Question: Let be a C*-algebra that is complemented in its bidual by a *-homomorphism, that is, there exists a *-homomorphism such that for all . Is a von Neumann algebra?

The converse is true: Let be a von Neumann algebra. Then has a (unique) isometric predual . Let be the natural inclusion of the Banach space in its bidual. We naturally identify the dual of with , and the dual of with . Then the transpose is a *-homomorphism that complements in .

Question: Let be a -factor. Is the unitary group contractible when equipped with the strong operator topology?

Background/Motivation: By Kuiper’s theorem, if is an infinite-dimensional Hilbert space, then the unitary group of the -factor is contractible in the norm topology. This was generalized by Breuer [1] to (certain) and 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 is not contractible. Similarly, the unitary group of a factor is not contractible in the norm topology since its fundamental group does not vanish – in fact, it was shown in [3] that for every factor .

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 is contractible in the strong operator topology if is a separable factor such that the associated factor admits a trace scaling one-parameter group of automorphisms. This includes all McDuff factors ( is McDuff if for the hyperfinite factor ) and all factors that satisfy , where 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.

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, and should be switched in the last sentence) the problem is:

Let be a topological manifold, and let be a continuous function. Let denote the subgroup of homeomorphisms that satisfy . Let be another manifold that is not homeomorphic to . Does there exist a continuous function such that is not isomorphic to ?

This is one of the few problems from the Scottish book that are still open. In modern terminology, the problem is:

Let and be Banach spaces, and let be a bijective map with the following property: For every there exists such that for the sphere , the restriction is isometric. Does it follow that is isometric?

It is noted in the Scottish Book that the answer is „yes“ whenever is continuous, which is automatic if is finite-dimensional, or if has the property that for any two elements satisfying and there exists such that .

Given groups and , let us say that the pair has property (*) if any two injective homomorphisms are conjugate if (and only if) they are pointwise conjugate.

Problem 1: Describe the class of groups such that has (*) for every group .

Problem 2: Describe the class of groups such that has (*) for every group .

Definitions: Given a group , two elements are conjugate if there exists such that . Two homomorphisms are pointwise conjugate if is conjugate to in for every . (Thus, for each there exists such that .) Further, and are conjugate if there exists such that for every . If and 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 have the following property: Pairs and in are conjugate (there exists such that and ) if (and only if) for every word in two noncommuting variables the elements and are conjugate. Using property (*) formulated above, Problem 30 asks to determine which groups have the property that satisfies (*) for every subgroup of that is generated by two elements. The above Problem 2 is a more general (and possibly more natural) version of this problem.

An automorphism is class-preserving if is conjuagte to for every . If has (*), then every class-preserving automorphism of 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 nor to . Note also that contains all abelian groups and that contains all cyclic groups.

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.

Question 1: If is a *-subalgebra of bounded linear operators on a separable Hilbert space such that is purely infinite as a ring, is the norm-closure 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 is a unital, simple, purely infinite ring. In the converse direction, it seems natural to ask:

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

Definitions: The relation on a ring is defined by setting if there exist such that ; see Definition 2.1 in [1]. A ring is purely infinite if no quotient of is a division ring, and if any satisfy if (and only if) ; see Definition 3.1 in [1]. (Here, denotes the two-sided ideal generated by , that is, .)

Given a C*-algebra , the Cuntz subequivalence relation is defined by setting if there exist sequences and in such that . A C*-algebra is purely infinite if it admits no nonzero one-dimensional representations and if any satisfy if and only if belongs to , the closed, two-sided ideal generated by ; see Definition 4.1 in [2]. (The definition in [2] only considers positive elements in , 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 , let denote its Pedersen ideal (the minimal dense ideal in ). By Proposition 8.5 in [1], if is purely infinite as a ring, then 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.

Given a C*-algebra and a closed, two-sided ideal , is the image of the normal elements in under the quotient map a closed subset of ?

Equivalently, if is a sequence of normal elements in that converge to , and if each admits a normal lift in , does 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 -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 in a C*-algebra is universally liftable if for every C*-algebra and every surjective *-homomorphism there exists a normal element with . One can show that a normal element is universally liftable if and only if there exists a projective C*-algebra , a normal element and a *-homomorphism with . (A C*-algebra is projective if for every C*-algebra , every closed, two-sided ideal , and every *-homomorphism there exists a *-homomorphism such that , where 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 , is the set of universally liftable normal elements in closed? Is it open (relative to the set of normal elements)?

Question: Is there a projective C*-algebra and a normal element such that for every C*-algebra , a normal element is universally liftable if and only if there exists a *-homomorphism with ?

Do simple, -stable, stably projectionless C*-algebras have stable rank one?

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

Background/Motivation: Rørdam showed in Theorem 6.7 in [2] that every unital, simple, finite -stable C*-algebra has stable rank one. Using this, one can show that every simple, finite, -stable C*-algebra that is not stably projectionless has stable rank one. Indeed, one may first reduce to the separable case (to show that a given element with is approximated by invertible elements, consider any separable, simple, -stable sub-C*-algebra of that is not stably projectionless and that contains .) Let be a nonzero projection. By Brown’s stabilization theorem, we have . Then the hereditary sub-C*-algebra is separable, unital, simple, finite and -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 has stable rank one.

Thus, a simple, -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, -stable, stably projectionless C*-algebra has stable rank one.

Let be a simple, -stable, stably projectionless C*-algebra. By Corollary 3.2 in [1], almost has stable rank one, that is, every hereditary sub-C*-algebra satisfies . In particular, every element in can be approximated by invertible elements in . To show that has stable rank one, one would need to show that every element in is approximated by invertibles. In Theorem 6.13 in [4] it is shown that has stable rank at most two, that is, the tuples in such that is invertible are dense in . One can also show:

Simple, -stable, stably projectionless C*-algebras have general stable rank one.

Here, the general stable rank of a unital C*-algebra , denoted , is the least integer such that acts transitively on for all . (We use to denote the set of invertible elements in the matrix algebra . Further, denotes the set of -tuples that generate as a left ideal, that is, such that is invertible.) For a nonunital C*-algebra . In general, one has , as noted in Theorem 3.3. of the overview article [5]. If is unital, then the action of , the connected component of the unit in , on has open orbits. It follows that the orbits of the action of on are also open (and hence also closed).

Let be a simple, -stable, stably projectionless C*-algebra. Let us prove that . By Theorem 6.13 in [4], we have , and so . To verify that , we need to show that acts transitively on . So let with and . First, we find such that with and . (If , we use the identity matrix; if , we use the flip matrix; and if and we use the upper-triangular matrix with diagonal entries and upper right entry .) Given , we use that to approximate by some invertible such that . We note that is in the orbit of , that is, there exists such that . Then . Since the orbits of the action of on are closed, it follows that for some . Finally, since is finite, it follows that .

We remark that implies , but not conversely in general.

(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 , the following conditions are equivalent:

has finite nuclear dimension.

is -stable, that is, where denote the Jiang-Su algebra.

has strict comparison of positive elements, which means that the Cuntz semigroup of is almost unperforated: if and satisfy for some , then .

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 is regular if it is -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 (-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, , for every the algebra 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 is nonregular. Villadsen showed that for every there exists a simple AH-algebra with stable rank . He also showed that his algebra with stable rank has either real rank or .

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 always either or ? Is this connected to the number of tracial states on ?

Elliott suggests that should correspond to a unique (or very few) tracial state on , while should correspond to a large tracial simplex of . It is known that for every C*-algebra . Moreover, if is a compact, Hausdorff space, then and . 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 -group, the simplex of tracial states, and the pairing between traces and . However, it does not encode the -group. One should therefore consider classification by the pair .

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

Given a separable, infinite-dimensional Hilbert space , what is the real rank of the minimal tensor product ?

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 number (its real rank) . 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 and , the product theorem for covering dimension shows that under certain weak assumptions (for example, both spaces are metric, or compact) we have . One might therefore expect that , but it was shown in [2] that this does not hold: there exists C*-algebras and such that

Later, Osaka showed in [3] that one can even take . Indeed, it is well known that von Neumann algebras have real rank zero, and hence so does . Osaka showed that does not have real rank zero. This raises the question of what the real rank of is, see Question 3.3 in [3].

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.