our definition of the group ${\mathbb{L}}$, we can write $${\vec y}-{\vec x}=\sum_{i=1}^n\ell_i{\vec x}_i+\ell{\vec c}$$ for $0\le\ell_i<p_i$ and $-d\le \ell\le d$. By (a), we have $e_i(P_i(\ell_i{\vec x}_i))={{\mathcal O}}$. Furthermore, using (\[pixi\]), we see that $e_i(P_i(\ell{\vec c}))={{\mathcal O}}(\ell)$. Thus $$e(P({\vec y}-{\vec x}))=(e_1(P_1(\ell_1{\vec x}_1))\otimes_{{{\mathcal O}}}\cdots\otimes_{{{\mathcal O}}}e_n(P_n(\ell_n{\vec x}_n))(\ell)={{\mathcal O}}(\ell).\qedhere$$

Now we are ready to prove Proposition \[ext vanishing\].

Let ${\vec x},{\vec y}\in[0,d{\vec c}]$. Since $({\vec x})$ is an autofunctor of ${\operatorname{\mathsf{mod}}}\Lambda$, we have $${\operatorname{Ext}\nolimits}^i_\Lambda(P({\vec x}),P({\vec y}))={\operatorname{Ext}\nolimits}^i_\Lambda(\Lambda e,P({\vec y}-{\vec x})).$$ By Lemmas \[cohomology\] and \[first entry 2\], we have $${\operatorname{Ext}\nolimits}^i_\Lambda(\Lambda e,P({\vec y}-{\vec x}))=H^i({\mathbb{P}}^d, e(P({\vec y}-{\vec x})))=H^i({\mathbb{P}}^d, {{\mathcal O}}(\ell))$$ for some $\ell$ with $-d\le \ell\le d$. This is zero by Theorem \[beilinson\] and so the proof is completed.

Proof of the generation condition
---------------------------------

The proof is broken up into two parts: first we will prove a seemingly weaker generation condition, and then show that in our case it is in fact sufficient.

\[generate\] If $X\in{\operatorname{\mathsf{mod}}}\Lambda$ satisfies ${\operatorname{Ext}\nolimits}^i_\Lambda(T,X)=0$ for all $i\ge0$, then $X=0$.

We call an idempotent $e'_i\in H^0({\mathbb{P}}^d, \Lambda_i)$ *standard* if it has one entry $1$ on the diagonal and all other entries are $0$.

Assume $X\in{\operatorname{\mathsf{mod}}}\Lambda$ satisfies ${\operatorname{Ext}\nolimits}^i_\Lambda(T,X)=0$ for all $i\ge0$. We need to show that $e'X=0$ for all idempotents $e'\in H^0({\mathbb{P}}^d, \Lambda)$ of 