OA-Problems – Hannes Thiel

Tensor products with pure C*-algebras

Von hannesthiel | August 4, 2024

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

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.

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

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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.

The Blackadar-Handelman conjectures

Von hannesthiel | November 1, 2021

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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]
B. Blackadar, D. Handelman, Dimension functions and traces on C∗-algebras, Journal of Functional Analysis. (1982) 297–340. https://doi.org/10.1016/0022-1236(82)90009-x.
[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). https://doi.org/10.1515/crelle.2008.062.
[3]
K. De Silva, A note on two Conjectures on Dimension funcitons of C*-algebras, ArXiv. (2016) 1601.03475.
[4]
F. Perera, The Structure of Positive Elements for C*-Algebras with Real Rank Zero, Int. J. Math. (1997) 383–405. https://doi.org/10.1142/s0129167x97000196.
[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 hannesthiel | Juli 18, 2021

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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 hannesthiel | Dezember 20, 2020

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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]
B. Blackadar, Shape theory for $C^*$ -algebras, MATH. SCAND. 56 (1985) 249–275. https://doi.org/10.7146/math.scand.a-12100.
[2]
D. Enders, Semiprojectivity for Kirchberg algebras, ArXiv Preprint ArXiv:1507.06091. (2015).
[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]
H. Thiel, Inductive limits of projective $C$ *-algebras, J. Noncommut. Geom. 13 (2020) 1435–1462. https://doi.org/10.4171/jncg/350.