What kind of additional properties and/or structures one needs to impose on the category of (commutative or noncommutative) monoids of some monoidal category so that one can recover the original monoidal category from this data?

What kind of additional properties and/or structures one needs to impose on a category to ensure that it is the category of monoids of some monoidal category?

The example I have in mind is the category of (commutative or noncommutative) C*-algebras (or von Neumann algebras). Can we obtain one of these categories as the category of monoids of some monoidal category?


2 Answers 2


Here is a characterization of categories of commutative monoids. I don't know the answer in the non-commutative case.

Let C be a category. Then C is the category of commutative monoids in some symmetric monoidal category if and only if C has finite coproducts.

For suppose that C = CMon(M) for some symmetrical monoidal category M = (M, @, I). Then one can show that the tensor product @ of M also defines a tensor product on C --- and that this is, in fact, binary coproduct in C. (Example: if M is the category of abelian groups then C is the category of commutative rings, and the tensor product of commutative rings is the coproduct.) Similarly, the unit object I of M is a commutative monoid in a unique way, and is in fact the initial object of C. So C has finite coproducts.

Conversely, suppose that C has finite coproducts. Then (+, 0) defines a symmetric monoidal structure on C, and with respect to this structure, every object of C is a commutative monoid in a unique way. Thus, C = CMon(C).

  • $\begingroup$ Notice that these constructions are not inverse. I don't think you can recover M from CMon(M), can you? $\endgroup$ Commented Oct 30, 2009 at 1:43
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    $\begingroup$ Certainly not without some extra structure, e.g., CMon(Set) is equivalent to CMon(CMon(Set)) (where I give everything the cartesian monoidal strcture). $\endgroup$ Commented Oct 30, 2009 at 5:15
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    $\begingroup$ These constructions aren't inverse, no, but one is adjoint to the other. $\endgroup$ Commented Oct 30, 2009 at 12:57

I may be misunderstanding you, but your second question seems different from the first. In the first, you are presupposing your category is the category of monoids for at least one monoidal category; in the second, you are asking if this is the case.

I'm not an expert in the relevant fields, but I think the second question has been looked at by some category theorists (Egger?) and it's either false or tricky. You need to build in an involution, for a start. If you only want operator algebras, then I think results of Blecher et al tell us that a monoid in the category of operator spaces equipped with Haagerup tensor product can always be realized as an operator algebra, and every operator algebra arises in this way.

We definitely seem to need to look for monoids in categories of operator spaces rather than mere Banach spaces - there is some old work of Carne which to some extent shows that the usual tensor norms of Banach space theory cannot characterize operator algebras (i.e. if @ denotes one of these reasonable norms, then there is no theorem to say: A is an operator algebra if and only if there exists an associative product A @ A --> A).

  • $\begingroup$ Yes, there actually are two questions. I edited the question to reflect this. I suspected that operator spaces might play a role here. What is the reference for Blecher's result? I looked through Egger's home page, but could not find anything related. $\endgroup$ Commented Oct 29, 2009 at 22:03
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    $\begingroup$ I might have misunderstood the category-theoretic side of things, but the result I was thinking of can be found as Theorems 6.1 and 6.2 of Pisier's Introduction to Operator Space Theory - which might be Googleable. Pisier goes on to give an alternative proof, or at least an alternative spin on those before, so it may be worth tracking down a copy of that part of the book. $\endgroup$
    – Yemon Choi
    Commented Oct 30, 2009 at 2:28

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