Understanding the bidual of a $C^*$-algebra as a $C^*$-algebra

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

• 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... Jul 17 '15 at 16:19
• you are right, $A$ instead of $C$ is much better. 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 $$\pi_u(A)''$$.

• 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). Jul 18 '15 at 12:43
• Note that, unless $A$ is finite-dimensional, it is never isomorphic to $\pi_u (A)''$. Not even when $A$ is a von Neumann algebra. Jul 22 '15 at 11:29
• oh sorry, you are right, I meant that i need the identification of $A^{**}$ with $\overline{\pi(A)}^{s.o.}$ as Banach spaces . Jul 22 '15 at 11:42