I used to participate in a seminar that taught students about foundations of non-commutative geometry. It isn’t very complicated to define a C*-module $\mathcal E$ (also known as C* Hilbert module) over a C*-algebra $\mathcal A$. This structure comprises the pairing, a 1½-linear mapping ${\mathcal E}\times{\mathcal E} \to {\mathcal A}$ with some properties analogous to ones of a Hilbert space. But the mess begins when one tries to define, in a similar fashion, a C*-bimodule; one needs it to define Morita equivalence of C*-algebras in a way other than category-theoretical. Namely, it is natural to expect from a ${\mathcal B},{\mathcal A}$-bimodule $\mathcal E$:

  • To be truly a bimodule, that means representations of $\mathcal B$ and $\mathcal A$ must commute, or the $b({\mathbf x}a) = (b{\mathbf x})a$ identity must be respected.
  • To have two pairings, one $\mathcal B$-valued and another $\mathcal A$-valued. Shouldn’t the definition be symmetric?
  • Two pairings should obey the exchange identity $\{{\mathbf x}|{\mathbf y}\}\,{\mathbf z} = {\mathbf x}\,({\mathbf y}|{\mathbf z})$ (that virtually defines one 2½-linear operation on $\mathcal E$, where $\mathbf y$ is the antilinear argument). It is the least obvious requirement, but it is necessary for correct behaviour of tensor product operations.

I presented these algebraic properties (and some other) at this HTML page as diagrams, to be more illustrative.

Alain Connes proposed to define a C*-bimodule over ${\mathcal B},{\mathcal A}$ as a representation of $\mathcal B$ in a C*-module over $\mathcal A$, i.e. a *-homomorphism from $\mathcal B$ to $\operatorname{End}_{\mathcal A} {\mathcal E}$, saying nothing about $\mathcal B$-valued pairing. The definition of $\operatorname{End}_{\mathcal A} {\mathcal E}$ implies commutation of representations. But it appears that this asymmetric definition is very weak on another side: examples, where any $\mathcal B$-valued pairing cannot be exchangeable with given $\mathcal A$-valued one, are almost trivial.

There is a book “Elements of Noncommutative Geometry” by José Gracia Bondía, Joseph C. Várilly, and Héctor Figueroa, where another definition is proposed: exchange identity is required, but authors missed the commutation of representations condition. A “C*-bimodule” in their sense is not proven to be a true bimodule. One can prove that if a “C*-bimodule” is full at least on one side, then $b({\mathbf x}a) = (b{\mathbf x})a$ follows from the definition. It appears that $b({\mathbf x}a) = (b{\mathbf x})a$ is necessary for many proofs in the book, but is it unknown whether it follows from it in the general case (there is neither counterexample nor an idea to prove it).

In the lectures 3 both exchange of pairings and $b({\mathbf x}a) = (b{\mathbf x})a$ were required for a C*-bimodule. But certain problem plagues even this definition. If we know that for any $b\in{\mathcal B}$ the respective left multiplication operator belongs to $\operatorname{End}_{\mathcal A} {\mathcal E}$ (that means it is bounded in the norm induced by the C*-$\mathcal A$-module structure), then we see a *-homomorphism of C*-algebras (that, as a general fact, must be bounded and have the norm 1 or 0), and subsequently can prove that left and right C*-module structures on $\mathcal E$ define the same norm (and hence, topology). Another approach is to define $\mathcal E$ as a topological vector space (and require continuity of both representations) that does the same thing, but it is unknown whether a plain vector space $\mathcal E$, without topology, has necessarily the same norm from two different structures of a C*-module. Again, counterexamples (satisfying aforementioned conditions) are unknown.

It seems that Morita equivalence, the main “user” of the concept of bimodule, is indifferent to some of these glitches because only few C*-bimodules can specify a Morita equivalence of two respective algebras. But the question persists: which should be the correct definition of a C*-bimodule?

  • $\begingroup$ Maybe this isn't the answer you want, but I once investigated C*-bimodules equipped with an involution which interacts with the bimodule structure in the natural way. (Namely, $(ax)^* = x^*a^*$ for $a \in \mathcal{A}$ and $x \in \mathcal{E}$.) My paper on them is here. It's not what you're asking about because these C*-bimodules are quite different from the C*-bimodules which arise in relation to Morita equivalence. $\endgroup$
    – Nik Weaver
    Aug 17, 2014 at 2:42
  • $\begingroup$ @Nik Weaver: yes, your bimodules seem to be a special case. One can’t introduce an involution in this way for a bimodule over two different algebras. But the latter is the most important thing for Morita equivalence (it doesn’t require ${\mathcal A},{\mathcal A}$-bimodules but tautological ones). $\endgroup$ Aug 17, 2014 at 5:20

1 Answer 1


As far as I'm concerned $C^*$ bimodules generally denote those you attributed to A.Connes. Such a bi-module define (by tensorization over B) a functor from $B$ $C^*$-modules to $A$ $C^*$-modules. Any $C^*$ functor between these categories which preserve orthogonal sums is of this form (essentially because of the stabilization theorem).

There is indeed a lot of them, they are more general than morphisms of $C^*$ algebra, but the general idea in non-commutative geometry is that we don't really have a good notion of "morphism" between non-commutative spaces, but $C^*$ bi - module form a really nice notion of "correspondences" between non-commutative space.

The relation with the bi-modules that arise in the theory of Morita equivalence (which I will call equivalence bimodule) is quite simple:

A $B,A$ bi-module $H$ induce an equivalence of category between $C^*$ A module and $C^*$ B module if and only if it is an equivalence bimodule.

In this case $B$ is exactly the algebra of compact operator of the $A$-module $H$ (that is the closure of the spam of the operator of the form $|x\rangle \langle y|$ for $x,y \in H$ and the $B$ valued parring is simply $[x,y]=|x\rangle \langle y|$.

I don't have access right now to the book you mentioned, and I don't read Russian, so I can't really comment on the other two definitions, but it seem to me that (unless they contains a small mistake) they should by equivalent to the previously mentioned definition (either of bimodule or of equivalence bimodule). Also note that equivalence bimodules are supposed to be full, which might answer a part of your questions.

Take a look at Bruce Blackadar's "Operator algebras" chapter II.7. It contains all the details about this.

  • $\begingroup$ Thanks. Could you write proper $\langle$angle brackets$\rangle$ for the bra-ket notation, not ugly substitution characters that cause erroneous spacing in LaTeX? Also, a hyphen spelled after the “bi” prefix is an extremely rare case (or a substandard spelling). $\endgroup$ Aug 17, 2014 at 10:16

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