# Traciality of compressions of von Neumann algebras

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! :)

• For II$_1$-factors this follows from uniqueness of the trace Feb 1 '19 at 13:57
• Could you please elaborate on this?
– Alvo
Feb 1 '19 at 14:51

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 $$\phi_2(AB)=\tfrac1{\phi_1(p)}\,\phi_1(pAp\,pBp)=\tfrac1{\phi_1(p)}\,\phi_1(pBp\,pAp)=\phi_2(BA).$$

• 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! :)
– Alvo
Feb 1 '19 at 14:46
• 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 $A,B\in \mathcal A$ Feb 1 '19 at 14:56
• Thank you. I changed the definition of $\phi_2$ accordingly. This should be fine now, right?
– Alvo
Feb 1 '19 at 14:59
• Yes, $\ \ \ \ \ \$ Feb 1 '19 at 15:03