I have encountered the definition of a UHF-algebra by an infinite tensor product.
And I think I explained it to myself incorrectly.
I would try to explain how I used to think about it by an example of the CAR-algebra $M_{2^\infty}$.
The infinite tensor product $\otimes_{n=1}^\infty M_2(\Bbb{C})$ is defined to be the closure of the elements of the form $A_1\otimes...\otimes A_j\otimes 1\otimes 1\otimes....$ where $A_i\in M_2(\mathbb{C})$ for $1\leq i\leq j$.
Now, let $\varphi_{n+1,n}:M_{2^n}\to M_{2^{n+1}}$ be defined by $\varphi_{n+1,n}(A)=\begin{bmatrix}A & 0\\0 & A \end{bmatrix}$ and $\varphi^n:M_{2^n}(\Bbb{C})\to \varinjlim M_{2^j}(\Bbb{C})$ be the natural maps. Then I tend to think of the above element $A_1\otimes...\otimes A_j\otimes 1\otimes 1\otimes....$ in the inductive limit $(M_{2^n}(\Bbb{C}), \varphi_{n+1,n})$ as identified with $\varphi^j(A_1\otimes...\otimes A_j)$ where $A_1\otimes...\otimes A_j\in M_{2^j}(\Bbb{C})$ as usual.
Why do I think that this is not true:
I'm reading the notes "Separable Exact $C^∗$-Algebras Embed Into the Cuntz Algebra" written by Paul Skoufranis, and there he proves the following proposition:
Proposition 6.6 (The Rokhlin Property of the Bernoulli Shift):
Let $\sigma$ be the one-sided Bernoulli shift on the $2^\infty$-UHF algebra $A = \otimes_{j=1}^\infty M_2(\Bbb{C})$; that is $\sigma(A_1 ⊗A_2 ⊗···)=I⊗A_1 ⊗A_2 ⊗···$.
For each $k \in \Bbb{N}$ let $A_k := \otimes_{j=1}^k M_2(\Bbb{C}) ⊆ A$ (which is a unital $C^∗$-subalgebra). Then for each $ε > 0$ and for each $r ∈ \Bbb{N}$ there exists a $k ∈ \Bbb{N}$ and projections $P_0,P_1,...,P_{2r−1},P_{2r} = P_0$ in $A_k$ such that $\sum_{j=1}^{2^r} P_j = I$ and $||\sigma(P_j)−P_{j+1}||<\epsilon$ for $j=0,1,...,2r −1$.
Well, if my explanations above were OK then the Bernoulli Shift is just the identity map, right?
So, there must be something incorrect in my above explanations, and I don't know how I should correct it.
Thank you for any help.