# modules over bound quiver algebra and bound quiver representations

Let $(Q,I)$ be a bound quiver over a fixed field $k$. Denote by $\mathrm{Mod}A$ (resp. $\mathrm{mod}A$) the category  of all (resp. all finite-dimensional) (right) modules over algebra  $A$ and by $\mathrm{REP}_{Q,I}$ (resp. $\mathrm{rep}_{Q,I}$) the category of all (resp. all finite-dimensional, (see this entry (http://planetmath.org/QuiverRepresentationsAndRepresentationMorphisms) for details) bound representations.

We will also allow $I=0$ (which is an admissible ideal only if lengths of paths in $Q$ are bounded, in particular when $Q$ is finite and acyclic). In this case bound representations are simply representations.

If $Q$ is a connected and finite quiver, $I$ and admissible ideal in $kQ$ and $A=kQ/I$, then there exists a $k$-equivalence of categories

 $F:\mathrm{Mod}A\to\mathrm{REP}_{Q,I}$

which restricts to the equivalence of categories

 $F^{\prime}:\mathrm{mod}A\to\mathrm{rep}_{Q,I}.$

Sketch of the proof. We will only define functor  $F$ and its quasi-inverse  $G$. For proof that $F$ is actually an equivalence please see [1, Theorem 1.6] (this not difficult, but rather technical proof).

Let $e_{a}$ be a stationary path in $a\in Q_{0}$ and put $\epsilon_{a}=e_{a}+I\in A$. Now if $M$ is a module in $\mathrm{Mod}A$, then define a representation

 $F(M)=(M_{a},M_{\alpha})$

by putting $M_{a}=M\epsilon_{a}$ ($M$ is a right module over $A$). Now for an arrow $\alpha\in Q_{1}$ define $M_{\alpha}:M_{s(\alpha)}\to M_{t(\alpha)}$ by putting $M_{\alpha}(x)=x\overline{\alpha}$, where $\overline{\alpha}=\alpha+I\in A$. It can be shown (see ) that $F(M)$ is a bound representation.

On module morphisms  $F$ acts as follows. If $f:M\to M^{\prime}$ is a module morphism, then define

 $F(f)=(f_{a})_{a\in Q_{0}}$

where $f_{a}:M_{a}\to M^{\prime}_{a}$ is a restriction  , i.e. $f_{a}(x)=f(x)$. It can be shown that $f_{a}$ is well-defined (i.e. $f_{a}(x)\in M^{\prime}_{a}$) and in this manner $F$ is a functor.

 $G(M)=\bigoplus_{a\in Q_{0}}M_{a}.$

Now we will define right $kQ$-module structure  on $G(M)$. For a stationary path $e_{a}$ in $a\in Q_{0}$ and for $x=(x_{a})\in G(M)$ put

 $x\cdot e_{a}=x_{a}.$

Now for a path $w=(a_{1},\ldots,a_{n})$ from $a$ to $b$ in $kQ$ we consider the evaluation map (see this entry (http://planetmath.org/RepresentationsOfABoundQuiver) for details) $f_{w}:M_{a}\to M_{b}$ and we put

 $(x\cdot w)_{c}=\delta_{bc}f_{w}(x_{a}),$

where $\delta_{bc}$ denotes the Kronecker delta. It can be shown that $G(M)$ is a $kQ$-module with the property that $G(M)I=0$. In particular $G(M)$ is a $kQ/I$-module.

Now, if $f=(f_{a}):M\to M^{\prime}$ is a morphism of representations then we define

 $G(f)=\bigoplus_{a\in Q_{0}}f_{a}:G(M)\to G(M).$

Also, it follows easily from definitions that both $F$ and $G$ take finite-dimensional objects to finite-dimensional.

It remains to show that these two functors are quasi-inverse. For the proof please see [1, Theorem 1.6]. $\square$

Corollary. If $Q$ is a finite, connected and acyclic quiver, then there exists an equivalence of categories $\mathrm{Mod}kQ\simeq\mathrm{REP}_{Q}$ which restricts to the equivalence of categories $\mathrm{mod}kQ\simeq\mathrm{rep}_{Q}$.

Proof. Since $Q$ is finite and acyclic, then the zero ideal   $I=0$ is admissible (because lengths of paths are bounded, so $R_{Q}^{m}=0$ for some $m\geqslant 1$, where $R_{Q}$ denotes the arrow ideal). The thesis follows from the theorem. $\square$

## References

Title modules over bound quiver algebra and bound quiver representations ModulesOverBoundQuiverAlgebraAndBoundQuiverRepresentations 2013-03-22 19:17:34 2013-03-22 19:17:34 joking (16130) joking (16130) 5 joking (16130) Theorem msc 14L24