The Blackadar-Handelman conjectures

Von | November 1, 2021

In ​[1]​, Blackadar and Handelman made two conjectures:

Conjecture 1: (Below Theorem I.2.4 in ​[1]​) Let A be a unital C*-algebra. Then the set \mathrm{LDF}(A) of lower-semicontinuous dimension functions is dense in the set \mathrm{DF}(A) of dimension functions.

Conjecture 2: (Below Theorem II.4.4 in ​[1]​) Let A be a unital C*-algebra. Then the compact, convex set \mathrm{DF}(A) is a Choquet simplex.

Here, a dimension function on a unital C*-algebra A is a map d\colon M_\infty(A)\to[0,\infty] that associates to every matrix over A a positive number and satisfying properties that generalize the classical properties of the rank of complex matrices:

  • If a and b are orthogonal matrices (that is, ab=a^*b=ab^*=a^*b^*=0), then d(a+b)=d(a)+d(b).
  • If a is Cuntz-dominated by b (that is, there exist sequences (r_n)_n and (s_n)_n such that \|a-r_nbs_n\|\to 0), then d(a)\leq d(b).
  • d(1)=1.

Such a dimension function is said to be lower-semicontinuous if it is lower-semicontinuous with respect to the norm-topology, that is, whenever (a_n)_n is a sequence in M_\infty(A) converging to some a, then d(a)\leq\liminf_n d(a_n). We equip \mathrm{DF}(A) with the topology of pointwise convergence. This gives \mathrm{DF}(A) the structure of a compact, convex set. We note that the subset \mathrm{LDF}(A)\subseteq\mathrm{DF}(A) is usually not closed (indeed, the conjecture is that it is dense). There is another (natural) topology on \mathrm{LDF}(A) giving it the structure of a compact, convex set that is even a Choquet simplex. Here, a compact, convex K set is a Choquet simplex if the set \mathrm{Aff}(K) of continuous, affine functions K\to\mathbb{R} satisfies Riesz interpolation, that is, given f_1,f_2,g_1,g_2\in\mathrm{Aff}(K) satisfying f_j\leq g_k for j,k\in\{1,2\} there exists h\in\mathrm{Aff}(K) such that f_j\leq h\leq g_k for j,k\in\{1,2\}. Choquet simplices have the property that every element can be represented in a unique way by a boundary measure. Important examples of Choquet simplices are the Bauer simplices: Given a compact, Hausdorff space X, the set M_1(X) of positive, Borel probability measures on X is a Choquet simplex with boundary \partial_e M_1(X)\cong X, and \mathrm{Aff}(M_1(X))\cong C(X,\mathbb{R}).

If A=C(X) is a commutative C*-algebra, then \mathrm{DF}(C(X)) naturally corresponds to the set of finitely-additive probability measures on X, while \mathrm{LDF}(C(X)) naturally corresponds to the set of (\sigma-additive) probability measures on X. For a general C*-algebra A, we therefore consider the dimension functions on A as „noncommutative, finitely-additive probability measures“, and similarly \mathrm{LDF}(A) are the „noncommutative probability measures“ on A. It is easy to see that the probability measures on a compact, Hausdorff space are dense in the set of finitely-additive probability measures (see the proof of Theorem I.2.4 in ​[1]​), and it is a classical result that the finitely-additive probability measures on a compact, Hausdorff space form a Choquet simplex. The Blackadar-Handelman conjectures predict that theses results generalize to the noncommutative setting.

The first Blackadar-Handelman conjecture has been verified in the following cases: if A is commutative (Theorem I.2.4 in ​[1]​); if A is simple, exact, stably finite and has strict comparison of positive elements (Theorem B and 6.4, and Remark 6.5 in ​​[2]​​); if A has finite radius of comparison (Theorem 3.3 in ​[3]​).

The second Blackadar-Handelman conjecture has been verified in the following cases: if A is commutative; if A is simple, exact, stably finite and \mathcal{Z}-stable (Theorem B in ​​[2]​); if A has real rank zero and stable rank one (Corollary 4.4 in ​​[4]​); in ​​[5]​, the assumption of real rank zero was removed from the result in ​​[4]​, thus verifying the second Blackadar-Handelman conjecture for all C*-algebras of stable rank one.

With view to the results in ​[5]​ it is natural to ask if the first Blackadar-Handelmann conjecture can be verified for all C*-algebras of stable rank one.

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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.

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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.

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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.

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    R.D. Mauldin, The Scottish Book, Springer International Publishing, 2015. https://doi.org/10.1007/978-3-319-22897-6.
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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.

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    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.
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    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?

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    B. Blackadar, The Homotopy Lifting Theorem for Semiprojective C^*-Algebras, MATH. SCAND. (2016) 291. https://doi.org/10.7146/math.scand.a-23691.