Lemma 11.18.1.5022159B8350 Put $G(v)=h(v,v)$ and $\rho =G^{-1}$ on $E^\times $. In a holomorphic frame that is unitary and Chern-normal at $x$, meaning that $M(x)$ is the identity and the first derivatives of $M$ vanish at $x$, the complex Hessian of $\rho $ at $(x,v)$ is
\[ \mathcal L_\rho (u,w) =\frac{h(R_ h(u,\bar u)v,v)}{G^2} -\frac{|w|_ h^2}{G^2} +\frac{2|h(w,v)|^2}{G^3}, \qquad u\in T_ x^{1,0}X,\quad w\in E_ x, \]
where $(u,w)$ are the horizontal and vertical components of a tangent vector. There are no mixed horizontal–vertical terms, and the vertical block has exactly one positive and $r-1$ negative eigenvalues.
Proof.
Such a frame is obtained by a constant change of frame making it unitary at $x$, followed by a holomorphic change of frame, linear in the coordinates, that cancels $\partial M$. Then $F_{\alpha \bar\beta }=-\partial _\alpha \partial _{\bar\beta }M$ at $x$. The horizontal first derivatives of $G$ and its mixed horizontal–vertical second derivatives vanish. The horizontal complex Hessian of $G$ is $-h(R_ h(u,\bar u)v,v)$, its vertical complex Hessian is $|w|_ h^2$, and its vertical first derivative in the direction $w$ is $h(w,v)$. Apply
\[ \partial \bar\partial (G^{-1}) =2G^{-3}\partial G\wedge \bar\partial G-G^{-2}\partial \bar\partial G. \]
The vertical block equals $G^{-1}>0$ at $w=v$ and equals $-|w|_ h^2/G^2$ when $w\perp _ h v$. These two subspaces are orthogonal for the vertical block, which proves the statement about its eigenvalues.
$\square$
Pome
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