I have a proof of something but it is very much provisional on some operator topology questions that I am looking for help with. I will state the problem generally but in fact what I am working with is $\mathcal{O}(S_N^+)\subset C_{\text{u}}(S_N^+)\subset C_{\text{u}}(S_N^+)^{**}$ where $S_N^+$ is the quantum permutation group of Wang on $N\geq 4$ symbols.
Let $\mathcal{A}$ be a unital $\mathrm{C}^*$-algebra generated by finitely many projections and $\mathcal{A}_0\subset \mathcal{A}$ a norm-dense *-algebra, generated by the same projections. We also have the bidual $\mathcal{A}^{**}$, and I understand that $\mathcal{A}\subset \mathcal{A}^{**}$ is weakly dense with respect to the weak operator topology on $\mathcal{A}^{**}$. Assume we have embeddings
$$\mathcal{A}_0\subset \mathcal{A}\subset \mathcal{A}^{**},$$ where $\mathcal{A}^{**}$ is the bidual.
Now the start of some confusion and perhaps an unsurmountable gap. We have $\mathcal{A}\subset \mathcal{A}^{**}$ is weakly dense. I would say, naively, that the norm closure of $\mathcal{A}$ is just $\mathcal{A}$, but perhaps that is for the norm on $\mathcal{A}$. I am guessing for the norm on $\mathcal{A}^{**}$, the norm closure of $\mathcal{A}$ is in fact $\mathcal{A}^{**}$. Now the question is: what kind of density does $\mathcal{A}_0$ have in $\mathcal{A}^{**}$ with respect to topologies on $\mathcal{A}^{**}$? I want some topology that plays nice both with biduals of *-homomorphisms, multiplication, and normal extensions of states on $\mathcal{A}$: oh and also with tensor products.
Let $T:\mathcal{A}\to \mathcal{A}\otimes_{\min}\mathcal{A}$ be a unital *-homomorphism. I understand that $T^{**}:\mathcal{A}^{**}\to (\mathcal{A}\otimes_{\min}\mathcal{A})^{**}$ is an ultraweakly continuous extension of $T$ which is also a *-homomorphism.
Let $p\in\mathcal{A}^{**}$ be a central projection such that the von Neumann algebra $\mathcal{A}^{**}p$ is finite dimensional. It is a sum of finitely many central projections $p=\sum_{i=1}^mq_i$, each $q_i$ in the strong closure of $\mathcal{A}_0$. So in fact I have a sequence in $\mathcal{A}_0$ that converges strongly to $p$. I want a topology $\tau$ on $\mathcal{A}^{**}$ such that
$$\text{exists a net } (p_{\lambda})\subset \mathcal{A}_0\text{ that converges to }p\qquad (1).$$
I also want $T^{**}$ to be continuous in this topology:
$$T^{**}(p)=\lim_\lambda T(p_\lambda)\qquad (2).$$
In addition $p$ is a meet of $N$ projections in $\mathcal{A}_0$ and the proof I have requires something like strong continuity for:
$$\begin{aligned}
T^{**}(p)=T^{**}(p_1\wedge\cdots\wedge p_N)&=T^{**}(p_1)\wedge\cdots \wedge T^{**}(p_N)
\\&=\underset{n\to\infty}{\operatorname{sot-lim}}\left(\left[(T(p_1))\cdots(T(p_N))\right]^n\right)\qquad (3)
\end{aligned}$$
I need to multiply by $1_{\mathcal{A}^{**}}\otimes q_i\in \mathcal{A}^{**}\overline{\otimes}\mathcal{A}^{**}$ in a continuous way (where $q_i\leq p$ is one of the sub-projections of $p$ (I presume there is some kind of standard von Neumann tensor product to use here)):
$$\begin{aligned}
T^{**}(p)(1_{\mathcal{A}^{**}}\otimes q_i)&=\underset{n\to\infty}{\operatorname{sot-lim}}\left(\left[(T(p_1))\cdots(T(p_N))\right]^n\right)[1_{\mathcal{A}^{**}}\otimes q_i]
\\& =\underset{n\to\infty}{\operatorname{sot-lim}}\left[\left(T(p_1)\cdots T(p_N)\right)^n(1_{\mathcal{A}^{**}}\otimes q_i)\right].\end{aligned}\qquad(4)$$
Now I have states $\rho_1$ and $\rho_2$ on $\mathcal{A}$. Let $\omega_1$ and $\omega_2$ be their extensions to a normal states on $\mathcal{A}^{**}$. I want these states to be continuous with respect to $\tau$ so that I can write:
$$\lim_\lambda\left[(\omega_1\otimes \omega_2)\left(T(p_\lambda)(p\otimes 1_{\mathcal{A}^*{**}})\right)\right]=(\omega_1\otimes \omega_2)\left[T^{**}(p)(p\otimes 1_{\mathcal{A}^{**}})\right]\qquad (5)$$
Question: Does there exist a topology on $\mathcal{A}^{**}$ that can achieve all of (1) to (5)?
I imagine that the various topologies I know a little about fall at one or other hurdle, but I am hoping in the zoo of operator algebra topologies there is something out there that might do the trick.
Best Answer
I think everything works; let me try to explain...
Firstly, as the inclusion $\newcommand{\A}{\mathcal{A}}\A\rightarrow\A^{**}$ is an isometry, the norm topologies are all the same. So $\A$ is closed in $\A^{**}$, the closure of $\A_0$ in $\A^{**}$ is just $\A$, and so forth.
Any of the weaker topologies we might put on $\A^{**}$ (thinking of this as a von Neumann algebra) are all weaker than the norm topology. So that $\A_0$ is dense in $\A$ means that as, e.g., $\A$ is $\sigma$-weakly dense in $\A^{**}$, also $\A_0$ is $\sigma$-weakly dense in $\A^{**}$.
It doesn't make sense to talk about the weak operator topology (WOT) on $\A^{**}$. The WOT is defined in terms the Hilbert space a von Neumann algebra acts on, and $\A^{**}$ isn't (a priori) acting on any Hilbert space.
However, the ``$\sigma$-topologies'' do make sense on an abstract von Neumann algebra as they only use the predual. The predual of $\A^{**}$ is just $\A^*$, and so we can speak in particular of the $\sigma$-weak topology on $\A^{**}$. This is the same as the weak$^*$-topology on $\A^{**}$, thinking of $\A^{**}$ as just being a Banach space.
My go to reference is Takesaki, Volume I, Chapter II.
If we use the $\sigma$-weak topology then everything you want will work. $T^{**}$ will be continuous for this topology. The product on $\A^{**} $ becomes separately continuous for this topology. The von Neumann tensor product is to use is just the standard one (Takesaki Chapter IV.5).
As the predual of $\A^{**}$ is $\A^*$, if you have a state (or just some functional) on $\A$ it is already a normal functional on $\A^{**}$ and so there is no need to take a normal extension. Thus certainly (5) holds, for example.