A group representation puts a symmetry action on a vector space. A quiver representation puts vector spaces at vertices and linear maps on arrows. The arrows need not be invertible, and the interesting structure can lie in how several maps fit together.
That change opens a different representation problem. We can study a single matrix up to independent source and target coordinates, a chain of maps, several parallel maps, or maps satisfying specified relations. Some of these problems have a finite list of indecomposable building blocks. Others already contain continuously varying families in tiny dimensions. The basic definitions and the finite-type classification are developed in the academic references at the end. [1–2]
This guide uses finite quivers and finite-dimensional complex vector spaces. It is a higher-Mathematics learning route, not a school syllabus claim. Begin with matrices, invariant subspaces and equivalence when the linear-algebra language is unfamiliar. The distinction between irreducible and indecomposable connects directly to the modular representation guide, although the categories are not identical.
The definition · One arrow · Path algebras · A complete chain calculation · Gabriel’s theorem · Practice and solutions · BTT Mathematics Hub
Why several individually simple maps can make a difficult problem
For one linear map, a suitable choice of bases can expose its rank immediately. For a chain of two maps, the middle vector space belongs to both maps. A basis change that simplifies the first also changes the second. We cannot independently optimise every matrix while pretending their common coordinates are unrelated.
Consider a sequence of measurements that are transferred between successive stages. The dimensions tell us how many coordinates exist at each stage. The ranks tell us how much each individual map carries forward. But a quantity surviving the first map may be killed by the second. To understand the sequence, we also need the rank of the composite.
Quiver representation theory treats the full arrangement as one object. Its equivalences must respect all arrows at once. Its subobjects must be closed under all relevant maps. Its decompositions must split the entire network of vector spaces, not merely one matrix selected from it.
What exactly is a quiver representation?
A quiver is a directed graph that may have multiple arrows between the same pair of vertices and may have loops. Write Q0 for the vertices and Q1 for the arrows. Each arrow a has a source s(a) and a target t(a).
A representation assigns a vector space Vi to every vertex i and a linear map Aa:Vs(a)→Vt(a) to every arrow. The tuple d=(dim Vi) is its dimension vector.
A dimension vector is not a complete description. The same dimensions can support zero maps, injective maps, surjective maps or many non-equivalent combinations. Likewise, the quiver itself specifies where maps are allowed, not what those maps are.
This is why a network drawing alone is not a quiver representation. We need the assigned vector spaces and maps. Conversely, a list of matrices without their sources, targets and shared vertices can omit essential information.
Equivalence allows one coordinate change at each vertex
Suppose V and W represent the same quiver. An isomorphism consists of invertible linear maps Pi:Vi→Wi satisfying Pt(a)Aa=BaPs(a) for every arrow. In coordinate matrices, this gives Ba=Pt(a)AaPs(a)−1.
The equation fixes our convention: the P maps translate vectors from the first representation to the second. A passive change-of-basis convention may write the inverses in the opposite positions, but the commuting equation determines what is intended.
The key constraint is consistency at a shared vertex. If three arrows enter or leave vertex i, they all use the same Pi. Changing it separately for each arrow would compare a different problem.
One arrow: rank completely classifies the representation
Take the quiver 1→2. A representation is one linear map A:Cm→Cn. Because the source and target have independent bases, Gaussian elimination puts A into a matrix with an r×r identity block and zeros elsewhere, where r=rank A.
Three building blocks suffice. S1 has C at vertex 1 and zero at vertex 2. S2 has zero at vertex 1 and C at vertex 2. I12 has C at both vertices and the identity arrow.
A ≅ rI12 ⊕ (m−r)S1 ⊕ (n−r)S2.
The formula can be derived directly. Choose r source vectors whose images form a basis of the image. Complete the source basis with a basis of the kernel. Complete the target basis with vectors outside the image. Each paired source and image vector gives I12; each kernel vector gives S1; each unused target vector gives S2.
Two maps with the same dimensions but different structures
Let m=2 and n=3. The map A(x,y)=(x,0,0) has rank 1, so its decomposition is I12⊕S1⊕2S2. The map B(x,y)=(x,y,0) has rank 2, giving 2I12⊕S2.
Both have dimension vector (2,3), but they are not isomorphic. Rank is preserved by invertible source and target transformations. No coordinate choice can turn a one-dimensional image into a two-dimensional image.
This is also a warning against using eigenvalues for the wrong problem. A rectangular arrow map has no ordinary characteristic polynomial. Even for a square arrow between two distinct vertices, the equivalence is independent left and right transformation, not similarity. Rank, rather than an eigenvalue list, is the right invariant for the one-arrow problem.
Subrepresentations must respect the arrows
A subrepresentation chooses a subspace Ui⊆Vi at every vertex such that Aa(Us(a))⊆Ut(a) for every arrow. It is not enough to select unrelated subspaces of the correct dimensions.
In I12, the choice U1=0 and U2=C is a subrepresentation: the zero source maps into the target. It is S2. The opposite choice U1=C and U2=0 is not a subrepresentation because the identity map sends C to C, not to zero.
That one-directional condition is a source of much of the theory’s structure. The direction of an arrow affects which subrepresentations exist, even when the dimensions at its two ends are the same.
A reducible representation that refuses to split
I12 has a nonzero proper subrepresentation S2. Its quotient is S1. Nevertheless it is not S1⊕S2: the direct sum has a zero arrow, whereas I12 has an identity arrow.
The sequence 0→S2→I12→S1→0 is therefore non-split. The middle object has two simple composition factors but is indecomposable. This conclusion follows from the actual arrow, not from a diagrammatic analogy.
Irreducible means no nonzero proper subrepresentation. Indecomposable means no direct-sum decomposition into two nonzero subrepresentations. These are different requirements. For quivers, even an elementary identity arrow demonstrates the difference.
Morphisms are compatible families of maps
A morphism T:V→W consists of maps Ti:Vi→Wi such that Tt(a)Aa=BaTs(a). The T maps need not be invertible. Once bases are fixed, these are linear equations in their entries.
For the one-arrow examples, Hom(S1,I12)=0. The commuting equation forces the source scalar to vanish. In the reverse direction, Hom(I12,S1) is one-dimensional: any source scalar works because the target of the morphism is zero.
Similarly Hom(S2,I12) is one-dimensional, while Hom(I12,S2)=0. These calculations display the injection and quotient in the non-split sequence. Equal dimensions of two Hom spaces cannot be assumed merely because their arguments have been reversed.
Endomorphisms detect splitting, not necessarily simplicity
An endomorphism of I12 is a scalar at each vertex, and the identity arrow forces those scalars to agree. Therefore End(I12)≅C. Its only idempotents are zero and one.
A nontrivial direct-sum splitting would create a nontrivial projection endomorphism. Since none exists, I12 is indecomposable. Yet it is not simple, as we already found the subrepresentation S2.
This example prevents an overextension of Schur’s lemma. A simple complex representation has scalar endomorphisms under the standard finite-dimensional hypotheses, but the converse need not hold in a non-semisimple representation category.
Path algebras package the whole arrangement
A path follows arrows in their permitted directions. Include a path of length zero ei at each vertex. The path algebra C[Q] has paths as a vector-space basis and multiplies them by concatenation when their endpoints match; an incompatible product is zero.
For the chain 1→2→3, call the arrows a:1→2 and b:2→3. We write ba for the path that first follows a and then b. The six basis elements are e1,e2,e3,a,b,ba.
Our convention gives e2a=a=ae1. The opposite vertex idempotents give zero products. The identity of the path algebra is e1+e2+e3, not one selected vertex idempotent.
A representation becomes a module on V1⊕V2⊕V3: vertex idempotents project onto their spaces, arrows act by the assigned maps, and longer paths act by composites. Conversely, the spaces eiM recover the vertex spaces from a module. This standard equivalence connects quiver representations with associative-algebra representation theory. [1]
A finite quiver need not have a finite-dimensional path algebra
The chain has finitely many paths because no directed cycle can be traversed repeatedly. By contrast, one vertex with one loop has paths of lengths 0,1,2,… . Its path algebra is C[t], which is infinite-dimensional as a vector space.
Representations of that loop assign a single operator A:V→V. Because source and target are the same vertex, the allowed coordinate change is similarity P A P−1. We have recovered the ordinary classification problem for an operator, including eigenvalues and Jordan blocks.
The graph explains why the one-loop and one-arrow problems use different matrix equivalences. One arrow has independent endpoint bases; one loop has one shared basis.
Three vertices: information can begin, persist and end
Consider the equioriented chain 1→2→3. For 1≤i≤j≤3, define Iij to have a one-dimensional space at vertices i through j, zero elsewhere, and identity maps between the consecutive nonzero spaces.
There are six such interval representations: I11, I22, I33, I12, I23 and I13. Each describes a one-dimensional component that begins at i and remains present through j.
For a finite chain, every finite-dimensional representation decomposes into these interval types. This is the type-A part of the theory explained in Ringel’s account. [2] The next example makes that theorem concrete rather than using the word interval as a substitute for a calculation.
Work a complete example from maps to intervals
Take V1=C³, V2=C⁴ and V3=C³. Define A(x,y,z)=(x,y,0,0) and B(u,v,w,t)=(u,w,t). Then rank A=2, rank B=3 and rank BA=1.
Track the standard basis vectors. The first vector of V1 passes to the first vector of V2, then to the first vector of V3. This gives one I13. The second source vector passes to the second middle vector and is then killed, giving one I12.
The third and fourth middle vectors are not images of A, but B sends them to the second and third target vectors. They give two I23 components. The third source vector is killed immediately by A, giving one I11.
V ≅ I13 ⊕ I12 ⊕ 2I23 ⊕ I11.
Check each vertex: 1+1+0+1=3 at the first; 1+1+2+0=4 at the second; 1+0+2+0=3 at the third. The decomposition accounts for every coordinate and every map.
Recover the same answer from ranks alone
Let the dimensions be n1,n2,n3, the arrow ranks be a and b, and the composite rank be c. If mij denotes the multiplicity of Iij, then
m13 = c m12 = a - c m23 = b - c m11 = n1 - a m33 = n3 - b m22 = n2 - a - b + c
Only I13 contributes to the composite rank, explaining the first equation. Both I12 and I13 contribute to rank A; both I23 and I13 contribute to rank B. The remaining equations count dimensions left over at each vertex.
Substitute (n1,n2,n3)=(3,4,3), a=2, b=3, c=1. We obtain m13=1, m12=1, m23=2, m11=1 and the other two multiplicities zero, agreeing with the basis-level construction.
Why separate arrow ranks are insufficient
Use dimension vector (1,2,1) and A(z)=(z,0). First choose B1(x,y)=x. Then B1A is the identity and has rank 1. The decomposition is I13⊕I22.
Instead choose B2(x,y)=y. Now B2A=0. The decomposition becomes I12⊕I23. Both arrows in both representations have rank 1, and the dimension vectors match.
The representations are nevertheless non-isomorphic because the composite rank is invariant under compatible coordinate changes. The example isolates the missing information: not just how much each map carries, but whether the image of the first lies in the kernel of the second.
Longer chains and a rank reconstruction formula
For 1→2→···→n, let r(i,j) be the rank of the composite from vertex i to vertex j, with r(i,i)=dim Vi. An interval Iab contributes one to r(i,j) exactly when a≤i≤j≤b.
Subtracting these cumulative counts twice gives the multiplicity of the interval starting at i and ending at j:
m(i,j)=r(i,j)−r(i−1,j)−r(i,j+1)+r(i−1,j+1).
Use zero for r(0,j) and r(i,n+1). The formula is an inclusion–exclusion calculation on the intervals: first remove those that began earlier, then those that continue later, and restore those subtracted twice.
This interpretation underlies the barcode language for finite one-parameter sequences of vector spaces. It does not imply that every directed network admits a barcode of the same kind. The shape of the index structure is a mathematical hypothesis, not merely a visual preference.
Gabriel’s theorem identifies the finite-type boundary
For a finite connected quiver over an algebraically closed field, the ordinary quiver representation category has only finitely many indecomposable isomorphism classes exactly when its underlying graph is a Dynkin diagram of type A, D or E. In this finite-type case, the indecomposable dimension vectors are the positive roots of the associated root system, with one indecomposable per positive root. [2]
The theorem concerns all finite-dimensional indecomposables, not just representations with a preselected dimension vector. For a disconnected quiver, one treats connected components separately. Parallel arrows and loops are not silently ignored when determining whether the underlying graph is one of the ordinary ADE diagrams.
The orientation changes the actual maps and subrepresentation relationships, but not whether an ADE graph has finite representation type. This is a precise connection between directed linear algebra and root-system geometry.
Read the six A3 roots as six representations
For the chain A3, the six positive-root dimension vectors are (1,0,0), (0,1,0), (0,0,1), (1,1,0), (0,1,1) and (1,1,1). They are exactly the six interval types we constructed.
The first three are simple vertex representations. The last three are indecomposable but not simple under the orientation 1→2→3. Thus Gabriel’s correspondence is with indecomposables, not with irreducibles alone.
The root-language connection also appears in Lie algebra representation theory, but the labels have different roles. A positive root here labels an indecomposable quiver dimension vector; a dominant highest weight labels a finite-dimensional irreducible module in the semisimple Lie setting. The shared geometry does not make these categories identical.
A quadratic form gives a useful check
The quiver quadratic form is q(d)=Σidi²−Σa:i→jdidj. For A3 it becomes d1²+d2²+d3²−d1d2−d2d3.
For (1,1,1), the value is 3−2=1, consistent with the interval I13. For the full worked dimension vector (3,4,3), it is 9+16+9−12−12=10. That vector is not a positive root of A3, so it cannot be the dimension vector of an indecomposable in this Dynkin setting.
The number q(d) alone is not a universal simplicity test for arbitrary quivers or quivers with relations. Its interpretation depends on the representation category. Our explicit interval decomposition is stronger evidence than a quadratic-form value detached from its hypotheses.
Two parallel arrows already create infinitely many types
The Kronecker quiver has two arrows from vertex 1 to vertex 2. Give both vertices the one-dimensional space C. The two arrow maps are then scalars a and b.
A basis change multiplies both scalars by the same nonzero ratio. Thus pairs (1,λ), with λ∈C, are pairwise non-isomorphic. There is also the type (0,1), often viewed as the missing point at infinity. Their parameter space is the projective line.
Each nonzero pair is indecomposable. A nontrivial decomposition of dimension vector (1,1) would have to be S1⊕S2, whose two arrow maps are both zero. Therefore a nonzero pair cannot split that way.
This explicit family explains why the parallel arrows matter. The diagram has only two vertices and the spaces have dimension one, yet there are infinitely many indecomposable isomorphism classes. Small dimensions do not guarantee a finite classification.
Relations change the question again
Sometimes we require specified paths to agree or vanish. In the chain 1→2→3, imposing ba=0 requires BA=0 in every representation. Algebraically, we replace the path algebra by its quotient by the ideal generated by ba.
The interval I13, whose two arrows are identities, is then forbidden because its composite is nonzero. The representation I12⊕I23 is allowed. The relation has changed the category, not merely selected a different coordinate basis.
Likewise, one loop with the relation t²=0 allows operators N satisfying N²=0, restricting their Jordan blocks to sizes at most two. Gabriel’s ordinary-quiver statement must not be treated as a complete classification theorem for every quotient by arbitrary relations.
Computation should preserve the shared-coordinate constraints
To compute Hom(V,W), introduce an unknown matrix Ti at every vertex and write the commuting equation for every arrow. Solve the resulting linear system. An isomorphism requires a solution in which every Ti is invertible.
Computing an endomorphism space is therefore a linear problem. Finding a nontrivial idempotent inside it, which would reveal a splitting, imposes the additional equation T²=T. That step is not automatically the same linear calculation.
For chains, rank computations and interval formulas provide a particularly direct route. For more complicated quivers, a dimension vector and a few ranks may leave substantial ambiguity. Report the invariants actually computed rather than calling a partial collection a complete classification.
Practice: maps, decompositions and counterexamples
1. One-arrow decomposition
A:C⁴→C³ has rank 2. Decompose its quiver representation. Solution: 2I12⊕2S1⊕S2. At the source the dimensions sum to 2+2=4, and at the target to 2+1=3. The two identity components account for the rank.
2. A subrepresentation that does not exist
Why is C→0 not a subrepresentation of the identity arrow C→C? Solution: the identity sends the proposed source subspace C onto C, which is not contained in the proposed target zero space. Dimensions alone do not establish a subrepresentation.
3. The zero-arrow contrast
What is C→C with zero arrow? Solution: it is S1⊕S2. It has the same dimension vector as I12, but the arrow rank is zero instead of one, so the two representations are not isomorphic.
4. Path count
How many basis paths does 1→2→3 have? Solution: three length-zero paths, two arrows and one length-two path, for a total of six. The answer would change for a loop, because arbitrarily long paths would become possible.
5. Composite rank determines the long interval
For a three-vertex chain, what is the multiplicity of I13? Solution: rank BA. Among the six interval types, only I13 contributes a nonzero composite from the first vertex to the third.
6. A second complete chain example
The dimensions are (2,3,2), arrow ranks are 1 and 2, and composite rank is 1. Find all multiplicities. Solution: m13=1, m12=0, m23=1, m11=1, m33=0 and m22=1. Thus the representation is I13⊕I23⊕I11⊕I22, which reconstructs all dimensions and ranks.
7. Detect impossible rank data
Can dimensions (1,1,1), arrow ranks 1 and 1, and composite rank 0 occur? Solution: no. Both arrows would be nonzero maps between one-dimensional spaces and hence invertible, so the composite rank would be 1. The multiplicity formula also gives m22=1−1−1+0=−1, exposing the inconsistency.
8. Kronecker parameters
Are the arrow pairs (1,2) and (1,3) isomorphic over C? Solution: no. An isomorphism multiplies both entries by one common nonzero scalar. Matching the first entry forces that scalar to be 1, which cannot change 2 to 3.
9. A relation excludes a component
Under the relation BA=0, can I13 occur as a direct summand of the three-vertex chain? Solution: no. On that summand the composite would be the identity. Direct-sum decomposition cannot cancel this identity against a separate summand.
10. Scalar endomorphisms do not prove simplicity
Give a representation with End(V)=C that is not simple. Solution: I12. Its endomorphisms must use one common scalar at both vertices, yet it contains S2 as a proper nonzero subrepresentation.
11. Loop versus arrow
Why is rank not a complete invariant for one loop? Solution: the one vertex permits only similarity, not independent row and column transformations. For example, the one-dimensional operators 1 and 2 both have rank 1 but are not similar. Their eigenvalues differ.
12. Count A3 indecomposables
How many indecomposable types occur for 1→2→3? Solution: six, one for each interval [i,j] with 1≤i≤j≤3. Only three of them are simple; the other three show that counting simples alone misses the representation category’s direct-sum building blocks.
Teach the shared structure, not just the individual matrices
Begin with one arrow and ask the learner to construct its decomposition using kernel and image bases. Then present two chain examples with the same arrow ranks but different composite ranks. Ask for the precise invariant that distinguishes them before revealing the interval labels.
Next use I12 to separate subrepresentation from direct summand. Require the learner to test the arrow condition and explain why the candidate complement fails. This small example makes later extension language concrete.
Only after these calculations introduce Gabriel’s theorem. The theorem then answers a question the learner has already encountered: when does the supply of indecomposable building blocks remain finite? Finish with the two-arrow Kronecker family, where a single parameter shows exactly why the finite list disappears.
Mathematical references and scope
[1] Pavel Etingof and collaborators, Introduction to Representation Theory, for quivers, path algebras and the module viewpoint. [2] Claus Michael Ringel, The Representation Theory of Dynkin Quivers: Three Contributions, for Dynkin quivers, interval representations and Gabriel’s finite-type theorem. The explicit rank calculations, Hom calculations and parameter comparisons in this guide are worked in full so that the abstract statements can be tested on concrete examples.
Continue Representation Mathematics — Batch 03
Compare quiver decomposition with Finite-Group Fourier Analysis, where the complex group algebra is semisimple. Study combinatorial labels in Symmetric Group Representations and Young Tableaux. Follow the extraction of unchanged polynomial quantities in Invariant Theory, Reynolds Operators and Molien Series. Return to the BTT Mathematics Learning Hub.