15.19.23 Curve areas and radial tube lengths83F8EFD496FF

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 $c$. $\square$