Pome

Proposition 15.19.264A271B6E1DDB

Proposition 15.19.26.4A271B6E1DDB Let $g_ j$ be smooth Hermitian metrics on $U_{\delta _0}$. Suppose their restrictions to $C$ satisfy

\[ g_ j|_ C\leq M h,\qquad A_ j=\operatorname {Area}_{g_ j|_ C}(C)\longrightarrow 0 \]

for a fixed $M>0$. Suppose also that there are $B>0$ and $0<\alpha \leq 1$ such that, for every unit fiber vector $v$ over any $x\in C$, the ray $\gamma _{x,v}(s)=sv$ satisfies

\[ g_ j(\gamma _{x,v}'(s),\gamma _{x,v}'(s)) \leq B s^{2\alpha -2}, \qquad 0<s<\delta _0, \]

uniformly in $j,x,v$. Then

\[ \operatorname {diam}^{\mathrm{int}}(U_\delta ,g_ j) \leq \frac{2\sqrt B}{\alpha }\delta ^\alpha +K_ h\sqrt{A_ j\bigl (1+\log (M/A_ j)\bigr )}, \qquad 0<\delta <\delta _0. \]

In particular,

\[ \lim _{\delta \downarrow 0}\limsup _{j\to \infty } \operatorname {diam}^{\mathrm{int}}(U_\delta ,g_ j)=0. \]

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

\[ \sqrt B\int _0^ r s^{\alpha -1}\, \mathrm{d}s =\frac{\sqrt B}{\alpha }r^\alpha . \]

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. $\square$

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