translation quiver

Let $Q=(Q_0,Q_1,s,t)$ be a locally finite quiver without loops. Recall that a loop is an arrow $\alpha$ such that $s(\alpha )=t(\alpha )$. Let $X,Y\subseteq Q_0$.

Definition 1. A pair $(Q,\tau )$ is said to be a translation quiver iff the following holds:

1. $\tau :X\to Y$ is a bijection;

2. If $x\in X$ and $y\in x^{-}$ is a direct predecessor of $x$, then the number of arrows from $y$ to $x$ is equal to the number of arrows from $\tau (x)$ to $y$.

If $(Q,\tau )$ is a translation quiver then we will say that $\tau (x)$ exists if $x\in X$ and $\tau (x)$ does not exist (or it is not defined) if $x\notin X$.

Definition 2. If $(Q,\tau )$ is a translation quiver, then a pair $(Q^{\prime },{\tau }^{\prime })$ is called a translation subquiver if it is a translation quiver, $Q^{\prime }$ is a full subquiver (http://planetmath.org/SubquiverAndImageOfAQuiver) of $Q$ and ${\tau }^{\prime }(x)=\tau (x)$ whenever $x$ is a vertex in $Q^{\prime }$ such that $\tau (x)$ exists and belongs to $Q^{\prime }$.

Example. Let $Q$ be the following quiver:

If we put $X=\{2,3,4,6,8\}$, $Y=\{1,2,3,5,6\}$ and

$$\tau (2)=1;\tau (3)=2;\tau (4)=3;$$

$$\tau (6)=5;\tau (8)=6;$$

then the pair $(Q,\tau )$ is a translation quiver and

is its translation subquiver, where ${\tau }^{\prime }(6)=5$.

Remark. It is common to write translation quivers as in example. This means that $Q$ is ,,oriented” to the right and in rows we have vertices such that ,,jumping” two places to the left gives us $\tau$ of this vertex. Note that in the example the vertex $7$ is not written in the same row as $9$ because $\tau (7)$ is not $9$ (indeed, $\tau (7)$ is not defined).