Minimal Tensor product of von Neumann algebras Let $M$ and $N$ be two von Neumann algebras. A natural way to define $M\overline{\otimes}N$ is the von Neumann algebra generated by the algebraic tensor product $M\otimes N$ in $B(H\otimes K)$ where $M\subseteq B(H)$ and $N\subseteq B(K).$ Is this tensor product same as the minimal $C^*$-algebraic tensor product of $M$ and $N$? If so how to prove this? What is the definition of minimal infinite $C^*$-algebraic tensor product of von Neumann algebras $(M_n,\tau_n)_{n\geq 1},$ where $\tau_n$'s are normal faithful tracial state?
 A: As Adrián remarked in the comments, the answer is negative. Let me give a somewhat inexplicit explanation of this. It actually fails even in the abelian case. Indeed, let $M=N = L^{\infty}[0,1]$. Note that we have a multiplication map $m: L^{\infty}[0,1]\otimes L^{\infty}[0,1] \to L^{\infty}[0,1]$ (on the algebraic tensor product) given by $f\otimes g \mapsto fg$. We can view each element of $L^{\infty}[0,1]\otimes L^{\infty}[0,1]$ as an element of the von Neumann algebraic tensor product $L^{\infty}[0,1]\overline{\otimes} L^{\infty}[0,1]\simeq L^{\infty}[0,1]^2$, i.e. a function of two variables. The multiplication map in this setting looks like this: we have a function $\sum_{i=1}^{k} f_i(x)g_{i}(y)$ and map it to the function $\sum_{i=1}^{k} f_i(x)g_i(x)$. So, if $F\in L^{\infty}[0,1]\otimes L^{\infty}[0,1]$, then $m(F)(x) = F(x,x)$. The multiplication map extends to the minimal tensor product -- this can be checked by hand.
On the other hand, if the minimal and von Neumann algebraic tensor products agreed in this case, we would have a multiplication map $m: L^{\infty}[0,1]^2\to L^{\infty}[0,1]$, which takes acts as restriction to the diagonal. Since the diagonal is of measure zero, there is no way this map can be well defined.
As for the definition of the infinite tensor product, you do the following. Use the trace $\tau_n$ to obtain a GNS representation of $M_n$ on $L^{2}(M_n)$ equipped with a cyclic vector $\Omega_n$. Now form the the infinite tensor product of Hilbert spaces $L^{2}(M_n)$ with respect to the stabilising sequence $(\Omega_n)$. Now the infinite (algebraic) tensor product acts on the resulting Hilbert space, so we may close it to obtain a von Neumann algebra.
