Exercise 6.3 (Millman & Parker, Elements of Differential Geometry). Let $$X_N = N - \langle N, n \rangle n $$ be the tangential component of the normal vector $N$ of a unit speed curve $\gamma$ on a surface $M$. Prove that the following are equivalent:

  1. $X_N = 0$.
  2. $\gamma$ is a geodesic.
  3. $X_N$ is parallel along $\gamma$.

I'm having trouble proving (3) implies (1) or (2). Any ideas? I think I can do the other implications..

(1 -> 2): If $X_N = 0$, then $N = \langle N, n \rangle n$ so that $$\kappa_g := \langle \gamma'', S\rangle = \kappa \langle N, S \rangle = \kappa \langle \langle N, n \rangle n, S \rangle = 0. $$

(2 -> 1): If $\kappa_g = 0$, then $\langle N, S \rangle = 0$. Since $\{n, T ,S\}$ form an orthonormal basis, $$X_N = \langle N, T\rangle T+ \langle N, S\rangle S = 0.$$

(1 -> 3): If $X_N = 0$, then $\frac{dX_N}{dt} = 0$ is clearly perpendicular to the surface $M$.

In my book's notation, $T = \gamma'$, $N= T' / \kappa$, $n$ is the unit normal to the surface, $S = n \times T$, and a vector field $X$ is said to be parallel along $\gamma$ if $dX/dt$ is perpendicular to $M$.


Since parallel translation preserves angles, if $X_N$ is parallel along $\gamma$, so is $T$. That's one of the definitions of a geodesic.

  • $\begingroup$ Ohh I think I got it.. If $X_N$ is parallel along $\gamma$, then $\langle X_N', T \rangle + \langle X_N, T' \rangle = \langle X_N, T \rangle' = 0$ implies $T'$ is perpendicular to $X_N\in T_PM$, i.e. $T$ parallel along $\gamma$, and so $\gamma$ is a geodesic! $\endgroup$ – Alain Jun 5 '16 at 20:51

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