ii) Let H = ker(f), which is a closed subgroup of R. Hence H = {0}, R or αZ for some α > 0. If H = {0}, f becomes a group isomorphism from (R, +) to (S1, ·), a contradiction since S1 has non-trivial elements of ﬁnite order. If H = R, take y = 0. Now suppose that H = αZ, α > 0. Then f is 1-1 on (0, α) and f([0, α]) must be connected. So f maps [0, α) continuously and bijectively onto S1. Moreover, f must take {kα/n : 0 ≤ k ≤ n − 1} onto the nth roots of unity since f is a homomorphism. First assume that f is orientation-preserving. Then f(kα/n) = e2πik/n = e2πi(kα/n)(1/α) for n ∈ N and 0 ≤ k < n − 1. Since f(x + mα) = f(x) for m ∈ Z, it follows that f(x) = e2πix(1/α) for every x ∈ Q, and then for every x ∈ R by the continuity of f. Thus we take y = 1/α. If f is orientation-reversing, by a similar argument, y = −1/α. (iii) First argument: if H = ker(f), then H is a closed subgroup of S1; now make use of Exercise- 40. If H = {1}, then f is a continuous bijection of S1 and hence is a homeomorphism. Since f is a group isomorphism, f must act as a permutation on {mth roots of unity} for each m ∈ N. Conclude that f = ±Id. So take n = ±1. If H = S1, take n = 0. If H = {mth roots of unity}, then f maps the arc [1, e2πi/m) bijectively and continuously onto S1. Depending upon f preserves or reverses orientation, try to conclude that n = m or n = −m. Second argument: If g : R → S1 is g(x) = e2πix, then f ◦ g is a character of R and hence by (ii), there is y ∈ R such that f(g(x)) = f(e2πix) = e2πixy for every x ∈ R. Since g(x + 1) = g(x), we have e2πi(x+1)y = e2πixy for every x ∈ R or e2πiy = 1, which gives y ∈ Z. That is, f(a) = ay for a ∈ S1. Take n = y. ¤

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