Mixing
Traciality of compressions of von Neumann algebras
$begingroup$
Let $phi_1$ be a linear functional on a von Neumann algebra $mathcal{A}.$ (I need the result in particular for $Pi_1$-factors), satisfying traciality. With "traciality" I mean the following: For $A,Binmathcal{A}$ we have $$phi_1(A B)=phi_1(B A).$$ Let further $pinmathcal{A}$ be a projection and $M_p:mathcal{A}rightarrow p mathcal{A} p$ be the compression of $mathcal{A}$ with respect to $p.$ Define a linear functional $phi_2$ on $mathcal{A}$ by $phi_2 (A):=phi_1 (p A p)$ for $Ainmathcal{A}.$
My question is: Does the traciality of $phi_1$ imply that $$phi_2 (A B) = phi_2(B A)?$$ Do I need additional assumptions for this to hold? Does anyone know a source where this might be covered?
Thank you very much! :)
operator-theory operator-algebras trace von-neumann-algebras
asked Feb 1 at 13:48
AlvoAlvo
375
$endgroup$
add a comment |
$begingroup$
Let $phi_1$ be a linear functional on a von Neumann algebra $mathcal{A}.$ (I need the result in particular for $Pi_1$-factors), satisfying traciality. With "traciality" I mean the following: For $A,Binmathcal{A}$ we have $$phi_1(A B)=phi_1(B A).$$ Let further $pinmathcal{A}$ be a projection and $M_p:mathcal{A}rightarrow p mathcal{A} p$ be the compression of $mathcal{A}$ with respect to $p.$ Define a linear functional $phi_2$ on $mathcal{A}$ by $phi_2 (A):=phi_1 (p A p)$ for $Ainmathcal{A}.$
My question is: Does the traciality of $phi_1$ imply that $$phi_2 (A B) = phi_2(B A)?$$ Do I need additional assumptions for this to hold? Does anyone know a source where this might be covered?
Thank you very much! :)
operator-theory operator-algebras trace von-neumann-algebras
asked Feb 1 at 13:48
AlvoAlvo
375
$endgroup$
$begingroup$
For II$_1$-factors this follows from uniqueness of the trace
$endgroup$
– Nick Bottom
Feb 1 at 13:57
$begingroup$
Could you please elaborate on this?
$endgroup$
– Alvo
Feb 1 at 14:51
add a comment |
$begingroup$
Let $phi_1$ be a linear functional on a von Neumann algebra $mathcal{A}.$ (I need the result in particular for $Pi_1$-factors), satisfying traciality. With "traciality" I mean the following: For $A,Binmathcal{A}$ we have $$phi_1(A B)=phi_1(B A).$$ Let further $pinmathcal{A}$ be a projection and $M_p:mathcal{A}rightarrow p mathcal{A} p$ be the compression of $mathcal{A}$ with respect to $p.$ Define a linear functional $phi_2$ on $mathcal{A}$ by $phi_2 (A):=phi_1 (p A p)$ for $Ainmathcal{A}.$
My question is: Does the traciality of $phi_1$ imply that $$phi_2 (A B) = phi_2(B A)?$$ Do I need additional assumptions for this to hold? Does anyone know a source where this might be covered?
Thank you very much! :)
operator-theory operator-algebras trace von-neumann-algebras
asked Feb 1 at 13:48
AlvoAlvo
375
$endgroup$
Let $phi_1$ be a linear functional on a von Neumann algebra $mathcal{A}.$ (I need the result in particular for $Pi_1$-factors), satisfying traciality. With "traciality" I mean the following: For $A,Binmathcal{A}$ we have $$phi_1(A B)=phi_1(B A).$$ Let further $pinmathcal{A}$ be a projection and $M_p:mathcal{A}rightarrow p mathcal{A} p$ be the compression of $mathcal{A}$ with respect to $p.$ Define a linear functional $phi_2$ on $mathcal{A}$ by $phi_2 (A):=phi_1 (p A p)$ for $Ainmathcal{A}.$
My question is: Does the traciality of $phi_1$ imply that $$phi_2 (A B) = phi_2(B A)?$$ Do I need additional assumptions for this to hold? Does anyone know a source where this might be covered?
Thank you very much! :)
operator-theory operator-algebras trace von-neumann-algebras
operator-theory operator-algebras trace von-neumann-algebras
asked Feb 1 at 13:48
AlvoAlvo
375
asked Feb 1 at 13:48
AlvoAlvo
375
edited Feb 1 at 14:57
Alvo
asked Feb 1 at 13:48
AlvoAlvo
375
asked Feb 1 at 13:48
AlvoAlvo
375
asked Feb 1 at 13:48
AlvoAlvo
375
375
$begingroup$
For II$_1$-factors this follows from uniqueness of the trace
$endgroup$
– Nick Bottom
Feb 1 at 13:57
$begingroup$
Could you please elaborate on this?
$endgroup$
– Alvo
Feb 1 at 14:51
add a comment |
$begingroup$
For II$_1$-factors this follows from uniqueness of the trace
$endgroup$
– Nick Bottom
Feb 1 at 13:57
$begingroup$
Could you please elaborate on this?
$endgroup$
– Alvo
Feb 1 at 14:51
$begingroup$
For II$_1$-factors this follows from uniqueness of the trace
$endgroup$
– Nick Bottom
Feb 1 at 13:57
$begingroup$
For II$_1$-factors this follows from uniqueness of the trace
$endgroup$
– Nick Bottom
Feb 1 at 13:57
$begingroup$
Could you please elaborate on this?
$endgroup$
– Alvo
Feb 1 at 14:51
$begingroup$
Could you please elaborate on this?
$endgroup$
– Alvo
Feb 1 at 14:51
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First of all, you have no relation between $phi_1$ and $phi_2$ so no, of course there is no implication.
But, more importantly, the relation you want does not even hold for $phi_1$. For instance in $M_2(mathbb C)$ (but you can easily lift this example to any II$_1$-factor), let
$$
p=begin{bmatrix} 1&0\0&0end{bmatrix}, A=begin{bmatrix} 1& 2\3&4end{bmatrix}, B=begin{bmatrix} 5&6\7&8end{bmatrix}.
$$
Then
$$
operatorname{Tr}(pABp)=19ne23=operatorname{Tr}(pBAp).
$$
In the case where $phi_2(A)=tfrac1{phi_1(p)}phi_1(pAp)$ (the factor to normalize $phi_2$ so that it is unital if $phi_1$ is) then yes, $phi_2$ is a trace on $pmathcal Ap$. If $A,Bin pmathcal Ap$, then $A=pAp$, $B=pBp$, so
$$
phi_2(AB)=tfrac1{phi_1(p)},phi_1(pAp,pBp)=tfrac1{phi_1(p)},phi_1(pBp,pAp)=phi_2(BA).
$$
answered Feb 1 at 14:00
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Thank you very much! Yes, you are completely right. I meant to define $phi_2$ on $mathcal{A}$ by $phi_2 (A) := phi_1 (pAp)$ for $Ainmathcal{A}.$ The way I wrote it did not constiute any connection between the two traces. Thanks again! :)
$endgroup$
– Alvo
Feb 1 at 14:46
1
$begingroup$
Then $phi_2$ is a trace, but you still need to be careful with the way you wrote the equality. It's true if $A,Bin pmathcal Ap$, but not in general if $A,Bin mathcal A$
$endgroup$
– Martin Argerami
Feb 1 at 14:56
$begingroup$
Thank you. I changed the definition of $phi_2$ accordingly. This should be fine now, right?
$endgroup$
– Alvo
Feb 1 at 14:59
$begingroup$
Yes, $ $
$endgroup$
– Martin Argerami
Feb 1 at 15:03
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First of all, you have no relation between $phi_1$ and $phi_2$ so no, of course there is no implication.
But, more importantly, the relation you want does not even hold for $phi_1$. For instance in $M_2(mathbb C)$ (but you can easily lift this example to any II$_1$-factor), let
$$
p=begin{bmatrix} 1&0\0&0end{bmatrix}, A=begin{bmatrix} 1& 2\3&4end{bmatrix}, B=begin{bmatrix} 5&6\7&8end{bmatrix}.
$$
Then
$$
operatorname{Tr}(pABp)=19ne23=operatorname{Tr}(pBAp).
$$
In the case where $phi_2(A)=tfrac1{phi_1(p)}phi_1(pAp)$ (the factor to normalize $phi_2$ so that it is unital if $phi_1$ is) then yes, $phi_2$ is a trace on $pmathcal Ap$. If $A,Bin pmathcal Ap$, then $A=pAp$, $B=pBp$, so
$$
phi_2(AB)=tfrac1{phi_1(p)},phi_1(pAp,pBp)=tfrac1{phi_1(p)},phi_1(pBp,pAp)=phi_2(BA).
$$
answered Feb 1 at 14:00
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Thank you very much! Yes, you are completely right. I meant to define $phi_2$ on $mathcal{A}$ by $phi_2 (A) := phi_1 (pAp)$ for $Ainmathcal{A}.$ The way I wrote it did not constiute any connection between the two traces. Thanks again! :)
$endgroup$
– Alvo
Feb 1 at 14:46
1
$begingroup$
Then $phi_2$ is a trace, but you still need to be careful with the way you wrote the equality. It's true if $A,Bin pmathcal Ap$, but not in general if $A,Bin mathcal A$
$endgroup$
– Martin Argerami
Feb 1 at 14:56
$begingroup$
Thank you. I changed the definition of $phi_2$ accordingly. This should be fine now, right?
$endgroup$
– Alvo
Feb 1 at 14:59
$begingroup$
Yes, $ $
$endgroup$
– Martin Argerami
Feb 1 at 15:03
add a comment |
$begingroup$
First of all, you have no relation between $phi_1$ and $phi_2$ so no, of course there is no implication.
But, more importantly, the relation you want does not even hold for $phi_1$. For instance in $M_2(mathbb C)$ (but you can easily lift this example to any II$_1$-factor), let
$$
p=begin{bmatrix} 1&0\0&0end{bmatrix}, A=begin{bmatrix} 1& 2\3&4end{bmatrix}, B=begin{bmatrix} 5&6\7&8end{bmatrix}.
$$
Then
$$
operatorname{Tr}(pABp)=19ne23=operatorname{Tr}(pBAp).
$$
In the case where $phi_2(A)=tfrac1{phi_1(p)}phi_1(pAp)$ (the factor to normalize $phi_2$ so that it is unital if $phi_1$ is) then yes, $phi_2$ is a trace on $pmathcal Ap$. If $A,Bin pmathcal Ap$, then $A=pAp$, $B=pBp$, so
$$
phi_2(AB)=tfrac1{phi_1(p)},phi_1(pAp,pBp)=tfrac1{phi_1(p)},phi_1(pBp,pAp)=phi_2(BA).
$$
answered Feb 1 at 14:00
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
$begingroup$
Thank you very much! Yes, you are completely right. I meant to define $phi_2$ on $mathcal{A}$ by $phi_2 (A) := phi_1 (pAp)$ for $Ainmathcal{A}.$ The way I wrote it did not constiute any connection between the two traces. Thanks again! :)
$endgroup$
– Alvo
Feb 1 at 14:46
1
$begingroup$
Then $phi_2$ is a trace, but you still need to be careful with the way you wrote the equality. It's true if $A,Bin pmathcal Ap$, but not in general if $A,Bin mathcal A$
$endgroup$
– Martin Argerami
Feb 1 at 14:56
$begingroup$
Thank you. I changed the definition of $phi_2$ accordingly. This should be fine now, right?
$endgroup$
– Alvo
Feb 1 at 14:59
$begingroup$
Yes, $ $
$endgroup$
– Martin Argerami
Feb 1 at 15:03
add a comment |
$begingroup$
First of all, you have no relation between $phi_1$ and $phi_2$ so no, of course there is no implication.
But, more importantly, the relation you want does not even hold for $phi_1$. For instance in $M_2(mathbb C)$ (but you can easily lift this example to any II$_1$-factor), let
$$
p=begin{bmatrix} 1&0\0&0end{bmatrix}, A=begin{bmatrix} 1& 2\3&4end{bmatrix}, B=begin{bmatrix} 5&6\7&8end{bmatrix}.
$$
Then
$$
operatorname{Tr}(pABp)=19ne23=operatorname{Tr}(pBAp).
$$
In the case where $phi_2(A)=tfrac1{phi_1(p)}phi_1(pAp)$ (the factor to normalize $phi_2$ so that it is unital if $phi_1$ is) then yes, $phi_2$ is a trace on $pmathcal Ap$. If $A,Bin pmathcal Ap$, then $A=pAp$, $B=pBp$, so
$$
phi_2(AB)=tfrac1{phi_1(p)},phi_1(pAp,pBp)=tfrac1{phi_1(p)},phi_1(pBp,pAp)=phi_2(BA).
$$
answered Feb 1 at 14:00
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
First of all, you have no relation between $phi_1$ and $phi_2$ so no, of course there is no implication.
But, more importantly, the relation you want does not even hold for $phi_1$. For instance in $M_2(mathbb C)$ (but you can easily lift this example to any II$_1$-factor), let
$$
p=begin{bmatrix} 1&0\0&0end{bmatrix}, A=begin{bmatrix} 1& 2\3&4end{bmatrix}, B=begin{bmatrix} 5&6\7&8end{bmatrix}.
$$
Then
$$
operatorname{Tr}(pABp)=19ne23=operatorname{Tr}(pBAp).
$$
In the case where $phi_2(A)=tfrac1{phi_1(p)}phi_1(pAp)$ (the factor to normalize $phi_2$ so that it is unital if $phi_1$ is) then yes, $phi_2$ is a trace on $pmathcal Ap$. If $A,Bin pmathcal Ap$, then $A=pAp$, $B=pBp$, so
$$
phi_2(AB)=tfrac1{phi_1(p)},phi_1(pAp,pBp)=tfrac1{phi_1(p)},phi_1(pBp,pAp)=phi_2(BA).
$$
answered Feb 1 at 14:00
Martin ArgeramiMartin Argerami
129k1184185
edited Feb 1 at 14:55
answered Feb 1 at 14:00
Martin ArgeramiMartin Argerami
129k1184185
answered Feb 1 at 14:00
Martin ArgeramiMartin Argerami
129k1184185
answered Feb 1 at 14:00
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
$begingroup$
Thank you very much! Yes, you are completely right. I meant to define $phi_2$ on $mathcal{A}$ by $phi_2 (A) := phi_1 (pAp)$ for $Ainmathcal{A}.$ The way I wrote it did not constiute any connection between the two traces. Thanks again! :)
$endgroup$
– Alvo
Feb 1 at 14:46
1
$begingroup$
Then $phi_2$ is a trace, but you still need to be careful with the way you wrote the equality. It's true if $A,Bin pmathcal Ap$, but not in general if $A,Bin mathcal A$
$endgroup$
– Martin Argerami
Feb 1 at 14:56
$begingroup$
Thank you. I changed the definition of $phi_2$ accordingly. This should be fine now, right?
$endgroup$
– Alvo
Feb 1 at 14:59
$begingroup$
Yes, $ $
$endgroup$
– Martin Argerami
Feb 1 at 15:03
add a comment |
$begingroup$
Thank you very much! Yes, you are completely right. I meant to define $phi_2$ on $mathcal{A}$ by $phi_2 (A) := phi_1 (pAp)$ for $Ainmathcal{A}.$ The way I wrote it did not constiute any connection between the two traces. Thanks again! :)
$endgroup$
– Alvo
Feb 1 at 14:46
1
$begingroup$
Then $phi_2$ is a trace, but you still need to be careful with the way you wrote the equality. It's true if $A,Bin pmathcal Ap$, but not in general if $A,Bin mathcal A$
$endgroup$
– Martin Argerami
Feb 1 at 14:56
$begingroup$
Thank you. I changed the definition of $phi_2$ accordingly. This should be fine now, right?
$endgroup$
– Alvo
Feb 1 at 14:59
$begingroup$
Yes, $ $
$endgroup$
– Martin Argerami
Feb 1 at 15:03
$begingroup$
Thank you very much! Yes, you are completely right. I meant to define $phi_2$ on $mathcal{A}$ by $phi_2 (A) := phi_1 (pAp)$ for $Ainmathcal{A}.$ The way I wrote it did not constiute any connection between the two traces. Thanks again! :)
$endgroup$
– Alvo
Feb 1 at 14:46
$begingroup$
Thank you very much! Yes, you are completely right. I meant to define $phi_2$ on $mathcal{A}$ by $phi_2 (A) := phi_1 (pAp)$ for $Ainmathcal{A}.$ The way I wrote it did not constiute any connection between the two traces. Thanks again! :)
$endgroup$
– Alvo
Feb 1 at 14:46
1
1
$begingroup$
Then $phi_2$ is a trace, but you still need to be careful with the way you wrote the equality. It's true if $A,Bin pmathcal Ap$, but not in general if $A,Bin mathcal A$
$endgroup$
– Martin Argerami
Feb 1 at 14:56
$begingroup$
Then $phi_2$ is a trace, but you still need to be careful with the way you wrote the equality. It's true if $A,Bin pmathcal Ap$, but not in general if $A,Bin mathcal A$
$endgroup$
– Martin Argerami
Feb 1 at 14:56
$begingroup$
Thank you. I changed the definition of $phi_2$ accordingly. This should be fine now, right?
$endgroup$
– Alvo
Feb 1 at 14:59
$begingroup$
Thank you. I changed the definition of $phi_2$ accordingly. This should be fine now, right?
$endgroup$
– Alvo
Feb 1 at 14:59
$begingroup$
Yes, $ $
$endgroup$
– Martin Argerami
Feb 1 at 15:03
$begingroup$
Yes, $ $
$endgroup$
– Martin Argerami
Feb 1 at 15:03
add a comment |
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$begingroup$
For II$_1$-factors this follows from uniqueness of the trace
$endgroup$
– Nick Bottom
Feb 1 at 13:57
$begingroup$
Could you please elaborate on this?
$endgroup$
– Alvo
Feb 1 at 14:51