11.12 Polar norms of semipositive Finsler metricsC40892E1279D

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. $\square$

Proof. The ratio is invariant under complex rescaling of $v$, so the maximum can be taken over the compact unit sphere in a local frame, which proves continuity. Positive upper and lower bounds for $G$ on that sphere compare $Q$ with a Hermitian squared dual norm; this gives positivity off zero and continuity at zero, and complex homogeneity is clear. Moreover,

Each term satisfies the triangle inequality in $\xi $, and so does their supremum.

On a coordinate ball, set $G_\varepsilon =\mathrm e^{-\varepsilon |z|^2}G$. By lemma 11.12.1, its Kobayashi curvature is strictly positive, and its polar square is $Q_\varepsilon =\mathrm e^{\varepsilon |z|^2}Q$. At a point $(x_0,\xi _0)$ with $\xi _0\ne 0$, choose a maximizing line and a vector $v_0$ on it. The lemma, applied to $G_\varepsilon $, gives a holomorphic section $\sigma $ such that

with equality at $(x_0,\xi _0)$. The pairing $\xi (\sigma (z))$ is nonzero near that point, so the right-hand side is pluriharmonic there. Restriction to sufficiently small holomorphic disks centered at the point therefore gives the submean inequality for $\log Q_\varepsilon $. This argument applies at every point off the zero section, and together with continuity it shows that $\log Q_\varepsilon $ is plurisubharmonic there.

As $\varepsilon \downarrow 0$, the functions $\log Q_\varepsilon =\log Q+\varepsilon |z|^2$ converge locally uniformly off the zero section. Passing to the limit in the submean inequalities on disks proves that $\log Q$ is plurisubharmonic off the zero section. Since the exponential is increasing and convex, Jensen's inequality shows that $Q=\exp (\log Q)$ is plurisubharmonic there. At a point of the zero section, $Q=0$ and $Q\geq 0$, so every centered disk satisfies the submean inequality there. Together with continuity and the submean inequalities at nonzero points, this proves plurisubharmonicity on the whole total space of $E^*$. $\square$

Proof. The real fiber Hessian of $G$ is bounded above by $2L$ times $k$ on the compact unit sphere bundle of $k$, for some $L>0$, and homogeneity extends this bound to every nonzero vector. Also, $G$ is $C^1$ at zero with derivative zero, since its value is $O(|v|^2)$ and its first fiber derivative is $O(|v|)$, uniformly over the base. Integrating the Hessian estimate along a segment gives

a segment passing through zero is handled by splitting it there and using the continuity of the first derivative.

Complex homogeneity and completing the square on each complex line give the identity

The maximum exists because $G(v)\geq a|v|_ k^2$ for a uniform constant $a>0$. Fix a maximizer $v_0$ for $\xi _0$. Differentiation, including the case $\xi _0=v_0=0$, gives $\mathrm dG(v_0)=2\operatorname {Re}\xi _0$. Evaluate the expression in (11.12.5) for $\xi _0+\zeta $ at $v_0+w$ and apply (11.12.4). The terms involving $\xi _0(w)$ cancel, and we obtain

Taking the supremum over $w$ yields

Set $c=L^{-1}$ and $\Phi =Q-c|\cdot |_{k^*}^2$. After subtracting the square, this inequality gives a real affine function on the whole dual fiber that lies below $\Phi $ and equals $\Phi $ at $\xi _0$. Hence $\Phi $ is the supremum of such affine functions, and therefore convex. Neither uniqueness of the maximizer nor real convexity of $G$ has been used. $\square$

Proof. By proposition 11.12.2, the polar square is a positive continuous plurisubharmonic complex homogeneous function of degree two, and by lemma 11.12.3 it has the strong convexity in the fibers required in proposition 11.10.5. Applying proposition 11.11.2 to $F=E^*$, whose dual $F^*=E$ is ample, gives the positive smooth strictly plurisubharmonic homogeneous function $H$ on $E^*$. Hence proposition 11.10.5 applies. Its proof uses theorem 11.10.4, which preserves strong convexity in the fibers, and the square root of the resulting function is the asserted smooth norm. $\square$

Proof. Apply corollary 11.12.6 to $E=F^*$. Such a metric would produce a smooth complex norm on $F$ with plurisubharmonic square, contrary to theorem 11.9.4. $\square$