11.21 Homogeneous plurisubharmonic functions and Hopf complements2ADE48A8B892

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. $\square$

Proof. Put $G(v)=h(v,v)$. At $x\in B$, take a holomorphic frame that is unitary and Chern-normal at $x$, and write a tangent vector to $F$ at $v\in F_ x$ as $(u,w)$, with $u\in T_ x^{1,0}B$ and $w\in F_ x$. The calculation in the proof of lemma 11.18.1 gives

with zero mixed block. Thus $G$ is smooth, plurisubharmonic and positive off the zero section, and it satisfies $G(cv)=|c|^2G(v)$. Apply proposition 11.21.1. $\square$

Proof. In parallel holomorphic unitary frames, the squared norm is $G(v)=\sum _ j|v_ j|^2$, which is plurisubharmonic on the total space. Since the transition matrices are constant and unitary, $G$ is globally defined; it is smooth, positive on $F^\times $ and homogeneous of degree two under complex scalars. Apply proposition 11.21.1. $\square$

Proof. Let $\alpha $ be the form on $X_ q$ induced by $\mathrm{dd}^{\mathrm{c}}\log |z|^2$; it descends because the contraction changes $\log |z|^2$ by a constant. For $u\in T_ z^{1,0}(\mathbb C^ n\setminus \{ 0\} )$, its coefficient form is, up to the positive factor $1/(2\pi )$,

where $\langle \cdot ,\cdot \rangle $ is the Euclidean Hermitian inner product. By the Cauchy–Schwarz inequality, it is positive semidefinite with kernel $\mathbb Cz$, hence of rank $n-1$. The curvature $-\mathrm{dd}^{\mathrm{c}}\log |z|^{2b}$ of the metric $G$ is $-b\alpha $. The integral of $\alpha \wedge \gamma ^{n-1}$ is strictly positive, so the degree has the sign of $-b$, which gives the three stated cases. The map $z\mapsto (z/|z|,\log |z|)$ identifies the covering space with $S^{2n-1}\times \mathbb R$, and the contraction translates the second coordinate by $\log q$. Hence $X_ q$ is diffeomorphic to $S^{2n-1}\times S^1$, and $H^2_{\mathrm{dR}}(X_ q,\mathbb R)=0$ since $n\geq 2$.

If $|a|<1$, then $b<0$, so every nonzero fiber vector gives $n-1$ positive curvature directions. Applying theorem 11.18.2 with $r=1$ and $s=n-1$, we obtain $2$-convexity. For the lower bound, consider the map $X_ q\to \mathbb P^{n-1}$, $[z]\mapsto [z]$, whose fibers are compact embedded elliptic curves $C\simeq \mathbb C^*/q^{\mathbb Z}$. On a lifted fiber $z=\zeta z_0$, the logarithm of the metric coefficient is $b\log |\zeta |^2+b\log |z_0|^2$, which is pluriharmonic in $\zeta $. Thus $E_ a|_ C$ has flat Chern connection. The complement $U_{E_ a|_ C}$ is a closed complex submanifold of $U_{E_ a}$. A $1$-convex exhaustion function of $U_{E_ a}$ would restrict to an exhaustion function of this submanifold that is strictly plurisubharmonic outside a compact set. This contradicts corollary 11.21.3, since the restricted bundle has base dimension one and rank one. The phase of $a$ affects the gluing but not these metric and curvature calculations.

If $|a|\geq 1$, then $b\geq 0$. On the covering space of $E_ a^\times $, the function

is plurisubharmonic: the first summand has semipositive Levi form and the second is pluriharmonic for $v\ne 0$. Consequently

The descended function $G$ is smooth, positive and homogeneous of degree two under complex scalars. Apply proposition 11.21.1 with rank one and total dimension $n+1$. $\square$

Proof. Let $\alpha $ be the form induced by $\mathrm{dd}^{\mathrm{c}}\log |z|^2$. By the calculation in example 11.21.4, it is semipositive of rank $n-1$, with radial kernel. The direct sum metric has diagonal curvature, and the normalized curvature evaluated at $v$ is

If every $|a_ j|<1$, then every $b_ j<0$. The coefficient is then strictly positive for $v\ne 0$, so theorem 11.18.2 with $s=n-1$ shows that $U_ F$ is $(r+1)$-convex. On any embedded elliptic fiber $C$ of $X_ q\to \mathbb P^{n-1}$, each line metric has flat Chern connection by the logarithmic calculation in example 11.21.4, and so does their direct sum. Hence corollary 11.21.3 gives least index $r+1$ for $U_{F|_ C}$. This is a closed complex submanifold of $U_ F$, being its inverse image over $C$, and lemma 11.17.3 gives the lower bound.

If every $|a_ j|\geq 1$, then every $b_ j\geq 0$. For $v_ j\ne 0$, the function $|z|^{2b_ j}|v_ j|^2$ is plurisubharmonic by the calculation with the logarithm and the exponential in example 11.21.4. It is also smooth at $v_ j=0$, because $z\ne 0$, and by continuity its complex Hessian is nonnegative there as well. The sum $G$ is therefore plurisubharmonic on the covering space of the total space and descends to $F$. It is positive on $F^\times $ and homogeneous of degree two under complex scalars, so proposition 11.21.1 applies. $\square$