12.19 Vanishing lower mass108F40B2DFAA

Proof. For fixed $\delta $, choose a constant $M_\delta >0$ with $B_\delta \leq M_\delta \omega $; this shows that $C_\delta \leq M_\delta $. Suppose instead that $C_{\delta _ l}\leq C$ along a sequence $\delta _ l\to 0$. Then $\int _ X(C\omega -B_{\delta _ l})\wedge \Theta \geq 0$ for every admissible $\Theta $, and in particular for the squares of Gauduchon metrics. By theorem 11.6.1, there are quasi-plurisubharmonic functions $v_ l$ such that

By theorem 5.25.2, a subsequence converges in $L^1$ to a $C\omega $-plurisubharmonic function $v$, not identically $-\infty $. Since $\delta ^{-2}p(\zeta /\delta )\sigma _\zeta $ is an approximate identity in the variable $\zeta $,

On a coordinate ball where $\chi =1$, add to $v$ a smooth strictly plurisubharmonic function whose complex Hessian dominates $C\omega $. The resulting plurisubharmonic function has a current that dominates $[\zeta =0]$, so its Lelong number is at least one at every point of an open subset of this hypersurface. Smooth corrections do not change Lelong numbers, so the same holds for $v$. Siu's analyticity theorem, recalled in 5.34.2, shows that $Z=\{ \nu (v,\cdot )\geq 1\} $ is an analytic subset of $X$. It is a proper subset because $v$ is not identically $-\infty $.

Let $\widetilde Z$ be the inverse image of $Z$ in $\Omega $. It contains an open subset of $H_0\cap \Omega $, where $H_0=\{ w_0=z_0\} $. This surface is a projective plane with two points removed, so it is connected and irreducible, and the identity principle gives $H_0\cap \Omega \subset \widetilde Z$. Since $\widetilde Z$ is invariant under $g$, also

Now use the chart $z_1=1$ near the point $[0:1:1:0]\in \Omega $. There $H_ k$ is the graph $z_0=q^{-k}w_0$. For each fixed $(w_0,w_1)$ near $(1,0)$, every local holomorphic defining function of $\widetilde Z$ vanishes at the distinct values $z_0=q^{-k}w_0$, which tend to zero as $k\to +\infty $. The one-variable identity theorem makes each defining function vanish identically on a polydisk. Thus $\widetilde Z$ contains a nonempty open subset of the connected manifold $\Omega $, contradicting properness. Hence no subsequence of $C_\delta $ stays bounded as $\delta \downarrow 0$. $\square$

Proof. On the parameter supports, $|a-bu|\geq 3/4-2/16=5/8$. Hence the matrix $\left(\begin{smallmatrix}1& u\\ b& a\end{smallmatrix}\right)$ is uniformly invertible, and the projective line (12.19.9) lies in $\Omega $. Distinct deck translates of this line are disjoint: equality at a point fixes $[z_0:z_1]$, and invertibility of the matrix then forces the deck scalar to be one. Thus each map is an embedding onto a compact rational curve in $X$.

The evaluation map from $\mathbb P^1$ times the open parameter domain to $X$ is a holomorphic submersion. Indeed, after fixing a projective gauge, variations of $u$ and $a$ span the two $w$-coordinates at points with $z_1\ne 0$, variations of $\tau $ and $b$ span them at points with $z_1=0$, and the coordinate on $\mathbb P^1$ supplies the remaining direction. For $\varepsilon >0$, the average is the integral along the fibers of this submersion of a smooth compactly supported parameter density, which proves smoothness. As an average of closed positive integration currents, $T_{j,\varepsilon }$ is closed and semipositive.

All curve parameters lie in one compact subset of the domain where the matrix is invertible, so the $\omega $-areas of the curves are uniformly bounded. The weights in $\tau $ and $u$ have mass one, and the weight in $(a,b)$ has a fixed finite mass. This proves the degree bound.

When $u=0$, the curve is given in the product chart by $\zeta =\tau $ and $y=ax+b$. Integrating in $b$ and $\tau $ yields (12.19.11); the $(1,1)$-factor there is the integration current of the graph of $y=ax+b$. On $\{ |x|,|y|\leq 1/4\} $, the density $f(a,y-ax)$ is bounded below by a positive constant for $a$ in a fixed small disk about $1$. The covectors $\mathrm{d}y-a\, \mathrm{d}x$, for $a$ in that disk, have no common nonzero annihilator, so integrating their squared absolute values and using compactness of the unit sphere in the $(x,y)$-directions gives the asserted lower bound.

For general $u$, the curve is given by

On a fixed compact subset of the chart, $1+ux$ stays away from zero for all sufficiently small $u$. Solving for the parameters gives $\tau =\zeta -\log (1+ux)$ and $b=(1+ux)y-ax$. Regarding $\rho _ j$ as a smooth function on $E$, this change of variables also applies near the boundary of its support. The Jacobians, the density coefficients and any fixed number of their derivatives depend smoothly on $u$. Averaging against the approximate identity in $u$ therefore gives local $C^\infty $ convergence. This convergence is needed only for each fixed $j$; it is not uniform in $j$. $\square$

Proof. Write the cofactor matrix of $\widehat\Theta _ j$ in block form, with a $2\times 2$ Hermitian block $A_ j$ in the $(x,y)$-directions and a column $\kappa _ j\in \mathbb C^2$:

The cofactor matrix of $T_{j,0}$ has the form $\operatorname {diag}(B_ j,0)$. Let $I_2$ denote the $2\times 2$ identity matrix. The lower bound for $K_ f$ and the definition of $\rho _ j$ give a constant $c_0>0$ such that $B_ j\geq c_0\delta _ j^{-2}I_2$ on $V_ j$. Since $Q_ j>0$, the Schur complement $A_ j-\kappa _ j\kappa _ j^*/\widehat F_ j$ is positive definite, and

by monotonicity of the determinant on positive Hermitian matrices. Combining the root formula with (12.19.8), we obtain

The smaller region $W_ j=\{ |x|,|y|<1/8,\ |\zeta |<\delta _ j/8\} $ has $\mathrm{d}V_0$-volume equal to a fixed positive constant times $\delta _ j^2$. Consequently

For this fixed $j$, lemma 12.19.10 gives smooth convergence of $T_{j,\varepsilon }$ to $T_{j,0}$ on the closure of $W_ j$. The cofactor matrix of $\widehat\Theta _ j+T_{j,0}$ is positive definite there, so the integrand of the root volume depends continuously on these smooth coefficients. We can therefore choose $0<\varepsilon _ j\leq 1/16$ so small that

Set $\Gamma _ j=\widehat\Theta _ j+T_{j,\varepsilon _ j}$. It is smooth, strictly positive and $\mathrm{dd}^{\mathrm{c}}$-closed. Its degree is bounded by (12.19.7) and lemma 12.19.10, whereas its root volume is at least $c_1j^2/2$. $\square$

of theorem 12.19.1. Fix a Hermitian form $h$ on $X$, and choose $A>0$ with $h\leq A\omega $. Let $\eta _ j$ be the positive root of $\Gamma _ j$, so that $\eta _ j^2=\Gamma _ j$. These metrics are Gauduchon, and

Set $\mu _ j=\eta _ j^3/R_ j$ and solve the smooth equation in the background $h$:

Since $D_{\mu _ j}(\eta _ j)=R_ j^{2/3}$, using $\eta _ j$ as a test in theorem 12.2.3 gives

The potentials $u_ j$ are smooth, hence bounded, so $v_-(h)=0$. $\square$

Proof. In dimension one the mass is $\int _ Xh>0$, and the surface case is corollary 12.18.1. In higher dimensions we use the monotonicity and the homogeneity of degree $n$ of $v_-$, which follow directly from (12.15.5), to pass from a split metric to every Hermitian metric.

For $n=3$ this is theorem 12.19.1. For $n>3$, take the product of the threefold $X$ with $\mathbb P^{n-3}$, let $\kappa $ be a Fubini–Study form on the projective factor, and consider the split metric $h_0=\omega +\kappa $, with pullbacks understood. For the potentials $u_ j$ given by the theorem for the background $\omega $,

For another Hermitian form $h$, choose $A$ with $h\leq Ah_0$. Monotonicity and homogeneity give $v_-(h)\leq A^ n v_-(h_0)=0$. $\square$

Proof. Take the product of the threefold above with the Hopf threefold of section 23.1, with a split metric. Potentials pulled back from the first factor make the masses tend to zero, while the smooth potentials of theorem 23.1.1, pulled back from the second factor, make them tend to infinity. In each case the product integral is $\binom 63$ times the product of the two factor masses. The metric comparison in the preceding proof handles the lower mass for every Hermitian form, and the metric independence in proposition 13.14.15 handles the upper mass. Products with projective spaces give every higher dimension by the same expansion of wedge powers. $\square$

Proof. Although $t$ is defined only on the covering space $\Omega $, both $\mathrm{d}t$ and $\mathrm d^{\mathrm c}t$ descend to $X$. The Fubini–Study potentials $\log \| z\| ^2$ and $\log \| w\| ^2$ give $\mathrm{dd}^{\mathrm{c}}t=\beta -\alpha $. The forms $\alpha ,\beta $ are closed, and $\alpha ^2=\beta ^2=0$ since each is pulled back from a curve. As $D^2=0$, we obtain

In the last identity, the term involving $D$ contains $\mathrm{d}t$ twice, and $(\alpha +\beta )\wedge (\beta -\alpha )=0$. Thus $\omega $ is balanced, hence Gauduchon, and the formula for $\mathrm{dd}^{\mathrm{c}}\omega $ follows from that for $\mathrm{dd}^{\mathrm{c}}D$.

On a fiber, write $\zeta =a+\mathrm{i}b$, where both real periods are $2\pi $. There $t=2a$ plus a constant, so

Since $\omega ^3=6\alpha \wedge \beta \wedge D$, fiber integration and the normalization of the Fubini–Study forms give $\int _ X\omega ^3=24\pi $.

The identity (12.16.4) for Gauduchon metrics on threefolds, with $\mathrm{dd}^{\mathrm{c}}\omega =2\alpha \wedge \beta $, gives (12.19.19). Its last integral is nonnegative. Thus every smooth strictly $\omega $-plurisubharmonic potential has mass at most $24\pi $, and the constant potential attains this value. Regularization of bounded potentials, as in the proof of theorem 12.15.4, gives the same supremum over bounded potentials, so $v_+(\omega )=24\pi $. The metric independence in proposition 13.14.15 then gives the bounded mass property of $X$. The assertion about the lower mass is theorem 12.19.1. $\square$

Proof. Write $p=f'$, an $L$-periodic function with mean zero. The chain rule gives

Positivity on the vertical line implies $p'>-1$. On the complex plane $\ker \partial t$, the form $D$ vanishes and the projection to the base is an isomorphism. For each value of $t$ modulo $L$, every base point has a point above it with that value. Hence, writing $\alpha _ B,\beta _ B$ for the two Fubini–Study forms on $B$, the form $(1-p(s))\alpha _ B+(1+p(s))\beta _ B+\mathrm{dd}^{\mathrm{c}}v$ is positive on $B$ for every $s$. Integration over the two rulings gives $1-p>0$ and $1+p>0$. Thus $|p|<1$ and $p'>-1$; when $v=0$, these inequalities also suffice for strict positivity.

In a logarithmic fiber coordinate $x+\mathrm{i}y$, both real periods are $2\pi $, $t=2x$ plus a constant, and $D|_ E=\pi ^{-1}\mathrm{d}x\wedge \mathrm{d}y$. Hence $\int _ E F(t)D=\int _0^ L F(s)\, \mathrm{d}s$ for every $L$-periodic function $F$. Only the derivative of $u$ along the fibers contributes to (12.19.19), and it equals $p(t)$ times the derivative of $t$. Fiber integration therefore gives

We now bound $\int _0^ Lp^2$. Suppose first that $p\not\equiv 0$, and put $m_+=\max p>0$, $m_-=-\min p>0$ and $r=m_++m_-<2$. The function $G(s)=(m_+-s)(s+m_-)$ is nonnegative on $[-m_-,m_+]$. Choose an interval $I$ of length at most $L$ that runs from a maximum point of $p$ forward to the next minimum point. Since $p'\geq -1$, we have $G(p)\geq -G(p)p'$. Using that $p$ has mean zero, we obtain

As $m_+m_-\leq r^2/4$, it follows that

where the last inequality holds because the middle expression is strictly increasing in $r\in (0,2]$, as $L>2$. The inequality $\int _0^ Lp^2<L-4/3$ holds trivially when $p=0$. Equation (12.19.21) now gives the strict lower bound $8$.

To prove sharpness, put $a=(L-2)/2$ and define an $L$-periodic function $p_0$ by

It has mean zero, $|p_0|\leq 1$ and $\int _0^ L p_0^2\, \mathrm{d}s=L-4/3$. Its distributional derivative is $-1$ on the descending interval and zero on the two constant intervals, and it has a positive atom where the period closes up. Thus $p_0'\geq -1$ as distributions on the circle. Let $\rho _\varepsilon $ be a nonnegative smooth periodic approximate identity of integral one, and set $p_\varepsilon =(1-\varepsilon )(p_0*\rho _\varepsilon )$ for $0<\varepsilon <1$. Then $p_\varepsilon $ is smooth, has mean zero, and

Its primitive $f_\varepsilon $ is smooth and periodic, so $u_\varepsilon =f_\varepsilon (t)$ is admissible. By (12.19.21), its mass tends to $8$. $\square$

Proof. By proposition 12.19.18, the nonzero smooth semipositive $(2,2)$-form $\alpha \wedge \beta =\mathrm{dd}^{\mathrm{c}}(\omega /2)$ is $\mathrm{dd}^{\mathrm{c}}$-exact. If a Hermitian metric $\eta $ satisfied $\mathrm{dd}^{\mathrm{c}}\eta =0$, Stokes' theorem would give

a contradiction. The first inequality holds because $\eta $ is positive in the vertical direction at every point.

Similarly, if a closed positive current $T$ satisfied $T\geq \varepsilon \omega $ for some $\varepsilon >0$, positivity and Stokes' theorem for currents would give

Hence there is no closed Kähler current. $\square$