I have a lot of problems trying to understand the double dual of a $C^*$-algebra. Let $A$ be a $C^*$-algebra, I read that if you endow the bidual Banach space $A^{**}$ of $A$ with the weak-*topology, then $A$ is dense in $A^{**}$ with respect to the weak-*topology. More precisely the range of $A$ under the canonical embedding $i:A\to A^{**},\; a\mapsto i(a)$, $i(a)(f)=f(a)$ for all $f\in A^*$, is dense in $A^{**}$ with respect to the weak-*topology. First question is, how to prove it, or do you have a reference?
Then you can extend the $C^*$-norm, the involution and the multiplication of $A$, and you obtain that $A^{**}$ is a $C^*$-algebra.

My second problem is, I need the fact that $\overline{\pi(A)}^{s.o.}$ can be identified with $A^{**}$. With the s.o.-closure I mean the strong operator topology -closure of $\pi(A)$ and $\pi$ is a faithful, non-degenerate *-representation of $A$. My question is: what kind of identification is this, is $\overline{\pi(A)}^{s.o.}\cong A^{**}$ as Banach spaces, Banach algebras or as $C^*$-algebras? I only know that $\overline{\pi(A)}^{s.o.}$ is a Von Neumann algebra. Here a proof is not necessary, but I'm interested in the isomorphism $\overline{\pi(A)}^{s.o.}\to A^{**}$.

I didn't find literature about this topic which is understandable for me and therefore I have lot of problems to understand it. I appreciate your help. Regards

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    $\begingroup$ This is probably petty, but do you mind if I edit your post so that the $C^*$-algebra is $A$, instead of $C$? It's a bit jarring because, once $C^{**}$ is a double dual, the eye tries to read $C^*$ as a single-dual, but... $\endgroup$
    – Mike F
    Jul 17 '15 at 16:19
  • 1
    $\begingroup$ you are right, $A$ instead of $C$ is much better. $\endgroup$
    – banach-c
    Jul 17 '15 at 18:26

The fact that $A$ is weak$^*$-dense in $A^{**}$ is basic functional analysis. I will be surprised if there is a functional analysis book that doesn't contain this result.

For your second question, it is not true as you stated it: $\pi$ cannot be any faithful representation but it is rather a very special one, the universal representation.

The way it works is like this: let $S\subset A^*$ be the state space of $A$, i.e. $$S=\{f\in A^*: \ f\geq0,\ \|f\|=1\}.$$ Then you construct the universal representation in this way: let $H_u=\bigoplus_{f\in S} H_f$ and $\pi_u=\bigoplus_{f\in S}\pi_f:A\to B(H_u)$, where for each $f\in S$ the pair $(H_f,\pi_f)$ is the one given by the GNS construction. So we can consider the von Neumann algebra $\pi_u(A)''$, which is known as the enveloping von Neumann algebra of $A$.

Now the key result is this (see for example Takesaki I, III.2.4):

Theorem. There exists a unique isometry $\pi:A^{**}\to\pi_u(A)''$ that is a homeomorphism with respect to the $\sigma(A^{**},A^*)$-topology on $A^{**}$ and the $\sigma$-weak topology on $\pi_u(A)''$.

In summary, the double dual is just that, the double dual; but because it has a canonical identification (isometric in norm, homeomorphic from the weak$^*$-topology to the $\sigma$-weak topology) with the enveloping von Neumann algebra, we can think about it as a von Neumann algebra. Or, even simpler, whenever someone in the context of C$^*$-algebras says "$A^{**}$", what they really mean is $\pi_u(A)''$.

  • $\begingroup$ okay, thanks for your answer. To my first question, you are right, this is Goldstine's theorem. I totally forgot that.To my second question: Thanks, it is good to know. With regard to the background of my question, I need the identification of $A$ with the s.o.-closure of $\pi_u(A)$. But I can combine this result in Takesaki's book with the fact that $\pi_u(A)''$ is s.o.closure of $\pi_u(A)$ (Von Neumann bicommutant theorem). Or I will ask the professor what to do at this particular point (I need everything for my bachelor's thesis). $\endgroup$
    – banach-c
    Jul 18 '15 at 12:43
  • $\begingroup$ Note that, unless $A $ is finite-dimensional, it is never isomorphic to $\pi_u (A)''$. Not even when $A $ is a von Neumann algebra. $\endgroup$ Jul 22 '15 at 11:29
  • $\begingroup$ oh sorry, you are right, I meant that i need the identification of $A^{**}$ with $\overline{\pi(A)}^{s.o.}$ as Banach spaces . $\endgroup$
    – banach-c
    Jul 22 '15 at 11:42

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