Three-dimensional subadjoint nonvanishing from a cubic Euler polynomial

Put $P=\mathrm{ceil}(-\mathrm{\Delta })$. Since the nonexceptional coefficients of Δ are nonnegative and all are below one, P is effective integral exceptional. Choose $\delta >0$ so small that (a−1)L−δT is ample and $\mathrm{\Gamma }=\mathrm{\Delta }+P+\delta T$ is effective SNC with every coefficient below one. Such a choice is possible by the explicit perturbation hypothesis and the finite coefficient gaps.

For every integer t≥−1, the Cartier divisor $N_t=P+tL$ has $N_t-K_V-\mathrm{\Gamma }$ ${\sim }_Q$ (a+t)L+M−δT.

This is the sum of the ample (a−1)L−δT and the curve-nef (t+1)L+M. It is ample by the actual nef-plus-ample supplier. Smooth SNC log-adjoint vanishing therefore gives $H^i(V,O_V(P+tL))=0$ for all $i>0$ and every integer t≥−1. The strict bound $a>1$ is used exactly for $t=-1$.

Normality of W and exceptionality of P give ${\nu }_{\ast }{O}_V(P)=O_W$ by the codimension-one regular-function criterion. Projection formula identifies $H^0(V,P+tL)=H^0(W,tA)$ for every integer t. Since W is integral proper and A ample, $\chi (V,P)=1$, $\chi (V,P-L)=0$, $\chi (V,P+L)=h^0(W,A)$.

The negative-degree H⁰ vanishes because an inverse ample bundle has no nonzero section. These identities use direct smooth vanishing, not a descent assertion about all higher direct images of $O_V$.

Write $p(t)=\chi (V,O_V(P+tL))=c_3{t}^3+c_2{t}^2+c_{1}t+c_0$. HRR on the smooth projective V, obtained from actual GRR to a point, gives $c_2=(2P-K_V)L^2\mathrm{/}4$.

Indeed the coefficient of t² in exp(P+tL)td(V) is $L^2(P+t{d}_1)\mathrm{/}2$, and $t{d}_1=-K_V\mathrm{/}2$. Neither $t{d}_2$ nor $t{d}_3$ is needed. Since P is exceptional and L is pulled back from W, $P{L}^2=0$: the image of each exceptional divisor has dimension at most one, so two general ample representatives pulled back from W have zero intersection with it.

Intersecting the exact subadjoint relation with L² gives $-K_V{L}^2=a{A}^3+\mathrm{\Delta }{L}^2+M{L}^2\ge a{A}^3$.

The first remaining term is $A^2{\nu }_{\ast }\mathrm{\Delta }\ge 0$ by effectivity. The second is nonnegative because M and L are nef; equivalently approximate L by ample divisors and intersect M with general complete-intersection curves. Therefore $c_2\ge a{A}^3\mathrm{/}4>0$.

The preceding vanishing identities give $c_0=p(0)=1$ and $p(-1)=0$. Hence $-c_3+c_2-c_1+1=0$, or $c_3+c_1=c_2+1$. Consequently $h^0(W,A)=p(1)=2+2c_2\ge 2+(a\mathrm{/}2)A^3>0$.

This proves the stated bound. It does not require a general high-index nonvanishing theorem or positivity of $c_2(T_W)$, and preserves the actual hypothesis $a>1$ rather than the weaker $a>0$ surface scope.