24.4.7 Primitive elliptic bundles148E1DFBEDF1

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 $K_ X\simeq \pi ^*K_ B$. $\square$

Proof. For $p=0$ the inequality holds because $F(M,T)>0$. Otherwise set $a=\max p>0$, $-b=\min p<0$ and $r=a+b$. Integrating $-p'<1$ over a forward interval $I$ from a maximum to the next minimum gives $r<|I|\leq T$; also $r<2M$, since $|p|<M$. The function $G(s)=(a-s)(s+b)$ is nonnegative on the range of $p$. Since $p$ has mean zero and $-p'<1$, integration over $I$ gives

As $ab\leq r^2/4$, the average of $p^2$ is at most $r^2/4-r^3/(6T)$. This expression increases strictly for $0<r<T$, and its value at $r=\min \{ 2M,T\} $ is $F(M,T)$. Since $r<\min \{ 2M,T\} $, the strict inequality follows.

For sharpness, consider the following $T$-periodic profiles $p_0$. When $T\geq 2M$, let $p_0$ be constant equal to $M$ and to $-M$ on two intervals of length $(T-2M)/2$ each, joined by a descent of slope $-1$ and length $2M$, and close the period by an upward jump. When $T\leq 2M$, let $p_0(s)=T/2-s$ for $0\leq s<T$, with its upward jump at the end of the period. In both cases $p_0$ has mean zero, $|p_0|\leq M$, $p_0'\geq -1$ in the sense of distributions, and the average of $p_0^2$ is $F(M,T)$; both constructions apply when $T=2M$. Convolution with a nonnegative periodic approximate identity preserves the mean and the bound on the distributional derivative, and multiplication by $1-\varepsilon $ makes both inequalities strict. As the convolution scale and $\varepsilon $ tend to zero, the resulting smooth functions converge to $p_0$ in $L^2$, which proves that the supremum is $F(M,T)$. $\square$

Proof. Put $p=f'$. The chain rule gives

Positivity on the vertical line implies $p'>-1$. Over every point of $B$, the function $t$ takes every value modulo $T$, so restriction to $\ker \partial t$ shows that $\eta +pc+\mathrm{dd}^{\mathrm{c}}v$ is a Kähler form on $B$ for every value of $p$. Its square has integral

since primitivity removes the mixed term and Stokes' theorem removes the exact terms. Hence $|p|<\kappa ^{-1/2}$.

Only the vertical derivative enters (24.4.9), because $-\pi ^*(c^2)$ has top degree in the base directions. For a $T$-periodic function $F_0$, integration over a fiber gives $\int _ E F_0(t)D=\int _0^ T F_0(s)\, \mathrm{d}s$, and therefore

Now apply lemma 24.4.10 with $M=\kappa ^{-1/2}$. The formula for $F$ gives $F(M,T)<M^2$, which proves positivity of the bound.

For $v=0$, diagonalizing $c$ with respect to $\eta $ shows that the horizontal positivity condition is exactly $|p|<M_0$; together with $p'>-1$ it also suffices for positivity of the full form. Moreover $\kappa $ is the $\eta ^2$-average of $\lambda ^2$, so $\kappa \leq M_0^{-2}$. Lemma 24.4.10 with $M=M_0$ therefore proves the claimed positive lower bound and its nonattainment. The smooth approximating functions in its proof have mean zero and hence periodic primitives $f_\varepsilon $, and the potentials $f_\varepsilon (t)$ prove that the bound is sharp. $\square$

Proof. In a hyperbolic plane of the K3 lattice, choose a basis $e,f$ with $e^2=f^2=0$ and $e\cdot f=1$, and put $\ell =e-2f$. This vector is primitive and has square $-4$. Choose a period in $\ell ^\perp $ outside the hyperplanes perpendicular to integral vectors not in $\mathbb Z\ell $. Such a period exists by Baire's theorem: the period domain in $\ell ^\perp $, of signature $(3,18)$, is nonempty and each excluded hyperplane is proper. The period theorem and the description of integral $(1,1)$-classes give the asserted Picard group. An irreducible curve would have nonzero class, because its Kähler area is positive, and hence square $-4m^2<-2$ for some integer $m\neq 0$, contradicting adjunction. Thus there are no curves.

The orthogonal complement of $\ell $ in $H^{1,1}(B,\mathbb R)$ has signature $(1,18)$. Choose a class of positive square there, with sign chosen so that it lies in the component of the positive cone containing the Kähler classes. Since there are no rational curves, the Kähler-cone theorem shows that this class is Kähler; choose $\eta $ in it. Proposition 24.4.8 then gives the assertions about the metric and the canonical bundle.

If a divisor $Z\subset X$ dominated $B$, the push-forward of its integration current would be $\pi _*[Z]=m$ for a positive generic degree $m$. Stokes' theorem and the projection formula would then give

Thus no divisor dominates $B$. The image of any other irreducible divisor would be a curve or a point; the base has no curves, and a fiber has dimension one, so both alternatives are impossible. Likewise, the image of an irreducible compact curve is a point, and the curve is therefore an entire elliptic fiber. $\square$

Proof. Since $\eta \wedge c=0$, the form $g\eta \wedge D$ is closed. The chain rule gives

the $a'c\wedge \eta ^2$ term having excess base degree. Thus $\mathrm{dd}^{\mathrm{c}}\Gamma =0$ forces $a''=0$, and periodicity makes $a$ constant. In a positive coframe with $\eta =\sigma _1+\sigma _2$, $D=\sigma _3$, the complementary coefficient matrix of $\Gamma $ is $\operatorname {diag}(g,g,2a)$. Hence positivity is exactly $a,g>0$. The displayed $\gamma $ is positive and squares to $\Gamma $. Multiplication and fiber integration give

For fixed $d_\Gamma >0$ set $s=d_\Gamma /(TA)$. Then $0<a<s$, $\bar g=s-a$, and $\sqrt a(s-a)$ has its unique maximum at $a=s/3$. Substitution proves the bound and its equality condition. $\square$