Theorem 11.18.2B93CB03C68B7

Proof. By lemma 11.18.1, the horizontal block of $\mathcal L_\rho $ has at least $s$ positive eigenvalues and the vertical block has one. Their direct sum thus has at least $s+1$ positive eigenvalues at every point of $E^\times $.

Choose $\epsilon >0$ and a smooth function $\chi :[0,+\infty )\to [0,1]$ equal to one on $[0,\epsilon ]$ and to zero on $[2\epsilon ,+\infty )$. Define

and extend $\psi $ by zero on the divisor at infinity $\mathbb P(E)$. This extension is smooth. Indeed, near a point $[0:v]$ with $v\ne 0$, use a local representative $[u:v]$ in a projective chart. On its intersection with the affine chart, $G(v/u)=h(v,v)/|u|^2$ tends to infinity as $u\to 0$, so $\chi (G)$ vanishes in a neighborhood of the point.

The function $\psi $ tends to infinity near the removed zero section. Since the ambient projective bundle is compact, every sublevel set of $\psi $ is compact in $U_ E$: it is closed there and cannot accumulate on the removed section. Put $D=\{ G<\epsilon \} \subset E$, including the zero section, and $K=\mathbb P(\mathcal O_ X\oplus E)\setminus D$. Then $K$ is a compact subset of $U_ E$, and $\psi =\rho $ on $U_ E\setminus K$. Thus the exhaustion $\psi $ has at least $s+1$ positive Levi eigenvalues outside a compact set. In dimension $n+r$, this is $(n+r-s)$-convexity. $\square$