$i\neq j$, $0< a_i<p_i$ and $0< a_j<p_j$ then $${\operatorname{Hom}\nolimits}_\Lambda(P(a_i{\vec x}_i),P(a_j{\vec x}_j))=
    H^0({\mathbb{P}}^d, P(a_j{\vec x}_j-a_i{\vec x}_i))=H^0({\mathbb{P}}^d,{{\mathcal O}}(-1))=0$$ whilst $${\operatorname{Hom}\nolimits}_\Lambda(P(a_i{\vec x}_i),P(\left( a_i+1 \right){\vec x}_i)=H^0({\mathbb{P}}^d, {{\mathcal O}})=1.$$ Hence the endomorphism algebra is given by the following quiver with relations: $$\xymatrix{
        &P({\vec x}_1)\ar[r]|{x_1}&P(2{\vec x}_1)\ar[r]|{x_1}&\dots&\dots\ar[r]|(.3){x_1}&
        P((p_1-1){\vec x}_1)\ar[ddr]|{x_1}\\
        &P({\vec x}_2)\ar[r]|{x_2}&P(2{\vec x}_2)\ar[r]|{x_2}&\dots&\dots\ar[r]|(.3){x_2}
        &P((p_2-1){\vec x}_2)\ar[dr]|{x_2}\\
        P\ar[uur]|{x_1}\ar[ur]|{x_2}\ar[dr]|{x_n}&\vdots&\vdots&\vdots&\vdots&\vdots&P({\vec c})\\
        &P({\vec x}_n)\ar[r]|{x_n}&P(2{\vec x}_n)\ar[r]|{x_n}&\dots&\dots\ar[r]|(.3){x_n}&P((p_n-1){\vec x}_n)\ar[ur]|{x_n}
    }$$ To see the relations, note firstly that by writing down our order we have implicitly chosen an $\eta_i\in H^0({\mathbb{P}}^1,{{\mathcal O}}(L_i))$ for all $1\leq i\leq n$. If $n\geq 3$ then necessarily for all $i\geq 3$ we have $\eta_i=\ell_i(\eta_1,\eta_2)$ for some functional $\ell_i$. The relations are thus $$x_{i}^{p_i}=\ell_i(x_1^{p_1},x_2^{p_2}),\quad{\rm for}\;
i\geq 3.$$

In this example we would like to show the relationship between a GL weighted ${\mathbb{P}}^1_{T_0:T_1}$ and the corresponding GL order on ${\mathbb{P}}^1$. Here, ${{\mathcal O}}={{\mathcal O}}_{{\mathbb{P}}^1}$ and we choose $\lambda_1=(1:0),\lambda_2=(0:1), \lambda_3=(1:1)$ be three points on ${\mathbb{P}}^1$. Consider $$\Lambda=
    \begin{bmatrix}
        {{\mathcal O}}&{{\mathcal O}}(-\lambda_1)\\
        {{\mathcal O}}&{{\mathcal O}}\end{bmatrix}\otimes
\begin{bmatrix}
        {{\mathcal O}}&{{\mathcal O}}(-\lambda_2)\\
        {{\mathcal O}}&{{\mathcal O}}\end{bmatrix}\otimes
\begin{bmatrix}
        {{\mathcal O}}&{{\mathcal O}}(-\lambda_3)\\
        {{\mathcal O}}&{{\mathcal O}}\end{bmatrix}$$ In this case, ${\mathbb{L}}={\langle {\vec x}_1,{\vec x}_2,{\vec x}_3,{\vec c}\rangle}/(2{\vec x}_1=2{\vec x}_2=2{\vec x}_3={\vec c})$ acts on ${{\operatorname{\mathsf{mod}}}\Lambda}$ where the action ${\vec x}_1$ is given by: $$\begin{aligned}
-\otimes_\Lambda I_1&=-\otimes_\Lambda
\left(\begin{bmatrix}
    {{\mathcal O}}&{{\mathcal O}}\\
    {{\mathcal O}}(\lambda_1)&{{\mathcal O}}\end{bmatrix}\otimes
\begin{bmatrix}
    {{\mathcal O}}&{{\mathcal O}}(-\lambda_2)\\
    