## Is there a trivial construction of the trace on the Jones basic construction?

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.

Question: What is the shortest known proof of the existence of the trace on the basic construction for a von Neumann subalgebra inclusion?

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, semifinite trace. If someone could indicate how to do this, I'd very much appreciate it.

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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).

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You didn't miss anything, and have answered the question I (am embarrassed to) have asked. What I meant to ask is if there is a short and easy explicit construction of the trace, and verification that it is, in fact, a normal, semifinite trace. I would still very much like to see this, if you should know how to do it. – Jon Bannon Apr 2 2011 at 11:09
I think your question is: how to construct a semifinite trace on $B'$ if $B\subset B(H)$ is a subalgebra for which some vector $\xi$ gives a tracial state $x\mapsto \langle x\xi,\xi\rangle$ (think for example of $B=\mathbb{C}\subset B(\mathbb{C}^N$ with trace vector $e_1=(1,0,...)$). You need to find isometries $v_i\in B'$ with $\sum v_iv_i^*=1$ and $v_i^* v_i\leq P$, where $P$ is the projection onto the closure of $B\xi\subset H$. Then $Tr(x)=\sum \langle v_i^* x v_i \xi ,\xi \rangle$ is the semifinite trace you want. This is a classical construction going back to Murray-von Neumann. – Dima Shlyakhtenko Apr 2 2011 at 16:39
Thanks, Dima. I'll give it another try! (Trying to "learn and relearn" the subject.) – Jon Bannon Apr 2 2011 at 19:16