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The finite representation type of a quiver condition means there are only finitely many isomorphism classes of indecomposable finite-dimensional $kQ$-modules. Equivalently, every finite-dimensional module is a direct sum of members of a finite list of indecomposables.

For the $E_6$ case, we give the reflection argument underlying the Gabriel theorem, rather than using its conclusion as the proof. Since the underlying graph is a tree, every orientation is acyclic. At a sink $i$, writefor the sum of the incoming arrow maps. If an indecomposable $V$ is not the simple $S_i$, then $h$ is surjective: a nonzero complementary subspace to its image at the sink would be a direct summand supported only at $i$. The Bernstein–Gelfand–Ponomarev reflection functor reverses the arrows and replaces $V_i$ by $\ker h$, with reversed maps the component projections. The source reflection replaces that kernel by the cokernel of its inclusion into $⨁V_j$, recovering $V_i$. The inclusion is injective, so the reflected representation has no direct summand $S_i$ at the new source. If it decomposed, applying the inverse would decompose $V$. Thus reflection preserves indecomposability except that it annihilates $S_i$.

On a dimension vector of a quiver representation, it acts by the integral simple reflectionLet $B=2I-\mathrm{Adj}(E_6)$. These reflections preserve the positive-definite quadratic form $d^{\mathsf{T}}Bd$. Positivity can be checked without a classification theorem: label the graph by the chain $1-2-3-4-5$ and attach vertex $6$ to $3$. The leading principal minors are $2,3,4,5,6,3$, all positive.

Choose a topological ordering so arrows point from smaller to larger indices. Reflect successively at sinks $6,5,\dots ,1$. This restores the orientation after one full cycle. The transformation of dimension vectors is the Coxeter element $C=s_1{s}_2\cdots s_6$. If $U$ has diagonal entries one and entry $-1$ above the diagonal at an arrow, direct multiplication of the reflection matrices givesThus $C$ has no fixed vector. It is an integral isometry of a positive definite form, so it has finite order $h$: the image of each lattice basis vector lies in a finite set of vectors of the same norm, giving only finitely many possible isometries. Since $C^h=I$ and $I-C$ is invertible,If an indecomposable survived all $h$ reflection cycles, its dimension vectors $d,Cd,\dots ,C^{h-1}d$ would all be nonzero and componentwise nonnegative, contradicting their zero sum. Therefore it reaches a simple at one of the finitely many reflection steps and is then annihilated. Applying the inverse reflections uniquely reconstructs it from that simple. There are at most $6h$ such stopping positions, provingThis supplies the requested finite-type proof sketch.

A vertex $i$ of a quiver is a sink when no arrow starts at $i$. A representation of a quiver assigns a vector space $V_j$ to every vertex and a linear map $V_a:V_j\to V_{\ell }$ to every arrow $a:j\to \ell$. A morphism $f:V\to W$ is a family of linear maps $f_j:V_j\to W_j$ such that $f_{\ell }{V}_a=W_a{f}_j$ for every arrow. The quiver has finite representation type when it has only finitely many isomorphism classes of indecomposable finite-dimensional representations.

At a sink $i$, the Bernstein–Gelfand–Ponomarev reflection functor replacesand reverses the arrows ending at $i$; the new arrow maps are the kernel inclusion followed by the coordinate projections. Every representation is a direct sum of copies of the simple representation $S_i$ and a representation for which the displayed incoming map is surjective. On the latter representations, reflection at the resulting source, using the corresponding cokernel, is inverse up to natural isomorphism. Thus reflection gives a bijection between indecomposable representations other than $S_i$ on the two sides. Adding the one omitted simple representation on each side proves that reversing all arrows into a sink preserves finite representation type.

For the four-arrow star $Q_1$, take $V_1=\cdots =V_4=k$, $V_5=k^2$, and let the four arrows have imagesAn endomorphism must preserve the first two lines, so its map on $V_5$ is diagonal. Preserving the third forces its diagonal entries to agree, and then all vertex maps are multiplication by that same scalar. The endomorphism ring is therefore $k$, so this representation is a brick module and hence indecomposable. Any isomorphism between parameters preserves the first three labelled lines; the induced projective linear transformation is therefore the identity, and the fourth line gives $\lambda =\mu$. Since $k$ is infinite, this is an infinite family of pairwise nonisomorphic indecomposables. Hence $Q_1$ is not of finite representation type. Repeatedly applying the reflection result to sinks or, dually, to sources shows that every orientation obtained by reversing some of its arrows also has infinite representation type.