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1 Construction of Haar Measure 1 Construction of Haar Measure Definition 1.1. A family G of linear transformations on a linear topological space X is said to be equicontinuous on a subset K of X if for every neighborhood V of the origin in X there is a neighborhood U of the origin such that the following condition holds if k1, k2 ∈ K and k1 − k2 ∈ U, then G(k1 − k2) ⊆ V that is T (k1 − k2) ∈ V for all T ∈ G. 1 Theorem 1.2 (Kakutani). Let K be a compact, convex subset of a locally convex linear topological space X, and let G be a group of linear mappings which is equicontinuous on K and such that G(K) ⊆ K. Then there exists a point p ∈ K such that T (p) = p ∀T ∈ G Proof. By Zorn’s lemma, K contains a minimal non-void compact convex subset K1 such that G(K1) ⊆ K1. If K1 contains just one point then the proof is complete. If this is not the case, the compact set K1 − K1 contains some point other than the origin. 2 Thus, there exists a neighborhood V of the origin such that V¯ 6⊇ K1 − K1. There is a convex neighborhood V1 of the origin such that αV1 ⊆ V for |α| ≤ 1. By the equicontinuity of G on the set K1, there is a neighborhood U1 of the origin such that if k1, k2 ∈ K1 and k1 − k2 ∈ U1 then G(k1 − k2) ⊆ V1. 3 Because each T ∈ G is invertible, T maps open sets to open sets (open mapping theorem) and T (A ∩ B) = TA ∩ TB for any sets A, B. Since T is linear, T convex-hull(A) = convex-hullT (A) for any set A. Because G is a group, G(GA) = GA for any set A. 4 Thus U2 := convex-hull(GU1 ∩ (K1 − K1)) = convex-hull(G(U1 ∩ (K1 − K1))) ⊆ V1 is relatively open in K1 − K1 and satisfies GU2 = U2 6⊇ K1 − K1. By continuity, GU¯2 = U¯2. Define ∞ > δ := inf{a : a > 0, aU2 ⊇ K1 − K1} ≥ 1 and U := δU2. For each 0 < < 1, (1 + )U ⊇ K1 − K1 6⊆ (1 − )U.¯ 5 −1 The family of relatively open sets {2 U + k}, k ∈ K1, is a covering of −1 −1 K1. Let {2 U + k1,..., 2 U + kn} be a finite sub-covering and let −1 p = (k1 + . kn)/n. If k is any point in K1, then ki − k ∈ 2 U for some 1 ≤ i ≤ n. Since ki − k ∈ (1 + )U for all i and all > 0, we have 1 p ∈ 2−1U + (n − 1) · (1 + )U + k. n 1 1 For = 4(n−1) , we have p ∈ (1 − 4n )U + k for each k ∈ K1. Let \ 1 K = K ∩ (1 − )U¯ + k 6= ∅. 2 1 4n k∈K1 6 1 ¯ Because (1 − 4n )U 6⊇ K1 − K1, we have K2 6= K1. The closed set K2 is clearly convex. Further since T (aU¯) ⊆ aU¯ for T ∈ G, we have T (aU¯ + k) ⊆ aU¯ + T k for all T ∈ G, k ∈ K1. Recalling TK1 = K1 for T ∈ G, we find that GK2 ⊆ K2, which contradicts the minimality of K1. 7 Theorem 1.3 (Haar Measure). Let G be a compact group. Let C(G) be the space of continuous maps from G to C. Then, there is a unique linear form m : C(G) −→ C having the following properties: 1. m(f) ≥ 0 for f ≥ 0 (m is positive). 2. m(11)= 1 (m is normalized). 3. m(sf) = m(f) where sf is defined as the function −1 sf(g) = f(s g) s, g ∈ G (m is left invariant). 4. m(fs) = m(f) where fs(g) = f(gs) for s, g ∈ G (m is right invariant). 8 Proof. For f ∈ C(G), let Cf denote the convex hull of all left translates of f. The elements of Cf are finite sums of the form: X X g(x) = aif(six) ai > 0, ai = 1 finite finite Clearly ||g|| = max{|g(x)| : x ∈ G} ≤ ||f|| Thus all sets Cf (x) = {g(x): g ∈ Cf } are bounded and relatively compact in C. Since G is compact, f is uniformly continuous, namely for all > 0, ∃ a neighborhood V = V of the identity element e ∈ G such that: y−1x ∈ V ⇒ |f(x) − f(y)| < 9 Since (s−1y)−1s−1x = y−1x, we also have −1 |sf(y) − sf(x)| < whenever y x ∈ V Since the functions g are convex combinations of functions of the form sf, |g(y) − g(x)| < whenever y−1x ∈ V Thus the set Cf is equicontinuous. By Ascoli’s theorem Cf is relatively compact in C(G). Define the compact convex set Kf = C¯f in C(G). The compact group G acts by left translations (isometrically) on C(G) and leaves Cf and hence Kf invariant. By Kakutani’s Theorem 1.2, there is a fixed point g of this action G in Kf . Such a fixed point satisfies by definition −1 sg = g (∀s ∈ G) ⇒ g(s ) = sg(e) = c (∀s ∈ G) for some constant c. 10 By the definition of the set Kf , given any > 0 there exists a finite set {s1, . sn} in G and ai > 0 such that n n X X ai = 1 and c − aif(six) < (∀x ∈ G) (1.1) 1 1 We first show that there is only one constant function Kf . Start the same construction as above, only now using right translations of f (e.g. we can apply the preceding construction to the opposite group G0 of G , 0 −1 0 or the function f = f(x )), obtaining a relatively compact set Cf with 0 0 compact convex closure Kf containing a constant function c . It will be 0 enough to show c = c .(all constants c in Kf must be equal to one 0 0 chosen constant c of Kf and conversely.) 11 There is certainly a finite combination of right translates which is close to c0 namely 0 X X |c − bjf(xtj)| < ( for some tj ∈ G, bj > 0 with bj = 1) Let us multiply this inequality by ai and put x = si to get 0 X |c ai − aibjf(sitj)| < ai (1.2) Summing over i, we obtain 0 X X X |c ai − aibjf(sitj)| < ai = (1.3) i,j 12 Operating symmetrically on Equation (1.1) (multiplying by bj, putting x = tj and summing over j), we find: X |c − aibjf(sitj)| < (1.4) i,j Subtracting (or adding) Equation (1.3) from (1.4) we get |c − c0| < 2. Since was arbitrary this completes the proof. From now on the constant c in Kf will be denoted by m(f). It is the only constant function which can be approximated arbitrarily close with convex combinations of left or right translates of f. 13 The following properties are obvious: • m(11)= 1 since Kf = {1} if f = 1. • m(f) ≥ 0 if f ≥ 0. • m(af) = am(f) for any a ∈ C (since Kaf = Kf ). • m(sf) = m(f) = m(fs) (by uniqueness) The proof will be complete if we show that m is additive (hence linear). Let us take f, g ∈ C(G) and start with Equation (1.1) above with c = m(f). Further let X h(x) = aig(six) Since h ∈ Cg, we certainly have Ch ⊆ Cg whence Kh ⊆ Kg. But the set Kg contains only one constant: m(h) = m(g). 14 We can write X |m(h) − bjh(tjx)| < P for finitely many suitable tj ∈ G and bj > 0 with bj = 1. Using the definition of h and m(h) = m(g), this implies X |m(g) − aibjg(sitjx)| < (1.5) i,j However multiplying Equation (1.1) by bj and replacing x by tjx and summing over j we find X |m(f) − aibjf(sitjx)| < (1.6) i,j 15 Adding Equation (1.5) and (1.6), this implies X |m(f) + m(g) − aibj(f + g)(sitjx)| < 2 i,j Thus the constant m(f) + m(g) is in Kf+g. However note that the only constant in this compact convex set is m(f + g). This completes the proof. 16 1.1 Exercises Exercise 1.4. Let m be the normalized Haar measure of a compact group G. For f ∈ C(G) or L1(G) show that m(f) = m(f˜) where the function f˜ is defined as the function f˜(x) = f(x−1). This equality is usually written as Z Z f(x)dx = f(x−1)dx G G Hint: Observe that f → m(f˜) is a Haar measure on G and use the uniqueness part of the Theorem on Haar measures, Theorem 1.3 17 Before stating the next exercise we need a definition Definition 1.5 (Semidirect products). Let L be a group and assume it contains a normal subgroup G and a subgroup H such that GH = L and G ∩ H = {e}. That is, suppose one can select exactly one element h from each coset of G so that {h} forms a subgroup H. If H is also normal then L is isomorphic with the direct product G × H. If H fails to be normal, we can still reconstruct L if we know how the inner automorphisms ρh behave on G. Namely for xj ∈ G and hj ∈ H (j = 1, 2), we have: −1 (x1h1)(x2h2) = x1h1x2h1 h1h2 = (x1ρh1 (x2)) h1h2 18 The construction just given can be cast in an abstract form. Let G and H be groups and suppose there is a homomorphism h → τh which carries H onto a group of automorphisms of G, namely τh ◦ τh0 = τhh0 0 for h, h ∈ H.
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