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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].
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 …
1
vote
Accepted
Certain groupoid and its $C^{*}$ algebra
I will explain how I ended up with the description given in my comment above.
Let $\{1, \dots ,n \}$ be your finite subset of $\mathbb{R}$ (only their order is important, we do not really care about …
3
votes
$C^*$-algebra which is not von Neumann, but satisfies the property that ever self-adjoint el...
Every monotone closed $C^*$-algebras has this property. And they are not all von-Neumann algebras. Although the counter-exemple are rather enormous objects.
In fact as commutative sub-algebra of $AW^ …
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 $c$ …
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 …
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 …
1
vote
Accepted
On the second dual of $C[0,1]$
It is easy to see that $\int \psi d\delta_t$ where $\delta_t$ is the Dirac mass at $t$ is $\psi(\{t\})$
So you are starting from a $T \in C([0,1])^{**}$ and you are attaching to it the function $t \m …
2
votes
C*-bimodules: the mess with definitions
As far as I'm concerned $C^*$ bimodules generally denote those you attributed to A.Connes. Such a bi-module define (by tensorization over B) a functor from $B$ $C^*$-modules to $A$ $C^*$-modules. Any …
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 …
2
votes
Comparision of two $C^{*}$ algebras associated to a non vanishing vector field on a compact ...
It depends if you use the monodromy groupoid or the holonomy groupoid (Is there a more canonical choice for this construction ? ).
Basically, the monodromy groupoid is exactly the same as the action …
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 …
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)= …
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 …
2
votes
1
answer
220
views
Non-perfect type one C^*-algebra, and a lemma in Fourier analysis
I would like to know if the following is true :
Let $\mathcal{H}$ be the complex Hilbert space $L^2([0,1])$ for the Lebesgue measure.
Let $q$ be the orthogonal projection on the subspace of $\mathcal{ …
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 …