# A question on K- theory of non commutative $C^\star$ algebra

Edit: According to the comment of Andre Henriques I revise the question:

What is an example of a noncommutative unital $C^\star$ algebra $A$, which is not Morita equivalent to a commutative algebra, such that for all unital subalgebra $B$ of $A$, $K_{0}(B)$ has $\mathbb{Z}$ as a summand? This question is motivated by this post and the fact that commutative algebras and their matrix algebras satisfies the above property.

• Euhh... 2x2 matrices? – André Henriques Nov 30 '14 at 20:05
• @AndréHenriques thank you for the comment. I revise my question. – Ali Taghavi Nov 30 '14 at 20:19
• Why would it be interesting if such an object existed? and why would it be interesting if no such objects existed? – Yemon Choi Nov 30 '14 at 20:22
• @YemonChoi this would be a possible characterization of commutative algebra in term of $K$-theory, up to Morita equivalent. – Ali Taghavi Nov 30 '14 at 20:24
• Here's the simplest example of a $C^*$-algebra that isn't Morita equivalent to a commutative $C^*$-algebra: functions from $[0,1]$ to 2x2-matrices, that take diagonal values at $0$. I think that it already provides a counterexample to your "characterization of commutative algebras". – André Henriques Nov 30 '14 at 22:09

The question only requires regarding K$_0$ as an abelian group (it has a natural pre-ordering too), which makes it easy to construct simple (in the technical sense) examples.
Let $A$ be a simple infinite dimensional unital AF C*-algebra, whose K$_0$ group is free (as an abelian group) [lots of examples exist; see any basic work on AF algebras]. If $B$ is a unital subalgebra, it is stably finite (as $A$ is stably finite). Hence the image of $K_0(B)$ in $K_0(A)$ is nonzero (actually, we knew this anyway, since the free module on one generator over $B$ has nonzero image in K$_0(A)$). The image of K$_0(B)$ is a nonzero subgroup of the free abelian group K$_0(A)$, so is itself free, and thus the map to its image splits. So $\bf Z$ is a direct summand of $K_0(B)$. [Even though $K_0(B) \to K_0(A)$ need not be one to one!]