infinite dimensional representation of a $C^*$ algebra [closed]
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When $H$ is separable infinite dimensional,$K(H)$ has no finite representation.Do there exist other non-unital $C^*$ algebras such that their representations cannot be finite dimensional?
operator-theory operator-algebras c-star-algebras
asked Dec 19 '18 at 18:21
mathrookiemathrookie
939512
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closed as off-topic by Saad, metamorphy, José Carlos Santos, Namaste, Paul Frost Dec 26 '18 at 23:26
This question appears to be off-topic. The users who voted to close gave this specific reason:
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When $H$ is separable infinite dimensional,$K(H)$ has no finite representation.Do there exist other non-unital $C^*$ algebras such that their representations cannot be finite dimensional?
operator-theory operator-algebras c-star-algebras
asked Dec 19 '18 at 18:21
mathrookiemathrookie
939512
$endgroup$
closed as off-topic by Saad, metamorphy, José Carlos Santos, Namaste, Paul Frost Dec 26 '18 at 23:26
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, metamorphy, José Carlos Santos, Namaste, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
When $H$ is separable infinite dimensional,$K(H)$ has no finite representation.Do there exist other non-unital $C^*$ algebras such that their representations cannot be finite dimensional?
operator-theory operator-algebras c-star-algebras
asked Dec 19 '18 at 18:21
mathrookiemathrookie
939512
$endgroup$
When $H$ is separable infinite dimensional,$K(H)$ has no finite representation.Do there exist other non-unital $C^*$ algebras such that their representations cannot be finite dimensional?
operator-theory operator-algebras c-star-algebras
operator-theory operator-algebras c-star-algebras
asked Dec 19 '18 at 18:21
mathrookiemathrookie
939512
asked Dec 19 '18 at 18:21
mathrookiemathrookie
939512
asked Dec 19 '18 at 18:21
mathrookiemathrookie
939512
asked Dec 19 '18 at 18:21
mathrookiemathrookie
939512
asked Dec 19 '18 at 18:21
mathrookiemathrookie
939512
939512
closed as off-topic by Saad, metamorphy, José Carlos Santos, Namaste, Paul Frost Dec 26 '18 at 23:26
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, metamorphy, José Carlos Santos, Namaste, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Saad, metamorphy, José Carlos Santos, Namaste, Paul Frost Dec 26 '18 at 23:26
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, metamorphy, José Carlos Santos, Namaste, Paul Frost
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
add a comment |
1 Answer
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Yes. Any simple infinite-dimensional C$^*$-algebra (that is, "most" of them in the sense that these are precisely the ones people study a lot) does not admit a finite-dimensional representation.
answered Dec 19 '18 at 18:52
Martin ArgeramiMartin Argerami
129k1184185
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Of course. Say $M_2 (mathbb C)oplus C_0 (mathbb R) $.
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– Martin Argerami
Dec 26 '18 at 15:39
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Yes. Any simple infinite-dimensional C$^*$-algebra (that is, "most" of them in the sense that these are precisely the ones people study a lot) does not admit a finite-dimensional representation.
answered Dec 19 '18 at 18:52
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Of course. Say $M_2 (mathbb C)oplus C_0 (mathbb R) $.
$endgroup$
– Martin Argerami
Dec 26 '18 at 15:39
add a comment |
$begingroup$
Yes. Any simple infinite-dimensional C$^*$-algebra (that is, "most" of them in the sense that these are precisely the ones people study a lot) does not admit a finite-dimensional representation.
answered Dec 19 '18 at 18:52
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Of course. Say $M_2 (mathbb C)oplus C_0 (mathbb R) $.
$endgroup$
– Martin Argerami
Dec 26 '18 at 15:39
add a comment |
$begingroup$
Yes. Any simple infinite-dimensional C$^*$-algebra (that is, "most" of them in the sense that these are precisely the ones people study a lot) does not admit a finite-dimensional representation.
answered Dec 19 '18 at 18:52
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
Yes. Any simple infinite-dimensional C$^*$-algebra (that is, "most" of them in the sense that these are precisely the ones people study a lot) does not admit a finite-dimensional representation.
answered Dec 19 '18 at 18:52
Martin ArgeramiMartin Argerami
129k1184185
answered Dec 19 '18 at 18:52
Martin ArgeramiMartin Argerami
129k1184185
answered Dec 19 '18 at 18:52
Martin ArgeramiMartin Argerami
129k1184185
answered Dec 19 '18 at 18:52
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
$begingroup$
Of course. Say $M_2 (mathbb C)oplus C_0 (mathbb R) $.
$endgroup$
– Martin Argerami
Dec 26 '18 at 15:39
add a comment |
$begingroup$
Of course. Say $M_2 (mathbb C)oplus C_0 (mathbb R) $.
$endgroup$
– Martin Argerami
Dec 26 '18 at 15:39
$begingroup$
Of course. Say $M_2 (mathbb C)oplus C_0 (mathbb R) $.
$endgroup$
– Martin Argerami
Dec 26 '18 at 15:39
$begingroup$
Of course. Say $M_2 (mathbb C)oplus C_0 (mathbb R) $.
$endgroup$
– Martin Argerami
Dec 26 '18 at 15:39
add a comment |
