# Module category

I am currently reading about module categories and have been not very successful. In this case a module category is a category with an action of a monoidal category. (more information on nLab)

In specific my task is to find a module category $$M$$ over $$G$$-vect ($$G$$-graded vector spaces), s.th. the Functor category $$Fun(M, M) \cong G$$-rep (representations of $$G$$), where G is a group.

Unfortunately I don't have any clue how that could look like. It would already be helpful if someone knew any example of a module category! (So that I can get a feeling for what this category could look like) If anyone knows more about this specific example, it would be of course even better!

Thanks in advance!

• Right, that is what I would call a representation of the category (at least with suitable additional structure, such as being additive, but that seems given here). I don't have an idea for the specific thing you are looking for, but a place to start with examples is the category acting on itself in the obvious way. – Tobias Kildetoft Dec 13 '18 at 10:55
• Actually come to think of it, that example might be exactly what you are looking for, though I didn't work out the details. – Tobias Kildetoft Dec 13 '18 at 10:57
• So first you need to figure out what a functor from this category to itself looks like (I assume that these are supposed to be functors compatible with the action of the category?) – Tobias Kildetoft Dec 13 '18 at 13:58
• Yes. The idea to keep in mind is that such a functor will be determined on objects by what it does to a suitable collection of objects, and similar for morphisms. In this case, if you know what it does to the $1$-dimension vector space concentrated in some degree $g\in G$ for all $g\in G$ then you also know what it does to all objects by additivity. So what about morphisms? – Tobias Kildetoft Dec 13 '18 at 14:13
• Actually, I may have been keeping the easy example of $\mathbb{Z}$-graded spaces a bit too hard in my mind, since there you see how to get everything from what it does to just the $1$-dimensional space in degree $1$, which gives you a vector space for each functor, and the action of $\mathbb{Z}$ should come from the morphisms. But I am now less sure if this all works out as nicely for arbitrary groups. – Tobias Kildetoft Dec 13 '18 at 14:15