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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.

metricforbees
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rkmath
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  • I have been wondering about the same question to get the hang of conditional expectation. In the C*-algebras case, you can use Tomiyama's theorem to say that a conditional expectation is just a contractive projection. This is much easier to work with. – Panini Jun 09 '23 at 12:24
  • @AksharaPrasad Is it possible that you had a question on MSE about this recently? I wanted to answer this, but could not find it anymore, so I posted my answer here. – J. De Ro Jun 09 '23 at 13:23
  • @J.DeRo Yes, I did. My question was closed because it was already answered at https://math.stackexchange.com/questions/3691011/conditional-expectations-on-c-algebras – Panini Jun 09 '23 at 14:29

2 Answers2

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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.

Adrián González Pérez
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  • Thank you. Can you please elaborate me what you mean by 'dualizing the inclusions' in the last paragraph.? What goes wrong in the case of a general von Neumann algebra(not finite)? – rkmath Sep 06 '18 at 15:05
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    If you have a bounded map $T:X \to Y$ between Baach spaces you can defince $T^:Y^\ast \to X^\ast$ in the natural way by $T^(\lambda)(x) = \lambda(T(x))$, that what i mean by "dualizing" (I am using that $(L^1)^\ast = L^\infty$) – Adrián González Pérez Sep 06 '18 at 17:52
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    In the case of non-finite algebras you risk that the restriction of the trace to the smaller algebra may not be semifinite. You can already observe that in the case of $\mathbb{R}$ with the Lebesgue measure $m$. Take the sub-$\sigma$-algebra $\Sigma_0$ generated by $(\infty,0]$ and $[0,\infty)$. The measure $m{|}_{\Sigma_0}$ is identically $\infty$ and so the inclusion of the $\Sigma_0$-measurable functions do not extend to $L^1$. – Adrián González Pérez Sep 06 '18 at 17:55
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    I get a little confused here.

    I think the definition for conditional expectation in probability theory should mean a map $E:L^1(X,\Sigma)\to L^1(X.\Sigma_0)$, then if we dualize this map we get $E^*:L^\infty(X,\Sigma_0)\to L^\infty(X.\Sigma)$, which is not a conditional expectation for von Neumann algebras.

    And if we want to get a conditional expectation in von Neumann algebra sense, we need to dualize the inclusion map. It is exactly what you did in this answer.

    So I think the conditional expectation are not the same in probability theory and von Neumann algebra sense?

     – Yanyu Oct 26 '22 at 05:29
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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!

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 $\omega \circ \mathbb{E}[-|\mathcal{G}]= \omega$.

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 $E(X) = \mathbb{E}[X|\mathcal{G}]$. $\quad \square$

J. De Ro
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