 Solution

Let $\kappa =R/12$, which is constant by the preceding argument. Substitution into geodesic deviation givesusing $g(T,T)=-1$ and the established orthogonality. With $V=DS$, metric compatibility yields $\dot{f}=2g(S,V)$ andThereforeBoth $S$ and $DS$ are orthogonal to $T$, so their squared norms are nonnegative. The minus sign in this conserved deviation energy in constant curvature reflects $D^{2}S=+\kappa S$, rather than an ordinary oscillator when $R>0$.