# 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 $l\geqslant 1$ from $x$ to $y$ in $Q$ is a sequence of arrows $(a_{1},\ldots,a_{l})$ such that

 $s(a_{1})=x;\ \ \ t(a_{l})=y;$
 $t(a_{i})=s(a_{i+1})$

for any $i=1,2,\ldots,l-1,l$.

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

If $a=(a_{1},\ldots,a_{l})$ and $b=(b_{1},\ldots,b_{k})$ 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

 $a\circ b=(a_{1},\ldots,a_{l},b_{1},\ldots,b_{k}).$

Note, that the length of $a\circ b$ is a sum of lengths of $a$ and $b$. Also a path $a=(a_{1},\ldots,a_{l})$ 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 PathAlgebraOfAQuiver 2013-03-22 19:16:19 2013-03-22 19:16:19 joking (16130) joking (16130) 4 joking (16130) Definition msc 14L24