OA-Problems – Seite 2

C*-algebras complemented in their biduals

Von hannesthiel | 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 hannesthiel | Oktober 7, 2020

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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]
M. Breuer, On the homotopy type of the group of regular elements of semifinite von Neumann algebras, Math. Ann. (1970) 61–74. https://doi.org/10.1007/bf01350761.
[2]
J. Brüning, W. Willgerodt, Eine Verallgemeinerung eines Satzes von N. Kuiper, Math. Ann. (1976) 47–58. https://doi.org/10.1007/bf01354528.
[3]
H. Araki, M.-S.B. Smith, L. Smith, On the homotopical significance of the type of von Neumann algebra factors, Commun.Math. Phys. (1971) 71–88. https://doi.org/10.1007/bf01651585.
[4]
S. Popa, M. Takesaki, The topological structure of the unitary and automorphism groups of a factor, Commun.Math. Phys. (1993) 93–101. https://doi.org/10.1007/bf02100051.

Scottish Book Problem 166

Von hannesthiel | Oktober 7, 2020

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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 hannesthiel | Oktober 7, 2020

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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 hannesthiel | Oktober 7, 2020

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Given groups $H$ and $G$ , let us say that the pair $(H,G)$ has property (*) if any two injective homomorphisms $\alpha_0,\alpha_1\colon H\to G$ are conjugate if (and only if) they are pointwise conjugate.

Problem 1: Describe the class $C_1$ of groups $H$ such that $(H,G)$ has (*) for every group $G$ .
Problem 2: Describe the class $C_2$ of groups $G$ such that $(H,G)$ has (*) for every group $H$ .

Definitions: Given a group $G$ , two elements $a,b\in G$ are conjugate if there exists $x\in G$ such that $a=xbx^{-1}$ . Two homomorphisms $\alpha_0,\alpha_1\colon H\to G$ are pointwise conjugate if $\alpha_0(a)$ is conjugate to $\alpha_1(a)$ in $G$ for every $a\in H$ . (Thus, for each $a\in H$ there exists $x_a\in G$ such that $\alpha_0(a)=x_a\alpha_1(a)x_a^{-1}$ .) Further, $\alpha_0$ and $\alpha_1$ are conjugate if there exists $x\in G$ such that $\alpha_0(a)=x\alpha_1(a)x^{-1}$ for every $a\in H$ . If $\alpha_0$ and $\alpha_1$ are conjugate, then they are locally conjugate. Property (*) records that the converse holds (for injective homomorphisms).

Motivation: Problem 30 of the Scottish Book [1] (which is still open) asks to determine which groups $G$ have the following property: Pairs $(a,a')$ and $(b,b')$ in $G$ are conjugate (there exists $x\in G$ such that $a=xbx^{-1}$ and $a'=xb'x^{-1}$ ) if (and only if) for every word $w$ in two noncommuting variables the elements $w(a,a')$ and $w(b,b')$ are conjugate. Using property (*) formulated above, Problem 30 asks to determine which groups $G$ have the property that $(H,G)$ satisfies (*) for every subgroup $H$ of $G$ that is generated by two elements. The above Problem 2 is a more general (and possibly more natural) version of this problem.

An automorphism $\alpha\colon G\to G$ is class-preserving if $\alpha(a)$ is conjuagte to $a$ for every $a\in G$ . If $(G,G)$ has (*), then every class-preserving automorphism of $G$ is inner. The study of groups with (or without) outer class-preserving automorphisms has a long history; see for instance [2] and [3]. There exist finite groups with outer class-preserving automorphisms. In particular, there exist finite groups that belong neither to $C_1$ nor to $C_2$ . Note also that $C_2$ contains all abelian groups and that $C_1$ contains all cyclic groups.

[1]
R.D. Mauldin, The Scottish Book, Springer International Publishing, 2015. https://doi.org/10.1007/978-3-319-22897-6.
[2]
C.-H. Sah, Automorphisms of finite groups, Journal of Algebra. (1968) 47–68. https://doi.org/10.1016/0021-8693(68)90104-x.
[3]
M.K. Yadav, Class preserving automorphisms of finite p-groups: a survey, in: C.M. Campbell, M.R. Quick, E.F. Robertson, C.M. Roney-Dougal, G.C. Smith, G. Traustason (Eds.), Groups St Andrews 2009 in Bath, Cambridge University Press, 2007: pp. 569–579. https://doi.org/10.1017/cbo9780511842474.019.

Purely infinite rings and C*-algebras

Von hannesthiel | September 23, 2020

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Question 1: If $A_0\subseteq\mathcal{B}(H)$ is a *-subalgebra of bounded linear operators on a separable Hilbert space $H$ such that $A_0$ is purely infinite as a ring, is the norm-closure $\overline{A_0}$ purely infinite as a C*-algebra?

This is Problem 8.4 in [1]. As noted in Problem 8.6 in [1], this is even unclear if $A_0$ is a unital, simple, purely infinite ring. In the converse direction, it seems natural to ask:

Question 2: Given a purely infinite C*-algebra $A$ , is there a dense *-subalgebra $A_0\subseteq A$ that is purely infinite as a ring?

Definitions: The relation $\precsim$ on a ring $R$ is defined by setting $x\precsim y$ if there exist $a,b\in R$ such that $x=ayb$ ; see Definition 2.1 in [1]. A ring $R$ is purely infinite if no quotient of $R$ is a division ring, and if any $x,y\in R$ satisfy $x\precsim y$ if (and only if) $x\in RyR$ ; see Definition 3.1 in [1]. (Here, $RyR$ denotes the two-sided ideal generated by $y$ , that is, $RyR=\{a_1yb_1+\ldots+a_nyb_n : n\geq 1, a_i,b_i\in R\}$ .)

Given a C*-algebra $A$ , the Cuntz subequivalence relation $\precsim$ is defined by setting $x\precsim y$ if there exist sequences $(a_n)_n$ and $(b_n)_n$ in $A$ such that $\lim_{n\to\infty} \| x - a_nyb_n \| = 0$ . A C*-algebra $A$ is purely infinite if it admits no nonzero one-dimensional representations and if any $x,y\in A$ satisfy $x\precsim y$ if and only if $x$ belongs to $\overline{\mathrm{span}}AyA$ , the closed, two-sided ideal generated by $y$ ; see Definition 4.1 in [2]. (The definition in [2] only considers positive elements in $A$ , but it equivalent to the definition given here.)

Background: By Proposition 3.17 in [1], if a C*-algebra is purely infinite as a ring, then it is purely infinite as a C*-algebra. The converse does probably not hold (Remark 3.18 in [1]), which is why we ask Question 2 above. A unital, simple C*-algebra is purely infinite as a C*-algebra if and only if it is purely infinite as a ring. Thus, Question 2 has a positive answer in this case.

Given a C*-algebra $A$ , let $\mathrm{Ped}(A)\subseteq A$ denote its Pedersen ideal (the minimal dense ideal in $A$ ). By Proposition 8.5 in [1], if $\mathrm{Ped}(A)$ is purely infinite as a ring, then $A$ is purely infinite. Thus, in this particular instance, Question 1 has a positive answer.

[1]
G. Aranda Pino, K.R. Goodearl, F. Perera, M. Siles Molina, Non-simple purely infinite rings, American Journal of Mathematics. (2010) 563–610. https://doi.org/10.1353/ajm.0.0119.
[2]
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 hannesthiel | September 22, 2020

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Given a C*-algebra $A$ and a closed, two-sided ideal $I\subseteq A$ , is the image of the normal elements in $A$ under the quotient map $A\to A/I$ a closed subset of $A/I$ ?

Equivalently, if $(x_n)_n$ is a sequence of normal elements in $A/I$ that converge to $x$ , and if each $x_n$ admits a normal lift in $A$ , does $x$ admit a normal lift?

This question is raised in Example 6.5 in [1]. It is equivalent to the question of whether the commutative C*-algebra of continuous functions on the disc vanishing at zero is $\ell$ -closed in the sense of Definition 6.1 in [1].

Of related interest is the class of normal elements that are universally liftable: We say that a normal element $a$ in a C*-algebra $A$ is universally liftable if for every C*-algebra $B$ and every surjective *-homomorphism $\pi\colon B\to A$ there exists a normal element $b\in B$ with $\pi(b)=a$ . One can show that a normal element $a\in A$ is universally liftable if and only if there exists a projective C*-algebra $P$ , a normal element $b\in P$ and a *-homomorphism $\varphi\colon P\to A$ with $\varphi(b)=a$ . (A C*-algebra $P$ is projective if for every C*-algebra $D$ , every closed, two-sided ideal $I\subseteq D$ , and every *-homomorphism $\psi\colon P\to D/I$ there exists a *-homomorphism $\tilde{\psi}\colon P\to D$ such that $\pi\circ\tilde{\psi}=\psi$ , where $\pi\colon D\to D/I$ is the quotient map.) Indeed, for the forward direct, one uses that every C*-algebra is the quotient of a projective C*-algebra, and for the converse direction one applies the definition of projectivity. In particular, every normal element in a projective C*-algebra is universally liftable.

Question: Given a C*-algebra $A$ , is the set of universally liftable normal elements in $A$ closed? Is it open (relative to the set of normal elements)?

Question: Is there a projective C*-algebra $P$ and a normal element $b\in P$ such that for every C*-algebra $A$ , a normal element $a\in A$ is universally liftable if and only if there exists a *-homomorphism $\varphi\colon P\to A$ with $\varphi(b)=a$ ?

[1]
B. Blackadar, The Homotopy Lifting Theorem for Semiprojective $C^*$ -Algebras, MATH. SCAND. (2016) 291. https://doi.org/10.7146/math.scand.a-23691.

Simple, Z-stable, projectionless C*-algebras

Von hannesthiel | September 22, 2020

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Do simple, $\mathcal{Z}$ -stable, stably projectionless C*-algebras have stable rank one?

Definitions: A C*-algebra $A$ is said to be $\mathcal{Z}$ -stable if it tensorially absorbs the Jiang-Su algebra $\mathcal{Z}$ , that is, $A\cong\mathcal{Z}\otimes A$ . Further, a simple C*-algebra is projectionless if it contains no nonzero projections, and it is stably projectionless if $A\otimes\mathcal{K}$ is projectionless. (In the nonsimple case, one should require that no quotient of $A$ contains a nonzero projection – see [1].) A unital C*-algebra $A$ is said to have stable rank one if the invertible elements in $A$ are dense. A nonunital C*-algebra $A$ has stable rank one if its minimal unitization $\widetilde{A}$ does.

Background/Motivation: Rørdam showed in Theorem 6.7 in [2] that every unital, simple, finite $\mathcal{Z}$ -stable C*-algebra has stable rank one. Using this, one can show that every simple, finite, $\mathcal{Z}$ -stable C*-algebra $A$ that is not stably projectionless has stable rank one. Indeed, one may first reduce to the separable case (to show that a given element $x_0+\lambda 1\in\widetilde{A}$ with $x_0\in A$ is approximated by invertible elements, consider any separable, simple, $\mathcal{Z}$ -stable sub-C*-algebra of $A$ that is not stably projectionless and that contains $x$ .) Let $p\in A\otimes\mathcal{K}$ be a nonzero projection. By Brown’s stabilization theorem, we have $A\otimes\mathcal{K}\cong (p(A\otimes\mathcal{K})p)\otimes\mathcal{K}$ . Then the hereditary sub-C*-algebra $p(A\otimes\mathcal{K})p$ is separable, unital, simple, finite and $\mathcal{Z}$ -stable (by Corollary 3.2 in [3]) and therefore has stable rank one. Since stable rank one is invariant under stable isomorphism, it follows that $A$ has stable rank one.

Thus, a simple, $\mathcal{Z}$ -stable C*-algebra that is not stably projectionless has stable rank one if and only if it is (stably) finite. A simple, stably projectionless C*-algebra is automatically stably finite, and it is therefore natural to expect that every simple, $\mathcal{Z}$ -stable, stably projectionless C*-algebra has stable rank one.

Let $A$ be a simple, $\mathcal{Z}$ -stable, stably projectionless C*-algebra. By Corollary 3.2 in [1], $A$ almost has stable rank one, that is, every hereditary sub-C*-algebra $B\subseteq A$ satisfies $B\subseteq\overline{\mathrm{Gl}(\widetilde{B})}$ . In particular, every element in $A$ can be approximated by invertible elements in $\widetilde{A}$ . To show that $A$ has stable rank one, one would need to show that every element in $\widetilde{A}$ is approximated by invertibles. In Theorem 6.13 in [4] it is shown that $A$ has stable rank at most two, that is, the tuples $(x,y)$ in $(\widetilde{A})^2$ such that $x^*x+y^*y$ is invertible are dense in $(\widetilde{A})^2$ . One can also show:

Simple, $\mathcal{Z}$ -stable, stably projectionless C*-algebras have general stable rank one.

Here, the general stable rank of a unital C*-algebra $A$ , denoted $\mathrm{gsr}(A)$ , is the least integer $n\geq 1$ such that $\mathrm{Gl}_m(A)$ acts transitively on $\mathrm{Lg}_m(A)$ for all $m\geq n$ . (We use $\mathrm{Gl}_m(A)$ to denote the set of invertible elements in the matrix algebra $M_m(A)$ . Further, $\mathrm{Lg}_m(A)$ denotes the set of $m$ -tuples $(a_1,\ldots,a_m)\in A^m$ that generate $A$ as a left ideal, that is, such that $a_1^*a_1+\ldots+a_m^*a_m$ is invertible.) For a nonunital C*-algebra $\mathrm{gsr}(A):=\mathrm{gsr}(\widetilde{A})$ . In general, one has $\mathrm{gsr}(A)\leq\mathrm{sr}(A)+1$ , as noted in Theorem 3.3. of the overview article [5]. If $A$ is unital, then the action of $\mathrm{Gl}_m^0(A)$ , the connected component of the unit in $\mathrm{Gl}_m(A)$ , on $\mathrm{Lg}_m(A)$ has open orbits. It follows that the orbits of the action of $\mathrm{Gl}_m(A)$ on $\mathrm{Lg}_m(A)$ are also open (and hence also closed).

Let $A$ be a simple, $\mathcal{Z}$ -stable, stably projectionless C*-algebra. Let us prove that $\mathrm{gsr}(A)=1$ . By Theorem 6.13 in [4], we have $\mathrm{sr}(A)\leq 2$ , and so $\mathrm{gsr}(A)\leq 3$ . To verify that $\mathrm{gsr}(A)\leq 2$ , we need to show that $\mathrm{Gl}_2(\widetilde{A})$ acts transitively on $\mathrm{Lg}_2(\widetilde{A})$ . So let $x=(\lambda_1+a_1,\lambda_2+a_2)\in\mathrm{Lg}_2(\widetilde{A})$ with $\lambda_1,\lambda_2\in\mathbb{C}$ and $a_1,a_2\in A$ . First, we find $u\in\mathrm{Gl}_2(\mathbb{C})\subseteq\mathrm{Gl}_2(\widetilde{A})$ such that $ux=(b_1,\kappa_2+b_2)$ with $\kappa_2\in\mathbb{C}$ and $b_1,b_2\in A$ . (If $\lambda_1=0$ , we use the identity matrix; if $\lambda_2=0$ , we use the flip matrix; and if $\lambda_1\neq 0$ and $\lambda_2\neq 0$ we use the upper-triangular matrix with diagonal entries $1$ and upper right entry $-\lambda_1/\lambda_2$ .) Given $\varepsilon>0$ , we use that $A\subseteq\overline{\mathrm{Gl}(\widetilde{A})}$ to approximate $b_1$ by some invertible $\kappa_1+c\in\widetilde{A}$ such that $\| x - u^{-1}(\kappa_1+c,\kappa_2+b_2) \|<\varepsilon$ . We note that $(\kappa_1+c,\kappa_2+b_2)$ is in the orbit of $(1,0)$ , that is, there exists $v\in\mathrm{Gl}_2(\widetilde{A})$ such that $(\kappa_1+c,\kappa_2+b_2)=v(1,0)$ . Then $\| x - u^{-1}v(1,0) \|<\varepsilon$ . Since the orbits of the action of $\mathrm{Gl}_2(\widetilde{A})$ on $\mathrm{Lg}_2(\widetilde{A})$ are closed, it follows that $x=w(1,0)$ for some $w\in\mathrm{Gl}_2(\widetilde{A})$ . Finally, since $\widetilde{A}$ is finite, it follows that $\mathrm{gsr}(A)=1$ .

We remark that $\mathrm{sr}(A)=1$ implies $\mathrm{gsr}(A)=1$ , but not conversely in general.

[1]
L. Robert, Remarks on Z-stable projectionless C*-algebras, Glasgow Math. J. (2015) 273–277. https://doi.org/10.1017/s0017089515000117.
[2]
M. Rørdam, The stable and the real rank of ${\mathcal Z}$ -absorbing C*-algebras, Int. J. Math. (2004) 1065–1084. https://doi.org/10.1142/s0129167x04002661.
[3]
A.S. Toms, W. Winter, Strongly self-absorbing $C^{*}$ -algebras, Trans. Amer. Math. Soc. (2007) 3999–4029. https://doi.org/10.1090/s0002-9947-07-04173-6.
[4]
L. Robert, L. Santiago, A revised augmented Cuntz semigroup, ArXiv:1904.03690. (2019). https://arxiv.org/abs/1904.03690v1.
[5]
B. Nica, Homotopical stable ranks for Banach algebras, Journal of Functional Analysis. (2011) 803–830. https://doi.org/10.1016/j.jfa.2011.03.001.

Nonregular, simple, nuclear C*-algebras

Von hannesthiel | September 7, 2020

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(Based on the talk „Thoughts on the classification problem for amenable C*-algebras“ of George Elliott, 30. June 2020, at the Zagreb Workshop on Operator Theory.)

Let us consider the class of unital, separable, simple, nuclear C*-algebras. The Toms-Winter conjecture predicts that for such an algebra $A$ , the following conditions are equivalent:

$A$ has finite nuclear dimension.
$A$ is $\mathcal{Z}$ -stable, that is, $A\cong\mathcal{Z}\otimes A$ where $\mathcal{Z}$ denote the Jiang-Su algebra.
$A$ has strict comparison of positive elements, which means that the Cuntz semigroup of $A$ is almost unperforated: if $x$ and $y$ satisfy $(n+1)x\leq ny$ for some $n\in\mathbb{N}$ , then $x\leq y$ .

As of today, it is known that (1) and (2) are equivalent and that (2) implies (3). Further, it is known that (3) implies (2) under certain additional assumptions.

As long as the Toms-Winter conjecture is not completely verified, the second condition seems most natural, and we say that $A$ is regular if it is $\mathcal{Z}$ -stable. By the spectacular recent breakthrough in the Elliott classification program, it is known that every regular C*-algebras that satisfy the Universal Coefficient Theorem (UCT) are classified by the Elliott invariant ( $K$ -theory and tracial data).

We have an obvious dichotomy: a unital, separable, simple, nuclear C*-algebra is either regular or nonregular. It is easy to find regular algebra. Indeed, since the Jiang-Su algebra is self-absorbing, that is, $\mathcal{Z}\otimes\mathcal{Z}\cong\mathcal{Z}$ , for every $A$ the algebra $\mathcal{Z}\otimes A$ is automatically regular. Examples of nonregular C*-algebras were constructed by Villadsen, Toms and Rørdam.

Question 1:
Are the regular or the nonregular C*-algebras generic?

A regular C*-algebra is either purely infinite or stably finite. A simple, unital, purely infinite C*-algebra has infinite stable rank. Rørdam showed that a stably finite, regular C*-algebra has stable rank one. Thus, a simple C*-algebra with finite stable rank $\geq 2$ is nonregular. Villadsen showed that for every $n\in\{1,2,\ldots,\infty\}$ there exists a simple AH-algebra with stable rank $n$ . He also showed that his algebra with stable rank $n$ has either real rank $n-1$ or $n$ .

Question 2:
Does there exist a nonregular C*-algebra of real rank zero?

Question 3:
What is the relation between the stable rank and real rank of a stably finite, simple (nuclear) C*-algebra? Is the real rank of $A$ always either $\mathrm{sr}(A)$ or $\mathrm{sr}(A)-1$ ? Is this connected to the number of tracial states on $A$ ?

Elliott suggests that $\mathrm{rr}(A)=\mathrm{sr}(A)-1$ should correspond to a unique (or very few) tracial state on $A$ , while $\mathrm{rr}(A)=\mathrm{sr}(A)$ should correspond to a large tracial simplex of $A$ . It is known that $\mathrm{rr}(B)\leq 2\mathrm{sr}(B)-1$ for every C*-algebra $B$ . Moreover, if $X$ is a compact, Hausdorff space, then $\mathrm{rr}(C(X))=\dim(X)$ and $\mathrm{sr}(C(X))=\lfloor \frac{\dim(X)}{2} \rfloor +1$ . Thus, the stable rank of a commutative C*-algebra is roughly half of its real rank.

Question 4:
Can the recent classification of regular (separable, simple, nuclear) C*-algebras be extended to include (some) nonregular C*-algebras by using a stronger invariant?

A natural candidate for a stronger invariant would be the Cuntz semigroup. The Cuntz semigroup encodes the $K_0$ -group, the simplex of tracial states, and the pairing between traces and $K_0$ . However, it does not encode the $K_1$ -group. One should therefore consider classification by the pair $(\mathrm{Cu}(-),K_1(-))$ .

Elliott suggests to consider the following testcase: Let $A$ and $B$ be simple inductive limits of matrix algebras over the Hilbert cube. Then $K_1(A)=K_1(B)=0$ , and the question becomes: Are $A$ and $B$ isomorphic whenever their Cuntz semigroups are isomorphic? If $\mathrm{Cu}(A)\cong\mathrm{Cu}(B)$ , and if $A$ is regular, then the Cuntz semigroup of $B$ is `regular‘ (in the sense of Winter) and it follows that both algebras are regular. Then, since $A$ and $B$ also satisfy the UCT, one can indeed deduce from the classification result that $A\cong B$ .

Real rank of B(H) tensor B(H)

Von hannesthiel | Juni 20, 2020

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Given a separable, infinite-dimensional Hilbert space $\Huge{H}$ , what is the real rank of the minimal tensor product $\mathcal{B}(H)\otimes \mathcal{B}(H)$ ?

Background/Motivation: The real rank is a noncommutative dimension theory that was introduced by Brown and Pedersen in [1]. It associates to each C*-algebra $A$ a number (its real rank) $\mathrm{rr}(A)\in\{0,1,\ldots,\infty\}$ . The lowest and most interesting value is zero. One can think of C*-algebras of real rank zero as zero-dimensional, noncommutative spaces. (There are other noncommutative dimension theories, and they lead to different concepts of zero- or low-dimensional noncommutative spaces.) For topological spaces $X$ and $Y$ , the product theorem for covering dimension shows that under certain weak assumptions (for example, both spaces are metric, or compact) we have $\dim(X\times Y)\leq\dim(X)+\dim(Y)$ . One might therefore expect that $\mathrm{rr}(A\otimes B)\leq\mathrm{rr}(A)+\mathrm{rr}(B)$ , but it was shown in [2] that this does not hold: there exists C*-algebras $A$ and $B$ such that

$\mathrm{rr}(A)=\mathrm{rr}(B)=0, \text{ yet } \mathrm{rr}(A\otimes B)>0.$

Later, Osaka showed in [3] that one can even take $A=B=\mathcal{B}(H)$

. Indeed, it is well known that von Neumann algebras have real rank zero, and hence so does $\mathcal{B}(H)$

. Osaka showed that $\mathcal{B}(H)\otimes\mathcal{B}(H)$

does not have real rank zero. This raises the question of what the real rank of $\mathcal{B}(H)\otimes\mathcal{B}(H)$

is, see Question 3.3 in [3].

[1]
L.G. Brown, G.K. Pedersen, C*-algebras of real rank zero, Journal of Functional Analysis. 99 (1991) 131–149. https://doi.org/10.1016/0022-1236(91)90056-b.
[2]
K. Kodaka, H. Osaka, Real Rank of Tensor Products of C*-Algebras, Proceedings of the American Mathematical Society. 123 (1995) 2213–2215. https://doi.org/10.2307/2160959.
[3]
H. Osaka, Certain C*-algebras with non-zero real rank and extremal richness, MATH. SCAND. 85 (1999) 79. https://doi.org/10.7146/math.scand.a-13886.