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 $k Q$ 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 $k Q$ and it can be easily checked that $k Q$ becomes an associative algebra in this manner called the path algebra of $Q$ over $k$ .