Lemma 11.12.182E28EBCB904

Proof. Since the vertical Hessian is positive, the Levi-orthogonal complement $H_{G,v}$ of the vertical tangent space at $v$ projects isomorphically onto the base tangent space. The Levi form of $G$ on $H_{G,v}$ is the Schur complement $S$, which is negative definite when $\mathcal K_ G>0$. For the vertical radial vector $R=v_ i\partial _{v_ i}$, complex homogeneity gives

for every tangent vector $a$, where $\mathcal L_ G$ is the Levi form. Since $R$ is vertical, the second identity implies $\partial G|_{H_{G,v}}=0$.

Choose a holomorphic section $\sigma _0$ through $v_0$ whose tangent space at $x_0$ is $H_{G,v_0}$. In coordinates centered at $x_0$, the function $G\circ \sigma _0$ has zero first derivative and Taylor expansion

where $P$ is a homogeneous holomorphic quadratic polynomial. The identity $\partial _ vG(v_0)(v_0)=G(v_0)>0$ gives a vertical vector $b$ with $\partial _ vG(v_0)(b)=1$. Set $\sigma =\sigma _0-bP$ in the local frame. This changes only the pure quadratic terms and cancels $P+\bar P$. The remaining quadratic form is negative definite, and therefore dominates the cubic remainder on a sufficiently small ball. The section stays nonzero there and satisfies the asserted inequality.

For the last assertion, differentiate $\mathrm e^{-\varepsilon \varphi }G$ in the Schur complement. Euler's identities give $G_{\bar j}=v_ iG_{i\bar j}$ and $G_\alpha =G_{\alpha \bar j}\bar v_ j$. Hence the terms containing $\partial \varphi $, including its square, cancel between the base Hessian and the mixed block product, leaving

Dividing by $G_\varepsilon $ gives the curvature formula. The vertical Hessian of $G_\varepsilon $ is the positive multiple $\mathrm e^{-\varepsilon \varphi }(G_{i\bar j})$, so $G_\varepsilon $ is again strongly pseudoconvex. $\square$