= Solution
For a <connected space>[connected] <Riemannian manifold> $(M,g)$, its <Riemannian distance> is
$$
d_g(p,q)=\inf_\gamma L_g(\gamma),
\qquad
L_g(\gamma)=\int_a^b|\dot\gamma(t)|_g\,dt,
$$
where the <infimum> is over the piecewise smooth curves from $p$ to $q$. Connectedness of a <smooth manifold> implies path connectedness, so this set of curves is nonempty.
The <Gauss lemma> says that the differential of $\exp_p$ preserves the radial inner product: for $v,w\in T_pM$,
$$
g_{\exp_p(v)}\bigl((d\exp_p)_v v,(d\exp_p)_v w\bigr)=g_p(v,w).
$$
Consequently radial <geodesic>[geodesics] from $p$ are orthogonal to the images of tangent vectors to spheres centred at the origin in $T_pM$. In a sufficiently small <normal neighbourhood> of $p$, this implies
$$
d_g(p,\exp_p v)=|v|_g:
$$
every competing curve has length at least the total variation of its radial coordinate, and the radial geodesic has that length.
The axioms $d_g(p,q)\geq0$, symmetry, and the <triangle inequality> follow directly from length and concatenation. Certainly $d_g(p,p)=0$. If $q\ne p$, choose a normal ball $B_g(p,r)$ that does not contain $q$. Every curve from $p$ to $q$ first meets its boundary, and its initial part has length at least $r$ by the Gauss lemma. Hence $d_g(p,q)\geq r>0$. Thus $d_g(p,q)=0$ if and only if $p=q$.
Back to article page
