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Let $(M, H, H_+,J)$ be a standard form of a von Neumann algebra $M$ acting on a complex Hilbert space $H$ endowed with a self-dual cone $H_+$. Is it true that $$\langle x,yz\rangle=\langle zx,y\rangle$$ whenever $x,y,z$ are real?

The motivation for this question is that the above property holds on non-commutative $L^2$-spaces (with a trace).

I have tried looking at the literature but did not find this stated anywhere. For example, I have already looked at Chapter IX of Takesaki.

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  • $\begingroup$ What are $x,y,z$? If the are elements of $H$, you cannot multiply them (and end up in $H$). If they are elements of $M$, you cannot take their inner product before fixing an embedding of $M$ into $H$. $\endgroup$
    – MaoWao
    Commented Jul 30, 2023 at 4:58
  • $\begingroup$ @MaoWao Scusi. What if $x,y\in H$ but $z\in M$ (say, a difference of projections)? Being a standard form, there already had to be an embedding of $M$ in $H$, correct? $\endgroup$
    – Guest
    Commented Jul 30, 2023 at 9:41
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    $\begingroup$ The definition of standard forms does not include an embedding of $M$ into $H$. Even when you fix a non-tracial normal state (or n.s.f. weight), there is a whole zoo of embeddings of $M$ into $H$. Btw, I do not think your identity holds even in finite von Neumann algebras: The left side is $\tau(xyz)$, the right side is $\tau(xzy)$. $\endgroup$
    – MaoWao
    Commented Jul 30, 2023 at 11:57

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