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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].
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 obstruc …
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 …
10
votes
Accepted
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 …
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 …
8
votes
Accepted
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 …
8
votes
Accepted
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 …
6
votes
Accepted
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 …
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. …
5
votes
Accepted
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)= …
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 $c$ …
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 …
5
votes
Accepted
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 …
4
votes
0
answers
134
views
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 …
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 …
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 …