Category of G-Set and Category of R-Module: A Uniform Approach via Category Theory

IntroductionCategory of G-ObjectCategory of left R-ModuleConnection between Rep_k(G) and R-Mod

Introduction

Category theory offers a unifying framework that reveals connections between classical algebraic structures. By viewing group actions and module structures through the lens of functors and natural transformations, we gain both conceptual clarity and their connection.

In what follows, we demonstrate how $G$$G$-sets and $R$$R$-modules can be elegantly described as functor categories. This perspective reveals that $G$$G$-equivariant maps and module homomorphisms are simply natural transformations, suggesting that seemingly different algebraic structures share fundamental categorical properties.

Category of G-Object

Let $G$$G$ be a group and $B(G)$$B(G)$ be the one object category with respect to $G$$G$. i.e. The object is $\ast$$*$ (It could be anything, nobody cares) and the morphism set is $G$$G$.

Let $\mathcal{C}$$\mathcal{C}$ be a category, then let us consider $\mathrm{Fun}(B(G),\mathcal{C})$$\mathrm{Fun}(B(G),\mathcal{C})$.

It is easy to see that for a functor $H\in \mathrm{Fun}(B(G),\mathcal{C})$$H \in \mathrm{Fun}(B(G),\mathcal{C})$, it should map $\ast$$*$ to an object $F(\ast )$$F(*)$ in $\mathcal{C}$$\mathcal{C}$, and for the morphism, the functor provides a group homomorphism from $G$$G$ to ${\mathrm{Aut}}_{\mathcal{C}}(F(\ast ))$$\mathrm{Aut}_\mathcal{C}(F(*))$.

By definition, the $G$$G$-equivariant map is just the natural transformation:

$$\begin{array}{}\text{(1)}& \begin{array}{ccc}F(\ast )& \stackrel{F(g)}{\to }& F(\ast )\\ \eta \downarrow & & \downarrow \eta & \\ G(\ast )& \underset{G(g)}{\overset{}{\to }}& G(\ast )\end{array}\end{array}$$

Hence the composition of $G$$G$-equivariant maps is still a $G$$G$-equivariant map. The isomorphism of $G$$G$-Object is just the natural isomorphism.

Let $\mathcal{C}=\mathrm{Set}$$\mathcal C=\mathrm{Set}$, then we get the category of $G$$G$-$\mathrm{Set}$$\mathrm{Set}$. If $\mathcal{C}=k$$\mathcal C=k$-$\mathrm{Vect}$$\mathrm{Vect}$, then we get ${\mathrm{Rep}}_k(G)$$\mathrm{Rep}_k(G)$.

Category of left R-Module

Let us consider something interesting, I claim that category of $R\text{-Mod}$$R\text{-Mod}$ is $\text{Add}(B(R),\mathrm{Ab})$$\text{Add}(B(R),\mathrm{Ab})$.

Here $B(R)$$B(R)$ is a one object category and $\mathrm{End}(\ast )=R$$\mathrm{End}(*)=R$.

The reason we need additive functors is because we need $F:\mathrm{End}(\ast )\to {\mathrm{End}}_{\mathrm{Ab}}(F(R))$$F:\mathrm{End}(*)\to\mathrm{End}_{\mathrm{Ab}}(F(R))$ to be a ring homomorphism, therefore we have $F(r+s)(a)=(F(r)+F(s))(a)=F(r)(a)+F(s)(a)$$F(r+s)(a)=(F(r)+F(s))(a)=F(r)(a)+F(s)(a)$, which is the distributive law.

The morphism between modules is a natural transformation between two additive functors:

$$\begin{array}{}\text{(2)}& \begin{array}{ccc}F(\ast )& \stackrel{F(r)}{\to }& F(\ast )\\ \eta \downarrow & & \downarrow \eta & \\ G(\ast )& \underset{G(r)}{\overset{}{\to }}& G(\ast )\end{array}\end{array}$$

By definition $\eta$$\eta$ is a morphism in $\mathrm{Ab}$$\mathrm{Ab}$, hence we have $\eta (a+b)=\eta (a)+\eta (b)$$\eta(a+b)=\eta(a)+\eta(b)$. The diagram commute tells us $\eta (rm)=r\eta (m)$$\eta(rm)=r\eta(m)$.

Remark You could replace $\mathrm{Ab}$$\mathrm{Ab}$ by any pre-additive cat. For example, if you consider $\text{Add}(B(R),\mathrm{TopAb})$$\text{Add}(B(R),\mathrm{TopAb})$, then you will get Category of topological left $R$$R$-Module. You can also consider $\mathrm{Add}(B(R),{\mathrm{Ch}}_R),\mathrm{Ban},\mathrm{Hill}$$\mathrm{Add}(B(R),\mathrm{Ch}_R),\mathrm{Ban},\mathrm{Hill}$...

Connection between Rep_k(G) and R-Mod

We know that the group algebra functor $k[-]$$k[-]$ is the left adjoint of the unit functor, hence we have

$$\begin{array}{}\text{(3)}& {\mathrm{Hom}}_{k\text{-}\mathrm{alg}}(k[G],{\mathrm{End}}_k(V))\cong {\mathrm{Hom}}_{\mathrm{Grp}}(G,\mathrm{GL}(V))\end{array}$$

This sugget you to see the isomorphism between ${\mathrm{Rep}}_k(G)$$\mathrm{Rep}_k(G)$ and $k[G]-$$k[G]-$Mod.