In general, an AF-algebra can has some different Bratteli diagrams. We can add some identical arrows to make the Bratteli diagrams different, but it is too trivial, are any good examples?

For which AF-algebras, its Bratteli diagram can be unique decided? I think the UHF algebras should be. How about others?

A same question is at here: https://math.stackexchange.com/questions/654472/bratteli-diagram-decided-by-af-algebras

  • 2
    $\begingroup$ I don't quite understand what you call "different" Bratteli diagrams. As you point out, they can "look" different but nevertheless yield the same AF-algebra. There is a way to define an equivalence relation among Bratelli diagrams such that two diagrams are equivalent if and only if the associated AF-algebras are isomorphic. If you're interested in UHF-algebras, there is a way to express the CAR algebra as a (non-trivial) direct limit of algebras of the form $M_{2^n}\oplus M_{2^n}$ (as opposed to the usual $M_{2^n}$). I can give you the details if you are interested. $\endgroup$ – Eusebio Gardella Jan 31 '14 at 22:17
  • $\begingroup$ By "give you the details" I mean "write a proper answer" and not just a comment. $\endgroup$ – Eusebio Gardella Jan 31 '14 at 22:18
  • $\begingroup$ I do not find the exact definition for the Bratteli diagram, so we can choose a suitable equivalence, I think the isomorphism as a minimum arrow diagrams is suitable, here minimum means no identical arrows. Maybe you have other natural choise. $\endgroup$ – Strongart Feb 3 '14 at 11:38

Right from the beginning of Bratteli diagrams (Bratteli; also Elliott), it was known, and also evident, that isomorphism of the diagrams (equivalent to Morita equivalence of the corresponding C* algebras) was given by intertwining sequences of maps (for isomorphism of unital AF algebras, the intertwining sequence has to be "pointed", that is preserving the designated order unit). The Bratteli diagram is thus nowhere near unique for every fixed choice of AF algebra.

As a particular (very special) case, stationary diagrams yielding Morita equivalence (isomorphism) arise from shift equivalence (pointed) of powers of the matrices.

| cite | improve this answer | |
  • 1
    $\begingroup$ I do not know much about Morita equivalence, as I know, we have A is Morita equivalence to B iff A⊙K=B⊙K, K is compact operator algebras. Maybe you can show me how does it effect to the Bratteli diagrams or give me some refenences. $\endgroup$ – Strongart Feb 3 '14 at 11:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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