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 $\underline{\mathrm{Hom}}(V, V)$?

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    $\begingroup$ 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. $\endgroup$ – Жека Feb 19 at 10:46
  • $\begingroup$ Thanks for your fast reply, I will look into it but are looking forward to your answer! Thanks a lot :) $\endgroup$ – 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.

  • $\begingroup$ 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! $\endgroup$ – P. Schulze Feb 20 at 6:54
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    $\begingroup$ $F$ 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 $F(V_g)\simeq k$. $\endgroup$ – Жека Feb 20 at 11:26
  • $\begingroup$ The functor $\textrm{Vect}\to kG-\textrm{mod}_\mathcal{C}$ which sends $V\to kG\otimes V$ is an equivalence. Under that equivalence $F(V_g)\simeq (kG)\otimes V_g$. $\endgroup$ – Жека Feb 20 at 11:29
  • $\begingroup$ I guess the basic (and rather trivial) fact one needs to check is $V_g\otimes kG\simeq kG$. $\endgroup$ – Жека Feb 20 at 11:30
  • $\begingroup$ 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 $F$? $\endgroup$ – P. Schulze Feb 21 at 10:16

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