The normal curvature is bounded by the principal curvatures.

Question

Let the inclusion $i:S\subset\mathbb R^3$ be an immersion of a surface $S$, and let $N:S\to \mathbb R^3$ be a local Gauss map. Let $a:I\to S$ be an arc length parametrized curve, with $a(0)=p$ and $a'(0)=v$.

The second fundamental form of $S$ at $(v,v)$ is $W_p(v,v)=-<N_p(v),v>=-<N_p(v),a'(0)>$.

$$dN_p(v)=v(N)=\frac{d}{dt}|_0N(a(t))=N_x(p)v_1+N_y(p)v_2+N_z(p)v_3$$

Therefore, $W_p(v,v)=<N_x(p)v_1+N_y(p)v_2+N_z(p)v_3,a'(0)>$.

The curvature of $a$ at $p$ is $k_a(p)=\sqrt{<a''(0),a''(0)>}$. How can we see that $k_a(p)=W_p(v,v)$?

Obviously, the other possible local Gauss map gives the opposite value for $W_p(v,v)$. How can we remedy this phenomenon?

Motivation:

The formula $k_a(p)=W_p(v,v)$, shows that the normal curvature of $S$ in the direction $v$ is a value of $W$. Hence, the normal curvature would be bounded by the principle curvatures.

Answer 1

First, let $S$ be a regular surface of $\mathbb{R}^3$with Gauss map $N$. Let $c\colon I:=(-\varepsilon,\varepsilon)\subset \mathbb{R}\longrightarrow S$ be a curve parametrized by arc length with $c(0)=p\in S$ and $c'(0)=v\in T_pS$. Then we have the following: $$II_p(v)=-\langle dN_p(v),v\rangle=-\langle dN_{c(0)}(c'(0)),c'(0)\rangle=-\langle (N\circ c)'(0)),c'(0)\rangle.\ (\dagger)$$ But, $\langle (N\circ c)(s)),c'(s)\rangle=0,\ \forall s\in I$. Therefore by differentiating and evaluating at $s=0$ we obtain $$\langle (N\circ c)(0)),c''(0)\rangle=-\langle (N\circ c)'(0)),c'(0)\rangle.$$ Thus, from $(\dagger)$ we obtain $$II_p(v)=\langle (N\circ c)(0)),c''(0)\rangle=k_n(c(0),c'(0))=k_n(p,v),$$ where $k_n(p,v)$ is the normal curvature at $p$ in the direction $v$. Now, BE CAREFUL: the curvature of $c$ at $p$ doesnt coincide in general with the normal curvature $k_n(p,v)$ (recall the relation $k^2=k_n^2+k_g^2$ where $k,k_n\ \&\ k_g$ are the curvature, the normal curvature and the geodesic curvature of $c$. Thus, you can obtain $k\geq |k_n|$). Let me outline you two facts:

1. At a point $p$ you have $$\mathrm{max}_{v\in T_pS,\ ||v||=1}k_n(p,v)=k_1(p)$$ and $$\mathrm{min}_{v\in T_pS,\ ||v||=1}k_n(p,v)=k_2(p)$$
2. If the Gaussian curvature at a point $p$ is positive then you can easily check that the following inequality holds: $$k(p)\geq\mathrm{min}\{|k_1(p)|,|k_2(p)|\}.$$ Hope i helped!