Geodesic curvature and projection onto the tangent plane

Question

Here is exercise 10 from section 4-4 outin do Carmos's Differential Geometry of Curves and Surfaces:

Show that the geodesic curvature of an oriented curve $C\subset S$ at a point $p\in C$ is equal to the curvature of the plane curve obtained by projecting $C$ onto the tangent plane $T_p(S)$ along the normal to the surface at $p$.

As a hint, he writes:

Apply the relation $k_g^2 + k_n^2 = k^2$ and the Meusnier theorem to the projecting cylinder.

I can't see how the Meusnier theorem applies here, since it deals with curves that are on the same surface. Besides, what cylinder is he talking about?

Answer 1

Let $\alpha$ be a unit-speed parametrization of $C$ with $\alpha(0) = p$. The projection of $\alpha$ in $T_pS$ along $N(p)$ is given by $$\beta(s) = \alpha(s)+\langle p-\alpha(s),N(p)\rangle N(p),$$whence $\beta'(s) = \alpha'(s) - \langle \alpha'(s),N(p)\rangle N(p)$ and $\beta''(s) = \alpha''(s) - k_n(\alpha'(s)) N(p)$. For $s=0$ we obtain $$\beta'(0) = \alpha'(0) \quad\mbox{and}\quad \beta''(0) = \frac{D\alpha'}{{\rm d}s}(0).$$Computing the curvature of $\beta$ at $0$ we have $$\kappa_\beta(0) = \frac{\|\beta'(0) \times \beta''(0)\|}{\|\beta'(0)\|^3} = \left\|\frac{D\alpha'}{{\rm d}s}(0)\right\| \sin \theta_1,$$where $\theta_1 = \angle(\alpha'(0), (D\alpha'/{\rm d}s)(0))$. On the other hand, the geodesic curvature of $\alpha$ at $p$ is given by: $$k_g(p) = \left\langle \frac{D\alpha'}{{\rm d}s}(0), N(p) \times \alpha'(0)\right\rangle = \left\|\frac{D\alpha'}{{\rm d}s}(0)\right\|\| N(p)\times \alpha'(0)\|\cos \theta_2 = \left\|\frac{D\alpha'}{{\rm d}s}(0)\right\|\cos \theta_2 $$since $N(p)$ and $\alpha'(0)$ are orthogonal unit vectors, where $\theta_2 = \angle((D\alpha'/{\rm d}s)(0),N(p)\times \alpha'(0))$. Since $\theta_1+\theta_2 = \pi/2$, the result follows.