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Proposition 12.22.58AB4DFD2719F

Proposition 12.22.5.8AB4DFD2719F Let $\pi :Y\to X$ be a modification of connected compact complex manifolds of dimension $n\geq 2$, and let $\theta $ be a smooth nef background on $X$. Then $\pi ^*\theta $ is nef. For fixed Hermitian forms $h_ X,h_ Y$ and smooth positive probability forms $\mu _ X,\mu _ Y$, there are constants $C,K>0$, independent of $\epsilon $, such that

2D24229F8C67 12.22.6
\begin{equation} \mathcal V_{\mu _ Y}(\pi ^*\theta +\epsilon h_ Y) \leq C\mathcal V_{\mu _ X}(\theta +K\epsilon h_ X) \qquad (\epsilon >0). \end{equation}

Consequently

\[ \underline a(\pi ^*\theta )\geq \underline a(\theta ),\qquad \overline a(\pi ^*\theta )\geq \overline a(\theta ). \]

In dimension one a modification is an isomorphism and preserves both orders.

Proof. Step 1: nefness. Choose $C'$ with $\pi ^*h_ X\leq C'h_ Y$. For $\epsilon >0$, pull back a smooth representative $\theta +\mathrm{dd}^{\mathrm{c}}u_\tau \geq -\tau h_ X$ with $\tau <\epsilon /C'$; this gives a positive representative of $\pi ^*\theta +\epsilon h_ Y$. Thus $\pi ^*\theta $ is nef, and both volumes in (12.22.6) are positive.

Step 2: regularized tests. The ratio $\pi ^*\mu _ X/\mu _ Y$ is smooth, nonnegative and bounded on the compact manifold $Y$, so there is $c>0$ with $\mu _ Y\geq c\pi ^*\mu _ X$. Fix a smooth Gauduchon metric $\gamma _0$ on $Y$, given by theorem 11.4.2. For an arbitrary Gauduchon metric $\gamma $ on $X$, write $\Psi =\gamma ^{n-1}$, and for $\delta >0$ define $\Gamma _\delta $ by

\[ \Gamma _\delta ^{n-1}=\pi ^*\Psi +\delta \gamma _0^{n-1}. \]

The right side is a strictly positive $(n-1,n-1)$-form, and such a form has a unique smooth positive $(n-1)$st root, obtained by solving for the Hermitian matrix of the root from the positive cofactor matrix. Since the right side is $\mathrm{dd}^{\mathrm{c}}$-closed, $\Gamma _\delta $ is a smooth Gauduchon metric on $Y$.

Step 3: the numerator. For $\epsilon >0$, lemma 12.22.3 gives

C7A393B9C4B8 12.22.7
\begin{equation} \lim _{\delta \downarrow 0}\int _ Y(\pi ^*\theta +\epsilon h_ Y) \wedge \Gamma _\delta ^{n-1} =\int _ X\theta \wedge \Psi +\epsilon \int _ Yh_ Y\wedge \pi ^*\Psi \leq \int _ X(\theta +K\epsilon h_ X)\wedge \Psi . \end{equation}

The first term in the middle is nonnegative because $\theta $ is nef, and the second is positive because $\pi ^*\Psi $ is positive on the set where $\pi $ is biholomorphic, so the limit is positive.

Step 4: the denominator. We claim that

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\begin{equation} \lim _{\delta \downarrow 0}D_{\mu _ Y}(\Gamma _\delta ) =\int _ Y(\pi ^*\gamma ^ n)^{(n-1)/n}\mu _ Y^{1/n} \geq c^{1/n}D_{\mu _ X}(\gamma )>0. \end{equation}

To justify the limit on the exceptional locus, represent $\Gamma _\delta ^{n-1}$ by its positive cofactor matrix. The density of $(\Gamma _\delta ^ n)^{(n-1)/n}$ is a fixed dimensional constant times the $1/n$th power of the determinant of that matrix, and this expression extends continuously to positive semidefinite matrices. Since $Y$ is compact, dominated convergence applies. The inequality follows from $\mu _ Y\geq c\pi ^*\mu _ X$ and the change of variables for the degree-one map $\pi $.

Step 5: conclusion. Test the infimum defining $\lambda _{\mu _ Y}(\pi ^*\theta +\epsilon h_ Y)$ with $\Gamma _\delta $, use (12.22.7) and (12.22.8), and let $\delta \downarrow 0$. For every Gauduchon metric $\gamma $ on $X$ this gives

\[ \lambda _{\mu _ Y}(\pi ^*\theta +\epsilon h_ Y) \leq c^{-1/n} \frac{\int _ X(\theta +K\epsilon h_ X)\wedge \gamma ^{n-1}}{D_{\mu _ X}(\gamma )}. \]

Taking the infimum over $\gamma $ and then the $n$th power proves (12.22.6) with $C=c^{-1}$.

Dividing the logarithm of (12.22.6) by $\log \epsilon <0$ reverses the inequality. The bounded error $\log C$ and the rescaling $\epsilon \mapsto K\epsilon $ do not affect lower and upper limits, as in the proof of proposition 12.20.1. This gives $\underline a(\pi ^*\theta )\geq \underline a(\theta )$ and $\overline a(\pi ^*\theta )\geq \overline a(\theta )$. Finally, a proper holomorphic map of degree one between compact Riemann surfaces is an isomorphism, which settles the case $n=1$. $\square$

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