Lemma 23.1.7.6674C9FC8075 For such a function, at $\xi \ne 0$ the inequality $\omega +\mathrm{dd}^{\mathrm{c}}V>0$ is equivalent to
814E85A56944
23.1.8
\begin{equation} 1+V_{w\bar w}>0,\qquad \mu +V_ x>0,\qquad Q[V]:=V_{xx}+\mu - \frac{|\mu +V_{x\bar w}|^2}{1+V_{w\bar w}}>0. \end{equation}
At $\xi =0$, the inequality $\omega +\mathrm{dd}^{\mathrm{c}}V>0$ is equivalent to $1+V_{w\bar w}>0$.
Proof.
In these coordinates,
\[ s=w+\bar w+\log (1+\mathrm{e}^ x),\qquad \partial s=\mathrm{d}w+\mu \partial x, \]
and
\[ 2\pi \alpha =\mathrm i\mu \partial \bar\partial x +\mathrm i\mu (1-\mu )\partial x\wedge \bar\partial x. \]
Choose horizontal vectors of unit length for $2\pi \alpha $, one orthogonal to the radial direction and the other
\[ e_ r=\frac{\xi _1\partial _{\xi _1}+\xi _2\partial _{\xi _2}}{\sqrt{\mu (1-\mu )}}. \]
Adjoin the vertical vector $\partial _ w$. In this frame the matrix of $2\pi (\omega +\mathrm{dd}^{\mathrm{c}}V)$ is
\[ \begin{pmatrix} (\mu +V_ x)/\mu
& 0
& 0
\\ 0
& (\mu +V_{xx})/[\mu (1-\mu )]
& (\mu +V_{x\bar w})/\sqrt{\mu (1-\mu )}
\\ 0
& (\mu +V_{wx})/\sqrt{\mu (1-\mu )}
& 1+V_{w\bar w}
\end{pmatrix}. \]
Taking the Schur complement of the last diagonal entry gives (23.1.8). At $\xi =0$, the derivatives of $V$ in the base directions vanish and $\partial s=\mathrm{d}w$, which gives the last assertion.
$\square$
Pome
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