23.1 Unbounded mass on Hopf manifolds142249962360

Proof. Since $\beta $ restricts to $2\mathrm{d}A_ E$ on each fiber,

By (11.1.6) and (23.1.3), followed by an integration by parts against the closed form $\alpha ^2$,

Wedging with $\alpha ^2$ removes every component of $\mathrm{d}V\wedge \mathrm d^{\mathrm c}V$ except the vertical one, which is $\mathrm i|V_ w|^2\mathrm{d}w\wedge \mathrm{d}\bar w/(2\pi )=2|V_ w|^2\mathrm{d}A_ E$. Integration over the fibers proves the formula. $\square$

Proof. In these coordinates,

and

Choose horizontal vectors of unit length for $2\pi \alpha $, one orthogonal to the radial direction and the other

Adjoin the vertical vector $\partial _ w$. In this frame the matrix of $2\pi (\omega +\mathrm{dd}^{\mathrm{c}}V)$ is

Taking the Schur complement of the last diagonal entry gives (23.1.8). At $\xi =0$, the derivatives of $V$ in the base directions vanish and $\partial s=\mathrm{d}w$, which gives the last assertion. $\square$

Proof. Integrating the nonconstant Fourier modes gives

This proves smoothness, reality and (23.1.13). The least nonzero value of $\pi ^2|\nu _{p,q}|^2$ is $1/4$, and any fixed number of differentiations introduces coefficients of polynomial growth in $|\nu _{p,q}|$. Factoring out $\mathrm{e}^{-t/4}$ for $t\geq 1$ therefore leaves a convergent Gaussian series, which proves (23.1.16).

To prove (23.1.14), fix a lift of $w$ and give $\Lambda $ the probability weights

Logarithmic differentiation of the Gaussian sum yields

and one more derivative gives

The expression in parentheses is a variance and is nonnegative.

Finally, Parseval's identity gives, with $\mathcal E_\eta (t)$ denoting the energy in (23.1.15),

The Gaussian formula at the origin implies

To see this, isolate the zero lattice vector and use $\mathrm{e}^{-|\lambda |^2/(2t)}\leq \mathrm{e}^{-1/(4t)}\mathrm{e}^{-|\lambda |^2/4}$ for $\lambda \ne 0$. The error term is integrable near $t=0$, and $\mathcal E_\eta (1)=O(\eta ^2)$. Integrating the derivative between $t$ and $1$ proves (23.1.15). $\square$

Proof. Choose a nearest lattice point $\lambda _0$, and put $d=|w-\lambda _0|$ and $s_\lambda =|w-\lambda |^2-d^2\geq 0$. The distance $d$ is bounded by the diameter of a fundamental rectangle. By the triangle inequality, $s_\lambda \geq |\lambda -\lambda _0|^2/4-C$, and outside a fixed finite set of $\lambda $ we have $s_\lambda \geq |\lambda -\lambda _0|^2/8$. Consequently,

The probability formula in the previous proof gives, with $G=|t\partial _{\bar w}\log k_ t|$,

If $d\geq 1/4$, this is at most $Cd$. If $d<1/4$, the nonzero lattice distances are at least one and

Summing the resulting Gaussian bound gives $G\leq C(d+\mathrm{e}^{-c/t})$ in all cases. When $G>2C\mathrm{e}^{-c/t}$, we have $d\geq G/(2C)$, and the bound $tk_ t\leq C\mathrm{e}^{-d^2/t}$ gives (23.1.18). In the remaining case $G^2/t$ is bounded, and (23.1.18) follows from $tk_ t\leq C$ after increasing $C_0$. $\square$

Proof. If $\sigma ^2t\geq 1/100$, the bound $tk_ t\leq C_0$ gives (23.1.20) after enlarging $C$. Assume therefore that $\sigma ^2t<1/100$. If $|\zeta -(1-\eta )^{-1}|\geq 1/4$, the subtracted expression is at least $h/[100(1+h)]$. Since $\log (1/(\eta t))>2$, the definition of $\sigma $ gives

Multiplication by $k_ t/(1+h)$ shows that the left side of (23.1.20) is nonpositive.

Otherwise $|\zeta |\geq 3/4$, so

Discarding the nonnegative square and applying lemma 23.1.17 gives (23.1.20) in this case as well. $\square$

Proof. We first derive an exact formula for $Q[W]$ on $[0,b]$, then bound its negative part separately on $[0,a^{-1}]$ and on $[a^{-1},b]$, integrate these bounds, and finally examine the ends of the radial line.

Step 1: the Schur complement on $[0,b]$. Write $F=F_\eta $ for the heat potential. On $[0,b]$ we have $q\geq 1$, hence $\vartheta (q)=1$ and $W=F(t(x),w)$. Set

The heat equation gives $F_ t=h-\eta $, and the chain rule gives

Substituting these in $Q[W]$, using $h_ t=h((\log k_ t)_{w\bar w}+|\partial _{\bar w}\log k_ t|^2)$, and completing the square, we obtain the exact identity

By (23.1.14), the part $\sigma ^2th[1+t(\log k_ t)_{w\bar w}]$ is nonnegative.

Step 2: the initial interval $[0,a^{-1}]$. Here

Moreover $q_0-1/2\leq q\leq q_0$ and $t\leq \varepsilon \mathrm{e}^{q_0}$, and the latter tends to zero faster than any power of $\eta $. In particular,

At each point where $\chi (ax)>0$, apply lemma 23.1.19 with $\kappa \chi (ax)^2$ in place of $\kappa $; its constant is uniform in $x$. The positive square in (23.1.27) then absorbs the negative term $-\kappa \chi ^2tk_ t$ coming from $\sigma _ xtF_ t$. The remaining part of $\sigma _ xtF_ t$, which involves $\chi '$, is at least $-\eta $, because $F_ t\geq -\eta $. The other negative terms in (23.1.27) are $O(\eta )$, and so is the exponential error of lemma 23.1.19, since $\sigma ^2t\leq \eta ^2$. Thus $d_\eta \leq C\eta $ on this interval. At points where $\chi (ax)=0$, the smooth nonnegative function $\chi $ has a minimum, so $\chi '(ax)=0$ and directly

since $\mu \leq \mu (a^{-1})<1-\eta $ for small $\eta $.

Step 3: the interval $[a^{-1},b]$. Here the formulas simplify to

For $t\leq 1$, lemma 23.1.19 and (23.1.27) give

Indeed, $-\eta \sigma ^2t=-2\kappa q^2t$ and $\sigma _ xtF_ t=-\kappa tk_ t+\kappa t$; these are the two terms for which the scale was chosen. When $t\geq 1$, we insert the exponential derivative estimates (23.1.16) directly in the definition of $Q$. The denominator stays above $1-\eta $, and the chain rule gives

The perturbing derivatives are bounded by constants times $\eta \mathrm{e}^{-t/4}$, $\eta \sigma t \mathrm{e}^{-t/4}$ and $\eta [(|\sigma _ x|+\sigma ^2)t+\sigma ^2t^2]\mathrm{e}^{-t/4}$. Since $q,t\geq 1$ here, each is bounded by a constant times the right side of (23.1.30). Since $1+W_{w\bar w}\geq 1-\eta $, subtracting the two quotients gives

The square of the mixed derivative is at most $C\kappa \eta q^2t^2\mathrm{e}^{-t/2}$, so it satisfies the same bound. Since $Q[0]>0$, this proves (23.1.30).

Step 4: integration of the negative part. The constant $2\eta $ and the initial interval contribute $O(\eta b)=o(1)$, by (23.1.23). In the small-time region, $q\geq \sqrt L$ and $t=\eta ^{-1}\mathrm{e}^{-q^2}$. Therefore

For the exponential term the integral is at most

The interval $[\sqrt L,2\sqrt L]$ has length $O(\sqrt L)$. For $q\geq 2\sqrt L$ and sufficiently small $\eta $, $\eta ^2\mathrm{e}^{q^2}/q^2\geq \mathrm{e}^{q^2/4}$. The integrand on this remaining interval is bounded by $\exp (-\mathrm{e}^{q^2/4}/(C\kappa ))$, which is integrable on $[1,+\infty )$. The prefactor is $O(\sqrt\eta )$, and the initial interval contributes $O(\sqrt{\eta L})$, so the full expression is $o(1)$. In the large-time region, (23.1.28) gives

This proves the integral assertion on $[0,b]$.

Step 5: the radial slope. The negative part of $W_ x$ is uniformly small as well. For $t\leq 1$, we have $F_ t\geq -\eta $ and $0\leq \sigma \leq 2aq$, so

The last inequality follows because $t\sqrt{L+\log (1/t)}$ is increasing on $(0,1]$ when $L>1$. For $t\geq 1$ and $q\geq 1$, use $|W_ x|\leq C\sqrt\eta \, qt \mathrm{e}^{-t/4}$ and $q\leq \sqrt L$.

Step 6: the ends. It remains to consider the two ends, including the cutoff region. If $b\leq x\leq x_*$, then $0\leq q\leq 1$ and $1/(\mathrm{e}\eta )\leq t\leq 1/\eta $. The derivatives of the cutoffs and scale have at most polynomial growth in $\eta ^{-1}$. Thus (23.1.16) implies

But $Q[0]=\mu (1-\mu )\geq \mathrm{e}^{-x_*}/4$ on this interval, and (23.1.23) shows that the error is $o(\mathrm{e}^{-x_*})$. Hence $Q[W]>0$ there. For $x\geq x_*$, $W=0$ and positivity is immediate. For $x\leq 0$, $W_ x=0$ and

since $\mu \leq 1/2$. On $[-2,-1]$, the bounds $1/9\leq \mu \leq 3/10$ and $1-\eta \geq 9/10$ give $Q[W]\geq 2/27$. Finally, $Q[W]$ is continuous, with positive denominator, so the maximum of its negative part over the compact fiber $E$ is continuous in $x$. This proves the assertions about support and continuity, and completes the proof. $\square$

of theorem 23.1.1. Choose a smooth function $r_\eta \geq 0$ supported in $(-1/2,b+1/2)$ with

Such a majorant exists by uniform approximation of the compactly supported continuous function $d_\eta $, followed by adding a positive bump on $[0,b]$; for each fixed $\eta $ the approximation error can be chosen small enough for the integral bound. Fix a smooth $\psi \geq 0$ supported in $(-2,-1)$ with integral one and set

The negative part of $g_\eta ''$ lies to the left of its positive part, and their integrals agree. Consequently,

Thus $g_\eta $ vanishes for $x\leq -2$ and is constant for $x\geq b+1/2$, and $V_\eta $ has the properties at the two ends required for smooth extension to $X$.

Since $g_\eta $ is independent of $w$, the Schur complement changes by exactly its second derivative:

This is positive on $[0,b]$ because $r_\eta >d_\eta $ there. On $(-2,-1)$ it is at least $2/27-m_\eta \| \psi \| _\infty >0$ for small $\eta $. Elsewhere $Q[W]>0$, $\psi =0$ and $r_\eta \geq 0$. The vertical coefficient remains above $1-\eta $ by (23.1.24). For the remaining coefficient, if $x\leq -2$, both $W_ x$ and $g_\eta '$ vanish; if $-2\leq x\leq 0$, then $W_ x=0$ and $\mu \geq \mu (-2)>0$, while $|g_\eta '|\leq m_\eta =o(1)$; and if $x\geq 0$, then $\mu \geq 1/2$ and $W_ x\geq -o(1)$. In all cases $\mu +(V_\eta )_ x>0$ for sufficiently small $\eta $. Therefore lemma 23.1.7 gives $\omega +\mathrm{dd}^{\mathrm{c}}V_\eta >0$ on $X$.

On the fixed ball $B=\{ |\xi |^2<\mathrm{e}^{-3}\} \Subset U_0$, one has $V_\eta =F_\eta (\varepsilon ,w)$, including at $\xi =0$. Thus

uniformly for $y\in B$. Since $\int _ B\omega _{\mathrm{FS}}^2>0$, the mass identity (23.1.5) shows that $\int _ X(\omega +\mathrm{dd}^{\mathrm{c}}V_\eta )^3\to +\infty $, and the theorem follows by taking $V_ j=V_{\eta _ j}$ along any sequence $\eta _ j\downarrow 0$. $\square$

Proof. Choose $n-2$ horizontal angular vectors, one horizontal radial vector and the vertical vector $\partial _ w$. In the normalization of the proof of lemma 23.1.7, each angular entry is $A=(\mu +V_ x)/\mu $, and the radial–vertical block is

This follows by differentiating $s=w+\bar w+\log (1+\mathrm e^ x)$: the radial part of $\partial \bar\partial x$ is zero, whereas its angular part contributes $V_ x/\mu $. The Schur complement of the last entry gives the stated positivity criterion. At $\xi =0$ the same endpoint argument as in dimension three applies.

Set $d=n-1$. The radial Fubini–Study probability density is $d\mu ^{d-1}\mathrm d\mu =d\mu ^ d(1-\mu )\mathrm dx$; it follows from the polar-coordinate determinant of $\mathrm{dd}^{\mathrm{c}}\log (1+|\xi |^2)$ and its integral is one. The background block $K_0$ has determinant one. Using the fiber mass two, therefore,

For $n=3$ this is $12\mu ^2(1-\mu )A\det K_ V$. Comparing the two expressions proves (23.1.34). $\square$

Proof. Use exactly the functions $V_\eta =W+g_\eta $ constructed in the proof of theorem 23.1.1, including the heat scale $\varepsilon =\eta ^{1/\eta ^2}$. Their endpoint properties give smooth potentials on every $H_ n$, and lemma 23.1.33 gives strict positivity. The slope bounds in lemma 23.1.25 and $-m_\eta \leq g_\eta '\leq 0$ imply

for all sufficiently small $\eta $.

We must first localize the divergent threefold mass to this region. On $x<0$, the heat potential is $W=F_\eta (\varepsilon ,w)$ and has no $x$ derivatives. The determinant formula in the preceding proof and $\int _ E F_{w\bar w}\, \mathrm dA_ E=0$ give

Here $0<\mu +g_\eta '\leq \mu $, and $g_\eta ''=r_\eta -m_\eta \psi $ with $\int r_\eta =m_\eta $ and $\int \psi =1$. Hence

No bound on the amplitude of the fiber Hessian is required; its mean has been integrated exactly. Since the total threefold mass tends to infinity, its mass on $x\geq 0$ tends to infinity. The dimensional comparison now gives

Proof. For a Hermitian form $h$, compactness gives $h\geq c\omega _ n$ for some $c>0$. The potentials $cV_\eta $ satisfy $h+\mathrm{dd}^{\mathrm{c}}(cV_\eta )\geq c(\omega _ n+\mathrm{dd}^{\mathrm{c}}V_\eta )>0$, and positivity of determinants bounds their masses below by $c^ n$ times the divergent masses in theorem 23.1.35. $\square$

Proof. First suppose that $u$ is smooth and $\omega +\mathrm{dd}^{\mathrm{c}}u>0$. Averaging the positive forms $\tau _ a^*(\omega +\mathrm{dd}^{\mathrm{c}}u)$ gives $\omega +\mathrm{dd}^{\mathrm{c}}Au>0$. Since $Au$ is smooth and constant on the fibers, $Au=\pi ^*v$ for a smooth function $v$ on $\mathbb P^2$. At each point, the kernel of $\partial s$ maps isomorphically onto the tangent space of the base, and $\beta $ vanishes on it. Restricting $\omega +\mathrm{dd}^{\mathrm{c}}Au=\pi ^*(\omega _{\mathrm{FS}}+\mathrm{dd}^{\mathrm{c}}v)+\beta $ to this kernel therefore shows that $\omega _{\mathrm{FS}}+\mathrm{dd}^{\mathrm{c}}v>0$.

For a smooth function $f:E\to \mathbb R$, the nonconstant Fourier characters in (23.1.11), indexed by $(p,q)\in \mathbb Z^2\setminus \{ (0,0)\} $, have eigenvalues $\pi ^2p^2+q^2/4\geq 1/4$ for $-\partial _ w\partial _{\bar w}$. Parseval's identity therefore gives

Applying this inequality on every fiber, integrating against $\omega _{\mathrm{FS}}^2$ and using the mass identity (23.1.5), we obtain (23.1.39).

For bounded $u$, choose uniformly bounded smooth functions $u_ j\downarrow u$ with $\omega +\mathrm{dd}^{\mathrm{c}}u_ j>0$ by theorem 5.32.1. Their averages decrease pointwise to $Au$ by dominated convergence. The corresponding smooth base potentials $v_ j$ decrease to a bounded $\omega _{\mathrm{FS}}$-plurisubharmonic function $v$, so $Au=\pi ^*v$. Dominated convergence gives $u_ j\to u$ and $Au_ j\to Au$ in $L^2(X,\nu )$, while bounded Bedford–Taylor continuity gives convergence of the total masses. Thus (23.1.39) passes to $u$.

If the mass is $6$, the inequality gives $u=Au$ almost everywhere. Since both functions are bounded and quasi-plurisubharmonic, lemma 5.24.4 gives equality everywhere. Conversely, if $u=\pi ^*v$ with bounded $v\in \mathrm{PSH}(\mathbb P^2,\omega _{\mathrm{FS}})$, choose smooth $v_ j\downarrow v$ with $\omega _{\mathrm{FS}}+\mathrm{dd}^{\mathrm{c}}v_ j>0$ by theorem 5.32.1. The pullbacks decrease to $u$ and are uniformly bounded. They are strictly $\omega $-plurisubharmonic because the base forms are positive and $\beta $ is positive on the vertical line. Their vertical derivatives vanish, so (23.1.5) gives mass $6$ for each of them. Bounded Bedford–Taylor continuity gives the same mass for $u$.

It remains to prove that the constant $2/3$ is optimal. Choose a nonzero real $\rho \in C_ c^\infty (U_0)$. For $\delta \in \mathbb R$, the function

extends by zero to a smooth function on $X$. For sufficiently small nonzero real $\delta $, it satisfies $\omega +\mathrm{dd}^{\mathrm{c}}u>0$. Its fiber average is zero, and its only fiber frequencies are $(p,q)=(0,\pm 1)$, with eigenvalue $1/4$. Hence equality holds in the fiber inequality, and therefore in (23.1.39), with nonzero left-hand side. No smaller coefficient can hold. $\square$

Proof. The lower bound $v_-(\omega )\geq 6$ follows from (23.1.39), and it is attained by constants. Theorem 23.1.1 gives $v_+(\omega )=+\infty $. $\square$

Proof. Set $m(x)=\int _ EV\, \mathrm dA_ E$ and $e(x)=\int _ E|V_ w|^2\, \mathrm dA_ E$. Periodic integration by parts gives

The mean of every fiber derivative vanishes. Expanding $\Delta _ V$ yields

Thus $f(x)=\log (1+\mathrm e^ x)+m(x)-e(x)/2$ is convex.

Both $m$ and $e$ are smooth global functions on the projective base: local fiber coordinates change by holomorphic translations, preserving fiber Haar area, the vertical derivative and the Dirichlet integral. They are $U(2)$-invariant. Smoothness and radial invariance at $\xi =0$ give $m',e'\to 0$ at the negative end. At the positive end use a normal coordinate $z_0/z_1$ to the hyperplane $\{ z_0=0\} $ on the base. The invariant base functions are constant along this hyperplane and smooth across it; their $x$ derivatives tend to zero. This does not require that the pointwise potential be constant on its elliptic fibers at that end. Consequently $f'(-\infty )=0$ and $f'(+\infty )=1$. Integrating the nonnegative $f''$ proves the flux identity.

For each fixed $w$, $R_ V\geq 0$ implies $V_{xx}\geq -1$ on all of $\mathbb R$. Hence for every real $h$,

Taking $h=V_ x(x,w)$ and using $V\leq 0$ proves the derivative bound. Finally $\mu \leq 1$ proves the bound for $A_ V$. $\square$

Proof. The compact group $G=U(3)\times (\mathbb R/\mathbb Z)$ acts transitively and holomorphically on $X$ by unitary transformations and positive scalar multiplication. It preserves $\omega $. Choose a nonnegative smooth central approximate identity $\rho _\epsilon $ on $G$, supported near its identity, and put

Convolution on a compact homogeneous space is smooth. Averaging preserves the current inequality and the bounds, and centrality makes convolution commute with the $U(2)$ action. Transporting the bounded tests defining capacity shows that $\operatorname {Cap}_\omega $ is $G$-invariant.

Suppose $L>0$. By quasi-continuity, choose an open set $O$ of arbitrarily small capacity such that $u$ is continuous on $X\setminus O$. When $g$ is sufficiently close to the identity, $u(gp)$ and $u(p)$ are uniformly close whenever both points lie outside $O$. Thus, outside $O$ and for small kernel support,

Markov's inequality, followed by capacity invariance, gives

where the supremum runs over $v\in \mathrm{PSH}(X,\omega )$. Letting the capacity of $O$ tend to zero proves the claim. If $L=0$, $u=0$ and no approximation is needed. $\square$

Proof. For bounded $u\leq 0$, apply lemma 23.1.46. The bounded capacity continuity of products and quasi-continuous coefficients in section 5.27, section 5.29 passes the total estimate (23.1.45) to $u$. For the tail at $T>0$, use continuous depth cutoffs majorizing $\mathbf1_{\{ -u>T\} }$ and supported in $\{ -u>T-\delta \} $, where $0<\delta <T$. The same product continuity gives a bound by $12Q_{T-\delta }(W)$. Continuity of $(1+\sqrt{2t})W(t)$ gives $Q_{T-\delta }(W)\to Q_ T(W)$. For $T=0$, use the total estimate.

For unbounded $u$, the canonical truncations $u^ K=\max (u,-K)$ remain bounded and invariant. On $\{ u>-K\} $ their measures agree with the finite-level restrictions defining $\mu _ u$. These restrictions increase to $\mu _ u$, so monotone convergence proves (23.1.48).

Splitting at depth $T$ and rescaling $u^ T$ into a capacity test gives, for every Borel set $B$,

Indeed $\omega +\mathrm{dd}^{\mathrm{c}}u^ T\leq (1+T)(\omega +\mathrm{dd}^{\mathrm{c}}(u^ T/(1+T)))$, and the latter potential has range of length at most one. If $\sqrt t\, W(t)\to 0$, then $Q_ T(W)\to 0$. Taking first $T$ large and then the capacity small proves uniform absolute continuity.

Suppose $u_ j\to u$ in capacity in $\mathcal R$. For fixed $K$, the truncations $u_ j^ K$ converge in capacity to $u^ K$, and their weighted bounded products converge weakly. Outside $\{ u\leq -K\} $, locality identifies $\nu _{W,u^ K}$ with $\nu _{W,u}$. Both measures give this exceptional set mass at most $12Q_{K-1}(W)$ when $K>1$, by their tail estimates; this also includes any truncation mass on $\{ u=-K\} $. Hence

uniformly over $u\in \mathcal R$. Letting $K\to +\infty $ removes the truncations and proves the claimed continuity. $\square$

Proof. For the upper bound use $k_ r\geq 0$ on $(0,1]$ and (23.1.16). The nonnegative Fourier coefficients imply $k_ r(w)\leq k_ r(0)$, and Gaussian summation gives $k_ r(0)\leq 2/r+C r^{-1}\mathrm e^{-1/(2r)}$ for $r\leq 1$. Its integrable error proves the lower bound. On the indicated annulus the nonzero-lattice terms and their derivatives are exponentially small relative to the zero-lattice term, proving the last two identities. For the first, split the defining integral at a fixed small $r_0$. On $[t,r_0]$ the nonzero-lattice contributions have bounded integral, and the difference between $(2/r)\mathrm e^{-|w|^2/r}$ and $2/r$ has integral at most $2|w|^2/t\leq 8$. The remaining integral is uniformly bounded. $\square$

Proof. Fix $\kappa $ as in lemma 23.1.19, also with $\kappa <1/100$. Choose a small fixed $c>0$ later and set

Use the cutoffs $\chi $ and $\vartheta $ from the earlier heat-time construction. Then $Z_\eta =F_\eta (\varepsilon ,\cdot )$ for $x\leq 0$, $Z_\eta =0$ for $x\geq x_*$, and

The larger value of $q_0$ changes the integrated positivity defect; we estimate it at this scale. Write $r_-=\max (-r,0)$ for the negative part of a real number $r$.

On $[0,b]$ we have $\vartheta =1$. Write $F=F_\eta $ and put $\sigma =(\log t)_ x$, $h=\eta k_ t$ and $\zeta =\sigma t\partial _{\bar w}\log k_ t/\mu $. The identity (23.1.27) applies to $Z_\eta $ on this interval. The contribution involving $h[1+t(\log k_ t)_{w\bar w}]$ is nonnegative. On $[0,a^{-1}]$,

This upper bound for $t$ is smaller than every power of $\eta $. Thus $\kappa q\chi 't\leq \eta $ and $\sigma ^2t\leq \eta ^2$. Use (23.1.20) with $\kappa \chi ^2$ where $\chi >0$, and $F_ t\geq -\eta $ for the $\chi '$ term. It follows that $[Q[Z_\eta ]]_-\leq C\eta $. Where $\chi =0$ also $\chi '=0$, and $Q[Z_\eta ]=\mu -\mu ^2/(1-\eta +h)>0$, since $x\leq a^{-1}=o(1)$.

On $[a^{-1},b]$, $\sigma =2aq$, $\sigma _ x=-\kappa /\eta $, $\mathrm dx=-\mathrm dq/a$. For $t\leq 1$, (23.1.27) and (23.1.20) give

For $t\geq 1$, (23.1.16) gives instead $[Q[Z_\eta ]]_-\leq C\kappa q^2t^2\mathrm e^{-t/4}$. Put $L=\log (1/\eta )$. Integrating the $q^2t$ term uses $\int _{\sqrt L}^\infty q^2\mathrm e^{-q^2}\, \mathrm dq\leq C\sqrt L\eta $. For the exponential term split at $2\sqrt L$: the initial interval has length $O(\sqrt L)$, and on the rest $\eta ^2\mathrm e^{q^2}/q^2\geq \mathrm e^{q^2/4}$ for small $\eta $. Its remaining integral is uniformly bounded even with upper endpoint infinity. For large time use $\mathrm dx=\sqrt{\eta /(2\kappa )}\, \mathrm dt/(tq)$, $q\leq \sqrt L$, and the integrability of $t\mathrm e^{-t/4}$. Including the constant term and initial interval yields

All constants here are independent of $q_0,c,\eta $ with fixed $\kappa $. The bound is small after choosing $c$, although it need not tend to zero.

Also $\sup [(Z_\eta )_ x]_-=O(\sqrt{\eta L})$: for $t\leq 1$ use $F_ t\geq -\eta $ and $t\sqrt{L+\log (1/t)}\leq \sqrt L$; for $t\geq 1$ use (23.1.16). On $[b,x_*]$, $0\leq q\leq 1$ and $(\mathrm e\eta )^{-1}\leq t\leq \eta ^{-1}$, so differentiated cutoffs and (23.1.16) give

Choose $c$ so that $\ell <1/(10\mathrm e)$. Since $Q[0]=\mu (1-\mu )\geq \mathrm e^{-x_*}/4$ here, this implies $Q[Z_\eta ]>0$ on the cutoff interval. For $x\leq 0$ use $Q[Z_\eta ]\geq \mu (1-\mu /(1-\eta ))>0$, and for $x\geq x_*$ use $Z_\eta =0$. On $[-2,-1]$ the same formula gives $Q[Z_\eta ]\geq 2/27$.

The continuous defect $d_\eta (x)=\max _ w[Q[Z_\eta ]]_-$ is therefore supported in $[0,b]$. Choose a smooth nonnegative $r_\eta $, supported in $(-1/2,b+1/2)$, with $r_\eta >d_\eta $ on $[0,b]$ and $m_\eta :=\int r_\eta \leq \int d_\eta +\eta $. Uniform approximation and a positive bump give this choice for each fixed $\eta $; no uniform derivative bound is needed. Fix a smooth probability bump $\psi $ supported in $(-2,-1)$ and put

The support order implies $-m_\eta \leq g_\eta '\leq 0$; $g_\eta =0$ at the negative end and is constant at the positive end. Furthermore $g_\eta \leq 0$ and $|g_\eta |\leq m_\eta (b+3)=O_ c(\eta ^{-1})$. Shrink the fixed $c$ so that (23.1.53) ensures

for all sufficiently small $\eta $. Now $Q[V_\eta ]=Q[Z_\eta ]+r_\eta -m_\eta \psi >0$ everywhere: use the majorant on $[0,b]$, the margin $2/27$ on $(-2,-1)$, and the established positivity elsewhere. The vertical coefficient is above $1-\eta $. For the horizontal coefficient, derivatives vanish for $x\leq -2$; on $[-2,0]$ use $\mu \geq \mu (-2)>m_\eta $; on $[0,\infty )$ use $\mu \geq 1/2$ and $(Z_\eta )_ x\geq -o(1)$. Thus (23.1.8) gives a smooth globally strictly positive potential.

On $x<-3$, $V_\eta =F_\eta (\varepsilon ,\cdot )$ has fiber mean zero, so $0\leq \sup _ XV_\eta \leq C\eta $ by lemma 23.1.51 and $g_\eta \leq 0$. Set $u_\eta =V_\eta -\sup _ XV_\eta $. Where the cutoff is nonzero $t\geq \varepsilon $, giving

Multiplication by a cutoff between zero and one preserves this lower bound for $Z_\eta $, including where $F_\eta $ is positive.

Mass and depth on the concentration region. Let $K_\eta $ consist of all radial angular directions and

using the small lift near zero in $E$. This lies in the linear transport region, with $\vartheta =1$, $x\asymp _ c\eta ^{-1}\to \infty $, $\sigma =2aq\asymp _ c\eta ^{-2}$, and superexponentially small $t$. Lemma 23.1.51 gives, uniformly,

In (23.1.27) the first $h$ contribution is nonnegative, $\sigma _ xtF_ t=-\kappa t(k_ t-1)\geq -3\kappa $, and the completed square tends to one. Its other negative terms tend to zero. Thus $Q[Z_\eta ]\geq 1/4$ for small $\eta $. On the core $\psi =0$ and $g_\eta ''=r_\eta \geq 0$, so $Q[V_\eta ]\geq 1/4$. Also

Indeed $(Z_\eta )_ x=\eta \sigma (tk_ t-t)$ and $\mu +g_\eta '>0$ here; the vertical inequality follows from $1-\eta +\eta k_ t$.

In (23.1.43), $\Delta _{V_\eta }=(1+(V_\eta )_{w\bar w})Q[V_\eta ]$. The fiber annulus has area $3t/2$. Integrating the preceding lower bounds and using $\sigma \, \mathrm dx=-2q\, \mathrm dq$ yields

The annular value estimate also gives

As $\eta q_0^2=c\eta ^{-2}+\eta \log (1/\eta )$, this is bounded above and below by fixed positive multiples of $\eta ^{-2}$ on the core. Together with (23.1.54), this proves the asserted mass and depth bounds. All these potentials are bounded: truncations beyond their individual depth equal them everywhere, so the defining extended product equals (23.1.43). $\square$

Proof. On $K_\eta $, monotonicity of $W$ gives

Every sufficiently large $T$ equals $B\eta ^{-2}$ for some sufficiently small $\eta $, proving the necessary bound. Also $K_\eta \subset \{ -u_\eta >A\eta ^{-2}/2\} $, so uniform deep tails force this lower bound to tend to zero. $\square$

Proof. First select the correction with additional local control. Put

Then $t(x_{\rm pre})=\eta ^3$, $x_{\rm pre}\to \infty $, and $b-x_{\rm pre}=O(\sqrt{\eta L})\to 0$. For a fixed large $K$ the preceding defect estimates give $d_\eta \leq (K/2)\eta $ on $[0,x_{\rm pre}]$: on the linear portion $q^2t\leq 4L\eta ^3$, and the exponential error is $O(\exp (-1/(C\kappa \eta ^2L)))=o(\eta )$. On the initial portion use the previously proved $C\eta $ bound. The integral of $d_\eta $ on $[x_{\rm pre},b]$ tends to zero by the same integrated estimates and this interval's length.

Choose a fixed left cutoff, zero for $x\leq -1/2$ and one for $x\geq -1/4$, and a translated right cutoff, one for $x\leq x_{\rm pre}$ and zero for $x\geq x_{\rm pre}+1/4$. Let $r^0_\eta $ be $K\eta $ times their product. It strictly exceeds $d_\eta $ on $[0,x_{\rm pre}]$. The continuous function $e_\eta =\max (d_\eta -r^0_\eta ,0)$ is supported in $[x_{\rm pre},b]$. Add a smooth nonnegative majorant $s_\eta >e_\eta $ on that interval, supported in $(x_{\rm pre}-1/4,b+1/2)$, with $\int s_\eta \leq \int e_\eta +\eta $. Set $r_\eta =r^0_\eta +s_\eta $. Then

After shrinking the fixed $c$, this is bounded by a fixed $m_0$ within all the positivity tolerances above. The conclusions of theorem 23.1.52 therefore hold for this choice, with fixed $A,B,D$. On every fixed compact interval $I\Subset (0,\infty )$, for sufficiently small $\eta $ we have $r_\eta =K\eta $ and exactly

Here $b_0,b_1$ depend only on the fixed left cutoff. Globally $0\leq -g_\eta (x)\leq m_0(x+2)$ for $x\geq 0$.

Let $G(w)=-\int _0^\infty (k_ r(w)-1)\, \mathrm dr$ for $w\ne 0$ on $E$. Gaussian integration gives the uniform bound

Indeed the nearest Gaussian has integral bounded by $C+2\log ^+(1/\operatorname {dist}_ E(w,0)^2)$ on $(0,1]$; the other lattice terms have bounded integral and large times decay. The same bound applies after the cutoff. On a fixed $I\Subset (0,\infty )$, $\vartheta =1$ for small $\eta $ and $t_\eta (x)\leq \exp (-c/(2\eta ^3))$. Every fixed $x$ derivative of $t_\eta $ has this bound times a polynomial in $\eta ^{-1}$. On compact fiber sets away from zero,

To verify derivatives, subtract $G$ from $G_ t=F_\eta (t,\cdot )/\eta $: $G_ t-G=\int _0^ t(k_ r-1)\, \mathrm dr=-t+O(\exp (-d^2/(Ct)))$ when $d$ stays positive. Differentiated Gaussian errors have only polynomial factors in $t^{-1}$, and repeated $x$ differentiation preserves their decay. Together with (23.1.57) and $0\leq \sup V_\eta \leq C\eta $, this gives uniform $C^\ell $ bounds for $u_\eta $ on each such fixed compact product set, for all sufficiently small $\eta $. The threshold may depend on the set and derivative order.

Fix $a_ j=2^{-j-2}$, $j\geq 1$, so $\sum a_ j=1/4$. Choose scales $\eta _ j\downarrow 0$ recursively and write $u_ j=u_{\eta _ j}$, $K_ j=K_{\eta _ j}$. Require the minimum $x$ on $K_ j$ to exceed every earlier $x_{*,i}$ by two. Choose a relatively compact product neighborhood $O_ j$ of $K_ j$, with positive $x$ interval and fiber closure away from zero, on which the earlier $u_ i$ are constant. The $K_ j$ are pairwise disjoint. With $E_ j=\sum _{i<j}a_ i\| u_ i\| _\infty $, impose

For each earlier $O_ i$, require the preceding uniform derivative bounds to hold for $u_ j$ through order $j$. All are finite constraints at stage $j$, achievable by making $\eta _ j$ smaller. They ensure that the future series $\sum _{k>i}a_ ku_ k$ converges in $C^\infty $ on compact subsets of $O_ i$: for every fixed derivative order the sufficiently late terms have a uniform bound there, and $\sum a_ k<\infty $.

These constraints also give $\sum _{k>j}a_ k\eta _ k\leq a_ j\eta _ j$. On $K_ j$ the coordinate and distance estimates are

Using (23.1.58), the global slope bound on $g_ k$, and the supremum constants, the future depth is at most

The last term is at most a fixed multiple of $a_ j\eta _ j^{-2}$ since $\eta _ j\leq a_ j$. Including the earlier and current terms, there is a fixed $C_*>0$, independent of the scales, such that

Fix this constant before the scale choice. Since $\sqrt T W(T)$ is unbounded, at each stage choose arbitrarily large $T_ j$ such that

and so small a scale that all finite preceding constraints hold. Such values occur arbitrarily far out because $W$ is finite on compact intervals. Thus the recursive choice is possible.

The smooth nonpositive partial sums $v_ N=\sum _{j\leq N}a_ ju_ j$ decrease, and

By theorem 5.25.2, $\int _ X(-v_ N)\, \mathrm dV\leq C\sum _ j a_ j<+\infty $. Thus corollary 5.25.3 gives a $\omega $-plurisubharmonic limit $v$ with $\omega +\mathrm{dd}^{\mathrm{c}}v\geq 3\omega /4$. Its supremum is finite and nonpositive.

On $O_ j$ the earlier terms are constant and the future series converges smoothly. Thus $v$ is smooth there and $\omega +\mathrm{dd}^{\mathrm{c}}v\geq a_ j(\omega +\mathrm{dd}^{\mathrm{c}}u_ j)$ pointwise. The determinant inequality for positive Hermitian matrices gives on $K_ j$

On these sets the extended product is the smooth product: on a compact neighborhood of $K_ j$ all sufficiently deep canonical truncations equal $v$. Set $U=v-\sup _ Xv$. Since $\sup v\leq 0$, $-U\leq -v$, and its local currents are unchanged. Monotonicity of $W$ and (23.1.59) give

The disjoint $K_ j$ prove infinite total mass for this single potential. They lie in its smooth locus, so no mass is assigned to poles. $\square$

Proof. Use the correction chosen in the proof of theorem 23.1.56. Its plateau and its late supplement satisfy

Therefore $m_\eta \to m:=K\ell >0$. Shrink the fixed $c$ so that $m$ satisfies all the established positivity margins. On each fixed finite $x$ interval, $r_\eta $ and all its derivatives tend to zero; the late supplement escapes the interval. Thus $g_\eta $ converges smoothly there to

It is zero for $x\leq -2$ and equals $-m(x-\mu _\psi )$ for $x\geq -1$. The same margins give $\mu +g'>0$ and $g''+\mu (1-\mu )\geq 0$ at finite $x$. Across $H$, its extension has logarithmic singularity of coefficient $m$: a local defining function $\zeta $ gives $g=m\log |\zeta |^2+O(1)$. Its upper semicontinuous extension is $\omega $-plurisubharmonic, has supremum zero and belongs to $\mathcal R$.

The bound (23.1.58) and $F_\eta \leq C\eta $ imply that, on compact base sets away from $H$, each fixed-depth exceptional set lies in an exponentially shrinking tube about the local holomorphic zero section $w=0$. These tubes have capacity tending to zero. Indeed, on a relatively compact coordinate chart, $\log |w|^2$ is locally integrable and plurisubharmonic; the weighted Chern–Levine–Nirenberg estimate of section 5.27 bounds the relative capacity of $\{ \log |w|^2<-T\} $ by $C/T$. A finite cover and the local capacity comparison transfer this estimate to $\operatorname {Cap}_\omega $.

The corrections converge uniformly on compact base sets away from $H$, including $\xi =0$, and the supremum normalization is $O(\eta )$. Since $g$ has its logarithmic pole on $H$, the estimate (5.29.4) gives neighborhoods of $H$ with arbitrarily small capacity. Removing one such neighborhood proves $u_\eta \to g$ in capacity.

Assume $\sqrt t\, W(t)$ is bounded but does not tend to zero. Choose the preceding sequence of cores with weighted mass bounded below by $c_0>0$. The proved upper bound makes their full weighted measures uniformly finite, so a subsequence has a weak cluster measure $\nu $. For each fixed closed neighborhood $H_ r$ of $H$, all sufficiently late cores lie in $H_ r$. Portmanteau gives $\nu (H_ r)\geq c_0$; decreasing continuity of the finite measure gives $\nu (H)\geq c_0$. But $\nu _{W,g}(H)=0$ by its defining non-pluripolar truncation. $\square$

Proof. The sufficient conditions are proposition 23.1.47. The necessary uniform mass and tail conditions follow from corollary 23.1.55, whose smooth potentials are invariant. If $\sqrt t\, W(t)$ is unbounded, theorem 23.1.56 disproves universal individual finiteness and prevents a map into finite measures.

For uniform absolute continuity, the concentration regions of theorem 23.1.52 lie in $\{ x\geq c_1/\eta \} $ for a fixed $c_1>0$, hence in shrinking neighborhoods of $H$. The globally defined $h_ H=\log (|z_0|^2/|z|^2)$ has supremum zero and $\omega +\mathrm{dd}^{\mathrm{c}}h_ H=\beta +[H]\geq 0$. Its sublevel estimate (5.29.4) shows that these neighborhoods have capacity tending to zero. If $\limsup _{t\to +\infty }\sqrt t\, W(t)>0$, choose the scales so that $D\eta ^{-1}W(B\eta ^{-2})$ stays bounded below. The weighted masses on these regions then violate uniform absolute continuity. Finally, lemma 23.1.61 disproves capacity-to-weak continuity whenever $\sqrt t\, W(t)$ is bounded but does not tend to zero. $\square$

Proof. Apply theorem 23.1.62 to $\sqrt t\, W_{p,a}(t)\sim t^{1/2-p}(\log t)^{-a}$. A smooth decreasing modification on a compact interval changes none of these conditions. $\square$

Proof. The Gauduchon identity $\mathrm{dd}^{\mathrm{c}}\omega ^2=0$ gives $\int _ X\omega ^2\wedge \mathrm{dd}^{\mathrm{c}}f=0$, and Stokes' formula gives $\int _ X(\mathrm{dd}^{\mathrm{c}}f)^3=0$. Expanding the three integrals with these identities proves the asserted values, and proposition 23.1.4 gives $d\geq 0$. The identity of measures

has four positive terms. After subtracting $\omega ^3+\nu $, the total masses of the remaining terms bound the total variation by $18t+18t+t(d+18t^2)+td+6t^3$, as asserted. $\square$

Proof. Fix a sequence $\varepsilon _ j$ as in the definition. Take a sequence $f_ k=t_ ku_ k$ realizing $\sigma $ as in (23.1.66), so that $f_ k$ has mass $6+m$, and put $H_ k=\mathrm{dd}^{\mathrm{c}}f_ k+t_ k\omega $. By lemma 23.1.68, the positive measures $\nu _ k=3\omega \wedge H_ k^2$ have masses $m+18t_ k^2$ and converge weakly to $\sigma $.

The idea is to replace $f_ k$ by a normalized sum of $N$ random translates. First suppose that $\rho $ has density at most $L$ relative to $h_ G$. Choose $N$ independent random group elements $g_1,\ldots ,g_ N$ with law $\rho $, and set

The quadratic measure of $W$ is

Averaging $u_ k\circ g$ against $h_ G$ gives a constant, hence $\int _ Gg^*H_ k\, \mathrm{d}h_ G=t_ k\omega $. Consequently $B=\int _ Gg^*H_ k\, \mathrm{d}\rho \leq Lt_ k\omega $, and independence gives

For any real $\varphi \in C(X)$ with $\| \varphi \| _\infty \leq 1$, the variance of $D(\varphi )$ is at most $(m+18t_ k^2)^2/N$.

For general $\rho $, convolve it with the normalized restrictions of $h_ G$ to shrinking neighborhoods of the identity, which have positive Haar measure. This gives probability measures $\rho _ j\rightharpoonup \rho $ with bounded densities $L_ j$. Fix a countable dense set $(\varphi _ l)$ in the real unit ball of $C(X)$ and numbers $0<\delta _ j<\min (1,m/4)$ tending to zero. Choose $N_ j$ large enough that

Next choose $k_ j$ large enough that $t_{k_ j}<1$, $\tau _ j=t_{k_ j}\sqrt{N_ j}<\min (\varepsilon _ j/4,1/j,1/4)$, $18N_ jt_{k_ j}^2<m/4$, and the expected cross-term mass is less than $\delta _ j/4$. Chebyshev's inequality for the first $j$ test functions and Markov's inequality for $C(X)$ show that some choice of $N_ j$ group elements satisfies

We may also arrange $k_ j\to \infty $. For each fixed test function $\varphi $, the family $\{ \varphi \circ g^{-1}:g\in G\} $ is compact in $C(X)$, so a finite uniform net and the bounded masses of $\nu _ k$ show that $\int \varphi \circ g^{-1}\, \mathrm{d}\nu _ k\to \int \varphi \circ g^{-1}\, \mathrm{d}\sigma $ uniformly in $g$. It follows that $\mathcal A_{\rho _ j}\nu _{k_ j}\rightharpoonup \mathcal A_\rho \sigma $. The dense test functions and the bounded masses now imply $D_ j\rightharpoonup \mathcal A_\rho \sigma $, while $C_ j\to 0$ in total variation.

By the integral identity of lemma 23.1.68,

Put $\lambda _ j=\sqrt{m/d_ j}$, so $\lambda _ j\to 1$ and $\lambda _ j<2$. The average $a_ j=N_ j^{-1}\sum _ a u_{k_ j}\circ g_ a$ is smooth and $\omega $-plurisubharmonic. With

the potential $r_ jU_ j$ differs from $\lambda _ jW_ j$ by a constant and has exact mass $6+m$. Its positive quadratic measure is $\lambda _ j^2(D_ j+C_ j)$. The total-variation estimate of lemma 23.1.68, with parameter $r_ j$, proves the required weak limit. $\square$

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. $\square$

Proof. Put $m=b-V_\Omega >0$. For necessity, let $\sigma $ be a shrinking defect for $(\Omega ,b)$. Proposition 13.15.20, applied with $h=\Omega $, gives a positive closed limit $K$ of the squares of the intermediate forms with $\sigma =3\Omega \wedge K$, and $\alpha \wedge K=0$ as in the proof of lemma 23.1.72. The local coefficient and Fourier argument of that proof gives $K=\pi ^*\tau $ for a positive measure $\tau $ on $\mathbb P^2$. Consequently

and $\mu $ has mass $m$.

Conversely, fix such a $\mu $ and set $\tau =\mu /(3A_\Omega )$. By theorem 23.1.74, choose a sequence $f_ j=t_ ju_ j$ realizing the shrinking defect $\mathcal L(6\tau )$ for $\omega $. The positive forms $H_ j=\mathrm{dd}^{\mathrm{c}}f_ j+t_ j\omega $ have trace mass $O(t_ j)$, and $H_ j^2\rightharpoonup \pi ^*\tau $ by lemma 23.1.72. Choose $C\geq 1$ with $\omega \leq C\Omega $, and put

The trace mass of $G_ j$ is $O(t_ j)$, and $G_ j^2-H_ j^2$ has trace mass $O(t_ j^2)$. Thus $G_ j^2\rightharpoonup \pi ^*\tau $ and $3\Omega \wedge G_ j^2\rightharpoonup \mathcal L_\Omega \mu $.

Unlike $\omega $, the form $\Omega $ need not be Gauduchon, so the mass polynomial may have a linear term. Define

These estimates follow by expanding $\mathrm{dd}^{\mathrm{c}}f_ j=G_ j-Ct_ j\Omega $ and using the preceding trace bounds. Stokes' formula removes the cubic integral, so

For large $j$, $q_ j>0$. The positive root

gives $M_\Omega (\lambda _ jf_ j)=b$. The normalized function $U_ j=u_ j/C$ is $\Omega $-plurisubharmonic, and $w_ j=\lambda _ j f_ j=r_ jU_ j$ with $r_ j=\lambda _ jCt_ j\to 0$.

To identify the full weak limit, first note that expanding $G_ j=\mathrm{dd}^{\mathrm{c}}f_ j+Ct_ j\Omega $ gives

Hence $G_ j^3\to 0$ in total variation. The positive expansion

converges weakly to $\Omega ^3+\mathcal L_\Omega \mu $, since the second and fourth terms tend to zero in total variation. Given prescribed bounds $\varepsilon _ j$, pass to a sufficiently late subsequence with $r_ j<\varepsilon _ j$ at every index. This preserves the normalization, the exact mass and the weak limit.

The fiber probabilities $h_ p^\Omega $ have smooth densities relative to $h_ p$ that are bounded above and below by positive constants, uniformly in $p$. Thus $\mathcal L_\Omega \mu $ and $\mathcal L\mu $ are bounded by fixed positive multiples of each other, and the assertions about support, point atoms and pluripolar sets follow from theorem 23.1.74. The global functions with fiber poles also remain available: if $\Omega \geq c\omega $, then $c\psi \in \mathrm{PSH}(X,\Omega )$.

Finally, for smooth $u\in \mathrm{PSH}(X,\omega )$ and $c>0$ with $\Omega \geq c\omega $,

Since $v_+(\omega )=+\infty $ by corollary 23.1.40 and the supremum may be computed over smooth potentials (proposition 13.14.15), it follows that $v_+(\Omega )=+\infty $. $\square$