# 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. 1.

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

2. 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 $x\not\in X$.

Definition 2. If $(Q,\tau)$ is a translation quiver, then a pair $(Q^{\prime},\tau^{\prime})$ is called a 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:

 $\xymatrix{1\ar[rd]&&2\ar[rd]&&3\ar[rd]&&4\\ &5\ar[rd]\ar[ru]&&6\ar[r]\ar[ru]&7\ar[r]&8\ar[ru]&\\ &&9\ar[ru]}$

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

 $\xymatrix{&2\ar[rd]&\\ 5\ar[ru]\ar[rd]&&6\\ &9\ar[ru]&}$

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 TranslationQuiver 2013-03-22 19:17:53 2013-03-22 19:17:53 joking (16130) joking (16130) 5 joking (16130) Definition msc 14L24