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Let $A$ be a $C^\ast$-algebra. Consider the canonical embedding $A\to A^{**},\; a\mapsto i(a)$, such that $i(a)(a^*)=a^*(a)$ for all $a\in A$. Here is $A^{**}$ considered as a Banach space. It's well known that $i(A)$ is weak$*$-dense in $A^{**}$. I want to endow $A^{**}$ with an inner multiplication, a norm and an involution map, such that $A^{**}$ is a $C^\ast$-algebra. The canonical embedding $i$ is not multiplicative in general. Therefore I want to know: Does it make sense, if one defines the multiplication and an involution on $A$ as follows: $i(a)i(b):=i(ab)$ for all $a,b\in A$ and $i(a)^*:=i(a^*)$ for all $a\in A$ and then you extend continuously on $A^{**}$?

Regards

Sabrina Gemsa
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1 Answers1

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I'm not sure what you mean by "the canonical embedding is not multiplicative" if you're only taking $A^{**}$ as a Banach space. A natural way to give $A^{**}$ multiplicative structure is via the Arens product, which does make the canonical embedding multiplicative. So this shows that your prescription of importing a product into $A^{**}$ using $i$ and extending by continuity does work. The analogous statements for the adjoint operation are also easy.

Since you're asking about the C*-algebra structure of $A^{**}$, are you aware that it is naturally $*$-isomorphic to the weak* completion of $A$ in its universal representation? This construction is called the universal enveloping von Neumann algebra and the statement of isometry is the Sherman-Takeda theorem.

Nik Weaver
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  • thanks a lot!I'm not familiar with the Arens product until now but I will change it. I first wanted to use a different approach for bidual c*algebras. And didn't know the Sherman-Takeda-theorem until now too, but good to know! – Sabrina Gemsa Aug 05 '15 at 18:32
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    I should mention that the Sherman-Takeda identification provides another way of giving $A^{**}$ multiplicative structure. I actually prefer this method to the Arens product definition, which I can never keep straight ... – Nik Weaver Aug 05 '15 at 18:32