### Differential geometry notes 4

Formulae of Frenet: These describe the derivatives of the moving trihedron of a curve in terms of the moving trihedron itself:

$\begin{bmatrix} \mathbf{\dot{t}} \\ \mathbf{\dot{p}} \\ \mathbf{\dot{b}} \end{bmatrix}$ $= \begin{bmatrix} 0 & \kappa & 0 \\ -\kappa & 0 & \tau \\ 0 & -\tau & 0 \end{bmatrix}$ $\begin{bmatrix} \mathbf{t} \\ \mathbf{p} \\ \mathbf{b} \end{bmatrix}$

Note that the coefficients inside the skew-symmetric matrix need not be constant, hence the system described by this equation is not necessarily linear.

Definition 16.1: A vector $\mathbf{d}$ is called a rotation vector of a rotation if it has the following properties:

1. $\mathbf{d}$ has the direction of the axis of rotation.
2. The sense of $\mathbf{d}$ is such that the rotation has the clockwise sense if one looks from the initial point of $\mathbf{d}$ to its terminal point.
3. The magnitude $|\mathbf{d}|$ of $\mathbf{d}$ equals the angular velocity $\omega$ of the rotation, that is, the velocity of points at distance 1 from the axis of rotation.

Theorem 16.2: The rotation vector of the trihedron of a curve $C: \mathbf{x}(s)$ of class $r \geq 3$ with non-vanishing curvature, when a point moves along $C$ with constant velocity 1 is given by the expression:

$\mathbf{d} = \tau \mathbf{t} + \kappa \mathbf{b}$

This vector is called the vector of Darboux.

The curves traced out by the terminal points of $\mathbf{t}(s)$, $\mathbf{b}(s)$ and $\mathbf{p}(s)$ on the unit sphere as the trihedron of a curve moves are called the tangent, binormal and principal normal indicatrix respectively. The linear elements of these curves $ds_T$, $ds_P$ and $ds_B$ are given by:

$ds_T^2 = \kappa^2 ds^2$

$ds_P^2 = (\tau^2 + \kappa^2) ds^2$

$ds_B^2 = \tau^2 ds^2$

Hence, we can write $ds_P^2 = ds_T^2 + ds_B^2$ (Equation of Lancret).

Shape of a curve in the neighborhood of any of its points: We can Taylor expand the representation of the curve around $s = 0$ (note that the zero point can be chosen arbitrarily):

$\mathbf{x}(s) = \mathbf{x}(0) + \sum_{v = 1}^3 \frac{s^v}{v!} \frac{d^v \mathbf{x}(0)}{ds^v} + o(s^3)$

We have $\mathbf{\dot{x}} = \mathbf{t}$, $\mathbf{\ddot{x}}=\mathbf{\dot{t}} = \kappa \mathbf{p}$, and $\mathbf{\dddot{x}} = \mathbf{\ddot{t}}= \dot{\kappa}\mathbf{p} + \kappa \mathbf{\dot{p}} = \dot{\kappa} \mathbf{p} - \kappa^2 \mathbf{t} + \kappa \tau \mathbf{b}$. We can choose the coordinate system such that $\mathbf{t}(0)=(1,0,0)$, $\mathbf{p}(0)=(0,1,0)$, $\mathbf{b}(0)=(0,0,1)$.

Plugging these expressions in the Taylor expansion, we get:

$x_1(s) = s - \frac{\kappa_0^2}{3!}s^3 + o(s^3)$

$x_2(s) = \frac{\kappa_0}{2}s^2 + \frac{\dot{\kappa_0}}{3!}s^3 + o(s^3)$

$x_3(s) = \frac{\kappa_0 \tau_0}{3!}s^3 + o(s^3)$

where $\kappa_0$ and $\tau_0$ are the values of the curvature and torsion at $s=0$. If we keep only the leading terms, we get:

$\mathbf{x}(s) \approx (s, \frac{\kappa_0}{2} s^2, \frac{\kappa_0 \tau_0}{6}s^3)$              ($\kappa_0 >0$, $\tau_0 \neq 0$)