An AF-algebra is a $C^* $-algebra which is the inductive limit of an inductive sequence of finite-dimensional $C^*$-algebras.

Elliott's theorem concerning the classification of AF-algebras says that two AF-algebras $\mathfrak{A}$ and $\mathfrak{B}$ are isomorphic as $C^* $-algebras if and only if $(K_0(\mathfrak{A}),K_0(\mathfrak{A})^+,\Gamma(\mathfrak{A}))$ and $(K_0(\mathfrak{B}),K_0(\mathfrak{B})^+,\Gamma(\mathfrak{B}))$ are isomorphic as scaled ordered groups, where $\Gamma(\mathfrak{A})$ denotes the dimension range, i.e. the elements of $K_0(\mathfrak{A})^+$ given as equivalence classes of projections in $\mathfrak{A}$.

Although this result is interesting and beautiful on its own, I would like to know whether there are interesting applications that can be understood by and might be interesting for students who are familiar with basic K-theory for $C^* $-algebras. Of course I'm also interested in more advanced applications or situations where Elliott's theorem provides insights which are hard to obtain otherwise.


This is more an example than an application, but, among many other non-obvious ordered groups that arise from AF algebras is the group of polynomials with integer coefficients, with positivity meaning having non-zero positive values on the open unit interval (unless the polynomial is the zero polynomial). As shown by Renault, this ordered group arises from Pascal´s triangle, interpreted as a Bratteli diagram! (As an application of this, it is possible to derive the solution to the classical Hausdorff moment problem.)

Another interesting example is the subgroup of the plane consisting of elements with rational coordinates, with positivity determined by the interior of a cone in the plane, the boundary lines of which do not necessarily have rational slope. When the cone is a half-plane, the ordered group is closely related to the continued fraction expansion of the slope of the (single) boundary line.

  • $\begingroup$ Thank you very much! These examples are very interesting indeed. $\endgroup$ – lvb Apr 15 '11 at 22:42

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.