Example 15.19.2711425F03EC01

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 $c$. $\square$