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 $q>0$.
$\square$
Pome
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