7
$\begingroup$

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.

$\endgroup$

1 Answer 1

11
$\begingroup$

Here I list some facts that may be useful for building your intuition:
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".

$\endgroup$
5
  • 1
    $\begingroup$ Do you know negative or positive counterparts of points 2,3,4 for $*$-isomorphism? $\endgroup$ Apr 4, 2016 at 18:58
  • 3
    $\begingroup$ 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). $\endgroup$
    – truebaran
    Apr 4, 2016 at 19:04
  • 6
    $\begingroup$ 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. $\endgroup$ Apr 4, 2016 at 19:10
  • $\begingroup$ @truebaran I see. I get the point that Morita equivalence is a good equivalence for NC-spaces, thanks a lot! Though I would also like to know what do you gain/lose when you consider other equivalences (e.g. Prugovecki)? $\endgroup$ Apr 4, 2016 at 20:01
  • 1
    $\begingroup$ Could you say something about the notion of equivalence in the sense of Prugovecki? I'm not familiar with this notion. $\endgroup$
    – truebaran
    Apr 7, 2016 at 23:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.