# representations of a bound quiver

Let $(Q,I)$ be a bound quiver (http://planetmath.org/AdmissibleIdealsBoundQuiverAndItsAlgebra) over a field $k$.

Let $\mathbb{V}$ be a representation of $Q$ over $k$ composed by $\{f(q)\}_{q\in Q_{1}}$ a family of linear maps. If

 $w=(\alpha_{1},\ldots,\alpha_{n})$

is a path in $Q$, then we have the evaluation map

 $f_{w}=f(\alpha_{n})\circ f(\alpha_{n-1})\circ\cdots\circ f(\alpha_{2})\circ f(% \alpha_{1}).$

For stationary paths we define $f_{e_{x}}:V_{x}\to V_{x}$ by $f_{e_{x}}=0$. Also, note that if $\rho$ is a relation (http://planetmath.org/RelationsInQuiver) in $Q$, then

 $\rho=\sum_{i=1}^{m}\lambda_{i}\cdot w_{i}$

where all $w_{i}$’s have the same source and target. Thus it makes sense to talk about evaluation in $\rho$, i.e.

 $f_{\rho}=\sum_{i=1}^{n}\lambda_{i}\cdot f_{w_{i}}.$

In particular

 $f_{\rho}:V_{s(w_{i})}\to V_{t(w_{i}})$

is a linear map.

Recall that the ideal $I$ is generated by relations (see this entry (http://planetmath.org/PropertiesOfAdmissibleIdeals)) $\{\rho_{1},\ldots,\rho_{n}\}$.

Definition. A representation $\mathbb{V}$ of $Q$ over $k$ with linear mappings $\{f(q)\}_{q\in Q_{1}}$ is said to be bound by $I$ if

 $f_{\rho_{i}}=0$

for every $i=1,\ldots,n$.

It can be easily checked, that this definition does not depend on the choice of (relation) generators of $I$.

The full subcategory of the category of all representations which is composed of all representations bound by $I$ is denoted by $\mathrm{REP}(Q,I)$. It can be easily seen, that it is abelian.

Title representations of a bound quiver RepresentationsOfABoundQuiver 2013-03-22 19:16:51 2013-03-22 19:16:51 joking (16130) joking (16130) 6 joking (16130) Definition msc 14L24