OA-Problems – Hannes Thiel

Primitive, factor and prime ideals

Von hannesthiel | April 13, 2026

Every primitive ideal of a C*-algebra is closed and prime. A long-standing problem of Dixmier asked whether the converse holds, and this was settled by Weaver in 2003, when he gave the first example [1] of a prime C*-algebra that is not primitive. Between primitive ideals and closed prime ideals lies the class of factor ideals, and Weaver’s example also shows that there exist closed prime ideals that are not factor ideals. This leads to the following question:

Is every factor ideal in a C*-algebra primitive?

Background/Motivation: Given a C*-algebra $A$ , a factor representation is a representation $\pi\colon A \to B(H)$ such that the generated von Neumann algebra $\pi(A)''$ is a factor, that is, $Z(\pi(A)'')=\mathbb{C}1$ . One can further distinguish the type of the factor. In particular, an irreducible representation is a type I factor representation. A (two-sided) ideal $I \subseteq A$ is:

primitive if $I = \ker(\pi)$ for some irreducible representation $\pi\colon A \to B(H)$ ;
factor if $I = \ker(\pi)$ for some factor representation $\pi \colon A \to B(H)$ ;
prime if for any ideals $J,K \subseteq A$ with $JK \subseteq I$ we have $J \subseteq I$ or $K \subseteq I$ .

Since irreducible representations are factor representations, we see that every primitive ideal is a factor ideal. Let us see that every factor ideal is a closed prime ideal.

First, if $p,q$ are nonzero projections in a factor $M$ , then $pMq \neq {0}$ . Indeed, projections in a factor are totally ordered with respect to Murray–von Neumann subequivalence, so we may assume that $p \precsim q$ . Then there exists $v \in M$ such that $p = vv^*$ and $v^*v \leq q$ , which implies that $0 \neq v = pvq \in pMq$ . Next, if $x,y$ are nonzereo elements in a factor $M$ , then using Borel functional calculus for $xx^*$ we can find $a \in M$ such that $p:=xx^*a$ is a nonzero projection, and similarly there is $b \in M$ such that $q := by^*y$ is a nonzero projection. It follows that $\{0\} \neq pMq = xx^*aMby^*y \subseteq xMy$ .

Now let $I$ be a factor ideal, that is, $I = \ker(\pi)$ for a representation $\pi \colon A \to B(H)$ such that $M := \pi(A)''$ is a factor, and let $J,K \subseteq A$ be ideals with $JK \subseteq I$ . Then $\pi(J)\pi(A)\pi(K) = \{0\}$ , which implies that $\pi(J) M \pi(K) = \{0\}$ . Then $\pi(J)=\{0\}$ or $\pi(K)=\{0\}$ , and so $J \subseteq I$ or $K \subseteq I$ , showing that $I$ is a prime ideal.

Letting $\mathrm{Prim}(A)$ , $\mathrm{Fac}(A)$ and $\mathrm{Prime}(A)$ denote the set of primitive, of factor, and of closed prime ideals of $A$ , respectively, we thus have the following inclusions:

$\mathrm{Prim}(A) \subseteq \mathrm{Fac}(A) \subseteq \mathrm{Prime}(A).$

There are three important cases in which closed prime ideals are automatically primitive, and hence $\mathrm{Prim}(A) = \mathrm{Fac}(A) = \mathrm{Prime}(A)$ .

Separable C*-algebras. Dixmier [2] showed that every closed prime ideal of a separable C*-algebra is primitive, see also Corollary II.6.5.15 in [3]. To see this, let $A$ be a separable C*-algebras, and let $I \in \mathrm{Prime}(A)$ . Passing to the quotient $A/I$ , we may assume that $A$ is prime (i.e. $\{0\}$ is a prime ideal) and we need to show that $A$ is primitive (i.e. admits a faithful, irreducible representation). We consider the primitive ideal space $X := \mathrm{Prim}(A)$ equipped with the hull-kernel topology, that is, a subset $F \subseteq X$ is closed if and only if $F=\{J \in X : K \subseteq J\}$ for some closed ideal $K$ (namely $K = \bigcap F$ ). Then $X$ is irreducible in the sense that if $F_1,F_2 \subseteq X$ are closed subsets with $X \subseteq F_1 \cup F_2$ , then $X \subseteq F_1$ or $X \subseteq F_2$ . Indeed, the closed ideals $K_1 := \bigcap F_1$ and $K_2 := \bigcap F_2$ satisfy

$K_1K_2 \subseteq K_1 \cap K_2 \subseteq \bigcap X = \{0\}$

since $X \subseteq F_1 \cup F_2$

. Since $\{0\}$

is prime, we get $K_1\subseteq \{0\}$

or $K_2 \subseteq \{0\}$

, which gives $X \subseteq F_1$

or $X \subseteq F_2$

Now let $U \subseteq X$ be a nonempty open subset. Then $U$ is dense, since $X \subseteq (X \setminus U) \cup \overline{U}$ and $X \nsubseteq X \setminus U$ , hence $X \subseteq \overline{U}$ . In general, the primitive ideal space of a C*-algebra is a Baire space (the intersection of a countable family of dense open sets is dense). Now, if $A$ is separable, then $X = \mathrm{Prim}(A)$ is second-countable. Let $U_1,U_2,\dots$ be a basis of nonempty open subsets of $X$ . Using that $X$ is a Baire space, the intersection $G:=\bigcap_n U_n$ is dense in $X$ , and in particular it is not empty. Pick $J \in G$ , which is a primitive ideal of $A$ . Then $J$ is contained in every nonempty open subset of $X$ , and hence $\{J\}$ is dense in $X$ . Then

$X = \overline{\{J\}} = \{P\in X : J \subseteq P\},$

which means that every primitive ideal of $A$

contains $J$

. But the intersection of all primitive ideals of $A$

is $\{0\}$

, and thus $J=\{0\}$

, showing that $\{0\}$

is a primitive ideal.

Von Neumann algebras. Every closed prime ideal in a von Neumann algebra (and more generally, in every AW*-algebra) is primitive. To see this, let $P$ be a closed, prime ideal in an AW*-algebra $A$ . Let $Z$ denote the center and consider $P \cap Z$ . Then $M := P \cap Z$ is a closed, prime ideal of $Z$ . Since closed, prime ideals in commutative C*-algebras are maximal, it follows that $M$ is a maximal ideal of $Z$ . Using that $M$ belongs to the center and applying Cohen’s factorization theorem, it follows that the closed ideal of $A$ generated by $M$ is equal to $AM$ . Then the quotient $A/AM$ is an AW* factor, and we have $AM \subseteq P$ . It was shown in Theorem 6 in [4] that the closed ideals of an AW* factor are well-ordered by inclusion. Since every closed ideal of a C*-algebra is an intersection of primitive ideals, it follows that every (proper) closed ideal of an AW* factor is primitive (see Corollary 7 in [4]). In particular, the (proper) closed ideal $P/AM$ of $A/AM$ is primitive, and it follows that $P$ is a primitive ideal of $A$ .

Type I C*-algebras. Kaplansky showed in Lemma 7.4 in [5] that every prime C*-algebra of type I is primitive. Since type I passes to quotients, we get that closed prime ideals in type I C*-algebras are primitive.

A C*-algebra has type I if and only if its bidual $A^{**}$ is a type I von Neumann algebra. Further, a C*-algebra is type I if and only if every of its factor representations is type I. Thus, a C*-algebra is not type I if and only if it has factor representations of type II or type III. Sakai showed that type III is always realized, that is, a C*-algebra is not type I if and only if it has a factor representation of type III. See IV.1.1.3 and IV.1.5.8 in [3]. For separable C*-algebras, the analog for type II is also known: A separable C*-algebra is not of type I if and only if it admits a factor representation of type II. The nonseparable case seems to be still open:

Does every C*-algebra that is not of type I admit a factor representation of type II?

One can further distinguish type $\mathrm{II}_1$ and type $\mathrm{II}_\infty$ . A non-type I C*-algebra need not have factor representations of type $\mathrm{II}_1$ . For example, simple purely infinite C*-algebras do not admit factor representations of type $\mathrm{II}_1$ , since this would entail the existence of a tracial state. In fact, a unital, simple, non-elementary C*-algebra admits a factor representations of type $\mathrm{II}_1$ if and only it has a tracial state. Simple, purely infinite C*-algebras can have factor representations of type $\mathrm{II}_\infty$ . For the Calkin algebra, this was shown in [6,7].

[1] Weaver. A prime C*-algebra that is not primitive. J. Funct. Anal. 203 (2003), 356–361.

[2] Dixmier. C*-algebras. North-Holland Math. Library, Vol. 15, 1977.

[3] Blackadar. Operator algebras. Theory of C*-algebras and von Neumann algebras. Encyclopaedia Math. Sci., 122, Springer 2006.

[4] Saitô, Wright. Ideals of factors. Rend. Circ. Mat. Palermo (2) 55 (2006), 360–368.

[5] Kaplansky. The structure of certain operator algebras. Trans. Amer. Math. Soc. 70 (1951), 219–255.

[6] Anderson, Bunce. A type II-infty factor representation of the Calkin algebra [Amer. J. Math. 99]

[7] Anderson. Extreme points in sets of positive linear maps on B(H). J. Functional Analysis 31 (1979), 195–217.

Pure group C*-algebras

Von hannesthiel | Februar 15, 2026

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Question: Is $C^*_r(G)$ pure if (and only if) $G$ is non-amenable?

Here, $C^*_r(G)$ denotes the reduced group C*-algebra of a discrete group $G$ . A C*-algebra $A$ is said to be \emph{pure} if it is Jiang-Su stable at the level of Cuntz semigroups, in the sense that

$\operatorname{Cu}(A) \cong \operatorname{Cu}(A)\otimes \operatorname{Cu}(\mathcal{Z}).$

Equivalently, the Cuntz semigroup $\operatorname{Cu}(A)$

is almost unperforated and almost divisible (see Theorem 7.3.11 in [1]). This notion was introduced by Winter in his seminal work on regularity properties of simple, nuclear C-algebras [2]. More recently, it was shown in [3,4] that pure C*-algebras form a robust class: they are closed under ideals, quotients, extensions, and inductive limits.

Jiang-Su stability implies purity, but reduced group C*-algebras are rarely Jiang-Su stable. Another important source of examples comes from Robert’s notion of \emph{self-less} C*-algebras [5], which are also pure.

If $G$ is amenable, then $C^*_r(G)$ admits a quotient isomorphic to $\mathbb{C}$ , arising from the trivial representation. This provides an obstruction to purity. In particular, if $C^*_r(G)$ is pure, then $G$ must be non-amenable.

There is some evidence for the converse: A recent breakthrough by Amrutam, Gao, Kunnawalkam Elayavalli, and Patchell showed that $C^*_r(G)$ is self-less (and hence pure) whenever $G$ is an acylindrically hyperbolic group with trivial finite radical and rapid decay. The assumption of rapid decay was later removed by Ozawa [7], and the expectation is that all C*-simple groups (that is, groups for which $C^*_r(G)$ is simple) have self-less and thus pure reduced group C*-algebra.

Things are more complicated for non-amenable groups for which $C^*_r(G)$ is not simple. A first step was obtained by extending the results of [6,7] to the twisted case, which implies that $C^*_r(G)$ is pure for every acylindrically hyperbolic group [8.9]. One might also expect that a minimal tensor product $A \otimes B$ is pure whenever $A$ or $B$ is pure (see this problem), which would handle groups like $\mathbb{F}_2 \times \mathbb{Z}$ , or more generally $G \times H$ for $G$ an acylindrically hyperbolic group and $H$ any other group. Update (February 2026): Seth and Vilalta [10] showed that $A \otimes B$ is pure whenever $A$ is simple and pure and $B$ is an ASH algebra. This handles in particular $\mathbb{F}_m \times \mathbb{Z}^n$ with $m \geq 2$ , and more generally groups of the form $G \times H$ with $G$ an acylindrically hyperbolic group (so that $C^*_r(G)$ is a finite direct sum of simple, pure C*-algebras) and $H$ virtually abelian (so that $C^*_r(H)$ is subhomogeneous).

A necessary condition for pureness is that the C*-algebra is nowhere scattered, meaning that none of its ideal-quotients are elementary (that is, isomorphic to the compact operators on some Hilbert spaces). It is known that $C^*_r(G)$ has no finite-dimensional irreducible representations (and hence no elementary quotients) whenever $G$ is non-amenable. However, it is unknown if $C^*_r(G)$ is nowhere scattered whenever $G$ is non-amenable.

[1] R. Antoine, F. Perera, H. Thiel. Tensor products and regularity properties of Cuntz semigroups. Mem. Amer. Math. Soc. 251 (2018).

[2] W. Winter. Nuclear dimension and Z-stability of pure C*-algebras. Invent. Math. 187 (2012), 259-342.

[3] R. Antoine, F. Perera, H. Thiel, E. Vilalta. Pure C*-algebras. preprint arXiv:2406.11052 (2024).

[4] F. Perera, H. Thiel, E. Vilalta. Extensions of pure C*-algebras. preprint arXiv:2506.10529 (2025).

[5] L. Robert. Selfless C*-algebras. Adv. Math. 478 (2025), Article ID 110409, 28 p.

[6] T. Amrutam, D. Gao, S. Kunnawalkam Elayavalli, G. Patchell. Strict comparison in reduced group C*-algebras. Invent. Math. 242 (2025), 639-657.

[7] N. Ozawa. Proximality and selflessness for group C*-algebras. preprint arXiv:2508.07938 (2025).

[8] S. Raum, H. Thiel, E. Vilalta. Strict comparison for twisted group C*-algebras. arXiv:2505.18569 (2025).

[9] F. Flores, M. Klisse, M. Cobhthaigh, M. Pagliero. Pureness and stable rank one for reduced twisted group C*-algebras of certain group extensions. preprint arXiv:2601.19758 (2026).

[10] A. Seth, E. Vilalta. Continuous functions over a pure C*-algebra. preprint arXiv:2602.14809 (2026).

Automatic continuity of homomorphisms

Von hannesthiel | September 17, 2025

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Introduction: The study of automatic continuity for homomorphisms (multiplicative, linear maps) $\varphi\colon A \to B$ between Banach algebras has a long history. In 1960, Bade and Curtis [1] proved the existence of discontinuous homomorphisms between commutative Banach algebras in ZFC, showing that automatic continuity results can only be achieved by imposing additional hypotheses on the algebras $A$ , $B$ or on the map $\varphi$ .

A natural class of Banach algebras are unital, commutative C*-algebras, that is, the algebras $C(X)$ of continuous functions on a compact, Hausdorff space $X$ . Using that every character (that is, homomorphism $A \to \mathbb{C}$ ) of a Banach algebra $A$ is contractive, it follows that every homomorphism $A \to C(X)$ is contractive. The situation is completely different if instead we assume that the domain is $C(X)$ : In 1976 Dales and Esterle [2] independently produced discontinuous homomorphisms $C(X) \to B$ for every infinite, compact, Hausdorff space $X$ , assuming the continuum hypothesis (CH). This suggested to consider the axiom „no discontinuous homomorphisms“

(NDH) Every homomorphism $C(X) \to B$ is continuous (for every compact, Hausdorff space $X$ and every Banach algebra $B$ ).

Cuntz [3] proved that continuity of linear maps out of a C*-algebra reduces to the commutative case: A linear map $\varphi\colon A\to X$ from a C*-algebra $A$ to a Banach space $X$ is continuous if and only if the restriction of $\varphi$ to every commutative sub-C*-algebra is continuous. It follows that (NDH) is equivalent to:

(NDH‘) Every homomorphism from a C*-algebra to a Banach algebra is continuous.

[This paragraph was corrected in February 2026. Special thanks to Ilijas Farah for the helpful feedback.] There are models of ZFC in which (NDH) holds [4]. In particular, Solovay described such a model in an unpublished note (see also Section 3 of [8]), and another model was constructed by Woodin [9]. Consequently, (NDH) is independent of ZFC: it fails under (CH), as shown by Esterle and Dales, but holds in certain models of ZFC.

This raises the question if one can impose further assumptions to show automatic continuity. A positive result was obtained by Sakai [5], who showed that every surjective homomorphism between C*-algebras is automatically continuous; see Lemma 4.1.12 in [5]. The following open questions ask if the result of Sakai can be generalized by removing the assumption on the target algebra (Question 1), by removing the assumption on the map (Question 2) or by relaxing the assumptions on target algebra and map (Question 3).

Question 1: Is every surjective homomorphism $A \to B$ from a C*-algebra $A$ onto a Banach algebra $B$ automatically continuous?

Question 2: Is every homomorphism $A \to B$ between C*-algebras automatically continuous?

Question 3: Is every homomorphism $A \to B$ with dense range from a C*-algebra $A$ to a semisimple Banach algebra $B$ automatically continuous?

The answers to all questions is known to be positive if $B$ is commutative: For Question 1 this was shown by Laursen (Theorem 4 in [6]), and for Questions 2 and 3 it follows using that characters are contractive. For further partial results see [7].

[1] Bade, Curtis. Homomorphisms of commutative Banach algebras. Am. J. Math. 82 (1960), 589-608.

[2] Dales, Esterle. Discontinuous homomorphisms from C(X). Bull. Am. Math. Soc. 83 (1977), 257-259.

[3] Cuntz. On the continuity of semi-norms on operator algebras. Math. Ann. 220 (1976), 171-183.

[4] Dales, Woodin. An introduction to independence for analysts. Cambridge University Press (1987).

[5] Sakai. C*-algebras and W*-algebras. Springer-Verlag (1971).

[6] Laursen. Continuity of homomorphisms from C*-algebras into commutative Banach algebras. J. Lond. Math. Soc., II. Ser. 36 (1987), 165-175.

[7] Runde. An epimorphism from a C*-algebra is continuous on the center of its domain. J. Reine Angew. Math. 439 (1993), 93-102.

[8] I. Farah. Embedding partially ordered sets into ww. Fund. Math. 151 (1996), 53-95.

[9] W.H. Woodin. A discontinuous homomorphism from C(X) without CH. J. Lond. Math. Soc., II. Ser. 48 (1993), 299-315.

Tensor products with pure C*-algebras

Von hannesthiel | August 4, 2024

1 Kommentar

Question: Is the minimal tensor product $A \otimes B$ of two C*-algebras pure whenever one of them is?

Update (February 2026): A partial positive answer has been given by Seth and Vilalta [6]. In particular, $A \otimes B$ is pure whenever $A$ is simple and pure and $B$ is an ASH algebra.

Background: Following Winter [1], a C*-algebra $A$ is said to be pure if it has very good comparison and divisibility properties, specifically its Cuntz semigroup is almost unperforated and almost divisible. This notion is closely related to that of $\mathcal{Z}$ -stability, where $\mathcal{Z}$ denotes the Jiang-Su algebra, and a C*-algebra $A$ is said to be $\mathcal{Z}$ -stable if $A \cong A \otimes \mathcal{Z}$ . Rørdam [2] showed that every $\mathcal{Z}$ -stable C*-algebra is pure, and the Toms-Winter conjecture predicts that a separable, unital, simple, nuclear C*-algebra is $\mathcal{Z}$ -stable if (and only if) it is pure. Moreover, it was shown in [4] that a C*-algebra $A$ is pure if and only if $\mathrm{Cu}(A) \cong \mathrm{Cu}(A) \otimes \mathrm{Cu}(\mathcal{Z})$ , that is, its Cuntz semigroup tensorially absorbs the Cuntz semigroup of $\mathcal{Z}$ . One may therefore think of pureness as $\mathcal{Z}$ -stability at the level of the Cuntz semigroup.

Using that $\mathcal{Z}$ is tensorially self-absorbing (even strongly self-absorbing in the sense of [3]), one sees that the minimal tensor product $A \otimes B$ is $\mathcal{Z}$ -stable whenever $B$ is $\mathcal{Z}$ -stable. Indeed, one has

$(A \otimes B) \otimes \mathcal{Z} \cong A \otimes (B \otimes \mathcal{Z}) \cong A \otimes B.$

This raises the question above, which is largely unexplored. For the special case that $A=C(X)$ and $B$ is pure, this was asked by Chris Phillips after my talk on pure C*-algebras in Shanghai 2024. For the case $A=C([0,1])$ and $B$ a simple, pure C*-algebra with stable rank one and vanishing $K_1$ -group, a positive answer to the question can likely be deduced from Corollary 2.7 in [5].

[1] W. Winter. Nuclear dimension and Z-stability of pure C*-algebras. Invent. Math. 187 (2012), 259-342.
[2] M. Rørdam. The stable and the real rank of Z-absorbing C*-algebras. Internat. J. Math. 15 (2004), 1065-1084.
[3] A. Toms, W. Winter. Strongly self-absorbing C*-algebras. Trans. Amer. Math. Soc. 359 (2007), 3999-4029.
[4] R. Antoine, F. Perera, H. Thiel. Tensor products and regularity properties of Cuntz semigroups. Mem. Amer. Math. Soc. 251 (2018).
[5] R. Antoine, F. Perera, L. Santiago. Pullbacks, C(X)-algebras, and their Cuntz semigroup. J. Funct. Anal. 260 (2011), 2844-2880.

[6] A. Seth, E. Vilalta. Continuous functions over a pure C*-algebra. preprint arXiv:2602.14809 (2026).

Commutators and square-zero elements in C*-algebras

Von hannesthiel | April 14, 2024

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Given elements $b$ and $c$ in a ring, the element $[b,c] := bc-cb$ is called an (additive) commutator. Given a C*-algebra $A$ , we use $[A,A]$ to denote the additive subgroup (equivalently, the linear subspace) generated by the set of additive commutators in $A$ . Note that $[A,A]$ is not necessarily a closed subspace. Given additive subgroups $V$ and $W$ , it is customary to use $[V,W]$ to denote the additive subgroup generated by the set $\{[v,w] : v\in V, w \in W\}$ .

An element $a$ in a ring is a square-zero element if $a^2=0$ . Given a square-zero element $x$ in a C*-algebra $A$ , we consider the polar decomposition $x=v|x|$ in the bidual $A^{**}$ . Then $|x|^{1/2}$ and $|x|^{1/2}v$ belong to $A$ , and we have

$x = [|x|^{1/2},|x|^{1/2}v] \in [A,A].$

More generally, Robert showed in Lemma 2.1 in [1] that every nilpotent element in $A$ belongs to $[A,A]$ . For $k\geq 2$ , we use $N_k(A)$ to denote the set of $k$ -nilpotent elements in $A$ , and we use $N_k(A)^+$ to denote the additive subgroup of $A$ generated by $N_k(A)$ . We thus have $N_k(A)^+ \subseteq [A,A]$ . This raises the following questions:

Question 1: (Robert, Question 2.5 in [1]) Is $[A,A] = N_2(A)^+$ ?

A positive answer to the above question is known in many cases, in particular if $A$ is unital and admits no characters (one-dimensional irreducible representations), by Theorem 4.2 in [1]. Further, it is known that $[A,A]$ is always contained in the closure of $N_2(A)^+$ .

In Theorem 4.2 in [2], it is shown that $N_2(A)^+ = [N_2(A)^+,N_2(A)^+]$ , that is, every square-zero element is a sum of commutators of square-zero elements, and every commutator of square-zero elements is a sum of square-zero elements. This raises the closely related question if every commutator in a C*-algebra is a sum of commutators of commutators:

Question 2: (Question 3.5 in [2]) Is $[A,A]=[[A,A],[A,A]]$ ?

A positive answer to Question 1 entails a positive answer to Question 2. Indeed, if a C*-algebra $A$ satisfies $[A,A]=N_2(A)^+$ , then

$[[A,A],[A,A]] = [N_2(A)^+,N_2(A)^+] = N_2(A)^+ = [A,A].$

[1]
L. Robert, On the Lie ideals of $\Cstar$ -algebras, J. Operator Theory (2016) 387–408. https://doi.org/10.7900/jot.2015may17.2070.
[2]
E. Gardella, H. Thiel, Prime ideals in C*-algebras and applications to Lie theory, Proc. Amer. Math. Soc. (to Appear) 1 (2024) 9.

Von Neumann’s problem for II-1 factors

Von hannesthiel | Dezember 8, 2023

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In the 1920s, von Neumann introduced the notion of amenability for groups, and he showed that a group is nonamenable whenever it contains the free group $\mathbb{F}_2$ . The question of whether this characterizes (non)amenability became known as von Neumann’s problem, and it was finally answered negatively by Olshanskii in 1980: There exist nonamenable groups that do contain no subgroup isomorphic to $\mathbb{F}_2$ .

The group von Neumann algebra $L(G)$ is the von Neumann subalgebra of bounded operators on $\ell^2(G)$ generated by the left regular representation of $G$ on $\ell^2(G)$ . There is a notion of amenability for von Neumann algebras, and Connes showed that $G$ is amenable if and only if $L(G)$ is. Further, if $H$ is a subgroup of $G$ , then $L(H)$ naturally is a sub-von Neumann algebra of $L(G)$ . Thus, if $G$ contains $\mathbb{F}_2$ , then $L(G)$ contains the free group factor $L(\mathbb{F}_2)$ as a sub-von Neumann algebra. The analog of von Neumann’s problem in this setting remains open:

Question: If $G$ is a nonamenable group, does $L(G)$ contain $L(\mathbb{F}_2)$ ?

The C*-algebraic version of von Neumann’s problem also remains open:

Question: If $G$ is a nonamenable group, does the reduced group C*-algebra $C^*_{\text{red}}(G)$ contain $C^*_{\text{red}}(\mathbb{F}_2)$ ?

Background: A (discrete) group is said to be amenable if it admits a finitely additive, left invariant probability measure. To quote Brown-Ozawa, there are approximately $10^{10^{10}}}$ different characterizations of amenability for groups [1].

[1]
N. Brown, N. Ozawa, 𝐶*-Algebras and Finite-Dimensional Approximations, Graduate Studies in Mathematics. (2008). https://doi.org/10.1090/gsm/088.

Nonseparable, exact C*-algebras

Von hannesthiel | Dezember 8, 2023

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It is known that every nuclear C*-algebra is exact, and that exactness passes to sub-C*-algebras. It follows that every sub-C*-algebra of a nuclear C*-algebra is exact. For separable C*-algebras, the converse holds. In fact, Kirchberg’s $\mathcal{O}_2$ -embedding theorem shows that a separable C*-algebra $A$ is exact if and only if it embeds into the Cuntz algebra $\mathcal{O}_2$ (which is nuclear), Theorem 6.3.11 in [1].

At the Kirchberg Memorial Conference in Münster, July 2023, Simon Wassermann asked if this result can be generalized to the nonseparable case:

Question: Does every (nonseparable) exact C*-algebra embed into a nuclear C*-algebra?

[1]
M. Rørdam, E. Størmer, Classification of Nuclear C*-Algebras. Entropy in Operator Algebras, Springer Berlin Heidelberg, 2002. https://doi.org/10.1007/978-3-662-04825-2.

Traces on purely infinite C*-algebras

Von hannesthiel | Oktober 26, 2022

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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]
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.
[2]
E. Kirchberg, M. Rørdam, Infinite Non-simple C*-Algebras: Absorbing the Cuntz Algebra O∞, Advances in Mathematics. (2002) 195–264. https://doi.org/10.1006/aima.2001.2041.

Commutators in factors

Von hannesthiel | Oktober 8, 2022

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

Update (August 2024): A positive answer to the questions was recently given In the following article:

S. Wen, J. Fang, Z. Yao. A stronger version of Dixmier’s averaging theorem and some applications. J. Funct. Anal. 287 (2024).

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:

$\tau(a) = 0$ for every tracial functional $\tau \colon A \to \mathbb{C}$ .
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. [Update: As noted above, in a recent article by S. Wen, J. Fang, and Z. Yao it is shown that every trace-zero element in a $\mathrm{II}_1$ factor is a commutator.]

[1]
H. Wielandt, �ber die Unbeschr�nktheit der Operatoren der Quantenmechanik, Math. Ann. (1949) 21–21. https://doi.org/10.1007/bf01329611.
[2]
N. Ozawa, Dixmier approximation and symmetric amenability for C*-algebras, J. Math. Sci. Univ. Tokyo. 20 (2013) 349–374.
[3]
H. Halpern, Commutators in properly infinite von Neumann algebras, Trans. Amer. Math. Soc. (1969) 55–73. https://doi.org/10.1090/s0002-9947-1969-0251546-8.
[4]
K. Dykema, A. Skripka, On single commutators in II $_{1}$ –factors, Proc. Amer. Math. Soc. (2012) 931–940. https://doi.org/10.1090/s0002-9939-2011-10953-5.

Are maximal ideals in C*-algebras closed?

Von hannesthiel | Oktober 8, 2022

0 Kommentare

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]
M.C. García, H.G. Dales, Á.R. Palacios, Maximal left ideals in Banach algebras, Bull. London Math. Soc. (2019) 1–15. https://doi.org/10.1112/blms.12290.