Example of norm separable c-star algebras [closed]
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I want to know enough examples of norm separable $C^{*}$-algebras which are neither finite dimensional nor commutative.
operator-algebras c-star-algebras von-neumann-algebras
asked Dec 12 '18 at 11:14
mathlovermathlover
130110
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closed as off-topic by Saad, Martin R, Brahadeesh, Ben, Cesareo Dec 14 '18 at 10:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Martin R, Brahadeesh, Ben, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I want to know enough examples of norm separable $C^{*}$-algebras which are neither finite dimensional nor commutative.
operator-algebras c-star-algebras von-neumann-algebras
asked Dec 12 '18 at 11:14
mathlovermathlover
130110
$endgroup$
closed as off-topic by Saad, Martin R, Brahadeesh, Ben, Cesareo Dec 14 '18 at 10:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Martin R, Brahadeesh, Ben, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
I want to know enough examples of norm separable $C^{*}$-algebras which are neither finite dimensional nor commutative.
operator-algebras c-star-algebras von-neumann-algebras
asked Dec 12 '18 at 11:14
mathlovermathlover
130110
$endgroup$
I want to know enough examples of norm separable $C^{*}$-algebras which are neither finite dimensional nor commutative.
operator-algebras c-star-algebras von-neumann-algebras
operator-algebras c-star-algebras von-neumann-algebras
asked Dec 12 '18 at 11:14
mathlovermathlover
130110
asked Dec 12 '18 at 11:14
mathlovermathlover
130110
asked Dec 12 '18 at 11:14
mathlovermathlover
130110
asked Dec 12 '18 at 11:14
mathlovermathlover
130110
asked Dec 12 '18 at 11:14
mathlovermathlover
130110
130110
closed as off-topic by Saad, Martin R, Brahadeesh, Ben, Cesareo Dec 14 '18 at 10:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Martin R, Brahadeesh, Ben, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, Martin R, Brahadeesh, Ben, Cesareo Dec 14 '18 at 10:23
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Saad, Martin R, Brahadeesh, Ben, Cesareo
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Your desire is somewhat vague since you do not write down your motivation.
Did you already browse through K. R. Davidson's book
$,C^*$-Algebras by Example ? (*)
It's worthwhile!
At least kinda start of a list, with
$mathsf H$ being an infinite-dimensional separable Hilbert space:
Compact operators $:mathcal K(mathsf H)$
Cuntz algebras $:mathcal O_n$ with $,nin{2,3,dots,42,dots ,infty}$
$dots$
* Fields Institute Monograph Volume 6, American Mathematical Society, 1996
answered Dec 12 '18 at 12:40
HannoHanno
2,213427
$endgroup$
add a comment |
$begingroup$
Any C$^*$-algebra which is generated by a finite or a countable set will be separable. Including the examples by Hanno, here is a very incomplete list:
$K(H)$
Cuntz and Cuntz-Krieger algebras
For any countable non-abelian group $Gamma$, the reduced C$^*$-algebra of the group, $C_r^*(G)$ (namely the norm closure of the span of $lambda(G)subset B(ell^2(G))$, where $lambda$ is the left regular representation). This includes for instance $C_r^*(mathbb F_n)$, where $mathbb F_n$ are the free groups.
For any countable non-abelian group $Gamma$, the universal C$^*$-algebra of the group, $C^*(G)$. This differs from the above whenever $G$ is non-amenable.
Irrational rotation algebras ($A_theta$ is the universal C$^*$-algebra generated by unitaries $u,v$ such that $uv=e^{2pi itheta}vu$)
AF C$^*$-algebras. That is, direct limits of finite-dimensional C$^*$-algebras. This class is huge, as can be seen by considering their Bratelli diagrams
Tensor products of the above
Reduced free products of the above
$c_0$-sums of the above
C$^*$-subalgebras of the above
answered Dec 12 '18 at 15:42
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
1
$begingroup$
Another remark: Separability passes to subalgebras. In particular lots of (necessarily exact and quasidiagonal) algebras are subalgebras of separable AF-algebras.
$endgroup$
– user42761
Dec 13 '18 at 7:42
$begingroup$
Indeed. $ $
$endgroup$
– Martin Argerami
Dec 13 '18 at 14:43
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Your desire is somewhat vague since you do not write down your motivation.
Did you already browse through K. R. Davidson's book
$,C^*$-Algebras by Example ? (*)
It's worthwhile!
At least kinda start of a list, with
$mathsf H$ being an infinite-dimensional separable Hilbert space:
Compact operators $:mathcal K(mathsf H)$
Cuntz algebras $:mathcal O_n$ with $,nin{2,3,dots,42,dots ,infty}$
$dots$
* Fields Institute Monograph Volume 6, American Mathematical Society, 1996
answered Dec 12 '18 at 12:40
HannoHanno
2,213427
$endgroup$
add a comment |
$begingroup$
Your desire is somewhat vague since you do not write down your motivation.
Did you already browse through K. R. Davidson's book
$,C^*$-Algebras by Example ? (*)
It's worthwhile!
At least kinda start of a list, with
$mathsf H$ being an infinite-dimensional separable Hilbert space:
Compact operators $:mathcal K(mathsf H)$
Cuntz algebras $:mathcal O_n$ with $,nin{2,3,dots,42,dots ,infty}$
$dots$
* Fields Institute Monograph Volume 6, American Mathematical Society, 1996
answered Dec 12 '18 at 12:40
HannoHanno
2,213427
$endgroup$
add a comment |
$begingroup$
Your desire is somewhat vague since you do not write down your motivation.
Did you already browse through K. R. Davidson's book
$,C^*$-Algebras by Example ? (*)
It's worthwhile!
At least kinda start of a list, with
$mathsf H$ being an infinite-dimensional separable Hilbert space:
Compact operators $:mathcal K(mathsf H)$
Cuntz algebras $:mathcal O_n$ with $,nin{2,3,dots,42,dots ,infty}$
$dots$
* Fields Institute Monograph Volume 6, American Mathematical Society, 1996
answered Dec 12 '18 at 12:40
HannoHanno
2,213427
$endgroup$
Your desire is somewhat vague since you do not write down your motivation.
Did you already browse through K. R. Davidson's book
$,C^*$-Algebras by Example ? (*)
It's worthwhile!
At least kinda start of a list, with
$mathsf H$ being an infinite-dimensional separable Hilbert space:
Compact operators $:mathcal K(mathsf H)$
Cuntz algebras $:mathcal O_n$ with $,nin{2,3,dots,42,dots ,infty}$
$dots$
* Fields Institute Monograph Volume 6, American Mathematical Society, 1996
answered Dec 12 '18 at 12:40
HannoHanno
2,213427
edited Dec 12 '18 at 15:13
answered Dec 12 '18 at 12:40
HannoHanno
2,213427
answered Dec 12 '18 at 12:40
HannoHanno
2,213427
answered Dec 12 '18 at 12:40
HannoHanno
2,213427
2,213427
add a comment |
add a comment |
$begingroup$
Any C$^*$-algebra which is generated by a finite or a countable set will be separable. Including the examples by Hanno, here is a very incomplete list:
$K(H)$
Cuntz and Cuntz-Krieger algebras
For any countable non-abelian group $Gamma$, the reduced C$^*$-algebra of the group, $C_r^*(G)$ (namely the norm closure of the span of $lambda(G)subset B(ell^2(G))$, where $lambda$ is the left regular representation). This includes for instance $C_r^*(mathbb F_n)$, where $mathbb F_n$ are the free groups.
For any countable non-abelian group $Gamma$, the universal C$^*$-algebra of the group, $C^*(G)$. This differs from the above whenever $G$ is non-amenable.
Irrational rotation algebras ($A_theta$ is the universal C$^*$-algebra generated by unitaries $u,v$ such that $uv=e^{2pi itheta}vu$)
AF C$^*$-algebras. That is, direct limits of finite-dimensional C$^*$-algebras. This class is huge, as can be seen by considering their Bratelli diagrams
Tensor products of the above
Reduced free products of the above
$c_0$-sums of the above
C$^*$-subalgebras of the above
answered Dec 12 '18 at 15:42
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
1
$begingroup$
Another remark: Separability passes to subalgebras. In particular lots of (necessarily exact and quasidiagonal) algebras are subalgebras of separable AF-algebras.
$endgroup$
– user42761
Dec 13 '18 at 7:42
$begingroup$
Indeed. $ $
$endgroup$
– Martin Argerami
Dec 13 '18 at 14:43
add a comment |
$begingroup$
Any C$^*$-algebra which is generated by a finite or a countable set will be separable. Including the examples by Hanno, here is a very incomplete list:
$K(H)$
Cuntz and Cuntz-Krieger algebras
For any countable non-abelian group $Gamma$, the reduced C$^*$-algebra of the group, $C_r^*(G)$ (namely the norm closure of the span of $lambda(G)subset B(ell^2(G))$, where $lambda$ is the left regular representation). This includes for instance $C_r^*(mathbb F_n)$, where $mathbb F_n$ are the free groups.
For any countable non-abelian group $Gamma$, the universal C$^*$-algebra of the group, $C^*(G)$. This differs from the above whenever $G$ is non-amenable.
Irrational rotation algebras ($A_theta$ is the universal C$^*$-algebra generated by unitaries $u,v$ such that $uv=e^{2pi itheta}vu$)
AF C$^*$-algebras. That is, direct limits of finite-dimensional C$^*$-algebras. This class is huge, as can be seen by considering their Bratelli diagrams
Tensor products of the above
Reduced free products of the above
$c_0$-sums of the above
C$^*$-subalgebras of the above
answered Dec 12 '18 at 15:42
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
1
$begingroup$
Another remark: Separability passes to subalgebras. In particular lots of (necessarily exact and quasidiagonal) algebras are subalgebras of separable AF-algebras.
$endgroup$
– user42761
Dec 13 '18 at 7:42
$begingroup$
Indeed. $ $
$endgroup$
– Martin Argerami
Dec 13 '18 at 14:43
add a comment |
$begingroup$
Any C$^*$-algebra which is generated by a finite or a countable set will be separable. Including the examples by Hanno, here is a very incomplete list:
$K(H)$
Cuntz and Cuntz-Krieger algebras
For any countable non-abelian group $Gamma$, the reduced C$^*$-algebra of the group, $C_r^*(G)$ (namely the norm closure of the span of $lambda(G)subset B(ell^2(G))$, where $lambda$ is the left regular representation). This includes for instance $C_r^*(mathbb F_n)$, where $mathbb F_n$ are the free groups.
For any countable non-abelian group $Gamma$, the universal C$^*$-algebra of the group, $C^*(G)$. This differs from the above whenever $G$ is non-amenable.
Irrational rotation algebras ($A_theta$ is the universal C$^*$-algebra generated by unitaries $u,v$ such that $uv=e^{2pi itheta}vu$)
AF C$^*$-algebras. That is, direct limits of finite-dimensional C$^*$-algebras. This class is huge, as can be seen by considering their Bratelli diagrams
Tensor products of the above
Reduced free products of the above
$c_0$-sums of the above
C$^*$-subalgebras of the above
answered Dec 12 '18 at 15:42
Martin ArgeramiMartin Argerami
127k1182182
$endgroup$
Any C$^*$-algebra which is generated by a finite or a countable set will be separable. Including the examples by Hanno, here is a very incomplete list:
$K(H)$
Cuntz and Cuntz-Krieger algebras
For any countable non-abelian group $Gamma$, the reduced C$^*$-algebra of the group, $C_r^*(G)$ (namely the norm closure of the span of $lambda(G)subset B(ell^2(G))$, where $lambda$ is the left regular representation). This includes for instance $C_r^*(mathbb F_n)$, where $mathbb F_n$ are the free groups.
For any countable non-abelian group $Gamma$, the universal C$^*$-algebra of the group, $C^*(G)$. This differs from the above whenever $G$ is non-amenable.
Irrational rotation algebras ($A_theta$ is the universal C$^*$-algebra generated by unitaries $u,v$ such that $uv=e^{2pi itheta}vu$)
AF C$^*$-algebras. That is, direct limits of finite-dimensional C$^*$-algebras. This class is huge, as can be seen by considering their Bratelli diagrams
Tensor products of the above
Reduced free products of the above
$c_0$-sums of the above
C$^*$-subalgebras of the above
answered Dec 12 '18 at 15:42
Martin ArgeramiMartin Argerami
127k1182182
edited Dec 13 '18 at 14:43
answered Dec 12 '18 at 15:42
Martin ArgeramiMartin Argerami
127k1182182
answered Dec 12 '18 at 15:42
Martin ArgeramiMartin Argerami
127k1182182
answered Dec 12 '18 at 15:42
Martin ArgeramiMartin Argerami
127k1182182
127k1182182
1
$begingroup$
Another remark: Separability passes to subalgebras. In particular lots of (necessarily exact and quasidiagonal) algebras are subalgebras of separable AF-algebras.
$endgroup$
– user42761
Dec 13 '18 at 7:42
$begingroup$
Indeed. $ $
$endgroup$
– Martin Argerami
Dec 13 '18 at 14:43
add a comment |
1
$begingroup$
Another remark: Separability passes to subalgebras. In particular lots of (necessarily exact and quasidiagonal) algebras are subalgebras of separable AF-algebras.
$endgroup$
– user42761
Dec 13 '18 at 7:42
$begingroup$
Indeed. $ $
$endgroup$
– Martin Argerami
Dec 13 '18 at 14:43
1
1
$begingroup$
Another remark: Separability passes to subalgebras. In particular lots of (necessarily exact and quasidiagonal) algebras are subalgebras of separable AF-algebras.
$endgroup$
– user42761
Dec 13 '18 at 7:42
$begingroup$
Another remark: Separability passes to subalgebras. In particular lots of (necessarily exact and quasidiagonal) algebras are subalgebras of separable AF-algebras.
$endgroup$
– user42761
Dec 13 '18 at 7:42
$begingroup$
Indeed. $ $
$endgroup$
– Martin Argerami
Dec 13 '18 at 14:43
$begingroup$
Indeed. $ $
$endgroup$
– Martin Argerami
Dec 13 '18 at 14:43
add a comment |
