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I would appreciate some help showing the following statement.

Let $\omega: [0,1] \rightarrow S^2$ be a smooth curve with velocity vector $V = \omega'$, speed $v = |V|$ and Frenet frame $\{T,N\}$.

Then for all vector fields $W: S^2 \rightarrow TS^2$ along the curve $\omega$ we have:

$$R(W,V)V = v^2 \langle W , N \rangle N $$

where $R$ is the Riemannian curvature tensor on $S^2$.

If $\omega$ were arc-length parametrized / unit-length, I suppose the terms $v^2$ could join the left side because of the normalization of $N = \frac{|T'|}{T'}$, but this isn't the case here.

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  • $\begingroup$ No, the $v^2$ is there because $V=vT$. What definition of the Frenet frame are they using? We're not thinking of $S^2\subset\Bbb R^3$ and of the Frenet frame of a curve in $\Bbb R^3$, I presume? $\endgroup$ – Ted Shifrin Jan 29 at 19:52
  • $\begingroup$ @TedShifrin I think they are but I’m not entirely sure. If it helps: Later calculations are made representing the vectorfield $W$ (as above) in terms of $W = fT + gN$. Furthermore for $\gamma $ with $\gamma‘ = \omega$ they get the Frenet Frame $\{T_\gamma, N_\gamma, B_\gamma \}$ such that $T_\gamma = \omega, N_\gamma = T, B_\gamma$ as expected. $\endgroup$ – Nhat Jan 30 at 12:42
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Yes, I think the authors' use of the term Frenet frame is a bit misleading — I personally would call this a Darboux frame. Regardless, by $N$ they mean the normalized covariant derivative of $T$.

As I already commented, it suffices to show that $R(W,T)T = \langle W,N\rangle N$ (since $V = vT$). Note that $R(T,T)T = 0$, so, writing $W=aT+bN$, we have $R(W,T)T = bR(N,T)T \overset{(*)}{=} bN = \langle W,N\rangle N$, as required. The equality (*) follows from the fact that the sphere has constant sectional curvature $1$.

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