ON A CLASS OF IDEALS OF THE TOEPLITZ ALGEBRA ON THE BERGMAN SPACE

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1 ON A CLASS OF IDEALS OF THE TOEPLITZ ALGEBRA ON THE BERGMAN SPACE TRIEU LE Abstract Let T denote the full Toeplitz algebra on the Bergman space of the unit ball B n For each subset G of L, let CI(G) denote the closed twosided ideal of T generated by all T f T g T gt f with f, g G It is known that CI(C(B n)) = K - the ideal of compact operators and CI(C(B n) L ) = T Despite these extreme cases, there are subsets G of L so that K CI(G) T This paper gives a construction of a class of such subsets 1 INTRODUCTION For any integer n 1, let C n denote the Cartesian product of n copies of C For z = (z 1,, z n ) and w = (w 1,, w n ) in C n, we write z, w = z 1 w z n w n and z = z z n 2 for the inner product and the associated Euclidean norm Let B n denote the open unit ball which consists of all z C n with z < 1 Let S n denote the unit sphere which consists of all z C n with z = 1 For any subset V of B n we write cl(v ) for the closure of V as a subset of C n with respect to the Euclidean metric We write B n for the closed unit ball which is also cl(b n ) Let C(B n ) (respectively, C(B n )) denote the space of all functions that are continuous in the open unit ball (respectively, the closed unit ball) Let ν denote the Lebesgue measure on B n normalized so that ν(b n ) = 1 Let L 2 = L 2 (B n, dν) and L = L (B n, dν) The Bergman space L 2 a is the subspace of L 2 which consists of all analytic functions The normalized reproducing kernels for L 2 a are of the form k z (w) = (1 z 2 ) (n+1)/2 (1 w, z ) n 1, z, w < 1 We have k z = 1 and g, k z = (1 z 2 ) (n+1)/2 g(z) for all g L 2 a, z B n The orthogonal projection from L 2 onto L 2 a is given by g(w) (P g)(z) = (1 z, w ) n+1 dν(w), g L2, z B n B n For any f L the Toeplitz operator T f : L 2 a L 2 a is defined by T f h = P (fh) for h L 2 a We have f(w)h(w) (11) (T f h)(z) = dν(w) (1 z, w ) n+1 B n for h L 2 a and z B n For all f L, T f f and Tf = T f In contrast with Toeplitz operators on the Hardy space of the unit sphere, there are functions f L so that T f < 2000 Mathematics Subject Classification Primary 47B35; Secondary 47B47 1

2 2 TRIEU LE f Since T f is an integral operator by equation (11), we see that T f is compact if f vanishes almost everywhere in the complement of a compact subset of B n Let B(L 2 a) be the C algebra of all bounded linear operators on L 2 a Let K denote the ideal of B(L 2 a) that consists of all compact operators The full Toeplitz algebra T is the C subalgebra of B(L 2 a) generated by {T f : f L } For any subset G of L, let I(G) denote the closed two-sided ideal of T generated by all T f with f G Let CI(G) denote the closed two-sided ideal of T generated by all commutators [T f, T g ] = T f T g T g T f with f, g G A result of L Coburn [1] in the 70 s showed that CI(C(B n )) = K In 2004 D Suárez [5] showed that CI(L ) = T for the case n = 1 This result has been generalized by the author [2] to all n 1 In fact, we are able to show that CI(G) = T for certain subsets G of L We can take G = {f C(B n ) L : f vanishes on B n \E} where 0 < ν(e) can be as small as we please We can also take G = {f L : f vanishes on B n \E} where E is a closed nowhere dense subset of B n with 0 < ν(e) as small as we please From these results, one may be interested in the question: is there any subset G of L so that K CI(G) T? The purpose of this paper is to show that there are infinitely many such subsets Our main result is the following theorem Theorem 11 To every closed subset F of S n, there is a subset G F of L so that the following statements hold true: (1) I(G ) = K and CI(G Sn C(B n )) = T (2) If F 1, F 2 are closed subsets of S n and F 1 F 2 then G F1 G F2 (3) If F 1, F 2 are closed subsets of S n and F 2 \F 1 then we have CI(G F2 C(B n ))\I(G F1 ) In particular, if F S n then K CI(G F C(B n )) I(G F ) T In Section 2 and Section 3, we provide some preliminaries and basic results In Section 4, we give a proof for Theorem 11 2 PRELIMINARIES For any z B n, let ϕ z denote the Mobius automorphism of B n that interchanges 0 and z For any z, w B n, let ρ(z, w) = ϕ z (w) Then ρ is a metric on B n (called the pseudo-hyperbolic metric) which is invariant under the action of the automorphism group Aut(B n ) of B n For any w B n, we have ρ(z, w) 1 as z 1 These properties and the following inequality can be proved by using the identities in [4, Theorem 222] See [2, Section 2] for more details For any z, w, u B n, (21) ρ(z, w) ρ(z, u) + ρ(u, w) 1 + ρ(z, u)ρ(u, w) For any a in B n and any 0 < r < 1, let B(a, r) = {z B n : z a < r} and E(a, r) = {z B n : ρ(z, a) < r} Inequality (21) shows that if z, w B n so that E(z, r 1 ) E(w, r 2 ) for some 0 < r 1, r 2 < 1 then ρ(z, w) ρ(z, u) + ρ(u, w) 1 + ρ(z, u)ρ(u, w) < r 1 + r r 1 r 2 (where u is any element in E(z, r 1 ) E(w, r 2 ))

3 IDEALS OF THE TOEPLITZ ALGEBRA 3 This implies that if ρ(z, w) r 1 + r r 1 r 2 then E(z, r 1 ) E(w, r 2 ) = Lemma 21 For any 0 < r < 1 and ζ S n there is an increasing sequence {t m } m=1 (0, 1) so that t m 1 as m and E(t k ζ, r) E(t l ζ, r) = for all k l Proof We will construct the required sequence {t m } m=1 by induction We begin by taking any t 1 in (0, 1) Suppose we have chosen t 1 < < t m so that 1 j 1 < t j < 1 for all 1 j m and E(t k ζ, r) E(t l ζ, r) = for all 1 k < l m, where m 1 Since ρ(tζ, t j ζ) 1 as t 1 for all 1 j m, we can choose a t m+1 with max{t m, 1 (m + 1) 1 } < t m+1 < 1 and ρ(t m+1 ζ, t j ζ) > 2r for all 1 j m 1 + r2 It then follows that E(t m+1 ζ, r) E(t j ζ, r) = for all 1 j m Also, since 1 m 1 < t m < 1 for all m, t m 1 as m Lemma 22 For any 0 < r < 1 and any ɛ > 0, there is a δ depending on r and ɛ so that for all ζ S n and all a B n with a ζ < δ, we have E(a, r) {z B n : z ζ < ɛ} As a consequence, if b B n and ζ S n so that E(b, r) {z B n : z ζ < δ} then b ζ < ɛ Proof From [4, Section 227], for any a 0, E(a, r) {z B n : P z c 2 r 2 s 2 + Qz 2 r 2 s < 1}, z, a where P z = a, Qz = z P z, c = 1 r2 1 a 2 a, a 1 r 2 a and s = a 2 1 r 2 It then a 2 follows that E(a, r) B(c, r s) Since a c = r 2 s a r s, we get B(c, r s) B(a, 2r s) Hence E(a, r) B(a, 2r s) Note that the inclusion certainly holds true for a = 0 (in this case s = 1) Now suppose a ζ < δ Then a ζ a ζ > 1 δ Hence, So for any z E(a, r), s = 1 a 2 1 r 2 a 2 1 a 2 2(1 a ) 1 r2 1 r 2 < 2δ 1 r 2 z ζ z a + a ζ 2r s + δ < 2r Choosing δ so that 2r 2δ 1 r 2 + δ 2δ + δ < ɛ, we then have the first conclusion of the 1 r2 lemma Now suppose a, b B n so that a ζ < δ and a E(b, r) Then since b E(a, r), the first conclusion of the lemma implies b ζ < ɛ For any z B n, the formula U z (f) = (f ϕ z )k z, f L 2, defines a bounded operator on L 2 It is well-known that U z is a unitary operator with L 2 a as a reducing subspace and U z T f Uz = T f ϕz on L 2 a for all z B n and all f L See, for example, [3, Lemmas 7 and 8]

4 4 TRIEU LE Lemma 23 For any sequence {z m } m=1 B n with z m 1, U zm weak operator topology of B(L 2 a) 0 in the Proof Since Span({k z : z B n }) is dense in L 2 a, it suffices to show that for all z, w in B n, we have lim U z m k z, k w = 0 m Fix such z and w For each m 1, U zm k z, k w = (1 w 2 ) (n+1)/2 (U zm k z )(w) = (1 w 2 ) (n+1)/2 k z (ϕ zm (w))k zm (w) ( (1 w 2 )(1 z 2 )(1 z m 2 ) ) (n+1)/2 = ( (1 ϕzm (w), z )(1 w, z m ) n+1 Since ϕ zm (w), z z and w, z m w, we obtain ( (1 w 2 )(1 z 2 )(1 z m 2 ) ) (n+1)/2 U zm k z, k w ( ) n+1 (1 z )(1 w ) It then follows that lim U z m k z, k w = 0 m Lemma 24 Let {z m } m=1 B n so that z m 1 as m non-zero positive operator on L 2 a Suppose A = m=1 U zm SU z m Let S be any exists in the strong operator topology and is a bounded operator on L 2 a Then there is a constant c > 0 and an f L 2 a so that AU zm f c > 0 for all m Proof Since S is non-zero and positive, there is an f L 2 a with f = 1 so that Sf, f > 0 For each m 1, AU zm f, U zm f U zm SU z m U zm f, U zm f SU z m U zm f, U z m U zm f Sf, f Since U zm f = 1 it follows that AUz m f Sf, f > 0 for all m 3 BASIC RESULTS The first result in this section shows that for a certain class of subsets G of L, I(G) possesses a special property This property will later help us distinguish I(G 1 ) and I(G 2 ) for G 1 G 2 Proposition 31 Let W be a subset of B n and let F = cl(w ) S n Let f be in L so that f vanishes almost everywhere in B n \W Let g 1,, g l be any functions in L Let {z m } m=1 be any sequence in B n so that z m 1 and z m w ɛ > 0 for all w F, all m 1, where ɛ is a fixed constant Then the sequence {T f T g1 T gl U zm } m=1 converges to 0 in the strong operator topology of B(L 2 a) Consequently, if we set G = {f L : f vanishes almost everywhere in B n \W } then for any T I(G), T U zm 0 in the strong operator topology of B(L 2 a)

5 IDEALS OF THE TOEPLITZ ALGEBRA 5 Proof Let V 1 = { z 1 : z w < ɛ/3 for some w F } and V 2 = { z 1 : z w < ɛ/2 for some w F } Let η be a continuous function on B n so that 0 η 1, η(z) = 1 if z cl(v 1 ) and η(z) = 0 if z / V 2 Let Z = cl(w ) (B n \V 1 ) Then Z B n and Z is compact with respect to the Euclidean metric We have Z S n = cl(w ) S n (B n \V 1 ) = F (B n \V 1 ) = Thus Z is a compact subset of B n Since the function f(1 η) vanishes almost everywhere in (B n \W ) cl(v 1 ) which contains B n \Z, the operator T f(1 η) is compact Since η is continuous on B n, the operators T g T 1 η T g(1 η) and T 1 η T g T g(1 η) are compact for all g L (see [1]) So we have (31) T f T g1 T gl = T f T g1 T gl T η + T f T g1 T gl T 1 η = T f T g1 T gl T η + T f(1 η) T g1 T gl + K 1 = T f T g1 T gl T η + K, where K 1 is a compact operator and K = T f(1 η) T g1 T gl + K 1 is also a compact operator For any h L 2 a L and any m 1 we have T η U zm h 2 ηu zm h 2 (U zm h)(z) 2 dν(z) V 2 = h(ϕ zm (z))k zm (z) 2 dν(z) V 2 h 2 k zm (z) 2 dν(z) V 2 Let V 3 = { z 1 : z w < ɛ for some w F } Since the map (z, w) 1 z, w is continuous and does not vanish on the compact set cl(v 2 ) (B n \V 3 ), there is a δ > 0 so that 1 z, w δ for all z cl(v 2 ) and w (B n \V 3 ) For each m 1, z m (B n \V 3 ), so for all z V 2, Hence we have k zm (z) (1 z m 2 ) (n+1)/2 1 z, z m n+1 (1 z m 2 ) (n+1)/2 δ n+1 T η U zm h h ν(v2 ) (1 z m 2 ) (n+1)/2 δ n+1 This implies T η U zm h 0 as m Since L 2 a L is dense in L 2 a and T η U zm T η 1 for all m, we conclude that T η U zm 0 in the strong operator topology of B(L 2 a) So T f T g1 T gl T η U zm 0 in the strong operator topology of B(L 2 a) Also by Lemma 23, U zm 0 in the weak operator topology, so KU zm 0 in the strong operator topology for any compact operator K Combining these facts with (31), we conclude that T f T g1 T gl U zm 0 in the strong operator topology of B(L 2 a) The following proposition was proved by Suárez for the case n = 1 (see [5, Proposition 29]) The case n 2 is similar and can be proved with the same

6 6 TRIEU LE method The point is that for all n 2, the metric ρ and the reproducing kernel functions have all the properties needed for Suárez s proof See [2, Section 2] for more details Proposition 32 Let 0 < r < 1 and {w m } m=1 be a sequence in B n so that E(w k, r) E(w l, r) = for all k l For each m N, let c 1 m,, c l m, a m, b m, d 1 m,, d k m L be functions of norm 1 that vanish almost everywhere on B n \E(w m, r) Then T c 1 m T c l m (T am T bm T bm T am )T d 1 m T d k m belongs to CI(L ) m N Remark 33 In the proof of Proposition 32, we work only with Toeplitz operators with symbols in the set G which consists of functions of the form f m, where F is a subset of N and f is one of the symbols c 1,, c l, a, b, d 1,, d k So in the conclusion of the proposition, we may replace CI(L ) by CI(G) m F 4 PROOF OF THE MAIN THEOREM We are now ready for the proof of Theorem 11 Fix 0 < r < 1 Let W = E(0, r) For any closed non-empty subset F of S n, let W F = E(tζ, r) 0<t<1 ζ F It is clear that W Sn = B n We always have F cl(w F ) S n We will show that in fact F = cl(w F ) S n Suppose ζ cl(w F ) S n For any m 1, applying Lemma 22 with ɛ = m 1, we get a δ m > 0 so that if E(b, r) V m, where V m = {z B n : z ζ < δ m }, then b ζ < m 1 Since W F V m, there is 0 < t m < 1 and ζ m F so that E(t m ζ m, r) V m Hence t m ζ m ζ < m 1 So t m ζ m ζ 0 as m Since t m = t m ζ m ζ = 1 as m, we get ζ m = t 1 m (t m ζ m ) ζ in the Euclidean metric as m This implies that ζ F Thus, cl(w F ) S n F and hence, cl(w F ) S n = F Now define G F = {f L : f vanishes almost everywhere in B n \W F } It is clear that if F 1 F 2 then W F1 W F2, hence G F1 G F2 Since T f is compact for all f G, I(G ) = K Since G Sn = L, we have CI(G Sn C(B n )) = CI(L C(B n )) = T Now suppose F 1 and F 2 are two closed subsets of S n so that F 2 \F 1 Let ζ F 2 \F 1 From Lemma 21 there is a sequence {t m } m=1 (0, 1) with t m 1 and E(t k ζ, r) E(t l ζ, r) = for all k l Let z m = t m ζ for all m 1 Since z m ζ 0 and ζ / F 1 which is a closed subset of S n, there is an ɛ > 0 so that z m w ɛ for all w F 1, all m 1 Since cl(w F1 ) S n = F 1, Proposition 31 shows that T U zm 0 in the strong operator topology for all T I(G F1 ) Take f to be any continuous function supported in E(0, r) such that [T f, T f ] 0 Any function of the form f(z) = z 1 η( z /r) where η is non-negative, continuous and supported in [0, 1] with η > 0 will work Let S = [T f, T f ] 2 then S is a non-zero,

7 IDEALS OF THE TOEPLITZ ALGEBRA 7 positive operator on L 2 a Define T = U zm SUz m m=1 By Lemma 24, there is a constant c > 0 and an h L 2 a so that T U zm h c for all m This implies that T is not in I(G F1 ) For each m, U zm [T f, T f ]Uz m = U zm T f T f Uz m U zm T f T f Uz m = T f ϕzm T f ϕzm T f ϕzm T f ϕzm = [T f ϕzm, T f ϕzm ] So U zm SUz m = [T f ϕzm, T f ϕzm ] 2 Hence T = [T f ϕzm, T f ϕzm ] 2 Since each f ϕ zm is continuous and supported in {w B n : ϕ zm (w) < r} = E(z m, r) and E(z k, r) E(z l, r) = for all k l, Proposition 32 and Remark 33 show that T CI(G F2 C(B n )) So T CI(G F2 C(B n ))\I(G F1 ) Acknowledgements This work was done when the author was a graduate student at the State University of New York at Buffalo References [1] Lewis A Coburn, Singular integral operators and Toeplitz operators on odd spheres, Indiana Univ Math J 23 (1973/74), MR (48 #957) [2] Trieu Le, On the commutator ideal of the Toeplitz algebra on the Bergman space of the unit ball in C n, J Operator Theory, to appear [3] Young J Lee, Pluriharmonic symbols of commuting Toeplitz type operators on the weighted Bergman spaces, Canad Math Bull 41 (1998), no 2, MR (99b:47035) [4] Walter Rudin, Function theory in the unit ball of C n, Grundlehren der Mathematischen Wissenschaften [Fundamental Principles of Mathematical Science], vol 241, Springer-Verlag, New York, 1980MR (82i:32002) [5] Daniel Suárez, The Toeplitz algebra on the Bergman space coincides with its commutator ideal, J Operator Theory 51 (2004), no 1, MR (2005b:47060) m=1 Department of Mathematics, University of Toledo, Toledo, OH address: trieule2@utoledoedu