The point in which one starts to talk about noncommutative geometry is the Gelfand Najmark theorem. It establishes an equivalence of the catgeories of commutative (non)unital $C^*$-algebras and the opposite category of the (locally) compact Hausdorff spaces. However the theory goes much much further: from the noncommutative analog of the measure theory (von Neumann algebras) to spectral triples which are noncommutative generalisations of (spinor) Riemanian manifolds. The paradigm of the noncommutative geometry is to look at the "space" through some (posiibly noncommutative) algebra which serves as the space of "functions on the space". However I have never heard about the precise definition of noncommutative space in such a way that these spaces forms a category. From the other hand, I've read that $\mathbb{C}$ and $M_n(\mathbb{C})$ and more general the algebra of compact operators should decribe the same "space" while many invariants (in the spirit of homology theories) for these algebras are the same. It suggest that the naive definition of isomorphism between "noncommutative spaces" defined as the isomorphism of the corresponding $C^*$-algebras seems to be to strong. So what is the most resonable way to define objects and morphisms of noncommutative (say topological) spaces?

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    $\begingroup$ Note that there are notions of equivalence for noncommutative algebras that restrict to isomorphism for commutative algebras, for example Morita equivalence. $\endgroup$ Mar 1, 2014 at 17:50
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    $\begingroup$ The sense in which it makes sense to think of $\mathbb{C}$ and $M_n(\mathbb{C})$ as the same is that the categories of modules are the same (this is what Morita equivalence is about). So this equivalence is more like homotopy equivalence than isomorphism. $\endgroup$ Mar 1, 2014 at 19:09
  • $\begingroup$ Proposals for how to define categories of spectral triples have been being advanced by Bertozzini & Conti & Lewkeeratiyutkul for a while now, see at ncatlab.org/nlab/show/spectral+triple#BCL05 $\endgroup$ Mar 2, 2014 at 1:51

1 Answer 1


If you want Morita equivalences to be the natural notion of equivalence, on the purely algebraic level you can work with the bimodule $2$-category. This is the $2$-category whose

  • objects are rings,
  • morphisms $R \to S$ are $(R, S)$-bimodules $_RM_S$ (composition is given by tensor product),
  • $2$-morphisms are given by $(R, S)$-bimodule homomorphisms.

Any $2$-category has a notion of equivalence, and in this $2$-category equivalence is precisely Morita equivalence. (You can also think of this $2$-category as the $2$-category whose objects are the categories of right $R$-modules over rings $R$, whose morphisms are cocontinuous functors between module categories, and whose $2$-morphisms are natural transformations between these. This is $2$-equivalent to the above by the Eilenberg-Watts theorem.)

There should be a version of this $2$-category in the C*-world although I don't know what the precise definitions are.

In the above definition we can replace rings with $\text{Ab}$-enriched categories and bimodules with $\text{Ab}$-enriched profunctors. We can also replace the bimodule categories above with derived bimodule categories; this is the natural setting for defining things like Hochschild (co)homology, which is actually derived Morita invariant, not just Morita invariant. Combining these two, we can talk about dg-categories and dg-bimodules between them; some authors use dg-categories as the objects of noncommutative (derived? algebraic?) geometry. See, for example, the nLab.

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    $\begingroup$ That is right. In the realm of C*-algebras it is however more natural to work with the so-called strong Morita equivalence introduced by Rieffel (see for instance his paper Morita equivalence for $C^\ast$-algebras and $W^\ast$-algebras). $\endgroup$ Mar 1, 2014 at 19:07

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