U.U.D.M. Project Report 2018:18
Examensarbete i matematik, 15 hp
Handledare: Martin Herschend
Examinator: Veronica Crispin Quinonez
Juni 2018
Department of Mathematics
Uppsala University
Homological Algebra for Quiver Representations
Mateusz Stroi ński
HOMOLOGICAL ALGEBRA FOR QUIVER REPRESENTATIONS
MATEUSZ STROIŃSKI
Abstract.
Quiver representations arise naturally in the study of representation theory for associative algebras. A par- ticularly simple case to consider is that of representations of finite acyclic quivers, which is the object of study of this thesis.
In this thesis we give an explicit presentation of some elementary constructions in the category of quiver representations, which then allows us to formulate and discuss some basic homological algebra within that category. The tools of homological algebra can then be applied to study the canonical problem of represen- tation theory: classification of the indecomposable representations.
Contents
1. Introduction 1
2. First definitions 1
3. Projective and Injective Representations 11
4. Irreducible morphisms and almost split sequences 23
5. Auslander-Reiten Translation Revisited 32
References 38
i
1. Introduction
Quivers, and in particular their representations, owe their significance to their role as an important tool in study of representations of finite-dimensional algebras (and more generally also Artinian algebras), especially due to the fundamental developments of that field made by Peter Gabriel (resulting in Gabriel’s famous theorem presented in [3]) and those made by Maurice Auslander and Idun Reiten, resulting in an entire field of study known as Auslander-Reiten theory (basic elements of which are introduced in the following text;
for a more detailed and advanced text we refer to [2] ).
The following text focuses on finite-dimensional representations of finite acyclic quivers, and hence treats them as an independent object of study rather than a tool used in study of other objects. The first part of the text introduces the category RepKQ of finite-dimensional representations of a given finite acyclic quiver together with its fundamental properties (in particular, RepKQ is a K-linear category), gives explicit descriptions of some essential categorical notions realized in quiver representations, such as direct sums and products, pullbacks and pushouts and also kernels and cokernels.
Second part introduces projective and injective representations, describes the fundamental family {P (i)}i∈Q0
of indecomposable projective representations, and similarly a family {I(i)}i∈Q0 of indecomposable injective representations, and then shows that those families generate the projective respectively injective represen- tations of given quiver Q, that is, given a projective representation P , P can be decomposed as a direct sum with summands in said family, P =L
i∈Q0(Lni
j=1P (i)), and similarly for injective representations. The extensive presentation of projective representations is followed up by a discussion of projective resolutions, concluding that every representation admits a projective resolution of length 1. Those facts then allow us to conclude that similar statements hold for injective representations of Q, when considered (due to a duality between RepKQ and RepKQop shown in the same part) as the projective representations of the opposite quiver Qop. The second part concludes with a brief introduction to the Auslander-Reiten translation - a map τ : RepKQ → RepKQ whose significance is revealed in the fourth and final part of the text.
The third part of the text gives a more concrete description of the relation between representation theory of quivers and that of finite-dimensional algebras, by presenting an equivalence of RepKQ and mod-KQ - the category of finite dimensional KQ-modules, where KQ denotes the path algebra of Q. It focuses more on homological algebra of those modules, introducing important notions such as almost split sequences and irreducible morphisms and their direct relation to the fundamental question of representation theory (here of finite-dimensional algebras, although for this text the path algebras of finite acyclic quivers, being a proper subset of finite-dimensional algebras, is the object of focus) - the classification of indecomposable representations.
This question is even more relevant to the fourth part of the text, which reveals some important properties of the aforementioned Auslander-Reiten translation, most notably it being a bijection between the set of indecomposable non-injective modules and that of indecomposable non-projective modules. It also introduces the Auslander-Reiten quiver Γ(mod-A) of mod-A (where A denotes a finite-dimensional algebra), which, using τ gives a visual representation of the structure of indecomposable modules of A, hence providing crucial information about the category mod-A itself.
Finally it should be noted that the first half of the text follows (roughly) chapters 1 and 2 of [7], whereas the latter half consists of parts of chapter IV of [1]. Hence the reader is referred to those texts for a more thorough exposition of the material presented here.
2. First definitions Definition 2.1. A quiver Q = (Q0, Q1, s, t) consists of
• Q0- a set of vertices
• Q1- a set of arrows
• s : Q1→ Q0 - a map mapping an arrow to its starting point
• t : Q1→ Q0 - a map mapping an arrow to its terminal point
Definition 2.2. M is a representation of Q over field K if and only if M = (Mi, ϕα)i∈Q0,α∈Q1, where Mi
is a K-vector space, and ϕα: Mi → Mj is a K-linear map, provided that s(α) = i, t(α) = j Throughout this text we assume the following:
1
• All representations considered are finite-dimensional, i.e. each Mi is finite-dimensional.
• All representations are over an algebraically closed field K (unless otherwise stated).
Definition 2.3. Let M = (Mi, ϕα)i∈Q0,α∈Q1, M0= (Mi0, ϕ0α)i∈Q0,α∈Q1 be two representations of quiver Q.
A morphism of representations f : M → M0 is a collection of linear maps (fi)i∈Q0 such that for each α ∈ Q1, the following diagram commutes:
Mi Mj
Mi0 Mj0
ϕα
fi fj
ϕ0α
In particular, f is an isomorphism if each fi is an isomorphism of vector spaces.
Finally, we denote the set of morphisms from M to M0 by Hom(M, M0)
Remark 2.4. Given representations M , N , P of quiver Q, and given morphisms g ∈ Hom(M, N ), f ∈ Hom(N, P ), let f ◦ g = (fi◦ gi)i∈Q0. Clearly f ◦ g is in Hom(M, P ) and thus representations over K of Q together with their morphisms constitute a category, denoted by RepKQ .
Proposition 2.5. Given two representations of Q, M, M0 ∈ RepKQ, the set of morphism Hom(M, M0) forms a vector space, i.e. we have Hom(M, M0) ∈ VectK.
Proof. Let f, g ∈ Hom(M, M0) We know that ∀i ∈ Q0: fi, gi∈ Hom(Mi, Mi0) ∈ V ectK and thus by setting f + g = (fi+ gi)i∈Q0 , kg = (kgi)i∈Q0(∀k ∈ K) we can satisfy all the vector space properties, provided that our morphisms are well defined, which is checked as follows:
ϕ0α◦ (fi+ gi) = ϕ0α◦ fi+ ϕ0α◦ gi = fj◦ ϕα+ gj◦ ϕα= (fj+ gj) ◦ ϕα and
ϕ0α◦ kfi= k(ϕ0α◦ fi) = k(fj◦ ϕα) = kfj◦ ϕα
Proposition 2.6. 0 = (0i, 0α) is a zero object of RepKQ i.e. it is both initial and terminal in RepKQ . Proof. Let M = (Mi, ϕα)i∈Q0,α∈Q1 ∈ RepKQ
0 is initial: ∀i ∈ Q0, 0iis initial in VectK i.e. ∃!ψ ∈ Hom(0i, Mi) = 0 and 0 ◦ ϕα= 0 = 0 ◦ 0, so the morphism is unique and well-defined.
0 is terminal: ∀i ∈ Q0, 0i is terminal in VectK i.e ∃!λ ∈ Hom(Mi, 0i) = 0 and ϕα◦ 0 = 0 = 0 ◦ 0, so the
morphism is unique and well-defined.
Definition 2.7. Let M = (Mi, ϕα)i∈Q0,α∈Q1 ∈ RepKQ and let M0 = (Mi0, ϕ0α)i∈Q0,α∈Q1 ∈ RepKQ. We define the direct sum of M and M0 as M ⊕ M0 = (Mi⊕ Mi0,hϕ
α 0 0 ϕ0α
i) Proposition 2.8. Let M, M0∈ RepKQ then M ⊕ M0 is:
a) The product MQ M0 of M and M0 in RepKQ b) The coproduct M` M0 of M and M0 in RepKQ
Proof. For both parts we use the fact that the notions of finite product and finite coproduct coincide in VectK, i.e for V, W ∈ VectK, we have V ⊕ W ' VQ W ' V ` W
First let N = (Ni, γα)i∈Q0,α∈Q1 ∈ RepKQ a) Let f ∈ Hom(N, M ), f0∈ Hom(N, M0).
Let π = (πi)i∈Q0 : M ⊕ M0 → M , where πi = [1 0]i, (i.e define π pointwise as projections from the direct sum onto the left summand), g = (gi)i∈Q0 = hf
i
fi0
i : N → M ⊕ M0 (i.e define g pointwise as morphisms induced by product property of M ⊕ M0). The following diagram then commutes for all i ∈ Q0 due to product property of M ⊕ M0:
2
Ni
(M ⊕ M0)i
Mi Mi0
f0 fi gi i
πi π0i
Moreover, the choice of g is unique as giis the unique morphism making said diagram commute. So all that needs to be checked is that π and g are well-defined, i.e for all α ∈ Q1 these diagrams commute:
(M ⊕ M0)i (M ⊕ M0)j
Mi Mj
ϕα 0
0 ϕ0α
πi πj
ϕα
Ni Nj
(M ⊕ M0)i (M ⊕ M0)j
γα
gi gj
ϕα 0
0 ϕ0α
And indeed we have
ϕα◦ πi= ϕα◦ [1 0]i= [ϕα0]j= [1 0]jhϕ
α 0 0 ϕ0α
i= πj◦hϕ
α 0 0 ϕ0α
i and also
hϕ
α 0 0 ϕ0α
i◦ gi=hϕ
α 0 0 ϕ0α
i◦hf
i
fi0
i
=hϕ
α◦fi
ϕ0α◦fi0
i
=hf
j◦γα
fj0◦γα
i
as both f and f0 are well-defined morphisms, and finallyhfj◦γα
fj0◦γα
i
= gj◦ γα
b) Let f ∈ Hom(M, N ), f0 ∈ Hom(M0, N ). Similarly to a), let ι = (ιi)i∈Q0, where ιi = [10]i (i.e define ι : M → M ⊕M0pointwise as inclusions of the left summand into the direct sum), g = (gi)i∈Q0: M ⊕M0→ N , where gi ' [fifi0] (i.e define g pointwise as morphisms induced by coproduct property of M ⊕ M0. Just like in the case of product, our desired diagram now commutes pointwise for all i ∈ Q0 and it also does so uniquely, all due to direct sum being the coproduct in VectK, and yet again all we need to check is that g and ι are well-defined. For ι we have:
hϕ
α 0 0 ϕ0α
i◦ ιi=hϕ
α 0 0 ϕ0α
i◦ [10]i= [ϕ0α] = [10]j◦ ϕα
and for g we proceed as follows:
γα◦ [fifi0] = [γα◦fiγα◦fi0] = [fjfj0] ◦hϕ
α 0 0 ϕ0α
i
= gj◦hϕ
α 0 0 ϕ0α
i
Definition 2.9. M ∈ RepKQ is indecomposable if and only if M 6= 0 and there are no non-zero represen- tations L, N such that M = L ⊕ N
Theorem 2.10 (Krull-Schmidt Theorem). Given a representation M ∈ RepKQ, M admits a unique (up to reordering of summands) direct sum decomposition into indecomposable summands, i.e M ' M1⊕ · · · ⊕ Mn
such that M1, . . . , Mn are indecomposable.
Proof. Theorem 1.2 in [7]
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Proposition 2.11. RepKQ has pullbacks, i.e., given the following diagram in RepKQ :
A B
D
f g
there is a representation X ∈ RepKQ together with morphisms ρA ∈ Hom(X, A), ρB ∈ Hom(X, B) consti- tuting a pullback of the diagram.
Lemma 2.12. The category R -Mod of left R-modules (for some ring R) has pullbacks.
Proof. Given a diagram as above in R -Mod, let X = {(x, y) ∈ A ⊕ B : f (x) = g(y)} = Ker(d), where d = (f, −g) : A ⊕ B → D, and let ρA, ρB be the restrictions of canonical projections πA, πB to X. Clearly g ◦ ρB = f ◦ ρA, so X completes the pullback square, thus it remains to show that the square is universal, i.e. given Y ∈ R -Mod, qA∈ Hom(Y, A), qB ∈ Hom(Y, B) such that f ◦ qA= g ◦ qB, ∃!h : Y → X such that qB= ρB◦ h, qA= ρA◦ h, which is illustrated by following diagram:
Y
X B
A D
qB
qA h
ρB
ρA g
f
We achieve that by letting h = (qA, qB) = [qqAB] (which is well-defined as f ◦ qA= g ◦ qB). Clearly we have ρA◦ h = ρA◦ (qA, qB) = qA, ρB◦ h = ρB◦ (qA, qB) = qB
so the diagram commutes. Uniqueness of h can be proven as following: given h0∈ Hom(Y, X), ρA◦h0= ρA◦h, ρB◦ h0= ρA◦ h and thus h0 = h, but both those conditions are necessary for h0to satisfy the diagram above,
and thus if h0 does so, we get h0= h.
This gives us a foundation to prove the same property for RepKQ just like we did with earlier statements: by borrowing the pointwise construction from VectK (= K -Mod) and endowing it with some canonical choice of arrow-wise structure so that all the morphisms are well defined.
Proof of proposition 2.11. Let X = (Xi,hϕα
Xi 0
0 ψαXi
i
)i∈Q0,α∈Q1(where A = (Ai, ϕα), B = (Bi, ψα)), where Xi is the pointwise pullback of
Ai Bi
Di
fi gi
Let ρA: X → A = (ρAi)i∈Q0 , ρB: X → B = (ρBi)i∈Q0. Then ρA∈ Hom(X, A) as ϕα◦ ρAi = ϕα◦ [1 0]i= [1 0]i ◦hϕ
αXi 0 0 ψαXi
i
. Analogous statement holds for ρB. Thus, X completes the commutative square.
We still need to prove that X does it in a universal manner. Let Y = (Yi, ωα)i∈Q0,α∈Q1 ∈ RepKQ, qA ∈ Hom(Y, A), qB ∈ Hom(Y, B) such that f ◦ qA= g ◦ qB Let h = (hi)i∈Q0 be the morphism defined pointwise by Xi being pullback pointwise (i.e. Xi’s universal property of pullback of diagram above in V ectK). It is clear that h is the only possible candidate, as for all i ∈ Q0, hi is the unique morphism making the diagram commute at i, which is necessary for h to make the corresponding diagram in RepKQ commute. Thus we only need to show that h is well-defined, i.e. h ∈ Hom(Y, X). Recall from Lemma2.12that hi= (qAi, qBi) = [qqAB] We need to show that given α : i → j ∈ Q1, we havehϕ
αXi 0 0 ψαXi
i◦ hi= hj◦ ωαIn particular, the following holds:
hϕ
αXi 0 0 ψαXi
i◦ hi=hϕ
αXi 0 0 ψαXi
i◦qAi
qBi =hϕ
αXi◦qAi 0 0 ψαXi◦qBi
i
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Now since qA∈ Hom(Y, A) we have that ϕα◦ qAi = qAj◦ ωαand that equality implies ϕαXi ◦qAi = qAj◦ ωα
A similar equality holds for B and ψ. Thus hϕ
αXi 0 0 ψαXi
i◦ hi=hϕ
αXi◦qAi 0 0 ψαXi◦qBi
i
=hq
Aj◦ωα qBj◦ωα
i
=hq
qAjBj
i◦ ωα= hj◦ ωα
Proposition 2.13. RepKQ has pushouts, i.e. given the following diagram in RepKQ :
D
A B
f g
There is a representation X ∈ RepKQ together with morphisms ιA ∈ Hom(A, X), ιB ∈ Hom(B, X) consti- tuting a pushout of the diagram.
Lemma 2.14. R -Mod has pushouts.
Proof. Consider a diagram as above in R -Mod. Let X = (A ⊕ B)/K, where K = {(f (d), g(−d) | d ∈ D} = (f, −g)(D). Let ι : A → A ⊕ B, ι0 : B → A ⊕ B be the usual inclusions, π : A ⊕ B → X be the canonical projection. Let γA= π ◦ ι, γB= π ◦ ι0. Clearly γA◦ f = γB◦ g as π ◦ ι ◦ f − π ◦ ι0◦ g = π ◦ (ι ◦ f − ι0◦ g) = 0 as Im(ι◦f −ι0◦g) = Ker(π). So the pushout square commutes. Now we need to show that it indeed is a pushout square, i.e. has the following universal property: given Y ∈ R -Mod, qA∈ Hom(A, Y ), qB∈ Hom(B, Y ) such that qA◦ f = qB◦ g, there is a unique morphism h ∈ Hom(X, Y ) such that β = h ◦ γA, β0 = h ◦ γB, which is illustrated by the diagram below:
D B
A X
Y
g
f γB
qB γA
qA h
Let ω = (qA, qB) i.e. the unique morphism such that ω ◦ ι = qA, ω ◦ ι0= qB. We have ω ◦ (ι ◦ f − ι0◦ g) = ω ◦ ιf − ω ◦ ι0◦ g = qA◦ f − qB◦ g = 0
so K ⊂ Ker(ω), and thus by factorization theorem there is a unique morphism η : X → Y such that η ◦ π = ω Thus η ◦ γA = η ◦ π ◦ ι = ω ◦ ι = qA and similarly for B, so the diagram commutes. η is also unique such morphism: if η0 ∈ Hom(X, Y ) makes the diagram commute, it must have η0 ◦ π ◦ ι = qA = η ◦ π ◦ ι, and η0◦ π ◦ ι0 = qB = η ◦ π ◦ ι0 so by universal property of coproducts we have η0◦ π = η ◦ π, and so η0= η as π
is epi.
Proof of proposition 2.13. Let X = (Xi, χα)i∈Q0,α∈Q1, where Xi is the pullback of Di
Ai Bi
fi gi
and χα: Xi→ Xj is defined by χα((a, b) + Ki) = (ϕα(a), ψα(b)) + Kj. (where A = (Ai, ϕα), B = (Bi, ψα)).
In analogy to the proof about pullbacks, we first show that γA= (γAi)i∈Q0 is a well-defined morphism:
χα◦ γAi(a) = χα◦ πi◦ ιi(a) = (ϕα(a), 0) + K¯ j = πj◦ ιj◦ ϕα= γAj ◦ ϕα
so γA∈ Hom(A, X), and similarly for B, and thus X completes the square. Let Y = (Yi, κα)i∈Q0,α∈Q1, qA= (qAi)i∈Q0, qB = (qBi)i∈Q0 such that qA◦ f = qB◦ g, i.e. Y completes the square. Let h : X → Y = (hi)i∈Q0
consist of morphisms induced pointwise by pushout property. Clearly, h is the only possible candidate for the induced pushout morphism. It remains to show that h is a well-defined morphism from X to Y , which it is if and only if for all arrows α ∈ Q1, α : i → j, the following equation is satisfied: κα◦ hi = hj◦ χα. We obtain that result as following: for qA∈ Hom(A, Y ) we have κα◦ qAi = qAj ◦ ϕα Now by the pointwise
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pushout property of h we know that qAi = hi◦ γAi and that qAj = hj◦ γAj. Now, by putting that to our former equation, we have the following: κα◦ hiγAi= hj◦ γAj◦ ϕαand finally, as γA∈ Hom(A, X), we have χα◦ γAi= γAj◦ ϕα we get
κα◦ hi◦ γAi = hj◦ χα◦ γAj ⇐⇒ κα◦ hi◦ πi◦ ιi= hj◦ χα◦ pij◦ ιj
and similarly for B we get κα◦ hi◦ πi◦ ι0i= hj◦ χα◦ pij◦ ι0j and thus by universal property of coproducts we get κα◦ hi◦ πi= hj◦ χα◦ πi so κα◦ hi= hj◦ χα as πi is epi.
We can use our construction to directly obtain the kernel and cokernel representations.
Definition 2.15. Let C be a category with a zero object 0. Let A, B ∈ C and f ∈ Hom(A, B). A pair (Ker(f ), ι) such that Ker(f ) ∈ C, ι ∈ Hom(Ker(f ), A) is said to be the kernel of f it it satisfies the following universal property:
Given D ∈ C and h ∈ Hom(D, A) such that f ◦ h = 0 (i.e. the unique morphism that factors through 0, that is f ◦ h = ¯0 ◦ ϕ, ϕ ∈ Hom(D, 0) with ¯0 being the unique morphism in Hom(0, B)), there is a unique morphism φ : D → Ker(f ) such that h = ι ◦ φ, i.e. the following diagram commutes:
D
Ker(f ) A B
∃!φ h
ι f
Proposition 2.16. In a category C with a zero object 0 and pullbacks, given f ∈ Hom(A, B), f has a kernel (X, γ) and it is given by the following pullback:
X 0
A B
γ 0
0 f
Proof. The universal property of kernels defined above is properly contained (as a diagram) in the following diagram induced by the universal property of pullbacks:
D
X 0
A B
0
h
∃!φ 0
γ 0
f
This yields the following definition:
Definition 2.17. Given A = (Ai, ψα)i∈Q0,α∈Q1, B ∈ RepKQ, f ∈ Hom(A, B) we define the kernel of f as the representation Ker(f ) ∈ RepKQ given by Ker(f ) = (Ker(fi), ψα Ker(fi))i∈Q0,α∈Q1, ι ∈ Hom(Ker(f ), A) = (ιi)i∈Q0, where ιi is the canonical inclusion of Ker(fi) into Ai in VectK
We treat cokernels similarly:
Definition 2.18. Let C be a category with a zero object 0. Let A, B ∈ C, f ∈ Hom(A, B). A pair (Coker(f ), π) such that Coker(f ) ∈ C, π ∈ Hom(B, Coker(f )) is said to be the cokernel of f if it satisfies the following universal property:
Given D ∈ C and a morphism h ∈ Hom(B, D) such that h ◦ f = 0, the following commutative diagram exists:
A B Coker(f )
D
f π
h
∃!φ
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Proposition 2.19. In a category C with a zero object 0 and pushouts, given f ∈ Hom(A, B), f has a cokernel (X, γ) and it is given by the following pushout:
A B
0 X
0 f
γ 0
Proof. Once again, the universal property of cokernels is properly contained in the following diagram induced by the universal property of pushouts:
A B
0 X
D
f
0 γ
h 0
0
∃!φ
This motivates the following definition:
Definition 2.20. Given representations A, B = (Bi, ψα)i∈Q0,α∈Q1, and a morphism f ∈ Hom(A, B) we define the cokernel of f as the representation Coker(f ) ∈ RepKQ given by Coker(f ) = (Coker(fi), χα), where χα(bi+ fi(Bi)) = ψα(bi) + fj(Bj) and π ∈ Hom(B, Coker(f )) = (πi)i∈Q0, where πi is the canonical projection from Bi onto Coker(fi)
Definition 2.21. A morphism of quiver representations f is:
a) mono if Ker(f ) = 0 b) epi if Coker(f ) = 0
c) an isomorphism if it is both epi and mono.
Definition 2.22. A representation L ∈ RepKQ is said to be a subrepresentation of M ∈ RepKQ if there is a monomorphism g : L → M .
Theorem 2.23 (First Isomorphism Theorem). Let A = (Ai, ϕα), B = (Bi, ψα∈ RepKQ, f ∈ Hom(A, B).
Then Coker(Ker(f ) ' Ker(Coker(f ))
Proof. Fortunately, the corresponding theorem for vector spaces equips us with an explicit isomorphism, i.e. Coker(Ker(fi)) = Ai/ Ker(fi) ' Im(fi) = Ker(Coker(fi)) With γi : Ai/ Ker(fi) −∼→ Im(fi) defined as follows: γi(xi+ Ker(fi)) = f (xi). By applying the definitions in said order we get that Coker(Ker(f )) = (Ai/ Ker(fi), χα) with χα(x + Ker(fi)) = ϕα(x) + Ker(fj), and Ker(Coker(f )) = (Im(fi), ψαIm(fi)) Now let γ = (γi)i∈Q0. Then γ is a morphism as for all α : i → j ∈ Q1 and all x + Ker(fi) ∈ Ai/ Ker(fi) we have (ψαIm(fi)◦γi)(x + Ker(fi) = (ψαIm(fi))(fi(x)) = (ψα◦ fi)(x) = (fj◦ ϕα)(x) = γj(ϕα(x) + Ker(fj)) = (γj◦ χα)(x + Ker(fi)) so (ψαIm(fi)) ◦ γi = γj◦ χα. Moreover, since γi is an isomorphism for all i ∈ Q0, its corresponding kernel and cokernel representations assign the 0 vector space to every i, and thus (as 0 is the zero object), they both are the zero representation, so γ is mono and epi, and thus an isomorphism.
It’s about time now to formally introduce the notion of image.
Definition 2.24. Given A −→ B in Repf KQ , we let Im(f ) ∈ RepKQ be the representation given by Im(f ) := Ker(Coker(f )).
Definition 2.25. A sequence · · ·−→ Mf0 1 f1
−→ M2 f2
−→ · · · in RepKQ is exact at Miif Im(fi−1) = Ker(fi), and is an exact sequence if it is exact for all i. In particular, exact sequences of the form 0 → L−→ Mf −→ N → 0g are said to be short exact.
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Definition 2.26.
a) f : L → M in RepKQ is called a section if there is h ∈ Hom(M, L) such that h ◦ f = 1L
b) g : M → N in RepKQ is called a retraction if there is h ∈ Hom(N, M ) such that g ◦ h = 1N
c) A short exact sequence 0 → L−→ Mf −→ N → 0 is said to split (equivalently, is said to be split exact) if fg is a section.
Proposition 2.27. Let 0 → L−→ Mf −→ N → 0 be a short exact sequence in Repg KQ . Then:
a) f is a section if and only if g is a retraction
b)If f is a section, then M ' L ⊕ N . Moreover, let η be the morphism induced by g being a retraction and let h be the morphism induced by f being a section. Let ιL, ιN be the canonical injections from L respectively M to L ⊕ N and let πL, πN be the canonical projections from L ⊕ N to L respectively N . Then the following diagrams are isomorphic:
0 L M N 0
f h
g η
0 L ιL L ⊕ N N 0
πL
πN
ιN
To prove that proposition we employ the strategy used earlier: we first prove the same result in category R -Mod to obtain candidate morphisms that satisfy our desired property pointwise, and then prove that those candidate morphisms indeed are morphisms.
Lemma 2.28. Let 0 → L−→ Mf −→ N → 0 be a short exact sequence in R -Mod. Then:g a) f is a section if and only if g is a retraction
b) If f is a section, then M ' L ⊕ N . Moreover, let η be the morphism induced by g being a retraction and let h be the morphism induced by f being a section. Let ιL, ιN be the canonical injections from L respectively M to L ⊕ N and let πL, πN be the canonical projections from L ⊕ N to L respectively N . Then the following diagrams are isomorphic:
0 L M N 0
f h
g η
0 L ιL L ⊕ N N 0
πL
πN ιN
Proof. a) " =⇒ ": If g was mono (it is epi due to exactness of the sequence) we would just set η = g−1, but that need not be the case, and although no elements in N have an empty preimage, some preimages need not be singletons, making the inverse mapping not well-defined. We thus "fix" that problem with diagram chasing. Let n ∈ N , m, m0 ∈ g−1(n). We have m − m0 ∈ Ker(g), so m − m0 ∈ Im(f ) and hence there is f−1(m − m0), which is unique as f is mono by exactness of the sequence. Now as f is a section, we have h ∈ Hom(M, L) such that h ◦ f = idL, and (h ◦ f )(f−1(m − m0)) = f−1(m − m0) = h(m − m0), so (f ◦ h)(m − m0) = f (f−1(m − m0)) = m − m0 and thus m − (f ◦ h)(m) = m0 − (f ◦ h)(m0) and g(m − (f ◦ h)(m)) = g(m) − (g ◦ f )(h(m) = g(m) as g ◦ f = 0. So, given n ∈ N , we let η(n) = m − (f ◦ h)(m) for any m ∈ g−1(n) which by above is well-defined and satisfies g ◦ η = idN And so g is a retraction.
" ⇐= ": in analogy to the construction above, given m ∈ M , and letting η ∈ Hom(N, M ) be the morphism induced by g being a retraction (i.e. g ◦ η = idN), we have g(m − (η ◦ g)(m)) = g(m) − (g ◦ η ◦ g)(m) = g(m) − (g ◦ η)(g(m) = g(m) − g(m) = 0 so we have m − (η ◦ g)(m) ∈ Ker(g) and thus there is a unique element f−1(m − (η ◦ g)(m)) and we let h(m) = f−1(m − (η ◦ g)(m)) and given l ∈ L we have (h ◦ f )(l) = f−1(f (l) − (η ◦ g ◦ f )(l)) = f−1(f (l)) = l. So h ◦ f = idL and we’re done.
Remark 2.29. The constructions given above commute, i.e. given a section f with h ∈ Hom(M, L) such that h ◦ f = idL, if we first construct a "retraction morphism" η from h as we did above, and then construct a "section morphism" h0 from η as above, we get h0= h, as
h0(m) = f−1(m − η(g(m))) = f−1(m − (g−1(g(m)) − (f ◦ h)(g−1(g(m)))) = f−1(m − m + (f ◦ h)(m)) = h(m)
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b) We prove this statement in several steps:
i) 0 ← L←− Mh ←− N ← 0 is a short exact sequenceη
Proof: idN = g ◦ η is mono, so η is mono, idL = h ◦ f is epi, hence h is epi. Now all that remains is to show Ker(h) = Im(η). Let m ∈ Im(η). Thus there is a unique element n such that m = η(n), so g(m) = (g ◦ η)(n) = n, and m ∈ g−1(n), so by earlier definition of η we have m = η(n) = m − (f ◦ h)(m), hence m = m − (f ◦ h)(m), so (f ◦ h)(m) = 0. Moreover, h(m) = 0 as f is mono, so Im(η) ⊆ Ker(h). Now let m ∈ Ker(h). Then h(m) = 0, so by the definition of h we get f−1(m − (η ◦ g)(m)) = 0, so m − (η ◦ g)(m) = 0 as f is mono, and thus m = η(g(m)). Hence m ∈ Im(η), so Ker(h) ⊆ Im(η), and thus Ker(h) = Im(η). Thus the sequence is exact.
ii) f ◦ h + η ◦ g = idM
Proof: Given m ∈ M we have
m − (f ◦ h + η ◦ g)(m) = m − (f ◦ h)(m) − (η ◦ g)(m)
= m − f (h(m) − (g−1(g(m)) − (f ◦ h)(g−1(g(m))))
= m − f (h(m)) − (m − f (h(m))) = 0 iii) M ' L` N with injections f, η
Claim: Given X ∈ R -Mod, ϕ ∈ Hom(L, X), ψ ∈ Hom(N, X), there is a unique γ ∈ Hom(M, X) such that the following diagram commutes:
L M N
X
f
ϕ γ
η
ψ
Proof of claim: Let γ = ϕ ◦ h + ψ ◦ g, so
γ ◦ η = ϕ ◦ h ◦ η + ψ ◦ g ◦ η = ψ ◦ g ◦ η as h ◦ η = 0, and g ◦ η = idN means that ψ ◦ g ◦ η = ψ, so γ ◦ η = ψ. Similarly,
γ ◦ f = ϕ ◦ h ◦ f + ψ ◦ g ◦ f = ϕ ◦ h ◦ f = ϕ
And so the diagram commutes. For uniqueness, let γ0such that γ0◦ η = ψ, γ0◦ f = ϕ. But f ◦ h + η ◦ g = idM, so
γ0 = γ0◦ (f ◦ h + η ◦ g) = (γ0◦ f ) ◦ h + (γ0◦ η) ◦ g = ϕ ◦ h + ψ ◦ g = γ iv) M ' LQ N with projections h, g
Claim: given a module X ∈ R -Mod, and morphisms of modules ϕ ∈ Hom(X, L), ψ ∈ Hom(X, N ), there is a unique γ ∈ Hom(X, M ) such that the following diagram commutes:
L M N
X
h g
ϕ γ
ψ
Proof of claim: let γ = η ◦ ψ + f ◦ ϕ. Then
g ◦ γ = g ◦ η ◦ ψ + g ◦ f ◦ ϕ = g ◦ η ◦ ψ = ψ and
h ◦ γ = h ◦ η ◦ ψ + h ◦ f ◦ ϕ = h ◦ f ◦ ϕ = ϕ For uniqueness, let γ0 such that h ◦ γ0= ϕ, g ◦ γ = ψ. But
γ0 = idM◦ γ0 = (f ◦ h + η ◦ g) ◦ γ0= f ◦ (h ◦ γ0) + η ◦ (g ◦ γ0) = f ◦ ϕ + η ◦ ψ = γ
Now we’re ready to prove the statement for RepKQ :
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Proof of proposition 2.27. Note that we assume that f is a section, and thus there is a morphism h = (hi)i∈Q0 ∈ Hom(M, L) such that h ◦ f = idL, and thus in particular, for all i ∈ Q0 we have hi◦ fi = idLi. Now we let η = (ηi)i∈Q0, where ηi is constructed from hi as above. Now, as mentioned earlier, all the desired properties stated in a) and b) will follow as long as η is a well-defined morphism. To show this we let L = (Li, ρα)i∈Q0,α∈Q1, M = (Mi, ϕα)i∈Q0,α∈Q1, and N = (Ni, ψα)i∈Q0,α∈Q1. We need to show that for all α : i → j ∈ Q1 we have ηj◦ ψα= ϕα◦ ηi. But since the properties proven earlier still hold pointwise, i.e.
for all i ∈ Q0, Mi' Li⊕ Ni, we can represent our morphisms with matrices in following manner:
Li Mi Ni
Lj Mj Ni
fi
ρα hi
ϕα gi ηi
ψα
fj
hj
gj
ηj
'
Li Li⊕ Ni Ni
Lj Lj⊕ Nj Ni
[10]i
ρα [ 1 0 ]i
ϕα [ 0 1 ]i
[01]i
ψα
[10]j
[ 1 0 ]j
[ 0 1 ]j
[01]j And in particular, ϕα=a b
c d, but since f is a morphism we have ϕα◦ fi= fj◦ ρα, so a b
c d [10]i= [10]jρα, so [ac] = [ρ0α], so a = ρα, c = 0, similarly for g we obtain gj◦ ϕα= ψα◦ gi and thus [0 ψα] = [c d]. Therefore we have d = ψα, and finally with help of h we get hj ◦ ϕα = ρα◦ hi, which means that [ρα0] = [a b], so b = 0, so that ϕα =hρ
α 0 0 ψα
i
Now we have ηj ◦ ψα= 0
ψα = hρ
α 0 0 ψα
i[01] = ϕα◦ ηi and so our candidate
morphism is well-defined, and the results follow.
Definition 2.30. Given a representation X, we define the covariant Hom-functor of X, Hom(X, −) : RepKQ → VectK by:
• Sending a representation M ∈ RepKQ to the vector space of morphisms Hom(X, M )
• Sending a morphism of representations f ∈ Hom(L, M ) to the K-linear map Hom(X, f ) : Hom(X, L) → Hom(X, M ) which, given a morphism of representations ϕ ∈ Hom(X, L), sends it to f ◦ ϕ ∈ Hom(X, M )
Proposition 2.31. A sequence
0 → L−→ Mf −→ Ng
in RepKQ is exact if and only if for all representations X ∈ RepKQ the following sequence is exact in VectK: 0 → Hom(X, L)−−−−−−→ Hom(X, M )Hom(X,f ) −−−−−−→ Hom(X, N )Hom(X,g)
In particular, the covariant Hom-functor Hom(X, −) is left exact.
Proof. " =⇒ " First, if ϕ ∈ Hom(X, L) such that Hom(X, f )(g) = f ◦ g = 0 then by universal property of kernels, ϕ = ι ◦ h for a unique h ∈ Hom(X, Ker(f )). But f is mono, so Ker(f ) = 0, so h = 0, and thus ϕ = 0 and so Hom(X, f ) is mono. Now we need to show Im(Hom(X, f )) = Ker(Hom(X, g)). Clearly, Im(Hom(X, f )) ⊂ Ker(Hom(X, g)) as g ◦ f = 0 means that Hom(X, g)(Hom(X, f )(ϕ)) = g ◦ f ◦ ϕ = 0 ◦ ϕ = 0.
We also have Ker(Hom(X, g)) ⊂ Im(Hom(X, f )), since ψ ∈ Ker(Hom(X, g)) ⇐⇒ g ◦ ψ = 0 and by universal property of kernels we thus get ∃!h ∈ Hom(X, Ker(g)) such that ψ = ι ◦ h, with ι being the kernel morphism.
But Ker(g) = Im(f ) and f is epi onto Im(f ), so there is a unique isomorphism ¯f : L−∼→ Im(f ) and ι ◦ ¯f = f , and finally ψ = f ◦ ¯f−1◦ h, so ψ is in Im(Hom(X, f )).
" ⇐= " First we show that Hom(X, f ) is mono implies that f is mono, arguing by contraposition: if f is not mono, then Ker(f ) 6= 0, ι 6= 0 (ι being the kernel morphism), but f ◦ ι = 0 = f ◦ 0, so Hom(X, f ) is not mono. Next we have Im(f ) ⊂ Ker(g) as Hom(X, g)(Hom(X, f )(idL)) = 0 = g ◦ f ◦ idL = g ◦ f , and finally Ker(Hom(X, g)) ⊂ Im(Hom(X, f )) as the kernel morphism of Ker(g), ιg : Ker(g) → M is in
Ker(Hom(X, g)), and thus factors through f .
Definition 2.32. Given a representation X, we define the contravariant Hom-functor of X, Hom(−, X) : RepKQ → VectK by:
• Sending a representation M ∈ RepKQ to the vector space of morphisms Hom(M, X)
• Sending a morphism of representations g ∈ Hom(M, N ) to the K-linear map Hom(f, X) : Hom(N, X) → Hom(M, X) which, given a morphism of representations ψ ∈ Hom(N, X), sends it to ψ ◦ g ∈ Hom(M, X)
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Proposition 2.33. A sequence
L−→ Mf −→ N → 0g
in RepKQ is exact if and only if for all representations X ∈ RepKQ the following sequence is exact in VectK: 0 → Hom(N, X)−−−−−−→ Hom(M, X)Hom(g,X) −−−−−−→ Hom(L, X)Hom(f,X)
In particular, the contravariant Hom-functor Hom(−, X) is left exact.
Proof. " =⇒ ": First, g epi means that Hom(g, X) is mono, since by universal property of cokernels, if φ ∈ Hom(N, X) such that Hom(g, X)(φ) = φ ◦ g = 0, φ = h ◦ π for a unique h ∈ Hom(Coker(g), X) (π being the cokernel projection), but Coker(g) = 0) means that h = 0, and hence φ = 0)
Secondly, Im(Hom(g, X)) ⊂ Ker(Hom(f, X) as Hom(f, X)(Hom(g, X)(φ)) = φ ◦ g ◦ f = φ ◦ 0 = 0.
And finally Ker(Hom(f, X)) ⊂ Im(Hom(g, X)) as φ ∈ Ker(Hom(f, X)) is equivalent to φ ◦ f = 0, which implies that there is a unique morphism h ∈ Hom(Coker(f ), X) such that φ = h ◦ π, π being the cok- ernel morphism of Coker(f ) (statement following from universal property of cokernels), and Coker(f ) ' Coker(Ker(Coker(f ))) =: A, since given the kernel morphism ι of Ker(Coker(f )), we have π ◦ ι = 0, and thus there is a unique morphism ψ ∈ Hom(A, Coker(f )) such that ψ ◦ ρ = π, ρ denoting the cokernel morphism of A, so φ = h ◦ ψ ◦ ρ, and h ◦ ψ acts as cokernel morphism, and is unique such: κ ∈ Hom(A, X) such that κ ◦ ρ = φ, so κ ◦ ρ = h ◦ ψ ◦ ρ , and thus κ = h ◦ ψ as ρ is epi, and the equality follows by earlier part of this proof. Now Coker(f ) = Coker(Ker(Coker(f ))) = Coker(Ker(g)) = Ker(Coker(g)) = Ker(0) = N , and thus φ factors through g.
" ⇐= ": First Hom(g, X) being mono implies that g is epi. We argue by contraposition: if g is not epi, then Coker(g) 6= 0 =⇒ π 6= 0, but Hom(g, X)(π) = π ◦ g = 0 = 0 ◦ g = Hom(g, X)(0) so Hom(g, X) is not mono.
Secondly Im(f ) ⊂ Ker(g) as Hom(f, X)(Hom(g, X)(idN)) = idN◦ g ◦ f = 0 =⇒ g ◦ f = 0
Thirdly Ker(g) ⊂ Im(f ): Let π denote the cokernel morphism of Coker(f ). By definition we have π ◦ f = 0, so π ∈ Ker(Hom(f, X)). Thus π ∈ Im(Hom(g, X)) and therefore there is a morphism ϕ ∈ Hom(N, Coker(f )) such that π = ϕ ◦ g. Now let ι denote the kernel morphism of Ker(g). We have π ◦ ι = ϕ ◦ g ◦ ι = ϕ ◦ 0 = 0
so Ker(g) ⊂ Ker(Coker(f )) = Im(f ).
Proposition 2.34. A sequence 0 → L −→ Mf −→ N → 0 in Repg KQ is split exact if and only if for all X ∈ RepKQ : 0 → Hom(X, L)−−−−−−→ Hom(X, M )Hom(X,f ) −−−−−−→ Hom(X, N ) → 0 is exact.Hom(X,g)
Proof. " =⇒ ": Since Hom(X, −) is left exact, to show exactness of the sequence, we only need to show that Hom(X, g) is epi. But the corresponding sequence in RepKQ is split exact, so g is a retraction, and thus there is a morphism η ∈ Hom(N, M ) such that g ◦ η = idN. Now let ϕ ∈ Hom(X, N ). We have ϕ = g ◦ η ◦ ϕ which means that Hom(X, g) is epi.
" ⇐= ": Here we need to both show that g is epi and also that the sequence splits.
We let X = N , and since Hom(X, g) is epi, we get that idN = g ◦ η for some η ∈ Hom(N, M ). But idN
being epi means that g is epi, and also g is a retraction, so the sequence splits. Proposition 2.35. A sequence 0 → L −→ Mf −→ N → 0 in Repg KQ is split exact if and only if for all X ∈ RepKQ : 0 → Hom(N, X)−−−−−−→ Hom(M, X)Hom(g,X) −−−−−−→ Hom(L, X) → 0 is exact.Hom(f,X)
Proof. " =⇒ ": all we need to show is that Hom(f, X) is epi. But the corresponding sequence in RepKQ splits, so f is a section, i.e. there is a morphism η ∈ Hom(M, L) such that η ◦f = idL. Given ϕ ∈ Hom(L, X), we have ϕ = ϕ ◦ η ◦ f , thus ϕ ∈ Im(Hom(f, X)) and hence Hom(f, X) is epi.
" ⇐= ": In analogy to the proof of the preceding proposition, idL= h ◦ f for some h ∈ Hom(M, L). Thus f
is mono and is a section, and the result follows.
3. Projective and Injective Representations
Definition 3.1. Given Q = (Q0, Q1, s, t), and vertices i, j ∈ Q0, a path from i to j is a sequence of arrows {αk}nk=1such that for all k, t(αk) = s(αk+1), and s(α1) = i, t(αn) = j
Definition 3.2. We say that a quiver is acyclic if and only if it contains no cycles, i.e. non-empty paths from i to i, for some i ∈ Q0
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Definition 3.3. Let Q = (Q0, Q1, s, t). i ∈ Q0 is a source if and only if there is no α ∈ Q1 such that i = t(α). Similarly, i is a sink if and only if there is no α ∈ Q1such that i = s(α).
Proposition 3.4. If Q is acyclic then:
a) There is a source in Q b) There is a sink in Q
Proof. a) Assume Q has no source. Then for all i ∈ Q0there is an arrow α ∈ Q1such that i = t(α). Thus if we start at some fixed vertex, we can go back to another vertex by that arrow and continue so indefinitely, inducing an infinite path in the quiver. But we assume (throughout) that our quiver is finite, so some vertices must have been visited multiple times, thus we can separate a subpath of our path that starts and ends at one of those vertices, and thus is a directed loop, contradicting our hypothesis.
b) Similarly, assume Q has no sink. Then, after choosing some i ∈ Q0, we can produce an infinite path by starting with arrow that has its source at i, and continuing with an arrow that has its source at the target of our arrow etc. And thus, just like in a), we obtain directed loops, leading us to contradiction.
From now on we assume that all quivers are acyclic, unless otherwise noted.
Definition 3.5. Let i ∈ Q0Define S(i), the simple representation at vertex i as (S(i)j, ϕα)j∈Q0,α∈Q1, where S(i)j= K if j = i and S(i)j = 0 else, and for all α ∈ Q1 we have ϕα= 0
Proposition 3.6. Given a representation M = (Mi, γα)i∈Q0,α∈Q1 ∈ RepKQ, M is simple (i.e. is not zero and has no non-zero subrepresentations) if and only if it is isomorphic to S(i) for some i ∈ Q0
Proof. " =⇒ ":
Choose a sink j in Q. We have two cases:
1) Mj6= 0 2) Mj= 0
For 1) We have a monomorphism ι : S(j) → M , ι = (ιi)i∈Q0 defined by ιj=
"1 0
...
0
#
, ιi= 0 for i 6= j, which is
well-defined, as the following square clearly commutes:
S(i)m S(i)j
Mm Mj
0
0 ιj
γα
For 2) We remove j and all the arrows into it (as there are no arrows from it to remove) from Q to obtain Q0 and a representation M0 of it, and consider the same problem for Q0. If we happen to be in case 1) this time and thus have a subrepresentation S(j0) of M0, it will also be a subrepresentation of M after we put j and its arrows back to our quiver, since we extend by 0 morphisms for both S(j0) and M0.
This procedure terminates after at most |Q0| steps, and the only way it could avoid case 1) in all steps is by all its vector spaces being 0, and thus M = 0, which is not allowed by the definition of a simple object.
" ⇐= ":
For N 6= 0 to be a subrepresentation of M ' S(i), i.e. have a monomorphism N → M , the morphism must be pointwise mono, so Nj= 0 for j 6= i and Nj' K ' S(i)i, and all the arrows being 0 now follows from 0
being a zero object in V ectK
Definition 3.7. Let i ∈ Q0. Define the projective representation at vertex i, P (i) = (P (i)j, ϕα)j∈Q0,α∈Q1, by P (i)j ' Kn, where n is the number of paths from i to j (from now on referred to as i-j paths), with a canonical basis formally consisting of those paths. Given α : j → k, the map ϕα is defined on that basis by ϕα: p → pα, where p is any of the i-j paths in the basis.
Proposition 3.8. For all i ∈ Q0, P (i) is a projective object of RepKQ , i.e. given M ,N ∈ RepKQ , g ∈ Hom(M, N ) epi, and h ∈ Hom(P (i), N ), there is a (not necessarily unique) morphism f ∈ Hom(P (i), M )
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