The Cuntz Comparison in the Standard 𝐂^∗-Algebra

Abstract

The Cuntz comparison, introduced by Cuntz in early 1978, associates each -algebra with an abelian semigroup which is an invariant for the classification of the nuclear -algebras and called the Cuntz semigroup. In this paper, we study the Cuntz comparison in the standard -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.

1. Introduction and the Statement of Results

Throughout this paper, let and be complex Hilbert spaces, let be the algebra of all bounded linear operators from into , and abbreviate to . For an operator , by ,  , and we denote the adjoint, the null space, and the range of , respectively. An operator is said to have finite rank if is finite, dimensional, and, in this case, we write . Denote by the ideal of all finite rank operators in . A standard -algebra acting on the Hilbert space is a -subalgebra of which contains the identity and the ideal . In this paper, we always assume that and are standard -algebras acting on and , respectively. Furthermore, we denote by the positive cone of all positive elements in .

In [1], Cuntz introduced a notion of the comparison for positive elements which extends the usual Murray-von Neumann comparison for projections in the -algebra. This comparison that we will denote by is nowadays called the Cuntz comparison.

Definition 1 (see [1]). Let . One writes , if there exists a sequence of elements in such that In this case, we say that is Cuntz subequivalent to . Furthermore, we say that is Cuntz equivalent to and write , if and .
The Cuntz comparison plays an important role in Elliott’s program for the classification of the nuclear separable simple -algebras. Indeed, the Cuntz comparison associates each -algebra with an abelian semigroup which is an invariant for the classification of the nuclear -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 -algebra.

In Section 2, we characterize the Cuntz comparison in terms of the dimension of the operator range. To classify -algebras via their Cuntz semigroups, one will prove the uniqueness and existence theorem for homomorphisms between -algebras. The uniqueness theorem says that if a semigroup map between the Cuntz semigroups of -algebras and is induced by two homomorphisms and between -algebras and , then for some unitary . 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 between and which preserves the Cuntz comparison. Recall that a map is said to be semilinear if is additive, and for all positive numbers and . Moreover, it is said to preserve the Cuntz comparison, if whenever . In Section 3, we present some results for the semilinear map which preserves the Cuntz comparison.

2. The Cuntz Comparison in the Standard -Algebra

In this section, we characterize the Cuntz comparison in terms of the dimension of the operator range.

Lemma 2. Let with and having finite rank. Then

Proof. Let . Then one has such that where is the rank- operator satisfying for all .
Since , there exists a sequence of elements in such that with respect to the norm topology on . It follows that with respect to the norm topology on . So the sequence is bounded, and thus, for each , the sequence is bounded. Now, one has a subsequence and a sequence of elements in such that, for each , with respect to the weak topology on . Consequently, with respect to the weak operator topology on . By formulas (5) and (7), Thus has finite rank, and .

If and there exists an element such that , then by Definition 1. The converse statement is not true in the general case (see [2]). But, we have the following.

Theorem 3. Let with at least one of them having finite rank. Then the following statements are equivalent: (a), (b), (c) for some finite rank operator .

Proof. It is clear that (c)(a).
(a)(b). Suppose that . By Lemma 2, the desired inequality clearly holds if has finite rank. Now, we suppose that is infinite. Then must have finite rank by the given assumption. Thus, the inequality holds too.
(b)(c). Suppose that . Let and so have finite rank . As ,   and thus we see that Pick a -dimensional subspace of and denote by the orthogonal projection of onto . Then has rank . Consequently, there is an invertible element such that It follows that And thus the rank- operator does the job.

Theorem 3 shows the relation between the Cuntz comparison and the dimension of the operator range. Moreover, we can characterize the rank- positive operator in terms of the Cuntz equivalence as follows.

Corollary 4. Let . The following statements are equivalent:(a),(b) for all with ,(c) for some with .

Proof. It is clear that (b)(c).
(a)(b). If and , then and by Theorem 3, and so by Definition 1.
(c)(a). Suppose that and . Then and by Definition 1. From Theorem 3, it follows that and . Thus, .

3. Preserver of the Cuntz Comparison

In this section, we focus our attention on the semilinear map which preserves the Cuntz comparison. The map is said to be semilinear if is additive, and for all positive numbers and . Moreover, it is said to preserve the Cuntz comparison, if whenever . And it is said to preserve the Cuntz comparison in both directions when if and only if . In a similar way, is said to preserve the Cuntz equivalence, if whenever . And it is said to preserve the Cuntz equivalence in both directions when if and only if .

For , we denote by the spectrum of as an element in the -algebra .

Lemma 5 (see [16]). Let and be unital -algebras with a common identity and norm such that . If , then .

Theorem 6. Let be a semilinear surjective transformation. The following statements are equivalent. (a) preserves the Cuntz comparison in both directions.(b) preserves the Cuntz equivalence in both directions.(c) for all .

Proof. (a)(b). Let . Then if and only if and , and, by (a), this is the case if and only if and , or, equivalently, if and only if .
(b)(c). By (b) and Corollary 4, one can conclude that, for any , This induces an injective map on , the set of all nonnegative integers, such that, for any nonnegative integer , Furthermore, since is surjective, so is . Thus is a bijective map. We claim that is indeed the identity map. Once the claim is proved, (c) of Theorem 6 clearly follows.
Suppose is not the identity map. Since is bijective, there exist and in such that Now take an operator , where denotes the set of all rank operators in . One can always find two operators and such that Since is additive and is a positive operator, we have Contradiction is reached.
(c)(a). Let . Then if and only if by Theorem 3, and this is the case if and only if by (c), which again is the case if and only if by Theorem 3.

Now, we give an explicit version of the surjective transformation which preserves the Cuntz comparison in both directions.

Proposition 7. Let be a surjective transformation. If is invertible and for all , then preserves the Cuntz comparison in both directions.

Proof. Let and . Then by Definition 1, one has a sequence of elements in such that . It follows that To prove , it remains to show that for all .
Since and are the linear spans of and , respectively, it is easy to see that for all . Moreover, since is invertible in and , we conclude by Lemma 5 that . Thus for all .
Conversely, let with . Then by Definition 1, one has a sequence of elements in such that . It follows that Now, to prove that , we show that for all .
Since is surjective, . Noting that is the linear span of , for all . Furthermore, since is invertible in and ,   by Lemma 5 again. So for all .

Proposition 7 shows that if has the explicit form as formula (17), then 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. Let and let be a map on such that Then it is easy to check that is a continuous semilinear surjective map and for all . From Theorem 6, it follows that preserves the Cuntz comparison in both directions.
Now, we show that does not have the explicit form as formula (17). Indeed, suppose on the contrary that there is a matrix satisfying for all . Then one has for all with and .
If we take and in (22), then we get and . If we take and in (22), we get and . Hence from formula (22), for all , which is impossible.

Conflict of Interests

The authors declare that there is no conflict of interests regarding the publication of this paper.

Acknowledgments

The authors thank Professor Chi-Kwong Li for drawing their attention to the subject of preserver of the Cuntz comparison and for many useful comments. The first author is supported by NSF of China (11371279). The third author is supported by NSF of China (11326107) and the Special Foundation for Excellent Young College and University Teachers (405ZK12YQ21-ZZyyy12021).