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:
$$\text{xymatrix}1\text{ar}[rd]\mathrm{\&}\mathrm{\&}2\text{ar}[rd]\mathrm{\&}\mathrm{\&}3\text{ar}[rd]\mathrm{\&}\mathrm{\&}4\mathrm{\&}5\text{ar}[rd]\text{ar}[ru]\mathrm{\&}\mathrm{\&}6\text{ar}[r]\text{ar}[ru]\mathrm{\&}7\text{ar}[r]\mathrm{\&}8\text{ar}[ru]\mathrm{\&}\mathrm{\&}\mathrm{\&}9\text{ar}[ru]$$ 
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
$$\text{xymatrix}\mathrm{\&}2\text{ar}[rd]\mathrm{\&}5\text{ar}[ru]\text{ar}[rd]\mathrm{\&}\mathrm{\&}6\mathrm{\&}9\text{ar}[ru]\mathrm{\&}$$ 
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).
Title  translation quiver 

Canonical name  TranslationQuiver 
Date of creation  20130322 19:17:53 
Last modified on  20130322 19:17:53 
Owner  joking (16130) 
Last modified by  joking (16130) 
Numerical id  5 
Author  joking (16130) 
Entry type  Definition 
Classification  msc 14L24 