fundamental relation among $e_{l,j}$’s.

\[easyConst\] If $e_{l,k}=0$ for all $k>j$, then $e_{l+1,k}=0$ for all $k>j-1$.

This follows immediately from the definition of $e_{l,j}$.

To reduce $e_{0,0}$ to $e_{0,0}'$, the following proposition is also fundamental.

\[secondconstraint\] If $\mathcal{E}(d_{\min})$ does not admit $G_0$ as a quotient. Then $e_{1,0}\geqq e_{0,0}$ and, in the standard resolution, we can replace $\mathcal{P}_0$ by $\mathcal{P}_0^0$ and $\mathcal{P}_1$ by $\mathcal{P}_1^0$, where $\mathcal{P}_0^0=\oplus_{j=1}^{m}G_{j}^{\oplus e_{0,j}}$ and $\mathcal{P}_1^0=G_0^{\oplus e_{1,0}-e_{0,0}}
\oplus(\oplus_{j=1}^{m-1}G_{j}^{\oplus e_{1,j}})$. In particular we see that $e_{0,0}'=0$.

Denote by $d_1$ the differential $\mathcal{P}_1\to \mathcal{P}_0$, and let $p_{l,0}:\mathcal{P}_l\to G_0^{\oplus e_{l,0}}$ be the projection. Then the composite $p_{0,0}\circ d_1$ factors through $p_{1,0}$: $p_{0,0}\circ d_1=d_{1,0}\circ p_{1,0}$ where $d_{1,0}:G_0^{\oplus e_{1,0}}\to G_0^{\oplus e_{0,0}}$ is the induced morphism. Suppose that $d_{1,0}$ is not surjective. Then we have a surjection $q:G_0^{\oplus e_{0,0}}\to G_0$ such that $q\circ d_{1,0}=0$. Since $(q\circ p_{0,0})\circ d_1=q\circ d_{1,0}\circ p_{1,0}=0$, there exists a surjection $\mathcal{E}(d_{\min})\to G_0$, which contradicts the assumption. Therefore $d_{1,0}$ is surjective and $e_{1,0}\geqq e_{0,0}$. Moreover $p_{0,0}\circ d_1
$ is surjective. Since $\operatorname{Ker}(p_{0,0})=
\mathcal{P}_0^0
$ and $\operatorname{Ker}(p_{0,0}\circ d_1)
=\mathcal{P}_1^0$, the desired replacement can be done in the standard resolution of $\mathcal{E}(d_{\min})$.

If $d_{\min}>0$, $G_0=\mathcal{O}$, and $\mathcal{E}$ is nef, then $\mathcal{E}(d_{\min})$ does not admit the sheaf $\mathcal{O}$ as a quotient.

Some easy constraints {#easyconstraints}
