\\
    \beta_4 & = e_1 + e_2 + e_3, & \beta_5 &= e_1 + e_2, & \beta_6 &= e_1,
    \end{aligned}$$ where one can check that $u<v$ implies $\langle \beta_u,\beta_v \rangle \geq 0$. Observe we can obtain another Reineke order by interchanging $e_2$ and $e_1+e_2+e_3$ above, since $\langle e_2,e_1+e_2+e_3 \rangle = 0$.

Given a dimension vector $\gamma \in D$, a choice of non-negative integers ${{\boldsymbol{\mathrm{m}}}}= (m_\beta)_{\beta\in\Phi_+}$ is called a *Kostant partition* of $\gamma$ if $$\gamma = \sum_{\beta\in\Phi_+} m_\beta\,\beta.$$ When this happens, we write ${{\boldsymbol{\mathrm{m}}}}{\vdash}\gamma$. Gabriel’s theorem [@pg1972] that the orbits are in one-to-one correspondence with Kostant partitions. In the sequel, we will denote the orbit corresponding to ${{\boldsymbol{\mathrm{m}}}}{\vdash}\gamma$ by $\Omega_{{{\boldsymbol{\mathrm{m}}}}}(Q) = \Omega_{{{\boldsymbol{\mathrm{m}}}}} \subset {\boldsymbol{\mathrm{Rep}}}_\gamma$. When $\Phi_+$ is in Reineke order $\beta_1 \prec \cdots \prec \beta_r$, we will abuse notation and write $m_u= m_{\beta_u}$.

\[ex:Orbit.A3\] For $Q$ the equioriented $A_3$ quiver of Example \[ex:Reineke.order.A3\], take the Reineke order in . Consider the Kostant partition with $m_1 = m_5 = 0$, $m_2=2$, and all other $m_u = 1$. For type $A$ quivers, one can draw a *lacing diagram* (originally due to [@saadf1980]), corresponding to the orbit $\Omega_{{\boldsymbol{\mathrm{m}}}}$. In this case, we have that $\Omega_{{\boldsymbol{\mathrm{m}}}}$ corresponds to the lacing diagram

\(11) at (0,1.5) [$\bullet$]{}; 