Understanding normal curvature

Question

I want to prove that for every compact surface $S\subset \mathbb{R}^3$ there must exist an elliptic point. The proof follows like this: consider the function $f:\mathbb{R}^3\to\mathbb{R}$ like $f(p)=||p||^2.$ Since S is compact, the restriction $f_{|_S}$ must attain a maximum, lets say for $p_0\in S$. Consider an unit tangent vector $v\in T_pS$ and a parametrized curve $\alpha:(-\epsilon,\epsilon)\to S$ such that $\alpha(0)=p_0$ and $\alpha'(0)=v$. Consider the function $h=f\circ \alpha:(-\epsilon,\epsilon)\to\mathbb{R}$, which attains a maximum at $t=0$. Since it's a maximum we have that $$h'(0)=2\langle \alpha'(0),\alpha(0)\rangle=0,$$ which implies that $\alpha(0)=p$ (seen as the vector $p-(0,0,0)$ of $\mathbb{R}^3$) is perpendicular to $S$ at the point $p_0$, so $p_0/||p_0||$ is an unit normal at this point. Also we have that $$h''(0)=2\langle \alpha''(0),\alpha(0)\rangle+2\langle \alpha'(0),\alpha'(0)\rangle=2(\alpha''(0)\cdot p_0+1)\leq 0\implies \alpha''(0)\cdot \frac{p_0}{||p_0||}\leq -\frac{1}{||p_0||}.$$ Now since $\alpha''(0)\cdot \frac{p_0}{||p_0||}$ is the normal curvature of $p$ in the direction of $v$, we have that for the principal directions the principal curvatures satisfie that $$k_1(p),k_2(p)<0\implies K(p)>0,$$ so $p$ is elliptic. The point i'm missing is when it says that $\alpha''(0)\cdot \frac{p_0}{||p_0||}$ is the normal curvature of $p$ in the direction of $v$. If $N$ is the gauss operator, the definition of normal curvature is $$k_n=-d_pN(v)\cdot v,$$ and i don't see how it relates to the expresion $\alpha''(0)\cdot \frac{p_0}{||p_0||}$. Any help will be very appreciated.

Answer 1

Recall that if $\alpha:(-\epsilon,\epsilon)\to {\mathbb R}^3$ is parametrized by arc length $s$, then $\alpha'(s)$ is the unit tangent, and $$ \| \alpha''(s)\| $$ is defined to be the curvature of $\alpha$.

Now if the curve $\alpha:(-\epsilon,\epsilon)\to S\subset {\mathbb R}^3$ lies on a surface $S$, still parametrized by arc length $s$, then $$ \alpha''(s)\cdot N_{\alpha(s)} $$ is defined to be the normal curvature of $\alpha$ on the surface. It is the projection of the curvature of $\alpha$ in the normal direction of the surface $S$.

You formula for normal curvature is obtained only after an "integration by parts calculation" calculation: $$ \alpha''(s)\cdot N_{\alpha(s)} = (\alpha'(s)\cdot N_{\alpha(s)})' - \alpha'(s)\cdot N_{\alpha(s)}' = -\alpha'(s)\cdot dN_{\alpha(s)}(\alpha'(s)). $$ Here $\ '=\frac{d}{ds}$, and note $\alpha'(s)\cdot N_{\alpha(s)}=0$. In particular, at $s=0$, the normal curvature is $$ \alpha''(0)\cdot N_{p_0} = -d_{p_0} N(v)\cdot v, $$ where $v=\alpha'(0)$, $p_0=\alpha(0)$. In the case at hand, $N_{p_0}=\frac{p_0}{\|p_0\|}$.