Proposition 11.21.11204B7FDAE31

Proof. Since every smaller convexity index would imply it, it suffices to exclude $(N-1)$-convexity. Suppose that a smooth exhaustion function $\psi $ of $U_ F$ has at least two positive Levi eigenvalues outside a compact set $K\subset U_ F$.

Fix a reference Hermitian norm on $F$. On its compact unit sphere bundle, $G$ has a positive minimum and a finite maximum, so by homogeneity $G$ is comparable to the squared reference norm uniformly on $F^\times $. Thus the level sets

are compact, and multiplication by $t$ maps $\Sigma _1$ onto $\Sigma _ t$. They are smooth hypersurfaces by the Euler identity: if $R_ v=v$ is the real radial vector at $v$, then $\mathrm dG(R_ v)=2G(v)>0$. Since $K$ stays away from the removed zero section, there is $\epsilon >0$ such that $\Sigma _ t$ is disjoint from $K$ for $0<t<\epsilon $.

Put

On any closed interval in $(0,\epsilon )$, the radial derivatives of the smooth function under the maximum are uniformly bounded over the compact level set, so $M$ is locally Lipschitz. Since $\psi $ is an exhaustion, $M(t)\to +\infty $ as $t\to 0$: otherwise maximizing points along a sequence $t\to 0$ would lie in one compact sublevel set while accumulating on the removed section. Choose $0<t_0<t_1<\epsilon $ with $M(t_0)>M(t_1)$. Since a Lipschitz function is absolutely continuous and is the integral of its derivative, there is a point $t\in (t_0,t_1)$ at which $M$ is differentiable with $M'(t)<0$.

Choose $v\in \Sigma _1$ attaining $M(t)$, and set $p=tv$. For both signs of $h$, the inequality

is an equality at $h=0$, so differentiability of $M$ gives $\frac{\mathrm d}{\mathrm dt}\psi ([1:tv])=M'(t)$. Since $\psi $ restricted to $\Sigma _ t$ attains its maximum at $p$ and $\mathrm dG\ne 0$, there is a real number $\lambda $ with $\mathrm d\psi =\lambda \, \mathrm dG$ at $p$. Evaluating on the radial direction gives

The function $\psi -\lambda G$ has zero differential at $p$ and nonpositive real Hessian on $T_ p\Sigma _ t$. Indeed, its restriction to $\Sigma _ t$ is $\psi $ minus a constant, which has a maximum at $p$, and since its differential vanishes at $p$, its second derivative along a curve in $\Sigma _ t$ is the Hessian evaluated on the tangent vector. On the complex tangent hyperplane $H=T_ p\Sigma _ t\cap \mathrm iT_ p\Sigma _ t$, this implies

The first inequality follows by adding the real Hessian inequalities for a tangent vector and its complex rotation; the second uses plurisubharmonicity of $G$ and $\lambda <0$. A two-dimensional complex subspace on which $\mathcal L_\psi $ is positive definite would meet the hyperplane $H$ nontrivially, so $\psi $ has at most one positive Levi eigenvalue at $p\notin K$. This contradiction excludes $(N-1)$-convexity. Together with theorem 11.18.2 for $s=0$, it proves the proposition. $\square$