Corollary 14.9.4.BFCF1480CC8B Let $G\in C^\infty (X,\mathbb R)$, let $L_0v=3\, \mathrm{dd}^{\mathrm{c}}v\wedge \omega ^2/\omega ^3$ be the linearized operator at $u=0$, and let $v$ be the unique smooth real function satisfying
\[ L_0v=G-\int _ XG\, \mathrm{d}\nu , \qquad \int _ Xv\, \mathrm{d}\nu =0. \]
Then
\[ \left.\frac{\mathrm{d}}{\mathrm{d}t}D(tG)\right|_{t=0}=0, \qquad \left.\frac{\mathrm{d}^2}{\mathrm{d}t^2}D(tG)\right|_{t=0} =2\int _ X|v_ w|^2\, \mathrm{d}\nu . \]
The second derivative vanishes if and only if $G$ is a smooth base pullback.
Proof.
For real $t$ near zero put $u_ t=u_{tG}$ and $b_ t=b(tG)$. By proposition 14.4.1, this is a smooth family, with $u_0=b_0=0$. Differentiating the logarithmic equation at $t=0$ gives $L_0\dot u_0=G+\dot b_0$. Since $\omega $ is Gauduchon, $\nu $ is the adjoint probability measure of $L_0$, and integration gives $\dot b_0=-\int _ XG\, \mathrm{d}\nu $. Hence $\dot u_0=v$; this also proves the existence and uniqueness of $v$.
Smooth dependence gives $(u_ t)_ w=t v_ w+O(t^2)$ uniformly on $X$, so (14.9.2) gives
\[ \mathrm{e}^{D(tG)}=1+t^2\int _ X|v_ w|^2\, \mathrm{d}\nu +O(t^3). \]
Taking logarithms proves the derivative formulas.
If the second derivative vanishes, then $v_ w=0$, so $v=\pi ^*q$ for a smooth real function $q$ on the base. Expanding $\omega ^2=\pi ^*\omega _{\mathrm{FS}}^2+2\pi ^*\omega _{\mathrm{FS}}\wedge \beta $ gives
\[ L_0(\pi ^*q)=\pi ^*\! \left( \frac{2\mathrm{dd}^{\mathrm{c}}q\wedge \omega _{\mathrm{FS}}}{\omega _{\mathrm{FS}}^2}\right). \]
Hence $G=L_0v+\int _ XG\, \mathrm{d}\nu $ is a base pullback. Conversely, when $G$ is a base pullback, proposition 14.9.1 gives $D(tG)=0$ for every real $t$.
$\square$
Pome
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