# path algebra of a quiver

Let $Q=({Q}_{0},{Q}_{1},s,t)$ be a quiver, i.e. ${Q}_{0}$ is a set of vertices, ${Q}_{1}$ is a set of arrows, $s:{Q}_{1}\to {Q}_{0}$ is a source function and $t:{Q}_{1}\to {Q}_{0}$ is a target function.

Recall that a path of length $l\u2a7e1$ from $x$ to $y$ in $Q$ is a sequence^{} of arrows $({a}_{1},\mathrm{\dots},{a}_{l})$ such that

$$s({a}_{1})=x;t({a}_{l})=y;$$ |

$$t({a}_{i})=s({a}_{i+1})$$ |

for any $i=1,2,\mathrm{\dots},l-1,l$.

Also we allow paths of length $0$, i.e. stationary paths.

If $a=({a}_{1},\mathrm{\dots},{a}_{l})$ and $b=({b}_{1},\mathrm{\dots},{b}_{k})$ are two paths such that $t({a}_{l})=s({b}_{1})$ then we say that $a$ and $b$ are compatibile and in this case we can form another path from $a$ and $b$, namely

$$a\circ b=({a}_{1},\mathrm{\dots},{a}_{l},{b}_{1},\mathrm{\dots},{b}_{k}).$$ |

Note, that the length of $a\circ b$ is a sum of lengths of $a$ and $b$. Also a path $a=({a}_{1},\mathrm{\dots},{a}_{l})$ of positive length is called a cycle if $t({a}_{l})=s({a}_{1})$. In this case we can compose $a$ with itself to produce new path.

Also if $a$ is a path from $x$ to $y$ and ${e}_{x},{e}_{y}$ are stationary paths in $x$ and $y$ respectively, then we define $a\circ {e}_{y}=a$ and ${e}_{x}\circ a=a$.

Let $kQ$ be a vector space with a basis consisting of all paths (including stationary paths). For paths $a$ and $b$ define multiplication^{} as follows:

If $a$ and $b$ are compatible, then put $ab=a\circ b$ and put $ab=0$ otherwise. This operation^{} extendes bilinearly to entire $kQ$ and it can be easily checked that $kQ$ becomes an associative algebra in this manner called the path algebra^{} of $Q$ over $k$.

Title | path algebra of a quiver |
---|---|

Canonical name | PathAlgebraOfAQuiver |

Date of creation | 2013-03-22 19:16:19 |

Last modified on | 2013-03-22 19:16:19 |

Owner | joking (16130) |

Last modified by | joking (16130) |

Numerical id | 4 |

Author | joking (16130) |

Entry type | Definition |

Classification | msc 14L24 |