Path Algebras

A quiver is a directed multigraph — vertices, and arrows between them, with loops and parallel arrows allowed. Its path algebra $kQ$ has a basis of all directed paths, one trivial path $e_v$ per vertex included, and the product is concatenation:

$$p \cdot q = \begin{cases} pq & \text{if } q \text{ starts where } p \text{ ends} \\ 0 & \text{otherwise}\end{cases}$$

Like the incidence algebra, most products are zero. Unlike every other family here, this one need not be finite-dimensional.

One loop and the algebra is infinite

$kQ$ is finite-dimensional exactly when $Q$ is acyclic. A single loop gives the paths $e, a, a^2, a^3, \ldots$ — the Jordan quiver, whose path algebra is the polynomial ring $k[x]$. There is no finite basis and no dimension, so Basis and AlgebraDimension have nothing to return and leave the call standing rather than enumerating forever.

#Acyclic or not

The linear quiver is acyclic and has a dimension; the Jordan quiver has a loop, so its dimension does not exist and the call stays put.

In$\mathrm{QuiverIsAcyclic}(\mathrm{LinearQuiver}(4))$Out$\mathrm{True}$In$\mathrm{QuiverIsAcyclic}(\mathrm{JordanQuiver})$Out$\mathrm{False}$In$\mathrm{AlgebraDimension}(\mathrm{PathAlgebra}(\mathrm{LinearQuiver}(4)))$Out$10$In$\mathrm{AlgebraDimension}(\mathrm{PathAlgebra}(\mathrm{JordanQuiver}))$Out$\mathrm{AlgebraDimension}(\mathrm{PathAlgebra}(\mathrm{JordanQuiver}))$

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<notatio-cell value="QuiverIsAcyclic(LinearQuiver(4))" data-allow-mismatch="children" /> <notatio-cell value="QuiverIsAcyclic(JordanQuiver)" data-allow-mismatch="children" /> <notatio-cell value="AlgebraDimension(PathAlgebra(LinearQuiver(4)))" data-allow-mismatch="children" /> <notatio-cell value="AlgebraDimension(PathAlgebra(JordanQuiver))" data-allow-mismatch="children" />

That distinction is the whole reason this family is worth having next to the others: every previous one was finite by construction, and this is the first place where "there is no answer" is a structural fact about the input rather than a missing feature.

Aₙ's path algebra is a chain's incidence algebra

The linear quiver $1 → 2 → \ldots → n$ has exactly one path from $i$ to $j$ when $i ≤ j$, and none otherwise. So its paths are the intervals of a chain, and $kA_n$ is the incidence algebra of Chain(n) — same dimension $\binom{n+1}{2}$, same product, two libraries describing one object from different directions.

#The same algebra, twice

Both are C(7,2) = 21.

In$\mathrm{AlgebraDimension}(\mathrm{PathAlgebra}(\mathrm{LinearQuiver}(6)))$Out$21$In$\mathrm{AlgebraDimension}(\mathrm{IncidenceAlgebra}(\mathrm{Chain}(6)))$Out$21$

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<notatio-cell value="AlgebraDimension(PathAlgebra(LinearQuiver(6)))" data-allow-mismatch="children" /> <notatio-cell value="AlgebraDimension(IncidenceAlgebra(Chain(6)))" data-allow-mismatch="children" />

The path count is verified a second way in the tests, against powers of the adjacency matrix: paths of length $ℓ$ are the entries of $A^ℓ$, and an acyclic quiver on $n$ vertices has none longer than $n-1$. Enumeration and linear algebra agree.

Composition, and zero

#Concatenate or annihilate

1→2 then 2→3 composes. The other order does not meet, so it is zero — not an error.

In$\mathrm{QuiverCompose}(\mathrm{LinearQuiver}(4), \mathrm{QuiverPath}(1, \bigl\lbrack0\bigr\rbrack), \mathrm{QuiverPath}(2, \bigl\lbrack1\bigr\rbrack))$Out$\mathrm{QuiverPath}(1, \bigl\lbrack0, 1\bigr\rbrack)$In$\mathrm{QuiverCompose}(\mathrm{LinearQuiver}(4), \mathrm{QuiverPath}(2, \bigl\lbrack1\bigr\rbrack), \mathrm{QuiverPath}(1, \bigl\lbrack0\bigr\rbrack))$Out$0$

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<notatio-cell value="QuiverCompose(LinearQuiver(4), QuiverPath(1, [0]), QuiverPath(2, [1]))" data-allow-mismatch="children" /> <notatio-cell value="QuiverCompose(LinearQuiver(4), QuiverPath(2, [1]), QuiverPath(1, [0]))" data-allow-mismatch="children" />

A path is written QuiverPath(start, [arrow indices]), so the trivial path at $v$ is QuiverPath(v, []). Those trivial paths are the local identities:

$$e_{\text{start}}\,p \;=\; p \;=\; p\,e_{\text{end}},$$

and together they sum to the identity of the algebra.

Parallel arrows are distinct

Quiver(n, [[from,to],…]) builds one explicitly, and arrows are indexed by position — so two arrows with the same endpoints are two different basis elements. The Kronecker quiver is the smallest example: two vertices, two parallel arrows, four paths.

#The Kronecker quiver

Four paths: two trivial, two arrows. The two arrows do not compose with each other, so their product is zero.

In$\mathrm{AlgebraDimension}(\mathrm{PathAlgebra}(\mathrm{Quiver}(2, \bigl\lbrack\bigl\lbrack1, 2\bigr\rbrack, \bigl\lbrack1, 2\bigr\rbrack\bigr\rbrack)))$Out$4$In$\mathrm{Basis}(\mathrm{PathAlgebra}(\mathrm{KroneckerQuiver}))$Out$\bigl\lbrack\mathrm{QuiverPath}(1, \bigl\lbrack\bigr\rbrack), \mathrm{QuiverPath}(1, \bigl\lbrack0\bigr\rbrack), \mathrm{QuiverPath}(1, \bigl\lbrack1\bigr\rbrack), \mathrm{QuiverPath}(2, \bigl\lbrack\bigr\rbrack)\bigr\rbrack$

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<notatio-cell value="AlgebraDimension(PathAlgebra(Quiver(2, [[1, 2], [1, 2]])))" data-allow-mismatch="children" /> <notatio-cell value="Basis(PathAlgebra(KroneckerQuiver))" data-allow-mismatch="children" />

Things worth knowing

Named quivers are recognised, not declared. JordanQuiver and KroneckerQuiver are bare symbols. Declaring them as nullary functions gave them the type () -> value, which then failed PathAlgebra's value parameter — an undeclared symbol types as unknown and passes. Same lesson as the named hypercomplex algebras.

Composition takes its quiver. A path does not know which quiver it belongs to, so QuiverCompose(quiver, p, q) names it rather than riding the shared ordered product.

Not built yet. Representations of a quiver — the actual subject, and where Gabriel's theorem lives: a connected quiver has finitely many indecomposable representations exactly when its underlying graph is a Dynkin diagram of type $A$, $D$ or $E$. Also quotients by an admissible ideal, which is how a cyclic quiver is made finite-dimensional in practice.