While reading about Morita equivalence in the category of $C^*$-algebras I met also the following notion: a correspondence between two $C^*$-algebras $A,B$ is a pair $(X,\varphi)$ where $X$ is a (right) $B$ Hilbert module and $\varphi:A \to \mathfrak{B}(X)$ is a homomorphism.
Question 1 In all definitions which I met I found that the only requirment for $\varphi$ is that it is a homomorphism but it is somehow natural to require that it preserves the involution. Do we assume this?

Identyfying isomorphic correspondences one can check that with the composition defined as tensor product we obtain the category: objects are $C^*$-algebras while morphism are isomorphism classes of correspondences. It is commonly stated that this definition is more flexible while it admits 'more' morphism: in fact given an ordinary morphism $f:A \to B$ one can define $X$ to be $B_B$ with the standard right Hilbert module structure and $\varphi(a)$ to be a multplication by $f(a)$.
Question 2 Is it true that given any two $C^*$-algebras $A,B$ one can always find a correspondence between them?
And finally I would like to know whether
Question 3 isomorphism in this category is precisely the same as Morita equivalence.

If $_AX_B$ is an equivalence $A,B$ bimodule then if we take a dual bimodule $_B\tilde{X}_A$ one can check the appropriate tensor products are isomorphic to $_AA_A$ and $_BB_B$-so we obtain isomorphisms in this category. I don't however see why the converse must be true.

EDIT: it seems that I was too quick claiming that I see that two Morita equivalent algebras $A,B$ are isomorphic in the category described above. It is always true that $_AX_B \otimes _B\tilde{X}_A \cong _AA_A$ but I don't see why $_AA_A$ is and identity morphism in this category. To be more precise: for each $C^*$-algebra $A$ we put $X$ to be $A$ with the natural action of $A$ from the right and from the left by multiplication and the product $\langle a,a' \rangle_A=a^*a'$ (left action is the same as the homomorphism $\varphi:A \to \mathfrak{B}(X)$ which here is defined via $\varphi(a)(a')=aa'$). Suppose now that another correspondence $(Y,\psi)$ from $A$ to $B$ is given: so $Y$ is a right $B$-Hilbert module and left $A$-module thanks to $\psi$. If we compose our correspondences we obtain the module $A \otimes_{\psi} X$ which has the structure of left $A$-module by $a \cdot (a' \otimes x)=(aa') \otimes x$ and right $B$-module by $(a \otimes x) \cdot b=a \otimes (x \cdot b)$. The $B$-valued inner product on $A \otimes_{\psi} X$ is defined by $\langle a \otimes x,a' \otimes x' \rangle_B=\langle x,\psi(\langle a,a' \rangle_A)x' \rangle_B$. Put $\alpha:A \otimes_{\psi} X \to X, \ \alpha(a \otimes x)=\psi(a)x$. (in purely algebraic seeting, when one deals with genuine unital algebras and unitary modules in the sense that the unit of the algebra act as identity this is a natural isomorphism). One easily checks that it is correctly defined: moreover if we assume that $\psi$ is a *-homomorphism (namely that $\psi$ preserves involution) one can check that $\alpha$ preserves the products $\langle \cdot,-\rangle$. But if we don't assume that $A$ is unital and $\psi$ is unital, I don't see how to conclude that $\alpha$ is an isomorphism. Similar difficulties we get if we compose with $_AA_A$ from the right.

MUCH LATER EDIT: This is rather funny coincidence: several days after I posed this question I found the following videos from the Trimester Programm on Noncommutative Geometry- here is the link for the first part https://www.youtube.com/watch?v=3ZGQxLy1ydk

Now some my questions which I posed previously (and some which originated while discussing this topic) are no longer current. Let me briefly explain some details: in fact in the definition of a correspondence one assumes that the underlying Hilbert module is nondegenerate, namely that $\psi(A)X$ is dense in $X$-this assumption allows us to prove that $A \otimes_{\psi} X \cong X$ via the map $\alpha(a \otimes x)=\psi(a)x$. However, in order to prove the similar condition from the right one doesn't need any additional assumption: the natural isomorphism $X \otimes_{\varphi} A \cong X$ where $\varphi$ is left multiplication is of the form $\beta(x \otimes a)=x \cdot a$ and now $X \langle X,X \rangle$ is always dense in $X$ so $X \cdot A$ is also dense. This proves (together with my previous remarks) that two Morita equivalent $C^*$-algebras are isomorphic in this category. Now, compatibility with the composition of ordinary morhpsims goes as I explained provided that our morphisms are nondegenerate: $f:A \to B$ is called nondegenerate if $f(A)B$ is dense in $B$. This assumption is somehow natural: the category of all $C^*$-algebras (possibly nonunital) with nondegenerate morphisms as arrows is equivalent to thwe opposite category of locally compact Hausdorff spaces and proper maps. If we relax the assumption that our morphisms are nondegenerate one produces from $f:A \to B$ the correspondence in which $X=\overline{f(A)B}$ which is a right Hilbert $B$-module (with standard inner product) and now the nondegeneracy of the left action follows. Finally one should also note the following: the most natural notion for arrows at the topological level is the notion of continuos map (between locally compact spaces-without the assumption of being proper). One can therefore ask which notion at the $C^*$-algebraic level corresponds to the notion of continuity. It appears that the appropriate notion is the following: the morphism $\varphi:A \to B$ is a nondegenerate *-homomorphism from $A$ to the algebra of multipliers $M(B)$ which is nondegenerate in the sense that $\varphi(A)B$ is dense in $B$. So in the data defining a correspondence there is a $*$-homomorphism $\varphi:A \to \mathfrak{B}(X)$ which is nondegenerate in the sense that $\varphi(A)X$ is dense in $X$. This is equivalent to declaring that $\varphi(A)\mathfrak{K}(X)$ is dense in $\mathfrak{K}(X)$ (the algebra of ''compact'' operators on $X$) and if we recall that $M(\mathfrak{K}(X))-\mathfrak{B}(X)$ we see that the map $\varphi$ being the part of the data defining correspondence is nothing more than morphism $A \to \mathfrak{K}(X)$ is the above sense.

I now suspect that given two arbitrary $C^*$-algebras one can always find a correspondence between them however I'm still interested in knowing some examples. Also, I still don't see how the isomorphism in this category implies Morita equivalence (I believe that it is true: I found some short note that this in fact is true, in Connes' Noncommutative Geometry).

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    $\begingroup$ In the nonunital case, one often imposes the additional condition on ${}_A E_B$ that $A E$ be dense in $E$; this should take care of the issue in your edit. $\endgroup$ Jan 3, 2015 at 23:34
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    $\begingroup$ In particular, observe that the “trivial line bundle” ${}_A A_A$ satisfies this condition precisely because $A$ admits an approximate unit. $\endgroup$ Jan 3, 2015 at 23:41
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    $\begingroup$ I think that the heart of the matter here is that Gelfand–Naimark duality for locally compact Hausdorff spaces is a subtle business. One way to cut the Gordian knot is Woronowicz's approach, which identifies a continuous map $f: X \to Y$ as inducing a nondegenerate $\ast$-homomorphism $f^t : C_0(Y) \to M(C_0(X))$, where $M(C_0(X))$ denotes the multiplier algebra of $C_0(X)$; indeed, if $\phi : A \to M(B)$ is a nondegenerate $\ast$-homomorphism, so that $\phi(A)B$ is dense in $B$, then the induced bimodule ${}_A B_B$ satisfies both nondegeneracy and fullness. $\endgroup$ Jan 4, 2015 at 19:19
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    $\begingroup$ See, for instance, mathoverflow.net/questions/82871/… and similar discussions on MathOverflow and Math.SE. I suppose the point is that naive $\ast$-homomorphisms $A \to B$ aren't necessarily all that natural in the nonunital case, when seen through the guiding lens of Gelfand–Naimark. $\endgroup$ Jan 4, 2015 at 19:20
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    $\begingroup$ @BranimirĆaćić +1 just because it warms my heart to see people discussing G-N duality and realizing that maybe one wants to consider continuous maps that aren't proper $\endgroup$
    – Yemon Choi
    Jan 11, 2015 at 16:33

1 Answer 1


(Apologies for resurrecting a year-old question. I thought it worth at least recording an answer to part 3, since I think this is something that comes up quite a lot.)

Question 1: As you've already discovered, the appropriate definition (in this context) is that $\varphi$ should be a nondegenerate $*$-homomorphism from $A$ into the adjointable operators on $X$.

Question 2: Yes. Take any correspondence $A\to \mathbb{C}$ (i.e. a nondegenerate $*$-representation of $A$ on a Hilbert space) and compose it with any correspondence $\mathbb{C}\to B$ (i.e. a right $B$-Hilbert module) to get a correspondence $A\to B$. The composition will be nonzero as long as both of the factors are.

Question 3: Yes: a $C^*$-correspondence is invertible if and only if it is an imprimitivity bimodule in the sense of Rieffel. For a short proof see:

Echterhoff, Kaliszewski, Quigg and Raeburn: A categorical approach to imprimitivity theorems for $C^*$-dynamical systems (Memoirs of the AMS 2006, arXiv:math/0205322, Lemma 2.4).

Different proofs can be found in:

Echterhoff, Kaliszewski, Quigg and Raeburn: Naturality and induced representations (Bulletin of the Australian Math. Soc. 2000, arXiv:math/0002039, Proposition 2.6)

Blecher: A new approach to Hilbert $C^*$-modules (Math. Ann. 1997, Theorem 5.5)

As you mention in the comments, for unital algebras one has the stronger result that $C^*$-algebra Morita equivalence coincides with the purely algebraic notion. For this, see:

Beer: On Morita equivalence of nuclear $C^∗$-algebras (J. Pure Appl. Algebra 1982, Theorem 1.8).

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    $\begingroup$ Nice answer. I for one appreciate it. $\endgroup$
    – Nik Weaver
    Feb 2, 2016 at 17:54

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