Theorem 10.5.4DBDF98C510DC

Proof. The non-nef locus. By the monotonicity in $\varepsilon $, $E_{\mathrm{nn}}(\alpha )$ is the union over positive integers $j,k$ of the sets $\{ x:\nu (V_{\theta +\omega /j},x)\geq 1/k\} $. Each of them is analytic by theorem 6.9.5, and proper, because it is contained in the pole set of $V_{\theta +\omega /j}$. If $\alpha $ is nef, then $\alpha +\varepsilon \{ \omega \} $ is Kähler for every $\varepsilon >0$, so $V_{\theta +\varepsilon \omega }$ is bounded by corollary 10.4.7 and all these Lelong numbers vanish. Conversely, suppose that $\nu (V_{\theta +\varepsilon \omega },x)=0$ for all $x\in X$ and $\varepsilon >0$. By corollary 6.9.3, $\mathcal I(mV_{\theta +\varepsilon \omega })$ is the unit ideal at every point, so the approximations $G_ m$ of lemma 6.17.1 for the background $\theta +\varepsilon \omega $ are smooth and satisfy $\theta +\mathrm{dd}^{\mathrm{c}}G_ m\geq -(\varepsilon +c/m)\omega $. Hence $\alpha $ is nef.

Realization of the non-Kähler locus. Let $\alpha $ be big. Kähler currents with analytic singularities exist in $\alpha $ by proposition 10.4.6, and their pole sets are closed analytic subsets of $X$. Since $X$ has a countable basis, countably many of the open complements of these pole sets already have the same union as all of them: for every element of a countable basis that is contained in one of these complements, choose one such complement. The intersection of the corresponding countably many pole sets $Z_1,Z_2,\ldots $ is therefore $E_{\mathrm{nK}}(\alpha )$. Let $\mathcal I_{Z_ k}$ be the ideal sheaf of $Z_ k$, which is coherent by theorem 9.3.10. The increasing sequence of coherent ideals $\mathcal I_{Z_1}+\cdots +\mathcal I_{Z_ k}$, whose zero sets are $Z_1\cap \cdots \cap Z_ k$, is stationary on the compact manifold $X$ by lemma 6.8.4. Hence $E_{\mathrm{nK}}(\alpha )=Z_1\cap \cdots \cap Z_ N$ for some $N$.

Let $T_ k=\theta +\mathrm{dd}^{\mathrm{c}}\varphi _ k$, $k\leq N$, be the corresponding Kähler currents, and choose $\delta >0$ with $T_ k\geq \delta \omega $ for all $k\leq N$. Then $\varphi =\max _ k\varphi _ k$ belongs to $\mathrm{PSH}(X,\theta -\delta \omega )$, it has analytic singularities by proposition 6.14.8, and its pole set is exactly $Z_1\cap \cdots \cap Z_ N=E_{\mathrm{nK}}(\alpha )$. Thus $\theta +\mathrm{dd}^{\mathrm{c}}\varphi $ is a Kähler current realizing $E_{\mathrm{nK}}(\alpha )$, which is a proper analytic subset because $\varphi $ is finite almost everywhere. If $\nu (\alpha ,x)>0$, then $\nu (\varphi ,x)\geq \nu (V_\theta ,x)=\nu (\alpha ,x)>0$ by minimality of $V_\theta $ and proposition 10.5.2, so $\varphi (x)=-\infty $; hence $E_{\mathrm{nn}}(\alpha )\subset E_{\mathrm{nK}}(\alpha )$.

If $E_{\mathrm{nK}}(\alpha )=\varnothing $, the potential $\varphi $ just constructed has analytic singularities and no poles, so it is bounded, and proposition 10.4.6 shows that $\alpha $ is Kähler. Conversely, if $\alpha $ contains a Kähler form $\theta +\mathrm{dd}^{\mathrm{c}}u$, then $u$ is smooth and has analytic singularities without poles, so $E_{\mathrm{nK}}(\alpha )=\varnothing $.

The identity (10.5.5). Suppose first that $x\notin E_{\mathrm{nK}}(\alpha )$. Then $\alpha $ is big, and there is a Kähler current $\theta +\mathrm{dd}^{\mathrm{c}}\varphi \geq \delta \omega $ with analytic singularities and $\varphi (x)>-\infty $; in particular $\varphi $ is bounded near $x$ and $\nu (\varphi ,x)=0$. For $0<\varepsilon <\delta $, the current $\theta -\varepsilon \omega +\mathrm{dd}^{\mathrm{c}}\varphi $ is positive and lies in $\alpha -\varepsilon \{ \omega \} $, which is therefore pseudo-effective, and the bound $\nu (\alpha -\varepsilon \{ \omega \} ,x)\leq \nu (\varphi ,x)=0$ shows that $x\notin E_{\mathrm{nn}}(\alpha -\varepsilon \{ \omega \} )$.

Conversely, suppose that $x\notin E_{\mathrm{nn}}(\alpha -\varepsilon \{ \omega \} )$ for some $\varepsilon >0$. By the convention in definition 10.5.1, the class $\alpha -\varepsilon \{ \omega \} $ is pseudo-effective. For $0<\eta <\varepsilon /2$ put $u=V_{\theta -(\varepsilon -\eta )\omega }$; then $\nu (u,x)\leq \nu (\alpha -\varepsilon \{ \omega \} ,x)=0$. By corollary 6.9.3, $\mathcal I(mu)$ is the unit ideal near $x$ for every $m$, and the approximation $G_ m$ of lemma 6.17.1 for $u$ and the background $\theta -(\varepsilon -\eta )\omega $ is smooth near $x$ and satisfies

For large $m$ this is a Kähler current in $\alpha $ with analytic singularities whose potential is finite at $x$. Hence $\alpha $ is big and $x\notin E_{\mathrm{nK}}(\alpha )$. This proves (10.5.5); the argument also shows that $\alpha $ is big as soon as the right-hand side of (10.5.5) is not all of $X$, so the formula is consistent with the convention $E_{\mathrm{nK}}(\alpha )=X$ for classes that are not big. $\square$