Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
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.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.