Pome

Proposition 12.35.9E45E9799C813

Proposition 12.35.9.E45E9799C813 The all-test infimum is attained and

DA1A40EA1D3A 12.35.10
\begin{equation} \mathcal V_\mu (\theta ;bD)=\mathcal V_\mu (\beta )=\frac{2(c-b)}{H}, \qquad \mathcal V_\mu (\theta )=\frac{2c}{H}. \end{equation}

Writing $V=2k/H$ and $g=Vm/(2a)$, a minimizing test is the closed positive form

\[ T=\frac ga w_1+w_2. \]

Proof. Since $\int gw=k$, solve $\mathrm{dd}^{\mathrm{c}}f=(g-k)w$ on the first sphere. Then $\Omega =\beta +\mathrm{dd}^{\mathrm{c}}f=gw_1+aw_2>0$ and $\Omega ^2=V\mu $. For every Gauduchon test, Stokes and the matrix arithmetic–geometric mean inequality give

\[ \int _ Y\beta \wedge T=\int _ Y\Omega \wedge T \geq \sqrt V\, D_\mu (T). \]

The displayed $T$ is closed, since $g/a$ depends only on the first factor, and $T=a^{-1}\Omega $, so equality holds pointwise. Explicitly,

\[ \int _ Y\beta \wedge T=2k, \qquad D_\mu (T)=\sqrt V\int \frac ma w=\sqrt V H. \]

This proves the infimum over all tests. The curvature of $D$ is $w_1$, so the divisorial residual is indeed $\beta $. Replacing $k$ by $c$ gives the formula for $\theta $. $\square$

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