Proposition 12.5.14BBB232A044C

Proof. The values of the derivative on all smooth $(1,1)$-forms determine the smooth form $\Xi _\theta $. Its positive root exists and is unique by the positive-root construction preceding the proof of theorem 11.6.1. If $D=D_\mu (\gamma _\theta )$, this root is $\widehat\gamma =D^{-1/(n-1)}\gamma _\theta $, and the homogeneity of $D_\mu $ gives $D_\mu (\widehat\gamma )=1$.

By theorem 12.2.3, the solution form $\Omega _\theta $ is conformal to $\widehat\gamma $, and its volume form is $\lambda _\mu (\theta )^ n\mu $. These two requirements determine the conformal factor pointwise and give the formula for $\Omega _\theta $. Thus the value and the derivative recover $\Omega _\theta $, a representative of the class of $\theta $. Without the value, the same formula recovers $\Omega _\theta $ up to a positive constant factor. Conversely, multiplying the class by a positive constant multiplies $\Omega _\theta $ by the same constant and leaves its minimizing Gauduchon ray, and hence $\Xi _\theta $, unchanged; so the derivative cannot determine more than the ray of the class.

For the surjectivity in the last assertion, take any Gauduchon metric $\gamma $ and set

This Hermitian form is unchanged by constant rescaling of $\gamma $ and satisfies $\Omega _\gamma ^ n=\mu $. The zero potential therefore solves the equation for the background $\Omega _\gamma $ with constant one, so its class has $\lambda _\mu =1$, and $\gamma $ is a minimizing Gauduchon metric by theorem 12.2.3. Injectivity follows from the recovery formula: two classes with $\lambda _\mu =1$ and the same Gauduchon ray have the same solution form. This proves the bijection. $\square$