# The category of finite dimensional right $KG$-modules is equivalent to the category of finite dimensional representations of a quiver $Q$

Let $$K$$ be an algebraically closed field and $$G$$ be a finite group such that $$|G|$$ is not divisible by the characteristic of $$K$$ (so that Maschke's theorem can be applied). Let $$Q$$ be the quiver consisting of $$n$$ vertices and no arrows, where $$n$$ is the number of distinct conjugacy classes of $$G$$. Consider the category $$\text{rep}(Q)$$ of finite dimensional representations of $$Q$$. How can we show that $$\text{rep}(Q)$$ is equivalent to the category of finite dimensional right $$KG$$-modules, where $$KG$$ is the group algebra? I know that $$\text{rep}(Q)$$ is equivalent to the category of finite dimensional right $$KQ$$-modules where $$KQ$$ is the quiver algebra (though I'm not sure that this fact is relevant). I can't see how to define a functor from $$\text{rep}(Q)$$ to $$\text{mod}$$-$$KQ$$. Thanks in advance.

• Since $Q$ has no arrows, you can parameterize the isomorphisms classes of objects of $\mathrm{rep}(Q)$ by $\mathbb{N}^n$; a representation here is just a choice of vector space at each node. This looks like a red flag to me. Can you really parameterize $KG$-modules up to isomorphism by $\mathbb{N}^n$? Jun 12, 2020 at 15:24
• It's been a while, but I recall being under the impression that this statement is true if you instead let $Q$ be the quiver with a single node, and taking $F_1 \hookrightarrow F_0 \twoheadrightarrow G$ be a presentations of $G$, giving $Q$ arrows corresponding to the generators of $F_0$ and relations corresponding to the generators of $F_1$. A group is just a category with a single object and all morphisms invertible, as the category theorists love to remind us ;) Jun 12, 2020 at 15:27
• @MikePierce From Maschke's theorem there are only $n$ number of isomorphism classes of simple $KG$-modules, say $M_1,\dots,M_n$. An object in $\text{rep}(Q)$ is collection of $n$ (finite-dimensional) $k$-vector spaces $(V_1,\dots,V_n)$. To define a functor $\text{rep}(Q)\to \text{mod}-KQ$, it seems natural to send this object to the direct sum $M_1^{k_1}\oplus \cdots \oplus M_n^{k_n}$ where $k_i$ is the dimension of $V_i$ over $k$. But I can't see how to send a morphism of $\text{rep}(Q)$ (which is a collection of $n$ number of $k$-linear maps) to a morphism in $\text{mod}-KQ$. Jun 13, 2020 at 1:03
• Have you looked at Morita theory and Morita equivalence? It should be possible to find a $(KG, KQ)$-bimodule $B$ such that the functor $\mathsf{Rep}_K(Q) \to \mathsf{Rep}_K(G)$ defined by $V \mapsto B \otimes_{KQ} V$ is the equivalence you are looking for. Jun 14, 2020 at 3:08
• This isn't true as stated. Is $K$ supposed to be assumed to be algebraically closed? Jun 15, 2020 at 3:31

The number $$n$$ of conjugacy classes does coincide with the number of isomorphism classes of irreducible, finite-dimensional $$G$$-representations, because $$K$$ is algebraically closed und Maschke’s theorem applies.¹ (Maybe we also need that $$\operatorname{char}(K) = 0$$?) Let $$S_1, \dotsc, S_n$$ be a set of representatives for the isomorphism classes of irreducible, finite-dimensional $$G$$-representations. Then by Maschke’s theorem every finite-dimensional $$G$$-representation is isomorphic to a representation of the form $$S_1^{r_1} \oplus \dotsb \oplus S_n^{r_n}$$ for some unique natural numbers $$r_1, \dotsc, r_n$$. Let $$\mathcal{A}$$ be the full subcategory of $$\mathbf{rep}_K(G)$$ whose objects are these direct sums. Then the inclusion from $$\mathcal{A}$$ to $$\mathbf{rep}_K(G)$$ is both fully faithful and essentially surjective, and therefore an equivalence.

We need to better understand morphisms in $$\mathcal{A}$$. For this we recall the matrix calculus for morphisms between direct sums:

Let $$R$$ be some ring and let $$M_1, \dotsc, M_s$$ and $$N_1, \dotsc, N_r$$ be $$R$$-modules.²

Let $$f_{ij} \colon M_j \to N_i$$ be a homomorphism. Then the map \begin{align*} f \colon M_1 \oplus \dotsb \oplus M_r &\to N_1 \oplus \dotsb \oplus N_s \,, \\ (m_1, \dotsc, m_r) &\mapsto \left( \sum_{j=1}^r f_{1j}(m_j) , \dotsc, \sum_{j=1}^r f_{sj}(m_j) , \right) \end{align*} is again a homomorphism. If we write the elements of $$M_1 \oplus \dotsb \oplus M_s$$ as column vectors $$\begin{bmatrix} m_1 \\ \vdots \\ m_s \end{bmatrix}$$ then we can write the homomorphism $$f$$ as a matrix $$f = \begin{bmatrix} f_{11} & \cdots & f_{1s} \\ \vdots & \ddots & \vdots \\ f_{r1} & \cdots & f_{rs} \end{bmatrix} \,.$$ The application of $$f$$ to an element of $$M_1 \oplus \dotsb \oplus M_s$$ is then just the usual matrix-vector-multiplication.

If on the other hand $$f$$ is any homomorphism from $$M_1 \oplus \dotsb \oplus M_s$$ to $$N_1 \oplus \dotsb \oplus N_r$$, then the compositions $$f_{ij} \colon M_j \hookrightarrow M_1 \oplus \dotsb \oplus M_s \xrightarrow{\;f\;} N_1 \oplus \dotsb \oplus N_r \twoheadrightarrow N_i$$ are again homomorphism.

These two constructions are mutually inverse, and result in a bijection between the set $$\operatorname{Hom}_R( M_1 \oplus \dotsb \oplus M_s, N_1 \oplus \dotsb N_r )$$ and the set of all matrices $$[ f_{ij} ]_{i,j}$$ consisting of homomorphisms $$f_{ij} \colon M_j \to N_i$$.

Given two such homomorphisms $$f = [f_{jk}]_{jk}$$ and $$g = [g_{ij}]_{ij}$$, their composition $$g \circ f$$ (if it is defined) can be computed via the usual rules of matrix multiplication, i.e. we have $$g \circ f = \left[ \sum_j g_{ij} \circ f_{jk} \right]_{ik} \,.$$ (This holds because the application of $$f$$ and $$g$$ can be computed via matrix-vector-multiplication.)

In our situation we have $$R = K[G]$$ and can apply Schur’s lemma to compute $$\operatorname{Hom}(S_i, S_j)$$ for any two indices $$i, j$$: We find that $$\operatorname{Hom}(S_i, S_j) = 0$$ for all $$i \neq j$$, and $$\operatorname{Hom}(S_i, S_i) = K$$ for every index $$i$$, because $$K$$ is algebraically closed. We hence find that every morphisms in $$\mathcal{A}$$ can be described by a block-diagonal matrix with entries in $$K$$, giving us a bijection $$\operatorname{Hom}_{\mathcal{A}} ( S_1^{s_1} \oplus \dotsb \oplus S_n^{s_n}, S_1^{r_1} \oplus \dotsb \oplus S_n^{r_n} ) \to \operatorname{M}(r_1 \times s_1, K) \times \dotsb \times \operatorname{M}(r_n \times s_n, K) \,.$$ We want to emphasize that this bijection does not rely on any additional choices. (One may call it “canonical”.) The composition of morphisms in $$\mathcal{A}$$ corresponds to the multiplication of block-diagonal matrices, which in turns corresponds under the above bijection to the componentwise multiplication.

We have now shown that the category $$\mathcal{A}$$ is isomorphic to the category $$\mathcal{B}$$ which is given as follows:

• The objects of $$\mathcal{B}$$ are tupels $$(r_1, \dotsc, r_n)$$ of natural numbers.
• The morphism sets of $$\mathcal{B}$$ are given by $$\operatorname{Hom}_{\mathcal{B}}( (s_1, \dotsc, s_n), (r_1, \dotsc, r_n) ) = \operatorname{M}(r_1 \times s_1, K) \times \dotsb \times \operatorname{M}(r_n \times s_n, K) \,.$$
• The composition of morphisms in $$\mathcal{B}$$ is given by $$(A_1, \dotsc, A_n) \circ (B_1, \dotsc, B_n) = (A_1 B_1, \dotsc, A_n B_n) \,.$$

Let us now consider the category $$\mathbf{rep}_K(Q)$$. Let $$\mathcal{C}$$ be the full subcategory of $$\mathbf{rep}_K(Q)$$ whose objects are those representations $$(V_1, \dotsc, V_n)$$ whose vector spaces $$V_i$$ are of the form $$V_i = K^{r_i}$$ for some natural numbers $$r_i$$. The inclusion from $$\mathcal{C}$$ to $$\mathbf{rep}_K(Q)$$ is fully faithful und essentially surjective, and therefore an equivalence.

A morphism in $$\mathcal{C}$$ from $$(K^{s_1}, \dotsc, K^{s_n})$$ to $$(K^{r_1}, \dotsc, K^{r_n})$$ is just an arbitrary tupel $$(f_1, \dotsc, f_n)$$ of linear maps $$f_i$$ from $$K^{s_i}$$ to $$K^{r_i}$$. (These linear maps don’t need to satisfy any compatability conditions because the quiver $$Q$$ doesn’t have any arrows.) Each linear map $$f_i$$ is then given by multiplication with a unique matrix of size $$r_i \times s_i$$ with coefficients in $$K$$. We see from this that the category $$\mathcal{C}$$ is also isomorphic to the category $$\mathcal{B}$$.

We have overall shown that $$\mathbf{rep}_K(G) \simeq \mathcal{A} \cong \mathcal{B} \cong \mathcal{C} \simeq \mathbf{rep}_K(Q)$$ and thus overall $$\mathbf{rep}_K(G) \simeq \mathbf{rep}_K(Q)$$.

¹ Every irreducible $$G$$-representation is automatically finite-dimensional because $$G$$ is finite.
² One can do the same construction in any additive category instead of just $$R\text{-}\operatorname{Mod}$$

• Thanks. It is very clear! Jun 16, 2020 at 2:46