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.

  • $\begingroup$ 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 $\mathbb{N}^n$? $\endgroup$ Jun 12, 2020 at 15:24
  • $\begingroup$ 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 ;) $\endgroup$ Jun 12, 2020 at 15:27
  • $\begingroup$ @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 $\text{mod}-KQ$. $\endgroup$
    – user302934
    Jun 13, 2020 at 1:03
  • $\begingroup$ 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. $\endgroup$
    – Joppy
    Jun 14, 2020 at 3:08
  • 1
    $\begingroup$ This isn't true as stated. Is $K$ supposed to be assumed to be algebraically closed? $\endgroup$ Jun 15, 2020 at 3:31

1 Answer 1


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 $R\text{-}\operatorname{Mod}$

  • $\begingroup$ Thanks. It is very clear! $\endgroup$
    – user302934
    Jun 16, 2020 at 2:46

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