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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.

Stephen Curry
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1 Answers1

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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.

Martin Argerami
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  • 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? – Stephen Curry Mar 10 '14 at 16:12
  • 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. – Martin Argerami Mar 10 '14 at 17:28
  • 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? – Stephen Curry Mar 11 '14 at 15:15
  • 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. – Martin Argerami Mar 11 '14 at 17:32
  • 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? – Shirly Geffen Oct 27 '16 at 16:32
  • On a very quick glance I cannot see that $|T^t|=|T|$, but it is obvious that $|T^t|\leq|T|$. It follows immediately that $|E^{**}|=|E|$. So you are right. – Martin Argerami Oct 27 '16 at 17:33
  • @MartinArgerami The induced multiplication on $A^{*}$ is actually the Arens products (both the Arens product agree on second dual of $C^$-algebras.) – NewB Feb 06 '23 at 21:36