How to construct an $AF$-algebra $N$ with $K_0(N)=\mathbb Q$?

Question

On the paper i'm reading, it says letting $N=\lim M_2\otimes ...\otimes M_{n!}$ will do.

I wonder why it doesn't simply let $N=\lim M_2\otimes ...\otimes M_n$. Take arbitary $p\in N\otimes M_n$, there is $p'\in M_{m!}\otimes M_n$ such that $|p'-p|<1/2$ so that $[p']=[p]$ in $K_0(N)$. Therefore, for any $r\in\mathbb N$, letting $rs>m$ and mapping $p'$ to $p'\otimes 1_{(rs)!/m!}\in M_{(rs)!}\otimes M_n$, we see $q=p\otimes 1_{(rs)!/m!/r}$ satisfies that $r[q]=[p]$. This shows $K_0(N)=\mathbb Q$.

Am I right?

Answer 1

You're right, if you take $A_n = \otimes_1^n M_i \simeq M_{n!}$ with connecting maps $A_n \ni a \mapsto a \otimes 1_{n+1} \in A_{n+1}$, then $$A = \underset{\to}{\lim} A_n $$ has $K_0(A) \simeq \mathbb{Q}$. Indeed, just use continuity of $K_0$. Applying $K_0$ to the sequence we get $$ \mathbb{Z} \to^2 \mathbb{Z} \to^3 \mathbb{Z} \to \cdots, $$ and its not too hard to convince yourself that the resulting limit is $\mathbb{Q}$. Note that the algebra $A$ is actually the universal UHF algebra in this case.

The same argument applies to more general inductive limits which give UHF algebras (see Direct limit of $\mathbb{Z}$-homomorphisms). As long as in the induced sequence on $K$-theory $$ \mathbb{Z} \to^{a_1} \mathbb{Z} \to^{a_2} \mathbb{Z} \to \cdots, $$ each prime divides an $a_i$ for some $i$, you should get all of $\mathbb{Q}$ as your limit. So the limit you described works as well (although I'm sure of why one might prefer one limit to another in this specific case). For example, you could take $\underset{\to}{\lim} M_2 \otimes M_3 \otimes \cdots \otimes M_{n!} \simeq M_{(n!)!}$, which intuitively is a subsequence of the one you wrote down. (I think I wrote a wrong limit before, edited for corectness)

There is also a classification of UHF algebras via their corresponding supernatural numbers, one can show that the supernatural numbers of the two sequences end up being the same.