15.19 Degenerating semipositive backgrounds28A495F29AAD

Proof. Write $\mu =f\, \mathrm{d}V$, where $f$ is smooth and positive. The strict barrier makes $\theta $ big. By lemma 15.3.1, there is a bounded $s\in \mathrm{PSH}(X,\theta )$ with $-1\leq s\leq 0$ and $\mathrm{MA}_\theta (s)\geq b\mu $ for some fixed $b>0$. If $c_\varepsilon <b$, positivity gives

for every constant $A$. Strict bounded domination would imply $u_\varepsilon \geq s+A$ for every $A$, which is absurd. Hence $c_\varepsilon \geq b$.

For an upper bound, choose a fixed Gauduchon form $\gamma $. The arithmetic–geometric mean inequality and Stokes give

The first integral is fixed and positive, giving $c_\varepsilon \leq B$.

For the lower bound on $u_\varepsilon $, note that all $u_\varepsilon $ are normalized $(\theta +\eta )$-plurisubharmonic functions. Uniform exponential integrability, lemma 5.26.2, gives $a>0$ such that $g_\varepsilon =\mathrm{e}^{-a u_\varepsilon }f$ has a common $L^2$ bound. Also $g_\varepsilon \geq f$. By theorem 15.3.3, the solutions

are uniformly bounded. Put $B_0=\max \{ 1,B\} $ and $z_\varepsilon =w_\varepsilon -a^{-1}\log B_0$. Then

The comparison principle for the exponential equation in the background $\theta +\varepsilon \eta $ gives $z_\varepsilon \leq u_\varepsilon \leq 0$, and $z_\varepsilon $ is bounded below uniformly in $\varepsilon $. $\square$

Proof. For the differentiated calculation write $U=u_\varepsilon /(2\pi )$ and $P=\rho /(2\pi )$, so that $\mathrm{dd}^{\mathrm{c}}u_\varepsilon =\mathrm{i}\partial \bar\partial U$. Put

The previous lemma and the smooth positive density give $0<d_1\leq D\leq d_2$. Choose a fixed $C_0$ so that $v=U-P+C_0>1$, and set

where $A$ will be fixed below. For each $\varepsilon $, $S$ is smooth on $X$ and $v$ tends to infinity at $Z$. Thus $H\to -\infty $ there, and $H$ attains its maximum on $X\setminus Z$. We bound this maximum independently of $\varepsilon $.

In local holomorphic coordinates, $h$ and $g$ denote the coefficient matrices of $\eta $ and $\omega _\varepsilon $ with respect to $\mathrm{i}\, \mathrm{d}z_ j\wedge \mathrm{d}\bar z_ k$, and commas denote coordinate derivatives, as in $h_{i\bar j,k}=\partial _ kh_{i\bar j}$ and $h_{i\bar j,\bar k}=\partial _{\bar k}h_{i\bar j}$. At a maximum point choose holomorphic coordinates in which

A linear change first normalizes $h$ and diagonalizes $g$. In these coordinates, the quadratic change $z_ i=w_ i+\frac12\sum _{j,k}a^ i_{jk}w_ jw_ k$, with symmetric coefficients chosen so that $a^ i_{ik}=-h_{i\bar i,k}$, then kills $h_{i\bar i,k}$. Its coefficients are uniformly bounded on a fixed finite cover. Write

The lowered torsion coefficients $T_{ij\bar k}$ and their first derivatives are uniformly bounded: the potential Hessian cancels, leaving only derivatives of $\theta +\varepsilon \eta $.

Let $\mu _{\mathrm{loc}}$ be the density of $\mu $ with respect to the coordinate volume form. Differentiating $S=h^{k\bar\ell }g_{k\bar\ell }$ and the equation $\log \det g=\log c_\varepsilon +\log \mu _{\mathrm{loc}}$ gives

The second derivatives of $g$ contribute $\sum _{i,j,k}|g_{i\bar j,k}|^2/(\lambda _ i\lambda _ j)$, up to $-CR-C$, by commuting the potential derivatives. The first derivatives of $h$ contribute $-2\operatorname {Re}\sum _{i,j,k}\lambda _ i^{-1}h_{j\bar k,\bar i}g_{k\bar j,i}$. Its terms with $j=k$ vanish. Replace $g_{k\bar j,i}$ by $g_{i\bar j,k}+T_{ik\bar j}$ and apply weighted Cauchy–Schwarz. This consumes exactly the terms of the preceding square with $j\ne k$ and costs at most $CRS+CR$. The second derivatives of $h^{-1}$ cost $CRS$. Finally the fixed density term $-C$ is absorbed using $SR\geq n^2$.

Since $\partial _ iS=\sum _ jg_{i\bar j,j}+E_ i$ at the maximum point, weighted Cauchy–Schwarz also gives

Indeed expand around $\partial S-E$ and discard the resulting term $-|E|_ g^2$. Combining with (15.19.5) yields

At the maximum, $\partial \log S=\gamma '\partial v$. Moreover (15.19.1) gives $\Delta _ gv\leq n-\delta R$, and $|E|_ g^2\leq CR$. Young's inequality with the positive coefficient $-\gamma ''=v^{-2}$ therefore gives

Since $\gamma '=A+v^{-1}$, choose $A$ with $\delta A-C\geq 1$. The resulting inequality is

All constants are independent of $\varepsilon $.

If the maxima of $H$ were unbounded, then at a sequence of maximizing points

Thus $S/v\to \infty $ and $S\to \infty $. Inequality (15.19.6) would imply $R\leq 2C$ along this sequence. But

a contradiction. Hence $H$ is uniformly bounded. On $K$, both $U$ and $P$ are bounded, so $S\leq C_ K$. The determinant lower bound then gives $\lambda _ i\geq d_1/C_ K^{n-1}$, completing the proof. $\square$

Proof. Monotonicity of the constants. Let $\alpha \leq \beta $ be Hermitian forms, and let $(u_\alpha ,c_\alpha )$ and $(u_\beta ,c_\beta )$ be the normalized smooth solutions of $(\alpha +\mathrm{dd}^{\mathrm{c}}u_\alpha )^ n=c_\alpha \mu $ and $(\beta +\mathrm{dd}^{\mathrm{c}}u_\beta )^ n=c_\beta \mu $. At a maximum point of $u_\alpha -u_\beta $,

so $c_\alpha \leq c_\beta $. Hence $c_\varepsilon $ decreases as $\varepsilon \downarrow 0$, and lemma 15.19.3 gives $c_0\geq b>0$. Write $c_\eta (\varepsilon )$ for the constant $c_\varepsilon $ defined with the perturbation $\eta $. For two perturbations with $M^{-1}\eta _1\leq \eta _2\leq M\eta _1$, the same comparison gives

Letting $\varepsilon \downarrow 0$ shows that $c_0$ does not depend on $\eta $.

Local estimates. The trace bound of lemma 15.19.4 gives uniform ellipticity on compact subsets of $X\setminus Z$. The smooth coefficients of $\theta +\varepsilon \eta $ and all derivatives of $\log \mu _{\mathrm{loc}}$ have common bounds there. Interior complex Evans–Krylov and Schauder estimates, as in the ellipticity argument of proposition 15.18.5, give uniform $C^ k$ bounds on smaller compact subsets for every $k$.

Convergence and the limiting equation. Normalized compactness in the fixed background $\theta +\eta $, followed by a diagonal extraction over an exhaustion of $X\setminus Z$, gives the stated convergence. The uniform bound $-C\leq u_\varepsilon \leq 0$ passes to the canonical limit, and normalized compactness preserves the supremum zero. Current convergence gives $\theta +\mathrm{dd}^{\mathrm{c}}u_0\geq 0$, and smooth convergence identifies the equation on $X\setminus Z$. The bounded non-pluripolar measure $\mathrm{MA}_\theta (u_0)$ and the smooth measure $\mu $ both vanish on the analytic set $Z$. This proves the equation on all of $X$. For closed $\theta $, bounded closed-background Stokes gives $\int _ X\mathrm{MA}_\theta (u_0)=\int _ X\theta ^ n$, proving the last assertion. $\square$

Proof. Smoothness and positivity off $E$ follow from the pullback and the biholomorphism. In one standard model put

The function $r^2$ is the sum of the squared absolute values of the four contracted coordinates. It is therefore globally defined on the bundle; $b$ is the pullback of the Fubini–Study form from its base. In the chart (15.19.9),

This is an inequality of currents across the zero section as well as a smooth inequality off it.

Ambient positivity gives $\theta \geq c\mathrm{dd}^{\mathrm{c}}r^2$ on a sufficiently small tube, for some $c>0$. The smooth form $\mathrm{dd}^{\mathrm{c}}r^2+b$ is positive definite there. Indeed the contraction is an immersion off the zero section, and on that section $\mathrm{dd}^{\mathrm{c}}r^2$ is positive in both fiber directions while $b$ is positive along the base. Compactness of the zero section and a smaller closed tube give a uniform lower bound by a positive multiple of $\eta $. Thus, for every fixed $a>0$,

near the zero section, with $\delta _ a>0$.

Choose disjoint model tubes with nonempty complement and numbers $0<t_1<t_2<1$ in each. Let $0\leq \chi \leq 1$ be a smooth cutoff of $r^2$, equal to one for $r^2\leq t_1$ and zero for $r^2\geq t_2$. Extend $\chi (r^2)\log r^2$ by zero outside its tube. On the compact transition annuli, the pullback $\theta $ is uniformly positive and the second derivatives of this cutoff potential are bounded. A sufficiently small positive rational $a$ therefore makes

satisfy $\theta +\mathrm{dd}^{\mathrm{c}}\rho \geq \delta \eta $ there and near every $E_ j$. On the remaining compact set, $\rho =0$ and $\theta $ is positive, so decreasing $\delta $ proves the global inequality. Near $E_ j$ the function is $a\log (|y_1|^2+|y_2|^2)+a\log (1+|z|^2)$, hence has analytic singularities with rational coefficient and pole set exactly $E_ j$. It is smooth elsewhere and nonpositive; its zero extension gives supremum zero. If $E$ is empty, take $\rho =0$ and use the positivity of $\theta $ on compact $X$. $\square$

Proof. Apply the theorem using lemma 15.19.10. Since $\pi $ contracts $E_ j$, the restriction of $\theta $ to it is zero. Stokes on the compact smooth curve gives $\int _{E_ j}\mathrm{dd}^{\mathrm{c}}u_\varepsilon =0$, proving the area identity. $\square$

Proof. The map is biholomorphic off $E$, and both smooth measures $\mu $ and $\theta ^3$ give $E$ zero mass. Hence $\nu _ Y=\pi _*\theta ^3$ is finite, $\mu _ Y$ has no node atom, and on the regular locus

In the chart (15.19.9), put $h=1+|z|^2$ and $s=|y_1|^2+|y_2|^2$, so $r^2=hs$. The fiber block of the complex Hessian of $r^2$ is $h$ times the identity, and its Schur complement is $s/h$. Consequently

where $\lambda _6$ is Euclidean volume in $(z,y_1,y_2)$. Ambient positivity bounds $\theta $ above and below by positive multiples of $\mathrm{dd}^{\mathrm{c}}r^2$ on a smaller tube. The smooth density of $\mu $ is bounded above and below there. On compact subcharts covering the base $\mathbb P^1$, it follows that

The other bundle chart gives the same comparison, since $r$ is the norm of the contracted coordinates.

On each bounded base subchart, $r$ is comparable to the Euclidean radius $R$ in the two complex fiber variables. Up to fixed positive constants, the contribution to $\int F^ p\, \mathrm{d}\nu _ Y$ is therefore

Two compact base subcharts give the upper estimate; a fixed open base subchart gives the lower estimate. The integral is finite precisely when $p<3$, and diverges logarithmically at $p=3$. Away from the finitely many nodes the density is smooth. The same comparisons give $\nu _ Y(\{ r<t\} )\asymp t^6$ near a node. The two bounds for $F$ sandwich $\{ F>T\} $ between such tubes of radii comparable to $T^{-1/2}$, proving the endpoint estimate. $\square$

Proof. Existence follows from corollary 15.19.12. For uniqueness we use an energy identity, which is available because $\theta $ is closed. Fix one cluster limit $u_0$ and put $A=\theta +\mathrm{dd}^{\mathrm{c}}u_0$. This is a positive closed current with bounded local potentials, smooth and strictly positive on $X\setminus E$ by the trace bounds. If $v$ is another bounded solution with scalar $c$, bounded closed-background Stokes gives $c=\int _ X\theta ^3=c_0$. Put $B=\theta +\mathrm{dd}^{\mathrm{c}}v$ and $w=v-u_0$. The bounded integration-by-parts calculus of section 5.27 gives

All three energies are nonnegative, so each of them vanishes, in particular the one with $A^2$. On every relatively compact chart of $X\setminus E$, the smooth positive factor $A^2$ controls the weak gradient of $w$, so $\mathrm{d}w=0$ there. The complement of a proper complex analytic subset in the connected manifold $X$ is connected. Thus $w$ is constant almost everywhere off $E$. Since $E$ has ordinary volume zero, canonical almost-everywhere equality of quasi-plurisubharmonic functions, lemma 5.24.3, gives $v=u_0+C$ on all of $X$. Normalization fixes $C=0$.

The compactness and smooth local estimates of theorem 15.19.7 now make every subsequential limit equal to $u_0$, proving convergence of the whole family. The equation and its unique normalized solution do not depend on $\eta $. Descent and the zero exceptional mass identify bounded solutions upstairs and downstairs, proving the final assertion. $\square$

Proof. Choose $\varepsilon >0$ with $2\varepsilon \geq \operatorname {dis}R$. Give the disjoint union $E\sqcup F$ its original distances within each space, and use the symmetric cross distance

It is finite and at least $\varepsilon $. The triangle inequalities involving one cross distance follow from the triangle inequality within the corresponding space and the infimum. For two cross distances, take $(x_0,y_0),(x_1,y_1)\in R$. The distortion bound gives

Taking the two infima gives $d_ E(x,x')\leq D(x,y)+D(y,x')$. The same argument with $E,F$ exchanged treats the other case. Hence $D$ is a metric extending both metrics. Each point has a correspondent at cross distance $\varepsilon $, so the Hausdorff distance between $E$ and $F$ in $(E\sqcup F,D)$ is at most $\varepsilon $. Letting $\varepsilon $ decrease to half the distortion proves the upper bound, including zero distortion.

For the lower bound, suppose two isometric images have Hausdorff distance $h$. For every $\delta >0$, two points in the first image have points in the second within $h+\delta $. Their distance is at most $\operatorname {diam}F+2h+2\delta $. Taking the supremum and then $\delta \downarrow 0$ gives $\operatorname {diam}E\leq \operatorname {diam}F+2h$. Exchanging the images and taking the infimum over embeddings proves the lower bound. $\square$

Proof. The neighborhoods can be chosen nested, with closures avoiding the other points of $S$. Every sequence of regular points converging to $s_ a$ is $d_0$-Cauchy: its tail lies in each prescribed neighborhood, whose punctured diameter tends to zero. Interleaving two such sequences shows that they define the same completion point, denoted by $\widehat s_ a$. Density of $Y^\circ $ guarantees the existence of these sequences.

For each prescribed positive diameter, remove sufficiently small neighborhoods of all the points of $S$. Their complement is a compact subset of $Y^\circ $ and hence is totally bounded for $d_0$. The removed punctured neighborhoods each have the prescribed small diameter. Thus $Y^\circ $ is totally bounded, and its completion is compact.

The points $\widehat s_ a$ are distinct from one another and from the regular points. To see this, fix $s_ a$ and a different point $y\in Y$. Choose an open neighborhood $V$ of $s_ a$ whose closure avoids $y$ and $S\setminus \{ s_ a\} $, and choose $W\Subset V$ containing $s_ a$. There is a smooth function $\chi $ on $Y^\circ $, equal to one on $W\setminus S$ and zero outside $V$, whose derivative has compact support in $Y^\circ $. Indeed the transition takes place in the compact regular set $\overline V\setminus W$, where smooth cutoffs exist. Therefore $|\mathrm{d}\chi |_{g_0}\leq C$ for some finite $C$. Integration along curves gives

The Lipschitz extension of $\chi $ to the completion has value one at $\widehat s_ a$ and zero at the regular point $y$, or at $\widehat s_ b$ if $y=s_ b$. This proves the assertion.

Define $\iota $ by the identity on the regular locus and by $\iota (s_ a)=\widehat s_ a$. The intrinsic distance induces the smooth topology at each regular point. At $s_ a$, adjoining $\widehat s_ a$ to a punctured neighborhood does not increase its diameter, so the diameter hypothesis proves continuity of $\iota $. Every completion point is represented by a Cauchy sequence in $Y^\circ $. Compact metrizability of $Y$ gives a subsequence converging topologically either to a regular point or to some $s_ a$; the preceding arguments give its metric limit. Hence $\iota $ is surjective as well as injective. A continuous bijection from compact $Y$ to Hausdorff $\widehat Y$ is a homeomorphism. $\square$

Proof. If $S$ is empty, uniform metric comparison on compact $X$ proves the distance convergence. Assume $N\geq 1$. Fix $\tau >0$. Choose pairwise disjoint neighborhoods $V_ a$ from the specified bases with

for all sufficiently large $j$, where $U_ a=\pi ^{-1}(V_ a)$. Pick $q_ a\in V_ a\setminus S$, and then choose open neighborhoods $W_ a$ of $s_ a$ with $\overline W_ a\subset V_ a$ and $q_ a\notin \overline W_ a$. The regular cores

are compact and identified by $\pi $. Uniform convergence and positivity of $g_0$ on this core give numbers $A_ j\geq 1$, with $A_ j\to 1$, such that lengths of curves contained in the cores compare in both directions by the factor $A_ j$.

Consider a piecewise smooth path in $Y^\circ $ joining two points of $K_ Y$. The preimages of the closed sets $\overline W_ a$ are compact subsets of its parameter interval. At its first visit to any such set, delete the portion from that visit through its last visit to the same set. Resume with the remaining tail. That set is never visited again; thus there are at most $N$ deleted portions. Each retained segment lies in $K_ Y$, and each deleted portion has endpoints in a single $\overline W_ a\subset V_ a$. The lifts of those endpoints can be joined in $U_ a$ with $g_ j$-length less than $\tau $, up to an arbitrarily small error. Applying this procedure to almost minimizing paths, comparing retained lengths, and letting the errors tend to zero gives, for $x,x'\in K_ X$,

Conversely start with an almost minimizing $g_ j$-path in $X$ and apply the same deletion procedure to the closed sets $\pi ^{-1}(\overline W_ a)$. Project its retained segments, which avoid $E$, to $K_ Y$. The two images of each deleted portion have completion distance less than $\tau $. The triangle inequality in $\widehat Y$ therefore gives

For any $x\notin K_ X$ with $\pi (x)\in W_ a$, replace $x$ by $p=\pi ^{-1}(q_ a)\in U_ a\cap K_ X$. Then $d_ j(x,p)<\tau $ and $\widehat d_0(G(x),G(p))<\tau $. Leave points of $K_ X$ fixed. Replacing both endpoints extends the two inequalities to all $x,x'\in X$, with the same error

If $D=\operatorname {diam}(\widehat Y,\widehat d_0)$, the first inequality gives $\operatorname {diam}(X,d_ j)\leq A_ jD+C_ j$. The positive and negative distance errors are consequently bounded by

respectively. Their upper limits are at most $(N+4)\tau $. Letting $\tau \downarrow 0$ proves uniform convergence.

The graph of $G$ is a correspondence between $X$ and $\widehat Y$, and its distortion $b_ j$ for the distances $d_ j$ and $\widehat d_0$ is the supremum in the statement. By lemma 15.19.20,

Proof. Replace $g$ by $g/M$. Its area is $a=A/M\leq 1$, and it is bounded above by $h$. We prove the estimate for this normalized metric.

Choose a finite cover by holomorphic coordinate disks $U_ i$ represented by $\{ |z|<1\} $, with each coordinate defined on the larger disk $\{ |z|<2\} $. Write $g=\rho _ i|\mathrm{d}z|^2$ there. There is a common $L>0$ such that $0<\rho _ i\leq L$ on all the closed unit disks, and

where $\lambda _2$ is Euclidean area. For $p,z\in U_ i$, let $\ell _ p(z)$ be the $g$-length of their straight coordinate segment, which stays in $U_ i$. Polar coordinates about $p$ and Fubini give

Indeed each ray meets the convex unit disk in an interval of length $R\leq 2$. Interchanging the radial integrals multiplies the length integrand at radius $s$ by $(R^2-s^2)/2\leq 2$.

For $0<a\leq 1/4$, put $t=\sqrt a$. The integral on the right, without its factor two, is bounded by

Thus for any $p,q\in U_ i$, averaging $\ell _ p(z)+\ell _ q(z)$ over $z\in U_ i$ gives a two-segment path of length at most $K\sqrt{a(1+\log (1/a))}$. The constant is common to the finite cover.

The overlap graph of this cover is connected because $C$ is connected. Choose points in consecutive overlaps along a simple path in this finite graph. The local estimates join any two points of $C$ through at most the number of covering disks, proving the global bound for $a\leq 1/4$. For $1/4\leq a\leq 1$, the comparison $g\leq h$ gives $\operatorname {diam}(C,g)\leq \operatorname {diam}(C,h)$, and increasing $K_ h$ gives the same estimate. Restoring the factor $M$ scales lengths by $\sqrt M$ and proves the assertion. $\square$

Proof. Choose a coordinate disk $\{ |z|<R\} $ with $0<R<1$, on whose closure $h=\lambda (z)|\mathrm{d}z|^2$ with $\lambda >0$ smooth. Choose a smooth cutoff $0\leq \chi \leq 1$, supported in this disk and equal to one for $|z|\leq R/2$, and a constant $c>0$ with $c\leq \min \lambda $. For sufficiently small $\varepsilon >0$, set

in the disk, and set $g_\varepsilon =\varepsilon ^2h$ outside it. The cutoff makes this a smooth positive metric, and both coefficients in its convex combination are at most one.

On $|z|\leq R/2$ its metric density is $c\varepsilon ^2/(\varepsilon ^2+|z|^2)$, whose area is

The remaining area is $O(\varepsilon ^2)$, proving the asserted area comparison. Any path from the origin to a point outside this smaller disk must first cross its boundary. Radial variation is bounded by Euclidean speed, so that initial portion has length at least

This gives the diameter lower bound. The upper bound follows from lemma 15.19.24, since $A_\varepsilon (1+\log (1/A_\varepsilon ))\asymp \varepsilon ^2\log ^2(1/\varepsilon )$. Finally rescale by

Its area is comparable to $1/\log (1/\varepsilon )$ and tends to zero, whereas its diameter is comparable to one. At the origin its ratio to $h$ tends to infinity, so it has no uniform upper bound. $\square$

Proof. A point of fiber radius $r<\delta $ joins its base point on the zero section along the radial segment, entirely inside $U_\delta $. Its length is at most

The metric is smooth at the zero section for each $j$; the integrable uniform bound controls the whole segment. Join the two base points by an almost minimizing path in $(C,g_ j|_ C)$. This path also lies in $U_\delta $, and up to an arbitrarily small excess its length is bounded by lemma 15.19.24. Together with the two radial segments, this gives the asserted bound for the intrinsic diameter. Since $A_ j(1+\log (M/A_ j))\to 0$, taking $j\to \infty $ and then $\delta \downarrow 0$ proves the final assertion. $\square$

Proof. Put $p=[1:0:0]$ and take the incidence surface

Off $p$, the second coordinate is $[z_1:z_2]$, so $\pi $ is biholomorphic there. Its exceptional curve is $E=\{ p\} \times \mathbb P^1$. In the affine chart $z_0=1$, a chart of $X$ is

The other chart is $([1:t'w':w'],[t':1])$, with $t'=1/t$ and $w'=tw$ on their overlap. These are the charts of the tautological line bundle $L=\mathcal O_{\mathbb P^1}(-1)$. They also show smoothness along $E$; smoothness elsewhere follows from the biholomorphism. Thus $X$ is a compact smooth complex surface. Its connected dense complement $X\setminus E$ makes it connected.

Let $q:L\to \mathbb P^1$ be the bundle projection. Give $L$ the norm induced by its tautological inclusion into $\mathbb P^1\times \mathbb C^2$. In the first chart its squared norm is

Fix an area-one Hermitian metric $h$ on $\mathbb P^1$. The Chern connection of the fiber norm defines a smooth semipositive Hermitian tensor $k$ on $L$ by the squared norm of the covariant fiber differential. Explicitly, with $\lambda =1+|t|^2$,

Under $t'=1/t$, $w'=tw$, the differential in this formula multiplies by $t$ and $\lambda $ divides by $|t|^2$, so the tensor is globally defined. It has complex rank one, is positive on nonzero vertical tangent vectors, and restricts to zero on $TE$. Consequently $q^*h+k$ is a smooth positive Hermitian metric on $L$.

Choose $0<\delta _0<1$ and extend this metric from $U_{\delta _0}=\{ r<\delta _0\} $ to a smooth Hermitian metric $g_*$ on $X$. Such an extension follows by a smooth cutoff supported in a slightly larger bundle tube and a convex combination with any global Hermitian metric. It agrees exactly with $q^*h+k$ on $U_{\delta _0}$.

Choose $0<\varepsilon _ j<1$ tending to zero, with $2\varepsilon _ j^2<\delta _0$, and a smooth function $0\leq \chi \leq 1$ on $[0,\infty )$, equal to zero on $[0,1]$ and to one on $[4,\infty )$. Set

on $U_{\delta _0}$, and set $g_ j=g_*$ elsewhere. The function $f_ j$ is smooth at $E$, is at least $\varepsilon _ j^2$, and equals one for $r\geq 2\varepsilon _ j^2$. Hence the metrics glue smoothly and are positive. They equal $g_*$ eventually on every compact subset of $X\setminus E$.

Let $\beta ,\kappa $ be the Hermitian forms of $q^*h,k$. Their complex rank one gives $\beta ^2=\kappa ^2=0$. On the tube,

Outside it the metrics coincide. This proves the global volume identity. On $E$, the restriction is $g_ j|_ E=\varepsilon _ j^2h$, so its area is $\varepsilon _ j^2$.

For any unit fiber vector $a$, a radial ray $\gamma (s)=sa$ has $k$-squared speed one and zero base component. Thus its $g_ j$-squared speed is $f_ j(s)^{-1}$. If $s<2\varepsilon _ j^2$, this is at most $\varepsilon _ j^{-2}\leq 2/s$; otherwise it is one, still at most $2/s$ for $s<\delta _0$. This is the uniform integrable radial bound with exponent $1/2$.

A radial segment of radius $r<\delta $ has length at most $r+2\varepsilon _ j$: its portion below $2\varepsilon _ j^2$ has speed at most $\varepsilon _ j^{-1}$ and length at most $2\varepsilon _ j$, and its remaining speed is one. Connect two points by their radial segments and an almost minimizing path on $E$. All three segments stay inside $U_\delta $, giving

Taking $j\to \infty $ before $\delta \downarrow 0$ proves the asserted shrinking condition. The affine balls

have precisely these tubes as their preimages.

Let $d_*$ be the global distance of the fixed smooth metric $g_*$. Choose distinct points $a,b\in E$, and put $c=d_*(a,b)>0$. Choose points $a_ r,b_ r$ of fiber norm $r>0$ over these base points. They tend to $a,b$ as $r\downarrow 0$. Every path in $Y\setminus \{ p\} $ lifts to a path in $X\setminus E$ with the same $g_0$-length as its $g_*$-length. Therefore

For any fixed $\delta >0$, these two images lie in $V_\delta $ when $r<\delta $. Taking the supremum and then $r\downarrow 0$ gives its punctured diameter at least $c$. $\square$