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 | |
Results tagged with c-star-algebras
Search options not deleted
user 24916
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].
6
votes
Almost idempotent approximate units in C*-algebras
Akemann has constructed a C*-algebra that does not contain an approximate unit of commuting elements. See Example 2.1 in
Akemann. Approximate units and maximal abelian C*-subalgebras. Pacific J. Math …
- 3,497
4
votes
Is the space of *-homomorphisms between two $C^*$-algebras locally path connected
If $A$ is semiprojective (the C*-analog of an absolute neighborhood retract, ANR), then your question also has a positive answer. The class of semiprojective C*-algebras includes also many non-commuta …
- 3,497
1
vote
All AI-algebras are AT-algebras
Contrary to what you said, the building block of an AI-algebra, namely $C([0,1],M_n)$, can naturally be embedded into the building block of an AT-algebra, namely $C(\mathbb{T},M_n)$, and not conversel …
- 3,497
9
votes
Is every maximal ideal in a C*-algebra always closed?
Theorem: Let $J\subseteq A$ be a maximal ideal. Then $J$ is hereditary (if $a\in A_+$ satisfies $a\leq b$ for some $b\in J_+$, then $a\in J_+$), strongly invariant (if $x^*x\in J_+$ then $xx^*\in J_+$ …
- 3,497
9
votes
0
answers
120
views
Real Rank of $M_n(A)$
The real rank for C*-algebras was defined by Brown-Pedersen in [1] as a noncommutative analog of covering dimension. Given a unital C*-algebra $A$, its real rank $\mathrm{rr}(A)$ is the smallest natur …
- 3,497
4
votes
Classification of $C^*$ algebras whose all non scalar elements have disconnected spectrum
If $K$ is compact, extremally disconnected and infinite, then for every compact subset $D\subseteq\mathbf{C}$ there exists $a\in C(K)$ with spectrum $\sigma(a)=D$.
[edit: Previously, $K$ was only ass …
- 3,497
8
votes
1
answer
280
views
Commutator ideal in nonunital C*-algebra
Let $A$ be a C*-algebra that has no one-dimensional irreducible representations, that is, there is no (closed) two-sided ideal $I\subseteq A$ such that $A/I\cong\mathbb{C}$.
Let $J$ denote the (not ne …
- 3,497
6
votes
Accepted
Finite codimensional subvector space of $C^{*}$ algebras which contains no invertible elements
Let $A$ be a unital C*-algebra.
As was already noted in the comments, we have $n(A)=1$ if and only if $A$ has a character.
Let $M_n=M_n(\mathbb{C})$.
It is easy to see that $n(M_n)\leq n$. Conversely …
- 3,497
4
votes
C$^*$-algebras isomorphic after tensoring
The answer to question 1 is `yes': Let $A$ and $B$ be any $C^*$-algebras. Let $N$ be a simple $C^*$-algebra of such high cardinality that it does not embed into either $A$ or $B$. Then take $C:=N\oplu …
- 3,497
8
votes
Which C*-algebras are complemented in their bidual?
(1) If $A$ is $1$-complemented in its bidual, then $A$ is an AW*-algebra.
Indeed, assume that $A$ is $1$-complemented in its bidual, via a contractive projection $p\colon A^{**}\to A$. By a theorem o …
- 3,497
3
votes
The closure of selfadjoint elements of an algebra whose spectrum consist of rational numbers
A unital C*-algebra has real rank zero if the self-adjoint elements with finite spectrum are norm-dense in the set of all self-adjoints. If a self-adjoint element has finite spectrum, then it is of th …
- 3,497
15
votes
2
answers
1k
views
Is a C*-algebra with an isomorphic predual a von Neumann algebra?
It is well-known that a C*-algebra $A$ is a von Neumann algebra if and only if it has an isometric predual, that is, if and only if there exists a Banach space $X$ such that $A$ is isometrically isomo …
- 3,497
2
votes
Accepted
Totally non hereditary $C^{*}$-subalgebras
So you want to consider a C*-algebra $A$ with the following property: Every sub-C*-algebra of $A$ is isomorphic to a hereditary sub-C*-algebra of $A$.
We can distinguish two cases:
If $A$ is finite …
- 3,497