Search Results
Search type | Search syntax |
---|---|
Tags | [tag] |
Exact | "words here" |
Author |
user:1234 user:me (yours) |
Score |
score:3 (3+) score:0 (none) |
Answers |
answers:3 (3+) answers:0 (none) isaccepted:yes hasaccepted:no inquestion:1234 |
Views | views:250 |
Code | code:"if (foo != bar)" |
Sections |
title:apples body:"apples oranges" |
URL | url:"*.example.com" |
Saves | in:saves |
Status |
closed:yes duplicate:no migrated:no wiki:no |
Types |
is:question is:answer |
Exclude |
-[tag] -apples |
For more details on advanced search visit our help page |
Algebras of operators on Hilbert space, $C^*-$algebras, von Neumann algebras, non-commutative geometry
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 …
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{ …
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 …
6
votes
Accepted
What is the Gelfand dual of an open surjection?
After more thought, I think the correct statement is the following:
Theorem: Let $\pi : X \to Y$ be a continuous map between compact Hausdorff spaces. Then the following condition are equivalents:
(a) …
7
votes
1
answer
230
views
Free extension of algebra for an operad
I fix $C$ a symmetric monoidal model category (with a cofibrant unit if it helps). I'm assuming that it is closed, or at least that the tensor product commutes to colimits in each variable.
If $X$ is …
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 …
12
votes
Accepted
Can one associate a "nice" topos to a von Neumann algebra?
(I'm going to be a bit informal to be able to go to the point relatively directly, but if you want more details on some specific aspect. I can try to add them)
Toposes are closely related to topologic …
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. …
2
votes
KMS-states of Bost-Connes type system
$\pi$ is an irreducible representation. So because of Schur lemma and the double commutant theorem the map $\pi:A \rightarrow B(H)$ is surjective.
Now the state $\phi$ is defined as a normale state o …
2
votes
Does every integer map generate a von Neumann algebra of type I?
I think I have an example:
Precisely, I will construct an integer function $m$ such that $M$ is bounded and the algebra $\mathcal{M}$ contains a corner which is the von Neuman algebra completion of a …
5
votes
Route to Alain Connes'work about classification of injective factors
I would strongly recommend to have a look (and to always keep available) to the books by Takesaki "The theory of Operator algebras" I,II and III
You might not need to know everything that is in these …
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 …
7
votes
1
answer
479
views
Two notions of bundle of C* algebras
One can find in the literature two notions of $C^*$-algebra over a topological space $X$.
The first is as the data of an open surjection $ \pi: B \rightarrow X$ together with the structure of a $C^*$ …
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 …
8
votes
1
answer
175
views
$\mathcal{O}_{\infty}$ and $\mathcal{Z}$ stable isomorphism as equivalence
If $O$ is a (strongly ?) self absorbing $C^*$-algebra, one has an equivalence relation on (separable) $C^*$-algebras: "being $O$-stably isomorphic" i.e. $A$ and $B$ are $O$-stably isomorphic if and on …