Definition 1. Let $B\subset A$ be C*-algebra. A projection from A onto B is a linear map $E: A \rightarrow B$ such that $E(b)=b$ for every $b\in B$. A conditional expectation from A onto B is a contractive completely positive projection $E$ from $A$ onto $B$ such that $E(bxb')=bE(x)b'$ for every $x\in A$ and $b, b'\in B$.

Theorem 2 (Tomiyama). Let $B\subset A$ be C*-algebra and $E$ be a projection from $A$ onto $B$. Then, the following are equivalent:

(1) $E$ is a conditional expectation;

(2) $E$ is contractive completely positive;

(3) $E$ is contractive.

Proof. We only have to prove that the last condition implies the first, so assume $E$ is contractive. Passing to double duals, we may assume that $A$ and $B$ are von Neumann algebras. We first prove that $E$ satisfies $E(bxb')=bE(x)b'$ for every $x\in A$ and $b, b'\in B$..........

My question: I can not understand why we can regard a C*-algebra as a von Neumann algebra? Could someone explain to me in detail? Many thanks.

  • $\begingroup$ Where can I find this theorem? $\endgroup$ – Mahmood Al Jul 12 '16 at 20:19
  • $\begingroup$ It appears as Theorem 1.5.10 in Brown-Ozawa. $\endgroup$ – Martin Argerami Oct 27 '16 at 17:38

You have canonical embeddings $A\hookrightarrow A^{**}$, $B\hookrightarrow B^{**}$. The embedding is given by $a(f)=f(a)$ for all $a\in A$, $f\in A^*$. Restriction allows us to embed $B^{**}$ into $A^{**}$.

Now, a map $E:A\to B$ like the conditional expectation admits a dual map $E^*:B^*\to A^*$ given by $(E^*f)(a)=f(E(a))$. And similarly we can obtain a double dual map $E^{**}:A^{**}\to B^{**}$. When restricted to $A$, the following happens; for $a\in A$, $g\in B^*$, $$ E^{**}(a)(g)=a(E^*g)=(E^*g)(a)=g(E(a))=(E(a))g. $$ This shows that $E^{**}|_A=E$. Since $\|E^{**}\|=\|E\|$, we get that $E^{**}$ is contractive, and now one performs that proof for $E^{**}$, $A^{**}$, $B^{**}$ to show that $E^{**}$ is completely positive. As $E$ is a restriction of $E^{**}$, it will also be completely positive.

| cite | improve this answer | |
  • $\begingroup$ Well, I still have some doubt. Although i know $A^{\ast \ast}$ is isometrically isomorphic to the enveloping von Neumann algebra of $A$, this isomorphism is just spatial isomorphism (in other words, $A^{**}$ do not have the algebraic structure). Why can we still regard $A^{**}$ as a von Neumann algebra in this proof? $\endgroup$ – Yan kai Mar 10 '14 at 16:12
  • $\begingroup$ The isometric isomorphism allows you to think of $E^{**}$ in the enveloping von Neumann algebra. So, speaking properly, one makes the proof for the inclusion of enveloping von Neumann algebras. $\endgroup$ – Martin Argerami Mar 10 '14 at 17:28
  • $\begingroup$ I suppose, even if it is a isometric isomorphism, there is no algebraic structure on $A^{\ast \ast}$, how can we regard $A^{**}$ as a von Neumann algebra? $\endgroup$ – Yan kai Mar 11 '14 at 15:15
  • $\begingroup$ When anyone says "regard $A^{**}$ as a von Neumann algebra, what they mean is "consider the enveloping von Neumann algebra". And actually $A^{**}$ does have an algebraic structure: the one induced by the isomorphism with the enveloping von Neumann algebra. $\endgroup$ – Martin Argerami Mar 11 '14 at 17:32
  • $\begingroup$ Does the weak-* density of $A$ in $A^{**}$ implies $||E^{**}||=||E||$? We don't know that $E$ is "weak$^*$ continuous"... Isn't it a general fact that if we have $T : X \to Y$ a bounded linear map between Banach spaces, then the induced map $T^t: Y^* \to X^*$ has the same norm? $\endgroup$ – Shirly Geffen Oct 27 '16 at 16:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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