Let $Q$ be the quiver with vertices $I = \{i_1,i_2,i_3,i_4,j\}$ and for each $k$ one arrow $i_k \to j$. Let $\Theta$ the function on $\Bbb Z I$ defined by $\Theta(i_k) = 0, \Theta(j) = 1$.
Proposition : A representation $X$ of $Q$ is $\Theta$-semistable if $\mu(U) \leq \mu(X)$ for all non-zero sub representation $U \subset X$, where we denote $\mu(M) = \Theta(\dim(M))/(\sum_k \dim (M_{i_k}) + \dim M_j)$.
I have now a problem with the simple case $i_k = 1$ and $j = 2$. In this case, Markus Reineke in his introductory lectures notes (page 19) claims that if three vectors are not proportional, then $X$ is semistable.
However it seems to me that there is no semistable representation : indeed if $X$ is a representation with at least a non-zero arrow, say $\alpha_1$ with image $v$, I think that $U_1 = k, U_2 = U_3 = U_4 = 0, U_j = k$ is a sub representation where the map $U_j \to X_j$ sends $1$ to $v$. But we have $\mu(U) = 1/2 > \mu(X) = 1/3$. Any explanations is appreciated thanks !
I beleive that you're right. Furthermore, a representation $X$ of that quiver can't be semistable so long as $X_j \neq 0$, since the representation $U$ with $U_{i_1} = U_{i_2} = U_{i_3} = U_{i_4} = 0$ and $U_j = X_j$ will be a subrepresentation of $X$.
Maybe this is a typo? There do appear to be a couple in the paper, relevantly Theorem 3.8 for example. Maybe Reineke meant the quiver with the same vertices but with arrows $j \to i_k$ instead? This would avoid my counterexample, and, after devoting a smidgeon of thought to it, makes what he's saying in that paragraph you referenced a bit more reasonable.