Proposition 11.23.12.F45FF03B558B Fix $p>1$. For each smooth strictly positive probability volume form $\nu $, put $\rho _\nu =\omega ^ n/(M\nu )$ and define
\[ \mathcal D_ p(\omega ,\nu ) :=\left.\frac{\mathrm{d}}{\mathrm{d}t}\right|_{t=0} \int _ X\left(\frac{(\omega +t\mathrm{dd}^{\mathrm{c}}\log \rho _\nu )^ n}{M\nu }\right)^ p\nu . \]
Then $\mathcal D_ p(\omega ,\nu )\leq 0$ for every such $\nu $ if and only if $\omega $ is Gauduchon. If $\omega $ is not Gauduchon, there is such a $\nu $ with $\mathcal D_ p(\omega ,\nu )>0$.
Proof.
The form in the definition is positive for sufficiently small $|t|$. The path $u_ t=t\log \rho _\nu $ starts at $u_0=0$ with the velocity $\log \rho _\nu $ prescribed by (11.23.7) at time zero. The proof of proposition 11.23.8 at a given time uses only the potential and its velocity at that time, so (11.23.10) holds for the derivative at $t=0$. If $\omega $ is Gauduchon, its first term vanishes and the derivative is nonpositive.
Conversely, suppose that $\lambda :=\mathrm{dd}^{\mathrm{c}}(\omega ^{n-1})\ne 0$. Write $\lambda =f\omega ^ n/M$, where $f$ is smooth and real. Stokes' theorem gives $\int _ Xf\omega ^ n=0$, whereas
\[ \int _ Xf\lambda =\frac1M\int _ Xf^2\omega ^ n>0. \]
For small $\varepsilon >0$, set
\[ c_\varepsilon =\frac1M\int _ X\mathrm{e}^{-\varepsilon f}\omega ^ n, \qquad r_\varepsilon =c_\varepsilon \mathrm{e}^{\varepsilon f}, \qquad \nu _\varepsilon =\frac{\omega ^ n}{M r_\varepsilon }. \]
Then $\nu _\varepsilon $ is a smooth strictly positive probability volume form, and $\rho _{\nu _\varepsilon }=r_\varepsilon $. Since $c_\varepsilon =1+O(\varepsilon ^2)$, $\int _ X\lambda =0$ and $\mathrm{d}r_\varepsilon =\varepsilon r_\varepsilon \mathrm{d}f$, the two terms in (11.23.10) give
\[ \mathcal D_ p(\omega ,\nu _\varepsilon ) =\frac{np}M\varepsilon \int _ Xf\lambda +O(\varepsilon ^2)>0 \]
for sufficiently small $\varepsilon >0$. This proves necessity and the last assertion.
$\square$
Pome
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