Commutants of Toeplitz operators with monomial symbols

0.08 MB·English

by  Akaki Tikaradze

COMMUTANTS OF TOEPLITZ OPERATORS WITH
MONOMIAL SYMBOLS
5
1 AKAKITIKARADZE
0
2
Abstract. In this note we describe the commutant of the multipli-
n
cation operator by a monomial in the Toeplitz algebra of a complete
a
J strongly pseudoconvexReinhardt domain.

0
1
Throughoutgiven an n-tuple of nonnegative integers α = (i ,··· ,i ) and
1 n
A] z = (z1,··· ,zn) ∈ Cn, we will denote monomial z1i1···znin by zα. Also, Z+
F will denote the set of all nonnegative integers.
. Recall that a boundeddomain Ω ⊂ Cn is said to be a complete Reinhardt
h
t domain if (z1,··· ,zn) ∈ Ω implies (a1z1,··· ,anzn) ∈ Ω for any complex
a numbers a ∈ D¯,1 ≤ i ≤ n, where D = {z ∈ C : |z| < 1} denotes the unit
m i
disk.
[
As usual, the Bergman space of square integrable holomorphic functions

1 on a bounded domain Ω ⊂ Cn is denoted by A2(Ω). Recall that given a
v
bounded measurable function f ∈ L∞(Ω), one defines the corresponding
3
0 ToeplitzoperatorTf :A2(Ω) → A2(Ω)withsymbolf asTf(φ) = P(fφ),φ ∈
3 A2(Ω),whereP : L2(Ω) → A2(Ω)istheorthogonalprojection. Ifthesymbol
2
f is holomorphic, then T is the multiplication operator on A2(Ω) with
0 f
. symbol f. Let T(Ω) denote the C∗-algebra generated by {Tg : g ∈ L∞(Ω)}.
1
We will refer to this algebra as the Toeplitz C∗-algebra of Ω.
0
5 GivenaToplitzoperatorTf,itisofgreatinteresttostudythecommutant
1 of T in the Toeplitz C∗-algebra of Ω.
f
v: To this end, in the case of the unit disk Ω = D in C, Z. Cuckovic [[Cu],
Xi Theorem 1.4] showed that the commutant of Tzk,k ∈N in the Toeplitz C∗-
algebra of D consists of the Toeplitz operators with bounded holomorphic
r
a symbols. Subsequently T. Le [[Le], Theorem 1.1] has generalized Cuckovic’s
result to the case of of the unit ball Ω = B ⊂ Cn and a monomial f =
n
zm1···zmn such that m > 0 for all 1 ≤ i ≤ n.
1 n i
In this note we extend Le’s result to the case of strongly pseudoconvex
completeReinhardtdomains. (Theorem0.2.) Moreover, ourproofissimpler
and computation free.
As in [Cu], [Le], the following result plays the crucial role in the proof of
Theorem 0.2.
Theorem 0.1. Let Ω ⊂ Cn be a bounded complete Reinhardt domain, and
let f = z1m1z2m2···znmn be a monomial such that mi > 0 for all 1 ≤ i ≤ n.
If S : A2(Ω) → A2(Ω) is a compact operator that commutes with T , then
f
S = 0.
1
2 AKAKITIKARADZE
We will need the following trivial
Lemma0.1. Letφ ∈L∞(Ω).Thenthereexistsǫ > 0suchthat |φ|m+1dµ ≥
Ω
ǫ |φ|mdµ, for all m ∈ N. R
Ω
R
Proof. Without loss of generality we may assume that µ(φ−1(0)) = 0. Put
Ω = |φ|−1((0,ǫ)). Choose ǫ > 0 so that µ(Ω ) ≤ 1µ(Ω). Then for all m ∈ N
ǫ ǫ 2
|φ|m+1dµ ≥ |φ|m+1dµ ≥ ǫ |φ|mdµ.
Z Z Z
Ω Ω\Ωǫ Ω\Ωǫ
But
|φ|mdµ ≤ ǫmµ(Ω )≤ |φ|mdµ.
ǫ
Z Z
Ωǫ Ω\Ωǫ
Therefore
1
|φ|m+1dµ ≥ ǫ |φ|mdµ.
Z 2 Z
Ω Ω
(cid:3)
Proof. (Theorem 0.1.) Let us write f = zτ,τ = (m ,··· ,m ). Since Ω is a
1 n
complete Reinhardt domain, it is well-known that monomials {zγ,γ ∈ Zn}
+
form an orthogonal basis of A2(Ω). Assume that there exits a nonzero
compact operator S : A2(Ω) → A2(Ω) that commutes with Tzτ. Thus
S(gzmτ) = S(g)zmτ, for all g ∈ A2(Ω),m ∈ N. Let α,β ∈ Zn be such
+
that hS(zα),zβi 6= 0. Write S(z ) = c zγ,c ∈ C. Thus c 6= 0. It
A2(Ω) α γ γ γ β
follows that for any m ∈ N P
kS(zα+mτ)k = kzmτS(zα)k ≥ |c |kzβ+mτk .
A2(Ω) A2(Ω) β A2(Ω)
Let β′ ∈ Nn,k ∈ N be such that zβzβ′ = zkτ. Such β′,k exist because
m > 0, for all 1 ≤ i ≤ m. Let ǫ′ > 0 be such that ǫ′||gzβ′|| ≤ ||g||
i A2(Ω) A2(Ω)
for all g ∈ A2(Ω). Then for ǫ = |c |ǫ′ > 0, we have
β
kS(zα+mτ)k ≥ ǫkz(m+k)τk ,m ≥ 0.
A2(Ω) A2(Ω)
By Lemma 0.1, there exists δ′ > 0 such that for all m ≥ 0
kz(m+k)τk ≥ δ′kzmτk .
A2(Ω) A2(Ω)
Put δ = ǫδ′. Combining the above inequalities we get that for all m ≥ 0
kS(zα+mτ)k ≥ δkzmτk .
A2(Ω) A2(Ω)
On the other hand since
||zα+mτ||A2(Ω) ≤ ||zα||L∞(Ω)||zmτ||A2(Ω),
we get that
kS(zα+mτ)kA2(Ω) ≥ δ ,m ∈ N.
kzα+mτkA2(Ω) kzαkL∞(Ω)
COMMUTANTS OF MONOMIALS 3
However, thesequence{ zα+mτ },m ∈ Nconvergesto0weakly. Thus
kzα+mτkA2(Ω)
compactness of S implies that
kS(zα+mτ)k
A2(Ω)
lim = 0,
m→∞ kzα+mτkA2(Ω)
a contradiction.
(cid:3)
In the proof of Theorem 0.2 we will use the Hankel operators. Recall that
given a function φ∈ L∞(Ω), the Hankel operator H :A2(Ω)→ L2(Ω) with
φ
symbol φ is defined by H (g) = φg−P(φg),g ∈ A2(Ω).
φ
The proof of the following uses Theorem 0.1 and is essentially the same
as in [[Cu], page 282].
Theorem 0.2. Let Ω ⊂ Cn be a bounded smooth strongly pseudoconvex
complete Reinhardt domain, and let f = zm1···zmn,m > 0 be a monomial.
1 n i
If S is an element of the Toeplitz C∗-algebra of Ω which commutes with T ,
f
then S is a multiplication operator by a bounded holomorphic function on Ω.
Proof. Recall that for any g ∈ L∞(Ω) and a holomorphic ψ ∈ A∞(Ω) we
have [T ,T ] = H ∗H . On the other hand H is a compact operator for all
g ψ ψ¯ g z¯i
1 ≤ i ≤ n, as easily follows from [Pe]. Thus, [T ,T ] is a compact operator
g zi
for all 1 ≤ i ≤ n,g ∈ L∞(Ω). This implies that the commutator [s,T ] is
zi
compact for any s ∈ T(Ω),1 ≤ i ≤ n. If S ∈ T(Ω) commutes with T , then
f
so do compact operators [S,T ],1 ≤ i ≤ n. Therefore, by Theorem 0.2 we
zi
have [S,T ] = 0,1 ≤ i ≤ n. This implies that S = T for some bounded
zi g
holomorphic g by [[SSU], proof of Theorem 1.4].
(cid:3)
References
[Cu] Z. Cuckovic, Commutants of Toeplitz operators on the Bergman space, Pacific J.
Math. 162 (1994), no. 2, 277–285.
[Le] T.Le,The commutants of certain Toeplitz operators on weighted Bergman spaces, J.
Math. Anal. Appl.348 (2008), no. 1, 1–11.
[Pe] M. Peloso, Hankel operators on weighted Bergman spaces on strongly pseudoconvex
domains, Illinois J. Math. 38 (1994), no. 2, 223–249.
[SSU] N.Salinas, A. Sheu,H. Upmeier, Toeplitz operators on pseudoconvex domains and
foliation C∗-algebras, Ann.Math. Vol. 130, No. 3 (1989), 531–565.
Department of Mathematics, Universiy of Toledo, Toledo, Ohio, USA
E-mail address: [email protected]