Pome

23.1.6 The radial–vertical block70014772CD9C

We look for potentials that depend on the base point only through its distance from the origin of the affine chart. In a suitable frame, the matrix of $\omega +\mathrm{dd}^{\mathrm{c}}V$ then consists of one tangential horizontal coefficient and a two-by-two radial–vertical block, whose Schur complement isolates the mixed derivative that must be controlled. Subscripts denote partial derivatives in the indicated variables.

For $\xi \ne 0$ introduce

\[ x=\log |\xi |^2,\qquad \mu =\frac{\mathrm{e}^ x}{1+\mathrm{e}^ x},\qquad \mu _ x=\mu (1-\mu ). \]

We use smooth real functions $V:\mathbb R\times E\to \mathbb R$, written $V=V(x,w)$, that are independent of $x$ for $x$ near $-\infty $ and constant in both variables for $x$ near $+\infty $. These conditions allow $V$ to be extended smoothly across $\xi =0$, and over the complement of the affine chart by the same constant. The region of the base where $V$ is nonconstant may depend on $V$.

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. $\square$

A rapid cutoff of a concentrated fiber potential can make the mixed derivative $V_{x\bar w}$ too large for this criterion. We therefore spread the concentration by the heat equation before cutting off, so that the fiber derivatives are already exponentially small in the cutoff region.

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