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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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Group 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 group 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 group 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 group 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 so I don't think you can actually do this.

I don't know if things work better in the category of topological vector spaces, but I believe 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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