Simple, Z-stable, projectionless C*-algebras

Von | September 22, 2020

Do simple, \mathcal{Z}-stable, stably projectionless C*-algebras have stable rank one?

Definitions: A C*-algebra A is said to be \mathcal{Z}-stable if it tensorially absorbs the Jiang-Su algebra \mathcal{Z}, that is, A\cong\mathcal{Z}\otimes A. Further, a simple C*-algebra is projectionless if it contains no nonzero projections, and it is stably projectionless if A\otimes\mathcal{K} is projectionless. (In the nonsimple case, one should require that no quotient of A contains a nonzero projection – see ​[1]​.) A unital C*-algebra A is said to have stable rank one if the invertible elements in A are dense. A nonunital C*-algebra A has stable rank one if its minimal unitization \widetilde{A} does.

Background/Motivation: Rørdam showed in Theorem 6.7 in ​[2]​ that every unital, simple, finite \mathcal{Z}-stable C*-algebra has stable rank one. Using this, one can show that every simple, finite, \mathcal{Z}-stable C*-algebra A that is not stably projectionless has stable rank one. Indeed, one may first reduce to the separable case (to show that a given element x_0+\lambda 1\in\widetilde{A} with x_0\in A is approximated by invertible elements, consider any separable, simple, \mathcal{Z}-stable sub-C*-algebra of A that is not stably projectionless and that contains x.) Let p\in A\otimes\mathcal{K} be a nonzero projection. By Brown’s stabilization theorem, we have A\otimes\mathcal{K}\cong (p(A\otimes\mathcal{K})p)\otimes\mathcal{K}. Then the hereditary sub-C*-algebra p(A\otimes\mathcal{K})p is separable, unital, simple, finite and \mathcal{Z}-stable (by Corollary 3.2 in ​[3]​) and therefore has stable rank one. Since stable rank one is invariant under stable isomorphism, it follows that A has stable rank one.

Thus, a simple, \mathcal{Z}-stable C*-algebra that is not stably projectionless has stable rank one if and only if it is (stably) finite. A simple, stably projectionless C*-algebra is automatically stably finite, and it is therefore natural to expect that every simple, \mathcal{Z}-stable, stably projectionless C*-algebra has stable rank one.

Let A be a simple, \mathcal{Z}-stable, stably projectionless C*-algebra. By Corollary 3.2 in ​[1]​, A almost has stable rank one, that is, every hereditary sub-C*-algebra B\subseteq A satisfies B\subseteq\overline{\mathrm{Gl}(\widetilde{B})}. In particular, every element in A can be approximated by invertible elements in \widetilde{A}. To show that A has stable rank one, one would need to show that every element in \widetilde{A} is approximated by invertibles. In Theorem 6.13 in ​[4]​ it is shown that A has stable rank at most two, that is, the tuples (x,y) in (\widetilde{A})^2 such that x^*x+y^*y is invertible are dense in (\widetilde{A})^2. One can also show:

Simple, \mathcal{Z}-stable, stably projectionless C*-algebras have general stable rank one.

Here, the general stable rank of a unital C*-algebra A, denoted \mathrm{gsr}(A), is the least integer n\geq 1 such that \mathrm{Gl}_m(A) acts transitively on \mathrm{Lg}_m(A) for all m\geq n. (We use \mathrm{Gl}_m(A) to denote the set of invertible elements in the matrix algebra M_m(A). Further, \mathrm{Lg}_m(A) denotes the set of m-tuples (a_1,\ldots,a_m)\in A^m that generate A as a left ideal, that is, such that a_1^*a_1+\ldots+a_m^*a_m is invertible.) For a nonunital C*-algebra \mathrm{gsr}(A):=\mathrm{gsr}(\widetilde{A}). In general, one has \mathrm{gsr}(A)\leq\mathrm{sr}(A)+1, as noted in Theorem 3.3. of the overview article ​[5]​. If A is unital, then the action of \mathrm{Gl}_m^0(A), the connected component of the unit in \mathrm{Gl}_m(A), on \mathrm{Lg}_m(A) has open orbits. It follows that the orbits of the action of \mathrm{Gl}_m(A) on \mathrm{Lg}_m(A) are also open (and hence also closed).

Let A be a simple, \mathcal{Z}-stable, stably projectionless C*-algebra. Let us prove that \mathrm{gsr}(A)=1. By Theorem 6.13 in ​​[4]​, we have \mathrm{sr}(A)\leq 2, and so \mathrm{gsr}(A)\leq 3. To verify that \mathrm{gsr}(A)\leq 2, we need to show that \mathrm{Gl}_2(\widetilde{A}) acts transitively on \mathrm{Lg}_2(\widetilde{A}). So let x=(\lambda_1+a_1,\lambda_2+a_2)\in\mathrm{Lg}_2(\widetilde{A}) with \lambda_1,\lambda_2\in\mathbb{C} and a_1,a_2\in A. First, we find u\in\mathrm{Gl}_2(\mathbb{C})\subseteq\mathrm{Gl}_2(\widetilde{A}) such that ux=(b_1,\kappa_2+b_2) with \kappa_2\in\mathbb{C} and b_1,b_2\in A. (If \lambda_1=0, we use the identity matrix; if \lambda_2=0, we use the flip matrix; and if \lambda_1\neq 0 and \lambda_2\neq 0 we use the upper-triangular matrix with diagonal entries 1 and upper right entry -\lambda_1/\lambda_2.) Given \varepsilon>0, we use that A\subseteq\overline{\mathrm{Gl}(\widetilde{A})} to approximate b_1 by some invertible \kappa_1+c\in\widetilde{A} such that \| x - u^{-1}(\kappa_1+c,\kappa_2+b_2) \|<\varepsilon. We note that (\kappa_1+c,\kappa_2+b_2) is in the orbit of (1,0), that is, there exists v\in\mathrm{Gl}_2(\widetilde{A}) such that (\kappa_1+c,\kappa_2+b_2)=v(1,0). Then \| x - u^{-1}v(1,0) \|<\varepsilon. Since the orbits of the action of \mathrm{Gl}_2(\widetilde{A}) on \mathrm{Lg}_2(\widetilde{A}) are closed, it follows that x=w(1,0) for some w\in\mathrm{Gl}_2(\widetilde{A}). Finally, since \widetilde{A} is finite, it follows that \mathrm{gsr}(A)=1.

We remark that \mathrm{sr}(A)=1 implies \mathrm{gsr}(A)=1, but not conversely in general.

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    L. Robert, Remarks on Z-stable projectionless C*-algebras, Glasgow Math. J. (2015) 273–277. https://doi.org/10.1017/s0017089515000117.
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    M. Rørdam, The stable and the real rank of  {\mathcal Z}-absorbing C*-algebras, Int. J. Math. (2004) 1065–1084. https://doi.org/10.1142/s0129167x04002661.
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    A.S. Toms, W. Winter, Strongly self-absorbing C^{*}-algebras, Trans. Amer. Math. Soc. (2007) 3999–4029. https://doi.org/10.1090/s0002-9947-07-04173-6.
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    L. Robert, L. Santiago, A revised augmented Cuntz semigroup, ArXiv:1904.03690. (2019). https://arxiv.org/abs/1904.03690v1.
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    B. Nica, Homotopical stable ranks for Banach algebras, Journal of Functional Analysis. (2011) 803–830. https://doi.org/10.1016/j.jfa.2011.03.001.

Nonregular, simple, nuclear C*-algebras

Von | September 7, 2020

(Based on the talk „Thoughts on the classification problem for amenable C*-algebras“ of George Elliott, 30. June 2020, at the Zagreb Workshop on Operator Theory.)

Let us consider the class of unital, separable, simple, nuclear C*-algebras. The Toms-Winter conjecture predicts that for such an algebra A, the following conditions are equivalent:

  1. A has finite nuclear dimension.
  2. A is \mathcal{Z}-stable, that is, A\cong\mathcal{Z}\otimes A where \mathcal{Z} denote the Jiang-Su algebra.
  3. A has strict comparison of positive elements, which means that the Cuntz semigroup of A is almost unperforated: if x and y satisfy (n+1)x\leq ny for some n\in\mathbb{N}, then x\leq y.

As of today, it is known that (1) and (2) are equivalent and that (2) implies (3). Further, it is known that (3) implies (2) under certain additional assumptions.

As long as the Toms-Winter conjecture is not completely verified, the second condition seems most natural, and we say that A is regular if it is \mathcal{Z}-stable. By the spectacular recent breakthrough in the Elliott classification program, it is known that every regular C*-algebras that satisfy the Universal Coefficient Theorem (UCT) are classified by the Elliott invariant (K-theory and tracial data).

We have an obvious dichotomy: a unital, separable, simple, nuclear C*-algebra is either regular or nonregular. It is easy to find regular algebra. Indeed, since the Jiang-Su algebra is self-absorbing, that is, \mathcal{Z}\otimes\mathcal{Z}\cong\mathcal{Z}, for every A the algebra \mathcal{Z}\otimes A is automatically regular. Examples of nonregular C*-algebras were constructed by Villadsen, Toms and Rørdam.

Question 1:

Are the regular or the nonregular C*-algebras generic?

A regular C*-algebra is either purely infinite or stably finite. A simple, unital, purely infinite C*-algebra has infinite stable rank. Rørdam showed that a stably finite, regular C*-algebra has stable rank one. Thus, a simple C*-algebra with finite stable rank \geq 2 is nonregular. Villadsen showed that for every n\in\{1,2,\ldots,\infty\} there exists a simple AH-algebra with stable rank n. He also showed that his algebra with stable rank n has either real rank n-1 or n.

Question 2:

Does there exist a nonregular C*-algebra of real rank zero?

Question 3:

What is the relation between the stable rank and real rank of a stably finite, simple (nuclear) C*-algebra? Is the real rank of A always either \mathrm{sr}(A) or \mathrm{sr}(A)-1? Is this connected to the number of tracial states on A?

Elliott suggests that \mathrm{rr}(A)=\mathrm{sr}(A)-1 should correspond to a unique (or very few) tracial state on A, while \mathrm{rr}(A)=\mathrm{sr}(A) should correspond to a large tracial simplex of A. It is known that \mathrm{rr}(B)\leq 2\mathrm{sr}(B)-1 for every C*-algebra B. Moreover, if X is a compact, Hausdorff space, then \mathrm{rr}(C(X))=\dim(X) and \mathrm{sr}(C(X))=\lfloor \frac{\dim(X)}{2} \rfloor +1. Thus, the stable rank of a commutative C*-algebra is roughly half of its real rank.

Question 4:

Can the recent classification of regular (separable, simple, nuclear) C*-algebras be extended to include (some) nonregular C*-algebras by using a stronger invariant?

A natural candidate for a stronger invariant would be the Cuntz semigroup. The Cuntz semigroup encodes the K_0-group, the simplex of tracial states, and the pairing between traces and K_0. However, it does not encode the K_1-group. One should therefore consider classification by the pair (\mathrm{Cu}(-),K_1(-)).

Elliott suggests to consider the following testcase: Let A and B be simple inductive limits of matrix algebras over the Hilbert cube. Then K_1(A)=K_1(B)=0, and the question becomes: Are A and B isomorphic whenever their Cuntz semigroups are isomorphic? If \mathrm{Cu}(A)\cong\mathrm{Cu}(B), and if A is regular, then the Cuntz semigroup of B is `regular‘ (in the sense of Winter) and it follows that both algebras are regular. Then, since A and B also satisfy the UCT, one can indeed deduce from the classification result that A\cong B.

Real rank of   B(H)\otimes B(H)

Von | Juni 20, 2020

Given a separable, infinite-dimensional Hilbert space \Huge{H}, what is the real rank of the minimal tensor product  \mathcal{B}(H)\otimes \mathcal{B}(H)?

Background/Motivation: The real rank is a noncommutative dimension theory that was introduced by Brown and Pedersen in ​[1]​. It associates to each C*-algebra A a number (its real rank) \mathrm{rr}(A)\in\{0,1,\ldots,\infty\}. The lowest and most interesting value is zero. One can think of C*-algebras of real rank zero as zero-dimensional, noncommutative spaces. (There are other noncommutative dimension theories, and they lead to different concepts of zero- or low-dimensional noncommutative spaces.) For topological spaces X andY, the product theorem for covering dimension shows that under certain weak assumptions (for example, both spaces are metric, or compact) we have \dim(X\times Y)\leq\dim(X)+\dim(Y). One might therefore expect that \mathrm{rr}(A\otimes B)\leq\mathrm{rr}(A)+\mathrm{rr}(B), but it was shown in ​[2]​ that this does not hold: there exists C*-algebras A and B such that

     $$\mathrm{rr}(A)=\mathrm{rr}(B)=0, \text{ yet } \mathrm{rr}(A\otimes B)>0.$$

Later, Osaka showed in ​[3]​ that one can even take A=B=\mathcal{B}(H). Indeed, it is well known that von Neumann algebras have real rank zero, and hence so does \mathcal{B}(H). Osaka showed that \mathcal{B}(H)\otimes\mathcal{B}(H) does not have real rank zero. This raises the question of what the real rank of \mathcal{B}(H)\otimes\mathcal{B}(H) is, see Question 3.3 in [3].

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    L.G. Brown, G.K. Pedersen, C*-algebras of real rank zero, Journal of Functional Analysis. 99 (1991) 131–149. https://doi.org/10.1016/0022-1236(91)90056-b.
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    K. Kodaka, H. Osaka, Real Rank of Tensor Products of C*-Algebras, Proceedings of the American Mathematical Society. 123 (1995) 2213–2215. https://doi.org/10.2307/2160959.
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    H. Osaka, Certain C*-algebras with non-zero real rank and extremal richness, MATH. SCAND. 85 (1999) 79. https://doi.org/10.7146/math.scand.a-13886.

Automorphisms of the Calkin algebra

Von | Juni 20, 2020

Does the Calkin algebra \mathcal{Q} admit an automorphism that induces the flip on  K_1(\mathcal{Q})?

Background/Motivation: Let H be a separable, infinite-dimensional Hilbert space. The Calkin algebra \mathcal{Q}:=\mathcal{B}(H)/\mathcal{K}(H) is the quotient of the bounded, linear operators on H by the closed, two-sided ideal of compact operators. The important problem of whether the Calkin algebra has outer automorphisms was eventually shown to be independent of the usual set theoretic axioms ZFC: Phillips and Weaver, ​[1]​, proved that the Continuum Hypothesis implies that \mathcal{Q} has many outer automorphism. This was complemented by a result of Farah, ​[2]​, who showed that the Open Coloring Axiom implies that all automorphisms of \mathcal{Q} are inner.

Given a C*-algebra A, every automorphism \alpha\colon A\to A induces a group automorphism K_1(A)\to K_1(A). It is known that K_1(\mathcal{Q})\cong\mathbb{Z}. Hence, an automorphism of \mathcal{Q} either induces the identity or the flip on K_1(\mathcal{Q}). Every inner automorphism of a C*-algebra acts trivially on K-theory. Therefore, the result of Farah shows that assuming the Open Coloring Axiom, every automorphism of \mathcal{Q} induces the identity on K_1(\mathcal{Q}). As it turns out, the outer automorphisms \alpha constructed by Phillips and Weaver are locally inner, that is, for every element a\in\mathcal{Q} there exists a unitary u\in\mathcal{Q} such that \alpha(a)=uau^*. (The unitary depends on the element.) Since the generator of K_1(\mathcal{Q}) is represented by a unitary in \mathcal{Q}, it follows that every locally inner automorphism of \mathcal{Q} induces the identity on K_1(\mathcal{Q}) as well. One may consider automorphisms of \mathcal{Q} that induce the flip on K_1(\mathcal{Q}) as ‚very outer‘, and the question is if such automorphisms exists (assuming some set-theoretic axioms).

  1. [1]
    N.C. Phillips, N. Weaver, The Calkin algebra has outer automorphisms, Duke Math. J. 139 (2007) 185–202. https://doi.org/10.1215/s0012-7094-07-13915-2.
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    I. Farah, All automorphisms of the Calkin algebra are inner, Ann. Math. 173 (2011) 619–661. https://doi.org/10.4007/annals.2011.173.2.1.