I am reading Finite von Neumann Algebras and Masas by Sinclair and Smith. I have some questions about the proof of Lemma 4.2.3 (i). First some definitions.

$N$ is a finite von Neumann algebra represented in standard form on $L^2(N)$, $B$ is a von Neumann subalgebra of $N$ and $\xi$ is the identity element in $L^2(N)$. Define $e_B$ to be the projection from $L^2(N)$ to $L^2(B)$ and the basic construction is the von Neumann algebra $\langle N, e_B \rangle := (N \cup \{e_B\})''$.

Lemma 4.2.3 (i) The $^*$-subalgebra $Ne_BN$ is weakly dense in $\langle N, e_B \rangle$.

Proof: It can be shown that $Ne_BN = \textrm{span}\{xe_By : x, y \in N\}$ is a $^*$-subalgebra of $\langle N, e_B \rangle$ and is non-unital in general. If $e$ is the projection onto the closed subspace $\overline{\textrm{span}}\{Ne_BNL^2(N)\}$ in $L^2(N)$, then $e=1$ since $N\xi \subset Ne_BNL^2(N)$. If $p$ is the identity of the weak closure of $Ne_BN$, then $\overline{\textrm{span}}\{Ne_BNL^2(N)\} \subset pL^2(N)$, and it follows that $p=1$. Thus we may apply the double commutant theorem to conclude that $$(Ne_BN)' = \{\{e_B\} \cup N\}' = \langle N, e_B \rangle'$$ and that $\langle N, e_B \rangle = (Ne_BN)''$ as required.

Overall, I am quite confused about the proof. I have the following questions:

  1. Why do we have $N\xi \subset Ne_BNL^2(N)$?
  2. Why do we have $\overline{\textrm{span}}\{Ne_BNL^2(N)\} \subset pL^2(N)$?
  3. How can we conclude that $(Ne_BN)' = \{\{e_B\} \cup N\}'$?

Edit: As Martin has pointed out, it doesn't make sense to say that $\xi$ is the identity element since $L^2(N)$ is a Hilbert space. What I really mean is that $\xi$ is the image of $1 \in N$ under the inclusion of $N$ in $L^2(N)$.


1 Answer 1


Clarification: $\xi$ is not "the identity element in $L^2(N)$", or at least this is a confusing way to say it. What $\xi$ is, is the element $1\in N$ seen as an element of $L^2(N)$.

  1. At the beginning of Section 4.2, the authors state

Whenever we consider a von Neumann subalgebra, we assume a common identity element.

So $1\in B$. Then $\xi=1\,e_B\,1\,\xi\in Ne_BN\,L^2(N)$.

  1. One clearly has that $Ne_BN\,L^2(N)\subset L^2(N)$, and hence $$ Ne_BN\,L^2(N)=pNe_BN\,L^2(N)\subset pL^2(N). $$

  2. If $T\in \{\{e_B\}\cup N\}'$, then $T$ commutes with every element of $N$ and with $e_B$, so $\{\{e_B\}\cup N\}'\subset (Ne_BN)'$. Conversely, because $1\in (Ne_BN)''$, from the Double Commutant Theorem we have that $(Ne_BN)''=\overline{Ne_BN}^{\rm sot}$. Then there exists a net $\{r_j\}\subset Ne_BN$ with $r_j\xrightarrow{\rm sot}1$. Fix $x\in N$. Then $xr_j\in Ne_BN$ and $xr_j\to x$, so $x\in \overline{Ne_BN}^{\rm sot}=(Ne_BN)''$. That is $N\subset (Ne_BN)''$. So $\{\{e_B\}\cup N\}\subset(Ne_BN)''$. Taking commutant, $ (Ne_BN)'\subset\{\{e_B\}\cup N\}'$.


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