Pome

Proposition 12.3.294AFD9E43DE0

Proposition 12.3.2.94AFD9E43DE0 The function $\lambda _\mu $ is smooth along smooth finite-dimensional families in $\mathscr P$. For a smooth real $(1,1)$-form $\chi $ and $n\geq 2$, its derivative at $\theta $ is

\[ (\mathrm{d}\lambda _\mu )_\theta (\chi ) =\frac{\displaystyle \int _ X\chi \wedge \gamma _\theta ^{n-1}}{D_\mu (\gamma _\theta )}, \]

where $\gamma _\theta $ is any minimizing Gauduchon metric. In dimension one the derivative is $\int _ X\chi $.

Proof. Choose a smooth potential $u_*$ with $\theta +\mathrm{dd}^{\mathrm{c}}u_*>0$. Backgrounds $\theta _ t$ near $\theta $ then have the positive representatives $\theta _ t+\mathrm{dd}^{\mathrm{c}}u_*$. The implicit-function argument in the proof of proposition 11.40.4 applies to this family and shows that the solution and its constant depend smoothly on the parameters. Adding $u_*$ to these solution potentials and fixing the mean gives the normalized solutions for $\theta _ t$, so $\lambda _\mu (\theta _ t)=c_{\theta _ t}(\mu )^{1/n}$ is smooth.

To compute the derivative, write $\lambda =\lambda _\mu (\theta )$ and $\Omega =\Omega _\theta $, and write the minimizing metric as $\gamma _\theta =\mathrm{e}^ g\Omega $ with $g$ smooth and real, as theorem 12.2.3 permits. Along $\theta _ t=\theta +t\chi $ for small real $t$, put $\Omega _ t=\theta _ t+\mathrm{dd}^{\mathrm{c}}u_ t$ and $\lambda _ t=\lambda _\mu (\theta _ t)$, and let dots denote derivatives at $t=0$. Differentiating $\Omega _ t^ n=\lambda _ t^ n\mu $ gives

\[ (\chi +\mathrm{dd}^{\mathrm{c}}\dot u)\wedge \Omega ^{n-1} =\lambda ^{n-1}\dot\lambda \, \mu . \]

We multiply by $\mathrm{e}^{(n-1)g}$, so that $\Omega ^{n-1}$ becomes $\gamma _\theta ^{n-1}$, and integrate. The term with $\mathrm{dd}^{\mathrm{c}}\dot u$ vanishes by the Gauduchon condition, and $\int _ X\chi \wedge \gamma _\theta ^{n-1}=\lambda ^{n-1}\dot\lambda \int _ X\mathrm{e}^{(n-1)g}\mu $. On the other hand, $\gamma _\theta ^ n=\mathrm{e}^{ng}\lambda ^ n\mu $, so that

\[ D_\mu (\gamma _\theta ) =\lambda ^{n-1}\int _ X \mathrm{e}^{(n-1)g}\mu . \]

Dividing the two identities proves the derivative formula. In dimension one the formula follows directly from the definition. $\square$

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