A Characterization of Finite EI Categories with Hereditary Category Algebras

Liping Li School of Mathematics, University of Minnesota, MN, 55455, USA

March 4, 2011

Abstract

In this paper we give an explicit algorithm to construct the ordinary quiver of a finite EI category for which the endomorphism groups of all objects have orders invertible in the field k. We classify all finite EI categories with hereditary category algebras, characterizing them as free EI categories (in a sense which we define) for which all endomorphism groups of objects have invertible orders. Some applications on the representation types of finite EI categories are derived.

1 Introduction

In this paper we study the representations of finite EI categories. They are small categories with finitely many morphisms in which every endomorphism is an isomorphism. This concept includes many structures such as finite groups, finite posets and free categories associated to finite quivers. A representation of C is a covariant functor from C to the category of finite dimensional vector spaces, and it may be regarded as the same thing as a finitely generated module for the category algebra.

We introduce the concepts of finite EI quivers and finite free EI categories. A finite EI quiver is a quiver equipped with two maps f and g. The map f assigns a finite group f(v) = G_{v} to each vertex v in the quiver and the map g assigns a (G_{w}, G_{v})-biset to each arrow from the vertex v to the vertex w in the quiver. Finite free EI categories are categories generated from finite EI quivers by a specific rule which generalizes the construction of a free category from a quiver [4]. We will show that they are characterized by a certain unique factorization property of the non-isomorphisms. In a certain sense a finite free EI category is the largest EI category which can be generated by a finite EI quiver.

For every finite EI category C, we will construct a finite free EI category Ĉ and a full functor F̂ : Ĉ → C such that F̂ is the identity map restricted to objects, endomorphisms and all non-isomorphisms which can not be expressed as composites of more than one non-isomorphism. This category Ĉ is unique up to isomorphism. We call Ĉ the free EI cover of C.

A finite EI category C determines a finite dimensional associative kalgebra kC with identity, called the category algebra (defined in section 2). Under the assumption that the endomorphism groups of all objects in C have orders invertible in k, we will describe an explicit algorithm to construct the ordinary quiver of the category algebra kC. This algorithm uses the representations of semisimple group algebras and their restrictions to subgroups. It is easier and more intuitive than the method by computing Exts, introduced in [6]. Our main results are described in the following theorems:

Theorem 1.1. Let C be a finite EI category for which the endomorphism groups of all objects have orders invertible in the field k. Then the quiver Q constructed by our algorithm is precisely the ordinary quiver of the category algebra kC. Moreover, kC has the same ordinary quiver as that of kĈ, the category algebra of the free EI cover Ĉ of C.

Theorem 1.2. Let C be a finite EI category. Then the category algebra kC is hereditary if and only if C is a finite free EI category satisfying the condition that the endomorphism groups of all objects have orders invertible in k.

These results are useful in determining the representation types of category algebras. Indeed, if C is a finite free EI category for which the endomorphism groups of all objects have orders invertible in k, then it has the same representation type as that of its ordinary quiver. Stated explicitly, kC is of finite (tame, resp.) representation type if and only if the underlying graph of its ordinary quiver is a disjoint union of Dynkin (Euclidean, resp.) diagrams. Otherwise, it is of wild type. For a general finite EI category C, kC is a quotient algebra of kĈ, the category algebra of its free EI cover Ĉ. Thus the finite representation type of kĈ implies the finite representation type of kC, no matter what the characteristic of k is. Moreover, if C has a full subcategory of infinite representation type, it is of infinite representation type as well. By studying those full subcategories which are finite free EI categories, for example the connected full subcategories with two objects, we can get certain information about the representation types of general finite EI categories.

Here is the layout of this paper: in section 2 we give some background on the representation theory of finite EI categories, and introduce the definitions of finite EI quivers and finite free EI categories. An equivalent definition by the unique factorization property of non-isomorphisms is proved, and we present some further basic properties.

From section 3 onwards we focus on the representations of finite EI categories for which all endomorphism groups of objects have orders invertible in k. In section 4 we give a detailed algorithm to construct an associated quiver Q for each finite EI category for which the endomorphism groups of all objects have orders invertible in the field k. Then we prove Theorem 1.1. In section 5 we prove Theorem 1.2, which classifies all finite EI categories with hereditary category algebras.

Even though we can study the representation type of C by constructing its ordinary quiver, it is more efficient sometimes to use certain simple criteria deduced from Theorem 1.1 and Theorem 1.2. In the last section we describe such criteria expressed in terms of full subcategories with two objects.

It is sometimes useful to know an explicit functor F from kC-mod, the category of all representations of a finite free EI category C, to kQ-mod, the category of all representations of the ordinary quiver Q of kC. Although our proof of Theorem 1.2 does not rely on this functor, we describe it in detail in the appendix. This functor is proved to be full, faithful and dense, and hence induces a Morita-equivalence between kC and kQ.

In this paper C is always a finite EI category with objects Ob(C) and morphisms Mor(C). We use Aut_{C}(x) (or End_{C}(x)) to denote the endomorphism group of a fixed object x in C and Hom_{C}(x, y) to denote the set of morphisms from an object x to another object y. The field k is supposed to be algebraically closed. All modules are finitely generated left modules. Composition of group actions, morphisms and maps is from right to left.

Acknowledgement: The author wants to express great appreciation to his advisor, Professor Peter Webb, who led the author into this area. Without his invaluable suggestions and help this paper would not have appeared. Professor Webb also contributed the concepts of finite EI quivers and finite free EI categories for this paper. These ideas are described in his unpublished notes.

2 Finite Free EI Categories

For the reader’s convenience, we include in this section some background on the representation theory of finite EI categories. Please refer to [7], [9], [10] for more details.

The category algebra kC of a finite EI category C is the k-space with basis the morphisms in C. Its product ∗ is determined by bilinearity from the product of the basis elements, given by the rule:

α ∗ β = \{ \begin{matrix}α ∘ β & if α and β can be composed \\ 0 & otherwise. \end{matrix}

Let R be representation of C defined in section 1. Then it assigns a vector space R(x) to each object x in C, and a linear transformation R(α) : R(x) → R(y) to each morphism α : x → y such that all composition relations of morphisms in C are preserved under R. Notice that R(x) has a kEnd_{C}(x)- module structure for every object x in C. Indeed, for each morphism g in End_{C}(x), R(g) is an automorphism of R(x). Thus we can define an action of g on R(x) by letting g · v = R(g)(v), for all v in R(x). A homomorphism φ : R_{1} → R_{2} of two representations is a natural transformation of functors. By Theorem 7.1 of [5], a representation of C is equivalent to a kC-module. Thus we don’t distinguish these two concepts throughout this paper.

A finite EI category C is said to be of finite (tame, wild, resp.) representation type if the category algebra kC is of finite (tame, wild, resp.) type. The category C is connected if for any two distinct objects x and y, there is a list of objects x = x_{0}, x_{1}, …, x_{n} = y such that either Hom_{C}(x_{i}, x_{i+1}) or Hom_{C}(x_{i+1}, x_{i}) is not empty, 0 ⩽ i ⩽ n − 1. Every finite EI category is a disjoint union of connected components, and each component is a full subcategory. If C = ⨆_{i=1}^{m} C_{i}, the category algebra kC has an algebra decomposition kC_{1} ⊕ … ⊕ kC_{m}. Moreover, if C and D are equivalent finite EI categories, kC-mod is equivalent to kD-mod by Proposition 2.2 in [7]. Thus it is sufficient to study the representations of connected, skeletal finite EI categories. We make the following convention:

Convention: All finite EI categories in this paper are connected and skeletal. Thus endomorphisms, isomorphisms and automorphisms coincide.

Under the hypothesis that C is skeletal, if x and y are two distinct objects in C with Hom_{C}(x, y) non-empty, then Hom_{C}(y, x) is empty. Indeed, if this is not true, we can take α ∈ Hom_{C}(y, x) and β ∈ Hom_{C}(x, y). The composite βα is an endomorphism of y, hence an automorphism. Similarly, the composite αβ is an automorphism of x. Thus both α and β are isomorphisms, so x is isomorphic to y. But this is impossible since C is skeletal and x ≠ y.

It is time to introduce the concept of finite free EI categories. Before giving a formal definition, we define finite EI quivers, which are finite quivers with extra structure.

Definition 2.1. A finite EI quiver Q̂ is a datum (Q_{0}, Q_{1}, s, t, f, g), where: (Q_{0}, Q_{1}, s, t) is a finite acyclic quiver with vertex set Q_{0}, arrow set Q_{1}, source map s and target map t. The map f assigns a finite group f(v) to each vertex v ∈ Q_{0}; the map g assigns an (f(t(α)), f(s(α)))-biset to each arrow α ∈ Q_{1}.

If f assigns the trivial group to each vertex in Q_{0} in the above definition, we obtain a quiver in the usual sense. In this sense, finite acyclic quivers are special cases of finite EI quivers.

Each finite EI quiver Q̂ = (Q_{0}, Q_{1}, s, t, f, g) determines a finite EI category CQ̂ in the following way: the objects in CQ̂ are precisely the vertices in Q_{0}. For a particular object v in CQ̂, we define End_{CQ̂}(v) = f(v), which is a finite group by our definition. It remains to define Hom_{CQ̂}(v, w) if v ≠ w are distinct vertices in Q_{0}, and the composition of morphisms.

Let [figure: v; γ; w] be a directed path from v to w. Then γ can be written uniquely as a composition of arrows, where v_{i} ∈ Q_{0} and α_{i} ∈ Q_{1} for i = 1, …, n.

\xymatrix{v = v_{0} \ar[r]^{α_{1}} & v_{1} \ar[r]^{α_{2}} & … \ar[r]^{α_{n}} & v_{n} = w}

Notice that g(α_{i}) is an (f(v_{i}), f(v_{i−1}))-biset, so we define:

H_{γ} = g(α_{n}) ×_{f(v_{n−1})} g(α_{n−1}) ×_{f(v_{n−2})} … ×_{f(v_{1})} g(α_{1}),

the biset product defined in [8]. Finally, Hom_{CQ̂}(f(v), f(w)) can be defined as ⨆_{γ} H_{γ}, the disjoint union of all H_{γ}, over all possible paths γ from x to y. In the case v = w we define Hom_{CQ̂}(v, v) = f(v).

Let α and β be two morphisms in CQ̂. They lie in two sets H_{γ_{1}} and H_{γ_{2}}, where γ_{1} and γ_{2} are two paths determined by α and β respectively, possibly of length 0. Their composite β ∘ α can be defined by the following rule: it is 0 if the composite γ_{2}γ_{1} is not defined in Q̂. Otherwise, the initial vertex v of γ_{2} is exactly the terminal vertex of γ_{1}. Since there is a natural surjective map p : H_{γ_{2}} × H_{γ_{1}} → H_{γ_{2}} ×_{f(v)} H_{γ_{1}}, we define β ∘ α = p(β, α), the image of (β, α) in H_{γ_{2}} ×_{f(v)} H_{γ_{1}}. This definition satisfies the associative rule, and in this way we get a finite EI category CQ̂ from Q̂.

Definition 2.2. A finite EI category C is a finite free EI category if it is isomorphic to the finite EI category CQ̂ generated from a finite EI quiver Q̂ by the above construction.

In practice it is inconvenient to check whether a finite EI category C is free or not by using the definition. Fortunately, there is an equivalent characterization built upon unfactorizable morphisms: the Unique Factorization Property (UFP).

Definition 2.3. A morphism α : x → z in C is unfactorizable if α is not an isomorphism and whenever it has a factorization as a composite

\xymatrix{x \ar[r]^{β} & y \ar[r]^{γ} & z}, then either β or γ is an isomorphism.

The reader may want to know the relation between the terminology unfactorizable morphism and the term irreducible morphism which is widely accepted and used, for example in [2] and [9]. Indeed, in this paper they coincide since we only deal with finite EI categories. But in a more general context, they are different, as we explain in the following example:

Example 2.4. Consider the following category C with two objects x ≇ y. The non-identity morphisms in C are generated by α : x → y and β : y → x with the only nontrivial relation being βα = 1_{x}. Then the morphisms in C are 1_{x}, 1_{y}, α, β and αβ. It is not a finite EI category since αβ ∈ End_{C}(y) is not an isomorphism. Then neither α nor β are irreducible morphisms since one of them is a split monomorphism and the other is a split epimorphism. However, the reader can check that they are unfactorizable morphisms. Furthermore, the algorithm constructing ordinary quivers of this category algebra (described in section 4) still works well. This example illustrates the reason that we introduce the notion unfactorizable morphisms: our algorithm works in a more general situation where unfactorizable morphisms do not coincide with irreducible morphisms any more.

[figure: x; 1_{x}; α; y; β; 1_{y}]

Note that the composite of an unfactorizable morphism with an isomorphism is still unfactorizable.

Proposition 2.5. Let α : x → y be an unfactorizable morphism. Then hαg is also unfactorizable for every h ∈ Aut_{C}(y) and every g ∈ Aut_{C}(x).

Proof. Fix a decomposition hαg = α_{1}α_{2}. Then we have α = (h^{−1}α_{1})(α_{2}g^{−1}). But α is unfactorizable, so by definition either one of h^{−1}α_{1} and α_{2}g^{−1} is an isomorphism. Without loss of generality, let h^{−1}α_{1} be an isomorphism. Then α_{1} is an isomorphism since h^{−1} is an automorphism.

Let C be a finite EI category. By the previous proposition, the set of unfactorizable morphisms from an object x to another object y is closed under the actions of Aut_{C}(x) and Aut_{C}(y). Choose a fixed representative for each (Aut_{C}(y),Aut_{C}(x))-orbit. Repeating this process for all pairs of different objects (x, y), we get a set A = {α_{1}, …, α_{n}} of orbit representatives. Elements in A are called representative unfactorizable morphisms.

We should point out here that each finite EI category C determines a finite EI quiver Q̂ in the following way: its vertices are objects in C; we put an arrow x :→ y in Q̂ for each representative unfactorizable morphism α : x → y in C. Thus the arrows biject with all representative unfactorizable morphisms α : x → y in C, or equivalently, all Aut_{C}(y)×Aut_{C}(x)-orbits of unfactorizable morphisms in C. The map f assigns End_{C}(x) to each object x; the map g assigns the (Aut_{C}(y), Aut_{C}(x))-biset where a representative unfactorizable morphism α : x → y lies to the corresponding arrow. Obviously, this finite EI quiver is unique up to isomorphism. We call this quiver the finite EI quiver of C.

Now suppose that C is a finite free EI category. It is possible that there is more than one finite EI quiver generating C, although they must have the same vertices. However, it is not hard to see that all those finite EI quivers are subquivers of the finite EI quiver of C.

All non-isomorphisms can be written as composites of unfactorizable morphisms.

Proposition 2.6. Let α : x → y be a morphism with x ≠ y. Then it has a decomposition \xymatrix{x = x_{0} \ar[r]^{α_{1}} & x_{1} \ar[r]^{α_{2}} & … \ar[r]^{α_{n}} & x_{n} = y} , where all α_{i} are unfactorizable.

Proof. If α is unfactorizable, we are done. Otherwise, α has a decomposition \xymatrix{x \ar[r]^{α_{1}} & x_{1} \ar[r]^{α_{2}} & y} where neither α_{1} nor α_{2} is an isomorphism. In particular, x_{1} is different from x and y. If both α_{1} and α_{2} are unfactorizable, we are done. Otherwise, assume α_{1} is not unfactorizable. Repeating the above process, we can get a decomposition \xymatrix{x \ar[r]^{α_{11}} & x_{11} \ar[r]^{α_{12}} & x_{1} \ar[r]^{α_{2}} & y}. With the same reasoning, x, x_{11}, x_{1}, y are pairwise different. Since there are only finitely many objects, this process ends after finitely many steps. Therefore we get a decomposition of α into unfactorizable morphisms.

For an arbitrary finite EI category C, the ways to decompose a nonisomorphism into unfactorizable morphisms need not to be unique. However, we can show that for finite free EI categories, this decomposition is unique up to a trivial relation, i.e., they satisfy the property defined below:

Definition 2.7. A finite EI category C satisfies the Unique Factorization Property (UFP) if whenever any non-isomorphism α has two decompositions into unfactorizable morphisms:

\xymatrix{x = x_{0} \ar[r]^{α_{1}} & x_{1} \ar[r]^{α_{2}} & … \ar[r]^{α_{m}} & x_{m} = y \\ x = x_{0} \ar[r]^{β_{1}} & y_{1} \ar[r]^{β_{2}} & … \ar[r]^{β_{n}} & y_{n} = y}

then m = n, x_{i} = y_{i}, and there are h_{i} ∈ Aut_{C}(x_{i}), 1 ⩽ i ⩽ n − 1 such that β_{1} = h_{1}α_{1}, β_{2} = h_{2}α_{2}h_{1}^{−1}, …, β_{n−1} = h_{n−1}α_{n−1}h_{n−2}^{−1}, β_{n} = α_{n}h_{n−1}^{−1}.

The UFP gives a characterization of finite free EI categories.

Proposition 2.8. A finite EI category C is free if and only if it satisfies the UFP.

Proof. Suppose C is a finite free EI category generated by a finite EI quiver Q̂ = (Q_{0}, Q_{1}, s, t, f, g). Let α : v → w be an arbitrary non-isomorphism. By the previous proposition α can be written as a composite of unfactorizable morphisms. Let α_{m} ∘ … ∘ α_{1} and β_{n} ∘ … ∘ β_{1} be two such decompositions of α. It is easy to see from definitions that an unfactorizable morphism in C lies in g(τ) for some unique arrow τ ∈ Q_{1}. Thus α_{m} ∘ … ∘ α_{1} and β_{n} ∘ … ∘ β_{1} determine two paths γ_{1} and γ_{2} in Q̂ from v to w. But α is contained in Hom_{C}(v, w) = ⨆_{γ} H_{γ}, the disjoint union taken over all possible paths from v to w, so γ_{1} must be the same as γ_{2}. Consequently, m = n, and α_{i} and β_{i} have the same target and source for 1 ⩽ i ⩽ n. By the definition of biset product, the fact

α_{n} ∘ (α_{n−1} … ∘ α_{1}) = β_{n} ∘ (β_{n−1} ∘ … ∘ β_{1})

in the biset product implies that there is an automorphisms g_{n−1} ∈ Aut_{C}(x_{n−1}) such that

α_{n} = β_{n}g_{n−1}, (α_{n−1} ∘ … ∘ α_{1}) = g_{n−1}^{−1}(β_{n−1} ∘ … ∘ β_{1}),

where x_{n−1} is the common target of α_{n−1} and β_{n−1}. By an easy induction on n, we show that {α_{i}}_{i=1}^{n} and {β_{j}}_{j=1}^{n} have the required relations in the previous definition. Thus C satisfies the UFP.

On the other hand, if C satisfies the UFP, we want to show that C is isomorphic to the finite free EI category CQ̂ generated from its finite EI quiver Q̂. Define a functor F : C → CQ̂ in the following way: First, F (x) = x for every object x in C since Ob(C) = Ob(CQ̂) by our construction. Furthermore, it is also clear that Aut_{C}(x) = AutCQ̂(x) for every object x, and the biset of unfactorizable morphisms from x to y in C is the same as that in CQ̂ for every pair of different objects x and y. Therefore we can let F be the identity map restricted to automorphisms and unfactorizable morphisms in C. Proposition 2.6 tells us that every non-isomorphism α in C is a composite α_{n} ∘ … ∘ α_{1} of unfactorizable morphisms, so F (α) can be defined as F (α_{n}) ∘ … ∘ F (α_{1}). By the UFP F is well-defined and is a bijection restricted to Hom_{C}(x, y) for each pair of distinct objects x and y. Consequently, F is a bijection from Mor(C) to Mor(CQ̂) and so is an isomorphism. This finishes the proof.

Finite free EI categories have a certain universal property which is stated in the following proposition:

Proposition 2.9. Let C be a finite EI category. Then there is a finite free EI category Ĉ and a full functor F̂ : Ĉ → C such that F̂ is the identity map restricted to objects, isomorphisms and unfactorizable morphisms. This finite free EI category is unique up to isomorphism.

We call Ĉ the free EI cover of C.

Proof. We mentioned before that every finite EI category C determines a finite EI quiver Q̂ (see the paragraphs before Proposition 2.6), hence a finite free EI category CQ̂ satisfying: Ob(C) = Ob(CQ̂); Aut_{C}(x) = AutCQ̂(x) for every object x; the biset of unfactorizable morphisms from x to y in C is the same as that in CQ̂ for every pair of different objects x and y.

Define a functor F̂ : CQ̂ → C in the following way: F is the identity map on objects, isomorphisms and unfactorizable morphisms. Now if δ : x → y is neither an isomorphism nor an unfactorizable morphism, it can be decomposed as the composite

\xymatrix{x = x_{0} \ar[r]^{β_{1}} & x_{1} \ar[r]^{β_{2}} & … \ar[r]^{β_{m}} & x_{m} = y,}

where each β_{i} is unfactorizable for 1 ⩽ i ⩽ m. Define

F̂(δ) = F̂(β_{m}) … F̂(β_{2})F̂(β_{1}).

F̂ is clearly well defined for all isomorphisms and unfactorizable morphisms. We want to verify that F̂ is well defined for factorizable morphisms as well. That is, if δ has another decomposition into unfactorizable morphisms

\xymatrix{x = x_{0} \ar[r]^{β′_{1}} & z_{1} \ar[r]^{β_{2}} & … \ar[r]^{β′_{n}} & z_{n} = y,}

then

F̂(β_{n})F̂(β_{n−1}) … F̂(β_{1}) = F̂(β′_{n}) … F̂(β′_{2})F̂(β′_{1}).

Since C satisfies the UFP, we have m = n and x_{i} = z_{i} for 1 ⩽ i ⩽ n, and β_{1} = h_{1}β′_{1}, β_{2} = h_{2}β′_{2}h_{1}^{−1}, …, β_{n−1} = h_{n−1}β′_{n−1}h_{n−2}^{−1}, β_{n} = β′_{n}h_{n−1}^{−1}, where h_{i} ∈ Aut_{C}(x_{i}). Thus:

F̂(β_{n})F̂(β_{n−1}) … F̂(β_{1})
= F̂(β′_{n}h_{n−1}^{−1})F̂(h_{n−1}β′_{n−1}h_{n−2}^{−1}) … F̂(h_{1}β′_{1})
= F̂(β′_{n})F̂(h_{n−1}^{−1})F̂(h_{n−1}) … F̂(h_{1}^{−1})F̂(h_{1})F̂(β′_{1})
= F̂(β′_{n}) … F̂(β′_{2})F̂(β′_{1}).

Therefore, F̂ is a well defined functor. It is full since all automorphisms and unfactorizable morphisms in C are images of F̂, and all other morphisms in C are their composites. Moreover, Ĉ is unique up to isomorphism since it is completely determined by objects, isomorphisms and unfactorizable morphisms in C.

It is well-known that every subgroup of a free group is still free. Finite free EI categories have a similar property.

Proposition 2.10. Let C be a finite EI category. Then C is a finite free EI category if and only if all of its full subcategories are finite free EI categories.

Proof. The if part is trivial since C is such a subcategory of itself. Now let D be a full subcategory of C. We want to show that D satisfies the UFP.

Take a factorizable morphism α in D and two decompositions of the following form:

\xymatrix{x = x_{0} \ar[r]^{α_{1}} & x_{1} \ar[r]^{α_{2}} & … \ar[r]^{α_{m}} & x_{m} = y \\ x = x_{0} \ar[r]^{β_{1}} & x′_{1} \ar[r]^{β_{2}} & … \ar[r]^{β_{n}} & x′_{n} = y}

where all α_{i} and β_{j} are unfactorizable in D, but possibly factorizable in C. Decomposing them into unfactorizable morphisms in C, we get two extended sequences of unfactorizable morphisms as follows:

\xymatrix{x = x_{0} \ar[r]^{δ_{1}} & w_{1} \ar[r]^{δ_{2}} & … \ar[r]^{δ_{r}} & w_{r} = y \\ x = x_{0} \ar[r]^{θ_{1}} & w_{1} \ar[r]^{θ_{2}} & … \ar[r]^{θ_{r}} & w_{r} = y}

Each pair of morphisms δ_{i} and θ_{i} have the same source and target since C is a finite free EI category. Moreover, there are h_{i} ∈ Aut_{C}(w_{i}), 1 ⩽ i ⩽ r−1, such that θ_{1} = h_{1}δ_{1}, θ_{2} = h_{2}δ_{2}h_{1}^{−1}, …, θ_{r−1} = h_{r−1}δ_{r−1}h_{r−2}^{−1}, θ_{r} = h_{r−1}^{−1}δ_{r}. If we can prove the fact that m = n and x_{1} = x′_{1}, …, x_{m} = x′_{n}, then the conclusion follows. Indeed, if this is true, say x_{1} = x′_{1} = w_{r_{1}}, …, x_{m} = x′_{m} = w_{r_{m}}, then we have β_{1} = h_{r_{1}}α_{1}, β_{2} = h_{r_{2}}α_{2}h_{r_{1}}^{−1}, …, β_{n} = α_{n}h_{r_{n}}^{−1}, which is exactly the UFP.

We show this fact by contradiction. Suppose that x_{1} = x′_{1}, x_{2} = x′_{2}, …, x_{i−1} = x′_{i−1}, and let x_{i} be the first object in the sequence different from x′_{i}. Notice both x_{m} … x_{1} and x′_{n} … x′_{1} are subsequences of the sequence w_{r} … w_{1}. As a result, x_{i} appears before x′_{i} in w_{r} … w_{1}, or after x′_{i}. Without loss of generality we assume that x_{i} is before x′_{i}. Let x_{i−1} = x′_{i−1} = w_{a}, x_{i} = w_{b} and x′_{i} = w_{c}. We must have a ≤ b < c. Furthermore, we check that

β_{i} = θ_{c} ∘ θ_{c−1} ∘ … ∘ θ_{a+1} = (θ_{c} ∘ … ∘ θ_{b+1}) ∘ (θ_{b} ∘ … ∘ θ_{a+1}),

with (θ_{c} ∘ … ∘ θ_{b+1}) contained in Hom_{C}(x_{i}, x′_{i}) and (θ_{b} ∘ … ∘ θ_{a+1}) contained in Hom_{C}(x_{i−1}, x_{i}). But D is a full subcategory, so β_{i} is the composite of two non-isomorphisms in D. This is a contradiction since we have assumed that β_{i} is unfactorizable in D.

The condition that D is a full subcategory of C is required. Consider the following two examples:

Example 2.11. Let C be the free category generated by the following quiver. Let D be the subcategory of C obtained by removing the morphism β from C. The category C is free, but the subcategory D is not free. Indeed, morphisms α, βα, γβ, γ are all unfactorizable in D. Thus the morphism γβα has two decompositions (γβ) ∘ α and (γ) ∘ (βα), which contradicts the UFP.

\xymatrix{• \ar[r]^{α} & • \ar[r]^{β} & • \ar[r]^{γ} & •}

Example 2.12. Let C be the following category, where g generates a cyclic group of order 2, interchanges β_{1} and β_{2}, and fixes α. Then β_{1}α = (β_{2}g)α = β_{2}(gα) = β_{2}α. It is not hard to check that C satisfies the UFP; but the subcategory formed by removing the morphism g from C doesn’t satisfies the UFP.

\xymatrix{1_{x} \ar[r]^{α} & ⟨g⟩ \ar[r]^{β_{1}} \ar[r]_{β_{2}} & 1_{z}}

Proposition 2.13. Let C be a finite EI category and Ĉ be its free EI cover. Then the category algebra kC is a quotient algebra of kĈ. In particular, if Ĉ is of finite representation type, so is C.

Proof. Since the functor F̂ : Ĉ → C is bijective on objects, by Proposition 2.2.3 in [10], it induces an algebra homomorphism from kĈ to kC by sending α ∈ Mor(Ĉ) to F̂(α) ∈ Mor(C). This homomorphisms is surjective since F̂ is full.

We want to clarify some confusion probably caused by the name “finite free EI categories”: a finite free EI category C is in general not a free category as the following example shows.

Example 2.14. Let C be a category with one object whose morphisms form a non-identity group. Then C is a finite free EI category by our definition. But it is not a free category since the morphism group is not a free monoid.

3 Representations of Free EI Categories

In this section we study the representations of finite free EI categories. First, let us introduce some notation.

A representative unfactorizable morphism α : x → y uniquely determines a subcategory D_{α} of C: Ob(D_{α}) = {x, y}, Aut_{D_{α}}(x) = Aut_{C}(x), Aut_{D_{α}}(y) = Aut_{C}(y), Hom_{D_{α}}(x, y) is the whole (Aut_{C}(y), Aut_{C}(x))-orbit where α lies. Obviously, D_{α} is a finite free EI category.

With the above setup, we can show that a representation of a finite free EI category C is determined by its local structures on subcategories D_{α} for all representative unfactorizable morphisms α. Stated formally, it is:

Proposition 3.1. Let C be a finite free EI category, and R be a rule assigning a kAut_{C}(x)-module to each object x and a linear transformation to each unfactorizable morphism in C. If R restricted to D_{α} is a representation of D_{α} for every representative unfactorizable morphism α in C, then R induces a unique representation R̃ of C such that R̃ restricted to each D_{α} is the same as R restricted to D_{α}.

Proof. Since we can define R̃(x) = R(x) for each object x in C, it suffices to define a linear map for each morphism in C such that the composition relations are preserved. If β is an automorphism or an unfactorizable morphism, it must belong to some subcategory D_{α} for a representative unfactorizable morphism α. Because R restricted to D_{α} is a representation, we then define R̃(β) = R(β).

Now suppose that β is factorizable and decompose it into unfactorizable morphisms of the following form

\xymatrix{x = x_{0} \ar[r]^{α_{1}} & x_{1} \ar[r]^{α_{2}} & … \ar[r]^{α_{m}} & x_{m} = y} .

Let R̃(α) = R(α_{m}) … R(α_{1}). We claim that it is well defined. That is: if α has two different decompositions

\xymatrix{x = x_{0} \ar[r]^{α_{1}} & x_{1} \ar[r]^{α_{2}} & … \ar[r]^{α_{m}} & x_{m} = y}

and

\xymatrix{x = x_{0} \ar[r]^{β_{1}} & y_{1} \ar[r]^{β_{2}} & … \ar[r]^{β_{n}} & y_{n} = y} ,

then it is true that R(α_{m}) … R(α_{1}) = R(β_{n}) … R(β_{1}).

Indeed, because C satisfies the unique factorization property, we have m = n, x_{i} = y_{i}, and β_{1} = h_{1}α_{1}, β_{2} = h_{2}α_{2}h_{1}^{−1}, …, β_{n−1} = h_{n−1}α_{n−1}h_{n−2}^{−1}, β_{n} = α_{n}h_{n−1}^{−1} for some h_{i} ∈ End_{C}(x_{i}), 1 ⩽ i ⩽ n − 1. Notice that each α_{i} and the corresponding β_{i} are unfactorizable, and lie in D_{α_{i}} since β_{i} and α_{i} are in the same orbit. The fact that R restricted to D_{α_{i}} is a representation of this subcategory implies R(h_{i})R(h_{i}^{−1}) = R(h_{i}h_{i}^{−1}) = 1. Thus:

R̃(β_{n}) … R(β_{2})R(β_{1})
= R(α_{n})R(h_{n−1}^{−1})R(h_{n−1})R(α_{n−1})R(h_{n−2}^{−1}) … R(h_{1})R(α_{1})
= R(α_{n}) … R(α_{1})
= R̃(α_{n} … α_{1}).

We proved that R̃ is indeed a representation of C. Moreover, R̃ restricted to any D_{α} is exactly the same as R restricted to this subcategory. The uniqueness is obvious since from the above process we conclude that R̃ is uniquely determined by its values on subcategories D_{α}, for all representative unfactorizable morphisms α in C.

Proposition 3.2. Let C be a finite free EI category. Let R_{1} and R_{2} be two representations of C. Then a family of kAut_{C}(x)-module homomorphisms {ϕ_{x} : R_{1}(x) → R_{2}(x) | x ∈ Ob(C)} is a kC-module homomorphism from R_{1} to R_{2} if and only if for each representative unfactorizable morphism α : x → y, we have ϕ_{y}R_{1}(α) = R_{2}(α)ϕ_{x}.

Proof. The only if part is trivial. For the other direction, let us consider unfactorizable morphisms first. If β : x → y is unfactorizable, it lies in the same orbit as a unique representative unfactorizable morphism α, i.e., β = hαg, where g ∈ Aut_{C}(x) and h ∈ Aut_{C}(y). Because ϕ_{x} is a kAut_{C}(x)- module homomorphism, and ϕ_{y} is a kAut_{C}(y)-module homomorphism, we have:

ϕ_{y}R_{1}(hαg) = ϕ_{y}R_{1}(h)R_{1}(α)R_{1}(g)
= R_{2}(h)ϕ_{y}R_{1}(α)R_{1}(g)
= R_{2}(h)R_{2}(α)ϕ_{x}R_{1}(g)
= R_{2}(h)R_{2}(α)R_{2}(g)ϕ_{x}
= R_{2}(hαg)ϕ_{x}.

That is, ϕ_{y}R_{1}(β) = R_{2}(β)ϕ_{x}.

If β is factorizable, let \xymatrix{x = x_{0} \ar[r]^{α_{1}} & x_{1} \ar[r]^{α_{2}} & … \ar[r]^{α_{m}} & x_{m} = y} be a decomposition of β into unfactorizable morphisms. Consider the following diagram:

\xymatrix{R_{1}(x_{0}) \ar[r]^{R_{1}(α_{1})} \ar[d]^{ϕ_{0}} & R_{1}(x_{1}) \ar[r]^{R_{1}(α_{2})} \ar[d]^{ϕ_{1}} & … \ar[r]^{R_{1}(α_{m})} & R_{1}(x_{m}) \ar[d]^{ϕ_{m}} \\ R_{2}(x_{0}) \ar[r]^{R_{2}(α_{1})} & R_{2}(x_{1}) \ar[r]^{R_{2}(α_{2})} & … \ar[r]^{R_{2}(α_{m})} & R_{2}(x_{m}).}

For i = 0, …, m − 1, we have ϕ_{i+1}R_{1}(α_{i+1}) = R_{2}(α_{i+1})ϕ_{i}. Thus:

ϕ_{m}R_{1}(β)
= ϕ_{m}R_{1}(α_{m})R_{1}(α_{m−1}) … R_{1}(α_{1})
= R_{2}(α_{m})ϕ_{m−1}R_{1}(α_{m−1}) … R_{1}(α_{1})
…
= R_{2}(α_{m})R_{2}(α_{m−1}) … R_{2}(α_{1})ϕ_{0}
= R_{2}(β)ϕ_{0}.

This finishes the proof.

Throughout the remaining of this section we use the following notation: α : x → y is a fixed representative unfactorizable morphism and D_{α} is the corresponding subcategory of C (see our notation before Proposition 3.1 for the definition of D_{α}). Let G and H be Aut_{C}(x) and Aut_{C}(y) respectively. Define: G_{0} = Stab_{G}(α), H_{0} = Stab_{H}(α), G_{1} = Stab_{G}(Hα), and H_{1} = Stab_{H}(αG). Obviously G_{1} = {g ∈ G : exists h ∈ H with αg = hα}, and H_{1} = {h ∈ H : exists g ∈ G with hα = αg}. The group G (H, resp.) acts transitively on Hom_{D_{α}}(x, y) if and only if H_{1} = H (G_{1} = G, resp.), bearing in mind that Hom_{D_{α}}(x, y) only has one orbit as a biset.

Lemma 3.3. With the above notation, G_{0} and G_{1} are subgroups of G, H_{0} and H_{1} are subgroups of H. Moreover, G_{0} ◁ G_{1}, H_{0} ◁ H_{1}, and G_{1}/G_{0} ≅ H_{1}/H_{0}.

Proof. This result is well-known to Bouc who described the structures of bisets in [3]. We give an elementary proof here. It is routine to check that G_{0} ⩽ G, G_{1} ⩽ G by definition. Similarly, H_{0} ⩽ H and H_{1} ⩽ H. It is also clear that G_{0} ⩽ G_{1} and H_{0} ⩽ H_{1}.

Let g_{0} ∈ G_{0}, g ∈ G_{1} and consider g^{−1}g_{0}g. Since g ∈ G_{1}, there is h ∈ H such that αg = hα. Then αg^{−1}g_{0}g = h^{−1}αg_{0}g = h^{−1}αg = h^{−1}hα = α. This means g^{−1}g_{0}g ∈ G_{0} and G_{0} ◁ G_{1}. Similarly, H_{0} ◁ H_{1}.

Now define a map ϕ : G_{1} ↦ H_{1}/H_{0} by the following rule: for every g ∈ G_{1}, there is some h ∈ H such that αg = hα. By the definition h is contained in H_{1}. Define ϕ(g) = h̄, the image of h in H_{1}/H_{0}. The reader can check that ϕ is well defined, surjective, and a group homomorphism. Moreover, the kernel of this map is exactly G_{0}. Thus G_{1}/G_{0} ≅ H_{1}/H_{0}.

Remark 3.4. Under the isomorphism given in this lemma, we identify the quotient group G_{1}/G_{0} with H_{1}/H_{0}, and the module k ↑_{G_{0}}^{G_{1}} ≅ k(G_{1}/G_{0}) with k ↑_{H_{0}}^{H_{1}} ≅ k(H_{1}/H_{0}).

From now on we insist:

Convention: All finite EI categories C we study satisfy that the endomorphism group of every object x in C has order invertible in k.

A representation R of D_{α} determines a linear map φ = R(α) from a kGmodule R(x) to a kH-module R(y). Notice that both R(x) and R(y) are semisimple by Maschke’s theorem in view of the previous convention.

Lemma 3.5. Let U be a simple summand of R(x) ↓_{G_{1}}^{G}. If φ(U) ≠ 0, then:

(1) U is a direct summand of k ↑_{G_{0}}^{G_{1}};

(2) the restricted map φ_{U} : U → φ(U) is a k(G_{1}/G_{0})-module isomorphism under the identification k ↑_{G_{0}}^{G_{1}} ≅ k ↑_{H_{0}}^{H_{1}};

(3) φ(U) is a direct summand of k ↑_{H_{0}}^{H_{1}}.

Proof. If φ(U) ≠ 0, there is some u ∈ V_{1} ⊆ U such that φ(u) = w ≠ 0, where V_{1} is a simple summand of U ↓_{G_{0}}^{G_{1}}. We claim that each g ∈ G_{0} fixes u, hence V_{1} ≅ k. Indeed, if there is g ∈ G_{0} such that gu ≠ u, then φ(u − gu) = φ(u) − (φg)(u) = 0 since g ∈ G_{0} fixes φ. But u − gu ≠ 0 is also in V_{1} and generates the simple kG_{0}-module V_{1}, too. Now the fact that G_{0} fixes φ and φ(u − gu) = 0 implies that V_{1} is sent to 0 by φ. In particular u is sent to 0. This is a contradiction. Thus V_{1} ≅ k | U ↓_{G_{0}}^{G_{1}}. But Hom_{kG_{1}}(k ↑_{G_{0}}^{G_{1}}, U) ≅ Hom_{kG_{0}}(k, U ↓_{G_{0}}^{G}) ≠ 0, so U | k ↑_{G_{0}}^{G_{1}}. This proves the first statement.

Take any g ∈ G_{1} and let ḡ be its image in G_{1}/G_{0} under the natural projection. Since U | k ↑_{G_{0}}^{G_{1}}, we know gu = ḡu for every u ∈ U. With the identification G_{1}/G_{0} ≅ H_{1}/H_{0}, ḡ is identified with a certain h̄ ∈ H_{1}/H_{0}, where h ∈ H_{1} acts on φ(U) in the same way as h̄ acts on it. But hα = αg ∈ Hom_{D_{α}}(x, y) and R is a representation of D_{α}, so we should have φ(gu) = hφ(u), that is: φ(ḡu) = h̄φ(u). Consequently, φ is a k(G_{1}/G_{0})-module homomorphism if we identify ḡ and h̄. Since U is simple, and φ is nonzero, it must be an isomorphism. This is the second statement.

On the other hand, every h ∈ H_{0} acts trivially on φ(u) = w since h fixes φ. Thus the kH_{0}-module generated by w is isomorphic to the trivial kH_{0}-module k, so k | φ(U) ↓_{H_{0}}^{H_{1}}. Again, Hom_{kH_{1}}(k ↑_{H_{0}}^{H_{1}}, φ(U)) ≅ Hom_{kH_{0}}(k, φ(U) ↓_{H_{0}}^{H_{1}}) ≠ 0, so φ(U) and k ↑_{H_{0}}^{H_{1}} have a common simple summand. By (2), φ(U) as the nontrivial homomorphic image of a simple module must be simple, too. Thus φ(U) | k ↑_{H_{0}}^{H_{1}}.

The conclusions in the previous lemma actually characterize linear maps R(α), where R is a representation of D_{α}.

Lemma 3.6. let φ : M → N be a linear map from a kG-module M to a kHmodule N such that for every simple summand U of M ↓_{G_{1}}^{G}, the restricted map φ_{U} : U → φ(U) is zero when U ∤ k ↑_{G_{0}}^{G_{1}} , and a k(G_{1}/G_{0})-module homomorphism otherwise. Then φ determines a representation R of D_{α} with R(x) = M, R(y) = N and R(α) = φ.

Proof. We define a functor R : D_{α} → k-Vec by letting R(x) = M, R(y) = N. If β is an automorphism of x or y, R(β) is already defined since M is a kAut_{D_{α}}(x)-module and N is a kAut_{D_{α}}(y)-module. Otherwise β in Hom_{D_{α}}(x, y) can be written as hαg, where h ∈ H, g ∈ G, α is the representative unfactorizable morphism determining D_{α}. In this case R(β) can be defined as hφg. The conclusion follows after we show that this definition is well-defined. That is, if h_{1}αg_{1} = h_{2}αg_{2}, then h_{1}φg_{1} = h_{2}φg_{2}.

Since M as a vector space is the direct sum of all the summands of M ↓_{G_{1}}^{G}, it is enough to show that for every summand U, and each u ∈ U, we have h_{1}φg_{1}(u) = h_{2}φg_{2}(u), or equivalently: h_{2}^{−1}h_{1}φ(u) = φ(g_{2}g_{1}^{−1}u). The fact that h_{1}αg_{1} = h_{2}αg_{2} implies h_{2}^{−1}h_{1}α = αg_{2}g_{1}^{−1}, hence h_{2}^{−1}h_{1} ∈ H_{1} and g_{2}g_{1}^{−1} ∈ G_{1}, so g_{2}g_{1}^{−1}u ∈ U and φ(g_{2}g_{1}^{−1}u) = φ_{U}(g_{2}g_{1}^{−1}u). Thus it is sufficient to prove that h_{2}^{−1}h_{1}φ_{U}(u) = φ_{U}(g_{2}g_{1}^{−1}u).

If φ_{U} = 0, we have φ_{U}(g_{2}g_{1}^{−1}u) = 0 = h_{2}^{−1}h_{1}0 = h_{2}^{−1}h_{1}φ_{U}(u) trivially. Otherwise, let h̄ be the projection of h_{2}^{−1}h_{1} into H_{1}/H_{0}, and ḡ be the projection of g_{2}^{−1}g_{1} into G_{1}/G_{0}. Then g_{2}^{−1}g_{1}u = ḡu, and h_{2}^{−1}h_{1}φ_{U}(u) = h̄φ_{U}(u) because U | k ↑_{G_{0}}^{G_{1}} and φ(U) | k ↑_{H_{0}}^{H_{1}}. Since ḡ is sent to h̄ by the isomorphism G_{1}/G_{0} ≅ H_{1}/H_{0}, and φ_{U} is a k(G_{1}/G_{0})-module homomorphism, we have: h_{2}^{−1}h_{1}φ_{U}(u) = h̄φ_{U}(u) = φ_{U}(ḡu) = φ_{U}(g_{2}^{−1}g_{1}u). Thus R is well defined, and the conclusion follows.

Remark 3.7. Let φ : M → N be a linear map from a kG-module M to a kHmodule N. Let U (T, resp.) be a simple summand of M ↓_{G_{1}}^{G} (N ↓_{H_{1}}^{H}, resp.). Take a fixed decomposition N = T ⊕ N′ of vector space and let p : N → T be the projection. Then φ is a direct sum of φ_{UT} s of the following form:

\xymatrix{φ_{UT} : U \ar@{^{(}->}[r] & M \ar[r]^{φ} & N \ar@{->>}[r]^{p} & T}

Lemma 3.5 and Lemma 3.6 tell us that φ is given by or gives a representation R of D_{α} if and only if the following is true: whenever φ_{UT} ≠ 0, then U | k ↑_{G_{0}}^{G_{1}}, T | k ↑_{H_{0}}^{H_{1}}, and φ_{UT} is a k(G_{1}/G_{0})-module isomorphism under the given identification.

4 Ordinary Quivers of Finite EI Categories

In this section we will construct the ordinary quiver Q for every finite EI category C under the hypothesis that the endomorphism groups of all objects in C have orders invertible in k. The algorithm used here to construct Q is expressed in terms of the simple modules of endomorphism groups of objects in C and their restrictions to subgroups. This is easier and more intuitive than the usual approach which uses the primitive idempotents and the radical of kC. Let us first introduce some preliminary results.

Let G be a finite group whose order is invertible in k and G_{1} ⩽ G be a subgroup. For every kG-module M, we denote the homogeneous space of a simple kG-module V in M by M(V), that is, the sum of all submodules of M isomorphic to V. If U is a simple summand of V ↓_{G_{1}}^{G}, we denote the homogeneous space of U in M(V) ↓_{G_{1}}^{G} by M(V, U). Since V ↓_{G_{1}}^{G} may contain more than one simple summand isomorphic to U, we choose a particular decomposition and index different isomorphic copies of U in V ↓_{G_{1}}^{G} by natural numbers s ∈ ℕ. That is:

V ↓_{G_{1}}^{G}= ⨁_{s=1}^{t} U_{s} ⊕ V′,

where each U_{s} is isomorphic to U and V′ has no summands isomorphic to U.

Now let M(V, U, s) be the sum of the s-th isomorphic copy of U in each V ↓_{G_{1}}^{G} of M(V) ↓_{G_{1}}^{G}, s ⩾ 1. Thus if

M(V) = V_{1} ⊕ V_{2} ⊕ … ⊕ V_{m}

is any decomposition of M(V) and θ_{i} : V → V_{i} is a kG-module isomorphism, and

V ↓_{G_{1}}^{G}= U_{1} ⊕ … ⊕ U_{t} ⊕ V′

is some fixed decomposition where U_{s} ≅ U for all 1 ⩽ s ⩽ t and V′ has no summands isomorphic to U as kG_{1}-modules, then M(V, U, s) is defined as θ_{1}(U_{s}) ⊕ θ_{2}(U_{s}) ⊕ … ⊕ θ_{m}(U_{s}). This definition of M(V, U, s) is dependent on the fixed decomposition of V ↓_{G_{1}}^{G}, but independent of the decomposition of M(V) up to an automorphism of M(V), or equivalently, independent of the particular isomorphisms θ_{i}. Indeed, if

M(V) = V_{1} ⊕ V_{2} ⊕ … ⊕ V_{m}

and

M(V) = W_{1} ⊕ W_{2} … ⊕ W_{n}

are two decompositions of M(V). Then we get two lists of isomorphisms {θ_{i}}_{i=1}^{m} and {ψ_{j}}_{j=1}^{n} with θ_{i}(V) = V_{i} and ψ_{j}(V) = W_{j}. By the Krull-Schmidt Theorem, we have m = n, and there is a permutation π of {1, …, n} together with a family of kG-module isomorphisms ϕ_{i} : V_{i} ≅ W_{π(i)}. Since both ψ_{i} and ϕ_{i}θ_{i} are isomorphisms from the simple kG-module V to the simple kG-module W_{i}, we know that ψ_{i} is a scalar multiple of ϕ_{i}θ_{i} by Schur’s lemma. Thus

ψ_{1}(U_{s}) ⊕ ψ_{2}(U_{s}) … ⊕ ψ_{m}(U_{s})
= ϕ_{1}(θ_{1}(U_{s})) ⊕ ϕ_{2}(θ_{2}(U_{s})) ⊕ … ⊕ ϕ_{m}(θ_{m}(U_{s}))
= ϕ(θ_{1}(U_{s}) ⊕ θ_{2}(U_{s}) ⊕ … ⊕ θ_{m}(U_{s}))

where ϕ is the direct sum of ϕ_{i}, an automorphism of M(V). Thus the automorphism ϕ sends the space M(V, U, s) determined by the first decomposition of M(V) to the space M(V, U, s) determined by the second decomposition of M(V).

Repeating the above process for all simple summands V of M, and all simple summands U of V ↓_{G_{1}}^{G}, we can decompose M as the direct sum of all subspaces of the form M(V, U, s). That is:

M = ⨁_{V |M} ⨁_{U|V ↓_{G_{1}}^{G}} ⨁_{s∈ℤ} M(V, U, s).

Notice that for each fixed V and U, there are only finitely many s such that M(V, U, s) ≠ 0. Thus this decomposition is well-defined. Let p_{V,U,s} : M → M(V, U, s) be the projection.

Lemma 4.1. With the above setup, every kG-module homomorphism ϕ : M_{1} → M_{2} as a linear map is the direct sum of all ϕ_{V,U,s}.

\xymatrix{ϕ_{V,U,s} : M_{1}(V, U, s) \ar@{^{(}->}[r] & M_{1} \ar[r]^{ϕ} & M_{2} \ar@{->>}[r]^{p_{V,U,s}} & M_{2}(V, U, s)}

Proof. Since M_{1} as a vector space is the direct sum of all subspaces of the form M_{1}(V, U, s), it suffices to show that every subspace M_{1}(V, U, s) is sent by ϕ into M_{2}(V, U, s). First, since ϕ is a module homomorphism, and both M_{1} and M_{2} are semisimple, then the homogeneous space M_{1}(V) ⊆ M_{1} is sent into the homogeneous space M_{2}(V) ⊆ M_{2}. If either M_{1}(V) or M_{2}(V) is 0, i.e., either M_{1} or M_{2} has no summand isomorphic to V, the conclusion holds trivially. Otherwise, take a particular simple summand V_{1} of M_{1}(V), and a particular simple summand V_{2} of M_{2}(V). The induced kG-module homomorphism from V_{1} to V_{2} by ϕ is a scalar multiplication, hence sends the s-th isomorphic copy of U in V_{1} ↓_{G_{1}}^{G} into the s-th isomorphic copy of U in V_{2} ↓_{G_{1}}^{G}. Consequently, M_{1}(V, U, s) is sent into M_{2}(V, U, s).

Now we construct the ordinary quiver Q for a finite EI category C. The detailed algorithm is as follows:

Step 1: The vertex set of Q is ⨆_{x∈Ob(C)} S_{x}, where S_{x} is a set of representatives of the isomorphism classes of simple kAut_{C}(x)-modules.

Step 2: Let α : x → y be a representative unfactorizable morphism. Then it determines uniquely:

• G = Aut_{C}(x), G_{0} = Stab_{G}(α), G_{1} = Stab_{G}(Hα);

• H = Aut_{C}(y), H_{0} = Stab_{H}(α), H_{1} = Stab_{H}(αG);

• {V_{1}, …, V_{m}}: the set of pairwise non-isomorphic simple kG-modules;

• {W_{1}, …, W_{n}}: the set of pairwise non-isomorphic simple kH-modules;

• {U_{1}, …, U_{r}}: the set of pairwise non-isomorphic simple summands of k ↑_{G_{0}}^{G_{1}}.

Step 3: For each particular simple kG-module V in {V_{1}, …, V_{m}} choose a decomposition V ↓_{G_{1}}^{G}≅ U_{1}^{e_{1}} ⊕ … ⊕ U_{r}^{e_{r}} ⊕ X where X has no summand isomorphic to any U_{i}. For each simple kH-module W in {W_{1}, …, W_{n}} choose a decomposition W ↓_{H_{1}}^{H}≅ U_{1}^{f_{1}} ⊕ … ⊕ U_{r}^{f_{r}} ⊕ Y such that Y has no summand isomorphic to any U_{i}. Then we put ∑_{i=1}^{r} e_{i}f_{i} arrows from the vertex V to the vertex W in Q.

Step 4: Repeat Steps 2-4 for all representative unfactorizable morphisms.

Remark 4.2. We can index an arrow in Q by a list (α, V, W, U, s, l): this arrow is induced by a representative unfactorizable morphism α : x → y; it starts from a simple kAut_{C}(x)-module V and ends at a simple kAut_{C}(y)- module W. This arrow is associated with the s-th isomorphic copy of U in V ↓_{G_{1}}^{G} and the l-th isomorphic copy of U in W ↓_{H_{1}}^{H}, where U is a common summand of V ↓_{G_{1}}^{G}, W ↓_{H_{1}}^{H} and k ↑_{G_{0}}^{G_{1}} under the given identification (G_{1}/G_{0}) ≅ (H_{1}/H_{0}). See Lemma 3.3, Lemma 3.5 and Lemma 3.6 for more details.

Although the subspaces M(V, U, s) of M and N(W, U, l) of N depend on the particular decomposition, the quiver we constructed by the above algorithm is independent of it. Indeed, we have:

e_{i} = dim_{k}Hom_{G_{1}}(U_{i}, V ↓_{G_{1}}^{G})

and

f_{i} = dim_{k}Hom_{H_{1}}(U_{i}, V ↓_{H_{1}}^{H})

since U_{i} can be viewed as both a kG_{1}-module and a kH_{1}-module by identifying G_{1}/G_{0} with H_{1}/H_{0}. Clearly e_{i} and f_{i} are invariant with respect to different decompositions.

The following two simple examples illustrate our construction.

Example 4.3. Let C be the following finite EI category where: G and L are cyclic groups of order 2 and 3 respectively; H and K are the symmetric groups on 3 letters; O_{1} is an (H, G)-biset with two morphisms generated by α, fixed by the trivial subgroup of G and the proper subgroup of order 3 in H; O_{2} is a (K, H)-biset of 6 morphisms generated by β, permuted regularly by H and K; O_{3} is a (L, H)-biset with one morphism δ, fixed by both H and L.

\xymatrix{G \ar[r]^{O_{1}} & H \ar[d]^{O_{3}} \ar[r]^{O_{2}} & K \\  & L}

The reader may check that C is in fact a finite free EI category. We have: kG ≅ k ⊕ S, the direct sum of two non-isomorphic one dimensional modules; kH ≅ kK ≅ k ⊕ ϵ ⊕ V_{2} ⊕ V_{2}, the direct sum of the trivial module k, the sign representation ϵ, and two isomorphic copies of 2-dimensional simple modules; kL ≅ k ⊕ ω ⊕ ω^{2}, the direct sum of three pairwise non-isomorphic one dimensional modules. Thus the associated quiver Q has 11 vertices.

To distinguish different vertices, we mark vertices corresponding to kGmodules with ⋆, vertices corresponding to kH-modules with ⋄, vertices corresponding to kK-modules with ∘, and vertices corresponding to kL-modules with •. Now we want to put arrows among them. It is not hard to see that C has 3 representative unfactorizable morphisms α, β and δ.

First, let us analyze α, which determines arrows from vertices marked by ⋆ to vertices marked by ⋄. Obviously, G_{0} = 1, G_{1} = G; H_{0} = C_{3} ⊴ H, H_{1} = H; and G_{1}/G_{0} ≅ H_{1}/H_{0} ≅ C_{2}. Thus α gives rise to no arrow ending at the vertex ⋄V_{2} since it is not a simple summand of k ↑_{H_{0}}^{H}. All other vertices are summands of k ↑_{G_{0}}^{G} and k ↑_{H_{0}}^{H}. There is one arrow from ⋆k to ⋄k since by identifying G_{1}/G_{0} and H_{1}/H_{0}, they are the trivial representations of this quotient group. Similarly, there is an arrow from ⋆S to ⋄ϵ since they both are the sign representations of this quotient group.

We omit the detailed analysis of β and δ. Finally, we get the associated quiver Q as below:

\xymatrix{ &  &  & ∘k & ∘ϵ & ∘V_{2} \\ •ω^{2} & •ω & •k & ⋄k \ar[l] \ar[u] & ⋄ϵ \ar[u] & ⋄V_{2} \ar[u] \\  &  &  & ⋆k \ar[u] & ⋆S \ar[u]}

Example 4.4. Let C be a finite EI category with objects x and y; H =Aut_{C}(y) is a copy of the symmetric group S_{3} on 3 letters; G =Aut_{C}(x) is cyclic of order 2; Hom_{C}(x, y) = S_{3} regarded as an (H, G)-biset where H acts from the left by multiplication, G acts from the right by multiplication after identifying G with a subgroup G^{†} of S_{3}; Hom_{C}(y, x) = ∅.

From the previous example, we find that Q has 5 vertices: ∘k and ∘S corresponding to x; •k, •V_{2} and •ϵ corresponding to y. We choose α = 1 ∈ S_{3} as the representative unfactorizable morphism and then G_{0} = 1, G_{1} = G, H_{0} = 1, H_{1} = G^{†}. As before, kG ≅ k ⊕ S, and kH ≅ k ⊕ ϵ ⊕ V_{2} ⊕ V_{2}. Moreover, k ↓_{H_{1}}^{H}≅ k, ϵ ↓_{H_{1}}^{H}≅ S, V_{2} ↓_{H_{1}}^{H}≅ S ⊕ k. Thus the quiver Q is as follows:

\xymatrix{ & ∘k \ar[dl] \ar[dr] &  & ∘S \ar[dl] \ar[dr] \\ •k &  & •V_{2} &  & •ϵ}

In the above examples we find that the associated quivers are acyclic. Moreover, if C is a finite free EI category, the underlying quiver of a finite EI quiver Q̂ generating C is a subquiver of Q. This is always true.

Proposition 4.5. Let C be a finite EI category for which the endomorphism groups of all objects have orders invertible in k. The associated quiver Q of C is acyclic. Moreover, if C is a finite free EI category generated by a finite EI quiver Q̂, then Q contains a subquiver isomorphic to the underlying quiver of Q̂.

Proof. It is obvious that the associated quiver Q of C contains no loops. Now let S_{1} → … → S_{n} = S_{1} be an oriented cycle with at least 2 vertices (therefore n ⩾ 3), where S_{i} is a simple summand of kAut_{C}(x_{i}), x_{i} ∈ Ob(C), 1 ⩽ i ⩽ n. Our construction shows that each arrow is induced by a unfactorizable morphism in C. Therefore we get a string of unfactorizable morphisms

\xymatrix{x_{1} \ar[r]^{α_{2}} & … \ar[r]^{α_{n−1}} & x_{n−1} \ar[r]^{α_{n}} & x_{n} = x_{1}} .

Notice that x_{2} ≠ x_{1} since α_{2} is unfactorizable, hence not an automorphism. But this implies that both Hom_{C}(x_{1}, x_{2}) and Hom_{C}(x_{2}, x_{1}) are non-empty, which is a contradiction (see the paragraphs before Definition 2.1), so the first statement is correct.

Now suppose that C is a finite free EI category generated by a finite EI quiver Q̂. We know that vertices in Q̂ are exactly objects in C, and each object x in C gives a unique vertex k_{x} in Q, the trivial kAut_{C}(x)-module. Furthermore, every arrow x → y in Q̂ corresponds to a unique representative unfactorizable morphism α : x → y, which in turn determines, by our construction, a unique arrow from k_{x} to k_{y} in Q. By identifying x with k_{x}, we get the second statement.

This associated quiver Q of C is actually the ordinary quiver of kC. To prove this, let us consider the radical of kC.

Proposition 4.6. Let C be a finite EI category for which the endomorphism groups of all objects have orders invertible in k. Then rad kC has as a basis the non-isomorphisms in C and rad kC/ rad^{2} kC has as a basis the images of all unfactorizable morphisms in C respectively.

Proof. Let Λ be the subspace of kC spanned by all non-isomorphisms in C. Clearly, Λ is an two-sided ideal of kC and it has the non-isomorphisms as a basis. Moreover, kC/Λ ≅ ⨁_{x∈Ob(C)} kEnd_{C}(x). Since the endomorphism groups of all objects in C have orders invertible in k, all groups algebras are semisimple. Thus kC/Λ is semisimple, and rad kC ⊆ Λ. On the other hand, because C has only finitely many distinct objects, Λ is nilpotent, so Λ ⊆ rad kC. This proves Λ = rad kC.

As an ideal of kC, rad^{2} kC contains all factorizable morphisms in C, and no unfactorizable morphism. Consequently, rad kC/ rad^{2} kC is spanned by the images of all unfactorizable morphisms in C. Actually, these images form a basis of rad kC/ rad^{2} kC.

Now we restate and prove Theorem 1.1.

Theorem 4.7. Let C be a finite EI category for which the endomorphism groups of all objects have orders invertible in the field k. Then the quiver Q constructed by our algorithm is precisely the ordinary quiver of the category algebra kC. Moreover, kC has the same ordinary quiver as that of kĈ, the category algebra of the free EI cover Ĉ of C.

Proof. Let Q′ be the ordinary quiver of kC. We only need to show that there is a bijection π from the vertices set of Q to the vertices set of Q′, and for every pair of vertices v and w in Q, the number of arrows from v to w is the same as that of the arrows from π(v) to π(w). By Corollary 4.5 of [7], a primitive idempotent of kC is exactly a primitive idempotent of kEnd_{C}(x) for some object x. Moreover, if e_{x} and e_{y} are two primitive idempotents of kC associated with objects x and y respectively, then kCe_{x} ≅ kCe_{y} if and only if x = y (since C is skeletal) and kEnd_{C}(x)e_{x} ≅ kEnd_{C}(x)e_{y} (see the paragraph before Proposition 4.3 of [7]). Thus the vertices of Q′, formed by isomorphism classes of indecomposable projective kC-modules, can be parametrized by isomorphism classes of indecomposable kEnd_{C}(x)-modules, where x ranges over the objects in C. But because every kEnd_{C}(x) is semisimple, indecomposable projective kEnd_{C}(x)-modules and simple kEnd_{C}(x)-modules coincide. Therefore Q and Q′ have the same vertices.

Now let e_{1} and e_{2} be two primitive idempotents of kC corresponding to objects x and y respectively. Let v = [kCe_{1}] and w = [kCe_{2}] be the corresponding vertices in Q′ (or Q since they have the same vertices). We know that the numbers of arrows in Q′ from v to w is the dimension of the k-space e_{2}(radkC/rad^{2}kC)e_{1}. Let α_{1}, α_{2}, …, α_{l} be all representative unfactorizable morphisms from x to y. We verify that

e_{2}(radkC/rad^{2}kC)e_{1} ≅ ⨁_{i=1}^{l} e_{2}(kHα_{i}G)e_{1}

by the previous proposition. Thus we only need to check that each representative unfactorizable morphism from x to y gives the same number of arrows from v to w as that given by our algorithm.

Take a particular representative unfactorizable α in the list α_{1}, α_{2}, …, α_{l}. Let us compute the dimension of e_{2}(kHαG)e_{1}. As we mentioned before, α determines groups H_{0} ⊴ H_{1} ⩽ H, G_{0} ⊴ G_{1} ⩽ G with H_{0}α = αG_{0} = α and H_{1}α = αG_{1}, and k(G_{1}/G_{0}) ≅ k(H_{1}/H_{0}). We identify these two modules and let {u_{1}, …, u_{n}} be a list of primitive idempotents of k(H_{1}/H_{0}) such that every simple sumand of k(H_{1}/H_{0}) is isomorphic to some k(H_{1}/H_{0})u_{i}, and k(H_{1}/H_{0})u_{i} ≇ k(H_{1}/H_{0})u_{j} if i ≠ j for 1 ⩽ i, j ⩽ n. Notice that all u_{i} are primitive idempotents of both kG_{1} and kH_{1} under the identification. Let U_{i} = k(H_{1}/H_{0})u_{i},

kGe_{1} ↓_{G_{1}}^{G}≅ U_{1}^{a_{1}} ⊕ … ⊕ U_{n}^{a_{n}} ⊕ U

and

kHe_{2} ↓_{H_{1}}^{H}≅ U_{1}^{b_{1}} ⊕ … ⊕ U_{n}^{b_{n}} ⊕ V

such that both U ≅ kG_{1}u and V ≅ kH_{1}v have no summand isomorphic to any U_{i}, where u and v are idempotents of kG and kH respectively. Then by using both the right and left module structure of kHαG we have:

e_{2}kHαGe_{1} ≅ Hom_{kH}(kHe_{2}, kHαGe_{1})

≅ Hom_{kH}(kHe_{2}, Hom_{kG}(e_{1}kG, kHαG))

≅ Hom_{kH}(kHe_{2}, Hom_{kG}(e_{1}kG, ⨁_{h∈H/H_{1}} khαG))

≅ Hom_{kH}(kHe_{2}, Hom_{kG}(e_{1}kG, kαG) ↑_{H_{1}}^{H})

≅ Hom_{kH}(kHe_{2}, Hom_{kG}(e_{1}kG, ⨁_{g∈G\G_{1}} kαgG_{1}) ↑_{H_{1}}^{H})

≅ Hom_{kH}(kHe_{2}, Hom_{kG}(e_{1}kG, kαG_{1} ↑_{G_{1}}^{G}) ↑_{H_{1}}^{H}) ≅ Hom_{kH}(kHe_{2}, Hom_{kG_{1}}(e_{1}kG ↓_{G_{1}}^{G}, kαG_{1}) ↑_{H_{1}}^{H})

≅ Hom_{kH_{1}}(kHe_{2} ↓_{H_{1}}^{H}, Hom_{kG_{1}}(e_{1}kG ↓_{G_{1}}^{G}, kαG_{1}))  (4.1)

The last two isomorphisms come from the Frobenius Reciprocity. Also notice that kαG_{1} = kH_{1}α is a (kH_{1}, kG_{1})-bimodule, which is isomorphic to k(G_{1}/G_{0}) as a right kG_{1}-module and is isomorphic to k(H_{1}/H_{0}) as a left kH_{1}-module. Moreover, the decomposition

kGe_{1} ↓_{G_{1}}^{G}≅ ⨁_{i=1}^{n}(kG_{1}u_{i})^{a_{i}} ⊕ kG_{1}u

as a left kG_{1}-module implies the decomposition

e_{1}kG ↓_{G_{1}}^{G}≅ ⨁_{i=1}^{n}(u_{i}kG_{1})^{a_{i}} ⊕ ukG_{1}

as a right kG_{1}-module. Thus:

Hom_{kG_{1}}(e_{1}kG ↓_{G_{1}}^{G}, kαG_{1}) ≅ Hom_{kG_{1}}(⨁_{i=1}^{n}(u_{i}kG_{1})^{a_{i}} ⊕ ukG_{1}, kαG_{1})

≅ Hom_{kG_{1}}(⨁_{i=1}^{n}(u_{i}kG_{1})^{a_{i}}, kαG_{1}) ≅ ⨁_{i=1}^{n}(kαG_{1}u_{i})^{a_{i}}

≅ ⨁_{i=1}^{n}(kH_{1}αu_{i})^{a_{i}} ≅ ⨁_{i=1}^{n} U_{i}^{a_{i}}

Consequently, we have:

R.H.S of (4.1) ≅ Hom_{kH_{1}}(kHe_{2} ↓_{H_{1}}^{H}, ⨁_{i=1}^{n} U_{i}^{a_{i}})

≅ Hom_{kH_{1}}(⨁_{i=1}^{n} U_{i}^{b_{i}} ⊕ V, ⨁_{i=1}^{n} U_{i}^{a_{i}})

≅ Hom_{kH_{1}}(⨁_{i=1}^{n} U_{i}^{b_{i}}, ⨁_{i=1}^{n} U_{i}^{a_{i}})

≅ ⨁_{i=1}^{n}(Hom_{kH_{1}}(U_{i}, U_{i}))^{a_{i}b_{i}}

The right hand side of the above identity has dimension ∑_{i=1}^{n} a_{i}b_{i}, which is the same as the number of arrows from the vertex v = [kCe_{1}] to the vertex w = [kCe_{2}] given by α in our construction (see Step 2 and Step 3 of our algorithm and Remark 4.2). This proves the first statement.

By Proposition 2.9, we know that C and its free EI cover Ĉ have the same objects. Moreover, for each pair of objects x and y, we have End_{C}(x) = EndĈ(x), End_{C}(y) = EndĈ(y), and the unfactorizable morphisms from x to y in C are the same as those in Ĉ. But from the above proof we find that these data completely determine their ordinary quivers. Thus kĈ and kC have the same ordinary quiver, as we claimed.

5 Hereditary Category Algebras

It is well known that the group algebra kG of a finite group G is hereditary if and only if the order of G is invertible in k, and the path algebra kQ of a finite acyclic quiver Q is always hereditary. Since a finite EI category C can be viewed as a combination of several finite groups and a finite acyclic quiver, it is convincing that a characterization of hereditary category algebras kC must be related closely to the following conditions: the endomorphism groups of all objects in C have orders invertible in k, and C is a finite free EI category. These two conditions actually characterize finite EI categories among hereditary category algebras and this is the content of Theorem 1.2.

Let us state some preliminary results which will be used in the proof of Theorem 1.2.

Lemma 5.1. Let α_{1} and α_{2} be two different unfactorizable morphisms in a finite free EI category C. Then as kC-modules, either kCα_{1} = kCα_{2}, or kCα_{1} ∩ kCα_{2} = 0.

Proof. For i = 1, 2, let B_{i} = {δα_{i} | δ ∈ Mor(C), s(δ) = t(α_{i})} be the set of all composites starting with α_{i}, where s(δ) and t(α_{i}) are the source of δ and the target of α_{i} respectively. Then B_{i} spans kCα_{i}. If we can that prove B_{1} and B_{2} are either the same, or have empty intersection, then the conclusion follows.

Suppose that B_{1} ∩ B_{2} is nonempty. Then we can find β ∈ B_{1} ∩ B_{2} and β can be expressed as δ_{1}α_{1} = δ_{2}α_{2}. But C is a finite free EI category and satisfies the UFP, so α_{1} and α_{2} have the same source x and the same target y. Moreover, there is an automorphism h ∈ End_{C}(y) such that α_{1} = hα_{2} and δ_{1} = δ_{2}h^{−1}. In particular, α_{1} ∈ B_{2} and α_{2} ∈ B_{1}. Thus B_{1} = B_{2}, which completes the proof.

Lemma 5.2. Let α : x → y be an unfactorizable morphism in a finite free EI category for which all endomorphisms groups of objects have orders invertible in k. Then the cyclic kC-module kCα is projective.

Proof. Let H = End_{C}(y) and H_{0} = Stab_{H}(α). Since the order of H is invertible in k, the subgroup H_{0} has invertible order in k, too. Thus we can define e = \frac{1}{|H_{0}|} ∑_{h∈H_{0}} h ∈ kH ⊆ kC. Since e is an idempotent of kC, it follows that kCe is a projective kC-module. We will show that kCα ≅ kCe as kC-modules.

Define a map φ : kCα → kCe by letting φ(rα) = re, where r ∈ kC. We claim that φ is a kC-module isomorphism.

First, we want to show that φ is well-defined. That is, if ∑_{i=1}^{m} a_{i}δ_{i}α = ∑_{j=1}^{n} b_{j}β_{j}α are two different expressions of a vector in kCα, where a_{i}, b_{j} ∈ k, δ_{i}, β_{j} ∈ Mor(C) are morphisms starting at y, then ∑_{i=1}^{m} a_{i}δ_{i}e = ∑_{j=1}^{n} b_{j}β_{j}e. This is equivalent to saying that if ∑_{i=1}^{m} a_{i}δ_{i}α = 0, then ∑_{i=1}^{m} a_{i}δ_{i}e = 0, where all δ_{i} are pairwise distinct morphisms starting at y.

Those δ_{i}α might not be all distinct. By changing the indices if necessary, we can suppose that δ_{1}α = … = δ_{i_{1}}α, δ_{i_{1}+1}α = … = δ_{i_{2}}α, and so on until δ_{i_{l−1}+1}α = … = δ_{i_{l}}α, where δ_{i_{1}}α, δ_{i_{2}}α, …, δ_{i_{l}}α are pairwise distinct and δ_{i_{l}} = δ_{m}. From the definition of the category algebra, δ_{i_{1}}α, …, δ_{i_{l}}α are in fact linearly independent. Thus we have:

∑_{i=1}^{m} a_{i}δ_{i}α = (a_{1}δ_{1}α + … + a_{i_{1}}δ_{i_{1}}α) + … + (a_{i_{l−1}+1}δ_{i_{l−1}+1}α + … + a_{i_{l}}δ_{i_{l}}α)
= (a_{1} + … + a_{i_{1}})δ_{i_{1}}α + … + (a_{i_{l−1}+1} + … + a_{i_{l}})δ_{i_{l}}α

Notice that (a_{1} + … + a_{i_{1}}) = … = (a_{i_{l−1}+1} + … + a_{i_{l}}) = 0 by the independence of the δ_{i_{j}}α, 1 ⩽ j ⩽ l. We want to show that each term (a_{i_{j−1}+1} + … + a_{i_{j}})δ_{i_{j}}α is sent to 0 by φ. Observe that since δ_{s}α = δ_{i_{j}}α for all 1 ⩽ s ⩽ i_{j}, by the UFP, there are h_{s} ∈ End_{C}(y) such that h_{s}α = α and δ_{s} = δ_{i_{j}}h_{s}^{−1}. In particular, h_{s} ∈ H_{0}. Thus:

φ(∑_{s=1}^{i_{j}} a_{s}δ_{s}α) = ∑_{s=1}^{i_{j}} a_{s}δ_{s}e = ∑_{s=1}^{i_{j}} a_{s}δ_{i_{j}}h_{s}^{−1}e

But all h_{s}^{−1} ∈ H_{0} fix e, i.e., h_{s}^{−1}e = e, so the right side of the above identity is actually (∑_{s=1}^{i_{j}} a_{s})δ_{i_{j}}e = 0. This shows that φ sends each term to 0, and hence is well-defined.

Now suppose φ(∑_{i=1}^{m} a_{i}δ_{i}α) = ∑_{i=1}^{m} a_{i}δ_{i}e = 0. Since e fixes α, i.e., eα = α, we have ∑_{i=1}^{m} a_{i}δ_{i}α = (∑_{i=1}^{m} a_{i}δ_{i}e)α = 0. Thus this map is injective.

It is clear that φ is surjective and is a kC-module homomorphism. In conclusion, φ is a kC-module isomorphism. The conclusion is proved.

Let us restate Theorem 1.2 and give a proof here.

Theorem 5.3. Let C be a finite EI category C. Then kC is hereditary if and only if C is a finite free EI category satisfying that the endomorphism groups of all objects have orders invertible in k.

Proof. If C has only one object, the conclusion holds obviously. Without loss of generality, we suppose that C has more than one object.

The if part. By Corollary 5.2 on page 17 of [1], it suffices to prove that Λ = rad kC is a projective kC-module. By Proposition 4.6, Λ has as a basis all non-isomorphisms in C. By Proposition 2.6, Λ as a kC-module is the sum of all submodules of the form kCα, where α ranges all unfactorizable morphisms in C. By Lemma 5.1, any two of them either coincide, or have a trivial intersection. Therefore, Λ is the direct sum of some of these submodules. But all these submodules are projective by Lemma 5.2, so Λ is also projective.

The only if part. Let C be a finite EI category such that kC is hereditary, we first show that all endomorphism groups of objects have orders invertible in k. If this is not true, then C has an object x whose endomorphism group G = End_{C}(x) has order not invertible in k. Let k_{x} be the simple kC-module which is k on x, and 0 on other objects. Let P → k_{x} be a minimal kCprojective resolution of k_{x}. By Lemma 5.1.1 of [10], it induces a projective resolution P(x) → k of kAut_{C}(x)-modules. Since Aut_{C}(x) has order not invertible in k, this induced projective resolution has infinite length, so P → k_{x} must be of infinite length. This is impossible since kC is hereditary. Consequently, all endomorphism groups of objects in C have orders invertible in k.

Next we prove that C is a finite free EI category. By Proposition 2.9, there is a full functor F̂ : Ĉ → C, where Ĉ is the free EI cover of C. Moreover, F̂ is the identity map restricted to objects, isomorphisms and unfactorizable morphisms. By Lemma 2.12, Ĉ gives a surjective algebra homomorphism ψ : kĈ → kC. Since Ĉ and C have the same objects, and the same endomorphism group for each object, we know from Proposition 4.3 of [7] that every simple kC-module can be viewed as a simple kĈ-module, giving all simple kĈ-modules. Moreover, kC and kĈ have the same ordinary quiver (Theorem 4.7). We know that kĈ is a hereditary algebra by the conclusion we just proved. Therefore, both kC and kĈ are hereditary algebras with the same simple modules and the same ordinary quiver. But for a hereditary algebra, these data completely determine the dimension of this algebra. Consequently, kC and kĈ have the same dimension, so ψ is an isomorphism and the functor F̂ : Ĉ → C must be bijective on morphisms. Therefore, F̂ is an isomorphism of categories. In conclusion, C is isomorphic to Ĉ, so is a finite free EI category.

Corollary 5.4. The category algebra kC of a finite EI category C is hereditary if and only if for all full subcategories D of C, the category algebras kD are hereditary.

Proof. One side is trivial since C is a full subcategory of itself. Now assume that kC is hereditary. By the previous theorem, C is a finite free EI category whose endomorphism groups of objects have orders invertible in k. If D is a full subcategory of C, then by Proposition 2.10, D is also a finite free EI category for which all endomorphism groups of objects obviously have orders invertible in k. By the above theorem again, kD is hereditary.

Corollary 5.5. Let C be a finite free EI category for which all endomorphism groups of objects have orders invertible in k. Then C is of finite (tame, resp.) representation type if and only if its ordinary quiver has underlying graph a disjoint union of Dynkin (Euclidean, resp.) diagrams. Otherwise, it has wild representation type.

Proof. The conclusion comes from the previous theorem and the classification of representation types of quivers.

As an application of this theorem, we assert that the categories shown in Example 4.3 and Example 4.4 have finite representation type over fields k whose characteristic is not 2 or 3.

6 Application to Representation Types

To determine the representation type of a finite EI category C is an interesting but challenging problem. In this section we only consider the finite EI categories C for which the endomorphism groups of all objects have orders invertible in k. Under this hypothesis, we can construct the ordinary quiver Q of kC according to the algorithm described in section 4. If furthermore C is a finite free EI category, its representation type is completely determined by Corollary 5.5. Otherwise, kC is Morita equivalent to kQ/I, where I is an nontrivial admissible ideal of kQ. Thus kC is of finite representation type if so is kQ, which is precisely Proposition 2.13 since kC is Morita equivalent to kQ.

It is well known that if a quiver Q has a subquiver of infinite representation type, Q is of infinite representation type as well. This conclusion holds for finite groups. Finite EI categories have a similar property:

Proposition 6.1. A finite EI category C is of infinite representation type if it has a full subcategory D of infinite representation type.

Proof. Let M be an indecomposable kD-module. By Theorem 4.4.2 of [10], the induced module M ↑_{D}^{C} is an indecomposable kC-module. Moreover, M is D-projective, i.e., M | M ↑_{D}^{C}↓_{D}^{C}. This implies that every indecomposable kD-module M is a direct summand of M̃ ↓_{D}^{C}, where M̃ is an indecomposable kC-module. If C is of finite representation type, it has only finitely many non-isomorphic indecomposable modules. But the restriction of these modules to D can produce only finitely many non-isomorphic indecomposable kD-modules, and hence D is of finite representation type, which is a contradiction!

Now let us consider full subcategories of C. The subcategories with one object always have finite representation type since their category algebras are precisely semisimple group algebras. They cannot provide us much information about the representation type of C. Without loss of generality we suppose that C has more than one object. Consider the connected subcategories with two objects. Since they are always finite free EI categories, we can completely determine their representation types by constructing the ordinary quivers. Fortunately, these categories give us many useful details about the representation type of C.

Let D be a connected full subcategory formed by two distinct objects x and y. From the previous proposition we know that if D is of infinite representation type, so is C. Let G = Aut_{D}(x), H = Aut_{D}(y).

Corollary 6.2. With the above notation, if Hom_{D}(x, y) has more than one orbit as an (H, G)-biset, then D, hence C are of infinite representation type.

Proof. Notice that every morphism in Hom_{D}(x, y) is unfactorizable. If it has more than one orbit, then D has at least two representative unfactorizable morphisms which are not in the same biset orbit. Take two representative unfactorizable morphisms α ≠ β lying in different biset orbits and consider the ordinary quiver Q of D. It contains two vertices k_{x} and k_{y}, the trivial representations of G and H respectively. Since both α and β give an arrow from k_{x} to k_{y} by our algorithm, there are at least two arrows from k_{x} to k_{y}. Thus Q contains multiple arrows and is of infinite representation type. By Corollary 5.5, D, hence C are of infinite representation type.

Remark 6.3. We comment that Proposition 6.1 and Corollary 6.2 are always true over all algebraically closed fields k, no matter what the characteristic of k is. We strengthen Corollary 6.2 under the hypothesis that all endomorphism groups have orders invertible in k.

By this corollary, we suppose that Hom_{D}(x, y) has only one orbit as an (H, G)-biset.

Corollary 6.4. With the above notation, if neither G nor H acts transitively on Hom_{D}(x, y), then D, hence C are of infinite representation type.

Proof. Since Hom_{D}(x, y) has only one orbit as an (H, G)-biset, it is generated by one morphism, say α. As before, we define G_{0} = Stab_{G}(α), H_{0} = Stab_{H}(α), G_{1} = Stab_{G}(Hα), H_{1} = Stab_{H}(αG). Because neither G nor H acts transitively on Hom_{D}(x, y), G_{1} is a proper subgroup of G, and H_{1} is a proper subgroup of H.

Let us consider the kG-module k ↑_{G_{1}}^{G}. We claim it has exactly one isomorphic copy of the trivial kG-module k as a direct summand. Indeed, Hom_{kG}(k, k ↑_{G_{1}}^{G}) ≅ Hom_{kG_{1}}(k ↓_{G_{1}}^{G}, k) = Hom_{kG_{1}}(k, k) ≅ k, which implies the claim. But the dimension of k ↑_{G_{1}}^{G} is strictly bigger than 1 since G_{1} is a proper subgroup of G, so it has another simple summand S not isomorphic to k. Again Hom_{kG_{1}}(k, S ↓_{G_{1}}^{G}) ≅ Hom_{kG}(k ↑_{G_{1}}^{G}, S) ≠ 0, and therefore S ↓_{G_{1}}^{G} has a trivial summand k. With the same reasoning, we can find a simple summand T | k ↑_{H_{1}}^{H} such that T is not isomorphic to the trivial representation k of H, but T ↓_{H_{1}}^{H} has a trivial summand k.

Consider the ordinary quiver Q of D. It contains at least four vertices: k_{x} and k_{y} which are the trivial modules of kG and kH respectively, S, and T. By our construction, the representative unfactorizable morphism α induces an arrow k_{x} → k_{y} and at least one arrow k_{x} → T since k_{x} ↓_{G_{1}}^{G}≅ k | T ↓_{H_{1}}^{H} under the identification. It also gives at least one arrow from S to k_{y} and at least one arrow from S to T. Thus we get a subquiver of Q, pictured below:

\xymatrix{k_{x} \ar[r] \ar[dr] & k_{y} \\ S \ar[r] \ar[ur] & T}

It is clear that the above subquiver is of infinite representation type, so Q is of infinite representation type as well. By Corollary 5.5, D, hence C are of infinite representation type.

Now without loss of generality we suppose that H acts transitively on Hom_{D}(x, y) (otherwise consider the opposite category C^{op}, which has the same representation type as C), so that G_{0} is a normal subgroup of G = G_{1}.

Corollary 6.5. Let S be a simple summand of k ↑_{H_{0}}^{H_{1}}. If S ↑_{H_{1}}^{H} is not multiplicity free, or has more than 3 summands, then D, hence C are of infinite representation type.

Proof. Notice that k ↑_{G_{0}}^{G}, after it is identified with k ↑_{H_{0}}^{H}, has a simple summand isomorphic to S. Denote this kG-module S′ to avoid confusion. Let T be a simple summand of S ↑_{H_{1}}^{H}. Since

Hom_{kH_{1}}(S, T ↓_{H_{1}}^{H}) ≅ Hom_{kH}(S ↑_{H_{1}}^{H}, T) ≠ 0,

S is a simple summand of T ↓_{H_{1}}^{H}. Moreover, S ↑_{H_{1}}^{H} has more than one summand isomorphic to T if and only if T ↓_{H_{1}}^{H} has more than one summand isomorphic to S.

Consider the ordinary quiver Q of D. Both S′ and T are vertices in Q since they are a simple kG-module and a simple kH-module respectively. If S ↑_{H_{1}}^{H} has more than one summand isomorphic to T, then T ↓_{H_{1}}^{H} has more than one summand isomorphic to S. Thus there are multiple arrows from the vertex S′ to the vertex T, and Q is of infinite representation type.

Now suppose S ↑_{H_{1}}^{H} has more than 3 summands. If it is not multiplicity free, we are done. Otherwise, all summands are pairwise non-isomorphic. Thus S ↑_{H_{1}}^{H} has at least four pairwise non-isomorphic simple summands, say T_{1}, T_{2}, T_{3} and T_{4}. They are different vertices in Q. By our construction, there are arrows from the vertex S′ to each of T_{i}, 1 ⩽ i ⩽ 4. As a result, the underlying graph of Q has a component whose underlying graph is not a disjoint union of Dynkin diagrams. Thus Q is of infinite representation type. The conclusion follows from Corollary 5.5.

We end this section with a finite EI category of infinite representation type.

Example 6.6. Let C be a finite EI category with: Ob(C) = {x, y}, Aut_{C}(x) = 1, Aut_{C}(y) is a copy of the symmetric group S_{3} on three letters, which acts on Hom_{C}(x, y) = S_{3} by multiplication from left, Hom_{C}(y, x) = ∅.

It is not hard to construct the ordinary quiver of this category, as shown below and it has infinite representation type.

\xymatrix{ & ∘V_{2} \\ ∘k & •k \ar[u] \ar[u] \ar[l] \ar[r] & ∘ϵ}

7 Appendix

In section 5 we proved that kC is hereditary if and only if C is a finite free EI category for which all endomorphism groups of objects have orders invertible in k. In this section we will construct a functor F : kC-mod → kQ-mod for such C, where Q is the ordinary quiver of kC constructed by our algorithm in section 4. This functor is faithful, dense and full, and hence induces a Morita equivalence between kC and kQ. This construction of F actually motivated Theorem 1.2, and gives a proof of its if part.

Definition of F on objects. We let R be a representation of C and show how to define a representation R′ of its ordinary quiver Q. Take a fixed vertex V in Q. By our construction, V is a simple kAut_{C}(x)-module for an object x in C. Let the homogeneous space R(x)(V) of V in the kEnd_{C}(x)-module R(x) be V^{a}, the direct sum of a copies of V. We then define R′(V) = k^{a}. In this way we assign a vector space to each vertex in Q. Repeating this process, R′ assigns a vector space to each vertex in Q.

Now we want to define a linear map for every arrow V → W in Q. By Remark 4.2, this arrow is indexed by a list (α, V, W, U, s, l) where α : x → y is an representative unfactorizable morphism. This morphism uniquely determines and is uniquely determined by the following data (see the notation before Lemma 4.1 and Remark 4.2):

• the groups G_{0} ⊴ G_{1} ⩽ G and H_{0} ⊴ H_{1} ⩽ H;

• a kG-module R(x) = M with M(V) ≅ V^{a}, and a kH-module R(y) = N with N(W) ≅ W^{b};

• the subspaces M(V, U, s) ⊆ M and N(W, U, l) ⊆ N.

The linear map φ = R(α) : R(x) → R(y) induces a linear map φ′, the composite of the following maps, where all inclusions and projections are defined as before:

\xymatrix{φ′ : M(V, U, s) \ar@{^{(}->}[r] & M \ar[r]^{φ} & N \ar@{->>}[r] & N(W, U, l).}

Notice that M(V, U, s) ≅ U^{a} since M(V) ≅ V^{a}, N(W, U, l) ≅ U^{b} since N(W) ≅ W^{b}. By Lemma 3.5 and Remark 3.7, the derived map from a particular isomorphic copy of U in M(V, U, s) into a particular isomorphic copy of U in N(V, U, l) is 0, or a k(G_{1}/G_{0})-module isomorphism. Both cases give a scalar multiplication λ since U is a simple k(G_{1}/G_{0})-module. Thus φ′ gives the following b × a block matrix B̃, where I is the dimU × dimU identity matrix. This matrix, in turn, gives us a b × a matrix B.

B̃ = [\begin{matrix}λ_{11}I & λ_{21}I & … & λ_{a1}I \\ λ_{12}I & λ_{22}I & … & λ_{a2}I \\ ⋯ & ⋯ & … & ⋯ \\ λ_{1b}I & λ_{2b}I & … & λ_{ab}I \end{matrix}]  ↔  B = [\begin{matrix}λ_{11} & λ_{21} & … & λ_{a1} \\ λ_{12} & λ_{22} & … & λ_{a2} \\ ⋯ & ⋯ & … & ⋯ \\ λ_{1b} & λ_{2b} & … & λ_{ab} \end{matrix}]

The matrix B provides the desired linear map θ from R′(x) = k^{a} to R′(y) = k^{b} with respect to the chosen bases.

Repeating the above process, we can define a linear map for each arrow in Q, hence a representation R′ of Q. Define F (R) = R′.

Definition of F on morphisms. Let π = {ϕ_{x} : x ∈ Ob(C)} be a homomorphism from a kC-module R_{1} to another kC-module R_{2}. We define a kQ-module homomorphism π′ from F (R_{1}) = R′_{1} to F (R_{2}) = R′_{2}.

Take an arbitrary vertex V in Q. Suppose R′_{1}(V) = k^{a} and R′_{2}(V) = k^{c}. By our construction, V uniquely determines an object x in C, and M_{1} = R_{1}(x) (M_{2} = R_{2}(x), resp.) satisfies M_{1}(V) ≅ V^{a} (M_{2}(V) ≅ V^{c}, resp.). Let G =

Aut_{C}(x). The composite map ϕ_{V} : \xymatrix{M_{1}(V) \ar@{^{(}->}[r] & M_{1} \ar[r]^{ϕ_{x}} & M_{2} \ar@{->>}[r] & M_{2}(V)} is a summand of ϕ_{x} since M_{1} and M_{2} are semisimple. This summand, with the following matrix representation C̃ (here I_{1} is the dimV × dimV identity matrix), gives a unique c × a matrix C, which defines a linear map ϕ′_{V} from R′_{1}(V) = k^{a} to R′_{2}(V) = k^{c} with respect to the chosen bases.

C̃ = [\begin{matrix}ϵ_{11}I_{1} & ϵ_{21}I_{1} & … & ϵ_{a1}I_{1} \\ ϵ_{12}I_{1} & ϵ_{22}I_{1} & … & ϵ_{a2}I_{1} \\ ⋯ & ⋯ & … & ⋯ \\ ϵ_{1c}I_{1} & ϵ_{2c}I_{1} & … & ϵ_{ac}I_{1} \end{matrix}]  ↔  C = [\begin{matrix}ϵ_{11} & ϵ_{21} & … & ϵ_{a1} \\ ϵ_{12} & ϵ_{22} & … & ϵ_{a2} \\ ⋯ & ⋯ & … & ⋯ \\ ϵ_{1c} & ϵ_{2c} & … & ϵ_{ac} \end{matrix}]

In this way we get π′ = {ϕ′_{V} : R′_{1}(V) → R′_{2}(V) | V is a vertex in Q}, a family of linear transformations. Define F (π) = π′. Now we need to verify that it is indeed a kQ-module homomorphism.

Remark 7.1. We constructed ϕ′_{V} from a summand ϕ_{V} of ϕ_{x}. Clearly, we can recover ϕ_{V} from ϕ′_{V}. In particular, ϕ_{V} = 0 ⇔ ϕ′_{V} = 0.

Take an arbitrary arrow V → W in Q and suppose that R′_{1}(V) = k^{a}, R′_{1}(W) = k^{b}, R′_{2}(V) = k^{c}, R′_{2}(W) = k^{d}. Let the linear maps θ_{1} and θ_{2} assigned to this arrow by R′_{1} and R′_{2} have the matrix representations B_{1} and B_{2}, and ϕ′_{V} , ϕ′_{W} ∈ F (π) = π′ have the matrix representations C_{x} and C_{y}:

B_{1} = [\begin{matrix}λ_{11} & λ_{21} & … & λ_{a1} \\ λ_{12} & λ_{22} & … & λ_{a2} \\ ⋯ & ⋯ & … & ⋯ \\ λ_{1b} & λ_{2b} & … & λ_{ab} \end{matrix}] B_{2} = [\begin{matrix}μ_{11} & μ_{21} & … & μ_{a1} \\ μ_{12} & μ_{22} & … & μ_{a2} \\ ⋯ & ⋯ & … & ⋯ \\ μ_{1b} & μ_{2b} & … & μ_{ab} \end{matrix}]

C_{x} = [\begin{matrix}ϵ_{11} & ϵ_{21} & … & ϵ_{a1} \\ ϵ_{12} & ϵ_{22} & … & ϵ_{a2} \\ ⋯ & ⋯ & … & ⋯ \\ ϵ_{1c} & ϵ_{2c} & … & ϵ_{ac} \end{matrix}] C_{y} = [\begin{matrix}η_{11} & η_{21} & … & η_{b1} \\ η_{12} & η_{22} & … & η_{b2} \\ ⋯ & ⋯ & … & ⋯ \\ η_{1d} & η_{2d} & … & η_{bd} \end{matrix}]

As we mentioned before, this arrow is indexed by a list (α, V, W, U, s, l). Let x and y be the source and target of α respectively. Then V and W are a simple kAut_{C}(x)-module and a simple kAut_{C}(y)-module respectively. Since π = {ϕ_{x} : x ∈ Ob(C)} is a kC-module homomorphism from R_{1} to R_{2}, the following diagram commutes:

\xymatrix{M_{1} \ar[r]^{φ_{1}} \ar[d]^{ϕ_{x}} & N_{1} \ar[d]^{ϕ_{y}} \\ M_{2} \ar[r]^{φ_{2}} & N_{2}.}  (7.1)

Obviously, ϕ_{x} sends M_{1}(V) into M_{2}(V), and ϕ_{y} sends N_{1}(W) into N_{2}(W). By Lemma 4.1, ϕ_{x} sends M_{1}(V, U, s) into M_{2}(V, U, s), and ϕ_{y} sends N_{1}(W, U, l) into N_{2}(W, U, l). Consequently, the above commutative diagram induces the following commutative diagram, where all inclusions and projections are defined in the usual sense:

\xymatrix{M_{1}(V, U, s) \ar@{^{(}->}[r] \ar[d]^{ϕ_{V,U,s}} & M_{1}(V) \ar@{^{(}->}[r] \ar[d]^{ϕ_{V}} & M_{1} \ar[r]^{φ} \ar[d]^{ϕ_{x}} & N_{1} \ar@{->>}[r] \ar[d]^{ϕ_{y}} & N_{1}(W) \ar@{->>}[r] \ar[d]^{ϕ_{W}} & N_{1}(W, U, l) \ar[d]^{ϕ_{W,U,l}} \\ M_{2}(V, U, s) \ar@{^{(}->}[r] & M_{2}(V) \ar@{^{(}->}[r] & M_{2} \ar[r]^{φ} & N_{2} \ar@{->>}[r] & N_{2}(W) \ar@{->>}[r] & N_{2}(W, U, l).}

In particular, the following diagram commutes:

\xymatrix{M_{1}(V, U, s) \ar[r]^{φ′_{1}} \ar[d]^{ϕ_{V,U,s}} & N_{1}(W, U, l) \ar[d]^{ϕ_{W,U,l}} \\ M_{2}(V, U, s) \ar[r]^{φ′_{2}} & N_{2}(W, U, l).}  (7.2)

But by our definition of the functor F, φ′_{1} and φ′_{2} induce θ_{1} and θ_{2}, the maps assigned to this fixed arrow by R′_{1} and R′_{2} respectively. They have the following matrix representation, where I is the dimU ×dimU identity matrix.

B̃_{1} = [\begin{matrix}λ_{11}I & λ_{21}I & … & λ_{a1}I \\ λ_{12}I & λ_{22}I & … & λ_{a2}I \\ ⋯ & ⋯ & … & ⋯ \\ λ_{1b}I & λ_{2b}I & … & λ_{ab}I \end{matrix}]  B̃_{2} = [\begin{matrix}μ_{11}I & μ_{21}I & … & μ_{a1}I \\ μ_{12}I & μ_{22}I & … & μ_{a2}I \\ ⋯ & ⋯ & … & ⋯ \\ μ_{1b}I & μ_{2b}I & … & μ_{ab}I \end{matrix}]

Also from the definition of F, ϕ_{V} and ϕ_{W} induces ϕ′_{V} and ϕ′_{W} respectively. They have the following matrix representations, where I_{1} is the dimV ×dimV identity matrix, I_{2} is the dimW × dimW identity matrix:

Ĉ_{x} = [\begin{matrix}ϵ_{11}I_{1} & ϵ_{21}I_{1} & … & ϵ_{a1}I_{1} \\ ϵ_{12}I_{1} & ϵ_{22}I_{1} & … & ϵ_{a2}I_{1} \\ ⋯ & ⋯ & … & ⋯ \\ ϵ_{1c}I_{1} & ϵ_{2c}I_{1} & … & ϵ_{ac}I_{1} \end{matrix}]  Ĉ_{y} = [\begin{matrix}η_{11}I_{2} & η_{21}I_{2} & … & η_{b1}I_{2} \\ η_{12}I_{2} & η_{22}I_{2} & … & η_{b2}I_{2} \\ ⋯ & ⋯ & … & ⋯ \\ η_{1d}I_{2} & η_{2d}I_{2} & … & η_{bd}I_{2} \end{matrix}]

It is easy to see from the above commutative diagram that ϕ_{V,U,s} and ϕ_{W,U,l} have the matrix representations C̃_{x} and C̃_{y}, with I_{1} and I_{2} replaced by I. Since the diagram (7.2) commutes, it must be true: C̃_{y}B̃_{1} = B̃_{2}C̃_{x}. But:

C_{y}B_{1} = B_{2}C_{x} if and only if C̃_{y}B̃_{1} = B̃_{2}C̃_{x},  (7.3)

that is: ϕ′_{W} θ_{1} = θ_{2}ϕ′_{V}, and the following diagram commutes. Since this arrow is arbitrarily chosen, we know π′ is indeed a kQ-module homomorphism.

\xymatrix{R′_{1}(V) = k^{a} \ar[r]^{θ_{1}} \ar[d]^{ϕ′_{V}} & R′_{1}(W) = k^{b} \ar[d]^{ϕ′_{W}} \\ R′_{2}(V) = k^{c} \ar[r]^{θ_{2}} & R′_{2}(W) = k^{d}}  (7.4)

Remark 7.2. The above proof actually implies that diagram (7.4) commutes if and only if diagram (7.2) commutes.

Therefore, F maps a kC-module homomorphism to a kQ-module homomorphism. F also preserves the homomorphism composition since matrix product preserves composition. Thus F is indeed a functor from Rep_{k}C to Rep_{k}Q.

Let us use Example 4.4 to show how a representation of C gives a representation of the associated quiver Q and vice versa.

Example 7.3. Let R be a representation of the finite EI category C shown in Example 4.4 with: R(x) = ⟨v_{1}⟩⊕⟨v_{2}⟩⊕⟨v_{3}⟩ ≅ k⊕k⊕S; R(y) = ⟨w_{1}⟩⊕⟨w_{2}⟩⊕ ⟨w_{3}, w_{4}⟩ ⊕ ⟨w_{5}, w_{6}⟩ ≅ k ⊕ ϵ ⊕ V_{2} ⊕ V_{2}. Moreover, since V_{2} ↓_{H_{1}}^{H}≅ k ⊕ S, we assume that w_{3} and w_{5} both generate the trivial submodules on restriction to H_{1}, and w_{4} and w_{6} both generates submodules isomorphic to S on restriction to H_{1}.

We already know from Example 4.4 that α = 1 ∈ S_{3} can be chosen as the unique representative unfactorizable morphism in C. Let φ = R(α). Then φ(v_{1}) and φ(v_{2}) are in the subspace of R(y) generated by w_{1}, w_{3}, w_{5}, and φ(v_{3}) lies in the subspace generated by w_{2}, w_{4}, w_{6}. Thus φ has the following matrix representation:

[\begin{matrix}λ_{11} & λ_{21} & 0 \\ 0 & 0 & λ_{32} \\ λ_{13} & λ_{23} & 0 \\ 0 & 0 & λ_{34} \\ λ_{15} & λ_{25} & 0 \\ 0 & 0 & λ_{36} \end{matrix}]

The induced representation R′ of Q is described below, where M_{1} =

[λ_{11} λ_{21}], M_{2} = [\begin{matrix}λ_{13} & λ_{23} \\ λ_{15} & λ_{25} \end{matrix}], M_{3} = [\begin{matrix}λ_{34} \\ λ_{36} \end{matrix}], and M_{4} = [λ_{32}].

\xymatrix{ & k^{2} \ar[dl]^{M_{1}} \ar[dr]^{M_{2}} &  & k \ar[dl]^{M_{3}} \ar[dr]^{M_{4}} \\ k &  & k^{2} &  & k}

The reader can easily recover R from R′.

Proposition 7.4. The above functor F we constructed is full, faithful and dense.

We give a lemma which will be used in the proof the this proposition.

Lemma 7.5. Let V = ⨁_{i=1}^{m} V_{i}, V′ = ⨁_{i=1}^{m} V′_{i}, W = ⨁_{i=1}^{n} W_{i} and W′ = ⨁_{i=1}^{n} W′_{i} be vector spaces. Let θ : V → V′, φ : V → W, ψ : W → W′ and φ′ : V′ → W′ be linear transformations of the following forms. Then the diagram D commutes if and only if all diagrams D_{ij} commute.

θ = [\begin{matrix}θ_{1} & 0 & … & 0 \\ 0 & θ_{2} & … & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & … & θ_{m} \end{matrix}] ψ = [\begin{matrix}ψ_{1} & 0 & … & 0 \\ 0 & ψ_{2} & … & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & … & ψ_{n} \end{matrix}]

φ = [\begin{matrix}φ_{11} & … & φ_{m1} \\ φ_{12} & … & φ_{m2} \\ ⋮ & ... & ⋮ \\ φ_{1n} & … & φ_{mn} \end{matrix}] φ′ = [\begin{matrix}φ′_{11} & … & φ′_{m1} \\ φ′_{12} & … & φ′_{m2} \\ ⋮ & ... & ⋮ \\ φ′_{1n} & … & φ′_{mn} \end{matrix}]

D = \xymatrix{V \ar[r]^{φ} \ar[d]^{θ} & W \ar[d]^{ψ} \\ V′ \ar[r]^{φ′} & W′} D_{ij} = \xymatrix{V_{i} \ar[r]^{φ_{ij}} \ar[d]^{θ_{i}} & W_{j} \ar[d]^{ψ_{j}} \\ V′_{i} \ar[r]^{φ′_{ij}} & W′_{j}}

Proof. This is just a calculation of the matrix product:

[\begin{matrix}ψ_{1} & 0 & … & 0 \\ 0 & ψ_{2} & … & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & … & ψ_{n} \end{matrix}] [\begin{matrix}φ_{11} & … & φ_{m1} \\ φ_{12} & … & φ_{m2} \\ ⋮ & ... & ⋮ \\ φ_{1n} & … & φ_{mn} \end{matrix}] = [\begin{matrix}φ′_{11} & … & φ′_{m1} \\ φ′_{12} & … & φ′_{m2} \\ ⋮ & ... & ⋮ \\ φ′_{1n} & … & φ′_{mn} \end{matrix}] [\begin{matrix}θ_{1} & 0 & … & 0 \\ 0 & θ_{2} & … & 0 \\ ⋮ & ⋮ & ... & ⋮ \\ 0 & 0 & … & θ_{m} \end{matrix}]

if and only if ψ_{j}φ_{ij} = φ′_{ij}θ_{i}. That is, D commutes if and only if all D_{ij} commute.

Now let us prove the previous proposition.

Proof. F is dense (or essentially surjective). First, we show the algorithm giving a representation of Q from a representation of C is invertible. That is: given a representation R′ of Q, we can define a representation R of C such that F (R) ≅ R′. By Proposition 3.1, it suffices to define a rule R which assigns a kAut_{C}(x)-module to each object x, a linear transformation to each unfactorizable morphism, so that R restricted to D_{α} is a representation of D_{α} for every representative unfactorizable morphism α (see the definition of D_{α} in section 3).

An object x gives a family of vertices {V_{1}, …, V_{m}} in Q. If R′(V_{i}) = k^{a_{i}}, then we define R(x) = M = V_{1}^{a_{1}} ⊕…⊕V_{m}^{a_{m}}. This definition gives a kAut_{C}(x)

-module for every object x in C. Now we construct a linear map R(α) : M = R(x) → N = R(y) for each representative unfactorizable morphism α. As we mentioned before, M as a vector space is the direct sum of all subspaces of the form M(V, U, s), and N as a vector space is the direct sum of all subspaces of the form N(W, T, l) (see the paragraph before Lemma 4.1). Thus it is enough to define linear maps from M(V, U, s) into N(W, T, l).

The morphism α determines groups G_{0} ⊴ G_{1} ⩽ G. By Remark 3.7, the derived map φ′ : M(V, U, s) → N(W, T, l) by φ should be a k(G_{1}/G_{0})-module homomorphism if U ≅ T | k ↑_{G_{0}}^{G_{1}} under the given identification G_{1}/G_{0} ≅ H_{1}/H_{0}; and 0 otherwise. If it is a k(G_{1}/G_{0})-module homomorphism, then U ≅ T and the list (α, V, W, U, s, l) uniquely determines an arrow from the vertex V to the vertex W. Let B = (λ_{ji})_{i=1,...,b}^{j=1,...,a} be the matrix representation of the linear map assigned to this arrow by R′. It provides a unique block matrix B̃ = (λ_{ji}I)_{i=1,...,b}^{j=1,...,a}, where I is the dimU × dimU identity matrix. This block matrix B̃ gives a linear map from M(V, U, s) to N(W, T, l) under the chosen decomposition of V ↓_{G_{1}}^{G}. All these maps determine a linear map from M to N, which can be defined as R(α). Repeating this process, we can define a linear map for each representative unfactorizable morphism.

By Lemma 3.6, the rule R we just defined is a representation of D_{α} while restricted to each D_{α}. By Proposition 3.1, R gives a representation of C. By abuse of notation, we denote it by R. It is direct to verify that F (R) is isomorphic to R′. Thus F is dense.

F is faithful. Let R_{1} and R_{2} be two kC-modules and π = {ϕ_{x} | x ∈ Ob(C)} be a kC-module homomorphism from R_{1} to R_{2}. Let V be a vertex in Q and x be the object in C corresponding to V. Let ϕ_{x} : M_{1} = R_{1}(x) → M_{2} = R_{2}(x) be the kAut_{C}(x)-module homomorphism in π, which is the direct sum of ϕ_{V} : M_{1}(V) → M_{2}(V) for all simple summands V of M_{1}. If F (π) = {ϕ′_{V} : V is a vertex in Q} is 0, then in particular ϕ′_{V} = 0 and therefore ϕ_{V} = 0 (see Remark 7.1). Since V is an arbitrary summand of M_{1}, we have ϕ_{x} = 0. Consequently, π is 0.

F is full. Let π′ = {ϕ′_{V} : V is a vertex in Q} be a kQ-module homomorphism from F (R_{1}) = R′_{1} to F (R_{2}) = R′_{2}. We can recover a kC-module homomorphism π = {ϕ_{x} | x ∈ Ob(C)} from π′ such that F (π) = π′. Indeed, to define ϕ_{x} : M_{1} = R_{1}(x) → M_{2} = R_{2}(x) for a fixed object x, it is enough to define ϕ_{V} : M_{1}(V) → M_{2}(V), where V is a simple summand of M_{1}. According to Remark 7.1, ϕ_{V} could be recovered from ϕ′_{V} ∈ π′. In this way we get from π′ a family of linear transformations π = {ϕ_{x} | x ∈ Ob(C)}. Clearly, F (π) and π′ have the same matrix representation, i.e., F (π) = π under the chosen decompositions and the chosen bases. The only thing which remains is to show that π is indeed a kC-module homomorphism. By Proposition 3.2, we only need to verify that diagram (7.2) commutes for every representative unfactorizable morphism α : x → y.

The kAut_{C}(x)-module M_{1} (M_{2}, resp.) is a direct sum of subspaces of the form M_{1}(V, U, s) (M_{2}(V, U, s), resp.). The kAut_{C}(y)-module N_{1} (N_{2}) is a direct sum of subspace of the form N_{1}(W, T, l) (N_{2}(W, T, l), resp.). By Lemma 7.1, it is enough to check that for an arbitrary M_{i}(V, U, s) and an arbitrary N_{i}(W, T, l), i = 1, 2, diagram (7.2) commutes. The morphism α determines groups G_{0} ⊴ G_{1} ⩽ G. Again by Lemma 3.5 and Remark 3.7, the maps φ′_{i} : M_{i}(V, U, s) → N_{i}(W, T, l) derived from φ_{i}, i = 1, 2, are 0 or k(G_{1}/G_{0})-module homomorphisms under the given identification. If they are 0, the diagram commutes trivially. Otherwise, U ≅ T | k ↑_{G_{0}}^{G_{1}}, and the list (α, V, W, U, s, l) determines an arrow. The linear maps assigned to this arrow by R′_{1} and R′_{2}, combined with ϕ′_{V} and ϕ′_{W} give the commutative diagram (7.4). By Remark 7.2 the diagram (7.2) also commutes. Thus π is indeed a kC-module homomorphism and F (π) = π′. That is, F is full.

We proved that F is dense, faithful and full. By Theorem 1 on page 91 of [4] F gives rise to a Morita equivalence between kQ and kC. This finishes the proof.

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