Local formula for distance from the centre of a normal neighbourhood

Put $R=\mathrm{\mid }v{\mathrm{\mid }}_{g_p}$ and $q={\exp }_p(v)$. By [F1], the radial segment $t\mapsto {\exp }_p(tv)$ has length $R$, so the infimum defining distance satisfies $d_g(p,q)\le R$.

Fix $s$ with $0<s<\rho$, and put $U_s={\exp }_p(B_s(0_p))$ and $K_s={\exp }_p({\bar{B}}_s(0_p))$. If $\dim M=n\ge 1$, instantiate one basis of $T_{p}M$ and apply [F2] to make it orthonormal. Its coordinate isometry identifies ${\bar{B}}_s(0_p)$ with the closed Euclidean $s$-ball, which is compact; continuity of the coordinate inverse and of ${\exp }_p$ makes $K_s$ compact. If $n=0$, the closed tangent ball is the singleton $\{0_p\}$ and the same conclusion is immediate. Since a manifold is Hausdorff, [F2] makes $K_s$ closed in $M$. Moreover $U_s$ is open, $U_s\subseteq K_s\subseteq U$, and injectivity of ${\exp }_p$ gives $K_s\setminus U_s={\exp }_p(\{w:\mathrm{\mid }w{\mathrm{\mid }}_{g_p}=s\})$.

We first bridge the two competitor classes in [F1]. Let $d:[u,w]\to U$ be any piecewise $C^1$ curve from $p$ to ${\exp }_p(z)$ and put $S=\mathrm{\mid }z{\mathrm{\mid }}_{g_p}>0$ and $h(t)=\mathrm{\mid }{\exp }_p^{-1}(d(t)){\mathrm{\mid }}_{g_p}$. For $0<\delta <S$, continuity and [F4] give a time with $h=\delta$; its level set is nonempty, closed in $[u,w]$, and compact, so [F4] gives its greatest time ${\tau }_{\delta }$. One has $h(t)>\delta$ on $({\tau }_{\delta },w]$, since a later value at most $\delta$ would cross the level again before $h(w)=S$. Thus $d$ avoids $p$ throughout $[{\tau }_{\delta },w]$. Subdivide this interval at its finitely many $C^1$ breakpoints. On each resulting piece, smoothness of $r$ away from $p$, the $C^1$ chain rule and the polar identity in [F4] give $\mathrm{\mid }{d}^{\mathrm{\prime }}{\mathrm{\mid }}_g^2=(h^{\mathrm{\prime }}{)}^2+\mathrm{\mid }{d}_{\perp }^{\mathrm{\prime }}{\mathrm{\mid }}_g^2$, hence $\mathrm{\mid }{d}^{\mathrm{\prime }}{\mathrm{\mid }}_g\ge h^{\mathrm{\prime }}$. Integrating on each nondegenerate piece using [F4], summing, and telescoping the radial increments yields $L_g(d)\ge L_g(d{\mathrm{\mid }}_{[{\tau }_{\delta },w]})\ge h(w)-h({\tau }_{\delta })=S-\delta \mathrm{.}$ This holds for every $\delta \in (0,S)$, so $L_g(d)\ge S$. For $S=0$ the same bound is simply nonnegativity of length. No derivative of $r$ at $p$ has been used.

Let $c:[a,b]\to M$ be any piecewise $C^1$ curve from $p$ to $q$. If its image is contained in $U$, step 1.3 gives $L_g(c)\ge R$. Suppose instead that it leaves $U$, and choose the explicit radius $s=(R+\rho )\mathrm{/}2$, so $R<s<\rho$. The set $E=c^{-1}(M\setminus U_s)$ is nonempty because a point outside $U$ is outside $U_s$; it is closed in $[a,b]$ because $U_s$ is open. By [F3], $E$ has a least element $\tau$. Since $c(a)=p\in U_s$ and $U_s$ is open, $\tau >a$.

For every $t<\tau$ one has $c(t)\in U_s\subseteq K_s$, while $c(\tau )\mathrm{\notin }{U}_s$. Continuity gives $c(\tau )\in \overline{U_s}$, and the closed set $K_s$ contains $U_s$, so $c(\tau )\in K_s\setminus U_s$. Thus $c(\tau )={\exp }_p(w)$ for a unique $w$ with $\mathrm{\mid }w{\mathrm{\mid }}_{g_p}=s$, and the prefix $c{\mathrm{\mid }}_{[a,\tau ]}$ is piecewise $C^1$ and lies in $K_s\subseteq U$. Applying the piecewise $C^1$ estimate in step 1.3 to this prefix gives $L_g(c)\ge L_g(c{\mathrm{\mid }}_{[a,\tau ]})\ge s>R\mathrm{.}$ Hence every competitor from $p$ to $q$ has length at least $R$.

Steps 2.1--3.1 cover respectively the competitors contained in $U$ and those leaving it, and step 3.1 gives the promised strict inequality in the latter case. Combining the lower bound for all competitors with step 1.1 gives $d_g(p,q)=R$, including $R=0$, where injectivity gives $q=p$ and the radial curve is constant.

For $0<s<\rho$, the equality just proved says $U_s=U\cap B_{d_g}(p,s)$: a point of $U$ has the unique form ${\exp }_p(w)$ and belongs to either side exactly when $\mathrm{\mid }w{\mathrm{\mid }}_{g_p}<s$. Empty $M$ supplies no centre; in dimension zero only $v=0$ occurs, and dimension one is already covered by the closed-interval Euclidean ball. The endpoints $\mathrm{\mid }v\mathrm{\mid }=\rho$ and $s=\rho$ are excluded by the open normal ball, while $v=0$ was handled in step 4.1. Assumption [A1] is used through [F1] for the radial upper bound and [F4] for the polar lower bound; the one orthonormal basis at the fixed point and the unique last-crossing and first-exit times require no additional choice. There is no iff claim.