Inductive limits of semiprojective C*-algebras

Von | Dezember 20, 2020

Question: Is every separable C*-algebra an inductive limit of semiprojective C*-algebras?

This question was first raised by Blackadar in ​[1]​. If we think of C*-algebras as noncommutative topological spaces, then semiprojective C*-algebras are noncommutative absolute neighborhood retracts (ANRs). It is a classical result from shape theory that every metrizable space X is homeomorphic to an inverse limit of metrizable ANRs X_n, that is, X\cong\varprojlim_n X_n. This means that C(X) is isomorphic to an inductive limit of the C(X_n), that is, C(X)\cong\varinjlim_n C(X_n). The question is if the noncommutative analog of this result also holds.

It has been verified for many classes of C*-algebras that they are inductive limits of semiprojective C*-algebras. For example, Enders showed in ​[2]​ that every UCT-Kirchberg algebra is an inductive limit of semiprojective C*-algebras. In ​[3]​, it was shown that the class of C*-algebras that are inductive limits of semiprojective C*-algebras is closed under shape domination, and in particular under homotopy equivalence. One deduces, for example, that if X is a contractible, compact, metrizable space, and if A is an inducitve limit of semiprojective C*-algebras, then so is C(X,A). It also follows that every contractible C*-algebra is an inductive limit of semiprojective C*-algebras – in fact, even of projective C*-algebras, as was shown in ​[4]​.

The commutative C*-algebra C(S^2) is probably the easiest C*-algebra where it is currently unknown if it is an inductive limit of semiprojective C*-algebras. Equivalently, it is unknown if C_0(\mathbb{R}^2) is an inductive limit of semiprojective C*-algebras. By Example 4.6 in ​[3]​, we know that the stabilization C_0(\mathbb{R}^2)\otimes\mathcal{K} is an inductive limit of semiprojective C*-algebras.

  1. [1]
    B. Blackadar, Shape theory for C^*-algebras, MATH. SCAND. 56 (1985) 249–275.
  2. [2]
    D. Enders, Semiprojectivity for Kirchberg algebras, ArXiv Preprint ArXiv:1507.06091. (2015).
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    H. Thiel, Inductive limits of semiprojective C^*-algebras, Advances in Mathematics. 347 (2019) 597–618.
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    H. Thiel, Inductive limits of projective C*-algebras, J. Noncommut. Geom. 13 (2020) 1435–1462.

C*-algebras complemented in their biduals

Von | Dezember 20, 2020

Question: Let A be a C*-algebra that is complemented in its bidual A^{**} by a *-homomorphism, that is, there exists a *-homomorphism \pi\colon A^{**}\to A such that \pi(a)=a for all a\in A. Is A a von Neumann algebra?

The converse is true: Let A be a von Neumann algebra. Then A has a (unique) isometric predual A_*. Let \kappa_{A_*}\colon A_*\to (A_*)^{**} be the natural inclusion of the Banach space A_* in its bidual. We naturally identify the dual of A_* with A, and the dual of (A_*)^{**} with A^{**}. Then the transpose \kappa_{A_*}^*\colon A^{**}=(A_*)^{***}\to (A_*)^*=A is a *-homomorphism that complements A in A^{**}.

Contractibility of unitary groups of II-1 factors

Von | Oktober 7, 2020

Question: Let M be a \mathrm{II}_1-factor. Is the unitary group \mathcal{U}(M) contractible when equipped with the strong operator topology?

Background/Motivation: By Kuiper’s theorem, if H is an infinite-dimensional Hilbert space, then the unitary group of the \mathrm{I}_\infty-factor \mathcal{B}(H) is contractible in the norm topology. This was generalized by Breuer ​[1]​ to (certain) \mathrm{I}_\infty and \mathrm{II}_\infty von Neumann algebras, and eventually Brüning-Willgerodt ​[2]​ showed that the unitary group of every properly infinite von Neumann algebra is contractible in the norm topology. This is no longer true for finite von Neumann algebras: The unitary group of a finite matrix algebra M_n(\mathbb{C}) is not contractible. Similarly, the unitary group of a \mathrm{II}_1 factor is not contractible in the norm topology since its fundamental group does not vanish – in fact, it was shown in ​[3]​ that \pi_1(\mathcal{U}(M))\cong\mathbb{R} for every \mathrm{II}_1 factor M.

As noted in the introduction of ​[4]​, the unitary group of every properly infinite von Neumann algebra is also contractible in the strong operator topology. This naturally leads to the above question, which was considered by Popa-Takesaki in ​[4]​. They showed that \mathcal{U}(M) is contractible in the strong operator topology if M is a separable \mathrm{II}_1 factor such that the associated \mathrm{II}_\infty factor M\bar{\otimes} \mathcal{B}(H) admits a trace scaling one-parameter group of automorphisms. This includes all McDuff factors (M is McDuff if M\cong M\bar{\otimes}\mathcal{R} for the hyperfinite \mathrm{II}_\infty factor \mathcal{R}) and all factors that satisfy M\cong M\bar{\otimes} L(\mathbb{F}_\infty), where L(\mathbb{F}_\infty) is the group von Neumann algebra of the free group on infinitely many generators.

  1. [1]
    M. Breuer, On the homotopy type of the group of regular elements of semifinite von Neumann algebras, Math. Ann. (1970) 61–74.
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    J. Brüning, W. Willgerodt, Eine Verallgemeinerung eines Satzes von N. Kuiper, Math. Ann. (1976) 47–58.
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    H. Araki, M.-S.B. Smith, L. Smith, On the homotopical significance of the type of von Neumann algebra factors, Commun.Math. Phys. (1971) 71–88.
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    S. Popa, M. Takesaki, The topological structure of the unitary and automorphism groups of a factor, Commun.Math. Phys. (1993) 93–101.

Scottish Book Problem 166

Von | Oktober 7, 2020

This is one of the few problems from the Scottish book that are still open. In slightly modernized form, and correcting the typo (in the book, f and f_0 should be switched in the last sentence) the problem is:

Let M be a topological manifold, and let f\colon M\to\mathbb{R} be a continuous function. Let G^M_f denote the subgroup of homeomorphisms T\colon M\to M that satisfy f\circ T=f. Let N be another manifold that is not homeomorphic to M. Does there exist a continuous function f_0\colon N\to\mathbb{R} such that G^M_f is not isomorphic to G^N_{f_0}?

Scottish Book Problem 155

Von | Oktober 7, 2020

This is one of the few problems from the Scottish book that are still open. In modern terminology, the problem is:

Let X and Y be Banach spaces, and let U\colon X\to Y be a bijective map with the following property: For every x_0\in X there exists \varepsilon>0 such that for the sphere S(x_0,\varepsilon) := \{ x\in X : \|x-x_0\|=\varepsilon \}, the restriction U|_{S(x_0,\varepsilon)} is isometric. Does it follow that U is isometric?

It is noted in the Scottish Book that the answer is „yes“ whenever U^{-1} is continuous, which is automatic if Y is finite-dimensional, or if Y has the property that for any two elements y_1,y_2\in Y satisfying y_2\neq 0 and \|y_1+y_2\|=\|y_1\|+\|y_2\| there exists \lambda\geq 0 such that y_1=\lambda y_2.

Conjugacy of pointwise conjugate homomorphisms

Von | Oktober 7, 2020

Given groups H and G, let us say that the pair (H,G) has property (*) if any two injective homomorphisms \alpha_0,\alpha_1\colon H\to G are conjugate if (and only if) they are pointwise conjugate.

Problem 1: Describe the class C_1 of groups H such that (H,G) has (*) for every group G.

Problem 2: Describe the class C_2 of groups G such that (H,G) has (*) for every group H.

Definitions: Given a group G, two elements a,b\in G are conjugate if there exists x\in G such that a=xbx^{-1}. Two homomorphisms \alpha_0,\alpha_1\colon H\to G are pointwise conjugate if \alpha_0(a) is conjugate to \alpha_1(a) in G for every a\in H. (Thus, for each a\in H there exists x_a\in G such that \alpha_0(a)=x_a\alpha_1(a)x_a^{-1}.) Further, \alpha_0 and \alpha_1 are conjugate if there exists x\in G such that \alpha_0(a)=x\alpha_1(a)x^{-1} for every a\in H. If \alpha_0 and \alpha_1 are conjugate, then they are locally conjugate. Property (*) records that the converse holds (for injective homomorphisms).

Motivation: Problem 30 of the Scottish Book ​[1]​ (which is still open) asks to determine which groups G have the following property: Pairs (a,a') and (b,b') in G are conjugate (there exists x\in G such that a=xbx^{-1} and a'=xb'x^{-1}) if (and only if) for every word w in two noncommuting variables the elements w(a,a') and w(b,b') are conjugate. Using property (*) formulated above, Problem 30 asks to determine which groups G have the property that (H,G) satisfies (*) for every subgroup H of G that is generated by two elements. The above Problem 2 is a more general (and possibly more natural) version of this problem.

An automorphism \alpha\colon G\to G is class-preserving if \alpha(a) is conjuagte to a for every a\in G. If (G,G) has (*), then every class-preserving automorphism of G is inner. The study of groups with (or without) outer class-preserving automorphisms has a long history; see for instance ​[2]​ and ​[3]​. There exist finite groups with outer class-preserving automorphisms. In particular, there exist finite groups that belong neither to C_1 nor to C_2. Note also that C_2 contains all abelian groups and that C_1 contains all cyclic groups.

  1. [1]
    R.D. Mauldin, The Scottish Book, Springer International Publishing, 2015.
  2. [2]
    C.-H. Sah, Automorphisms of finite groups, Journal of Algebra. (1968) 47–68.
  3. [3]
    M.K. Yadav, Class preserving automorphisms of finite p-groups: a survey, in: C.M. Campbell, M.R. Quick, E.F. Robertson, C.M. Roney-Dougal, G.C. Smith, G. Traustason (Eds.), Groups St Andrews 2009 in Bath, Cambridge University Press, 2007: pp. 569–579.

Purely infinite rings and C*-algebras

Von | September 23, 2020

Question 1: If A_0\subseteq\mathcal{B}(H) is a *-subalgebra of bounded linear operators on a separable Hilbert space H such that A_0 is purely infinite as a ring, is the norm-closure \overline{A_0} purely infinite as a C*-algebra?

This is Problem 8.4 in ​[1]​. As noted in Problem 8.6 in ​[1]​, this is even unclear if A_0 is a unital, simple, purely infinite ring. In the converse direction, it seems natural to ask:

Question 2: Given a purely infinite C*-algebra A, is there a dense *-subalgebra A_0\subseteq A that is purely infinite as a ring?

Definitions: The relation \precsim on a ring R is defined by setting x\precsim y if there exist a,b\in R such that x=ayb; see Definition 2.1 in ​[1]​. A ring R is purely infinite if no quotient of R is a division ring, and if any x,y\in R satisfy x\precsim y if (and only if) x\in RyR; see Definition 3.1 in ​[1]​. (Here, RyR denotes the two-sided ideal generated by y, that is, RyR=\{a_1yb_1+\ldots+a_nyb_n : n\geq 1, a_i,b_i\in R\}.)

Given a C*-algebra A, the Cuntz subequivalence relation \precsim is defined by setting x\precsim y if there exist sequences (a_n)_n and (b_n)_n in A such that \lim_{n\to\infty} \| x - a_nyb_n \| = 0. A C*-algebra A is purely infinite if it admits no nonzero one-dimensional representations and if any x,y\in A satisfy x\precsim y if and only if x belongs to \overline{\mathrm{span}}AyA, the closed, two-sided ideal generated by y; see Definition 4.1 in ​[2]​. (The definition in ​[2]​ only considers positive elements in A, but it equivalent to the definition given here.)

Background: By Proposition 3.17 in ​[1]​, if a C*-algebra is purely infinite as a ring, then it is purely infinite as a C*-algebra. The converse does probably not hold (Remark 3.18 in ​[1]​), which is why we ask Question 2 above. A unital, simple C*-algebra is purely infinite as a C*-algebra if and only if it is purely infinite as a ring. Thus, Question 2 has a positive answer in this case.

Given a C*-algebra A, let \mathrm{Ped}(A)\subseteq A denote its Pedersen ideal (the minimal dense ideal in A). By Proposition 8.5 in ​[1]​, if \mathrm{Ped}(A) is purely infinite as a ring, then A is purely infinite. Thus, in this particular instance, Question 1 has a positive answer.

  1. [1]
    G. Aranda Pino, K.R. Goodearl, F. Perera, M. Siles Molina, Non-simple purely infinite rings, American Journal of Mathematics. (2010) 563–610.
  2. [2]
    E. Kirchberg, M. Rordam, Non-simple purely infinite C*-algebras, American Journal of Mathematics. (2000) 637–666.

Liftable normal elements

Von | September 22, 2020

Given a C*-algebra A and a closed, two-sided ideal I\subseteq A, is the image of the normal elements in A under the quotient map A\to A/I a closed subset of A/I?

Equivalently, if (x_n)_n is a sequence of normal elements in A/I that converge to x, and if each x_n admits a normal lift in A, does x admit a normal lift?

This question is raised in Example 6.5 in ​[1]​. It is equivalent to the question of whether the commutative C*-algebra of continuous functions on the disc vanishing at zero is \ell-closed in the sense of Definition 6.1 in ​[1]​.

Of related interest is the class of normal elements that are universally liftable: We say that a normal element a in a C*-algebra A is universally liftable if for every C*-algebra B and every surjective *-homomorphism \pi\colon B\to A there exists a normal element b\in B with \pi(b)=a. One can show that a normal element a\in A is universally liftable if and only if there exists a projective C*-algebra P, a normal element b\in P and a *-homomorphism \varphi\colon P\to A with \varphi(b)=a. (A C*-algebra P is projective if for every C*-algebra D, every closed, two-sided ideal I\subseteq D, and every *-homomorphism \psi\colon P\to D/I there exists a *-homomorphism \tilde{\psi}\colon P\to D such that \pi\circ\tilde{\psi}=\psi, where \pi\colon D\to D/I is the quotient map.) Indeed, for the forward direct, one uses that every C*-algebra is the quotient of a projective C*-algebra, and for the converse direction one applies the definition of projectivity. In particular, every normal element in a projective C*-algebra is universally liftable.

Question: Given a C*-algebra A, is the set of universally liftable normal elements in A closed? Is it open (relative to the set of normal elements)?

Question: Is there a projective C*-algebra P and a normal element b\in P such that for every C*-algebra A, a normal element a\in A is universally liftable if and only if there exists a *-homomorphism \varphi\colon P\to A with \varphi(b)=a?

  1. [1]
    B. Blackadar, The Homotopy Lifting Theorem for Semiprojective C^*-Algebras, MATH. SCAND. (2016) 291.

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.

  1. [1]
    L. Robert, Remarks on Z-stable projectionless C*-algebras, Glasgow Math. J. (2015) 273–277.
  2. [2]
    M. Rørdam, The stable and the real rank of  {\mathcal Z}-absorbing C*-algebras, Int. J. Math. (2004) 1065–1084.
  3. [3]
    A.S. Toms, W. Winter, Strongly self-absorbing C^{*}-algebras, Trans. Amer. Math. Soc. (2007) 3999–4029.
  4. [4]
    L. Robert, L. Santiago, A revised augmented Cuntz semigroup, ArXiv:1904.03690. (2019).
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    B. Nica, Homotopical stable ranks for Banach algebras, Journal of Functional Analysis. (2011) 803–830.

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.