# Decomposition of normal curvature of curve in $\mathbb{R}^3$

Given curve $c$ in $\mathbb{R}^3$, there are $3$ 2-dimensional submanifolds $N_i$ containing $c$ s.t. they are ruled surfaces $$N_i(s,t)=c(t)+s V_i$$ where $\{V_i\}$ is an orthonormal set.

Clearly, $N_i$ has a intrinsic metric so that $c$ has a normal curvature $k_i$

Hence show that $\sum_i\ k_i\geq k\ \ast$

Proof : If there is a surface $S$ containing $c$, then there is two ruled surfaces $$S_N =(N,T:=c'),\ S_n=(n,T)$$ generated by a unit out normal vector $N$ and a unit vector $n\in T_{c'}S$, which is orthogonal to $c'$.

Hence $k =\sqrt{k_N^2+k_n^2}$ where $k_N,\ k_n$ is normal curvature on these surfaces.

How can we finish the proof ?

[Add] I will enumerate some facts :

(1) Convex simple closed curve in $\mathbb{R}^2$ has total curvature $2\pi$

(2) Total curvature of $c$ is a limit of sum of external angles in polygonal line $c_n$ where $c_n\rightarrow c$.

(3) $\ast$ can be restated in terms of total curvature.

(4) If $u, \ v$ is unit vector, then we fix $v$ and an orthonormal set $\{V_i\}$. Then we have three circles $C_i:=(V_i^\perp + u)\cap S^2(1)$ Hence we find an $u_i'\in C_i$ s.t. $|x-v|\leq |u_i'-v|$ for $x\in C_i$.

If $u_i',\ v$ has external angle $\theta_i$, then $\ast$ is equivlant to $\sum_i\ \theta_i\geq \pi-\angle\ (u,v)$.

(5) If $f:=\cos^{-1} : [-1,1]\rightarrow [0,\pi]$ is strictly decreasing, then note that $f| [-1,0]$ is convex and $f|[0,1]$ is concave.

Then $\ast$ is equivalent to $$f (Ax -\sqrt{(1-A^2)(1-x^2)} ) + f(By-\sqrt{(1-B^2)(1-y^2)} ) + f(Cz-\sqrt{(1-C^2)(1-z^2)} ) \leq 2\pi + f( u\cdot v)$$ where $u=(A,B,C),\ v=(x,y,z)$.