For which algebra $A$: $\mathrm{vect} \cong A-$mod

In general there is an equivalence between a module category (see Module Category nLab) over a monoidal category $$\mathcal{C}$$ and $$A-$$mod, where $$A$$ is an algebra over $$\mathcal{C}$$: $$\mathcal{M}_{\mathcal{C}} \cong A-mod$$ This is proven in Ostrik, Theorem 1 on page 10.

I now want to consider an explicit example. Let $$k$$ be a field. Let $$\mathcal{C}=G-\mathrm{vect}$$ be the category of $$G$$-graded vector spaces over $$k$$. And let $$\mathcal{M}_{G-\mathrm{vect}}$$ be the category $$\mathrm{vect}_{k}$$, all vector spaces over a field $$k$$. The module structure is given by the tensor product of vector spaces, where we use the forgetful functor on the $$G$$-graded vector spaces.

Now I want to find such an algebra $$A$$ as in the statement above. In the proof of Theorem 1 in Ostrik, $$A$$ is given as $$\underline{\mathrm{Hom}}(V, V)$$ for an arbitrary vector space $$V$$.

$$\underline{\mathrm{Hom}}(V, V)$$ is the representing object for the functor $$X \mapsto \mathrm{Hom}(X \otimes V, V)$$, with $$X \in \mathcal{C}$$=$$G-\mathrm{vect}$$.

Unfortunately I have no idea how to find such a representing object, especially since $$V$$ is arbitrary. Does anyone have an idea on how I could proceed to get such a $$V$$ and $$\underline{\mathrm{Hom}}(V, V)$$?

• I think that $A$ is the group algebra graded by the natural $G$-action. I will write a full answer later today when I have the time. – Жека Feb 19 at 10:46
• Thanks for your fast reply, I will look into it but are looking forward to your answer! Thanks a lot :) – P. Schulze Feb 19 at 10:47

Let $$\mathcal{M}$$ be a module category over $$\mathcal{C}$$ given by the forgetful functor $$F:\mathcal{C}\to \mathcal{M}=\textrm{Vect}$$. Choose an arbitrary irreducible object $$M\in\mathcal{M}$$.

Now we compute the inner hom $$\underline{Hom}(M,M)$$. For every $$V\in \mathcal{C}$$ we should have $$Hom_\mathcal{C}(V,\underline{Hom}(M,M))\simeq Hom_\mathcal{M}(V\otimes M,M).$$ For every $$g\in G$$ let $$V_g\in \mathcal{C}$$ be a $$1$$-dim vector space graded by $$g$$. The we have $$Hom_\mathcal{C}(V_g,\underline{Hom}(M,M))\simeq Hom_\mathcal{M}(F(V_g)\otimes M,M)\simeq Hom_\mathcal{M}(M,M)=k,$$ hence $$\underline{Hom}(M,M)$$ has each $$V_g$$ exactly one time. Therefore $$\underline{Hom}(M,M)\simeq \bigoplus\limits_{g\in G}{V_g}$$.

It is not hard to "guess" that the algebra structure given by composition is the group algebra structure on $$kG$$. Therefore $$\underline{Hom}(M,M)$$ is canonically isomorphic to $$kG$$.

Also one can manually check that for every $$kG$$-module in $$\mathcal{C}$$ is free, hence the category of $$kG$$-modules in $$\mathcal{C}$$ is equivalent to $$\textrm{Vect}$$.

Hope that helps.

• Thanks you for your answer! Could you maybe explain why $F(V_g) \otimes M \cong M$? I Suppose the question actually is, why $F(V_g) = k$... and then I don't quite get the conclusion: If I check that every $kG$- module in $\mathcal{C}$ is free (do have a hint on how to check this?), why do we get the equivalence to $\mathrm{vect}$? I am very sorry for my deficiency :( Thanks again! – P. Schulze Feb 20 at 6:54
• $F$ is the forgetful functor, $V_g$ is $1$-dimensional vector space graded by $g$, if we forget the grading then $F(V_g)$ is just $1$-dimensional vector space, hence $F(V_g)\simeq k$. – Жека Feb 20 at 11:26
• The functor $\textrm{Vect}\to kG-\textrm{mod}_\mathcal{C}$ which sends $V\to kG\otimes V$ is an equivalence. Under that equivalence $F(V_g)\simeq (kG)\otimes V_g$. – Жека Feb 20 at 11:29
• I guess the basic (and rather trivial) fact one needs to check is $V_g\otimes kG\simeq kG$. – Жека Feb 20 at 11:30
• Thanks again! In your second comment you say $F(V_g) \cong (kG) \otimes V_g$ - do you still mean $F$ as the forgetful functor or did you call the functor $Vect \rightarrow kG-mod$ also $F$? – P. Schulze Feb 21 at 10:16