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 $l⩾1$ 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