Dtjytyk

Are spiders unable to hurt humans, especially very small spiders?

Why are PDP-7-style microprogrammed instructions out of vogue?

How did passengers keep warm on sail ships?

Is 'stolen' appropriate word?

How to read αἱμύλιος or when to aspirate

How do you keep chess fun when your opponent constantly beats you?

What happens to a Warlock's expended Spell Slots when they gain a Level?

What is the padding with red substance inside of steak packaging?

Can each chord in a progression create its own key?

What aspect of planet Earth must be changed to prevent the industrial revolution?

Can I visit the Trinity College (Cambridge) library and see some of their rare books

Homework question about an engine pulling a train

How did the audience guess the pentatonic scale in Bobby McFerrin's presentation?

"is" operation returns false even though two objects have same id

Is it ok to offer lower paid work as a trial period before negotiating for a full-time job?

Do ℕ, mathbb{N}, BbbN, symbb{N} effectively differ, and is there a "canonical" specification of the naturals?

How do spell lists change if the party levels up without taking a long rest?

Circular reasoning in L'Hopital's rule

Did the UK government pay "millions and millions of dollars" to try to snag Julian Assange?

Working through the single responsibility principle (SRP) in Python when calls are expensive

Is every episode of "Where are my Pants?" identical?

What do I do when my TA workload is more than expected?

Windows 10: How to Lock (not sleep) laptop on lid close?

Can the Right Ascension and Argument of Perigee of a spacecraft's orbit keep varying by themselves with time?

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

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

846

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

846

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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

846

$endgroup$

share|cite|improve this question

edited Mar 22 at 15:01

sirjoe

asked Mar 22 at 13:02

sirjoesirjoe

846

share|cite|improve this question

edited Mar 22 at 15:01

sirjoe

asked Mar 22 at 13:02

sirjoesirjoe

846

asked Mar 22 at 13:02

sirjoesirjoe

846

asked Mar 22 at 13:02

sirjoesirjoe

846

1 Answer
1

active

oldest

votes

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 $mathbb{C}$.

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 $mathbb{C}$.

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

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 | 

Your Answer

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});

draft saved

draft discarded

Sign up or log in

StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});

Sign up using Google

Sign up using Facebook

Sign up using Email and Password

Post as a guest

Name

Email

Required, but never shown

StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3158113%2fwhen-is-the-set-of-positive-elements-in-a-c-algebra-a-totally-ordered-set%23new-answer', 'question_page');
}
);

Post as a guest

Name

Email

Required, but never shown

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 $mathbb{C}$.

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 $mathbb{C}$.

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

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 $mathbb{C}$.

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 $mathbb{C}$.

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

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 $mathbb{C}$.

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 $mathbb{C}$.

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

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