# Frenet Frame along a curve and Riemannian Curvature on $S^2$

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.

• 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? Jan 29 '19 at 19:52
• @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.
– Nhat
Jan 30 '19 at 12:42

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 $$1$$.