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Suppose, that $A$ is a $C^{\ast}$-algebra. It follows, according to Sakai's book, that double dual of $A$ is also $C^{\ast}$-algebra. I'm not quite sure if I understand the proof correctly. Author proves that $A^{\ast \ast}=\pi(A)''$ where $\pi$ is universal representation showing first that the corresponding preduals coincide. But preduals ale only Banach spaces, so the question rises: how do we define multiplication and involution in the second dual of $A$? I've heard that can be done in natural way (not using the universal representation), using so called Goldstein theorem but I don't know what precisely this theorem states. If we follow this 'natural' procedure via Goldstein theorem and simultaneously use universal representation we get that double dual of $A$ and $\pi(A)''$ are isomorphic as BANACH SPACES-but is it true that they are isomorhic as $C^{\ast}$-algebras (in general it seems that this need not to be true as $\ell^{\infty}$ and $L^{\infty}[0,1]$ are isomorhic as Banach spaces but not as $C^{\ast}$-algebras).


Thank You for Your answers-still there is one detail which I don't understand, namely-we have isomorphism of Banach spaces $T:A^{\ast \ast} \to \pi(A)''$, this isomorphism preserves multiplication when restricted to $A$ and $A$ is dense in $A^{\ast \ast}$ as well as $T(A)$ in $\pi(A)''$. Morever $T$ is $\ast$ weak continuous. But it seems to me that multiplication IS NOT continuos with respect to weak $\ast$ topology, so I'm not sure if we have enough information to conclude that $T$ preserves multiplication in general ( for example preserving $\ast$ structure can be proved via Goldstine-we define $x^{\ast}$ for $x \in A^{\ast \ast}$ as $\lim_{i}a_i^{\ast}$ where $a_i \to x$ is suitable net. As involution $\ast$ on $A$ is continuos with respect to $\ast$ weak topology, therefore it can be extended to whole $A^{\ast \ast}$ and $T$ preserves this operation-how to repeat this argument for the multiplication if we have lack of continuity?)

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    $\begingroup$ Firstly, I think the name is Goldstine. Secondly, what you are after are the Arens products on the dual of any Banach algebra. Unfortunately I am out of the office so cannot give a precise reference for the fact you need, but I suggest looking in volume 2 of Palmer's book on Banach algebras if you have access to a copy. $\endgroup$
    – Yemon Choi
    May 26 '12 at 22:01
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    $\begingroup$ Sorry, that should have been "...on the double dual of any Banach algebra" $\endgroup$
    – Yemon Choi
    May 26 '12 at 22:52
  • $\begingroup$ For future reference, the site interprets asterisks as italics, so use \ast instead. $\endgroup$ May 26 '12 at 23:43
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As Yemon says, it is the Arens product on the double dual of a Banach algebra that you're thinking of. Yes, $A^{**}$ and $\pi(A)''$ are isometrically isomorphic as Banach spaces and $*$-isomorphic as C*-algebras. This is surely in Sakai's book that you're already looking at.

A related question that you didn't quite ask is whether two C*-algebras which are isometrically isomorphic as Banach spaces must be $*$-isomorphic as C*-algebras. The answer is no --- there are C*-algebras which are not $*$-isomorphic to their opposite algebras (change the order of the product and leave everything else, including norm, alone). However, isometric C*-algebras will have the same state space and this implies that they are Jordan isomorphic, and roughly speaking, switching the order of the product is the worst thing that can happen. See the book State Spaces of Operator Algebras by Alfsen and Schultz.

The "right" version of this result is that two C*-algebras that are completely isometric must be $*$-isomorphic.

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I haven't thought about this in a long time. But it seems to me that the situation is like this:

The von Neumann algebra $\pi(A)''$ contains $A$ as a $\sigma$-weakly dense $\ast$-subalgebra. Then any normal linear functional on $M$ restricts to a bounded linear functional on $A$, so we have a map $\rho : \pi(A)''_* \to A^*$. This is a map of Banach spaces.

Then you use Kaplansky Density to show that $\rho$ is an isometry. And finally if you have a state on $A$, it is actually a vector state for $\pi(A)$, so it extends to a normal linear functional on $\pi(A)''$. Thus $\rho$ is an isomorphism of Banach spaces.

Then the dual map $\rho^* : A^{**} \to (\pi(A)''_*)^* \simeq \pi(A)''$ is also an isomorphism of Banach spaces. This allows us to define the algebraic operations in $A^{**}$ via transport of structure through the map $\rho^*$. You are asking how to define these operations intrinsically on $A^{**}$ without the use of the map $\rho^*$.

As you say, one way to understand this is to view $A$ as sitting weak-* densely in $A^{**}$ via Goldstine's Theorem. Then I guess you can try to extend the operations of $A$ to $A^{**}$ via continuity. Your question, then, is why does that definition agree with the operations on $A^{**}$ that we get from $\pi(A)''$?

I have not worked out the details, but it seems to me that it should work like this: we have an isomorphism of Banach spaces $\rho^* : A^{**} \to \pi(A)''$. But we know more, namely that this isomorphism comes from an isomorphism of preduals, so in fact this is an isomorphism of the weak-* topologies on these two spaces.

Both $A^{**}$ and $\pi(A)''$ contain $A$ as a weak-* dense subspace. The multiplication of $\pi(A)''$, when restricted to $A$, is exactly the original multiplication in $A$. And similarly for the multiplication of $A^{**}$ restricted to $A$ (since that's how we defined it). Thus we have a weak-* continuous isomorphism which is an isomorphism of algebras on a dense subspace, so it should be an isomorphism of von Neumann algebras.

There is another approach to defining an intrinsic multiplication on $A^{**}$, which is called Arens multiplication. I don't know much about it, but you can find the original paper "The adjoint of a bilinear operation" (Proceedings of the AMS 2, 1951) by Richard Arens freely available online.

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