# The Cuntz Comparison in the StandardC*-Algebra

Authors: Xiaochun Fang; Nung-Sing Sze; Xiao-Ming Xu
Journal: Abstract and Applied Analysis
 (2014)

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## Abstract

The Cuntz comparison, introduced by Cuntz in early 1978, associates eachC*-algebra with an abelian semigroup which is an invariant for the classification of the nuclear C*-algebras and called the Cuntz semigroup. In this paper, we study the Cuntz comparison in the standard C*-algebra. We characterize the Cuntz comparison in terms of the dimension of the operator range. Also, we consider the structure of the semilinear map which preserves the Cuntz comparison.


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## Body

## 1. Introduction and the Statement of Results

Throughout this paper, letH and K be complex Hilbert spaces, let B(H,K) be the algebra of all bounded linear operators from H into K, and abbreviate B(H,H) to B(H). For an operator T∈B(H,K), by T*,  N(T), and R(T) we denote the adjoint, the null space, and the range of T, respectively. An operator T∈B(H) is said to have finite rank if R(T) is finite, dimensional, and, in this case, we write rank⁡(T)=dim⁡(R(T)). Denote by F(H) the ideal of all finite rank operators in B(H). A standard C*-algebra acting on the Hilbert space H is a C*-subalgebra of B(H) which contains the identity I and the ideal F(H). In this paper, we always assume that A and B are standard C*-algebras acting on H and K, respectively. Furthermore, we denote by A+ the positive cone of all positive elements in A.In [1], Cuntz introduced a notion of the comparison for positive elements which extends the usual Murray-von Neumann comparison for projections in the C*-algebra. This comparison that we will denote by ≾ is nowadays called the Cuntz comparison.Definition 1 (see [1]).
LetA,B∈A+. One writes A≾B, if there exists a sequence (Xn)n=1∞ of elements in A such that
(1)A=lim⁡n→∞XnBXn*.
In this case, we say that A is Cuntz subequivalent to B. Furthermore, we say that A is Cuntz equivalent to B and write A~B, if A≾B and B≾A.
The Cuntz comparison plays an important role in Elliott’s program for the classification of the nuclear separable simpleC*-algebras. Indeed, the Cuntz comparison associates each C*-algebra with an abelian semigroup which is an invariant for the classification of the nuclear C*-algebras and called the Cuntz semigroup. Recently, it has been studied intensively by many authors (see [2–7]). In the present paper, we study the Cuntz comparison in the standard C*-algebra.In Section2, we characterize the Cuntz comparison in terms of the dimension of the operator range. To classify C*-algebras via their Cuntz semigroups, one will prove the uniqueness and existence theorem for homomorphisms between C*-algebras. The uniqueness theorem says that if a semigroup map between the Cuntz semigroups of C*-algebras A and B is induced by two homomorphisms α and β between C*-algebras A and B, then β=Adu∘α for some unitary u∈B. Motivated by the investigation of this uniqueness theorem and the extensive study of the preserver problems in matrix spaces or general operator algebras (see [8–15]), we discuss the structure of the semilinear map ϕ between A+ and B+ which preserves the Cuntz comparison. Recall that a map ϕ:A+→B+ is said to be semilinear if ϕ is additive, and ϕ(λA)=λϕ(A) for all positive numbers λ and A∈A+. Moreover, it is said to preserve the Cuntz comparison, if ϕ(A)≾ϕ(B) whenever A≾B. In Section 3, we present some results for the semilinear map which preserves the Cuntz comparison.


## 2. The Cuntz Comparison in the StandardC*-Algebra

In this section, we characterize the Cuntz comparison in terms of the dimension of the operator range.Lemma 2.
LetA,B∈A+ with A≾B and B having finite rank. Then
(2)rank⁡(A)≤rank⁡(B).Proof.
Letrank⁡(B)=k. Then one has e1,…,ek∈H such that
(3)B=∑i=1k‍ei⊗ei,
where ei⊗ei is the rank-1 operator satisfying (ei⊗ei)(h)=〈h,ei〉ei for all h∈H.
SinceA≾B, there exists a sequence {Xn}n=1∞ of elements in A such that
(4)A=lim⁡n→∞XnBXn*
with respect to the norm topology on B(H). It follows that
(5)A=lim⁡n→∞Xn(∑i=1k‍ei⊗ei)Xn*=lim⁡n→∞∑i=1k‍Xn(ei)⊗Xn(ei)
with respect to the norm topology on B(H). So the sequence {∑i=1k‍Xn(ei)⊗Xn(ei)}n=1∞ is bounded, and thus, for each i, the sequence {Xn(ei)}n=1∞ is bounded. Now, one has a subsequence {nj}j=1∞ and a sequence {e~1,e~2,…,e~k} of elements in H such that, for each i,
(6)lim⁡j→∞Xnj(ei)=e~i
with respect to the weak topology on H. Consequently,
(7)lim⁡j→∞∑i=1k‍Xnj(ei)⊗Xnj(ei)=∑i=1k‍e~i⊗e~i
with respect to the weak operator topology on B(H). By formulas (5) and (7),
(8)A=∑i=1k‍e~i⊗e~i.
Thus A has finite rank, and rank⁡(A)≤k=rank⁡(B).IfA,B∈A+ and there exists an element X∈A such that A=XBX*, then A≾B by Definition 1. The converse statement is not true in the general case (see [2]). But, we have the following.Theorem 3.
LetA,B∈A+ with at least one of them having finite rank. Then the following statements are equivalent: (a)
A≾B,(b)
dim⁡(R(A))≤dim⁡(R(B)),(c)
A=XBX* for some finite rank operator X∈F(H).Proof.
It is clear that (c)⇒(a).
(a)⇒(b). Suppose that A≾B. By Lemma 2, the desired inequality clearly holds if B has finite rank. Now, we suppose that dim⁡(R(B)) is infinite. Then A must have finite rank by the given assumption. Thus, the inequality holds too.
(b)⇒(c). Suppose that dim⁡(R(A))≤dim⁡(R(B)). Let A and so A1/2 have finite rank k. As B≥0,  dim⁡(R(B))=dim⁡(R(B1/2)) and thus we see that
(9)dim⁡(R(B1/2))≥k.
Pick a k-dimensional subspace H1 of R(B1/2) and denote by PH1 the orthogonal projection of H onto H1. Then PH1B1/2 has rank k. Consequently, there is an invertible element U∈A such that
(10)A1/2=UPH1B1/2.
It follows that
(11)A=UPH1BPH1U*.
And thus the rank-k operator X=UPH1 does the job.Theorem3 shows the relation between the Cuntz comparison and the dimension of the operator range. Moreover, we can characterize the rank-k positive operator in terms of the Cuntz equivalence as follows.Corollary 4.
LetA∈A+. The following statements are equivalent:(a)
rank⁡(A)=k,(b)
A~X for all X∈A+ with rank⁡(X)=k,(c)
A~X for some X∈A+ with rank⁡(X)=k.Proof.
It is clear that (b)⇒(c).
(a)⇒(b). If X∈A+ and rank⁡(X)=k, then A≾X and X≾A by Theorem 3, and so A~X by Definition 1.
(c)⇒(a). Suppose that A~X and rank⁡(X)=k. Then A≾X and X≾A by Definition 1. From Theorem 3, it follows that rank⁡(A)≤rank⁡(X) and rank⁡(X)≤rank⁡(A). Thus, rank⁡(A)=rank⁡(X)=k.


## 3. Preserver of the Cuntz Comparison

In this section, we focus our attention on the semilinear mapϕ:A+→B+ which preserves the Cuntz comparison. The map ϕ:A+→B+ is said to be semilinear if ϕ is additive, and ϕ(λA)=λϕ(A) for all positive numbers λ and A∈A+. Moreover, it is said to preserve the Cuntz comparison, if ϕ(A)≾ϕ(B) whenever A≾B. And it is said to preserve the Cuntz comparison in both directions when ϕ(A)≾ϕ(B) if and only if A≾B. In a similar way, ϕ is said to preserve the Cuntz equivalence, if ϕ(A)~ϕ(B) whenever A~B. And it is said to preserve the Cuntz equivalence in both directions when ϕ(A)~ϕ(B) if and only if A~B.ForA∈A, we denote by σA(A) the spectrum of A as an element in the C*-algebra A.Lemma 5 (see [16]).
LetA and B be unital C*-algebras with a common identity and norm such that A⊆B. If A∈A, then σA(A)=σB(A).Theorem 6.
Letϕ:F(H)+→F(K)+ be a semilinear surjective transformation. The following statements are equivalent. (a)
ϕ preserves the Cuntz comparison in both directions.(b)
ϕ preserves the Cuntz equivalence in both directions.(c)
rank⁡(ϕ(A))=rank⁡(A) for all A∈F(H)+.Proof.
(a)⇒(b). Let A,B∈F(H)+. Then A~B if and only if A≾B and B≾A, and, by (a), this is the case if and only if ϕ(A)≾ϕ(B) and ϕ(B)≾ϕ(A), or, equivalently, if and only if ϕ(A)~ϕ(B).
(b)⇒(c). By (b) and Corollary 4, one can conclude that, for any A,B∈F(H)+,
(12)rank⁡(A)=rank⁡(B)⟺rank⁡(ϕ(A))=rank⁡(ϕ(B)).
This induces an injective map f on N0, the set of all nonnegative integers, such that, for any nonnegative integer k∈N0,
(13)rank⁡(A)=k⟺rank⁡(ϕ(A))=f(k).
Furthermore, since ϕ is surjective, so is f. Thus f is a bijective map. We claim that f is indeed the identity map. Once the claim is proved, (c) of Theorem 6 clearly follows.
Supposef is not the identity map. Since f is bijective, there exist k and l in N0 such that
(14)k<l,f(k)>f(l).
Now take an operator B∈Fl(H)+, where Fl(H)+ denotes the set of all rank l operators in F(H)+. One can always find two operators B1∈Fk(H)+ and B2∈Fl-k(H)+ such that
(15)B=B1+B2.
Since ϕ is additive and ϕ(B2) is a positive operator, we have
(16)f(l)=rank⁡(ϕ(B))=rank⁡(ϕ(B1)+ϕ(B2))≥rank⁡(ϕ(B1))=f(k)>f(l).
Contradiction is reached.
(c)⇒(a). Let A,B∈F(H)+. Then A≾B if and only if rank⁡(A)≤rank⁡(B) by Theorem 3, and this is the case if and only if rank⁡(ϕ(A))≤rank⁡(ϕ(B)) by (c), which again is the case if and only if ϕ(A)≾ϕ(B) by Theorem 3.Now, we give an explicit version of the surjective transformation which preserves the Cuntz comparison in both directions.Proposition 7.
Letϕ:A+→B+ be a surjective transformation. If M∈B(H,K) is invertible and
(17)ϕ(A)=MAM*
for all A∈A+, then ϕ preserves the Cuntz comparison in both directions.Proof.
LetA,B∈A+ and A≾B. Then by Definition 1, one has a sequence (Xn)n=1∞ of elements in A such that A=lim⁡n→∞XnBXn*. It follows that
(18)ϕ(A)=MAM*=lim⁡n→∞MXnM-1MBM*M*-1Xn*M*=lim⁡n→∞(MXnM-1)ϕ(B)(MXnM-1)*.
To prove ϕ(A)≾ϕ(B), it remains to show that MXnM-1∈B for all n.
SinceA and B are the linear spans of A+ and B+, respectively, it is easy to see that MXnM*∈B for all n. Moreover, since MM* is invertible in B(K) and MM*=ϕ(IH)∈B+, we conclude by Lemma 5 that (MM*)-1∈B+. Thus MXnM-1=MXnM*(MM*)-1∈B for all n.
Conversely, letA,B∈A+ with ϕ(A)≾ϕ(B). Then by Definition 1, one has a sequence (Yn)n=1∞ of elements in B such that ϕ(A)=lim⁡n→∞Ynϕ(B)Yn*. It follows that
(19)A=M-1ϕ(A)M*-1=lim⁡n→∞M-1Ynϕ(B)Yn*M*-1=lim⁡n→∞(M-1YnM)B(M-1YnM)*.
Now, to prove that A≾B, we show that M-1YnM∈A for all n.
Sinceϕ is surjective, M-1B+M*-1=M-1ϕ(A+)M*-1=A+. Noting that B is the linear span of B+, M-1YnM*-1∈A for all n. Furthermore, since M-1M*-1 is invertible in B(H) and M-1M*-1=M-1IKM*-1∈A+,  M*M=(M-1M*-1)-1∈A+ by Lemma 5 again. So M-1YnM=(M-1YnM*-1)(M*M)∈A for all n.Proposition7 shows that if ϕ has the explicit form as formula (17), then ϕ preserves the Cuntz comparison in both directions. Unfortunately, the converse statement fails to hold. A counterexample, indebted to Professor Fangyan Lu, is presented as follows.Example 8.
LetA=M2(C) and let ϕ be a map on A+ such that
(20)ϕ((abb-d))=(dbb-a).
Then it is easy to check that ϕ is a continuous semilinear surjective map and rank⁡(ϕ(A))=rank⁡(A) for all A∈A+. From Theorem 6, it follows that ϕ preserves the Cuntz comparison in both directions.
Now, we show thatϕ does not have the explicit form as formula (17). Indeed, suppose on the contrary that there is a matrix M=(xuvy) satisfying
(21)ϕ(A)=MAM*
for all A∈A+. Then one has
(22)(xuvy)(abb-d)(x-v-u-y-)=(dbb-a)
for all a,b,d∈C with a,d≥0 and ad≥|b|2.
If we takeb=d=0 and a=1 in (22), then we get x=0 and |v|=1. If we take a=b=0 and d=1 in (22), we get y=0 and |u|=1. Hence from formula (22), uv-b-=b for all b∈C, which is impossible.



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