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Take sources $k,k,k$, sink $k^2$, and arrows sending $1$ to $e_1,e_2,e_1+e_2$. An endomorphism preserving these three lines must be scalar, so the representation has endomorphism ring $k$ and is a brick module. It is the only indecomposable of the three-subspace quiver with a two-dimensional sink.

The four-subspace quiver has four sources with arrows into one sink. Four one-dimensional sources mapped to the lines $k{e}_1$, $k{e}_2$, $k(e_1+e_2)$ and $k(\lambda e_1+e_2)$ in $k^2$ give a brick module for every $\lambda$. Preserving the first three lines forces scalar endomorphisms. Its Tits form of a quiver is zero, so the Ringel form gives one-dimensional self-extensions.

The four-subspace quiver has four one-dimensional sources and a two-dimensional sink. Its Tits form of a quiver at this dimension vector is $q=4\cdot 1^2+2^2-4(1\cdot 2)=0$.

An endomorphism comprises source scalars $a_1,\dots ,a_4$ and a sink matrix $T$. The first two columns $e_1,e_2$ force $T=\mathrm{diag}(a_1,a_2)$. The third column $e_1+e_2$ then forces $a_1=a_2=a_3$, making $T$ scalar. Since the fourth column $(\lambda ,1{)}^T$ is always nonzero, its scalar is also the same, for every $\lambda$.

Thus ${\mathrm{End}}_Q(V)=k$, so the representation is a brick module. The Ringel form givesThis conclusion also covers $\lambda =0,1$, where the fourth line repeats one of the earlier lines; the first three lines already force scalar endomorphisms.

For the Kronecker quiver representation with arrows $I$ and $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$, an endomorphism $(P,Q)$ satisfies $Q=P$ and $PN=NP$. Solving the second equation givesThus this representation is indecomposable but is not a brick: its endomorphism ring is a local endomorphism ring, while the nonzero endomorphism $N$ is nilpotent.

For a general indecomposable non-brick, the proof of Ringel lemma on bricks finds a proper indecomposable submodule with nonzero self-extensions. Repetition in strictly decreasing dimension reaches a brick module $Y\subset X$ with ${\mathrm{Ext}}_Q^1(Y,Y)\ne 0$. The linked proof supplies the minimal-rank, retraction and hereditary-extension steps.

Now assume the Tits form of a quiver is positive definite. If an indecomposable $X$ were not a brick, this $Y$ would give the contradictionHence $X$ is a brick. For its nonzero dimension vector $\mathbf{n}$, positivity and integrality then implyThus every indecomposable in this case is a rigid brick. This deduction uses the Ringel lemma on bricks and the Ringel form, without requiring the full Gabriel theorem.

A nonzero finite-dimensional representation is a brick module when its endomorphism ring is a division algebra. Over the algebraically closed field $k$, this means ${\mathrm{End}}_Q(X)=k$: for any endomorphism $f$, an eigenvalue $\lambda$ makes $f-\lambda I$ noninvertible, hence zero in a division algebra.

For a counterexample to the converse of “brick implies indecomposable”, take the one-loop representation $k^2$ with nilpotent Jordan block $N=\left(\begin{array}{cc}0& 1\\ 0& 0\end{array}\right)$. Its endomorphism ring is $k[N]\cong k[t]/(t^2)$, a local endomorphism ring of dimension two. It has no nontrivial idempotents, so the module is indecomposable, but it is not a brick.

For the one-arrow quiver, splitting the kernel, image and target complement decomposes any representation into copies of $k\to 0$, $0\to k$ and $k\stackrel{1}{\to }k$. Each has endomorphism ring $k$. Therefore every indecomposable of the one-arrow quiver is a brick.

An indecomposable of dimension vector $\mathbf{n}$ is a brick module and has $q_Q(\mathbf{n})=1$ by the previous parts. Its automorphism group of a quiver representation has dimension one. ThereforeIt follows that its orbit is open and dense in the irreducible quiver representation space. Two different indecomposables of the same dimension vector would give two disjoint nonempty open orbits. Nonempty open subsets of an irreducible space must intersect, so this is impossible. Hence an indecomposable is uniquely determined by its dimension vector, up to isomorphism. This is the positive definite Tits form indecomposable classification.

If $X$ is indecomposable, part (i) makes it a brick module. For its nonzero dimension vector of a quiver representation $\mathbf{n}$,The Tits form of a quiver takes integer values on integer vectors, so $q_Q(\mathbf{n})=1$ and $X$ has no self-extensions.

Conversely, suppose $\mathbf{n}$ has nonnegative integer entries and $q_Q(\mathbf{n})=1$. Choose a representation $M$ of that dimension vector whose orbit has maximal dimension. Such an orbit exists because dimensions are integers bounded by $\dim {\mathrm{Rep}}_Q(\mathbf{n})$. If $M=U\oplus V$ with nonzero $U,V$, a nonzero extension in either direction would, by part 5(c), produce a middle representation of the same dimension vector with larger orbit. Hence ${\mathrm{Ext}}_Q^1(U,V)={\mathrm{Ext}}_Q^1(V,U)=0$.

Writing $\mathbf{u}=\dim U$ and $\mathbf{v}=\dim V$, the Ringel form identity then givesThe last inequality uses positive integral values of $q_Q$ at both nonzero vectors. This contradiction makes $M$ indecomposable. Thus the indecomposable dimension vectors are exactly the positive roots $q_Q(\mathbf{n})=1$.

Use the allowed Ringel lemma on bricks in the following precise form: a finite-dimensional indecomposable quiver representation that is not a brick module contains a nonzero brick $B$ with ${\mathrm{Ext}}_Q^1(B,B)\ne 0$.

Suppose such a nonbrick indecomposable existed, and let $\mathbf{b}=\dim B$. The Tits form of a quiver is $q_Q(\mathbf{b})=⟨\mathbf{b},\mathbf{b}{⟩}_Q$, so the Ringel form identity would givePositive definiteness gives $q_Q(\mathbf{b})>0$ for nonzero $\mathbf{b}$, a contradiction. Therefore every indecomposable is a brick.

A nonzero finite-dimensional quiver representation is a brick module when ${\mathrm{End}}_Q(X)=k{1}_X$. If $X=U\oplus V$ with both summands nonzero, projection onto $U$ is an idempotent endomorphism different from zero and the identity. That is impossible in $k{1}_X$. Hence every brick is indecomposable.

For the requested converse counterexample, take the dual-number algebra $A=k[\epsilon ]/({\epsilon }^2)$ and its left regular module $A$. The algebra is local with maximal ideal $(\epsilon )$, and it is not simple because this is a nonzero proper ideal. The module is indecomposable: its endomorphism ring is $A^{\mathrm{op}}\cong A$, which has no nontrivial idempotents. But this ring has dimension two over $k$, so the regular module is not a brick. Thus indecomposable need not mean brick without the positive-definiteness hypothesis used below.

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.

Write the nonsplit short exact sequenceFor an endomorphism $g:M\to M$, the composite $qgj:L\to N$ vanishes because ${\mathrm{Hom}}_R(L,N)=0$. Hence $g$ restricts to an endomorphism $f$ of $L$ and induces an endomorphism $h$ of $N$, giving a commutative diagram of short exact sequences.

Because $L$ and $N$ are brick modules, each of $f,h$ is either zero or an isomorphism. If both are isomorphisms, the short five lemma makes $g$ an isomorphism. If both vanish, $g$ factors successively through $N$ and through $L$, hence through a map $N\to L$; this map is zero, so $g=0$.

The mixed cases would split the sequence. If $f$ is invertible and $h=0$, then $qg=0$, so $g=j\alpha$ for some $\alpha :M\to L$; the identity $\alpha j=f$ makes $f^{-1}\alpha$ a retraction of $j$. If $f=0$ and $h$ is invertible, then $gj=0$, so $g=\beta q$; the identity $q\beta =h$ makes $\beta h^{-1}$ a section of $q$. Both contradict nonsplitting. Thus every endomorphism of $M$ is zero or invertible, and $M$ is a brick.

No. Let $Q$ be the quiver $1\to 2$ over a field $k$, and take the representation of a quiverAn endomorphism is a pair of scalar maps $(a,b)$ satisfying $b=a$, so its endomorphism ring is $k$. Every nonzero endomorphism is therefore an isomorphism, making this representation a brick module. It nevertheless has the proper nonzero subrepresentation $0\to k$, so it is not an irreducible module.

Equivalently, this is a nonsimple module over the path algebra $kQ$ whose endomorphism ring is a division ring.

The Ringel lemma on bricks makes every indecomposable a brick module; positive definiteness and the Ringel form then force $q_Q=1$ and rigidity. Conversely, a maximal-orbit representation at a positive root cannot split: nonzero cross extensions increase orbit dimension, while vanishing cross extensions make the quadratic form of a split sum at least two. Open dense orbits give uniqueness. A compact bound on the integer roots gives finite representation type.

The Fitting lemma supplies a nonzero nilpotent endomorphism $f$ of a non-brick indecomposable $X$. Minimize its nonzero rank; nilpotence and minimality give $f^2=0$. Set $I=\mathrm{im}f\subset K=\ker f={⨁}_j{K}_j$. Choose a nonzero component $u:I\to K_j$. The square-zero endomorphism $X\stackrel{f}{\to }I\stackrel{u}{\to }{K}_j\hookrightarrow X$ has rank at least $\dim I$, so $u$ is injective.

If ${\mathrm{Ext}}^1(I,K_j)=0$, the projection $K\to K_j$ extends to $X\to K_j$ by the long exact sequence of Ext groups, giving a module retraction and contradicting indecomposability. Thus this extension group is nonzero. Since path algebras are hereditary, $u$ induces a surjection ${\mathrm{Ext}}^1(K_j,K_j)\twoheadrightarrow {\mathrm{Ext}}^1(I,K_j)$. Hence $K_j$ is a proper indecomposable submodule with self-extensions. Iterate until a brick module is reached; dimensions strictly decrease.

This minimal-rank argument is given in section 2 of William Crawley-Boevey's quiver lectures.

An indecomposable finite-dimensional quiver representation that is not a brick module contains a brick with nonzero self-extensions. The Ringel form then gives $q_Q(\dim B)=1-\dim {\mathrm{Ext}}_Q^1(B,B)\le 0$, impossible for a positive definite Tits form of a quiver.

The quadratic form is $q_Q(\mathbf{n})=⟨\mathbf{n},\mathbf{n}{⟩}_Q={\sum }_i{n}_i^2-{\sum }_{\rho :i\to j}{n}_i{n}_j$. It depends on the underlying graph rather than its orientation. For a representation it equals $\dim {\mathrm{End}}_Q(X)-\dim {\mathrm{Ext}}_Q^1(X,X)$. Its positive definiteness forces indecomposables to be rigid brick modules, using the Ringel lemma on bricks.