Theorem 5.39.6.21429B9E1BF3 Assume that $D(r_0/2)<+\infty $ and that $T$ is a smooth Hermitian form $\eta $ on $U\setminus A$ for a proper analytic subset $A\subset U$. Then every point of $U$ has a neighborhood $V\Subset U$ with constants $C,r_1>0$ such that, for $x,y\in V$,
\[ d_\eta ^ U(x,y)\leq C D(8|x-y|), \qquad x,y\notin A,\quad |x-y|<r_1. \]
The constants depend only on the charts and an upper bound for $|\theta |$; the dependence on $v$ enters only through $D$. All sequences in $V\setminus A$ converging to a given point of $A\cap V$ have a common limit in the metric completion. If these hypotheses hold on a finite cover of a compact connected manifold by charts, the metric completion of the regular locus is compact, and the identity of the regular locus extends to a continuous surjection of the manifold onto it. If $A$ is nonempty, the metric $\eta $ on the regular locus is incomplete.
Proof.
Write $d=2n$ and $\mu =T\wedge \omega _{\mathrm E}^{n-1}$. The same cutoff calculation as in theorem 5.39.1, with a constant subtracted from $v$, gives
\[ \mu (B(x,s))\leq C\bigl (s^ d+s^{d-2}m(2s)\bigr ). \]
This uses only the bound for $\theta $; closedness of $\theta $ is not needed. If $\tau $ is the Euclidean trace of $\eta $, then $\tau \, \mathrm{d}\lambda \leq C\mu $ on the regular locus. Polar integration bounds the average length of the segments from a regular point $x$ to the points of $B(x,R)$ by
\[ C\int _{B(x,R)}\sqrt{\tau (\zeta )} |\zeta -x|^{1-d}\, \mathrm{d}\lambda (\zeta ). \]
On an annulus of outer radius $s$, the Cauchy–Schwarz inequality and the trace bound make the contribution to this integral at most
\[ Cs^{1-d}s^{d/2} \bigl (s^ d+s^{d-2}m(2s)\bigr )^{1/2} \leq C\bigl (s+\sqrt{m(2s)}\bigr ). \]
For $s_ j=2^{-j}R$, monotonicity gives
\[ \sum _{j\geq 0}\sqrt{m(2s_ j)} \leq \frac1{\log 2}\int _0^{4R}\frac{\sqrt{m(t)}}t\, \mathrm{d}t. \]
Indeed, $\sqrt{m(2s_ j)}\log 2$ is at most the integral of $\sqrt{m(t)}/t$ over $[2s_ j,4s_ j]$, and these intervals have disjoint interiors; no doubling condition on $m$ is needed. The average length is therefore at most $CD(4R)$.
For $x,y\in V\setminus A$, average the lengths of the broken segments $[x,z]\cup [z,y]$ over $z$ in the ball of radius $|x-y|/2$ about the midpoint of $x$ and $y$, applying the preceding bound with $R=2|x-y|$ at each endpoint. Since the volume of this ball is comparable to $R^ d$, some broken path has length at most $CD(8|x-y|)$. As in the proof of theorem 5.39.1, almost every such path avoids $A$, and for small $V$ it stays inside $U$.
Since $D(r_0/2)<+\infty $, we have $D(r)\to 0$ as $r\to 0$. Hence any two regular sequences converging to the same point are Cauchy and have the same limit in the completion. The argument for the completion in the proof of theorem 5.39.1 applies with this modulus. A positive lower bound for $\eta $ near a regular point prevents a sequence converging to a point of $A$ from converging in the metric to the regular point. This also proves incompleteness when $A$ is nonempty.
$\square$
Pome
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