$\textrm{II}_1$ hyperfinite factor isomorphic to the infinite tensor power of matrix algebra Let $M$ be a uniformly hyperfinite factor of the type $\mathrm{II}_1$. It is stated in Power's paper Representations of Uniformly Hyperfinite Algebras and Their Associated von Neumann Rings that all such factors are isomorphic (and he refers to works of Murray and von Neumann).
Is it true that one can construct a sequence $\{M_k\}$ of commuting factors of type $\mathrm{I}_2$ in $M$ such that their infinite tensor product $\bigotimes_{k=1}^{\infty}M_k$ will coincide with $M$? Here we associate $M_k$ with the subalgebra $I\otimes\ldots\otimes I\otimes M_k\otimes I\otimes\ldots$, where $M_k$ is on the $k$-th place.
In other words, can we regard $M$ as the infinite tensor power of $\mathrm{Mat}(2,\mathbb{C})$? Any references would be appreciated.
 A: I'm not aware of a reference that does the construction explicitly (not saying there isn't). You can find generic details in Chapter XIV of Takesaki's Theory of Operator Algebras 3. Those details also cover those glossed over in the sketch below.
In any case, it is not hard to verify the construcion. Let
$$
A_n= \overbrace{M_2(\mathbb C)\otimes \cdots\otimes M_2(\mathbb C)}^{\text{$n$ times}}.
$$
We consider the obvious embedding $A_n\hookrightarrow A_{n+1}$ by using the first $n$ "factors" in $A_{n+1}$. This is an inductive system of C$^*$-algebras, so we can construct the inductive limit C$^*$-algebra
$$
A=\lim_{\to}A_n.
$$
Define on $A$ the state $\tau$ as the inductive limit of the traces. That is,
$$
\tau(x_1\otimes\cdots\otimes x_m)=\tau_2^\vphantom2(x_1)\cdots\tau_2^\vphantom2(x_m), 
$$
extended by linearity, where $\tau_2^\vphantom2$ is the normalized trace on $M_2(\mathbb C)$. There is no issue extending it to all of $A$, because $\tau$ is bounded on each $A_n$ (being the normalized trace, $|\tau(x)\|\leq\|x\|$ for all $x\in A_n$, for all $n$).
Now we do GNS for $A$ and $\tau$. Because $\tau$ is a tracial state, it extends to a tracial state on $\pi_\tau(A)''$. Indeed,
$$
\tau(x)=\langle x\,\Omega,\Omega\rangle,
$$
so it is defined everywhere. Because it is tracial in a dense subset of $\pi_\tau(A)''$, it is tracial on $\pi_\tau(A)''$. It is not too hard to show that $\pi_\tau(A)''$ is a factor, and so $\tau$ is faithful. Thus $\pi_\tau(A)''$ has a faithful trace. Being an infinite-dimensional factor with a faithful trace, it is a II$_1$-factor. It has a dense AFD C$^*$-algebra, so it is AFD, and thus isomorphic to the hyperfinite II$_1$.
The factor $\pi_\tau(A)''$ is often called an ITPF (Infinite tensor product factor) and is often denoted by
$$
\bigotimes_{k=1}^\infty M_2(\mathbb C).
$$

Clarification: naively, the natural approach would be to avoid the C$^*$-algebra and construct the tensor product von Neumann algebra directly on $\bigotimes_n\mathbb C^2$; this is actually the way it's done for finite tensor products. The problem is that it is not entirely obvious how to construct $\bigotimes_n\mathbb C^2$, or at least there are difficulties that do not show up in the finite product. Usually in infinite tensor product constructions one takes the elementary tensors to only be "finite" products, by filling the rest with identities, but there is no such thing in a Hilbert space. So one would be forced to take actual infinite elementary tensors, which requires that if $\xi=\xi_1\otimes\xi_2\otimes\cdots$ then $\prod_k\|\pi_k\|<\infty$. But now we have the problem that this product can be zero, which forces us to take a quotient and a completion to get an actual Hilbert space. I think that this program can be carried, but it is by no means as straightforward as constructing a finite tensor product.
