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It is known that every AI-algebra (i.e. inductive limit of interval algebras) is an AT-algebra (i.e. inductive limit of circle algebras)?

This seems a little bit odd because a building block of an AT-algebra can be embedded in a building block of an AI-algebra. Is there any example to clarify this?

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The way you phrase this question makes it seem like a (non-trivial!) exercise. If this is a known result then presumably you are able to look up the papers in which this is proved. On the other hand, if this is an open question then I think you should give more details about which cases you already know how to do, where you have tried looking, and so on. – Yemon Choi Jun 6 '13 at 23:01
It is a well-known result. Its proof is on page 57 of "Classification of Nuclear $C^{\ast}$-Algebras, Entropy in Operator Algebras, M. Rordam, E. Stormer". I am looking for an intuitive description or possibly an example. As I mentioned before, its converse makes more sense to me. – David Jun 7 '13 at 0:23
The way you phrase it is wrong. When you assume that the C*-algebras in question are unital and simple, then I guess that it is true. – Gabor Szabo Jul 4 '13 at 21:31

1 Answer 1

Contrary to what you said, the building block of an AI-algebra, namely $C([0,1],M_n)$, can naturally be embedded into the building block of an AT-algebra, namely $C(\mathbb{T},M_n)$, and not conversely.

Indeed let $\pi\colon\mathbb{T}\to[0,1]$ be a surjective, continuous map (e.g., identifying the north and south halfcircles). Then the map $$ C([0,1],M_n)\to C(\mathbb{T},M_n),\quad f\mapsto f\circ\pi $$ is an injective ${}^*$-homomorphism.

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But there are also continuous surjections from $[0,1]$ to $\mathbb T$. – Rasmus Bentmann Mar 3 at 13:57
Yes, I agree. But the homomorphism $C([0,1],M_n)\to C(\mathbb{T},M_n)$ feels to me 'more natural' than the map $C(\mathbb{T},M_n)\to C([0,1],M_n)$. For instance, the latter 'forgets' K-theory. – Hannes Thiel Mar 3 at 15:49

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