# Properties of Basic Construction in von Neumann Algebras

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 $$L^2(N)$$.

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 $$L^2(N)$$.
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, $$(Ne_BN)'\subset\{\{e_B\}\cup N\}'$$.