Let $M_{\mathbb{Q}}$ be the universal UHF-algebra and let $\mathcal{O}_{\infty}$ be the infinite Cuntz algebra. Let $A$ be a Kirchberg algebra that satisfies the UCT with $K_0(A) \cong \mathbb{Q}^n$ and $K_1(A) = 0$. By the Kirchberg-Phillips classification theorem the projections $p_i \colon \mathbb{Q}^n \to \mathbb{Q}$ induce homomorphisms $A \otimes \mathbb{K} \to M_{\mathbb{Q}} \otimes \mathcal{O}_{\infty} \otimes \mathbb{K}$, which combine to give a homomorphism $$ \psi \colon A \otimes \mathbb{K} \to (M_{\mathbb{Q}} \otimes \mathcal{O}_{\infty} \otimes \mathbb{K})^n $$ that is the identity on $K$-Theory. Since $A \otimes \mathbb{K}$ is simple, $\psi$ is injective.

Given an automorphism $\alpha \in Aut(A \otimes \mathbb{K})$, is there an extension of $\alpha$ to an endomorphism $\beta_{\alpha} \in End((M_{\mathbb{Q}} \otimes \mathcal{O}_{\infty} \otimes \mathbb{K})^n)$ with $\psi \circ \alpha = \beta_{\alpha} \circ \psi$? If yes, can $\beta_{\alpha}$ be chosen such that $\beta_{\alpha \circ \alpha'} = \beta_{\alpha} \circ \beta_{\alpha'}$?


This is definitely not an answer but too long for a comment.

The algebra on the right-hand side has a fairly simple primitive ideal space, and it may not be too hard to try to do what you want by hand, using Kirchberg's results on non-simple purely infinite classification. I'm not sure but you might be able to write the right-hand side algebra as some nice crossed product (using non-simple AF-algebras or non-pointwise outer automorphisms) that can help with this. Maybe a careful application of Kirchberg-Phillips will also work.

Whether you can choose $\beta$ to behave well under composition may turn out to be hard to prove. You are essentially asking whether actions on $K$-theory lift to actions on the algebra. Even on Kirchberg UCT algebras, this is tricky. In this paper http://arxiv.org/pdf/math/0608093.pdf Katsura does some things for some finite groups.

Hope this helps!

  • 1
    $\begingroup$ You surely know this, but Kirchberg's paper where he classifies non-simple purely inifinite algebras is in German. This shouldn't be a problem for you, though ;) $\endgroup$ Feb 7 '14 at 21:17

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.