most recent 30 from http://mathoverflow.net 2013-05-18T08:14:32Z http://mathoverflow.net/feeds/question/20418 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/20418/when-does-a-conditional-expectation-preserve-some-trace James Tener 2010-04-05T20:23:56Z 2010-04-18T01:19:23Z

<p>In developing a theory of index for inclusions of finite von Neumann algebras, several authors ([Kosaki, 1986], [Fidaleo &amp; Isola,1996], etc.) define the index of a conditional expectation of a von Neumann algebra M onto a vN-subalgebra N (here, a conditional expectation is a normal, faithful N-N bimodule map fixing the subalgebra pointwise). An inclusion is said to have finite index if there exists a conditional expectation that has finite index. However, in the case where M is finite we might be interested in restricting ourselves to the conditional expectations that preserve some trace on M.</p> <p>This leads us to the question: For a given (normal, faithful, finite) trace on M, Umegaki gives us a unique trace preserving conditional expectation E:M->N. Are there any nice necessary and sufficient conditions for a conditional expectation to arise in this manner? What if we allow the trace to be semifinite?</p> <p>Since subfactors give rise to more than one conditional expectation, it is certainly not the case that all conditional expectations come from traces. A necessary condition is that E(xy)=E(yx) whenever x or y is an element of the relative commutant $N^\prime \cap M$. This is not sufficient, however.</p>

http://mathoverflow.net/questions/20418/when-does-a-conditional-expectation-preserve-some-trace/21712#21712 Martin Argerami 2010-04-18T01:19:23Z 2010-04-18T01:19:23Z

<p>I would expect a general answer to be difficult, because the set of traces on your von Neumann algebra will depend a lot on the centre of the algebra. </p> <p>In the case of a factor, the question becomes how to tell if a given expectation is the one that commutes with the trace. Not checking all my facts very carefully, I think that in this case the necessary condition is also sufficient: that is, if $E(xy)=E(yx)$ whenever x is in $N^\prime \cap M$, then $E$ commutes with the trace. This follows from the fact that this condition is equivalent to the modular group of the expectation (see Combes-Delaroche, 1975) being trivial; and in the case of a factor, the modular group characterizes the expectation (Remark 4.12.b in Combes-Delaroche).</p>