representations of a bound quiver

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

Let $𝕍$ be a representation of $Q$ over $k$ composed by ${\{f(q)\}}_{q\in Q_1}$ a family of linear maps. If

$$w=({\alpha }_1,\mathrm{\dots },{\alpha }_n)$$

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

$$f_w=f({\alpha }_n)\circ f({\alpha }_{n-1})\circ \mathrm{\cdots }\circ f({\alpha }_2)\circ f({\alpha }_1).$$

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)) $\{{\rho }_1,\mathrm{\dots },{\rho }_n\}$.

Definition. A representation $𝕍$ 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 $i=1,\mathrm{\dots },n$.

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
Canonical name	RepresentationsOfABoundQuiver
Date of creation	2013-03-22 19:16:51
Last modified on	2013-03-22 19:16:51
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	6
Author	joking (16130)
Entry type	Definition
Classification	msc 14L24