42
votes

### What is the current status of the Kaplansky zero-divisor conjecture for group rings?

Apologies for the self-promotion, but there is now a counterexample to the unit conjecture (U) with $K=\mathbb{F}_2$ and virtually abelian $G = \langle a, b \,|\, (a^2)^b=a^{-2}, (b^2)^a=b^{-2} \...

23
votes

Accepted

### Why C*-algebras is not as popular as other areas of pure mathematics?

One way to tell how active a field is is by looking at what's appearing on the arXiv in that area. I think that will show you that operator algebra is a robust subject with a lot of activity.
In the ...

Community wiki

20
votes

Accepted

### Are algebraically isomorphic $C^*$-algebras $*$-isomorphic?

Answering the question in the body of the original post, which seems to be more restricted than the implicit question in the title of the post....
The answer is YES. See
L. Terrell Gardner, On ...

18
votes

Accepted

### Non-commutative duality I: Which C*-algebras are (isomorphic to a) convolution algebra?

The main obstruction to this kind of duality is not so much that not every $C^*$-algebra is a convolution algebra (though, at least if we don't use twisted convolution algebra, there are known ...

17
votes

Accepted

### Is this a characterization of commutative $C^{*}$ algebras?

Yes. I will show that any two positive elements of $A$ commute. Since every element is a linear combination of positive elements, this suffices.
Say $a$ and $b$ are positive. Then $a^{1/2}ba^{1/2} \...

17
votes

Accepted

### Maximal ideals of ultraproducts of full matrix algebras

I think Nik Weaver is right that the ideal mentioned is the unique maximal ideal.
This simultaneously answers both questions (since the quotient is clearly infinite dimensional). Let $\tau$ be the ...

14
votes

Accepted

### On equation $e^{xy-yx}=e^xe^ye^{-x}e^{-y}$ in $C^*$ algebras

Yes:
A $C^*$-algebra satisfies the identity $e^{[xy-yx]}=e^xe^ye^{-x}e^{-y}$ iff it is commutative.
This follows from two independent facts (I write $[x,y]=xy-yx$)
1) A (real/complex) unital ...

14
votes

Accepted

### Amenable action intuition

It is impossible to understand the motivation behind the definition of an amenable action without first understanding the definition of amenable groups, so let me first talk about groups (for ...

13
votes

### Which $\ast$-algebras are $C^\ast$-algebras?

Given an algebra $A$, one can ask whether it has a unit. If one exists, one then shows it is unique: $1_A = 1_A1_A' = 1_A'$. Thus being unital is a property of an algebra and not extra structure. ...

13
votes

Accepted

### For what kind of $C^*$ algebras does the inequality $\frac{(ab+ba)}{2}\leq\frac{ a^p}{p} +\frac{b^q }{q}$ hold for $a,b>0$?

Let me expand slightly on the comments I made above, and give the most general solution.
Clearly the inequality $\frac{ab + ba}{2} \leq \frac{a^2}{2} + \frac{b^2}{2}$ holds for all positive elements $...

13
votes

### Linear map between projective finitely generated Hilbert modules is adjointable

This is a comment on the definition of being "finitely generated".
There is a difference between algebraically finitely generated and topological finitely generated Hilbert $C^*$-modules.
...

12
votes

### In which sense the GNS-construction is a functor?

Although this response is a bit late, perhaps this perspective may help nonetheless. It only addresses the question about functoriality of the GNS construction.
The GNS construction is not quite a ...

12
votes

Accepted

### A non nuclear $C^*$ algebra $A$ for which the algebraic tensor product $A\otimes A$ admits a unique $C^*$ norm

Pisier https://arxiv.org/abs/1908.02705 very recently constructed a non-nuclear $C^\ast$-algebra $A$ with the weak expectation property (WEP) and the local lifting property (LLP). By a celebrated ...

12
votes

### Vector-Valued Stone-Weierstrass Theorem?

I think that you want something like this:
Let $E\to X$ be a (finite rank) vector bundle over a compact, Hausdorff topological space $X$, let $\mathcal{A}\subset C(X,\mathbb{R})$ be a subalgebra that ...

12
votes

### Non-commutative duality I: Which C*-algebras are (isomorphic to a) convolution algebra?

As already pointed out, Buss and Sims have found an example of a $C^*$-algebra which is not isomorphic to its opposite, and
hence it is not a groupoid $C^*$-algebra. However twisted groupoid $C^*$-...

12
votes

Accepted

### Faithful traces on quasi-diagonal C*-algebras

No, separable (unital) quasi-diagonal $C^\ast$-algebras do not necessarily admit a faithful tracial state. For instance, the $C^\ast$-algebra
\begin{equation}
A= \{ f\in C([0,1], \mathcal O_2) : f(0) \...

12
votes

### Simplicity of group $C^\ast$-algebra implies fullness of group-von Neumann algebra?

No, whenever $\Gamma$ is an infinite direct product of C$^*$-simple groups, we obtain a counterexample. For instance, taking $\Gamma = \mathbb{F}_2^{(\mathbb{N})}$ to be the direct sum of infinitely ...

11
votes

Accepted

### Kazhdan's property (T) vs. residual finiteness

Rufus Willett and Guoliang Yu, MR 3246936 Geometric property (T), Chin. Ann. Math. Ser. B 35 (2014), no. 5, 761--800. showed that if a finitely generated group is residually finite and finite ...

11
votes

Accepted

### Is a C*-algebra with an isomorphic predual a von Neumann algebra?

Via my colleague Garth Dales, some observations which answer your question in the negative, even in the abelian case:$\newcommand{\N}{{\mathbb N}}$
We know that $K$ is hyper-Stonean iff $C(K)$ is ...

11
votes

Accepted

### Generator of $K_0(C_0(\mathbb{C}))$

The group $K_0(C_0(\mathbb{C}))$ is generated by by the class $[p_{Bott}] - [1]$ where $p_{Bott} \in M_2(C_0(\mathbb{C})^\sim)$ is the so-called "Bott projection" given by
$$
p_{Bott}(z) = \frac{1}{1+...

11
votes

Accepted

### $*$-algebras, completions, and $K$-theory

Any infinite discrete group $\Gamma$ with Kazhdan's property (T) gives an example. Since it is not amenable, the full and reduced C*-algebras (which are both completions of the group algebra) do not ...

11
votes

Accepted

### The double dual of the unitization of a $C^*$-algebra

Believe it or not, these are $*$-isomorphisms as C${}^*$-algebras. If $J$ is a closed two-sided ideal of $B$ then $J^{**}$ is a weak* closed two-sided ideal of $B^{**}$, and every weak*-closed two-...

11
votes

### Two inequalities in $C^*$ algebras

As observed, the quadratic term may be equivalently removed from the inequality due to different homogeneity; then $x^*a^*ax+a^*x^*xa\leq a^*x^*ax+x^*a^*xa$ can be rewritten $[a,x]^*[a,x]\le0$, so ...

11
votes

Accepted

### Impact of annihilators in C*-algebras

An AW${}^*$-algebra is a C${}^*$-algebra which satisfies this condition for both right and left annihilators. So every AW${}^*$-algebra has your property, and any C${}^*$ algebra that is isomorphic to ...

10
votes

### Multiplier algebra of $A \otimes \mathcal{K}$

The fact stated in the answer by vap is proven in the paper "Multipliers of C*-algebras" by Akemann, Pedersen and Tomiyama (see Theorem 3.3, I guess). Moreover, they prove in Theorem 3.8 that ...

10
votes

Accepted

### Linear independency and compactness of the set of pure states of a $C^*$-algebra

As pointed out in the comments, the answer to question 1 is "no". More explicitly, if $v$ is any unit vector in $\mathbb{C}^2$ then the map $A \mapsto \langle Av,v\rangle$ is a pure state on $M_2$, ...

10
votes

### C*-algebras: Existence of an element inducing an injective map

In any Banach space $B$, if $S$ is a countable set of bounded linear maps, there is $a \in B$ such that $S\ni T \mapsto Ta \in B$ is injective.
This follows easily from the Baire Category Theorem: $B$...

10
votes

Accepted

### Behaviour of direct limit with matrices

You’re asking whether the functor $M_2$ on Banach spaces preserves colimits of direct sequences.
(In case you’re not familiar with the categorical terminology I’m using here, don’t be put off — it’s ...

10
votes

### What does it mean for a category to admit direct integrals?

This is a question that I have been working on recently together with Robert Furber and Bas Westerbaan. Let me sketch what we know so far, starting with the case of (infinite) direct sums, treated in ...

10
votes

### Which $\ast$-algebras are $C^\ast$-algebras?

A good way of figuring out whether a definition is good or not, is to see how it survives generalisations.
The notion of $C^*$-algebra admits a couple of interesting generalisations:
Real $C^*$-...

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