Haar Measure on Locally Compact Topological Groups

Home , Haar measure, Locally compact group

Abhishek Gupta February 2, 4, 2015

These are notes of my presentation delivered as a part of the course MTH 638: Abstract Harmonic Analysis at IIT Kanpur. Please send errors to [email protected].

The existence and uniqueness (i.e. upto scaling) of the Haar measure is proved. We follow the proof given by Weil and discussed in the book ‘Abstract Harmonic Analysis’ by G.B. Folland. The modular function is also discussed brieﬂy. By a locally compact group we shall mean a topological group which is locally compact and Hausdorﬀ as a topological space. Note that in such a space the Urysohn lemma applies which we shall use, often without explicit comment.

1 Haar Measure: its Existence and Uniqueness

Let G be a locally compact group. Cc(G) denotes the space of real valued compactly supported functions + on G and Cc (G) = {f ∈ Cc(G): f ≥ 0, f 6= 0}. Recall that a Radon measure on a topological space X is a Borel measure which is finite on compact subsets (compact subsets of Hausdorff spaces are Borel measurable), outer regular on all Borel subsets and inner regular on all σ-finite Borel sets.

Deﬁnition 1.1. A left(respectively right) Haar measure on a locally compact topological group G is a Radon measure µ such that µ(E) = µ(xE)(respectively µ(E) = µ(Ex)) for every Borel set E and ∀x ∈ G.

Given a left Haar measure µ, one can immediately see thatµ ˜ deﬁned byµ ˜(E) = µ(E−1) is a right Haar measure. Because of this, one can simply restrict attention to either left or right Haar measure, following the book we choose left. The modular function studies the failure of a left Haar measure to be right invariant and we shall look at this brieﬂy in the next section. We now recall the Reisz representation theorem(see pp. 212 of Folland’s Real Analysis for a proof):

Theorem 1.1. (Reisz representation theorem) If I is a positive linear functional on Cc(X), there is a R unique Radon measure µ on X such that I(f) = f dµ ∀f ∈ Cc(G) (in other words µ represents I).

1 R Cc(G) ⊆ L (µ) as continuous functions are Borel measurable and for f ∈ Cc(G), f dµ ≤ ||f||∞µ(supp(f)) < ∞ as f is compactly supported. Our strategy to prove the existence of a measure will be to utilize this theorem after producing a positive linear functional on Cc(G). To prove the left invariance of the measure we will use the following proposition:

R R + Proposition 1.1. A Radon measure µ is a left Haar measure ⇔ Ly dµ = f dµ, ∀f ∈ Cc (G)∀y ∈ G.

1 Proof. µy(E) = µ(yE) deﬁnes another Radon measure. For a characteristic function χ of a compact R −1 R set S we easily see that Lyχ dµ = µ(y S) = χ dµy, which implies this equality for simple functions + R by linearity of the Lebesgue integral. Thus, for any given φ ∈ C (G) the two sets { Lyf dµ : R c f is a simple function, f ≤ φ} and { f dµy : f is a simple function, f ≤ φ} are the same and so are R R + there supremums. Hence Lf dµ = f dµy ∀ f ∈ Cc (G). Now if µ is a Haar measure then µ = µy thus we have the required equality. Conversely given the R R integral equality, we have f dµ = f dµy, thus we have two measures representing the linear R functional Lf dµ hence by the uniqueness in Reisz representation theorem µ = µy. Thus µ is a left Haar measure. We note the following lemma proved in a previous lecture which we shall use:

Lemma 1.1. (Uniform Continuity Lemma) If f ∈ Cc(G) then f is both left and right uniformly continuous.

Theorem 1.2. Every locally compact group has a left Haar measure.

+ Pn Pn Proof. For f, φ ∈ Cc (G) define (f : φ) = inf{ j=1 cj|f ≤ j=1 cjLxj φ, for some xj’s}. This makes sense as we can cover support of f with translates of the open set {x ∈ G : φ(x) ≥ 1/2||φ||∞}. Because of compactness of support finitely many translates, say N, will suffice. Thus we have (f : φ) ≤ 2N||f||∞/||φ||∞. We have the following properties:

1.( f : φ) = (Lyf : φ)∀ y ∈ G

2.( f1 + f2 : φ) ≤ (f1 : φ) + (f2 : φ) 3.( cf : φ) = c(f : φ)∀ c > 0

4. f1 ≤ f2 ⇒ (f1 : φ) ≤ (f2 : φ)

5.( f : φ) ≥ ||f||∞/||φ||∞

+ 6.( f : φ) ≤ (f : ψ)(ψ : φ)∀ ψ ∈ Cc (G)

+ (f:φ) + Fix f0 ∈ C (G) and deﬁne Iφ = for f, φ ∈ C (G). By the above properties Iφ’s are left-invariant, c (f0:φ) c sub-additive, homogeneous of degree 1 and monotone(i.e. order preserving). The last property implies (f : φ) ≤ (f : f0)(f0 : φ) and (f0 : φ) ≤ (f0 : f)(f : φ). In other words,

−1 (f0 : f) ≤ Iφ(f) ≤ (f : f0)

+ Lemma 1.2. ∀f1, f2 ∈ Cc (G), > 0, ∃ a neighborhood V of 1, such that supp φ ⊆ V ⇒ Iφ(f1) + Iφ(f2) ≤ Iφ(f1 + f2) + .

+ Proof. Let g ∈ Cc (G) be such that g = 1 on supp f1 + f2. Define h = f1 + f2 + δg where δ is to be chosen later. Define hi = fi/h which is zero whenever the numerator is. Note that because of this definition we have h1 + h2 ≤ 1. By the preceding proposition we have a neighborhood V of 1 such that −1 + P |hi(x) − hi(y)| ≤ whenever y x ∈ V . Consider φ ∈ Cc (G) with supp φ ⊆ V . If h ≤ j cjLxj φ then:

X −1 X −1 fi(x) = h(x)hi(x) ≤ cjφ(xj x)h(x) ≤ cjφ(xj x)(hi(xj) + δ) j j

−1 because |hi(x) − hi(xj)| < δ whenever xj x ∈ supp φ. Since h1 + h2 ≤ 1 we have: X X X (f1 : φ) + (f2 : φ) ≤ cj(h1(xj) + δ) + cj(h2(xj) + δ) ≤ cj(1 + 2δ) j j j

2 taking the inﬁmum of right hand side and normalizing we get:

Iφ(f1) + Iφ(f2) ≤ (1 + 2δ)Iφ(h) ≤ Iφ(f1 + f2) + δ[2(Iφ(f1 + f2)) + (1 + 2δ)Iφ(g)]

+ Now note that for any k ∈ Cc (G) we have (k : φ) ≤ (k : f0)(f0 : φ) in other words Iφ(k) ≤ (k : f0). Thus choosing δ so that δ[2(f1 + f2 : f0) + (1 + 2δ)(g : f0)] < (which we can always do) ensures that

δ[2(Iφ(f1 + f2)) + (1 + 2δ)Iφ(g)] < and we are done.

−1 Now let X denote the interval [(f : f) , (f : f )] and X = Π + X . Hence, by Tychonoff f 0 0 f∈Cc (G) f theorem it is a compact Hausdorff space. For a neighborhood V of 1 denote by K(V ) the closure of the set {Iφ : supp φ ⊆ V }. Then K(V ) are all non-empty closed subsets of X, since given any open V we can construct a continuous function with support in V by the Urysohn Lemma. The family of closed n n subsets {K(V )} has the finite intersection property as K(∪i=1Vi) ⊆ ∪i=1K(Vi). Now by the following elementary fact: Proposition 1.2. A topological space is compact if and only if every family of closed subsets having finite intersection property has a nonempty intersection. we conclude that there is an element I in the intersection of all the K(V )’s. In other words given + + an > 0 and any f1, . . . , fn ∈ Cc (G) and any neighborhood V of 1 ∃φ ∈ Cc (G), suppφ ⊆ V such that |I(fi) − Iφ(fi)| < for all i. Then I has the following properties:

|Iφ(f + g) − Iφ(f) − Iφ(g)| + |Iφ(f) − I(f)| + |Iφ(g) − I(g)| + |I(f + g) − Iφ(f + g)| < 4 ∀ > 0

The triangle inequality then yields |I(f + g) − I(f) − I(g)| < 4 ∀ > 0, hence giving the desired result.

+ Thus I is a function on Cc (G) which is additive, left-invariant and homogenous of degree 1. Now we + extend I to whole of Cc(G) as follows. For f ∈ Cc(G) write f = g − h where g, h ∈ Cc (G). Define I(f) = I(g) − I(h). This is well defined because if f = g0 − h0 then g0 + h = g + h0 ⇒ I(g0) + I(h) = 0 0 0 I(g) + I(h ) ⇒ I(g ) − I(h ). It follows that I is a positive, left-invariant linear functional on Cc(G). Thus by the Reisz representation theorem it yields a Radon measure µ such that R f dµ = I(f). As R R the linear functional I satisfies I(Lyf) = I(f), we have Lyf dµ = f dµ. Thus by the previous proposition µ is a left Haar measure.

3 Note that we have used the axiom of choice in the form of Tychonoﬀ’s theorem in our construction. There are proofs that avoid this. We now prove uniqueness upto scaling.

Lemma 1.3. Given a Haar measure µ, we have, µ(U) > 0 for all non-empty open subsets and R f dµ > + 0∀f ∈ Cc (G) . Proof. If µ(U) = 0 then µ(xU) = 0∀x ∈ G. Covering any compact set K by ﬁnitely many such translates we have µ(K) = 0 for all compact subsets K. This forces µ(G) = 0 by inner regularity + which contradicts that µ 6= 0. Given f ∈ C (G), let U = {x ∈ G : f(x) > ||f||∞/2}. Then R c f dµ > ||f||∞µ(U)/2.

Proposition 1.3. If λ and µ are two left Haar measures on G the ∃c ∈ R>0 such that λ = cµ Proof. Given a left Haar measure µ, φ(f) = R f dµ defines a positive left-invariant linear functional on + R R Cc (G). Thus, given two left Haar measures λ, µ it is sufficient to show that f dµ = c f dλ∀f ∈ + R R + Cc (G). We show that the ratio f dµ/ f dλ = c∀f ∈ Cc (G)(the denominator is nonzero due to previous lemma). Now let V0 be a symmetric neighborhood of 1 with V0 compact. Such a set exists as we take some U neighborhood of 1 with U compact, then U ∩ U −1 works as closed subsets of compact + sets are compact. Let f, g ∈ Cc (G). We show that the ratio of two integrals for these functions is the same. If supp f, supp f denote compact sets containing the support of f, g then the functions f(xy) − f(yx) and g(xy) − g(yx) of x with y ∈ V0 are supported in the compact sets A = supp fV0 ∪ V0 supp f and B = supp gV0 ∪ V0 supp g respectively. To see this note that if x∈ / supp fV0 ∪ V0 supp f, then xy∈ / ( supp f)V0y ∪ V0( supp f)y. Also supp f ⊂ ( supp f)V0y ∪ V0( supp f)y, thus xy lies outside the support of f hence f(xy) vanishes. Similarly f(yx) also vanishes. Then by the uniform continuity we have that ∃ a neighborhood W of 1 such that |f(xy) − f(yx)| < for all x whenever y ∈ W . This W is obtained as follows: we get a W1 so that ||Ryf − f||∞ < /2 whenever y ∈ W1 and similarly W2 for ||f − Ly−1 f|| < /2. Add these two inequalities and use the triangle inequality, −1 thus W = W1 ∩ W2 works. By taking V1 = W ∩ W ∩ V0 we can replace W by V1 ⊆ V0 which is also symmetric. Similarly getting V2 and taking V = V1 ∩ V2 we get a symmetric neighborhood V ⊆ V0 such + that |f(xy) − f(yx)| < and |g(xy) − g(yx)| < for all x whenever y ∈ V . Choose h ∈ Cc (G) such −1 + that h(x) = h(x ) and supp h ⊆ V . Such an h exists because for an arbitrary j ∈ Cc (G) with supp j ⊆ V we can take h(x) = j(x) + j(x−1), since V is symmetric the support of h will again lie in V . Recall the Fubini theorem: Theorem 1.3. Suppose X and Y are measure spaces, and suppose that X × Y is given the maximal product measure. Fubini’s theorem states that if f(x, y) is X×Y integrable, meaning that it is measurable and Z |f(x, y)|d(x, y) < ∞ X×Y then Z Z Z Z Z f(x, y) dy dx = f(x, y) dx dy = f(x, y) d(x, y) X Y Y X X×Y In what follows, this theorem is applicable as all the integrals are over compact subsets which have finite measure. Then: Z Z ZZ h dµ f dλ = h(y)f(x)dµ(y)f dλ(x) ZZ = h(y)f(yx)dµ(y)f dλ(x) (using left invariance)

4 Z Z ZZ h dλ f dµ = h(x)f(y)dµ(y) dλ(x) ZZ = h(y−1x)f(y)dµ(y) dλ(x) (using left invariance) ZZ = h(x−1y)f(y)dµ(y) dλ(x) (using h(x) = h(x−1)) ZZ = h(y)f(xy)dµ(y) dλ(x) (left translating the integral on y by x−1)

Now subtracting these two equations gives: Z Z Z Z ZZ

h dµ f dλ − h dλ f dµ = h(y)(f(xy) − f(yx))dµ(y) dλ(x)

Z ≤ λ(A) h dµ where in the last step we have used the fact that f(xy) − f(yx) vanishes outside A, support of h is inside V and |f(xy) − f(yx)| < under these conditions. Similarly we have for g: Z Z Z Z Z

h dµ g dλ − h dλ g dµ ≤ λ(B) h dµ

Dividing these by R h dµ R f dµ and R h dµ R g dµ respectively and adding and then using the triangle inequality we have: R R f dλ f dλ λ(A) λ(B) − ≤ + R f dµ R f dµ R f dµ R g dµ Since > 0 is arbitrary the left hand side must be zero and we have our result.

2 Modular Function and Unimodular Groups

A left Haar measure λ is not right invariant in general, the modular function measures this failure. Given a left Haar measure λ, λx(E) := λ(Ex) also defines a left Haar measure as λx(yE) = λ((yE)x) = λ(y(Ex)) = λ(Ex) = λx(E). Thus, ∃∆(x) ∈ R× := R>0 such that λx = ∆(x)λ. As ∆(x) is clearly independent of the choice of λ(defined only upto a scalar multiple), we get a well-defined function ∆ : G → R×.

Deﬁnition 2.1. The function ∆ : G → R× is the called the modular function of G. If the modular function is the constant function mapping to 1, in other words if a left Haar measure is right invariant, then G is said to be unimodular. Proposition 2.1. The modular function is a continuous homomorphism to the multiplicative group of positive reals. For f ∈ L1(λ), Z Z −1 Ryf dλ = ∆(y ) f dλ

Proof. It is a homomorphism as:

∆(x)∆(y)λ = ∆(y)λx = (λx)y = λxy = ∆(xy)λ thus, ∆(x)∆(y) = ∆(xy). As RyχE(x) = χE(xy) = χEy−1 (x), we have: Z Z Z −1 −1 −1 χE(xy) dλ(x) = χEy−1 (x) dλ(x) = λ(Ey ) = ∆(y )λ(E) = ∆(y ) χE(x) dλ(x)

5 + Thus approximating f by simple functions yields the required equality. Fix a f ∈ Cc (G) then y 7→ Ryf + is a continuous function from G to Cc (G). To show the continuity of any map f : X → Y one can show that for any point x ∈ X the inverse image of an open neighbourhood of f(x) is again open in X. We shall do this. Let y ∈ G and consider an neighbourhood of Ryf. Let Rzf be another element of the image that lies in this neighbourhood. Then ||Ryf − Rzf||∞ < . We show that all such z ∈ G form an open set. By the deﬁnition of right uniform continuity we have that for any 0 > 0 there is 0 a neighbourhood V of 1 such that ||Rw(Rzf) − Rzf||∞ < for all w ∈ V . Choose α to be a real number in the non-empty open interval (||Ryf − Rzf||∞, ). So we have ||Ryf − Rzf||∞ < α and let 0 = − α. Then adding the inequalities ||Ryf − Rzf||∞ < α and ||Rw(Rzf) − Rzf||∞ < − α gives ||Rzf − Ryf||∞ + ||Rw(Rzf) − Rzf||∞ < . By triangle inequality ||Rwzf − Ryf||∞ < for all w ∈ V (remember that V was a neighborhood of 1), in other words ||Rxf − Ryf||∞ < for all x ∈ zV and zV is a neighborhood of z. Thus for every z in the preimage of an epsilon ball of y we have produced an open set that maps into that open ball. Hence the map G → Cc(G) taking an element to the right translation of a ﬁxed f is continuous. R R Now, since f 7→ f dλ is a continuous function Cc(G) → R, the composition y 7→ Ryf dλ is also R R continuous thus ∆ is continuous as ∆(x) = Rx−1 f dλ / f dλ. Abelian groups and discrete groups are unimodular. Compact groups and groups with compact abelianization are also unimodular as we show:

Proposition 2.2. G is unimodular if G or G/[G, G] is compact.

Proof. If G is compact then ∆(G) ⊆ R× is a compact subgroup and the only compact subgroup of R× is {1}. By the universal property of abelianization ∆ must factor through G/[G, G]: ∃ unique homomorphism ∆ : G/[G, G] → R× such that ∆ ◦ q = ∆, where q : G → G/[G, G] is the quotient map. As G/[G, G] is compact ∆ is the trivial homomorphism forcing ∆ to be the trivial homomorphism. We state without proof a useful result:

Theorem 2.1. Let λ be a left Haar measure and ρ be the associated right Haar measure deﬁned by ρ(E) = λ(E−1). Then Z Z f(x) dρ(x) = f(x)∆(x−1) dλ(x) E E for any Borel subset E.

Examples. We now see some examples of Haar measures on familiar groups.

1. The Haar measure on the groups (Rn, +) is just the Lebegue measure restricted to the Borel subsets.

× R 2. The group (R ), · has the measure given by µ(E) = E dµ(x)/|x| where µ is the Lebesgue measure. We only need to check that this is left invariant which is easy.

× 3. The Haar measure on the group G = (C ), · is given by the positive linear functional I : Cc(G) → R dxdy C: I(f) = f(x + iy) x2+y2 . We only need to check that this is a left-invariant functional. Since the integrand is continuous and has compact support, Riemann and Lebesgue integrals agree. So we can change variables and use the Jacobian. So this gives:

Z dxdy Z dxdy Z dx0dy0 Z dxdy f(z z) = f(x x−y y+i(xy +x y)) = f(x0+iy0) = f(z) 0 x2 + y2 0 0 0 0 x2 + y2 x02 + y02 x2 + y2

0 0 ∀z0 ∈ C, z0 = x0 + iy0. Here we have used the change of variable x = x0x − y0y, y = xy0 + x0y. So we have left invariance.

6 4. Using exactly the same idea as above we get that the measure on Gl(n, R) is given by Z dν(x) µ(E) = E |det(x)|

2 where ν is the Lebesgue measure on Rn . 5. Again using the same idea tells us that the Haar measure on the multiplicative group of non-zero Quaternions is given by Z dν(q) µ(E) = 4 E |q| 4 where q = q0 + iq1 + jq2 + kq3 and ν is the Lebesgue measure on R . a b 6. G = { |a, b ∈ , a 6= 0} is a group and is locally compact as a subset of × . One can 0 1 R R R check that dadb/a2 is a left Haar measure and dadb/|a| is a right Haar measure. It is also easy to check that these measures are not proportional so that this group is non-unimodular.

7. Qp is a locally compact group. If one normalizes the Haar measure so that µ(Zp) = 1 then m −m × R dµ(x) µ(a+p Zp) = p and the Haar measure on the multiplicative group Qp is given ν(E) = E |x| .