states that quivers with finite number of indecomposables corresponds to Dynkin quivers.
A quiver is of finite type iff for every dimension vector, the set of indeomposable representation with such dimension vector is finite.
If \(Q\) if of finite type, it would imply that there is finite number of isomorphism classes of representations of dimension vector \(v\) for any \(v\). So the group action on \(\mathrm{Rep}(Q)\) has finite number of orbits.
As a result, there must be a maximal orbit \(O\) which has maximal dimension \(\dim O = \dim \mathrm{Rep}(Q,v)\). So by orbit formula \(\dim O_x = \dim G - \dim G_x\) and the fact that \(K^\times \subset G_x\), we have
\[ \dim \mathrm{Rep}(Q,v) = \dim O = \dim G - \dim G_x \le \dim G - 1.\]
Since \(\dim \mathrm{Rep}(Q,v)\ge \dim O\), We always have
\[ \dim G_x \ge \langle v, v \rangle_Q.\]
Note that \(1\le \dim G - \dim \mathrm{Rep}(Q,v) = \sum v_i^2 - \sum v_s v_t = \langle v, v \rangle_Q\) is the Cartan form of \(Q\), therefore the quadratic form Tits form is positive definite, hence \(Q\) must be Dynkin.
In summary, a maximal dimension orbit exists for \(v\) would imply \(\langle v, v\rangle_Q \ge 1\).
Given a quiver \(Q\),
a sink is a point \(i\in Q_0\) such that arrows only go in.
a source is similar a point where arrows only go out.
For a sink \(i\in Q\), we define a converter \(s_i^+\) that sends \(Q\) to \(Q'\), which inverts all the arrow attaching \(i\), coverting a sink to a source. Similarly a map \(s_i^-\) converting a source to a sink.
If \(Q\) as undirected quiver, is a tree, then arrow orientations are connected with each other by successive applications of the two quiver converters.
The reflection functors are defined by
For a sink \(i\in Q\),
\[\Phi_i^+ : \mathrm{Rep}(Q) \to \mathrm{Rep}(s_i^+Q), \quad \begin{cases} V_i \mapsto \ker(\bigoplus_{k\to i} V_k \to V_i) \\ V_j\mapsto V_j\quad (j\neq i) \end{cases}\]
For a source \(i\in Q\),
\[\Phi_i^- : \mathrm{Rep}(Q) \to \mathrm{Rep}(s_i^-Q), \quad \begin{cases} V_i \mapsto \mathrm{coker}(V_i \to \bigoplus_{i\to k} V_k) \\ V_j\mapsto V_j\quad (j\neq i) \end{cases}\]