Let $N$ be a type $II\_{1}$-factor with trace $\tau$, and $B$ a von Neumann subalgebra. The existence of the semifinite trace on the Jones basic construction $\langle N, e\_{B} \rangle$ is reasonably easy to establish if $B$ is a subfactor of $N$, but appears not to be so easy in general.



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> Question: What is the shortest known proof of the existence of the trace on the basic construction for a *von Neumann subalgebra* inclusion?
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Two complete proofs of this appear in the excellent book *Finite von Neumann algebras and Masas* by Sinclair and Smith. I (and others far more adept than I) am curious if any other proofs of this exist in the literature.


**EDIT:** The previous question, as asked, is embarrassingly silly. What seems less than obvious is how to construct the trace and verify that what you have written down is actually a faithful, NORMAL, semifinite trace. If someone could indicate how to do this, I'd very much appreciate it.



Is there a trivial construction of the trace on the Jones basic construction?
Perhaps I am missing some hypothesis, but I think the proof is just about the same whether or not $B$ is a factor. Here is the proof from Jones' original paper, and I believe it does not use factoriality of B (and not even facoriality of $N$?)


**Lemma.** *Let $J:L^2(N)\to L^2(N)$ be the modular conjugation. Then $\langle N, e\_B\rangle$ is $J B' J$.* 


Indeed, clearly $N=JN'J \subset JB'J$ and $e\_B=Je\_BJ\subset JB'J$ since $L^2(B)$ is $B$-invariant; thus $\langle N,e\_B\rangle \subset JN'J$. On the other hand, $J\langle N, e\_B\rangle J ' \subset JNJ ' \cap \{e\_B\}' =\{e\_B\}'\cap N$ since $JNJ'=N$ and $Je\_BJ=e\_B$. If $x\in N$ commutes with $e\_B$ then, denoting by $1\in L^2(N)$ the trace vector, $x e\_B 1 = e\_B x 1 \in L^2(B)$ so that $x\in L^2(B)$. Thus $x=E\_B(x)$ (where $E\_B$ is the trace-preserving conditional expectation onto $B$) and so $x\in B$. Thus $\{e\_B'\}\cap N = B$ and as a result $J\langle N, e\_B\rangle J' \subset B$. By the bicommutant theorem you then get that $\langle N, e\_B\rangle \subset JB'J$.


Now, given the Lemma, it is clear that $JB'J$ has a semi-finite trace, since $B'$ has a semi-finite trace (since $B$ has a finite trace).


