Abstract:
Let $\mathscr M$ be a von Neumann algebra of operators in a Hilbert space $\mathscr H$, let $\tau$ be an exact normal semifinite trace on $\mathscr M$, and let $L_1(\mathscr M,\tau)$ be the Banach space of $\tau$-integrable operators. The following results are obtained.
If $X=X^*$, $Y=Y^*$ are $\tau$-measurable operators and $XY\in L_1(\mathscr M,\tau)$, then $YX\in L_1(\mathscr M,\tau)$ and $\tau(XY)=\tau(YX)\in\mathbb R$. In particular, if $X,Y\in\mathscr B(\mathscr H)^{\mathrm{sa}}$ and $XY\in\mathfrak S_1$, then $YX\in \mathfrak S_1$ and $\operatorname{tr}(XY) =\operatorname{tr}(YX)\in\mathbb R$. If $X\in L_1(\mathscr M,\tau)$, then $\tau(X^*)=\overline{\tau(X)}$.
Let $A$ be a $\tau$-measurable operator. If the operator $A$ is $\tau$-compact and $V\in\mathscr M$ is a contraction, then it follows from $V^*AV=A$ that $VA=AV$. We have $A=A^2$ if and only if $A=|A^*||A|$. This representation is also new for bounded idempotents in $\mathscr H$. If $A=A^2\in L_1(\mathscr M,\tau)$, then $\tau(A)=\tau(\sqrt{|A|}\mspace{2mu}|A^*|\sqrt{|A|}\mspace{2mu}) \in\mathbb R^+$. If $A=A^2$ and $A$ (or $A^*$) is semihyponormal, then $A$ is normal, thus $A$ is a projection. If $A=A^3$ and $A$ is hyponormal or cohyponormal, then $A$ is normal, and thus $A=A^*\in\mathscr M$ is the difference of two mutually orthogonal projections $(A+A^2)/2$ and $(A^2-A)/2$. If $A,A^2\in L_1(\mathscr M,\tau)$ and $A=A^3$, then $\tau(A)\in\mathbb R$.