# Most natural equivalence between $C^*$-algebras in NCG

I have listen or read that, in the context of noncommutative geometry, Morita equivalence is a more natural equivalence for $C^*$-algebras than $*$-isomorphism.

Can someone explain this sentence or know some text that could be helpful?

Does anybody know some comparisons of different $C^*$-algebras categories?

NOTE: I asked this same question en SE last week, it's still unanswered.

1. Two commutative Morita equivalent $C^*$-algebra are in fact $*$-isomorphic.
2 If $A$ is $C^*$-algebra and you take $B=M_n(A)$ then $A$ and $B$ are Morita equivalent.
3 Many invariants for $C^*$-algebras such as $K$-theory or Hochschild or cyclic homology are the same for Morita equivalent $C^*$-algebras.
4 Two Morita equivalent $C^*$-algebras have the same representation theory so from this point of view they should represent the same "noncommutative space".
• Do you know negative or positive counterparts of points 2,3,4 for $*$-isomorphism? – Melquíades Ochoa Apr 4 '16 at 18:58
• In 2. the answer is negative for $*$-isomorphism just because that if $A$ is commutative then $B$ is not (for $n>1$) so $A$ and $B$ can not be isomorphic. Obviously $*$-isomorphic $C^*$-algebras trivially have same invariants (K-theory, Hochschild homology). In point 4. "same representation theory" means that the category of representations are naturally equivalent (objects are representations and morphism are intertwiners between them). – truebaran Apr 4 '16 at 19:04
• One more crucial fact: the crossed product $C_0(X) \rtimes_r G$ of the $C^\ast$-algebra of a locally compact Hausdorff space by a free and proper action of a locally compact topological group is, in general, only Morita equivalent to the $C^\ast$-algebra of the quotient space $C_0(X/G)$. Indeed, $C_0(X) \rtimes_r G$ is necessarily noncommutative the moment $G$ is non-Abelian. – Branimir Ćaćić Apr 4 '16 at 19:10