Type III Von Neumann algebras and spectra of the modular operators.
$begingroup$
I´m studying the paper of Fredenhagen. There he said that he would prove that the algebra of local observables under certain conditions is of type III, by showing that all modular operators satisfy $spec(Delta_{mathcal{O}})=mathbb{R}_{+}$.
I'm just beginning to study von Neumann algebras and really don't know if it was a sufficient condition.
The definition I have for type III algebras is that there is only projectors of dimension $0$ or $infty$. My question is, how are this two assertions related?
Many thanks in advance.
operator-algebras c-star-algebras von-neumann-algebras quantum-field-theory
edited Dec 17 '18 at 14:33
Martin Argerami
127k1182183
asked Dec 15 '18 at 3:42
Gabriel PalauGabriel Palau
1106
$endgroup$
add a comment |
$begingroup$
I´m studying the paper of Fredenhagen. There he said that he would prove that the algebra of local observables under certain conditions is of type III, by showing that all modular operators satisfy $spec(Delta_{mathcal{O}})=mathbb{R}_{+}$.
I'm just beginning to study von Neumann algebras and really don't know if it was a sufficient condition.
The definition I have for type III algebras is that there is only projectors of dimension $0$ or $infty$. My question is, how are this two assertions related?
Many thanks in advance.
operator-algebras c-star-algebras von-neumann-algebras quantum-field-theory
edited Dec 17 '18 at 14:33
Martin Argerami
127k1182183
asked Dec 15 '18 at 3:42
Gabriel PalauGabriel Palau
1106
$endgroup$
$begingroup$
I cannot find anywhere in that paper saying that any algebra is not of type III. Could you be more specific?
$endgroup$
– Martin Argerami
Dec 17 '18 at 4:17
$begingroup$
Sorry, I was wrong. I have just edit the question. I should say "tipe III" instead of "not tipe III". The coment is in the abstract.
$endgroup$
– Gabriel Palau
Dec 17 '18 at 5:27
add a comment |
$begingroup$
I´m studying the paper of Fredenhagen. There he said that he would prove that the algebra of local observables under certain conditions is of type III, by showing that all modular operators satisfy $spec(Delta_{mathcal{O}})=mathbb{R}_{+}$.
I'm just beginning to study von Neumann algebras and really don't know if it was a sufficient condition.
The definition I have for type III algebras is that there is only projectors of dimension $0$ or $infty$. My question is, how are this two assertions related?
Many thanks in advance.
operator-algebras c-star-algebras von-neumann-algebras quantum-field-theory
edited Dec 17 '18 at 14:33
Martin Argerami
127k1182183
asked Dec 15 '18 at 3:42
Gabriel PalauGabriel Palau
1106
$endgroup$
I´m studying the paper of Fredenhagen. There he said that he would prove that the algebra of local observables under certain conditions is of type III, by showing that all modular operators satisfy $spec(Delta_{mathcal{O}})=mathbb{R}_{+}$.
I'm just beginning to study von Neumann algebras and really don't know if it was a sufficient condition.
The definition I have for type III algebras is that there is only projectors of dimension $0$ or $infty$. My question is, how are this two assertions related?
Many thanks in advance.
operator-algebras c-star-algebras von-neumann-algebras quantum-field-theory
operator-algebras c-star-algebras von-neumann-algebras quantum-field-theory
edited Dec 17 '18 at 14:33
Martin Argerami
127k1182183
asked Dec 15 '18 at 3:42
Gabriel PalauGabriel Palau
1106
edited Dec 17 '18 at 14:33
Martin Argerami
127k1182183
asked Dec 15 '18 at 3:42
Gabriel PalauGabriel Palau
1106
edited Dec 17 '18 at 14:33
Martin Argerami
127k1182183
edited Dec 17 '18 at 14:33
Martin Argerami
127k1182183
edited Dec 17 '18 at 14:33
Martin Argerami
127k1182183
127k1182183
asked Dec 15 '18 at 3:42
Gabriel PalauGabriel Palau
1106
asked Dec 15 '18 at 3:42
Gabriel PalauGabriel Palau
1106
asked Dec 15 '18 at 3:42
Gabriel PalauGabriel Palau
1106
1106
$begingroup$
I cannot find anywhere in that paper saying that any algebra is not of type III. Could you be more specific?
$endgroup$
– Martin Argerami
Dec 17 '18 at 4:17
$begingroup$
Sorry, I was wrong. I have just edit the question. I should say "tipe III" instead of "not tipe III". The coment is in the abstract.
$endgroup$
– Gabriel Palau
Dec 17 '18 at 5:27
add a comment |
$begingroup$
I cannot find anywhere in that paper saying that any algebra is not of type III. Could you be more specific?
$endgroup$
– Martin Argerami
Dec 17 '18 at 4:17
$begingroup$
Sorry, I was wrong. I have just edit the question. I should say "tipe III" instead of "not tipe III". The coment is in the abstract.
$endgroup$
– Gabriel Palau
Dec 17 '18 at 5:27
$begingroup$
I cannot find anywhere in that paper saying that any algebra is not of type III. Could you be more specific?
$endgroup$
– Martin Argerami
Dec 17 '18 at 4:17
$begingroup$
I cannot find anywhere in that paper saying that any algebra is not of type III. Could you be more specific?
$endgroup$
– Martin Argerami
Dec 17 '18 at 4:17
$begingroup$
Sorry, I was wrong. I have just edit the question. I should say "tipe III" instead of "not tipe III". The coment is in the abstract.
$endgroup$
– Gabriel Palau
Dec 17 '18 at 5:27
$begingroup$
Sorry, I was wrong. I have just edit the question. I should say "tipe III" instead of "not tipe III". The coment is in the abstract.
$endgroup$
– Gabriel Palau
Dec 17 '18 at 5:27
add a comment |
1 Answer
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$begingroup$
To try to make things simple (they are not; modular theory and Connes work on it are very very far from trivial), if $M$ is semifinite then the modular operator is the identity. That is, at least when talking about factors the modular operator is non-trivial only on factors of type III.
That said, the goal of the paper is not that much to show that the factors are to type III, but to show that the factors are of type III_1 (I'm not entirely sure if the algebras in the paper are factors, but at least in talking about factors I'm more sure I'm saying the right thing). This has to do with A. Connes classification.
Among many many other things, Connes proved that the set
$$
Gamma(M)=(0,infty)cap,bigcap{operatorname{sp}Delta_phi: phi text{ is a fns weight on }M}
$$
is a closed multiplicative group of $(0,infty)$. The only possibilities are
$Gamma(M)={1}$; if $M$ is also type III, we say that $M$ is of type III$_0$
$Gamma(M)={lambda^n: ninmathbb Z}$ for some $lambdain(0,1)$; we say that $M$ is of type III$_lambda$
$gamma(M)=(0,infty)$; we say that $M$ is of type III$_1$.
When $M$ is semifinite, you always have $Gamma(M)={1}$.
So if you can show that the spectrum of $Delta_phi$ is $(0,infty)$, then $M$ is of type III$_1$.
answered Dec 17 '18 at 5:41
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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votes
$begingroup$
To try to make things simple (they are not; modular theory and Connes work on it are very very far from trivial), if $M$ is semifinite then the modular operator is the identity. That is, at least when talking about factors the modular operator is non-trivial only on factors of type III.
That said, the goal of the paper is not that much to show that the factors are to type III, but to show that the factors are of type III_1 (I'm not entirely sure if the algebras in the paper are factors, but at least in talking about factors I'm more sure I'm saying the right thing). This has to do with A. Connes classification.
Among many many other things, Connes proved that the set
$$
Gamma(M)=(0,infty)cap,bigcap{operatorname{sp}Delta_phi: phi text{ is a fns weight on }M}
$$
is a closed multiplicative group of $(0,infty)$. The only possibilities are
$Gamma(M)={1}$; if $M$ is also type III, we say that $M$ is of type III$_0$
$Gamma(M)={lambda^n: ninmathbb Z}$ for some $lambdain(0,1)$; we say that $M$ is of type III$_lambda$
$gamma(M)=(0,infty)$; we say that $M$ is of type III$_1$.
When $M$ is semifinite, you always have $Gamma(M)={1}$.
So if you can show that the spectrum of $Delta_phi$ is $(0,infty)$, then $M$ is of type III$_1$.
answered Dec 17 '18 at 5:41
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
To try to make things simple (they are not; modular theory and Connes work on it are very very far from trivial), if $M$ is semifinite then the modular operator is the identity. That is, at least when talking about factors the modular operator is non-trivial only on factors of type III.
That said, the goal of the paper is not that much to show that the factors are to type III, but to show that the factors are of type III_1 (I'm not entirely sure if the algebras in the paper are factors, but at least in talking about factors I'm more sure I'm saying the right thing). This has to do with A. Connes classification.
Among many many other things, Connes proved that the set
$$
Gamma(M)=(0,infty)cap,bigcap{operatorname{sp}Delta_phi: phi text{ is a fns weight on }M}
$$
is a closed multiplicative group of $(0,infty)$. The only possibilities are
$Gamma(M)={1}$; if $M$ is also type III, we say that $M$ is of type III$_0$
$Gamma(M)={lambda^n: ninmathbb Z}$ for some $lambdain(0,1)$; we say that $M$ is of type III$_lambda$
$gamma(M)=(0,infty)$; we say that $M$ is of type III$_1$.
When $M$ is semifinite, you always have $Gamma(M)={1}$.
So if you can show that the spectrum of $Delta_phi$ is $(0,infty)$, then $M$ is of type III$_1$.
answered Dec 17 '18 at 5:41
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
To try to make things simple (they are not; modular theory and Connes work on it are very very far from trivial), if $M$ is semifinite then the modular operator is the identity. That is, at least when talking about factors the modular operator is non-trivial only on factors of type III.
That said, the goal of the paper is not that much to show that the factors are to type III, but to show that the factors are of type III_1 (I'm not entirely sure if the algebras in the paper are factors, but at least in talking about factors I'm more sure I'm saying the right thing). This has to do with A. Connes classification.
Among many many other things, Connes proved that the set
$$
Gamma(M)=(0,infty)cap,bigcap{operatorname{sp}Delta_phi: phi text{ is a fns weight on }M}
$$
is a closed multiplicative group of $(0,infty)$. The only possibilities are
$Gamma(M)={1}$; if $M$ is also type III, we say that $M$ is of type III$_0$
$Gamma(M)={lambda^n: ninmathbb Z}$ for some $lambdain(0,1)$; we say that $M$ is of type III$_lambda$
$gamma(M)=(0,infty)$; we say that $M$ is of type III$_1$.
When $M$ is semifinite, you always have $Gamma(M)={1}$.
So if you can show that the spectrum of $Delta_phi$ is $(0,infty)$, then $M$ is of type III$_1$.
answered Dec 17 '18 at 5:41
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
To try to make things simple (they are not; modular theory and Connes work on it are very very far from trivial), if $M$ is semifinite then the modular operator is the identity. That is, at least when talking about factors the modular operator is non-trivial only on factors of type III.
That said, the goal of the paper is not that much to show that the factors are to type III, but to show that the factors are of type III_1 (I'm not entirely sure if the algebras in the paper are factors, but at least in talking about factors I'm more sure I'm saying the right thing). This has to do with A. Connes classification.
Among many many other things, Connes proved that the set
$$
Gamma(M)=(0,infty)cap,bigcap{operatorname{sp}Delta_phi: phi text{ is a fns weight on }M}
$$
is a closed multiplicative group of $(0,infty)$. The only possibilities are
$Gamma(M)={1}$; if $M$ is also type III, we say that $M$ is of type III$_0$
$Gamma(M)={lambda^n: ninmathbb Z}$ for some $lambdain(0,1)$; we say that $M$ is of type III$_lambda$
$gamma(M)=(0,infty)$; we say that $M$ is of type III$_1$.
When $M$ is semifinite, you always have $Gamma(M)={1}$.
So if you can show that the spectrum of $Delta_phi$ is $(0,infty)$, then $M$ is of type III$_1$.
answered Dec 17 '18 at 5:41
Martin ArgeramiMartin Argerami
127k1182183
edited Jan 15 at 17:51
answered Dec 17 '18 at 5:41
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 17 '18 at 5:41
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 17 '18 at 5:41
Martin ArgeramiMartin Argerami
127k1182183
127k1182183
add a comment |
add a comment |
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$begingroup$
I cannot find anywhere in that paper saying that any algebra is not of type III. Could you be more specific?
$endgroup$
– Martin Argerami
Dec 17 '18 at 4:17
$begingroup$
Sorry, I was wrong. I have just edit the question. I should say "tipe III" instead of "not tipe III". The coment is in the abstract.
$endgroup$
– Gabriel Palau
Dec 17 '18 at 5:27