Just let me say first, I am no expert neither in $C^*$-algebras nor in $W^*$-algebras. But I came across the following question:
Let $A$ be a $C^*$-algebra. Then its bidual $A^{**}$ is also a $C^*$-algebra and thus a $W^*$-algebra since it obviously has a predual, namely $A^*$. My question is whether the ultra-weak topology $\sigma(A^{**}, A^*)$ coincides with the weak topology $\sigma(A^{**}, A^{***})$ on the closed unit ball $B_1$ of on $A^{**}$.
I think the statement is true, but I cannot find any reference for it. But I guess I just do not know the literature on that topic well enough. I would prefer a reference where it is stated in the above context and not under the identification that $A^{**}$ is isomorphic to some $L(H)$, where $H$ is some Hilbert space. However, I would be also glad with a reference in the latter setting. Thanks in advance!
This is never true if $A$ is infinite-dimensional. In this case, $A$ is a proper (closed) subspace of $A^{\ast\ast}$. Let $q\colon A^{\ast\ast}\to A^{\ast\ast}/A$ be the quotient map and $\psi\in (A^{\ast\ast}/A)^\ast$ a non-zero linear functional. Then $\phi=\psi\circ q$ is a non-zero bounded linear functional on $A^{\ast\ast}$ that vanishes on $A$.
By Goldstine's theorem, the unit ball $(A)_1$ is weak$^\ast$ dense in $(A^{\ast\ast})_1$. Let $x\in (A^{\ast\ast})_1$ such that $\phi(x)\neq 0$ and $(x_\alpha)$ a net in $(A)_1$ such that $x_\alpha\to x$ in $\sigma(A^{\ast\ast},A^\ast)$. Since $\phi(x_\alpha)=0\not\to \phi(x)$, the net $(x_\alpha)$ does not converge to $x$ in $\sigma(A^{\ast\ast},A^{\ast\ast\ast})$.