= Solution
The <Riemannian exponential map> at $p$ is $\exp_p(v)=\gamma_v(1)$, where $\gamma_v$ is the affinely parametrized <geodesic> with $\gamma_v(0)=p$ and $\dot\gamma_v(0)=v$. Its derivative at zero is the identity. The <inverse function theorem> therefore gives a <normal neighbourhood> $U=\exp_p(V)$, where $V\subset T_pM$ is an open star-shaped neighbourhood of zero on which $\exp_p$ is a <diffeomorphism>. Star-shaped means $tv\in V$ whenever $v\in V$ and $0\leq t\leq1$; it ensures the radial geodesic remains in $U$. This is distinct from a <convex normal neighbourhood>, which imposes a unique short <geodesic> between every pair of its points.
The <radial distance in a normal neighbourhood> and the unit <radial vector field in a normal neighbourhood> are defined, respectively, by
$$
r(q)=|v|,\qquad E_q=(d\exp_p)_v\frac{v}{|v|},\qquad v=\exp_p^{-1}(q),\quad q\ne p.
$$
The function $r$ is continuous at $p$ and smooth away from $p$, but is not differentiable at $p$ in positive dimension. Some conventions use the scaled radial field $W_q=(d\exp_p)_v v$ and the smooth radial function $r^2/2$ instead; the associated gradient identities will distinguish them.
The <Gauss lemma> states that for $v\in V$ and $w\in T_pM$,
$$
\boxed{g_{\exp_p(v)}\bigl((d\exp_p)_v v,(d\exp_p)_v w\bigr)=g_p(v,w).}
$$
To prove it, consider the <geodesic variation> $F(s,t)=\exp_p(t(v+sw))$ for $0\leq t\leq1$ and small $s$. Set $T=\partial_tF$ and $J=\partial_sF$. The <Levi-Civita connection> has zero <torsion tensor>, so $D_tJ=D_sT$. Each $t$-curve is a <geodesic>, hence $D_tT=0$, and its squared speed is the constant $|v+sw|^2$. At $s=0$,
$$
\frac{d}{dt}g(T,J)=g(T,D_sT)=\frac12\partial_s|T|^2=g_p(v,w).
$$
Since $J(0)=0$, integration gives $g(T,J)=t\,g_p(v,w)$. At $t=1$, $T=(d\exp_p)_v v$ and $J=(d\exp_p)_v w$, proving the <Gauss lemma>. Taking $w=v$ shows $|E|=1$; taking $w\perp v$ shows that $E$ is orthogonal to the images of tangent vectors to the tangent-space sphere, hence to the <geodesic spheres> in $U$.
For $u\in T_pM$ of unit length, write the unit-speed radial <geodesic> as $\gamma_u(t)=\exp_p(tu)$. Differentiating this expression and substituting $v=tu$ gives
$$
\boxed{E_{\gamma_u(t)}=\dot\gamma_u(t)\qquad(t>0).}
$$
To identify the <gradient>, write an arbitrary tangent vector at $q=\exp_p(v)$ as $\eta=(d\exp_p)_v w$. Differentiating $r(\exp_p(v))=|v|$ and using the <Gauss lemma>,
$$
dr_q(\eta)=\frac{g_p(v,w)}{|v|}=g_q(E_q,\eta).
$$
By the defining property of the <gradient>,
$$
\boxed{E=\operatorname{grad}r,\qquad W=rE=\operatorname{grad}(r^2/2).}
$$
The latter field extends smoothly across $p$ by $W_p=0$, because $r^2/2$ is the smooth quadratic function $|v|^2/2$ in <normal coordinates>.
The same calculation also proves local radial minimization: for a piecewise smooth <curve> in the normal neighbourhood, $|dr(\dot\alpha)|\leq|\dot\alpha|$, so its length from $p$ to $q$ is at least $r(q)$, attained by the radial <geodesic>. Take a small tangent ball with closure inside a larger normal ball. Any path leaving the small ball before reaching a point inside it has already accumulated length at least its radius. Therefore, for points in the small ball, $r(q)=d(p,q)$ even when competitors may leave the neighbourhood. This local argument does not assert that $r$ equals global <Riemannian distance> throughout every arbitrarily large star-shaped normal neighbourhood.
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