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When is the set of positive elements in a C* algebra a totally ordered set

strictly positive elements in $C^*$-algebraFinding Strictly Positive ElementsCharacterization of positive elements in unital C*-algebraA simple question about positive element in C*-algebraWhy does the order on positive elements respect the order on the norm?Positive elements in a Banach algebraWhy the self-adjointness condition for positivity of an element of a C*-algebra?positive elements in the unitization of a $C^*$-algebraPositive elements in * AlgebrasBounding norms with sums of positive elements

0

$begingroup$

As the title suggests I was wondering if there are any properties that ensure a C*-algebra has all positive elements comparable to each other. ( Recall $aleq b$ if $b-a$ is a positive element in the C* algebra)

c-star-algebras von-neumann-algebras

share|cite|improve this question

edited Mar 22 at 15:01

sirjoe

asked Mar 22 at 13:02

sirjoesirjoe

796

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is a full lattice ?
  $endgroup$
  – Epsilon
  Mar 22 at 13:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes it wasn't clear what I meant, I changed the title now
  $endgroup$
  – sirjoe
  Mar 22 at 15:02
  

  
  
  
  

add a comment | 

0

$begingroup$

As the title suggests I was wondering if there are any properties that ensure a C*-algebra has all positive elements comparable to each other. ( Recall $aleq b$ if $b-a$ is a positive element in the C* algebra)

c-star-algebras von-neumann-algebras

share|cite|improve this question

edited Mar 22 at 15:01

sirjoe

asked Mar 22 at 13:02

sirjoesirjoe

796

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What is a full lattice ?
  $endgroup$
  – Epsilon
  Mar 22 at 13:48
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes it wasn't clear what I meant, I changed the title now
  $endgroup$
  – sirjoe
  Mar 22 at 15:02
  

  
  
  
  

add a comment | 

$begingroup$

As the title suggests I was wondering if there are any properties that ensure a C*-algebra has all positive elements comparable to each other. ( Recall $aleq b$ if $b-a$ is a positive element in the C* algebra)

c-star-algebras von-neumann-algebras

share|cite|improve this question

edited Mar 22 at 15:01

sirjoe

asked Mar 22 at 13:02

sirjoesirjoe

796

$endgroup$

share|cite|improve this question

edited Mar 22 at 15:01

sirjoe

asked Mar 22 at 13:02

sirjoesirjoe

796

share|cite|improve this question

edited Mar 22 at 15:01

sirjoe

asked Mar 22 at 13:02

sirjoesirjoe

796

asked Mar 22 at 13:02

sirjoesirjoe

796

asked Mar 22 at 13:02

sirjoesirjoe

796

1 Answer
1

active

oldest

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2

$begingroup$

Here is a partial answer:

Let $A$ be a $C^*$-algebra. Assume that there is a projection $p$ such that $p not in 0,1$. Then $p$ and $1-p$ are not comparable : indeed, the spectrum of $p - (1-p)$ is $pm 1$ so $p - (1-p)$ is not positive, and the same goes for $1-p - p = 1 - 2p$.

EDIT:

Claim: It never happens if the algebra is unital and not $mathbbC$.

Let $A$ be a unital $C^*$-algebra of dimension greater or equal than $2$. Then there is $a in A$ such that the subalgebra generated by $a$ isn't isomorphic to $mathbbC$.

[EDIT2: We can assume that $a$ is normal. Indeed, if both $a+a^*$ or $a-a^*$ are elements of $mathbbC$, then $a$ is as well. So, $a+a^* not in mathbbC$ or $a-a^* not in mathbbC$, and each of these possibilities gives a normal element of $A setminus mathbbC$.]

Let us denote by $B$ the closure of this subalgebra. Then $B$ is commutative, since $a$ is assumed to be normal, so it is isomorphic to $C_0(X)$ where $X$ is a compact space with at least two points, $x$ and $y$. Since $X$ is normal, there is a positive continuous function $f$ such that $f(x) = 0$ and $f(y) = 1$, and there is a positive continuous function $g$ such that $g(x) = 1$ and $g(y) = 0$. Then $f$ and $g$ are not comparable.

share|cite|improve this answer

edited Mar 25 at 10:31

answered Mar 22 at 15:14

PlopPlop

485216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I guess in this case the C* algebra would need to be projectionles. so this would never hold for Von Neuman algebras. But I wonder if you have a totally ordered set in projectionless C* algebras e.g. the Jiang-Su algebra.
  $endgroup$
  – sirjoe
  Mar 22 at 15:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  About your edit: $B$ is not commutative unless $a$ is normal. That is not a serious restriction however, as $a+a^*$ and $a-a^*$ are normal.
  $endgroup$
  – s.harp
  Mar 24 at 15:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for spotting my mistake. I forgot the $*$ in $C^*$-algebra :D
  $endgroup$
  – Plop
  Mar 25 at 10:32
  

  
  
  
  

add a comment | 

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2

$begingroup$

Here is a partial answer:

Let $A$ be a $C^*$-algebra. Assume that there is a projection $p$ such that $p not in 0,1$. Then $p$ and $1-p$ are not comparable : indeed, the spectrum of $p - (1-p)$ is $pm 1$ so $p - (1-p)$ is not positive, and the same goes for $1-p - p = 1 - 2p$.

EDIT:

Claim: It never happens if the algebra is unital and not $mathbbC$.

Let $A$ be a unital $C^*$-algebra of dimension greater or equal than $2$. Then there is $a in A$ such that the subalgebra generated by $a$ isn't isomorphic to $mathbbC$.

[EDIT2: We can assume that $a$ is normal. Indeed, if both $a+a^*$ or $a-a^*$ are elements of $mathbbC$, then $a$ is as well. So, $a+a^* not in mathbbC$ or $a-a^* not in mathbbC$, and each of these possibilities gives a normal element of $A setminus mathbbC$.]

Let us denote by $B$ the closure of this subalgebra. Then $B$ is commutative, since $a$ is assumed to be normal, so it is isomorphic to $C_0(X)$ where $X$ is a compact space with at least two points, $x$ and $y$. Since $X$ is normal, there is a positive continuous function $f$ such that $f(x) = 0$ and $f(y) = 1$, and there is a positive continuous function $g$ such that $g(x) = 1$ and $g(y) = 0$. Then $f$ and $g$ are not comparable.

share|cite|improve this answer

edited Mar 25 at 10:31

answered Mar 22 at 15:14

PlopPlop

485216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I guess in this case the C* algebra would need to be projectionles. so this would never hold for Von Neuman algebras. But I wonder if you have a totally ordered set in projectionless C* algebras e.g. the Jiang-Su algebra.
  $endgroup$
  – sirjoe
  Mar 22 at 15:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  About your edit: $B$ is not commutative unless $a$ is normal. That is not a serious restriction however, as $a+a^*$ and $a-a^*$ are normal.
  $endgroup$
  – s.harp
  Mar 24 at 15:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for spotting my mistake. I forgot the $*$ in $C^*$-algebra :D
  $endgroup$
  – Plop
  Mar 25 at 10:32
  

  
  
  
  

add a comment | 

2

$begingroup$

Here is a partial answer:

Let $A$ be a $C^*$-algebra. Assume that there is a projection $p$ such that $p not in 0,1$. Then $p$ and $1-p$ are not comparable : indeed, the spectrum of $p - (1-p)$ is $pm 1$ so $p - (1-p)$ is not positive, and the same goes for $1-p - p = 1 - 2p$.

EDIT:

Claim: It never happens if the algebra is unital and not $mathbbC$.

Let $A$ be a unital $C^*$-algebra of dimension greater or equal than $2$. Then there is $a in A$ such that the subalgebra generated by $a$ isn't isomorphic to $mathbbC$.

[EDIT2: We can assume that $a$ is normal. Indeed, if both $a+a^*$ or $a-a^*$ are elements of $mathbbC$, then $a$ is as well. So, $a+a^* not in mathbbC$ or $a-a^* not in mathbbC$, and each of these possibilities gives a normal element of $A setminus mathbbC$.]

Let us denote by $B$ the closure of this subalgebra. Then $B$ is commutative, since $a$ is assumed to be normal, so it is isomorphic to $C_0(X)$ where $X$ is a compact space with at least two points, $x$ and $y$. Since $X$ is normal, there is a positive continuous function $f$ such that $f(x) = 0$ and $f(y) = 1$, and there is a positive continuous function $g$ such that $g(x) = 1$ and $g(y) = 0$. Then $f$ and $g$ are not comparable.

share|cite|improve this answer

edited Mar 25 at 10:31

answered Mar 22 at 15:14

PlopPlop

485216

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I guess in this case the C* algebra would need to be projectionles. so this would never hold for Von Neuman algebras. But I wonder if you have a totally ordered set in projectionless C* algebras e.g. the Jiang-Su algebra.
  $endgroup$
  – sirjoe
  Mar 22 at 15:25
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  About your edit: $B$ is not commutative unless $a$ is normal. That is not a serious restriction however, as $a+a^*$ and $a-a^*$ are normal.
  $endgroup$
  – s.harp
  Mar 24 at 15:32
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks for spotting my mistake. I forgot the $*$ in $C^*$-algebra :D
  $endgroup$
  – Plop
  Mar 25 at 10:32
  

  
  
  
  

add a comment | 

$begingroup$

Here is a partial answer:

Let $A$ be a $C^*$-algebra. Assume that there is a projection $p$ such that $p not in 0,1$. Then $p$ and $1-p$ are not comparable : indeed, the spectrum of $p - (1-p)$ is $pm 1$ so $p - (1-p)$ is not positive, and the same goes for $1-p - p = 1 - 2p$.

EDIT:

Claim: It never happens if the algebra is unital and not $mathbbC$.

Let $A$ be a unital $C^*$-algebra of dimension greater or equal than $2$. Then there is $a in A$ such that the subalgebra generated by $a$ isn't isomorphic to $mathbbC$.

[EDIT2: We can assume that $a$ is normal. Indeed, if both $a+a^*$ or $a-a^*$ are elements of $mathbbC$, then $a$ is as well. So, $a+a^* not in mathbbC$ or $a-a^* not in mathbbC$, and each of these possibilities gives a normal element of $A setminus mathbbC$.]

Let us denote by $B$ the closure of this subalgebra. Then $B$ is commutative, since $a$ is assumed to be normal, so it is isomorphic to $C_0(X)$ where $X$ is a compact space with at least two points, $x$ and $y$. Since $X$ is normal, there is a positive continuous function $f$ such that $f(x) = 0$ and $f(y) = 1$, and there is a positive continuous function $g$ such that $g(x) = 1$ and $g(y) = 0$. Then $f$ and $g$ are not comparable.

share|cite|improve this answer

edited Mar 25 at 10:31

answered Mar 22 at 15:14

PlopPlop

485216

$endgroup$

share|cite|improve this answer

edited Mar 25 at 10:31

answered Mar 22 at 15:14

PlopPlop

485216

answered Mar 22 at 15:14

PlopPlop

485216

answered Mar 22 at 15:14

PlopPlop

485216