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Lemma 17.5.33EB970BD6E881

Lemma 17.5.33.EB970BD6E881 Let $A$ be a closed subset of $X$ of Lebesgue measure zero in every chart, and let $S$ be a closed positive $(1,1)$-current on $X$ whose local potentials are locally bounded on $X\setminus A$. Let $\tau $ be a smooth strictly positive real $(n-1,n-1)$-form, and put $\nu =\omega \wedge \tau $. If

\[ \langle S^ n\rangle =m\nu \qquad \text{for some }m\geq 0, \]

then

0511A794D8D0 17.5.34
\begin{equation} \left(\int _ X S\wedge \tau \right) \left(\int _ X\langle S^{n-1}\rangle \wedge \omega \right) \geq \frac mn\left(\int _ X\nu \right)^2. \end{equation}

For $n=1$ the second factor is $\int _ X\omega $.

Proof. We prove a pointwise inequality for smooth forms, apply it to regularizations of the potentials of $S$ on small balls in $X\setminus A$, and globalize with a partition of unity.

Step 1: a pointwise inequality. Let $\sigma $ be a smooth positive definite real $(1,1)$-form. At a given point, choose linear coordinates on the tangent space that diagonalize the Hermitian matrices of $\omega $ and $\sigma $ simultaneously, so that $\omega =\sum _ j\gamma _ j$ and $\sigma =\sum _ j\lambda _ j\gamma _ j$ with $\lambda _ j>0$, where $\gamma _ j=\mathrm i\, \mathrm dz_ j\wedge \mathrm d\bar z_ j$, and put $\mathrm dV=\gamma _1\wedge \cdots \wedge \gamma _ n$. Since $\tau $ is strictly positive (section 5.3), the form $\tau -c\, \omega ^{n-1}$ is positive at this point for some $c>0$. The product of the strongly positive form $\gamma _ j$ with this positive form is a nonnegative multiple of $\mathrm dV$ (proposition 5.3.2), and $\gamma _ j\wedge \omega ^{n-1}=(n-1)!\, \mathrm dV$, so $\gamma _ j\wedge \tau \geq c(n-1)!\, \mathrm dV$; thus $\gamma _ j\wedge \tau =t_ j\, \mathrm dV$ with $t_ j>0$. Expanding the products, we obtain

\[ \frac{\sigma \wedge \tau }{\omega \wedge \tau } =\frac{\sum _ j\lambda _ jt_ j}{\sum _ jt_ j},\qquad \frac{\sigma ^{n-1}\wedge \omega }{\sigma ^ n} =\frac1n\sum _ j\lambda _ j^{-1}. \]

Since $\lambda _ j\sum _ k\lambda _ k^{-1}\geq 1$ for every $j$, the product of these two ratios is at least $1/n$. If $\sigma $ is only semipositive, we apply this to $\sigma +s\omega $ and let $s\downarrow 0$. Thus, for every smooth semipositive real $(1,1)$-form $\sigma $, as densities with respect to any fixed smooth volume form,

B73555CD9C1D 17.5.35
\begin{equation} (\sigma \wedge \tau )(\sigma ^{n-1}\wedge \omega ) \geq \frac1n\sigma ^ n\nu . \end{equation}

Step 2: regularization on a ball. Since the local potentials of $S$ are locally bounded on the open set $X\setminus A$, every point of $X\setminus A$ is the center of a coordinate ball $B\Subset X\setminus A$ on which $S=\mathrm{dd}^{\mathrm{c}}v$ with $v$ plurisubharmonic and bounded. Fix such a ball, put $\eta =\mathrm{dd}^{\mathrm{c}}|z|^2$ as in section 7.8, and write $\nu =g\eta ^ n$ with $g$ smooth and positive on $B$. Since $v$ is bounded on $B$, the non-pluripolar powers of $S$ on $B$ are the Bedford–Taylor powers of $v$ (see the remark following definition 13.2.4); in particular $(\mathrm{dd}^{\mathrm{c}}v)^ n=mg\eta ^ n$.

Let $v_\varepsilon =v*\rho _\varepsilon $ be the regularizations of $v$ by the radial mollifiers of section 2.5. On each ball $B'$ concentric with $B$ and of smaller radius, the functions $v_\varepsilon $ are defined for $\varepsilon $ smaller than the difference of the radii, and by proposition 5.9.2(2) they are smooth and plurisubharmonic on $B'$ and decrease to $v$ as $\varepsilon \downarrow 0$. As recalled at the beginning of section 7.8, $(\mathrm{dd}^{\mathrm{c}}v_\varepsilon )^ n=\det \mathcal H(v_\varepsilon )\, \eta ^ n$. The last assertion of proposition 7.8.3, applied on $B$ to $v$ and the continuous density $mg$, therefore gives

\[ (\mathrm{dd}^{\mathrm{c}}v_\varepsilon )^ n=F_\varepsilon \eta ^ n, \qquad F_\varepsilon =\det \mathcal H(v_\varepsilon )\geq \bigl ((mg)^{1/n}*\rho _\varepsilon \bigr )^ n. \]

The right side converges to $mg$ uniformly on compact subsets of $B$, by lemma 2.5.1(1) applied to the continuous function $(mg)^{1/n}$. Hence $\sqrt{F_\varepsilon g}$ is bounded below by functions converging uniformly to $\sqrt mg$ on compact sets, and for every nonnegative $\chi \in C^\infty _ c(B)$,

93DD9AEE16DC 17.5.36
\begin{equation} \varliminf _{\varepsilon \downarrow 0} \int \chi \sqrt{F_\varepsilon g}\, \eta ^ n \geq \sqrt m\int \chi \nu . \end{equation}

Moreover, for every $\chi \in C^\infty _ c(B)$, as $\varepsilon \downarrow 0$ along any sequence,

\[ \int \chi \, \mathrm{dd}^{\mathrm{c}}v_\varepsilon \wedge \tau \longrightarrow \int \chi \, S\wedge \tau , \qquad \int \chi \, (\mathrm{dd}^{\mathrm{c}}v_\varepsilon )^{n-1}\wedge \omega \longrightarrow \int \chi \, \langle S^{n-1}\rangle \wedge \omega . \]

For the first convergence, note that $\int \chi \, \mathrm{dd}^{\mathrm{c}}v_\varepsilon \wedge \tau =\int v_\varepsilon \, \mathrm{dd}^{\mathrm{c}}(\chi \tau )$, by the definition of $\mathrm{dd}^{\mathrm{c}}$ on currents (section 5.3); since $v_\varepsilon \to v$ in $L^1$ on $\operatorname{Supp}\chi $ (lemma 2.5.1(2)), this tends to $\int v\, \mathrm{dd}^{\mathrm{c}}(\chi \tau )=\int \chi \, S\wedge \tau $.

For the second convergence, choose a ball $B'$ as above containing $\operatorname{Supp}\chi $, and a sequence $\varepsilon _ j\downarrow 0$. By theorem 7.3.7, applied on $B'$ to the closed positive current $\omega $ of bidimension $(n-1,n-1)$ and to $n-1$ copies of the decreasing sequence $(v_{\varepsilon _ j})$, the currents $(\mathrm{dd}^{\mathrm{c}}v_{\varepsilon _ j})^{n-1}\wedge \omega $ converge weakly to the Bedford–Taylor product $(\mathrm{dd}^{\mathrm{c}}v)^{n-1}\wedge \omega $. By proposition 7.2.2(1) and (4), these products are the exterior products with $\omega $ of the smooth forms $(\mathrm{dd}^{\mathrm{c}}v_{\varepsilon _ j})^{n-1}$ and of the current $(\mathrm{dd}^{\mathrm{c}}v)^{n-1}$, which equals $\langle S^{n-1}\rangle $ on $B$. (For $n=1$ both sides of the second convergence equal $\int \chi \, \omega $.)

Step 3: globalization. Let $\chi $ be a smooth function with compact support in $X\setminus A$ and $0\leq \chi \leq 1$. Cover $\operatorname{Supp}\chi $ by finitely many balls $B_ l$ as in Step 2, and write $\chi =\sum _ l\chi _ l$ with nonnegative $\chi _ l\in C^\infty _ c(B_ l)$; for instance, multiply $\chi $ by a partition of unity with compact supports subordinate to the cover of $X$ by the balls $B_ l$ and $X\setminus \operatorname{Supp}\chi $ (proposition 5.2.6). In each $B_ l$ we use the notation of Step 2 with an index $l$ and put

\[ a_{l,\varepsilon }=\int \chi _ l\, \mathrm{dd}^{\mathrm{c}}v_{l,\varepsilon }\wedge \tau ,\qquad b_{l,\varepsilon }=\int \chi _ l\, (\mathrm{dd}^{\mathrm{c}}v_{l,\varepsilon })^{n-1}\wedge \omega . \]

Let $\alpha _{l,\varepsilon }$ and $\beta _{l,\varepsilon }$ be the densities of $\mathrm{dd}^{\mathrm{c}}v_{l,\varepsilon }\wedge \tau $ and $(\mathrm{dd}^{\mathrm{c}}v_{l,\varepsilon })^{n-1}\wedge \omega $ with respect to $\eta _ l^ n$. By (17.5.35) with $\sigma =\mathrm{dd}^{\mathrm{c}}v_{l,\varepsilon }$, $\alpha _{l,\varepsilon }\beta _{l,\varepsilon }\geq F_{l,\varepsilon }g_ l/n$, so the Cauchy–Schwarz inequality gives

\[ \int \chi _ l\sqrt{F_{l,\varepsilon }g_ l}\, \eta _ l^ n \leq \sqrt n\int \chi _ l\sqrt{\alpha _{l,\varepsilon }\beta _{l,\varepsilon }}\, \eta _ l^ n \leq \sqrt{n\, a_{l,\varepsilon }b_{l,\varepsilon }}. \]

Summing over $l$ and applying the Cauchy–Schwarz inequality to the finite sum $\sum _ l\sqrt{a_{l,\varepsilon }b_{l,\varepsilon }}$, we obtain

\[ \left(\sum _ l a_{l,\varepsilon }\right) \left(\sum _ l b_{l,\varepsilon }\right) \geq \frac1n\left(\sum _ l \int \chi _ l\sqrt{F_{l,\varepsilon }g_ l}\, \eta _ l^ n\right)^2. \]

Now let $\varepsilon \downarrow 0$ along a sequence. By Step 2, the two sums on the left converge to $\int _ X\chi \, S\wedge \tau $ and $\int _ X\chi \langle S^{n-1}\rangle \wedge \omega $. These are at most the corresponding integrals over $X$ without $\chi $, because $S\wedge \tau $ and $\langle S^{n-1}\rangle \wedge \omega $ are positive measures and $0\leq \chi \leq 1$. By (17.5.36) and $g_ l\eta _ l^ n=\nu $, the lower limit of the right side is at least $(m/n)(\int _ X\chi \nu )^2$. Therefore

\[ \left(\int _ X S\wedge \tau \right) \left(\int _ X\langle S^{n-1}\rangle \wedge \omega \right) \geq \frac mn\left(\int _ X\chi \nu \right)^2. \]

Finally, we remove the cutoff. Fix a metric $d$ inducing the topology of $X$ (lemma 5.2.7), and for $k\geq 1$ let $K_ k=\{ x\in X:d(x,A)\geq 1/k\} $, with $K_ k=X$ if $A=\varnothing $. These compact subsets of $X\setminus A$ increase to $X\setminus A$, because $A$ is closed. By proposition 5.2.6 there are smooth functions $\chi _ k\colon X\to [0,1]$ with compact support in $X\setminus A$ and $\chi _ k=1$ on $K_ k$. Since $\nu $ has a smooth density and $A$ has Lebesgue measure zero, $\nu (K_ k)\to \nu (X\setminus A)=\int _ X\nu $, so $\int _ X\chi _ k\nu \to \int _ X\nu $. Applying the last inequality to $\chi =\chi _ k$ and letting $k\to \infty $ proves (17.5.34). Both factors on its left are finite: the first is the pairing of a current with a smooth form on the compact manifold $X$, and the second is finite by theorem 13.5.4. $\square$

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