Theorem 44.2.8.label Let $(M, \tau)$ be a tracial von Neumann algebra, then:
- (1)
$L(M) = LC(M) = R(M)'$.
- (2)
$JL(M)J = R(M) = RC(M) = L(M)'$.
- (3)
In particular, for each $\xi \in L^{2}(M, \tau)$, $L_{\xi} \in B(L^{2}(M, \tau))$ if and only if $\xi \in \wh M$.
Proof. (1): By definition, $L(M) \subset LC(M)$. By Lemma 44.2.7, $LC(M) \subset RC(M)'$, so
\[L(M) \subset LC(M) \subset RC(M)' \subset R(M)'\]
and
\[R(M) \subset RC(M) \subset LC(M)' \subset L(M)'\]
Using the Bicommutant Theorem, it remains to show that $R(M)' \subset LC(M)$. To this end, let $T \in R(M)'$ and $\xi = T \wh 1$, then for each $x \in M$,
\[T\wh x = TR_{x} \wh 1 = R_{x} T \wh 1 = R_{x} \xi = \xi x = L_{\xi}^{0} \wh x\]
so $T|_{\wh M}= L_{\xi}^{0}$. By density of $\wh M$ in $L^{2}(M, \tau)$ and continuity of $T$, the closure of $L_{\xi}^{0}$ is $T$, hence $T = L_{\xi}$.$\square$
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