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For positive curvature, the reciprocal norm in lemma 11.18.1 is useful near a removed zero section. With negative curvature, the squared norm itself is plurisubharmonic. Adding its logarithm gives a weight whose Levi form controls both the base and the fiber directions.
In this subsection the base $X$ need not be compact. Suppose that it has a Kähler form $\omega $ and a smooth nonnegative plurisubharmonic exhaustion $\rho $. Here an exhaustion has compact sublevel sets; a constant function qualifies on a compact base. Let $\pi :E\to X$ be a holomorphic vector bundle of rank $r\geq 1$, with a smooth Hermitian metric $h$, and let $Z$ be its zero section. Use the unscaled curvature and coefficient forms $\alpha _{h,v}$ of 11.18, and assume
\[ \alpha _{h,v}\leq -|v|_ h^2\omega \qquad (x\in X,\ v\in E_ x). \]
This is a uniform negative curvature bound with respect to the chosen base form.
Proposition 11.19.1.FD9DEB3AA8D6 Put $U(v)=|v|_ h^2$, $q=U\mathrm e^ U$ and $\Psi =\rho \circ \pi +U$. Then $\Psi $ is a smooth plurisubharmonic exhaustion of $E$ and
\[ \Omega =\pi ^*\omega +2\pi \mathrm{dd}^{\mathrm{c}}U \]
is a Kähler form on $E$. The function $\log q$, extended by $-\infty $ on $Z$, is plurisubharmonic and satisfies
\[ 2\pi \mathrm{dd}^{\mathrm{c}}\log q\geq \Omega \quad \hbox{on }E\setminus Z. \]
Both $U$ and $q$ are strictly plurisubharmonic off $Z$. For the real radial vector field $\mathcal R$, whose flow is $v\mapsto \mathrm e^ s v$, one has $\mathcal Rq=2(1+U)q$. In particular $q$ has no critical points off $Z$.
Proof.
At $x$, choose a Chern-normal unitary holomorphic frame as in lemma 11.18.1. Write a tangent vector at $(x,v)$ as $(a,b)$, with $a$ horizontal and $b$ vertical. The first base derivatives and mixed base–fiber derivatives of $U$ vanish there. The same normal-frame calculation gives
\[ \begin{aligned} \mathcal L_ U(a,b) & =-h(R_ h(a,\bar a)v,v)+|b|_ h^2 \geq U\omega (a,\bar a)+|b|_ h^2,\\ \mathcal L_{\log U}(a,b) & =-\frac{h(R_ h(a,\bar a)v,v)}U +\frac{|b|_ h^2U-|h(b,v)|^2}{U^2} \geq \omega (a,\bar a)\quad (v\ne 0). \end{aligned} \]
The last vertical term is nonnegative by Cauchy–Schwarz. Here $\mathcal L$ is the unscaled complex Hessian, corresponding to $2\pi \mathrm{dd}^{\mathrm{c}}=\mathrm i\partial \bar\partial $. These formulas show that $U$ is plurisubharmonic, that $\Omega $ is positive definite even on $Z$, and that $2\pi \mathrm{dd}^{\mathrm{c}}(\log U+U)\geq \Omega $ off $Z$. The form $\Omega $ is closed because $\omega $ is closed. The first formula gives strict positivity for $U$ off $Z$; exponentiating $\log q$ gives it for $q$. Locally, $U$ is comparable to the squared Euclidean fiber norm, so $\log q$ tends to $-\infty $ at $Z$ and is locally bounded above. By lemma 5.24.7, the upper-limit extension is plurisubharmonic and has value $-\infty $ on $Z$.
The function $\Psi $ is plurisubharmonic. Its sublevel set of height $M$ lies over the compact set $\{ \rho \leq M\} $ and in the disk bundle $\{ U\leq M\} $ over that set, which is compact. The sublevel is closed, so $\Psi $ is an exhaustion. Finally, $\mathcal RU=2U$ gives the radial identity and excludes critical points where $q>0$.
$\square$
Corollary 11.19.2.029DAFB20616 For every $m>0$,
\[ \mathcal I(m\log q) =\mathcal I_ Z^{\max \{ \lfloor m\rfloor -r+1,0\} }. \]
In particular, if $k\geq 1$ is an integer and $m\geq r+k-1$, a holomorphic function that is locally square integrable against $q^{-m}$ times a smooth positive volume form vanishes to order at least $k$ along $Z$. If $m<r+k-1$, this conclusion fails for some local holomorphic functions.
Proof.
In a local bundle frame, $\log q=\log |v|^2+O(1)$ near $Z$. Apply lemma 5.37.1 in codimension $r$. Its radial integrability condition for a homogeneous fiber term of degree $a$ is $a+r>m$. For $m<r+k-1$, a monomial of degree $k-1$ satisfies this condition and has insufficient vanishing order. Smooth positive volume forms do not change local integrability.
$\square$
Corollary 11.19.3.6CD37D30AF59 The manifold $E\setminus Z$ admits a complete Kähler metric.
Proof.
Choose a smooth convex increasing function $\eta $ on $\mathbb R$ equal to $-\log (-t)$ for $t\leq -1$. On $E\setminus Z$ put
\[ \widehat\Omega =\Omega +2\pi \mathrm{dd}^{\mathrm{c}}(\Psi ^2) +2\pi \mathrm{dd}^{\mathrm{c}}\eta (\log q). \]
The added forms are nonnegative, so this is Kähler. Since $\Psi \geq 0$ is plurisubharmonic, $2\pi \mathrm{dd}^{\mathrm{c}}(\Psi ^2)\geq 2\mathrm i\partial \Psi \wedge \bar\partial \Psi $. Thus $\Psi $ has bounded gradient for $\widehat\Omega $, and a curve of finite length stays in a compact sublevel of $\Psi $ in $E$. The only possible escape from that compact set within $E\setminus Z$ is approach to $Z$. Where $\log q\leq -1$, set $\tau =\log (-\log q)$. Convexity and the chain rule give
\[ 2\pi \mathrm{dd}^{\mathrm{c}}\eta (\log q) \geq \mathrm i\partial \tau \wedge \bar\partial \tau . \]
Hence $\tau $ also has bounded gradient there, while $\tau \to +\infty $ on approach to $Z$. A finite-length curve cannot escape in this way either. The criterion that every escaping curve has infinite length proves completeness.
$\square$
For example, let $B\to X$ have a smooth Hermitian metric with positive unscaled curvature form $\omega $. The direct sum $E=(B^*)^{\oplus r}$ with its induced metric satisfies $\alpha _{h,v}=-|v|_ h^2\omega $. Thus the construction applies whenever this base also has the stated plurisubharmonic exhaustion, including a compact base.
If the base is connected, compact and has positive dimension, one may take $\rho =0$. Then $U$ is an exhaustion that is strictly plurisubharmonic off $Z$, but the total space $E$ is not Stein: every holomorphic function on $E$ restricts to a constant on the compact connected zero section and therefore fails to separate its distinct points. Strict positivity away from a compact analytic set does not remove this obstruction.
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