Geodesic curvature need not equal ambient curve curvature

Take the unit sphere $S^2\subset {\mathbb{R}}^3$ with its induced round metric and outward orientation, and let $\gamma (t)=(\cos t,\sin t,0)$ for $t\in \mathbb{R}$. Then ${\gamma }^{\mathrm{\prime }}=(-\sin t,\cos t,0)$ has norm $1$, while ${\gamma }^{\mathrm{\prime }\mathrm{\prime }}=(-\cos t,-\sin t,0)=-\gamma (t)$ has norm $1$ and is normal to $S^2$ because $T_{\gamma (t)}{S}^2=\gamma (t{)}^{\perp }$. Thus the ordinary space-curve curvature is $\mathrm{\parallel }{\gamma }^{\mathrm{\prime }\mathrm{\prime }}(t)\mathrm{\parallel }=1$.

Around any point of the equator use a longitude-latitude chart $X(u,v)=(\cos v\cos u,\cos v\sin u,\sin v)$ on a longitude interval and $\mathrm{\mid }v\mathrm{\mid }<\pi \mathrm{/}2$. Differentiating $X$ gives $g_{uu}={\cos }^{2}v$, $g_{uv}=0$, and $g_{vv}=1$. Formula [F2] therefore yields ${\mathrm{\Gamma }}^u{}_{uu}=0$ and ${\mathrm{\Gamma }}^v{}_{uu}=\sin v\cos v$. The curve from step 1.1 has coordinates $(u,v)=(t,0)$ on this chart, so its coordinate second derivatives vanish and both components of its covariant acceleration are zero at $v=0$. This calculation applies in a chart around every parameter value.

Hence $D_t{\gamma }^{\mathrm{\prime }}=0$ everywhere, so [F3] identifies $\gamma$ as an affinely parametrized geodesic. Its covariant acceleration is $A_{\gamma }=0$, and [F1] then gives $k_g=0$. Together with step 1.1 this gives $k_g=0\mathrm{\ne }1=\mathrm{\parallel }{\gamma }^{\mathrm{\prime }\mathrm{\prime }}\mathrm{\parallel }$, so the proposed universal equality is false. The witness is a single explicit curve and uses no choice principle.