OA-Problems – Hannes Thiel

Commutators and square-zero elements in C*-algebras

Von hannesthiel | April 14, 2024

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?

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.

[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.

C*-algebras complemented in their biduals

Von hannesthiel | Dezember 20, 2020

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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 $A^{**}$ .