Pome

Lemma 23.1.72D04CA3BB7868

Lemma 23.1.72.D04CA3BB7868 Let $f_ j=t_ ju_ j$, where $u_ j\in \mathrm{PSH}(X,\omega )$ are smooth, $\sup _ Xu_ j=0$ and $0<t_ j\to 0$. Suppose that $M_\omega (f_ j)=6+m$ and $(\omega +\mathrm{dd}^{\mathrm{c}}f_ j)^3\rightharpoonup \omega ^3+\sigma $. Then there is a unique positive measure $\mu $ on $\mathbb P^2$, of mass $m$, such that

\[ \sigma =\mathcal L\mu ,\qquad (\mathrm{dd}^{\mathrm{c}}f_ j+t_ j\omega )^2\rightharpoonup \tfrac 16\pi ^*\mu . \]

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. $\square$

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