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A C*-algebra is a complex Banach algebra together with an isometric antilinear involution satisfying (a b)* = b* a* and the C*-identity ‖a* a‖ = ‖a‖². Related tags: [banach-algebras], [von-neumann-algebras], [operator-algebras], [spectral-theory].

8 votes
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Residually finite-dimensional $C^*$-algebra

By "trace" I assume you mean tracial state, and in that case the answer is "not necessarily". A counter example is produced below. The goal is this: we construct an unital RFD $C^\ast$-algebra $B$ an …
Jamie Gabe's user avatar
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10 votes
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Endomorphisms of the Cuntz algebra

This is true: $\mathcal O_n$ is singly generated, i.e. there exists $x\in \mathcal O_n$ such that $C^\ast(x) = \mathcal O_n$. In particular, if $\phi, \psi \colon \mathcal O_n \to B$ are $\ast$-homomo …
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12 votes
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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) …
Jamie Gabe's user avatar
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3 votes
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Question on Cuntz' proof of Bott periodicity

The conclusion drawn in the book of Wegge-Olsen is wrong (explained below), but can, however, easily be tweaked to a correct proof. What is shown is that $j\circ q$ is homotopic to the identity on $\m …
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5 votes
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Is there a C*-algebra whose Pedersen ideal is not proper?

Examples include all non-unital algebraically simple $C^\ast$-algebras. By [Blackadar, Bruce E.; Cuntz, Joachim The structure of stable algebraically simple C∗-algebras. Amer. J. Math. 104 (1982), no. …
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  • 2,471
2 votes

Pure infiniteness of tensor product $C^\ast$-algebras

Kirchberg announced quite a lot of years ago that $A\otimes_{\min{}} B$ is strongly purely infinite if $A$ is strongly purely infinite and $B$ is exact. Unfortunately, I don't think this result has ev …
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6 votes
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Approximating a projection by a sum of elementary tensors with a certain property

Yes, this is possible, assuming that $A$ (or $B$) is nuclear. The same argument below (using that exact $C^\ast$-algebras are locally reflexive) also works if $A$ or $B$ is exact and the tensor produc …
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5 votes
Accepted

Primitive ideals of minimal tensor product

I expect that this answer is satisfactory, although it isn't a complete answer. This is really the best result one can hope for. For a $C^\ast$-algebra $A$ let $Prime(A)$ be the prime ideal space (def …
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8 votes
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Monotone approximation of elements in AF-algebras

No, this is not possible in general. $A^{LS}$ might have trivial intersection with a non-zero hereditary $C^\ast$-subalgebra of $A$, and thus any non-zero positive element in such a hereditary $C^\ast …
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5 votes
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Is $N_\phi = \{x \in E: \phi(\langle x,x\rangle)=0\}$ a Hilbert submodule of $E$?

It is not true. Take $B= M_2(\mathbb C)$ (with standard matrix units $e_{i,j}$), $E= B$ as a Hilbert $B$-module in the usual way, and let $\phi \in B^\ast$ be compression to the $(1,1)$-corner. Then …
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5 votes
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Two densely defined traces on a $C^*$-algebra coinciding on a dense subalgebra are equal

Yes, this is true. Let $a\in A_+$. By lower semicontinuity it suffices to show that $t_1((a-\delta)_+) = t_2((a-\delta)_+)$ for all $\delta>0$ (where $(a-\delta)_+$ is the positive part of $a-\delta 1 …
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7 votes
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Is the reduced group $C^*$-algebra quasidiagonal

This is true for countable discrete groups by the celebrated Quasidiagonality Theorem of Tikuisis-White-Winter (Quasidiagonality of nuclear C∗-algebras. Ann. of Math. (2) 185 (2017), no. 1, 229–284.)
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13 votes
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For what kind of $C^*$ algebras does the inequality $\frac{(ab+ba)}{2}\leq\frac{ a^p}{p} +\f...

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 $ …
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7 votes
Accepted

multiplier algebra of a simple $C^*$ algebra

If $A$ is a $\sigma$-unital, simple, non-unital $C^\ast$-algebra, then $M(A)$ is non-exact (in particular, it is non-nuclear). The following argument is modelled after Yemon's idea in the comments abo …
Jamie Gabe's user avatar
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7 votes

Morita-invertible C*-algebras

I know my answer is coming a bit late, but the answer to your question is: yes. If $A$ is a $C^\ast$-algebra, and there exists a $C^\ast$-algebra $B$ such that $A\otimes_\alpha B$ is strongly Morita …
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