Let $\mathcal{A}$ be a $\mathrm{C}^*$-algebra, and consider projections $p_1,\dots,p_N$ and $q_1,\dots,q_N$, all in $\mathcal{A}$. Using Halperin (I. Halperin, The product of projection operators. Acta Sci. Math. (Szeged) 23 (1962), 96–99), where $f_p=p_1 p_2\cdots p_N$ and $f_q:=q_1 q_2\cdots q_N$, both $(f_p^n)_{n\geq 1}$ and $(g_p^n)_{n\geq 1}$ converge $\sigma$-weakly in the bidual $\mathcal{A}^{**}$ to, respectively, $p:=p_1\wedge p_2\wedge \cdots p_N$ and $q:=q_1\wedge q_2\wedge \cdots q_N$.
Now, I have a term:
$$\lim_{n\to \infty}(f_p\otimes f_q)^n=\lim_{n\to \infty}(f_p^n\otimes f_q^n),$$
and I have asserted that this converges $\sigma$-weakly to:
$$\lim_{n\to \infty}(f_p^n\otimes f_q^n)=\lim_{n\to \infty}(f_p^n)\otimes \lim_{n\to \infty}(f_q^n)=p\otimes q.\qquad\qquad(*)$$
However, I have not justified why the limit can distribute, nor what tensor product I have with $p\otimes q$. The $f_p$ and $f_q$ actually live in a norm-dense $\ast$-algebra, $\mathcal{A}_0\subset\mathcal{A}$, where the tensor product is the algebraic tensor product, but it may well make as much sense to say that $f_p^n\otimes f_q^n$ lives in the minimal tensor product $\mathcal{A}\otimes\mathcal{A}$. I guess in general we cannot even say that $(\mathcal{A}\otimes \mathcal{A})^{**}\subseteq \mathcal{A}^{**}\overline{\otimes}\mathcal{A}^{**}$ (here I am thinking in terms of $\sigma$-weak closures of $\mathrm{C}^*$-algebras in biduals.
I guess we can say instead that we have
$$r_0=f_p\otimes f_q=(p_1\cdots p_N)\otimes (q_1\cdots q_N).$$
But then
$$r_0=(p_1\otimes q_1)(p_2\otimes q_2)\cdots (p_N\otimes q_N),$$
and this is a product of projections, and so:
$$\lim_{n\to \infty}(f_p\otimes f_q)^n=\lim_{n\to \infty}r_0^n,$$
and so by Halperin this converges strongly to:
$$r:=(p_1\otimes q_1)\wedge (p_2\otimes q_2)\wedge\cdots\wedge (p_N\otimes q_N).$$
But is $r=p\otimes q$?
I guess this is the same as asking, via Takeda–Sherman, if:
$$\bigcap_{i=1}^N\operatorname{ran}(p_i\otimes q_i)=\left(\bigcap_{i=1}^N\operatorname{ran}(p_i)\right)\otimes\left(\bigcap_{j=1}^N\operatorname{ran}(q_j)\right)$$
Question: Is ($\ast$) justifiable? With perhaps $p\otimes q$ the von Neumann tensor product, i.e. $p\otimes q\in \mathcal{A}^{**}\overline{\otimes}\mathcal{A}^{**}$?
Perhaps what could be helpful is that both $p$ and $q$ are central, and are both in fact the support projection of characters $\mathcal{A}\to\mathbb{C}$. That is, there are characters $\varphi_p,\,\varphi_q\to\mathbb{C}$ such that $p$ resp. $q$ are the smallest projections in the bidual such that the extensions satisfy $\varphi_p(p)=1$ and $\varphi_q(q)=1$. I am under the impression that this means these projections are minimal.
Best Answer
I hope the following answers (part of) your question.
If $M,N$ are von Neumann algebras and if we have bounded nets $\{m_i\}_i\subseteq M$ and $\{n_i\}_i\subseteq N$ such that $m_i \to m$ $\sigma$-weakly on $M$ and $n_i\to n$ $\sigma$-weakly on $N$, then it is true that $$m_i\otimes n_i\to m\otimes n$$ $\sigma$-weakly in $M \overline{\otimes} N$.
Proof: We may assume $M\subseteq B(H), N \subseteq B(K)$. Since any normal functional on $M \overline{\otimes}N$ is in the norm-closure of the linear span of the functionals $\omega_{\xi, \eta} \overline{\otimes} \omega_{\xi',\eta'}$, it suffices to show (via a density argument, using the fact that the nets are bounded) that $$(\omega_{\xi, \eta} \overline{\otimes} \omega_{\xi',\eta'})(m_i\otimes n_i)\to (\omega_{\xi, \eta} \overline{\otimes} \omega_{\xi',\eta'})(m\otimes n)$$ which is evident, given the assumption. Alternatively, if you know that on bounded subsets, the weak and the $\sigma$-weak topology coincide, you can simplify the above a bit.
In your situation, we can apply this to $M= N= \mathcal{A}^{**}$ to conclude that $$f_p^n \otimes f_q^n \to p\otimes q$$ in $\mathcal{A}^{**}\overline{\otimes}\mathcal{A}^{**}$.