Rieffel first studied Morita equivalence for $C^*$-algebras and von Neumann algebras in "Morita equivalence for C∗-algebras and W∗-algebras" Zbl 0295.46099 as a direct generalisation of the algebraic setup: the algebras have isomorphic categories of $*$-representations on Hilbert spaces. This seems okay for von Neumann algebras, but for $C^*$-algebras it reduces to the enveloping von Neumann algebras being Morita equivalent, and ends up giving a rather weak condition.

Later work by Rieffel and others seemed to clarify that the "correct" notion is what was initially termed strong Morita equivalence (perhaps these days no-one says "strong" any more). This means that there is an equivalence or imprimitivity bimodule, an object built using the language of Hilbert $C^*$-modules. Using the linking algebra construction, this is the same as the algebras being complementary full corners of a larger $C^*$-algebra. I like the presentation of Raeburn and Williams in "Morita equivalence and continuous-trace C∗-algebras." Zbl 0922.46050.

In a short survey paper, "Morita equivalence for operator algebras" Zbl 0541.46044 Rieffel notes that "strong" Morita equivalence implies usual (categorical) Morita equivalence, but has stronger properties, for example, preserving weak containment of representations.

My question is whether there is more recent work giving a more "categorical" interpretation of strong Morita equivalence?

This answer gives one possible answer, but without references.

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    $\begingroup$ Maybe correspondence categories is what you are after? Echterhoff has a nice survey article arxiv.org/abs/1006.4975 and in section 5 he describes a categorical approach developed by him and coauthors in 2000 and also developed independently by Landsman in 2001 $\endgroup$ Commented Jun 25 at 13:59
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    $\begingroup$ In a different direction from David Penneys's answer, there is a result of Blecher which expresses strong Morita equivalence as equivalence of certain "operator module categories" via completely contractive functors arxiv.org/abs/math/9906082 As you might expect, DB then extended this to strong Morita equivalence of operator algebras arxiv.org/abs/math/9906080 I suspect, but have not checked, that this would be written up in his book with Le Merdy. $\endgroup$
    – Yemon Choi
    Commented Jun 25 at 22:37
  • $\begingroup$ Thanks. In Caleb's comment, Prop 2.5.9 seems most relevant: there is a category, the "correspondence category", whose objects are $C^*$-algebras, and the isomorphisms in this category and exactly the equivalence bimodules. This seems in a similar direction to David's answer below. Yemon's comment seems closest to what I had in mind initially, in that this really does look at categories of certain modules, and asks for an equivalence of these categories. Thanks all! $\endgroup$ Commented Jun 26 at 11:28
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    $\begingroup$ @J.DeRo Doesn't Rieffel essentially do this? He showed that if $A,B$ are "weakly Morita equivalent" then the enveloping von Neumann algebras $A^{**}, B^{**}$ are "weakly Morita equivalent", but then this is the same as (strong) Morita equivalence, i.e. there is a (automatically normal) equivalence bimodule for $A^{**}$ and $B^{**}$. $\endgroup$ Commented Jun 26 at 15:10
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    $\begingroup$ @MatthewDaws Thanks. You are right. I was hoping for something that avoids the use of biduals. $\endgroup$
    – J. De Ro
    Commented Jun 26 at 15:12

2 Answers 2


I will have to look for a reference, but strong Morita equivalence is the notion of equivalence in the $\rm C^*$-2-category of $\rm C^*$-algebras, Hilbert $\rm C^*$-correspondences, and adjointable intertwiners.

In more detail, there is a weak 2-category (a.k.a. a bicategory) $\rm C^*$-Alg whose objects are $\rm C^*$-algebras (let's assume they are unital for ease of exposition) and whose 1-morphisms are right Hilbert $\rm C^*$-correspondences. This means a 1-morphism $X: A\to B$ is a Banach space equipped with a right $B$-valued inner product such that the norm on $X$ is equal to $\|\langle x|x\rangle_B\|_B^{1/2}$, together with a left $A$-action by adjointable operators. The 2-morphisms $X\Rightarrow Y$ are the adjointable $A-B$ bimodular operators. 1-Composition is the relative tensor product of correspondences. [I'm sure there is a better reference for all this, but I'm pulling all this from my joint article arXiv:2105.12010 since I know it best. It also includes many details on $\rm C^*$-2-categorie with the $\rm C^*$-algebra community specifically in mind.]

Now in a 2-category, an equivalence between 2 objects $A,B$ is a pair of 1-morphisms $X: A\to B$ and $Y: B\to A$ together with two 2-isomorphisms ${}_AX\otimes_B Y_A \Rightarrow 1_A$ and ${}_BY\otimes_A X_B\Rightarrow 1_B$. Here I am writing 1-composition left-to-right, suggestively using relative tensor product notation. When the 2-category is a $\rm C^*$-2-category, one further requires that these two 2-isomorphisms are unitary. One then checks that two (unital) $\rm C^*$-algebras are strongly Morita equivalent if and only if they are equivalent in the 2-category $\rm C^*$-Alg. It is straightforward to prove that an imprimitivity bimodule gives an equivalence in $\rm C^*$-alg, but the other direction is a bit more subtle, and requires me to discuss adjoints, unitary adjoints, and adjoint equivalences.

First, in a 2-category, there is a notion of left and right adjoint of a 1-morphism ${}_AX_B$, similar to the notion of adjoint functor. (Indeed, the notion of adjoint functor is exactly the notion of adjoint in the 2-category Cat.) We call a 1-morphism adjointable if it admits both a left and a right adjoint. The composite of two adjointable 1-morphisms is again adjointable, so we may restrict to a 2-subcategory whose 1-morphisms are adjointable. This 2-subcategory necessarily contains the equivalences, and it is a fact that every equivalence in a 2-category can be promoted to an adjoint equivalence, i.e., an equivalence which satisfies the extra zig-zag/snake equations.

In a $\rm C^*$-2-category, however, one should be careful to choose adjoints of 1-morphisms which are compatible with adjoints of 2-morphisms. It is at this stage that we are calling too many things "adjoints," which can lead to confusion. So in the rest of this answer, I'll use the term "dual" for an adjoint of a 1-morphism and use the term "adjoint" only for linear operators. The notion we are looking for here is that of unitary dual functor from arXiv:1808.00323. Typically, there are many that one can put on a given $\rm C^*$-2-category, and this article explores this in detail. In contrast, in the non-$\rm C^*$-setting, there is a contractible choice of dual functor; you can read the introduction to that paper if you're interested in that story.

Now, it is well-known by [MR2085108 = arXiv:math/0301259, MR1624182] that a Hilbert $\rm C^*$-correspondence is dualizable if and only if it is finitely generated projective (fgp) on both sides. Restricting to the fpg 1-morphisms in $\rm C^*$-Alg, it should be the case that there is a unitary dual functor where the dual is given by the contragredient bimodule and the evaluation map $\operatorname{ev}_X: {}_B\overline{X}\otimes_A X_B\to {}_BB_B$ is given by the $B$-valued inner product. At this step, I realize there is really something to check, as ${}_AX_B$ was only equipped with a right $B$-valued inner product, and not a left $A$-valued inner product. If my memory serves me correctly, I think we can equip a dualizable 1-morphisms ${}_AX_B$ with such a left $A$-valued inner product so that everything here works out; the references here are almost certainly [MR2085108 = arXiv:math/0301259, MR1624182].

Now once we have equipped the dualizable 2-subcategory of $\rm C^*$-Alg with this unitary dual functor, we see that an adjoint equivalence here should be an imprimitivity bimodule giving a strong Morita equivalence in the sense of Rieffel. There are some steps to check here, but I think it should all work out. The last thing to point out here is that in the $\rm C^*$-setting with a fixed unitary dual functor, it is still true that every equivalence can be promoted to an adjoint equivalence.

As you mention in the question, this gives a categorical proof that the categories of Hilbert space representations are equivalent. Observe that given a (unital) $\rm C^*$-algebra $A$, a Hilbert $\rm C^*$ $A-\mathbb{C}$ correspondence $H$ is indistinguishable from a Hilbert space $H$ equipped with a left $A$-action. Thus the $\rm C^*$ hom 1-category $\operatorname{Hom}(A\to \mathbb{C})$ is the $\rm W^*$-category $\operatorname{Rep}(A)$ of Hilbert space representations of $A$. Given an equivalence 1-morphism ${}_AX_B$, pre-composition with $X$ gives an equivalence of categories ${}_AX\otimes_B - : \operatorname{Rep}(B) \to \operatorname{Rep}(A)$.

However, we can also look at homs the other way. That is, the $\rm C^*$ hom 1-category $\operatorname{Hom}(\mathbb{C}\to A)$ is the $\rm C^*$ category $\operatorname{Mod}(A)$ of right Hilbert $\rm C^*$-modules. Again, given an equivalence 1-morphism ${}_AX_B$, post-composition with $X$ gives an equivalence of categories $-\otimes_AX_B: \operatorname{Mod}(A) \to \operatorname{Mod}(B)$.

While an equivalence $\operatorname{Rep}(A) \cong \operatorname{Rep}(B)$ is too weak to capture strong Morita equivalence, I suspect the equivalence $\operatorname{Mod}(A) \cong \operatorname{Mod}(B)$ fully captures the notion.

  • $\begingroup$ The weak $2$-category of $C^\ast$-algebras, Hilbert $C^*$-correspondences, and unitary intertwiners was considered in some detail by Buss–Meyer–Zhu, though they refer to Echterhoff–Kaliszewski–Quinn–Raeburn for the proof that equivalences in this weak 2-category are precisely strong Morita equivalence bimodules. $\endgroup$ Commented Jun 26 at 16:22

Given a $C^*$-algebra $C$, let us write $\operatorname{Mod}(C)$ for the category (without identities!) with as objects right Hilbert $C^*$-modules and as morphisms adjointable compact operators.

Proposition: Two $C^*$-algebras $C$ and $D$ are strongly $C^*$-Morita equivalent if and only if there is an equivalence $\operatorname{Mod}(D)\sim \operatorname{Mod}(C)$ through $*$-functors.

Proof. Suppose that there is an equivalence $$\operatorname{Mod}(D)\sim \operatorname{Mod}(C)$$ through $*$-functors $F: \operatorname{Mod}(D)\to \operatorname{Mod}(C)$ and $G: \operatorname{Mod}(C)\to \operatorname{Mod}(D)$.

It is an easy consequence of the $C^*$-identity that $*$-functors are norm-continuous on morphism spaces.

View $D\in \operatorname{Mod}(D)$ in the obvious way (and similarly for $C$). Put $$\mathscr{E}:= F(D), \quad \mathscr{F}:= G(C).$$

In order to conclude that $C,D$ are strongly Morita equivalent, it suffices to show two things (this is e.g. the definition of Morita equivalence in Lance's book on Hilbert $C^*$-modules, see chapter 7).

(1) $\mathcal{K}_C(\mathscr{E})\cong D$. But this is clear, since by fully-faithfulness of $F$ $$\mathcal{K}_C(\mathscr{E}) = \mathcal{K}_C(F(D)) \cong \mathcal{K}_D(D)\cong D.$$ (2) $\mathscr{E}$ is a full $C$-module. This is a little trickier to show. We start from the observation that $$[\mathcal{K}_D(D, \mathscr{F}) \mathcal{K}_D(\mathscr{F}, D)] = \mathcal{K}_D(\mathscr{F})$$ (this follows e.g. from the fact that $[\mathscr{F}D] = \mathscr{F}$ by use of an approximate unit for $D$). Consequently, since $D\cong GF(D)$, we also see that $$[\mathcal{K}_D(GF(D), \mathscr{F}) \mathcal{K}_D(\mathscr{F}, GF(D))] = \mathcal{K}_D(\mathscr{F}).$$ In other words, $$[G(\mathcal{K}_C(\mathscr{E}, C)) G(\mathcal{K}_C(C, \mathscr{E}))]= G(\mathcal{K}_C(C))$$ from which we deduce that $$[\mathcal{K}_C(\mathscr{E}, C)\mathcal{K}_C(C, \mathscr{E})]= \mathcal{K}_C(C)$$ which implies that $\mathscr{E}$ is full (since $[\langle \mathscr{E}, \mathscr{E}\rangle]$ is an ideal inside $C$).

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    $\begingroup$ Interesting. In A new approach to Hilbert C*-modules. Blecher proves Theorem 5.5 which is similar, but doesn't use compacts, but does assume strong continuity of the functors involved. A note added in proof (after the proof of the theorem) says that Skandalis showed him that such an equivalence automatically preserves compact maps, and then the result is "quite trivial". I wonder if (a) this is the argument he had in mind; and (b) what Skandalis's argument is... $\endgroup$ Commented Jun 27 at 14:08
  • $\begingroup$ @MatthewDaws What is the strong continuity of the functor? Compatibility with the strict (= strong*) topologies on adjointables? $\endgroup$
    – J. De Ro
    Commented Jun 27 at 18:23
  • $\begingroup$ Blecher actually works a little more generally, with all A-module maps (but remarks that you can restrict to the adjointables if you want). So he wants literal strong topology. But as all functors are assumed to be $*$-functors, won't you also get strict (=strong*) convergence as well? $\endgroup$ Commented Jun 28 at 9:14

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