QuiverCompose

Concatenate two paths — or zero, when $q$ does not start where $p$ ends. Most products in a path algebra are zero, and that is the structure rather than a failure.

WikipediaQuiver (mathematics)↗WikidataQ493980↗Encyclopedia of MathematicsQuiver↗nLabquiver↗

QuiverCompose(quiver, p, q)$p$ then $q$, or 0

Typed signatures

Type	Package	Applies
(expression<LinearQuiver> | expression<Quiver> | symbol, number, number) -> number	quiver	always

Domain: Path algebras

Details

• Associative, and the trivial paths act as local identities
• Zero is an honest answer here — it is returned as 0, not left symbolic
• The quiver is an argument because a path carries no reference to its own quiver

Examples

Basic #

$1\to2$ then $2\to3$

live ↗

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

Properties #

the ends do not meet

live ↗

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

a trivial path is a local identity

live ↗

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

About

Examples	3
Crosswalk	WikipediaWikidataEncyclopedia of MathematicsnLab

See also: QuiverPath, QuiverIsAcyclic