40
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 ...
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
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
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
Accepted
Can $C^*$-algebra of continuous functions on $R^n$ ($S^n$) be characterized alternatively?
Yes, it can be defined as the univeral commutative $C^*$-algebra with unit, generated by $n+1$ self adjoint elements $x_1,...,x_{n+1}$ subject to the relation $x_1^2+...+x_{n+1}^2=1$. Here universal ...
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 ...
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