Realizing Cuntz classes in commutative subalgebras

Von | Juli 18, 2021

Let A be a unital, simple C*-algebra of stable rank one. Does there exist a commutative sub-C*-algebra C(X)\subseteq A such that for every lower-semicontinuous function f\colon\mathrm{QT}_1(A)\to[0,1] there exists an open subset U\subseteq X such that f(\tau)=\mu_\tau(U) for \tau\in\mathrm{QT}_1(A)? Here, \mathrm{QT}_1(A) denotes the Choquet simplex of normalized 2-quasitraces on A (if A is exact, then this is just the Choquet simplex of tracial states on A), and \mu_\tau denotes the probability measure on X induced by the restriction of \tau to C(X).

More specifically, one may ask if this is always the case for a Cartan subalgebra of A.

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. https://doi.org/10.7146/math.scand.a-12100.
  2. [2]
    D. Enders, Semiprojectivity for Kirchberg algebras, ArXiv Preprint ArXiv:1507.06091. (2015).
  3. [3]
    H. Thiel, Inductive limits of semiprojective C^*-algebras, Advances in Mathematics. 347 (2019) 597–618. https://doi.org/10.1016/j.aim.2019.02.030.
  4. [4]
    H. Thiel, Inductive limits of projective C*-algebras, J. Noncommut. Geom. 13 (2020) 1435–1462. https://doi.org/10.4171/jncg/350.

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. https://doi.org/10.1007/bf01350761.
  2. [2]
    J. Brüning, W. Willgerodt, Eine Verallgemeinerung eines Satzes von N. Kuiper, Math. Ann. (1976) 47–58. https://doi.org/10.1007/bf01354528.
  3. [3]
    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. https://doi.org/10.1007/bf01651585.
  4. [4]
    S. Popa, M. Takesaki, The topological structure of the unitary and automorphism groups of a factor, Commun.Math. Phys. (1993) 93–101. https://doi.org/10.1007/bf02100051.

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. https://doi.org/10.1007/978-3-319-22897-6.
  2. [2]
    C.-H. Sah, Automorphisms of finite groups, Journal of Algebra. (1968) 47–68. https://doi.org/10.1016/0021-8693(68)90104-x.
  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. https://doi.org/10.1017/cbo9780511842474.019.

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. https://doi.org/10.1353/ajm.0.0119.
  2. [2]
    E. Kirchberg, M. Rordam, Non-simple purely infinite C*-algebras, American Journal of Mathematics. (2000) 637–666. https://doi.org/10.1353/ajm.2000.0021.

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. https://doi.org/10.7146/math.scand.a-23691.

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. https://doi.org/10.1017/s0017089515000117.
  2. [2]
    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.
  3. [3]
    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.
  4. [4]
    L. Robert, L. Santiago, A revised augmented Cuntz semigroup, ArXiv:1904.03690. (2019). https://arxiv.org/abs/1904.03690v1.
  5. [5]
    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.