Conditional Expectation on von Neumann algebras 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.
 A: 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.
