Proposition 24.4.88E5CAA6E5122

Proof. Write $a=\mathrm{d}t$ and $b=\mathrm d^{\mathrm c}t$. Then $\mathrm{d}a=0$, $\mathrm{d}b=\pi ^*c$, $\mathrm d^{\mathrm c}a=-\pi ^*c$ and $\mathrm d^{\mathrm c}b=0$. Consequently

so $\omega $ is balanced. Since $c$ is primitive, its eigenvalues relative to $\eta $ are $\lambda $ and $-\lambda $ for a function $\lambda \geq 0$, so $c^2=-\lambda ^2\eta ^2$. This proves the asserted sign of $\mathrm{dd}^{\mathrm{c}}\omega $.

To compute the volume, fix an elliptic fiber $E=\pi ^{-1}(b_0)$, where $b_0\in B$. In a logarithmic fiber coordinate $x+\mathrm{i}y$, the real periods are $\log q$ and $2\pi $, while $t$ equals $2x$ up to an additive constant and $D|_ E=\pi ^{-1}\mathrm{d}x\wedge \mathrm{d}y$. Thus $\int _ ED=T$. Since $\omega ^3=3\pi ^*(\eta ^2)\wedge D$, the volume formula follows.

For the mass identity, expand the cube in (24.4.9). The terms with one and with three factors $\mathrm{dd}^{\mathrm{c}}u$ integrate to zero, because $\omega ^2$ is closed and $\mathrm{dd}^{\mathrm{c}}u$ is exact. For the remaining term, put $S=\mathrm{dd}^{\mathrm{c}}\omega $. Integration by parts gives

where the last equality uses $\mathrm{d}S=0$. This proves the identity. Its last integral is nonnegative, and $u=0$ attains the upper bound $3TA$. Since smooth and bounded potentials have the same mass extrema, by theorem 12.15.4, this proves $v_+(\omega )=3TA$.

If $c_1(L)=0$ in real cohomology, then $c$ is exact and Stokes' theorem gives $\int _ Bc^2=0$. The pointwise identity $c^2=-\lambda ^2\eta ^2$ then forces $c=0$, and $\omega $ is closed. Otherwise $S=-\pi ^*(c^2)$ is nonzero, smooth and semipositive. For a pluriclosed metric $\gamma $, Stokes' theorem would give

a contradiction. Likewise, a closed positive current $R\geq \varepsilon \omega $ with $\varepsilon >0$ would give $0<\int _ XR\wedge S=\int _ X\omega \wedge \mathrm{dd}^{\mathrm{c}}R=0$. All these pairings are pairings of currents with smooth forms.

Finally, let $w$ be a local holomorphic fiber coordinate given by a frame of $L$, and let $\sigma $ be a local holomorphic two-form on $B$. The form $\pi ^*\sigma \wedge \mathrm{d}w/w$ is invariant under the quotient action. Changing the frame of $L$ adds to $\mathrm{d}w/w$ a horizontal one-form, whose wedge product with $\pi ^*\sigma $ vanishes. These local sections of $K_ X$ therefore have the transition functions of $\pi ^*K_ B$, which proves $K_ X\simeq \pi ^*K_ B$. $\square$