MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In developing a theory of index for inclusions of finite von Neumann algebras, several authors ([Kosaki, 1986], [Fidaleo & 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.

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?

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.

share|cite|improve this question
That was the intended title - thanks. – James Tener Apr 5 '10 at 22:27
up vote 3 down vote accepted

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.

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).

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.