Signs of geodesic curvature under reversals

Fix $s\in I$. The metric, curve, and Levi–Civita connection are unchanged when the surface orientation is reversed; by [F3] the quarter-turn becomes $-J$. By [F1] and [F2], $k_{-J,\gamma }(s)=g(A_{\gamma }(s),(-J)T(s))=-g(A_{\gamma }(s),JT(s))=-k_{J,\gamma }(s)\mathrm{.}$ This proves the surface-orientation reversal identity pointwise, including when the curvature is zero.

Fix $t\in -I$, and choose a chart containing $\gamma (-t)$. In its coordinates write $x(s)=x\circ \gamma (s)$ and $\bar{x}(t)=x\circ {\gamma }^{\mathrm{r}\mathrm{e}\mathrm{v}}(t)=x(-t)$. Then ${\dot{\bar{x}}}^i(t)=-{\dot{x}}^i(-t)$ and ${\ddot{\bar{x}}}^k(t)={\ddot{x}}^k(-t)$. Substitution into [F1]'s coordinate formula for covariant acceleration gives ${\bar{A}}^k(t)={\ddot{x}}^k(-t)+{\mathrm{\Gamma }}^k{}_{ij}(x(-t)){\dot{x}}^i(-t){\dot{x}}^j(-t)=A_{\gamma }^k(-t),$ because the two velocity signs cancel in the Christoffel term. The reversed unit tangent is $\bar{T}(t)=-T(-t)$, so [F2] yields $k_{J,{\gamma }^{\mathrm{r}\mathrm{e}\mathrm{v}}}(t)=g(A_{\gamma }(-t),-JT(-t))=-k_{J,\gamma }(-t)\mathrm{.}$ The coordinate calculation and inner product also hold with one-sided derivatives at included endpoints.

For both reversals, the acceleration in step 1.2 remains $A_{\gamma }(-t)$, while the new normal is $(-J)(-T(-t))=JT(-t)$. Hence $k_{-J,{\gamma }^{\mathrm{r}\mathrm{e}\mathrm{v}}}(t)=g(A_{\gamma }(-t),JT(-t))=k_{J,\gamma }(-t)\mathrm{.}$ Thus the two sign changes cancel. The calculations are pointwise and use no division, so zero curvature is included. The curve and both reversal maps are supplied explicitly; no choice is used.