Is there a trivial construction of the trace on the Jones basic construction? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T15:02:03Z http://mathoverflow.net/feeds/question/60296 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/60296/is-there-a-trivial-construction-of-the-trace-on-the-jones-basic-construction Is there a trivial construction of the trace on the Jones basic construction? Jon Bannon 2011-04-01T14:44:43Z 2011-04-02T11:26:17Z <p>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.</p> <blockquote> <p>Question: What is the shortest known proof of the existence of the trace on the basic construction for a <em>von Neumann subalgebra</em> inclusion?</p> </blockquote> <p>Two complete proofs of this appear in the excellent book <em>Finite von Neumann algebras and Masas</em> by Sinclair and Smith. I (and others far more adept than I) am curious if any other proofs of this exist in the literature.</p> <p><strong>EDIT:</strong> 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.</p> http://mathoverflow.net/questions/60296/is-there-a-trivial-construction-of-the-trace-on-the-jones-basic-construction/60349#60349 Answer by Dima Shlyakhtenko for Is there a trivial construction of the trace on the Jones basic construction? Dima Shlyakhtenko 2011-04-02T06:22:27Z 2011-04-02T06:22:27Z <p>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 $N$?)</p> <p><strong>Lemma.</strong> *Let $J:L^2(N)\to L^2(N)$ be the modular conjugation. Then $\langle N, e_B\rangle$ is $J B' J$.* </p> <p>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 $\langle N, e_B\rangle \subset JB'J$.</p> <p>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).</p>