I am studying the quasi-local algebra on Bratteli and Robinson Operator Algebras and Quantum Statistical Mechanics, Vol. I (see definition 2.6.3), but there is one thing that is not clear to me at the moment. Let's say that the quasi-local algebra is the norm-closure of $\mathcal{A} = \cup_{n=1}^{+\infty} \mathcal{A}_n $ where the set of $\mathcal{A}_n$ is an increasing net of C*-algebras. Given a set { $O_n \in \mathcal{A}_n $ }, I define the following limit: $\lim_{N \to +\infty} \sum_{n=1}^{N} O_n$. My questions are:
- Is the existence of that limit guaranteed by the structure of quasi-local algebra, or in any case I should check that the limit converges in a certain topology?
- Can I say that, for every element A in the quasi-local algebra, there is the identity: $ \lim_{N \to +\infty} [\sum_{n=1}^{N} O_n, A] = [ \lim_{N \to +\infty}\sum_{n=1}^{N} O_n, A]$?
In this case, I introduced the commutation between two elements of the algebra as [B,C] = BC - CB.
A series either converges, or it doesn't. That depends on what the concrete elements $O_n$ are. For a trivial case, if $O_n=1$ for all $n$, the series obviously does not converge.
For $\sum_{n=1}^\infty O_n$ to exist what you need is, as usual, that the sequence of partial sums are Cauchy. That is, you need that for every $\varepsilon>0$ there exists $n_0$ such that for all $m\geq n\geq n_0$ you have $$ \bigg\|\sum_{k=n+1}^m O_n\bigg\|<\varepsilon. $$ In this situation the sequence of partial sums is Cauchy, and as C$^*$-algebra is a complete normed space the sequence converges, and the series exists.
How to guarantee that a certain series converges, there is no general recipe. Absolute convergence implies convergence, so if you know that $\sum_n\|O_n\|<\infty$ then the series will converge. But there are series that converge and are not absolutely convergent.
As for the commutators, here the series obscures what's happening. You have a sequence $\{S_n\}$ such that $S_n\to S$. And you are asking if $$ \lim_n [S_n,A]=[S,A]. $$ Expanding, this is $$ \lim_n S_nA-AS_n=SA-AS. $$ And this is true because $$ \big(S_nA-AS_n\big)-\big(SA-AS\big)=(S_n-S)A-A(S_n-S). $$ So $$ \big\|\big(S_nA-AS_n\big)-\big(SA-AS\big)\big\| \leq 2\|S_n-S\|\,\|A\|. $$