Conditional Expectation on von Neumann algebras

Question

A linear map $\phi$ from a von Neumann algebra M to the subalgebra N is called a conditional expectation when $\phi$ has the following properties. 1)$\phi(I)=I$, 2) $\phi(x_{1}y x_{2})=x_{1}\phi(y)x_{2}$ whenever $x_{1},x_{2}\in N$ and $y\in M.$ Can anyone explain me why this map is called conditional expectation? How it is related to the classical case? Thanks in advance.

Answer 1

I assume, by the "classical case", you mean the conditional expectations used in probability theory.

Let $(\Omega, \Sigma, \mathbb P)$ be a probability space with $\Sigma$ its $\sigma$-algebra and let $\Sigma_0 \subset \Sigma$ be a sub-$\sigma$-algebra. You have a natural inclusion $$ L^\infty(\Omega, \Sigma_0) \subset L^\infty(\Omega, \Sigma)$$ Both are von Neumann algebras and the inclusion preserves the probability measure $\mathbb P$. Classical probability theory gives you a conditional expectation $$\mathbb E(\cdot | \Sigma_0): L^\infty(\Omega, \Sigma) \to L^\infty(\Omega, \Sigma_0).$$ This is a (weak-$\ast$ continuous) conditinal expectation in the sense of von Neumann algebras, i.e. it is unit-preserving and $$X \, \mathbb{E}( Y | \Sigma_0) \, Z = \mathbb{E}(X \, Y \, Z | \Sigma_0),$$ for every $Y, Z$ $\Sigma_0$-measurable.

If you want a hint on how to obtain the conditional expectation $\mathbb{E}( \cdot | \Sigma_0)$ just use that, since the inclusion above is $\mathbb P$-preserving, it extends to an inclusion $j: L^1(\Omega, \Sigma_0;\mathbb P) \to L^1(\Omega, \Sigma; \mathbb P)$. Dualizing that inclusion gives the conditional expectation. The same proof works in (finite) von Neumann algebras if the von Neumann sub-algebra is unital.

Answer 2

Another point of view: Fix a probability space $(\Omega, \mathcal{F}, \mathbb{P})$ and a sub-$\sigma$-algebra $\mathcal{G}\subseteq \mathcal{F}$. Then classical conditional expectation from probability theory gives us a map $$\mathbb{E}[-| \mathcal{G}]: L^1(\Omega, \mathcal{F}, \mathbb{P})\to L^1(\Omega, \mathcal{G}, \mathbb{P}).$$ Consider the following Lemma:

Lemma: If $X\in L^\infty(\Omega, \mathcal{F}, \mathbb{P})$, then $Z:= \mathbb{E}[X|\mathcal{G}] \in L^\infty(\Omega, \mathcal{G}, \mathbb{P})$.

Proof lemma: By assumption, there is $M\ge 0$ such that $|X| \le M$ a.e. Then $$|Z| = |\mathbb{E}[X|\mathcal{G}]]\le \mathbb{E}[|X||\mathcal{G}]\le M$$ a.e. which proves the lemma. $\quad \square$

Therefore, by restriction we get a map $$\mathbb{E}[-|\mathcal{G}]: L^\infty(\Omega, \mathcal{F}, \mathbb{P})\to L^\infty(\Omega, \mathcal{G}, \mathbb{P}).$$ It is a conditional expectation of von Neumann algebras!

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We can also prove a converse. First, note that $$\omega: L^\infty(\Omega, \mathcal{F}, \mathbb{P})\to \mathbb{C}: X\mapsto \int_\Omega X d\mathbb{P}$$ is a state with $\omega \circ \mathbb{E}[-|\mathcal{G}]= \omega$.

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We then have:

Proposition: If $$E: L^\infty(\Omega, \mathcal{F}, \mathbb{P})\to L^\infty(\Omega, \mathcal{F}, \mathbb{P})$$ is a conditional expectation with $\omega \circ E = \omega$, then $E= \mathbb{E}[-|\mathcal{G}].$

Proof: Let $A\in \mathcal{G}$. Then for $X\in L^\infty(\Omega, \mathcal{F}, \mathbb{P})$, we have $$\int_A E(X) d\mathbb{P}=\int_\Omega E(X)I_A d\mathbb{P}=\int_\Omega E(XI_A) d\mathbb{P}= \omega(E(XI_A)) = \omega(XI_A) = \int_A X d\mathbb{P}$$ where we used that $E$ is a module map. This means that $E(X) = \mathbb{E}[X|\mathcal{G}]$. $\quad \square$