# Toeplitz Operators Acting on True-Poly-Bergman Type Spaces of the Two-Dimensional Siegel Domain: Nilpotent Symbols

Authors: Yessica Hernández-Eliseo; Josué Ramírez-Ortega; Francisco G. Hernández-Zamora
Journal: Journal of Function Spaces
 (2021)

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## Abstract

We describe certainC∗-algebras generated by Toeplitz operators with nilpotent symbols and acting on a poly-Bergman type space of the Siegel domain D2⊂ℂ2. Bounded measurable functions of the form cImζ1,Imζ2−ζ12 are called nilpotent symbols. In this work, we consider symbols of the form aImζ1bImζ2−ζ12, where both limits lims→0+bs and lims→+∞bs exist, and as belongs to the set of piecewise continuous functions on ℝ¯=−∞,+∞ and having one-side limit values at each point of a finite set S⊂ℝ. We prove that the C∗-algebra generated by all Toeplitz operators Tab is isomorphic to CΠ¯, where Π¯=ℝ¯×ℝ¯+ and ℝ¯+=0,+∞.


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## Body

## 1. Introduction

In the study of Toeplitz operators, one of the common strategies consists in selecting a set of symbolsE⊂L∞ in such a way that the algebra generated by Toeplitz operators with symbols in E can be described up to isomorphism, say, with an algebra of continuous functions or finding its spectrum. In this paper, we study Toeplitz operators with nilpotent symbols and acting on a poly-Bergman type space of the Siegel domain D2⊂ℂ2. In [1–3], the authors have fully described all commutative C∗-algebras generated by Toeplitz operators with symbols invariant under the action of a maximal abelian subgroup of biholomorphisms and acting on the Bergman spaces of both the unit disk D and the Siegel domain Dn⊂ℂn. For the unit disk, they discovered three families of symbols associated to commutative C∗-algebras of Toeplitz operators, while for the Siegel domain, they found n+2 classes of symbols. Each class of symbols is invariant under the action of a maximal abelian group of biholomorphism. Certainly, one can use these classes of symbols to study Toeplitz operators acting on poly-Bergman type spaces of the unit disk or the Siegel domain.LetΠ=z=x+iy∈ℂ:y>0 be the upper half-plane. Toeplitz operators with vertical symbols, which depend on y=Imz, and acting on Bergman type spaces have been studied. In [4–9], the authors proved that the algebra generated by Toeplitz operators with vertical symbols and acting on the weighted Bergman space Aλ2Π is isometrically isomorphic to the algebra VSOℝ+ of all bounded functions that are very slowly oscillating on ℝ+. Taking vertical symbols having limit values at y=0 and y=∞, in [10, 11], the authors found that ℝ¯+=0,+∞ is the spectrum of the algebra generated by all Toeplitz operators on the true-poly Bergman space An2Π. Similar research was made for Toeplitz operators on poly-Bergman spaces with homogeneous symbols ([12, 13]). Other works about it were made in [14–16], where the authors studied Toeplitz operators acting on An2Π from the point of view of wavelet spaces. On the other hand, in [17, 18], the authors studied Toeplitz operators on the Fock space F12ℂ with radial and bounded horizontal symbols; they found that spectral functions are uniformly continuous with respect to an adequate metric. Taking horizontal symbols having one-side limits at x=±∞, in [19, 20], the authors studied Toeplitz operators acting on poly-Fock spaces Fk2ℂ; they found the spectrum of the C∗-algebra generated by such Toeplitz operators. Even though the authors described the C∗-algebras generated by all spectral functions, the spectrum of the algebras is not fully understood in some cases; for this reason, additional conditions on the symbols are imposed.In [1, 2], the authors made remarkable research on the study of Toeplitz operators acting on the Bergman space of the Siegel domain Dn⊂ℂn. In particular, they studied the C∗-algebra TNn generated by all Toeplitz operators with bounded nilpotent symbols, which are functions of the form aζ=aImζ1,⋯,Imζn−1,Imζn−ζ′2, where ζ′=ζ1,⋯,ζn−1. Let us denote this kind of symbols by Nn. Although TNn is commutative, it is too large, so it is impossible to figure out what its spectrum is. In particular, in [21, 22], the authors described the algebra generated by Toeplitz operators acting on the weighted Bergman space Aλ2D3 over three-dimensional Siegel domain D3 using nilpotent symbols of the form cy2gImζ3−ζ1,ζ22.The main purpose of the paper is to find the spectrum of the algebra generated by Toeplitz operators acting on the true-poly-Bergman type spaceAL2D2 over two-dimensional Siegel domain D2 by selecting a particular set of nilpotent symbols. In this sense, we just consider nilpotent symbols of the form aImζ1 and bImζ2−ζ12. This paper is organized as follows. In Section 2, we recall how poly-Bergman type spaces are defined for the Siegel domain, and how they can be identified with a L2-space through a Bargmann type transform. In Section 3, we introduce Toeplitz operators acting on AL2D2 with nilpotent symbols; we show that such Toeplitz operators are unitary equivalent to multiplication operators. In Section 4, we take symbols of the form bImζ2−ζ12 for which both limits lims→0+bs and lims→+∞bs exist; it is proved that the C∗-algebra generated by all Toeplitz operators Tb is isomorphic to Cℝ¯+, where ℝ¯+=0,+∞ is the one-point compactification of 0,+∞. In Section 5, we take nilpotent symbols of the form aImζ1, where a∈Cℝ¯, and ℝ¯=−∞,+∞ is the two-point compactification of ℝ; we prove that the C∗-algebra generated by all Toeplitz operators Ta is isomorphic to CΔ, where Δ=Π¯/ℝ¯×+∞ and Π¯=ℝ¯×ℝ¯+. In Section 6, we describe the C∗-algebra generated by all Toeplitz operators Td, where dImζ1∈PCℝ¯,S, and PCℝ¯,S is the set of all piecewise continuous functions on ℝ¯ having one-side limit values at each point of a finite set S⊂ℝ. Finally, in Section 7, we describe the C∗-algebra generated by all Toeplitz operators Tab=TaTb=TbTa.


## 2. Poly-Bergman Type Spaces of the Siegel Domain

In this section, we recall some results obtained in [23], which are needed in our research about Toeplitz operators. Each ζ∈ℂn will be represented as an ordered pair ζ=ζ′,ζn, where ζ′=ζ1,⋯,ζn−1∈ℂn−1. Besides, the Euclidean norm function will be denoted by ·. The Siegel domain is defined by
(1)Dn=ζ=ζ′,ζn∈ℂn−1×ℂ:Imζn−ζ′2>0.We will study Toeplitz operators acting on certain poly-Bergman type subspaces ofL2Dn,dμλ, where dμλζ=Imζn−ζ′2λdμζ, with λ>−1, and dμζ is the usual Lebesgue measure. Once and for all, L2X means L2X,dm, where X is any subset of a Euclidean space and dm is the Lebesgue area measure on X.For each multi-indexL=l1,⋯,ln∈ℕn, the poly-Bergman type space AλL2Dn is the closed subspace of L2Dn,dμλ consisting of all L-analytic functions, that is, all functions fζ satisfying the equations
(2)∂∂ζ¯m−2iζm∂∂ζ¯nlmf=0,1≤m≤n−1,∂∂ζ¯nlnf=0.In particular, forL=1,⋯,1, AλL2Dn is just the Bergman space. Likewise, the anti-poly-Bergman type space A~λL2Dn is defined to be the complex conjugate of AλL2Dn. Thus, we introduce true-poly-Bergman type spaces as follows:
(3)AλL2Dn=AλL2Dn⊖∑m=1nAλ,L−em2Dn,A~λL2Dn=A~λL2Dn⊖∑m=1nA~λ,L−em2Dn,where em=0,⋯,1,⋯,0 and the 1 is placed at the m-entry. We assume that AλL2Dn=0 whenever L∈ℤn\ℕn.In [23], the authors proved that L2Dn,dμλ equals to the direct sum of all the true-poly-Bergman type spaces:
(4)L2Dn,dμλ=⊕L∈ℕnAλL2Dn⨁⊕L∈ℕnA~λL2Dn.The authors also proved thatAλL2Dn is isomorphic and isometric to the tensor product
(5)L2ℝn−1⊗ℍl1−1⊗⋯⊗ℍln−1−1⊗L2ℝ+⊗Lln−1,where ℝ+=0,∞. Both ℍm and Lm are one-dimensional spaces defined below. Recall the Hermite and Laguerre polynomials:
(6)Hmy≔−1mey2dmdyme−y2,Lmλy≔eyy−λm!dmdyme−yym+λ,for m=0,1,2,⋯ Recall also the Hermite and Laguerre functions
(7)hmy=−1m2nπn!1/2Hmye−y2/2,ℓmλy=−1mcmLmλye−y/2,where cm=m!/Γm+λ+1 and Γ is the usual Gamma function. It is well known that hmm=0∞ and ℓmλm=0∞ are orthonormal bases for L2ℝ and L2ℝ+,yλdy, respectively. Finally, ℍm=spanhm and Lm=spanℓmλ.In this work, we restrict ourselves to the study of Toeplitz operators acting on the true-poly-Bergman type spaces over two-dimensional Siegel domainD2 with the Lebesgue measure dμ (λ=0). Henceforth, the space A0L2D2 will be simply denoted by AL2D2; similarly, ℓmy and Lmy stand for ℓm0y and Lm0y, respectively. The true-poly-Bergman type space AL2D2 can be identified with L2ℝ×ℝ+ through a Bargmann type transform ([23]), such identification fits to the study of Toeplitz operators with nilpotent symbols. Several operators are needed to define such identification. To begin with, we introduce the flat domain D=ℂ×Π, where Π=ℝ×ℝ+⊂ℂ. Then, D can be identified with D2 using the mapping
(8)κ:D∋w=w1,w2↦ζ=w1,w2+iw12∈D2.Thus, we have the unitary operatorU0:L2D2,dμ⟶L2D,dη given by
(9)U0fw=fκw,where dηw=dμw. Take w=w1,w2∈ℂ×Π, with wm=um+ivm and m=1,2. We identify w=u1+iv1,u2+iv2 with u1,v1,u2,v2. Then
(10)L2D,dη=L2ℝ,du1⊗L2ℝ,dv1⊗L2ℝ,du2⊗L2ℝ+,dv2.Introduce(11)U1=F⊗I⊗F⊗I,where F is the Fourier transform acting on L2ℝ by the rule
(12)Fgt=12π∫−∞∞gxe−itxdx.Consider now the following two mappings acting onD:
(13)ψ1:ξ=ξ1,t2+is2↦w=ξ1,t2+is22t2,ψ2:z=x1+iy1,z2↦ξ=x2x1+y1+i12x2−x1+y1,z2,where ξ=ξ1,ξ2, z=z1,z2∈D, ξm=tm+ism, and zm=xm+iym. Both functions ψ1 and ψ2 lead to the following unitary operators acting on L2D,dη:
(14)V1fξ=12t21/2fψ1ξ,V2gz=gψ2z.Henceforth,L=l1,l2=j,k.Theorem 1 (see [23]).
The operatorU=V2V1U1U0 is unitary and maps L2D2,dμ onto the space
(15)L2D,dη=L2ℝ,dx1⊗L2ℝ,dy1⊗L2ℝ,dx2⊗L2ℝ+,dy2.
For eachL=j,k∈ℕ2, the operator U restricted to AL2D2 is an isometric isomorphism onto the space
(16)HL+=L2ℝ⊗spanhj−1y1⊗L2ℝ+⊗spanℓk−1y2.Introduce the isometric linear embeddingR0L:L2ℝ×ℝ+⟶L2D defined by
(17)R0Lgx1,y1,x2,y2=χℝ+x2gx1,x2hj−1y1ℓk−1y2.Of course,HL+ is the range of R0L, and it is also the image of AL2D2 under U. Thus, the operator
(18)RL=R0L∗U:L2D2⟶L2ℝ×ℝ+,isometrically maps the true-poly-Bergman type space AL2D2 onto L2ℝ×ℝ+. Therefore, RLRL∗=I and RL∗RL=BL, where BL is the orthogonal projection from L2D2 onto AL2D2. In addition, the operator RL∗=U∗R0L plays the role of the Segal-Bargmann transform for the true-poly-Bergman type space AL2D2, where the adjoint operator R0L∗:L2D⟶L2ℝ×ℝ+ is given by
(19)R0L∗fx1,x2=∫ℝ∫ℝ+hj−1y1ℓk−1y2fx1,y1,x2,y2dy2dy1,with x1,x2∈ℝ×ℝ+.


## 3. Toeplitz Operators with Nilpotent Symbols

In this section, we study Toeplitz operators with nilpotent symbols and acting on the true-poly-Bergman type spaceAL2D2. In [3], the author has widely developed the theory of Toeplitz operators on the Bergman spaces, and the author’s techniques can be applied to the study of Toeplitz operators acting on AL2D2. To begin with, a function c∈L∞D2,dμ is said to be a nilpotent symbol if it has the form cζ1,ζ2=cImζ1,Imζ2−ζ12. Then, the Toeplitz operator acting on AL2D2, with nilpotent symbol cζ, is defined by
(20)Tcfζ=BLcfζ,where BL is the orthogonal projection from L2D2 onto AL2D2. The Bargmann-type operator RL identifies the space AL2D2 with L2ℝ×ℝ+, and it fits properly in the study of the Toeplitz operator Tc.Theorem 2.
Letc be a nilpotent symbol. Then, the Toeplitz operator Tc is unitary equivalent to the multiplication operator γcI=RLTcRL∗, where γc:ℝ×ℝ+⟶ℂ is given by
(21)γcx1,x2=∫ℝ∫ℝ+c−x1+y12x2,y22x2hj−1y12ℓk−1y22dy2dy1.Proof.
We have(22)RLTcRL∗=RLBLcIRL∗=RLRL∗RLcIRL∗=RLcIRL∗=R0L∗V2V1U1U0cIU0−1U1−1V1−1V2−1R0L.
Recall thatζ=κw=w1,w2+iw12, where w=w1,w2∈D and wm=um+ivm. For g∈L2D,
(23)U0cIU0−1gw=cκwU0−1gκw=cκwgw.
That is,U0cIU0−1=cκwI, where cκw=cv1,v2. It is easy to see that U1cv1,v2IU1−1=cv1,v2I,(24)V1cIV1−1=cs1,s22t2I,V2V1cIV1−1V2−1=c−x1+y12x2,y22x2I.
Thus(25)RLTcRL∗=R0L∗c−x1+y12x2,y22x2IR0L=γcx1,x2I,where γcx1,x2 is given in (21).By Theorem2, the C∗-algebra generated by all Toeplitz operators Tc is commutative (see [1, 3]), but its spectrum is difficult to figure out what it is. For this reason, we assume certain continuity conditions on the nilpotent symbols in order to describe the spectrum of the subalgebra generated by the Toeplitz operators. We will split our research into two cases concerning the symbols. Firstly, we study Toeplitz operators with symbols of the form bImζ2−ζ12, for which
(26)γbx1,x2=γbx2=∫ℝ+by22x2ℓk−1y22dy2.Secondly, we analyze Toeplitz operators with symbols of the formaImζ1, for which
(27)γax1,x2=∫ℝa−x1+y12x2hj−1y12dy1.As mentioned above, theC∗-algebra generated by all Toeplitz operators Ta is still complicated to be fully described despite its commutative property. Fortunately, the C∗-algebra generated by all operators Ta can be described when the symbols a are taken to be continuous on ℝ¯=−∞,+∞, where ℝ¯ is the two-point compactification of ℝ. Even more, the C∗-algebra of Toeplitz operators Ta can be still described for symbols having finitely many jump discontinuities, as shown in Section 6. Finally, we analyze Toeplitz operators with symbols of the form cζ1,ζ2=aImζ1bImζ2−ζ12.


## 4. Toeplitz Operators with SymbolsbImζ2−ζ12

In this section, we study theC∗-algebra generated by all Toeplitz operators Tb with symbols of the form bImζ2−ζ12, where by has limit values at y=0,+∞. Under this continuity condition, we will see that γb is continuous on Π¯≔ℝ¯×ℝ¯+, where ℝ¯+=0,+∞ is the two-point compactification of ℝ+=0,+∞. Apply the change of variable y2↦2x2y2 in the integral representation of γb, then
(28)γbx1,x2=γbx2=2x2∫ℝ+by2ℓk−12x2y22dy2.Actually,γb depends only on the variable x2 and is continuous on ℝ+ because of the continuity of ℓk−1y and the Lebesgue dominated convergence theorem.LetL0,+∞∞ℝ+ denote the subspace of L∞ℝ+ consisting of all functions having limit values at 0 and +∞. For b∈L0,+∞∞ℝ+, define
(29)b0≔limy→0+by,b∞≔limy→+∞by.It is worth mentioning thatγb was obtained in [10] as the spectral function of a Toeplitz operator acting on a true-poly-Bergman space of the upper half-plane. Thus, we have at least two scenarios in which γb appears as a spectral function.Lemma 3 (see [10]).
Letb∈L0,+∞∞ℝ+. Then, the spectral function γb satisfies
(30)b∞=limx2→0+γbx2,b0=limx2→+∞γbx2.According to Lemma3 and Theorem 4.8 in [10], we have the following.Theorem 4.
Forb∈L0,+∞∞ℝ+, the spectral function γbx2 is continuous on ℝ¯=0,+∞. The C∗-algebra generated by all functions γb, with b∈L0,+∞∞ℝ+, is isomorphic and isometric to the algebra C0,∞. That is, the C∗-algebra generated by all Toeplitz operators Tb, with bImζ2−ζ12∈L0,+∞∞ℝ+, is isomorphic to C0,∞, where the isomorphism is defined on the generators by
(31)Tb↦γb.Obviously, the spectral functionγbx2 is defined and continuous on Π¯, but it is constant along each horizontal straight line. Thus, γb is identified with a continuous function on the quotient space Π¯/ℝ¯, which is homeomorphic to ℝ¯+.


## 5. Toeplitz Operators with Continuous SymbolsaImζ1

In this section, we study theC∗-algebra generated by all Toeplitz operators Ta, where symbols aImζ1 are taken to be continuous on ℝ¯. Once again, such a C∗-algebra can be identified with the algebra of all continuous functions on a quotient space of Π¯. Henceforth, x1,x2 will denote points in Π¯ instead of intervals.It is fairly simple to see thatγa is continuous on Π. Take the change of variable y1↦2x2y1+x1 in the integral representation of γa, then
(32)γax1,x2=2x2∫ℝay1hj−12x2y1+x12dy1.The functionγa is continuous at each point x1,x2∈Π because of the continuity of hj−1 and the Lebesgue dominated convergence theorem. Next, we will prove that γa has one-side limit value at each point of ℝ×0. For a∈L∞ℝ, we introduce the notation
(33)a−=limy→−∞ayanda+=limy→+∞ay,if such limits exist.Lemma 5.
Leta∈L∞ℝ, and suppose that ay converges at y=±∞. Then, for each x0∈ℝ, the spectral function γa satisfies
(34)limx1,x2→x0,0γax1,x2=a−∫−∞x0hj−1y12dy1+a+∫x0∞hj−1y12dy1.Proof.
LetA denote the right-hand side of equality (34). Take ε>0. We will prove that there exist δ>0 such that γax1,x2−A<ε whenever x1−x0<δ and 0<x2<δ. Note that a−,a+≤a∞. Since ∫−∞∞hj−1y12dy1=1, there exists δ1>0 such that
(35)a∞∫−δ1+x0δ1+x0hj−1y12dy1<ε5.
Then(36)I≔γax1,x2−A=∫−∞∞a−x1+y12x2hj−1y12dy1−a−∫−∞x0hj−1y12dy1−a+∫x0∞hj−1y12dy1≤∫−∞−δ1+x0a−x1+y12x2−a−hj−1y12dy1+a−∫−δ1+x0x0hj−1y12dy1+a+∫x0δ1+x0hj−1y12dy1+∫−δ1+x0δ1+x0a−x1+y12x2hj−1y12dy1+∫δ1+x0∞a−x1+y12x2−a+hj−1y12dy1≤max−∞<y1<−δ1+x0a−x1+y12x2−a−+3ε5+maxδ1+x0<y1<∞a−x1+y12x2−a+.
We have assumed thatay converges at ±∞; then, there exists N>0 such that ay−a−<ε/5 and ay−a+<ε/5 for y>N. Let δ=minδ1/2,δ12/16N2. Then, we have 1/2x2−x1+y1>N if x1−x0<δ, 0<x2<δ, and y1−x0≥δ1. Thus,
(37)max−∞<y1<−δ1+x0a−x1+y12x2−a−<ε5,maxδ1+x0<y1<∞a−x1+y12x2−a+<ε5.
Finally, we conclude thatγax1,x2−A<ε whenever x1−x0<δ and 0<x2<δ.In general,γax1,x2 does not converge at each point x=±∞,+∞∈Π¯; however, γax1,x2 has limit values along the parabolas x2=αx12+1, with α>0. We will define a bijective mapping Φ:Π⟶Π so that ϕa=γa∘Φ−1 will be a continuous mapping on Π¯=ℝ¯×ℝ¯+ with the usual topology.


### 5.1. Modified Spectral Function forTa

LetΦ:Π⟶Π be the mapping
(38)Φx1,x2=x1,x2x12+1.It is easy to see thatΦ−1t1,t2=t1,t12+1t2. Concerning the spectral properties of Ta, the function ϕa≔γa∘Φ−1 is as important as γa, but ϕa behaves much better than γa, at least for a continuous on ℝ¯. From now on, we take ϕa as the spectral function for the Toeplitz operator Ta. A direct computation shows that
(39)ϕat1,t2=∫−∞∞a−t1+s12t2t12+1hj−1s12ds1.BothΦ and Φ−1 are continuous on Π. Besides, the spectral function ϕa=γa∘Φ−1 is continuous on Π because γa is. Since Φ−1t1,0=t1,0, we have ϕat1,0=γax1,0. By Lemma 5, ϕa is also continuous on ℝ×0.Theorem 6.
ForaImζ1∈Cℝ¯, the spectral function ϕa:Π⟶ℂ can be extended continuously to Π¯=ℝ¯×ℝ¯+.Proof.
Follows from Lemmas5 and 7–9.For any domainX⊂ℝm and a function φ:X⟶ℂ, we write φx0 to mean the limit value of φ at x0, even if x0 does not belong to X. For example, a+∞ means limx→+∞ax.Lemma 7.
Leta∈L∞ℝ, and suppose that ay converges at the points y=±∞. Then, ϕa satisfies
(40)limt1,t2→+∞,0ϕat1,t2=a−∞.
That is, forε>0, there exists δ>0 and N>0 such that ϕat1,t2<ε whenever 0<t2<δ and t1>N. Analogously,
(41)limt1,t2→−∞,0ϕat1,t2=a+∞.Proof.
Suppose thata−∞=0. Let ε>0. Since hj−1∈L2ℝ, there exists s0>0 such that
(42)a∞∫s0∞hj−1s12ds1<ε2.
Take into account∫−∞∞hj−1s12ds1=1 in the following computation
(43)ϕat1,t2=∫-∞∞a-t1+s12t2t12+1hj-1s12ds1≤∫-∞s0a-t1+s12t2t12+1hj-1s12ds1+∫s0∞a-t1+s12t2t12+1hj-1s12ds1≤max-∞<s1<s0a-t1+s12t2t12+1+ε2.
Sinceas converges to zero at −∞, there exists N1>0 such that as<ε/2 for −s>N1. Take δ=1/16N12. Then, we have 1/2t2>2N1 for 0<t2<δ. On the other hand, assume t1>s0 and −∞<s1<s0. Then
(44)t1−s1t12+1>t1−s0t12+1.
The right-hand side of this inequality converges to1 when t1 tends to +∞; thus, there exists N2>s0 such that t1−s0/t12+1>1/2 for t1>N2. Consequently,
(45)N1=2N112<12t2t1−s0t12+1<t1−s12t2t12+1.
For0<t2<δ and t1>N≔maxs0,N2, we have
(46)a−t1+s12t2t12+1<ε2.
We definea~s=as−a2 in the case a−∞≠0, where a2≔a−∞ is a constant. Note that a~s converges to zero at −∞, and ϕa1+a2=ϕa1+ϕa2 for any nilpotent symbols a1 and a2. Then
(47)limt1,t2→+∞,0ϕat1,t2=limt1,t2→+∞,0ϕa~+a2t1,t2=limt1,t2→+∞,0ϕa~t1,t2+a2∫−∞∞hj−1s12ds1=a−∞.
Finally, the limit ofϕat1,t2 at −∞,0 can be proved analogously.Lemma 8.
Lett0∈ℝ+. If a∈L∞ℝ is continuous at −1/2t0, then the spectral function ϕa satisfies
(48)limt1,t2→+∞,t0ϕat1,t2=a−12t0.
Analogously, ifa is continuous at 1/2t0, then
(49)limt1,t2→−∞,t0ϕat1,t2=a12t0.Proof.
Suppose thata converges to zero at −1/2t0. Let ε>0. Since hj−1∈L2ℝ, there exists s0>0 such that
(50)a∞∫−∞−s0hj−1s12ds1<ε3,a∞∫s0∞hj−1s12ds1<ε3.
Take into account∫−∞∞hj−1s12ds1=1 in the following computation
(51)ϕat1,t2≤∫-∞-s0a-t1+s12t2t12+1hj-1s12ds1+∫-s0s0a-t1+s12t2t12+1hj-1s12ds1+∫s0∞a-t1+s12t2t12+1hj-1s12ds1<2ε3+max-s0<s1<s0a-t1+s12t2t12+1.
Because of the continuity ofas at −1/2t0, there exists δ1>0 such that as<ε/3 for s−−1/2t0<δ1. Let us estimate the value of the argument of a:
(52)I≔12t2t12+1−t1+s1−−12t0≤−12t2+12t0t1t12+1+12t01−t1t12+1+s12t2t12+1.
Chooseδ>0 in such a way −1/2t2+1/2t0<δ1/3 for t2−t0<δ. Pick N1>0 such that 1−t1/t12+1<2t0δ1/3 whenever t1>N1. Now assume that t2−t0<δ and s1<s0. Then, 1/2t2<1/2t0+δ1/3. Thus, s1/2t2t12+1 converges to 0 when t1 tends to +∞. Therefore, there exists N>N1 such that s1/2t2t12+1<δ1/3 for t1>N. The additional condition t1>N implies
(53)a−t1+s12t2t12+1<ε/3.
Hence,ϕat1,t2<ε if t2−t0<δ and t1>N.
Ifa does not converge to zero at −1/2t0, then take the function a~s=as−a2 and proceed as in the proof of Lemma 7, where a2=a−1/2t0.
Finally, the justification of the limit ofϕat1,t2 at −∞,t0 can be done analogously.Lemma 9.
Leta∈L∞ℝ be continuous at 0∈ℝ. For t0∈ℝ¯, the spectral function ϕa satisfies
(54)limt1,t2→t0,+∞ϕat1,t2=a0.
Actually, we have uniform convergence ofϕat1,t2 at t0,+∞, that is, for ε>0, there exists N>0 such that ϕat1,t2−a0<ε for all t2>N and for all t1∈ℝ¯.Proof.
Suppose thata0=0. Let ε>0, and choose s0>0 such that equations (50) hold. Then
(55)ϕat1,t2≤∫−∞−s0a−t1+s12t2t12+1hj−1s12ds1+∫−s0s0a−t1+s12t2t12+1hj−1s12ds1+∫s0∞a−t1+s12t2t12+1hj−1s12ds1<2ε3+max−s0<s1<s0a−t1+s12t2t12+1.
By the continuity ofas at 0, there exists δ1>0 such that as<ε/3 for s<δ1. For −s0<s1<s0, we have
(56)−t1+s12t2t12+1≤12t2t1t12+1+s1t12+1<12t21+s0.
TakeN=1+s02/4δ12. The inequality t2>N implies 1/2t2<δ1/1+s0. Thus, if t2>N, t1∈ℝ¯, and −s0<s1<s0, then
(57)−t1+s12t2t12+1<δ1.
Consequently,ϕat1,t2<ε for all t2>N and t1∈ℝ¯.
Finally, in the casea0≠0, the proof can be carry out by considering the symbol a~s=as−a0.For each nilpotent symbolaImζ1∈Cℝ¯, the spectral function ϕa is continuous on Π¯ and is constant along ℝ¯×+∞. For this reason, the C∗-algebra generated by all spectral functions ϕa is not CΠ¯, but it coincides with the algebra of continuous functions on a quotient space of Π¯.


### 5.2. Toeplitz OperatorsTa with Continuous Symbols aImζ1

Introduce the quotient spaceΔ≔Π¯/ℝ¯×+∞. By Lemma 9, the spectral function ϕa can be identified with a continuous function on Δ, which will be also denoted by ϕa:Δ⟶ℂ. We establish now one of our main results in this work.Theorem 10.
TheC∗-algebra generated by all spectral functions ϕa, with aImζ1∈Cℝ¯, is isomorphic and isometric to the algebra CΔ. That is, the C∗-algebra generated by all Toeplitz operators Ta is isomorphic to CΔ, where the isomorphism is defined on the generators by the rule
(58)Ta↦ϕa.Proof.
The functionsϕa separate the points of Δ according to Lemmas 11–13 below. The Stone-Weierstrass theorem completes the proof.Lemma 11.
Lett1,t2 and y1,y2 be distinct points of Π¯\Π, where they do not belong simultaneously to ℝ¯×+∞. Then, there exists a∈Cℝ¯ such that ϕat1,t2≠ϕay1,y2.Proof.
First consider the nilpotent symbola1s=s/s2+1, which is continuous on ℝ¯. Note that
(i)
ϕa1±∞,t2=a1∓1/(2t2)=∓1/1+4t2for0<t2≤+∞,(ii)
ϕa1±∞,0=a1∓∞=∓1,(iii)
ϕa1t1,+∞=a10=0 for t1∈ℝ¯,(iv)
ϕa1t1,0=−∫−∞t1hj−1s12ds1+∫t1∞hj−1s12ds1 for t1∈ℝ.
From∫−∞∞hj−1s12ds1=1, we get
(59)ϕa1t1,0=2∫t1∞hj−1s12ds1−1.
This formula also says thatϕa1±∞,0=∓1. On the other hand, the Hermite function hj−1 is continuous, and it has just a finitely many roots. Hence, ϕa1t1,0 is monotonically decreasing with respect to t1. Thus, points in ℝ¯×0 are separated by ϕa1.
Recall thatℝ¯×+∞ is identified with one point in Δ so take −∞,+∞ as representative point of the equivalence class ℝ¯×+∞. For t2,t~2∈0,∞, the three points −∞,t2, −∞,+∞, and +∞,t~2 are separated by ϕa1±∞,t because of the injective property of 1/1+4t.
Consider now the nilpotent symbola2s=1/s2+1, which is continuous on ℝ¯. We have
(i)
ϕa2±∞,t2=a2∓1/(2t2)=4t2/(1+4t2)>0 for t2>0,(ii)
ϕa2t1,+∞=a20=1 for t1∈ℝ¯,(iii)
ϕa2t1,0=0 for all t1∈ℝ¯.
Thus,ϕa2 separates each point t1,0 from the points ±∞,t2 and t1,+∞.Now our aim is to separate the points ofΠ. Consider the following family of continuous functions:
(60)aαs=0si−∞≤s≤−α,1αs+1si−α≤s≤0,1si0≤s≤+∞.Then,ϕaαt1,t2=ψαt1,t2+φt1, where
(61)ψαt1,t2=∫t1−2αt2t12+1t1s1−t12αt2t12+1+1hj−1s12ds1,(62)φt1=∫t1∞hj−1s12ds1.Lemma 12.
Ift1,t2 and y1,y2 are distinct points in Π, then there exists α>0 such that ϕaαt1,t2≠ϕaαy1,y2, where aα is defined in (60).Proof.
At first suppose thatt1,t2 and y1,y2 satisfy t1<y1. Introduce k≔φt1−φy1>0. It is easy to see that ψαt1,t2 can be written as
(63)ψαt1,t2=∫−10s+1hj−1t1+2αt2t12+1s22αt2t12+1ds.
This integral representation allows us to prove easily that(64)limα→0+ψαt1,t2=0.
Takeα small enough that ψαt1,t2−ψαy1,y2<k/2. Then
(65)ϕaαt1,t2−ϕaαy1,y2=k+ψαt1,t2−ψαy1,y2≥k−ψαt1,t2−ψαy1,y2≥k−k2>0.
Now suppose thatt1=y1=t and t2<y2. Let α=1/2t2+1>0. Then
(66)ψαt,t2=∫t−t2ts1−tt2+1hj−1s12ds1,ψαt,y2=∫t−y2ts1−ty2+1hj−1s12ds1.
Since0<t2<y2, we have t−y2<t−t2. Besides, the following inequality holds
(67)0<s1−tt2+1<s1−ty2+1,for t−t2<s1<t. Hence, ψαt,t2<ψαt,y2. Finally, from φt1=φy1=φt, we get ϕaαt,t2<ϕaαt,y2.Lemma 13.
Lett1,t2∈Π and y1,y2∈Π¯\Π. Then, there exists α>0 such that ϕaαt1,t2≠ϕaαy1,y2, where aα is given in (60).Proof.
Taket1,t2∈Π, y1,0∈ℝ×0, and α>0. It is easy to see that ψaαt1,t2>0. Then, we have ϕaαy1,0=φy1≤φt1<ϕaαt1,t2 for t1≤y1. Now suppose that y1<t1. Let k≔φy1−φt1>0. The inequality ψαt1,t2<k holds for α>0 small enough. For such α,
(68)ϕaαt1,t2=ψαt1,t2+φt1<φy1=ϕaαy1,0.
This proves that all points ofΠ can be separated from points of ℝ×0. On the other hand,
(69)0<ϕaαt1,t2=ψαt1,t2+φt1≤∫t1−2αt2t12+1t1hj−1s12ds1+φt1=∫t1−2αt2t12+1∞hj−1s12ds1<1.
Fory2∈ℝ¯+, we have ϕaα−∞,y2=aα1/2y2=1; hence, ϕaαt1,t2≠ϕaα−∞,y2.
Finally, consider the functionas≔aαs−α. Then, ϕa+∞,y2=a−1/2y2=0, and consequently, ϕat1,t2≠ϕa+∞,y2.


## 5.1. Modified Spectral Function forTa

LetΦ:Π⟶Π be the mapping
(38)Φx1,x2=x1,x2x12+1.It is easy to see thatΦ−1t1,t2=t1,t12+1t2. Concerning the spectral properties of Ta, the function ϕa≔γa∘Φ−1 is as important as γa, but ϕa behaves much better than γa, at least for a continuous on ℝ¯. From now on, we take ϕa as the spectral function for the Toeplitz operator Ta. A direct computation shows that
(39)ϕat1,t2=∫−∞∞a−t1+s12t2t12+1hj−1s12ds1.BothΦ and Φ−1 are continuous on Π. Besides, the spectral function ϕa=γa∘Φ−1 is continuous on Π because γa is. Since Φ−1t1,0=t1,0, we have ϕat1,0=γax1,0. By Lemma 5, ϕa is also continuous on ℝ×0.Theorem 6.
ForaImζ1∈Cℝ¯, the spectral function ϕa:Π⟶ℂ can be extended continuously to Π¯=ℝ¯×ℝ¯+.Proof.
Follows from Lemmas5 and 7–9.For any domainX⊂ℝm and a function φ:X⟶ℂ, we write φx0 to mean the limit value of φ at x0, even if x0 does not belong to X. For example, a+∞ means limx→+∞ax.Lemma 7.
Leta∈L∞ℝ, and suppose that ay converges at the points y=±∞. Then, ϕa satisfies
(40)limt1,t2→+∞,0ϕat1,t2=a−∞.
That is, forε>0, there exists δ>0 and N>0 such that ϕat1,t2<ε whenever 0<t2<δ and t1>N. Analogously,
(41)limt1,t2→−∞,0ϕat1,t2=a+∞.Proof.
Suppose thata−∞=0. Let ε>0. Since hj−1∈L2ℝ, there exists s0>0 such that
(42)a∞∫s0∞hj−1s12ds1<ε2.
Take into account∫−∞∞hj−1s12ds1=1 in the following computation
(43)ϕat1,t2=∫-∞∞a-t1+s12t2t12+1hj-1s12ds1≤∫-∞s0a-t1+s12t2t12+1hj-1s12ds1+∫s0∞a-t1+s12t2t12+1hj-1s12ds1≤max-∞<s1<s0a-t1+s12t2t12+1+ε2.
Sinceas converges to zero at −∞, there exists N1>0 such that as<ε/2 for −s>N1. Take δ=1/16N12. Then, we have 1/2t2>2N1 for 0<t2<δ. On the other hand, assume t1>s0 and −∞<s1<s0. Then
(44)t1−s1t12+1>t1−s0t12+1.
The right-hand side of this inequality converges to1 when t1 tends to +∞; thus, there exists N2>s0 such that t1−s0/t12+1>1/2 for t1>N2. Consequently,
(45)N1=2N112<12t2t1−s0t12+1<t1−s12t2t12+1.
For0<t2<δ and t1>N≔maxs0,N2, we have
(46)a−t1+s12t2t12+1<ε2.
We definea~s=as−a2 in the case a−∞≠0, where a2≔a−∞ is a constant. Note that a~s converges to zero at −∞, and ϕa1+a2=ϕa1+ϕa2 for any nilpotent symbols a1 and a2. Then
(47)limt1,t2→+∞,0ϕat1,t2=limt1,t2→+∞,0ϕa~+a2t1,t2=limt1,t2→+∞,0ϕa~t1,t2+a2∫−∞∞hj−1s12ds1=a−∞.
Finally, the limit ofϕat1,t2 at −∞,0 can be proved analogously.Lemma 8.
Lett0∈ℝ+. If a∈L∞ℝ is continuous at −1/2t0, then the spectral function ϕa satisfies
(48)limt1,t2→+∞,t0ϕat1,t2=a−12t0.
Analogously, ifa is continuous at 1/2t0, then
(49)limt1,t2→−∞,t0ϕat1,t2=a12t0.Proof.
Suppose thata converges to zero at −1/2t0. Let ε>0. Since hj−1∈L2ℝ, there exists s0>0 such that
(50)a∞∫−∞−s0hj−1s12ds1<ε3,a∞∫s0∞hj−1s12ds1<ε3.
Take into account∫−∞∞hj−1s12ds1=1 in the following computation
(51)ϕat1,t2≤∫-∞-s0a-t1+s12t2t12+1hj-1s12ds1+∫-s0s0a-t1+s12t2t12+1hj-1s12ds1+∫s0∞a-t1+s12t2t12+1hj-1s12ds1<2ε3+max-s0<s1<s0a-t1+s12t2t12+1.
Because of the continuity ofas at −1/2t0, there exists δ1>0 such that as<ε/3 for s−−1/2t0<δ1. Let us estimate the value of the argument of a:
(52)I≔12t2t12+1−t1+s1−−12t0≤−12t2+12t0t1t12+1+12t01−t1t12+1+s12t2t12+1.
Chooseδ>0 in such a way −1/2t2+1/2t0<δ1/3 for t2−t0<δ. Pick N1>0 such that 1−t1/t12+1<2t0δ1/3 whenever t1>N1. Now assume that t2−t0<δ and s1<s0. Then, 1/2t2<1/2t0+δ1/3. Thus, s1/2t2t12+1 converges to 0 when t1 tends to +∞. Therefore, there exists N>N1 such that s1/2t2t12+1<δ1/3 for t1>N. The additional condition t1>N implies
(53)a−t1+s12t2t12+1<ε/3.
Hence,ϕat1,t2<ε if t2−t0<δ and t1>N.
Ifa does not converge to zero at −1/2t0, then take the function a~s=as−a2 and proceed as in the proof of Lemma 7, where a2=a−1/2t0.
Finally, the justification of the limit ofϕat1,t2 at −∞,t0 can be done analogously.Lemma 9.
Leta∈L∞ℝ be continuous at 0∈ℝ. For t0∈ℝ¯, the spectral function ϕa satisfies
(54)limt1,t2→t0,+∞ϕat1,t2=a0.
Actually, we have uniform convergence ofϕat1,t2 at t0,+∞, that is, for ε>0, there exists N>0 such that ϕat1,t2−a0<ε for all t2>N and for all t1∈ℝ¯.Proof.
Suppose thata0=0. Let ε>0, and choose s0>0 such that equations (50) hold. Then
(55)ϕat1,t2≤∫−∞−s0a−t1+s12t2t12+1hj−1s12ds1+∫−s0s0a−t1+s12t2t12+1hj−1s12ds1+∫s0∞a−t1+s12t2t12+1hj−1s12ds1<2ε3+max−s0<s1<s0a−t1+s12t2t12+1.
By the continuity ofas at 0, there exists δ1>0 such that as<ε/3 for s<δ1. For −s0<s1<s0, we have
(56)−t1+s12t2t12+1≤12t2t1t12+1+s1t12+1<12t21+s0.
TakeN=1+s02/4δ12. The inequality t2>N implies 1/2t2<δ1/1+s0. Thus, if t2>N, t1∈ℝ¯, and −s0<s1<s0, then
(57)−t1+s12t2t12+1<δ1.
Consequently,ϕat1,t2<ε for all t2>N and t1∈ℝ¯.
Finally, in the casea0≠0, the proof can be carry out by considering the symbol a~s=as−a0.For each nilpotent symbolaImζ1∈Cℝ¯, the spectral function ϕa is continuous on Π¯ and is constant along ℝ¯×+∞. For this reason, the C∗-algebra generated by all spectral functions ϕa is not CΠ¯, but it coincides with the algebra of continuous functions on a quotient space of Π¯.


## 5.2. Toeplitz OperatorsTa with Continuous Symbols aImζ1

Introduce the quotient spaceΔ≔Π¯/ℝ¯×+∞. By Lemma 9, the spectral function ϕa can be identified with a continuous function on Δ, which will be also denoted by ϕa:Δ⟶ℂ. We establish now one of our main results in this work.Theorem 10.
TheC∗-algebra generated by all spectral functions ϕa, with aImζ1∈Cℝ¯, is isomorphic and isometric to the algebra CΔ. That is, the C∗-algebra generated by all Toeplitz operators Ta is isomorphic to CΔ, where the isomorphism is defined on the generators by the rule
(58)Ta↦ϕa.Proof.
The functionsϕa separate the points of Δ according to Lemmas 11–13 below. The Stone-Weierstrass theorem completes the proof.Lemma 11.
Lett1,t2 and y1,y2 be distinct points of Π¯\Π, where they do not belong simultaneously to ℝ¯×+∞. Then, there exists a∈Cℝ¯ such that ϕat1,t2≠ϕay1,y2.Proof.
First consider the nilpotent symbola1s=s/s2+1, which is continuous on ℝ¯. Note that
(i)
ϕa1±∞,t2=a1∓1/(2t2)=∓1/1+4t2for0<t2≤+∞,(ii)
ϕa1±∞,0=a1∓∞=∓1,(iii)
ϕa1t1,+∞=a10=0 for t1∈ℝ¯,(iv)
ϕa1t1,0=−∫−∞t1hj−1s12ds1+∫t1∞hj−1s12ds1 for t1∈ℝ.
From∫−∞∞hj−1s12ds1=1, we get
(59)ϕa1t1,0=2∫t1∞hj−1s12ds1−1.
This formula also says thatϕa1±∞,0=∓1. On the other hand, the Hermite function hj−1 is continuous, and it has just a finitely many roots. Hence, ϕa1t1,0 is monotonically decreasing with respect to t1. Thus, points in ℝ¯×0 are separated by ϕa1.
Recall thatℝ¯×+∞ is identified with one point in Δ so take −∞,+∞ as representative point of the equivalence class ℝ¯×+∞. For t2,t~2∈0,∞, the three points −∞,t2, −∞,+∞, and +∞,t~2 are separated by ϕa1±∞,t because of the injective property of 1/1+4t.
Consider now the nilpotent symbola2s=1/s2+1, which is continuous on ℝ¯. We have
(i)
ϕa2±∞,t2=a2∓1/(2t2)=4t2/(1+4t2)>0 for t2>0,(ii)
ϕa2t1,+∞=a20=1 for t1∈ℝ¯,(iii)
ϕa2t1,0=0 for all t1∈ℝ¯.
Thus,ϕa2 separates each point t1,0 from the points ±∞,t2 and t1,+∞.Now our aim is to separate the points ofΠ. Consider the following family of continuous functions:
(60)aαs=0si−∞≤s≤−α,1αs+1si−α≤s≤0,1si0≤s≤+∞.Then,ϕaαt1,t2=ψαt1,t2+φt1, where
(61)ψαt1,t2=∫t1−2αt2t12+1t1s1−t12αt2t12+1+1hj−1s12ds1,(62)φt1=∫t1∞hj−1s12ds1.Lemma 12.
Ift1,t2 and y1,y2 are distinct points in Π, then there exists α>0 such that ϕaαt1,t2≠ϕaαy1,y2, where aα is defined in (60).Proof.
At first suppose thatt1,t2 and y1,y2 satisfy t1<y1. Introduce k≔φt1−φy1>0. It is easy to see that ψαt1,t2 can be written as
(63)ψαt1,t2=∫−10s+1hj−1t1+2αt2t12+1s22αt2t12+1ds.
This integral representation allows us to prove easily that(64)limα→0+ψαt1,t2=0.
Takeα small enough that ψαt1,t2−ψαy1,y2<k/2. Then
(65)ϕaαt1,t2−ϕaαy1,y2=k+ψαt1,t2−ψαy1,y2≥k−ψαt1,t2−ψαy1,y2≥k−k2>0.
Now suppose thatt1=y1=t and t2<y2. Let α=1/2t2+1>0. Then
(66)ψαt,t2=∫t−t2ts1−tt2+1hj−1s12ds1,ψαt,y2=∫t−y2ts1−ty2+1hj−1s12ds1.
Since0<t2<y2, we have t−y2<t−t2. Besides, the following inequality holds
(67)0<s1−tt2+1<s1−ty2+1,for t−t2<s1<t. Hence, ψαt,t2<ψαt,y2. Finally, from φt1=φy1=φt, we get ϕaαt,t2<ϕaαt,y2.Lemma 13.
Lett1,t2∈Π and y1,y2∈Π¯\Π. Then, there exists α>0 such that ϕaαt1,t2≠ϕaαy1,y2, where aα is given in (60).Proof.
Taket1,t2∈Π, y1,0∈ℝ×0, and α>0. It is easy to see that ψaαt1,t2>0. Then, we have ϕaαy1,0=φy1≤φt1<ϕaαt1,t2 for t1≤y1. Now suppose that y1<t1. Let k≔φy1−φt1>0. The inequality ψαt1,t2<k holds for α>0 small enough. For such α,
(68)ϕaαt1,t2=ψαt1,t2+φt1<φy1=ϕaαy1,0.
This proves that all points ofΠ can be separated from points of ℝ×0. On the other hand,
(69)0<ϕaαt1,t2=ψαt1,t2+φt1≤∫t1−2αt2t12+1t1hj−1s12ds1+φt1=∫t1−2αt2t12+1∞hj−1s12ds1<1.
Fory2∈ℝ¯+, we have ϕaα−∞,y2=aα1/2y2=1; hence, ϕaαt1,t2≠ϕaα−∞,y2.
Finally, consider the functionas≔aαs−α. Then, ϕa+∞,y2=a−1/2y2=0, and consequently, ϕat1,t2≠ϕa+∞,y2.


## 6. Toeplitz Operators with Piecewise Continuous SymbolsaImζ1

Take a finite subsetS=β0,⋯,βm⊂ℝ and let PCℝ¯,S be the set of functions continuous on ℝ¯\S and having one-side limit values at each point of S. In this section, we study the C∗-algebra generated by all Toeplitz operators Ta, where aImζ1∈PCℝ¯,S. Obviously, we have to study the algebra generated by the spectral functions ϕa. To begin with, take the indicator function χ+=χ0,+∞. Then, the spectral function ϕχ+ is continuous on Π¯\ℝ¯×+∞ according to Lemmas 5, 7, and 8. Actually,
(70)ϕχ+t1,t2=∫−∞∞χ0,+∞−t1+s12t2t12+1hk−1s12ds1=φt1,where φ is defined in (61). Hence, ϕχ+ is continuous on Π¯ because φ is. Of course, we have now a spectral function which is not constant along ℝ¯×+∞ anymore.For anya∈PCℝ¯,0, we have
(71)as=a~s+a0+−a0−χ+s,where a0− and a0+ are the one-side limits of a at 0, and a~s=as+a0−−a0+χ+s. This function has a removable discontinuity at 0; thus, ϕa is continuous on Π¯.Theorem 14.
TheC∗-algebra generated by all Toeplitz operators Ta, with aImζ1∈PCℝ¯,0, is isomorphic and isometric to CΠ¯. The isomorphism is defined on the generators by the rule
(72)Ta↦ϕa.Proof.
Ift1≠y1, the function ϕχ+ separates any two points t1,t2 and y1,y2∈Π¯. By Lemma 12, two points in Π can be separated by ϕaα, where aα is the nilpotent symbol given in (60). Further,
(73)ϕaαt1,0=φt1<ϕaαt1,t2<1=aα0=ϕaαt1,+∞.We continue our study by introducing another point of discontinuity. Take the indicator functionχβ≔χβ/2,+∞s, where β>0. We have
(74)ϕχβt1,t2=∫−∞∞χβ/2,+∞−t1+s12t2t12+1hk−1s12ds1=∫t1+βt2t12+1∞hk−1s12ds1.According to Lemmas5 and 7–9, the spectral function ϕχβ is continuous on Π¯, except at the point −∞,1/β2. For λ0∈ℝ, ϕχβ takes the constant value φλ0 along the curve t1+βt2t12+1=λ0. From this equation, we get
(75)t2=ft1=t1−λ02β2t12+1,t1≤λ0.The horizontal linet2=1/β2 is an asymptote of the graph of t2=ft1, and of course, ϕχβt1,ft1=φλ0 for t1≤λ0. Thus, the level curves of ϕχβ converge to the point −∞,1/β2; our aim is to separate them at −∞,1/β2 through a mapping Yβ in such a way that ϕχβ∘Yβ−1 is continuous on Π¯.Lemma 15.
Letg:ℝ¯⟶−1/2,1/2 be any bijective, smooth, and increasing function, with g0=0. Take β>0, and let Yβ be the function on Π defined by the rule
(76)Yβt1,t2=t1,t21+gλt1,t2,where λt1,t2=t1+βt2t12+1. Then, Yβ is an homeomorphism from Π onto itself, which can be continuously extended to Π¯\−∞,1/β2 with range Π¯\−∞×Jβ, where Jβ=y:.5/β2≤y≤1.5/β2.Proof.
The functionht1,t2≔1+gλt1,t2 has range contained in 1/2,3/2. Hence, YβΠ⊂Π. Now suppose that Yβt1,t2=Yβa,b. Then, t1=a and t2ha,t2=bha,b. The function ha,t2 is strictly increasing with respect to t2. Thus, t2>b implies that t2ha,t2>bha,b. Consequently, Yβ is injective. Let c,d be a point in Π. Consider the equation Yβt1,t2=c,d, which is equivalent to the system of equations t1=c, t2ht1,t2=d. Thus, we have to prove that t2hc,t2=d is solvable. The function t2hc,t2 is bijective from 0,+∞ into itself with respect to t2. Therefore, t2hc,t2=d has a unique solution. That is, Yβ:Π⟶Π is surjective. On the other hand, Yβ is smooth and so is Yβ−1 because of the Inverse Function Theorem.
The correspondence (76) also defines Yβ on ℝ×0 by Yβt1,0=t1,0. Actually, Yβ can be defined on Π¯\Π∪−∞,1/β2 according to the following limits:
(77)limt1,t2→t10,0Yβt1,t2=t10,0,t10∈ℝ¯,limt1,t2→+∞,t20Yβt1,t2=+∞,32t20,t20∈ℝ¯,limt1,t2→t10,+∞Yβt1,t2=t10,+∞,t10∈ℝ¯,limt1,t2→−∞,t20Yβt1,t2=−∞,12t20,0<t20<1/β2,limt1,t2→−∞,t20Yβt1,t2=−∞,32t20,t20>1/β2.
We will justify just the last limit. Fort1<0,
(78)λt1,t2=t1−1+βt21+1t12.
Ift20>1/β2 and t2 is close enough to t20, then λt1,t2 tends to +∞ when t1 tends to −∞. Thus, limt1,t2→−∞,t20gt1,t2=1/2. Consequently,
(79)limt1,t2→−∞,t20t2ht1,t2=32t20,that is, Yβt1,t2 converges to −∞,3t20/2 when t1,t2 tends to −∞,t20. A local analysis proves that Yβ−1 is continuous. This completes the proof.Lemma 16.
The functionϕχβ∘Yβ−1 is continuous on Π¯, and it separates the points in the line segment −∞×Jβ. For each continuous function ϕ:Π¯⟶ℂ, ϕ∘Yβ−1 is also continuous on Π¯ and has constant value along −∞×Jβ.With a discontinuityβ1=β, we define Φ1=Yβ. Introduce another point of discontinuity β2, with β1<β2. Let P2=−∞,0.5/β22. The function ϕχβ2∘Φ1−1 has a continuous extension to Π¯\P2, and its level curves Φ1t1,t1−λ2/(β22t12+1), t1>λ, converge to P2. As in Lemma 15, we can construct a mapping Φ2 that separates all these level curves, and ϕχβ2∘Φ1−1∘Φ2−1 is continuous on Π¯. Adding more discontinuity points β3,⋯,βm, with βj<βj+1, we can construct mappings Φ3,⋯,Φm in such a way that ϕχβm∘Φ1−1∘⋯∘Φm−1 is continuous on Π¯.LetTSm denote the C∗-algebra generated by all Toeplitz operators Ta with nilpotent symbols aImζ1∈PCℝ¯,Sm, where Sm=β0,⋯,βm. For simplicity in our explanation, we assume that β0=0 and Sm−1⊂Sm⊂ℝ+∪0. We will explain how the Toeplitz algebra TSm increases as Sm does. By Theorem 14, TS0 is isomorphic to CΠ¯, where the isomorphism is given on the generators by the rule
(80)Ψ0:TS0∋Ta↦ϕa∈CΠ¯.Consider now the algebraTS1, where S1=0,β1. By Lemma 16, ϕa∘Φ1−1 is continuous on Π¯ for every a∈PCℝ¯,S1. Then, the algebra TS1 is also isomorphic and isometric to CΠ¯, where the isomorphism is given by
(81)Ψ1:TS1∋Ta↦ϕa∘Φ1−1∈CΠ¯.At first sight, both algebrasTS0 and TS1 seem to have the same spectrum Π¯, but they do not; they are identified with CΠ¯ through different isomorphisms. Of course, TS0 is a subalgebra of TS1. If Ta∈TS0, then ϕa∘Φ1−1∈CΠ¯ and has constant value along the line segment I1=−∞×Jβ1. According to the isomorphism Ψ1, we can say that the spectrum of TS1 equals Π¯; meanwhile, the spectrum of TS1 is the quotient space Π¯/I1. This phenomenon persists as long as the set Smgrows.Theorem 17.
LetSm=0,β1,⋯,βm⊂ℝ+. Then, there exist bijective continuous functions Φj:Π⟶Π, j=1,⋯,m, such that ϕa∘Φ1−1∘⋯∘Φm−1 admits a continuous extension to Π¯ for each aImζ1∈PCℝ¯,Sm. The C∗-algebra generated by all Toeplitz operators Ta is isomorphic and isometric to CΠ¯. The isomorphism is defined on the generators by the rule
(82)Ψm:Ta↦ϕa∘Φ1−1∘⋯∘Φm−1.Note that for each piecewise continuous symbolaImζ1∈PCℝ¯,Sm, in general, the spectral function γa:Π⟶ℂ does not admit a continuous extension to Π¯, but γa∘Φ−1∘Φ1−1∘⋯∘Φm−1 does, which means that γa is uniformly continuous with respect to a new metric on Π; this metric is the pushforward of the usual metric using the mapping Φ−1∘Φ1−1∘⋯∘Φm−1.


## 7. Toeplitz Operators with SymbolsaImζ1bImζ2−ζ12

In this section, we describe theC∗-algebra generated by all Toeplitz operators with symbols of the form cζ1,ζ2=aImζ1bImζ2−ζ12, where as∈Cℝ¯, and bt∈L∞ℝ+ has limit values at t=0,+∞. For such a symbol c, we have that γc=γaγb, which means that Tc=TaTb=TbTa. Although γb belongs to CΠ¯, the spectral function
(83)γb∘Φ−1t1,t2=∫ℝ+by22t12+1t2ℓk−1y22dy2,is continuous on Π¯\P−,P+, where P−=−∞,0 and P+=+∞,0. Since the level curves of γbx1,x2 are the horizontal lines x2=μ, the level curves of γb∘Φ−1 are given by the equations t2=μ/t12+1, with μ∈ℝ+.Lemma 18.
Letf:0,+∞→0,1 be any bijective, smooth, and increasing function. Then, the function
(84)Θt1,t2=t1,t2+t12t12+1ft2t12+1,is an homeomorphism from Π onto itself, which can be continuously extended to Π¯\P−,P+ with range Π¯\I∞, where I∞=τ1,τ2:τ1=±∞and0≤τ2≤1. We have Θ±∞,t2=±∞,t2+1 for 0<t2<+∞, and Θ acts like the identity mapping at the rest of points in Π¯\Π∪P−,P+.Proof.
Similar to the proof of Lemma15.The image of the level curvet2=μ/1+t12 under Θ is the curve
(85)τ2=μτ12+1+fμτ12τ12+1.This means that the level curves ofγb∘Φ−1∘Θ−1 do not converge to a single point anymore.Lemma 19.
The functionγb∘Φ−1∘Θ−1 is continuous on Π¯, and for each continuous function ϕ:Π¯⟶ℂ, ϕ∘Θ−1 is also continuous on Π¯ and has constant value along each component of I∞.Theorem 20.
TheC∗-algebra generated by all Toeplitz operators Tab, with aImζ1∈PCℝ¯,0 and bt∈L∞ℝ+ having limits values at t=0,+∞, is isomorphic and isometric to CΠ¯. The isomorphism is defined on the generators by the rule
(86)Tab↦γab∘Φ−1∘Θ−1.For the Toeplitz operatorTχβ with symbol χβ=χβ/2,+∞Imζ1, there exists a mapping Θβ:Π⟶Π such that γχβ∘Φ−1∘Θ−1∘Θβ−1 admits a continuous extension to Π¯. The construction of Θβ is similar to the construction of Φβ given in Lemma 15, where one has to take into account the level curves of the spectral function γχβ∘Φ−1∘Θ−1, which converge to the point −∞,1+1/β2.Implicitly, we have considered several compactifications ofΠ associated to the C∗-algebras studied herein; each compactification depends on the kind of symbols. Take Q−=−∞,+∞ and Q+=+∞,+∞, let us explain the situation in the case of the algebra generated by the Toeplitz operators with symbols aImζ1∈Cℝ¯. Essentially, the corresponding compactification of Π is obtained from Π¯\Q−,Q+ by gluing a line segment at each corner Q− and Q+. Each spectral function γax1,x2 is continuous on Π¯\Q−,Q+ and has limit values when x1,x2 moves along the parabolas x2=μ+μx12 and tends to Q±. For a net Xλ tending to Q+, γaXλ converges if Xλ is eventually in gaps between two parabolas close enough from each other with respect to the parameter μ.



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