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One categorical definition of a group $G$ is that it is a category $C$ with a single object $X$ such that every morphism in the set $C(X,X)$ is invertible, i.e. such that $C(X,X)$ is precisely the set-theoretical group $G$. Using this definition, a representation of the group $G$ is a functor from $C$ to the category of vector spaces.

Now there is another categorical definition of a group, which is perhaps less strange, and perhaps more illuminating -- namely that a group $G$ is a group object in the category of sets. This definition has the advantage that a group object can be defined relative any category that admits finite products, so in particular it can be defined relative to, say, the category of topological spaces, thus giving us a topological group.

My question then concerns the relationship between the two definitions: namely, if we fix an arbitrary cartesian monoidal category $M$, then I believe that a group object $G$ in $M$ gives rise to an $M$-enriched category $C$ with a single object $X$ with $C(X,X)$ -- the group object $G$ in $M$. Is this correct?

Furthermore, does this mean that we can define some sort of categorical representations of a group object $G$ in a cartesian monoidal category $M$ as the functors from $C$ to some fixed $M$-enriched category? So in particular we can treat representations of topological groups from this perspective (since I think any category enriched over the category of sets is enriched over the category of topological spaces as the former embeds faithfully into the latter)?

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I think the main obstruction to your approach is that the category of topological spaces is not enriched over itself in an natural way: it fails to be cartesian closed. Moreover isn't a representation of a topological group supposed to be a continuous map $G \times V \to V$, where $V$ is a topological vector space? – Zhen Lin Mar 19 '12 at 19:43

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Edit, 3/25/15: In an earlier version of this answer I claimed that group objects can be defined in any monoidal category. In fact this isn't true; one of the axioms requires access to a diagonal map, and so it's standard to only define group objects in cartesian monoidal categories. But since talking about actions only uses the monoid structure this doesn't affect the rest of the answer.

(On the other hand, we can define a Hopf monoid in any monoidal category; this is a bimonoid with an antipode satisyfing the usual Hopf algebra axioms. In $(\text{Vect}, \otimes)$ this reproduces Hopf algebras, but in any cartesian monoidal category the comonoid structure is unique and given by the diagonal, so this reproduces group objects. This is all pretty much irrelevant to the point I'm making below though.)

Monoid objects can actually be defined in any monoidal category $(M, \otimes, I)$; in particular the monoidal product doesn't need to be the categorical product in $M$. For example, a monoid object in $(\text{Ab}, \otimes, 1)$ is just a ring (and note that the categorical product in $\text{Ab}$ is the direct product, not the tensor product). If $G$ is a monoid object in $M$, the correct notion of an action of $G$ then ought to be an object $S$ with a map $G \otimes S \to S$ satisfying certain axioms. This correctly reproduces the usual notion of a continuous action of a topological group on a topological space and does not require that $M$ is closed monoidal (the monoidal generalization of cartesian closed).

Certainly a monoid object in $M$ gives rise to an $M$-enriched category with a single object; this is a straightforward consequence of the definitions. So you can certainly consider enriched functors from such a monoid object to other $M$-enriched categories. But you need to be careful: for example, you'd like to consider $\text{Top}$ as a category enriched over $\text{Top}$, but $\text{Top}$ is not cartesian closed.

I don't know if things work better in the category of topological vector spaces, but, for example, the category $\text{Ban}$ of Banach spaces is enriched over itself and so in particular enriched over $\text{Top}$ so things ought to work out fine there. Certainly the category of finite-dimensional real vector spaces is enriched over itself and so in particular enriched over $\text{Top}$ in a more-or-less canonical way.

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As you said, one has to be very careful. This mathoverflow thread should serve as a word of warning. – t.b. Mar 20 '12 at 0:53
Right. A $\text{Top}$-enriched functor $G \to \text{Ban}$ gives you a norm-continuous representation and this is not necessarily what you want. – Qiaochu Yuan Mar 20 '12 at 2:31

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