Question about trace values of the hyperfinite factor I want to show that there exists a projection p in the hyperfinite factor $R= \bigotimes_{n = 1}^\infty \text{Mat}_{3 \times 3}(\mathbb{C})$ of trace equal to $\frac{1}{2}$. This seems simple in $\bigotimes_{n = 1}^\infty \text{Mat}_{2 \times 2}(\mathbb{C})$ and I know that it is a known theorem that such a projection exists but I would like to see how to exactly produce such a projection in this specific example.
Thank you in advance!
 A: Because of the form of the trace, any finite product (with finitely many factors distinct from the identity) will have trace with a denominator $3^n$ for some $n$. So a projection with trace $1/2$ will necessarily be a limit of those.
It is important to avoid the common mistake (don't ask me how I know) of thinking only in terms of elementary tensors.
Anyway, one can think of $R= \bigotimes_{n = 1}^\infty \text{Mat}_{3 \times 3}(\mathbb{C})$ as the inductive limit $M_3(\mathbb C)\subset M_9(\mathbb C)\subset\cdots$ with the embeddings
$$\tag1
A\longmapsto \begin{bmatrix} A&\\ &A\\ &&A\end{bmatrix}.
$$
If we let $k_n=\lfloor 3^n/2\rfloor$, then the numbers $k_n/3^n$ converge monotonically to $1/2$, and $k_{n+1}=3k_n+1$. We start with
$$
p_1=\begin{bmatrix} 1&0&0\\0&0&0\\0&0&0\end{bmatrix}.
$$
Given $p_n$, we form
$$
p_{n+1}=\begin{bmatrix} p_n\\ & p_n\\ &&p_n\end{bmatrix}+E_{3^{n+1},3^{n+1}}.
$$
What we were doing is switching one $0$ to a $1$ in the diagonal. Then
$$
\operatorname{tr}(p_{n+1})=3\operatorname{tr}(p_n)+\frac1{3^{n+1}}=\frac{k_{n+1}}{3^{n+1}}.
$$
We also have $p_n\leq p_{n+1}$, which immediately guarantees that the sequence is sot-convergent. By normality of the trace (or by the definition of the trace in the inductive limit), if $p=\lim p_n$ then
$$
\operatorname{tr}(p)=\lim_n\frac{k_n}{3^{n}}=\frac12.
$$
Of course, there are infinitely many other projections with trace $1/2$, so this construction is just a particular, more or less straightforward,  case.
