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.

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    $\begingroup$ Euhh... 2x2 matrices? $\endgroup$ Nov 30, 2014 at 20:05
  • $\begingroup$ @AndréHenriques thank you for the comment. I revise my question. $\endgroup$ Nov 30, 2014 at 20:19
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    $\begingroup$ Why would it be interesting if such an object existed? and why would it be interesting if no such objects existed? $\endgroup$
    – Yemon Choi
    Nov 30, 2014 at 20:22
  • $\begingroup$ @YemonChoi this would be a possible characterization of commutative algebra in term of $K$-theory, up to Morita equivalent. $\endgroup$ Nov 30, 2014 at 20:24
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    $\begingroup$ 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". $\endgroup$ Nov 30, 2014 at 22:09

1 Answer 1


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!]

And a simple infinite-dimensional C*-algebra is not Morita equivalent to a commutative one.

  • $\begingroup$ Prof. Handelman, Thank you very much for your very interesting answer. $\endgroup$ Dec 2, 2014 at 18:18

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