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closed ideal in a $C^*$- algebra

If a sub-C*-algebra does not contain the unit, is it contained in a proper ideal?Unital nonabelian banach algebra where the only closed ideals are $0$ and $A$Non-closed ideals in $C^*$-algebrasElement $a$ in a unital C$^*$-algebra $A$ with $overlinelangle arangle=A$, but $a$ not left-invertibledescription of an ideal generated by the projections in a $C^*$ algebrastate on a non-unital $C^*$ algebratracial state on a unital infinite dimensional simple $C^*$ algebracentral projectionsIdeal in a $C^*$ algebracommutator of $C^*$ algebra

0

$begingroup$

Suppose $A$ is a non-simple $C^*$-algebra, let $x_0$ be a nonzero element in $A$, and let $S=x_0y-yx_0:yin A$. If $I$ is the closed ideal generated by the set $S$. I think there is a possibility that $I=A$, but I cannot think of a concrete example.

operator-theory operator-algebras c-star-algebras

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edited Mar 14 at 18:37

Aweygan

14.6k21442

asked Mar 13 at 18:18

mathrookiemathrookie

914512

$endgroup$

add a comment | 

0

$begingroup$

Suppose $A$ is a non-simple $C^*$-algebra, let $x_0$ be a nonzero element in $A$, and let $S=x_0y-yx_0:yin A$. If $I$ is the closed ideal generated by the set $S$. I think there is a possibility that $I=A$, but I cannot think of a concrete example.

operator-theory operator-algebras c-star-algebras

share|cite|improve this question

edited Mar 14 at 18:37

Aweygan

14.6k21442

asked Mar 13 at 18:18

mathrookiemathrookie

914512

$endgroup$

add a comment | 

$begingroup$

Suppose $A$ is a non-simple $C^*$-algebra, let $x_0$ be a nonzero element in $A$, and let $S=x_0y-yx_0:yin A$. If $I$ is the closed ideal generated by the set $S$. I think there is a possibility that $I=A$, but I cannot think of a concrete example.

operator-theory operator-algebras c-star-algebras

share|cite|improve this question

edited Mar 14 at 18:37

Aweygan

14.6k21442

asked Mar 13 at 18:18

mathrookiemathrookie

914512

$endgroup$

share|cite|improve this question

edited Mar 14 at 18:37

Aweygan

14.6k21442

asked Mar 13 at 18:18

mathrookiemathrookie

914512

share|cite|improve this question

edited Mar 14 at 18:37

Aweygan

14.6k21442

asked Mar 13 at 18:18

mathrookiemathrookie

914512

edited Mar 14 at 18:37

Aweygan

14.6k21442

edited Mar 14 at 18:37

Aweygan

14.6k21442

asked Mar 13 at 18:18

mathrookiemathrookie

914512

asked Mar 13 at 18:18

mathrookiemathrookie

914512

1 Answer
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1

$begingroup$

If $A$ is a simple $C^*$-algebra with trivial center, then the ideal $I$ as you constructed must be the whole algebra. For we must have $I=A$ or $0$, and if $I=0$, then $x_0$ is in the center of $A$, in which case $x_0=0$.

An example of such an algebra is $K(H)$ for a separable infinite-dimensional Hilbert space $H$.

If you want the additional assumption that $A$ is non-simple, let $A=K(H)oplus K(H)$, and let $x_0=(x_1,x_1)$ for any nonzero element $x_1in K(H)$.

share|cite|improve this answer

edited Mar 14 at 18:39

answered Mar 13 at 21:12

AweyganAweygan

14.6k21442

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks,I mean a non-simple $C^*$ algebra.I have edited the question.
  $endgroup$
  – mathrookie
  Mar 14 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Alright, I will think of another example. Out of curiosity, where does this question come from?
  $endgroup$
  – Aweygan
  Mar 14 at 2:40
  

  
  
  
  

add a comment | 

Your Answer

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1

$begingroup$

If $A$ is a simple $C^*$-algebra with trivial center, then the ideal $I$ as you constructed must be the whole algebra. For we must have $I=A$ or $0$, and if $I=0$, then $x_0$ is in the center of $A$, in which case $x_0=0$.

An example of such an algebra is $K(H)$ for a separable infinite-dimensional Hilbert space $H$.

If you want the additional assumption that $A$ is non-simple, let $A=K(H)oplus K(H)$, and let $x_0=(x_1,x_1)$ for any nonzero element $x_1in K(H)$.

share|cite|improve this answer

edited Mar 14 at 18:39

answered Mar 13 at 21:12

AweyganAweygan

14.6k21442

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks,I mean a non-simple $C^*$ algebra.I have edited the question.
  $endgroup$
  – mathrookie
  Mar 14 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Alright, I will think of another example. Out of curiosity, where does this question come from?
  $endgroup$
  – Aweygan
  Mar 14 at 2:40
  

  
  
  
  

add a comment | 

1

$begingroup$

If $A$ is a simple $C^*$-algebra with trivial center, then the ideal $I$ as you constructed must be the whole algebra. For we must have $I=A$ or $0$, and if $I=0$, then $x_0$ is in the center of $A$, in which case $x_0=0$.

An example of such an algebra is $K(H)$ for a separable infinite-dimensional Hilbert space $H$.

If you want the additional assumption that $A$ is non-simple, let $A=K(H)oplus K(H)$, and let $x_0=(x_1,x_1)$ for any nonzero element $x_1in K(H)$.

share|cite|improve this answer

edited Mar 14 at 18:39

answered Mar 13 at 21:12

AweyganAweygan

14.6k21442

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks,I mean a non-simple $C^*$ algebra.I have edited the question.
  $endgroup$
  – mathrookie
  Mar 14 at 2:21
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Alright, I will think of another example. Out of curiosity, where does this question come from?
  $endgroup$
  – Aweygan
  Mar 14 at 2:40
  

  
  
  
  

add a comment | 

$begingroup$

If $A$ is a simple $C^*$-algebra with trivial center, then the ideal $I$ as you constructed must be the whole algebra. For we must have $I=A$ or $0$, and if $I=0$, then $x_0$ is in the center of $A$, in which case $x_0=0$.

An example of such an algebra is $K(H)$ for a separable infinite-dimensional Hilbert space $H$.

If you want the additional assumption that $A$ is non-simple, let $A=K(H)oplus K(H)$, and let $x_0=(x_1,x_1)$ for any nonzero element $x_1in K(H)$.

share|cite|improve this answer

edited Mar 14 at 18:39

answered Mar 13 at 21:12

AweyganAweygan

14.6k21442

$endgroup$

share|cite|improve this answer

edited Mar 14 at 18:39

answered Mar 13 at 21:12

AweyganAweygan

14.6k21442

answered Mar 13 at 21:12

AweyganAweygan

14.6k21442

answered Mar 13 at 21:12

AweyganAweygan

14.6k21442