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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 $A,B\in\mathcal{A}$ we have $$\phi_1(A B)=\phi_1(B A).$$ Let further $p\in\mathcal{A}$ 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 $A\in\mathcal{A}.$ 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! :)

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  • $\begingroup$ For II$_1$-factors this follows from uniqueness of the trace $\endgroup$ Feb 1 '19 at 13:57
  • $\begingroup$ Could you please elaborate on this? $\endgroup$
    – Alvo
    Feb 1 '19 at 14:51
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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 $$ p=\begin{bmatrix} 1&0\\0&0\end{bmatrix},\ \ A=\begin{bmatrix} 1& 2\\3&4\end{bmatrix},\ \ B=\begin{bmatrix} 5&6\\7&8\end{bmatrix}. $$ Then $$ \operatorname{Tr}(pABp)=19\ne23=\operatorname{Tr}(pBAp). $$


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 $p\mathcal Ap$. If $A,B\in p\mathcal Ap$, then $A=pAp$, $B=pBp$, so $$ \phi_2(AB)=\tfrac1{\phi_1(p)}\,\phi_1(pAp\,pBp)=\tfrac1{\phi_1(p)}\,\phi_1(pBp\,pAp)=\phi_2(BA). $$

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  • $\begingroup$ 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 $A\in\mathcal{A}.$ The way I wrote it did not constiute any connection between the two traces. Thanks again! :) $\endgroup$
    – Alvo
    Feb 1 '19 at 14:46
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    $\begingroup$ Then $\phi_2$ is a trace, but you still need to be careful with the way you wrote the equality. It's true if $A,B\in p\mathcal Ap$, but not in general if $A,B\in \mathcal A$ $\endgroup$ Feb 1 '19 at 14:56
  • $\begingroup$ Thank you. I changed the definition of $\phi_2$ accordingly. This should be fine now, right? $\endgroup$
    – Alvo
    Feb 1 '19 at 14:59
  • $\begingroup$ Yes, $\ \ \ \ \ \ $ $\endgroup$ Feb 1 '19 at 15:03

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