Tangent-angle formula for geodesic curvature

Along the curve, the product rule and [F1] give ${\mathrm{\nabla }}_{T}T=-{\theta }^{\mathrm{\prime }}\sin \theta E_1+\cos \theta \omega (T)E_2+{\theta }^{\mathrm{\prime }}\cos \theta E_2-\sin \theta \omega (T)E_1=({\theta }^{\mathrm{\prime }}+\omega (T))(-\sin \theta E_1+\cos \theta E_2)$. By [F3], the final vector in parentheses is $JT$.

By [F2] and [F4], $k_g=g({\mathrm{\nabla }}_{T}T,JT)$. By [F3] and step 1.1, $JT=-\sin \theta E_1+\cos \theta E_2$, whose squared norm is ${\sin }^2\theta +{\cos }^2\theta =1$. Step 1.1 therefore gives $k_g={\theta }^{\mathrm{\prime }}+\omega (T)$. If $\tilde{\theta }$ is another angle lift on the same connected subinterval, then $\tilde{\theta }-\theta$ is continuous and takes values in $2\pi \mathbb{Z}$, so it is constant and has zero derivative. Thus the formula is lift-independent. The same computation uses one-sided derivatives at included endpoints. If $k_g=0$, the equality still holds without division; if no curve is present, the universal assertion is vacuous. The argument uses only the supplied frame, curve and local lift, so it makes no choice from any nonempty family.