Traces on purely infinite C*-algebras

Von | Oktober 26, 2022

Does there exist a purely infinite C*-algebra that admits a tracial weight taking a finite, nonzero value?

Here a C*-algebra A is purely infinite if every element a \in A_+ is properly infinite (a\oplus a is Cuntz subequivalent to a in M_2(A)). This notion was introduced and studied by Kirchberg-Rørdam in ​[1]​ and ​[2]​. Further, a weight on a C*-algebra A is a map \varphi \colon A_+ \to [0,\infty] that is additive and satisfies \varphi(\lambda a)=\lambda\varphi(a) for all \lambda\in[0,\infty) and a\in A_+. A weight \varphi is tracial if \varphi(xx^*)=\varphi(x^*x) for all x\in A.

If I \subseteq A is a (not necessarily closed) two-sided ideal that is strongly invariant (for x\in A, we have xx^* \in I if and only if x^*x \in I), then the map \tau_I\colon A_+ \to [0,\infty] given by \tau_I(a)=0 if a\in I and \tau_I(a)=\infty otherwise, is a tracial weight. Note that these tracial weights are trivial in the sense that they only take the values 0 and \infty.

The question is if there exists a nontrivial tracial weight on a purely infinite C*-algebra.

It is well-known that every lower-semicontinuous tracial weight on a purely infinite C*-algebra is trivial. In particular, purely infinite C*-algebras do not admit tracial states. The question is about tracial weights that are not lower-semicontinuous. Sometimes, such tracial weights are called singular traces.

  1. [1]
    E. Kirchberg, M. Rordam, Non-simple purely infinite C*-algebras, American Journal of Mathematics. (2000) 637–666.
  2. [2]
    E. Kirchberg, M. Rørdam, Infinite Non-simple C*-Algebras: Absorbing the Cuntz Algebra O∞, Advances in Mathematics. (2002) 195–264.

Commutators in factors

Von | Oktober 8, 2022

Question: Is every element of trace zero in a \mathrm{II}_1 factor a commutator?

Background on commutators: It is a classical question to determine the elements in an algebra A (over a field k) that are commutators, that is, of the form [a,b] := ab-ba for some a,b \in A. A related (and often easier) question is to determine the commutator subspace \mathrm{span} [A,A], which is defined as the linear subspace of A generated by the set [A,A] := \{[a,b] : a,b \in A \} of commutators. (Warning: In the literature, [A,A] is often used to denote \mathrm{span} [A,A].)

A linear functional \tau \colon A \to k is said to be tracial if \tau(ab)=\tau(ba) for all a,b \in A; equivalently, \tau vanishes on \mathrm{span} [A,A]. It follows that an element in A belongs to \mathrm{span} [A,A] if and only if it vanishes under every tracial functional.

Thus, a commutator has to vanish under every tracial functional. Another obstruction occurs in the normed setting, since the unit of a normed algebra is not a commutator (although it may be a sum of two commutators). It follows that nonzero scalar multiples of the unit are not commutators either. A simple proof was given by Wielandt in ​[1]​. Since commutators are mapped to commutators in quotients, we obtain according obstructions quotients by closed, two-sided ideals.

To summarize, if A is a unital Banach algebra, and a \in [A,A], then:

  1. \tau(a) = 0 for every tracial functional \tau \colon A \to \mathbb{C}.
  2. for every closed ideal J \subseteq A (equivalently: every maximal ideal J \subseteq A), we do not have a + J = \lambda 1 + J for some \lambda \in \mathbb{C} \setminus \{0\}.

We will see below that these are the only obstructions for an element to be a commutator in the case of properly infinite factors, as well as type \mathrm{I} factors.

Let us specialize to the case that A is a unital C*-algebra. In this case, one can consider the space T(A) of tracial states on A. Since every tracial state is continuous, it vanishes on the closure of \mathrm{span} [A,A]. Further, it follows from Theorem 5 in ​[2]​ that

    \[\overline{\mathrm{span} [A,A]} = \bigcap_{\tau\in T(A)}\ker(\tau).\]

In many cases, \mathrm{span} [A,A] is a closed subspace. If it is not closed, then the C*-algebra admits non-continuous tracial functionals.

Commutators in von Neumann factors. The set of commutators have been completely characterized in properly infinite factors, and in factors of type \mathrm{I}_n. After presenting the results for these cases, we discuss some partial results for the case of a \mathrm{II}_1 factor.

Type \mathrm{I}_n: The matrix algebra M_n(\mathbb{C}) is simple and has a unique tracial state \tau. A matrix a \in M_n(\mathbb{C}) is a commutator if and only if \tau(a)=0. In particular, the set [M_n(\mathbb{C}, M_n(\mathbb{C}] of commutators agrees with the commutator subspace. More generally, this holds for every matrix algebra M_n(k) over a field, which was shown by Shoda in 1936 for the case of characteristic zero, and by Albert-Muckenhoupt in 1957 for arbitrary characteristic.

Types \mathrm{I}_\infty, \mathrm{II}_\infty and \mathrm{III}: Let M be a properly infinite factor. Then M has no tracial states, and M has a unique maximal ideal J. (If M=B(H) for some infinite-dimensional Hilbert space, then J is the closure of the ideal of operators whose closed range have dimension strictly less than that of H. In particular, if H is separable, then J is the ideal K(H) of compact operators on H.) Then

    \[[M,M] = \{ a \in M : a+J\neq \lambda 1 + J \text{ for every } \lambda\in\mathbb{C}\setminus\{0\}\},\]

that is, an element a\in M is a commutator if and only if its image in the simple C*-algebra M/J (if M=B(H) and H is separable, then this is the Calkin algebra B(H)/K(H)) is not a nonzero scalar multiple of the identity. This was shown by Halpern in ​[3]​. In particular, [M,M] is neither closed, nor a subspace, but every element in M is a sum of two commutators and thus \mathrm{span}[M,M]=M.

Type \mathrm{II}_1: Let M be a \mathrm{II}_1 factor. Then M is simple and has a unique tracial state \tau. It is expected that [M,M]=\{a\in M : \tau(a)=0\}. Partial results in this direction have been obtained by Dykema-Skripka in ​[4]​. In particular, every nilpotent element in M is a commutator, and every normal element with vanishing trace and with atomic spectral measure is a commutator.

  1. [1]
    H. Wielandt, �ber die Unbeschr�nktheit der Operatoren der Quantenmechanik, Math. Ann. (1949) 21–21.
  2. [2]
    N. Ozawa, Dixmier approximation and symmetric amenability for C*-algebras, J. Math. Sci. Univ. Tokyo. 20 (2013) 349–374.
  3. [3]
    H. Halpern, Commutators in properly infinite von Neumann algebras, Trans. Amer. Math. Soc. (1969) 55–73.
  4. [4]
    K. Dykema, A. Skripka, On single commutators in II_{1}–factors, Proc. Amer. Math. Soc. (2012) 931–940.

Are maximal ideals in C*-algebras closed?

Von | Oktober 8, 2022

Surprisingly, the following question is open:

Question: Are maximal ideals in C*-algebras closed?

Let A be a C*-algebra. By an ideal in A we mean a two-sided ideal I \subseteq A that is not necessarily closed. We say that I is a maximal ideal if the only ideals J \subseteq A satisfying I \subseteq J are I and A. An ideal I is proper if I \neq A.

If A is unital, then every maximal ideal is closed. Indeed, in this case, using that a proper ideal does not contain any invertible element of A, and using that the set of invertible elements is open, we see that if I \subseteq A is a proper ideal, then so is its closure \overline{I}. Thus, if I is maximal, then I = \overline{I}, and I is closed.

Further, it is known that every maximal left ideal in a C*-algebra is closed; see for example Proposition 3.5 in ​[1]​. (This even holds in Banach algebras admitting a bounded approximate unit.) In commutative C*-algebras, every maximal ideal is also a maximal left ideal and therefore closed.

Assume I is a non-closed, maximal ideal in a C*-algebra A (if it exists). Then I is dense and therefore contains the Pederson ideal of A. Further, the quotient A/I is a simple \mathbb{C}-algebra. It is easy to see that A/I is radical, that is, \mathrm{rad}(A/I)=A/I, because A/I has no nonzero maximal (modular) left ideals. One can show that I is hereditary (if 0 \leq x \leq y in A, and y belongs to I, then so does x), strongly invariant (if xx^* \in I, then x^*x \in I), and invariant under powers: if a \in I_+ and t > 0, then a^t \in I. For the arguments see this post in mathoverflow.

  1. [1]
    M.C. García, H.G. Dales, Á.R. Palacios, Maximal left ideals in Banach algebras, Bull. London Math. Soc. (2019) 1–15.

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.

  1. [1]
    B. Blackadar, D. Handelman, Dimension functions and traces on C∗-algebras, Journal of Functional Analysis. (1982) 297–340.
  2. [2]
    N.P. Brown, F. Perera, A.S. Toms, The Cuntz semigroup, the Elliott conjecture, and dimension functions on C*-algebras, Journal Für Die Reine Und Angewandte Mathematik (Crelles Journal). (2008).
  3. [3]
    K. De Silva, A note on two Conjectures on Dimension funcitons of C*-algebras, ArXiv. (2016) 1601.03475.
  4. [4]
    F. Perera, The Structure of Positive Elements for C*-Algebras with  Real Rank Zero, Int. J. Math. (1997) 383–405.
  5. [5]
    R. Antoine, F. Perera, L. Robert, H. Thiel, C*-algebras of stable rank one and their Cuntz semigroups, Duke Math. J. (2022) (to appear).

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.
  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.
  4. [4]
    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.
  2. [2]
    J. Brüning, W. Willgerodt, Eine Verallgemeinerung eines Satzes von N. Kuiper, Math. Ann. (1976) 47–58.
  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.
  4. [4]
    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.