PomeLemma 11.29.1.E495F7FE3CE6 The function $G_ h$ is strictly plurisubharmonic off the zero section if and only if $h$ is a smooth positive Hermitian metric with strictly Griffiths-negative curvature. Equivalently, its dual metric on $E$ is strictly Griffiths-positive.
Lemma 11.29.1E495F7FE3CE6
Proof. On vertical tangent vectors the Levi form is $h$ itself, so strict positivity forces $h>0$. At a fixed base point, choose a holomorphic frame in which the matrix $H$ of $h$ is the identity and has vanishing first derivatives. Then $R_ h(u,\bar u)=-H_{u\bar u}$ at this point. A tangent vector to $F$ has horizontal and vertical components $(u,w)$, and
\[ \mathcal L G_ h(u,w)=|w|^2+v^*H_{u\bar u}v =|w|^2-h(R_ h(u,\bar u)v,v). \]
For $v\ne 0$, this is positive for every nonzero $(u,w)$ precisely when the curvature evaluation is negative for every nonzero $u$. The curvature of the dual metric is the negative transpose of the original curvature, with respect to the dual Hermitian pairing, which gives the last assertion. $\square$
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