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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].

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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 obstruc …
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10 votes

Is this a functor on the category of $C^{*}$ algebras?

This paper proves that there is no functor from the category of C* algebra (and morphisms) to the category of locales/topological spaces/a lot of other things that extend the gelfand duality and send …
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10 votes
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A C*-algebra enjoying some different C*-norms

No that's not possible (except the trivial case). Any $*$-homomorphism between $C^*$-algebras is automatically contractive, and if it is injective then it is isometric. You can apply this to the ident …
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10 votes

Strong Morita Equivalence and Morphisms Between $ C^{*} $-Algebras

You cannot says anything about morphisms between $A$ and $B$ in general, but from the two bi-modules you can construct a third algebra $C$, such that both $A$ and $B$ embeds into $C$ with the embeddin …
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8 votes
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A $C^{*}$ algebra associated to a group

Because the group is compact one can assume the representation is isometric and the Haar measure is normalized. In this situation, $T$ is just the orthogonal projection on the space of $G$-invariant v …
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8 votes
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How the modular theory of von Neumann algebras, deal with generating C*-algebras?

All groups algebra are trivial examples because for them the modular time evolution is (can be chosen) trivial: the modular time evolution attached to a state is trivial if and only if the state is a …
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6 votes
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The Gelfand duality for pro-$C^*$-algebras

The answer is No. Rougly, because it is not a good idea to look at continuous $\mathbb{C}$ valued function on a space which is not completely Haussdorff as completely haussdorf is exactly the hypothes …
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6 votes

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

The following is an argument for showing that "having direct integral" is definitely not a property nor a "property-like-structure" of $W^*$-categegories, but a real, non-trivial additional structure. …
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5 votes
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Counterexample to Riesz representation for Hilbert modules

Take $A= \mathcal{C}([0,1])$ and $H$ the ideal of $A$ of functions that vanish at $0$. $H$ is a Hilbert $A$ module (as any ideal, with the natural multiplication of $A$ and the scalar product $(x,y)= …
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5 votes

Relation between norm of any element of $C^*$-algebra in terms of self adjoint elements

The norm of the self adjoint part is not sufficient: Even if $b$ and $c$ commutes (which should be the easiest case) then it corresponds (by looking at the commutative algebra generated by $b$ and $c$ …
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5 votes

unitization-process of unital- and non-unital $C^*$-algebras

The question is already answer but there is a point I want to add: Some time ago I wrote a paper about the Gelfand duality for non-unital algebra within constructive mathematics, my proof goes throug …
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5 votes
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A coproduct of $C^\ast$-algebras

Given two locally compact spaces $X$ and $Y$ then the product $X \times Y$ is an open subset inside $\overline{X} \times \overline{Y}$, where $\overline{X}$ and $\overline{X}$ are the one point compac …
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4 votes
0 answers
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References for a lemma about compact operators on a Hilbert module

I am looking for a reference for the following result: If $A$ and $B$ are C* algebras, $H$ is a right Hilbert $A$-modules, $\phi :A \rightarrow B$ is a morphism, and assume that there is a map $\eta …
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4 votes

A Possible characterization of F.D or AF commutative $C^{*}$ algebras

I don't know what are the motivation for the formulaton in terms of $C^*$ algebras, but you are essentially asking for hausdorff compact/locally compact spaces such all their compact/locally compact q …
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4 votes

Noncommutative version of Littlewood's First Principle

Here is one half of an answer, but it was too long for a comment. I'll edit if I manage to finish it... When you have a $C^*$ algebra $C$ included in a von Neumann algebra $B$ ( and you might want to …
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