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MAT 449 : Problem Set 2 MAT 449 : Problem Set 2 Due Thursday, September 27 More examples of Haar measures 1. Let G be a locally compact group, and let H be a closed subgroup of G. We write π for the quotient map from G to G=H. We denote by ∆G (resp. ∆H ) the modular function of G (resp. H), and we assume that ∆GjH = ∆H . We fix left Haar measures µG and µH on G and H. a) (1) Show that, for every compact subset K0 of G=H, there exists a compact subset K of G such π(K) = K0. 1 R b) (1) Let f 2 L (G). Show that the function G ! C, x 7! H f(xh)dµH (h) is invariant by right translations by elements of H. Hence it defines a function G=H ! C, that we will denote by f H . H c) (2) If f 2 Cc(G), show that f 2 Cc(G=H). H d) (2) Show that the map Cc(G) !Cc(G=H), f 7! f is surjective. (Hint : You may + use the fact that, for every compact subset K of G, there exists a function ' 2 Cc (G) such that '(x) > 0 for every x 2 K.) H R e) (2) If f 2 Cc(G) is such that f = 0, show that G f(x)dµG(x) = 0. (Hint : use a function in Cc(G=H) that is equal to 1 on π(supp(f)), and proposition 2.12 of the notes. (Sorry.)) f) (2) Show that there exists a unique regular Borel measure µG=H on G=H that is invariant by left translations by elements of G and such that, for every f 2 Cc(G), R R H we have G f(x)dµG(x) = G=H f (y)dµG=H (y). g) (1) If P is a closed subgroup of G such that π induces a homeomorphism P !∼ G=H, show that the inverse image of µG=H by this homeomorphism is a left Haar measure on P . h) (2) If P is a closed subgroup of G such that the map P × H ! G,(p; h) 7! ph is a homeomorphism, and if dµP is a left Haar measure on P , show that the linear R R functional Cc(G) ! C, f 7! H P f(ph)dµP (p)dµH (h) defines a left Haar measure on G. Solution. a) Let V be a compact neighborhood of 1 in G=H. Then π(V ) is a compact neighbor- 0 S 0 hood of π(1) in G=H. We have K ⊂ x2π−1(K0) π(xV ). As K is compact, we can 0 Sn −1 0 Sn find x1; : : : ; xn such that K ⊂ i=1 π(xiV ). Let K = π (K ) \ ( i=1 xiV ). Then Sn K is a closed subset of the compact set i=1 xiV , hence it is compact, and we have π(K) = K0. 1 b) Let x 2 H. Then, for every g 2 G, we have Z Z f(gxh)dµH (h) = f(gh)dµH (h) H H by the left invariance of µH . c) We need to show that f H is continuous and that it has compact support. Fix a symmetric compact neighborhood V0 of 1, and note that A := supp f [ V0(supp f) is compact. Let " > 0. As f is left uniformly continuous, there ex- ists a neighborhood V ⊂ V0 of 1 such that, for every x 2 G and every y 2 V , we have jf(yx) − f(x)j ≤ ". Then, for every x 2 G and every y 2 V , we have Z H H −1 jf (π(yx)) − f (x)j = j (f(yxh) − f(xh))dµH (h)j ≤ εµH (x A \ H); H because f(yxh) = f(xh) = 0 unless y 2 (x−1 supp f) [ (x−1y−1 supp f) ⊂ x−1A. As x−1A \ H is compact, it has finite measure, and the calculation above implies that f H is continuous at the point π(x). Now we show that f H has compact support. By definition of f H , we have f H (π(x)) = 0 if x 62 KH. So the support of f H is contained in π(KH) = π(K), hence it is compact. 0 d) Let g 2 Cc(G=H), and let K be its support. By question (a), there exists a compact 0 + subset K of G such that π(K) = K . Let ' 2 Cc (G) be such that '(x) > 0 for every x 2 K. We show that 'H (y) > 0 for every y 2 K0. Let y 2 K0, write y = π(x) with x 2 K. As '(x) > 0 and ' is continuous, we can find an open neighborhood V of 1 0 0 in G and a c 2 R>0 such that '(x ) ≥ c for every x 2 xV . In particular, Z Z H ' (y) = '(xh)dµH (h) ≥ '(xh)dµH (h) ≥ c · µH (U \ H) > 0 H H\V (as U \ H is a nonempty open subset of H, we have µH (U \ H) > 0). We define a function F : G ! C in the following way : ( g(π(x)) if 'H (π(x)) > 0 F (x) = 'H (π(x)) 0 otherwise. H Note that F is continuous on the open subsets U1 = fx 2 Gj' (π(x)) > 0g and −1 0 U2 = G − supp(g ◦ π) (on the second subset, it is identically zero). As U1 ⊃ π (K ) −1 0 and π (K ) = supp(g ◦ π), we have U1 [ U2 = G, the function F is continuous on G. H Finally, we take f = F'. Then f 2 Cc(G), and we just need to show that f = g. Let x 2 G. If 'H (π(x)) = 0, then f(xh) = 0 for every h 2 H, so f H (π(x)) = 0. We have seen that 'H takes positive values on K0 = supp(g), so we also have x 62 supp(g), i.e., g(x) = 0 = f H (x). Now assume that 'H (π(x)) > 0. Note that the function H ! C, h 7! F (xh) is constant. So Z H g(π(x)) H f (π(x)) = F (x) '(xh)dµH (h) = 'H (π(x)) ' (π(x)) = g(π(x)): H + + Finally, note that f 2 Cc (G) if g 2 Cc (G=H), and that we also proved along the H + + way that f 2 Cc (G=H) if f 2 Cc (G) (we proved this for '). 2 e) Let 2 Cc(G=H) be such that (y) = 1 for every y 2 π(supp f). By question (d), H there exists ' 2 Cc(G) such that ' = . We have Z Z H f(x)dµG(x) = f(x)' (π(x))dµG(x) G G Z = f(x)'(xh)dµG(x)dµH (h) G×H Z Z = ( f(x)'(xh)dµG(x))dµH (h) H G Z Z −1 −1 = (∆G(h) f(xh )'(x)dµG(x))dµH (h) H G Z Z −1 −1 = '(x)( ∆H (h) f(xh )dµH (h))dµG(x) G H Z Z = '(x)( f(xh)dµH (h))dµG(x) (by proposition 2.12 of the notes) G H = 0 (because f H = 0). R f) By question (e), the positive linear function Cc(G) ! C, f 7! G fdµG factors through H the linear map Cc(G) !Cc(G=H); f 7! f . By question (d) (and the remark at the end of its solution), it defines a positive linear functional Cc(G=H) ! C. By the Riesz representation theorem, this comes from a regular Borel measure µG=H on G=H. Unravelling the definition, we get, for every f 2 Cc(G), Z Z H fdµG = f dµG=H : G G=H By the left invariance of µG and question (d), we have, if f 2 Cc(G=H) and x 2 G, Z Z f(xy)dµG=H (y) = f(y)dµG=H : G=H G=H Using the uniqueness part of the Riesz representation theorem (as we did in class), we see that µG=H (xE) = µG=H (E) for every Borel subset E of G=H. ∼ g) Let ν be the inverse image of µG=H by the homeomorphism α : P ! G=H. It is a regular Borel measure because α is a homeomorphism. Also, note that α(xy) = xα(y) for every x 2 P (this is obvious on the definition of α). As µG=H is invariant by left translations by elements of P , so is ν. h) The hypothesis implies that π induces a homeomorphism P !∼ G=H, hence we get a left Haar measure ν on P as in question (g). By the uniqueness of left Haar measures, we have µP = cν for some c 2 R>0. Hence, for every f 2 Cc(G), Z Z Z Z f(ph)dµP (p)dµH (h) = c ( f(ph)dµH (h))dν(p) = H P P H Z Z H c f (y)dµG=H (y) = c f(x)dµG(x): G=H G R R So the functional f 7! H P f(ph)dµP (p)dµH (h) is positive and corresponds to the left Haar measure cµG on G. 3 2. Let G be a locally compact group. Let A and N be two closed subgroups of G such that A × N ! G,(a; n) 7! an is a homeomorphism and that A normalizes N (i.e. for every a 2 A and n 2 N, we have ana−1 2 N).
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