Let $X$ be a smooth projective variety, and assume its divisor class group is finite and free. Let $E_1,E_2,\ldots,E_n$ be line bundles on $X$. Define $L_k=E_1+\ldots E_k$, and let $Q$ be the endomorphism quiver of the collection $\{L_1,\ldots,L_n\}$. Note $Hom(L_i,L_j)=H^0(X,E_{i+1}\otimes\ldots\otimes E_j)$. Let $Q_1$ be the set of arrows in $Q$. Then there is a natural map $$\phi: k[x_a,a\in Q_1]/I\to Cox(X)$$ where $I$ is the ideal used to define the set of representations of $Q$ as an affine variety. This map is in general not injective.
Let $\vec{e_i}$ be the standard basis for the characters, i.e. $e_i=(0,\ldots,-1,1,\ldots,0)$ where the $-1$ shows up at the $i$-th position. Let $\chi=\sum_{i=1}^na_i\vec{e_i}$ where all $a_i\geq 0$. Let $(k[x_a,a\in Q_1]/I)_{\chi}$ be the $\chi$ graded piece. the we have induced map: $$\phi_{\chi}: (k[x_a,a\in Q_1]/I)_{\chi}\to H^0(X,E_1^{a_1}\otimes\ldots\otimes E_n^{a_n})$$
Note if $\chi=\vec{e_i}+\ldots\vec{e_j}$ for $i<j$, then $\phi_{\chi}$ is injective. In fact, it is an isomorphism by definition of the quiver.
My question is: Under what condition (on both the collection $\{E_i\}$ and $\chi$ )do we have injectivity of $\phi_{\chi}$?
I believe this is equivalent as asking if all the relations coming from characters of the type $\vec{e_i}+\ldots+\vec{e_j}$ is enough to generate all the relations on the right hand side.
