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.

Theorem. 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 ${ϵ}_a=e_a+I\in A$. Now if $M$ is a module in $\mathrm{Mod}A$, then define a representation

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

by putting $M_a=M{ϵ}_a$ ($M$ 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 [1]) 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 $f_a:M_a\to M_a^{\prime }$ is a restriction, i.e. $f_a(x)=f(x)$. It can be shown that $f_a$ is well-defined (i.e. $f_a(x)\in M_a^{\prime }$) and in this manner $F$ is a functor.

The inverse functor is defined on objects as follows: for a representation $(M_a,M_{\alpha })$ put

$$G(M)=\underset{a\in Q_0}{\oplus }{M}_a.$$

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 $w=(a_1,\mathrm{\dots },a_n)$ 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

$$G(f)=\underset{a\in Q_0}{\oplus }{f}_a:G(M)\to G(M).$$

It can be shown that $G(f)$ is indeed an $A$-homomorphism and that $G$ is a functor.

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]. $\mathrm{\square }$

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 $m⩾1$, where $R_Q$ denotes the arrow ideal). The thesis follows from the theorem. $\mathrm{\square }$

Title	modules over bound quiver algebra and bound quiver representations
Canonical name	ModulesOverBoundQuiverAlgebraAndBoundQuiverRepresentations
Date of creation	2013-03-22 19:17:34
Last modified on	2013-03-22 19:17:34
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	5
Author	joking (16130)
Entry type	Theorem
Classification	msc 14L24