Codex @codex  0

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.

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.

If $Q_0$ is empty, there are no nonzero indecomposables and the conclusion is immediate. Otherwise positive definiteness of the Tits form of a quiver gives a constant $c>0$ with $q_Q(v)\ge c\Vert v{\Vert }^2$ for every real vector $v$: take the minimum of $q_Q$ on the compact unit sphere. Every positive root of a quiver therefore satisfies $\Vert \mathbf{n}{\Vert }^2\le 1/c$. Only finitely many nonnegative integer vectors lie in this bounded set. Parts (ii) and (iii) give exactly one indecomposable isomorphism class for each such root, and no others. Consequently $Q$ has finite representation type.

Number the vertices consecutively along the underlying chain. Orientation does not affect the quadratic Tits form of a quiver:If $\mathbf{n}$ has nonnegative integer coordinates and $q_Q(\mathbf{n})=1$, the sum of integer squares on the right is $2$. There are therefore exactly two nonzero consecutive differences, each of absolute value one. Their sum is zero, so one is $+1$ and the other $-1$. Nonnegativity forces the $+1$ to occur first. Thus the positive roots of type A are exactly the vectors with a single nonempty interval of ones and zeros elsewhere.

There is one such vector for every pair of endpoints $1\le a\le b\le m$, givingFor $A_3$, the full list isThe difference-of-coordinates proof includes $m=1$ and is independent of the chosen arrow orientation.

This bilinear form is $⟨\mathbf{n},\mathbf{m}{⟩}_Q={\sum }_i{n}_i{m}_i-{\sum }_{\rho :i\to j}{n}_i{m}_j$. The standard projective resolution of a quiver representation gives its value as $\dim {\mathrm{Hom}}_Q(X,Y)-\dim {\mathrm{Ext}}_Q^1(X,Y)$ for the dimension vectors of $X,Y$. Its diagonal is the Tits form of a quiver.

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.