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I have two mapping tori $A_{n}$={$f\in C([0,1], M_{n}(\mathbb{R}))\mid f(1)=\alpha_{1}(f(0))$}, $B_{n}$={$f\in C([0,1], M_{n}(\mathbb{R}))\mid f(1)=\alpha_{2}(f(0))$} where $\alpha_{1}, \alpha_{2}$ are two inner automorphisms of $M_{n}(\mathbb{R})$. I need to pick the connecting maps in the inductive sequence such that the limit algebras become non-isomorphic. Any suggestions?

I already constructed some connecting maps that produce simple algebras with isomorphic $K_{0}$-groups. If I know how to produce non-isomorphic limit algebras, I can modify my connecting maps accordingly.

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Your algebras $A_n$ and $B_n$ are isomorphic (and thus isomorphic to $C(S^1,M_n)$). Due to a classification theorem of Elliott, this means that, if your limits are simple, then they are not isomorphic iff their Elliott invariant (ordered $K_0$-group, $K_1$-group, trace simplex, and pairing between traces and $K_0$) differs.

I'm not sure about the other constraints you have (and I'm not sure if you really want the fibres of $A_n$ to be $M_n$ instead of say $M_{k_n}$ for some integer $k_n$), but probably the $K_0$-groups are forced to be isomorphic (eg. if the connecting maps are unital). But, you should probably be able to get different $K_1$-groups.

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  • $\begingroup$ In my case, the two unitary matrices of the two inner automorphisms $\alpha_{1}$ and $\alpha_{2}$ belong to different components of $O_{n}$. $\endgroup$
    – David
    Commented May 11, 2013 at 14:39
  • $\begingroup$ I'm very sorry - I totally read over the part about real numbers. But this means that you have 8 K-groups to try to use to show that the limits are non-isomorphic. Do your limits have the same K-theory? $\endgroup$
    – Aaron Tikuisis
    Commented May 12, 2013 at 6:43

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