23.1.71 Fiber measures7802C5C9B69D

Proof. Put $H_ j=\mathrm{dd}^{\mathrm{c}}f_ j+t_ j\omega $. By proposition 13.15.20, applied with $h=\omega $ and $n=3$, every convergent subsequence of $H_ j^2$ has a positive closed limit $K$ with $\sigma =3\omega \wedge K$; moreover $\alpha \wedge K=0$, because $\alpha $ is closed and semipositive, so that $\alpha \wedge K$ is a positive measure with $\int _ X\alpha \wedge K=0$. In holomorphic coordinates $(\xi _1,\xi _2,w)$ on a local trivialization of the elliptic bundle, the coefficient matrix of $K$ on the three exterior pairs is positive semidefinite as a matrix of measures. To see this directly, note that positivity of the matrices of $H_ j^2$ passes to the limit when they are tested against constant vectors, and take Radon–Nikodym derivatives with respect to the diagonal trace. Since $\alpha $ is strictly positive in the horizontal directions, the identity $\alpha \wedge K=0$ makes both diagonal entries containing $\mathrm{d}w$ vanish. Positivity then makes their rows and columns vanish, so only the horizontal top-degree coefficient remains.

Since $K$ is closed, both real fiber derivatives of that coefficient vanish in the sense of distributions. Testing against a compactly supported base function times a nonconstant Fourier mode on $E=\mathbb C/(\mathbb Z+2\pi \mathrm{i}\mathbb Z)$ shows that every such mode vanishes. By density of trigonometric polynomials in $C(E)$, the coefficient is a base measure times the constant fiber area. Changes of bundle coordinates translate $w$ by a holomorphic function of the base and preserve that area. Thus $K=\pi ^*\tau $ for a positive measure $\tau $ on $\mathbb P^2$.

Since $\alpha \wedge K=0$, (23.1.65) gives $\sigma =3\beta \wedge \pi ^*\tau =\mathcal L(6\tau )$. Therefore $\mu =6\tau =\pi _*\sigma $ is uniquely determined by $\sigma $, and it has mass $m$. In particular, every subsequential limit of $H_ j^2$ is the same, which proves convergence of the entire sequence. $\square$

Proof. Start with smooth $f_ j=t_ ju_ j$ realizing the shrinking defect $(m/6)\omega ^3=\mathcal L(m\, \omega _{\mathrm{FS}}^2)$, and put $H_ j=\mathrm{dd}^{\mathrm{c}}f_ j+t_ j\omega $. By lemma 23.1.72, $H_ j^2\rightharpoonup (m/6)\alpha ^2$, while $H_ j$ has trace mass $O(t_ j)$. Fix $A\in GL(3,\mathbb C)$, regarded as an automorphism of $X$, and let $a$ be the induced projective transformation of $\mathbb P^2$. Choose $C_ A\geq 1$ with $A^*\omega \leq C_ A\omega $, and put

For this fixed $A$, the trace mass of $A^*H_ j$ is $O_ A(t_ j)$, and the positive difference $(H_ j^ A)^2-A^*(H_ j^2)$ has trace mass $O_ A(t_ j^2)$. Since $\alpha =\pi ^*\omega _{\mathrm{FS}}$, it follows that

The limiting measure has mass $m$, so the integral identity of lemma 23.1.68, applied with the parameter $C_ At_ j$, gives $d_ j^ A=M_\omega (f_ j\circ A)-6\to m$. For large $j$, multiply $f_ j\circ A$ by $\sqrt{m/d_ j^ A}\to 1$. The mass is then exactly $6+m$, and the weak limit is unchanged. The underlying normalized potential is $u_ j\circ A/C_ A\in \mathrm{PSH}(X,\omega )$, with parameter $C_ At_ j\sqrt{m/d_ j^ A}\to 0$. The total-variation estimate of lemma 23.1.68 therefore shows that $\mathcal L(m\, a^*\omega _{\mathrm{FS}}^2)$ is a shrinking defect; any prescribed bounds $\varepsilon _ j$ are met by passing to a sufficiently late subsequence.

Now take $A_ l=\operatorname {diag}(R_ l^{-1},1,1)$, $R_ l\to +\infty $. Since

and $a_ l^{-1}[q_0:q_1:q_2]=[R_ lq_0:q_1:q_2]\to [1:0:0]$ outside the hyperplane $\{ q_0=0\} $, which has zero volume, dominated convergence gives $a_ l^*\omega _{\mathrm{FS}}^2\rightharpoonup \delta _{[1:0:0]}$. By weak continuity of $\mathcal L$, the shrinking defects $\mathcal L(m\, a_ l^*\omega _{\mathrm{FS}}^2)$ converge to $mh_{[1:0:0]}$. For each $l$, choose one corrected potential for $A_ l$, sufficiently late in its sequence that its parameter is less than $\varepsilon _ l$ and its measure differs from the corresponding limit by less than $1/l$ on the first $l$ functions of a fixed dense sequence of test functions. This diagonal sequence has the exact mass $6+m$ and the limit $\omega ^3+mh_{[1:0:0]}$. Unitary automorphisms move this fiber to any prescribed fiber. $\square$

Proof. Put $m=b-6$. Necessity follows from lemma 23.1.72. Finite convex combinations of the measures $mh_ p$ are shrinking defects by lemma 23.1.73, proposition 23.1.70, the latter applied to an atomic measure $\rho $ supported on unitary maps that move a fixed fiber to the desired fibers. Approximate an arbitrary base measure of mass $m$ by finite atomic measures of the same mass, using finite partitions of diameter tending to zero. Weak continuity of $\mathcal L$ and the diagonal choice on dense test functions used in the preceding proof give a realizing sequence for every prescribed sequence $\varepsilon _ j$.

Every nonempty open subset of a fiber has positive Haar measure. Local trivializations therefore show that $\mathcal L\mu $ has support $\pi ^{-1}(\operatorname {supp}\mu )$, and the fiber measures have no point atoms. If $B$ is a pluripolar subset of the base, local plurisubharmonic functions whose pole sets contain $B$ pull back by the submersion to such functions on $X$. Hence $\pi ^{-1}(B)$ is pluripolar, and $\mathcal L\mu (\pi ^{-1}(B))=\mu (B)$; thus $\mathcal L\mu $ charges a pluripolar set whenever $\mu $ does.

Conversely, suppose that $\mu $ charges no pluripolar subset of the base. Cover a pluripolar set in $X$ by countably many local product charts, each carrying a plurisubharmonic function $q(\xi ,w)\not\equiv -\infty $ whose pole set contains the part of the set in that chart. For a relatively compact disk $D$ in the fiber direction, the function

is plurisubharmonic and not identically $-\infty $ on a smaller base chart. Indeed, local integrability of $q$ and Fubini's theorem show that $F$ is locally integrable, and in particular finite almost everywhere; a common upper bound and the reverse Fatou lemma give upper semicontinuity; and integrating the submean inequalities in the base directions gives the submean property. The restriction of $q$ to a fiber is either subharmonic with a pole set of zero area or identically $-\infty $. The fibers of the second kind lie over the pluripolar set $\{ F=-\infty \} $, which has zero $\mu $-measure. Integrating the Haar measures of the fibers against $\mu $ therefore shows that each local pole set has zero $\mathcal L\mu $-measure, and the countable cover proves the assertion for this pluripolar set. For arbitrary pluripolar subsets, the same argument applies to Borel pole sets containing them, with outer measure. Similarly, for the first implication, a Borel $\mathcal L\mu $-null set containing $\pi ^{-1}(B)$ has a measurable fiber-mass function equal to one on $B$, which gives the corresponding outer-measure statement. $\square$