A non-commutative Radon-Nikodym derivative. - MathOverflow

A non-commutative Radon-Nikodym derivative.

2010-05-01T01:57:37Z

In this classic paper, Sakai proves the following Radon-Nikodym theorem:

Let $M$ be a von Neumann algebra, and let $\phi$ and $\psi$ be two normal positive linear functionals on $M$. If $\psi \leq \phi$, then there is a positive operator $t_0\in M$ such that $0 \leq t_0 \leq 1$, and $\psi(x) = \phi(t_0 x t_0)$ for all $x \in M$.

The paper provides no uniqueness result. One would naively expect that any two such operators $t_0$ and $t_1$ would satisfy $\phi((t_1-t_0)^2)=0$. I can find no such statement in the literature. Is this true?

Please note that $\phi$ is not assumed to be faithful.

Answer by Jon for A non-commutative Radon-Nikodym derivative.

2010-05-01T04:12:33Z

Does this paper have any relevance?

http://arxiv.org/PS_cache/math-ph/pdf/0303/0303056v3.pdf

Answer by Dmitri Pavlov for A non-commutative Radon-Nikodym derivative.

2010-05-01T08:20:02Z

Such t_0 is unique if its support is at most p, where p is the support of ϕ. Note that we can replace t_0 by pt_0p and the support of pt_0p is at most p.

Without this additional condition t_0 is highly non-unique, because we can replace t_0 by t_0 + q, where q is an arbitrary self-adjoint element with support at most 1-p such that t_0 + q ≥ 0. Using simple algebraic manipulations one can show that all solutions can be obtained in this way.

See Lemma 15.4 (page 104) in Takesaki's book “Tomita's theory of modular Hilbert algebras and its applications”. Electronic version: http://gen.lib.rus.ec/get?md5=ACC2A399A5C65C5CB2CCEE7CBEB3FAC3 [Note that Takesaki implicitly assumes that φ_0 is faithful, hence you need to introduce an additional condition on the support of h.]

One would naively expect that any two such operators t_0 and t_1 would satisfy ϕ((t_1−t_0)^2)=0.

This is a trivial corollary of the above statement characterizing all possible solutions.