In baby do Carmo's book Differential Geometry of Curves and Surfaces, definition 8a states that a regular connected curve $C$ in a regular surface $S$ is called a geodesic if, for each point $p \in S$, the parameterization $\alpha(s)$ of a coordinate of $p$ by the arc length $s$ is a parameterized geodesic (it is geodesic for all $t$ in the interval domain $I$ of the curve $C$).

My question is about a comment in the book, right after this definition:

From a point of view exterior of the surface $S$, Definition 8a is equivalent to saying that $\alpha''(s) = kn$ (principal normal vector) is normal to the tangent plane, that is, parallel to the normal of the surface."

A regular curve is a geodesic iff this happens for every point in the curve.

I don't get it. Isn't the principal normal vector at a point of a curve always parallel to the normal of the surface? When calculating the second fundamental form of $\alpha'(0)$, we get $<N,kn> = k<N,n> = k_n$. I believe my confusion comes from always thinking about spheres and cylinders and planes, and so we would have $<N,n> = 1$. Please enlighten me on this.

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    $\begingroup$ Indeed, only for geodesics is it true that $N$ is a scalar multiple of $n$. In general you have $k^2 = k_n^2 + k_g^2$ and the curve is a geodesic precisely when $k_g = 0$. $\endgroup$ Nov 26 '19 at 21:47

Consider the parallel $\lambda=45^\circ$ on a globe.

enter image description here

It's a flat circle, so its normal vector also lies on the same plane. It's obvious that the normal vector of this curve isn't perpendicular to the surface of the sphere (it is angled at $45^\circ$). (At this lattitude, even at noon the sun is never directly above the head). This happens because it's not geodesic.

However, the equator is geodesic and its normal vector is also normal the sphere


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