Weingarten map

To study the curvature of the shape $S$ in a neighborhood of $x_0 = \xi(u_0, v_0)$, one can examine the variations of the normal $\vec{n} = \vec{\nu}(u,v)$ around $x_0$. In particular, one can derivate the normal $\vec{n}$ in the direction specified by the tangent vectors at $x_0$: this is called the Weingarten map (note that it is not its derivative by $\vec{\tau}$). Because the normal is unitary, it is orthogonal to its derivatives (implying the derivatives are tangent vectors). Let $C$ be a curve of cartesian equation $\Gamma$, contained in $S$ going though $x$, and $(T, \gamma)$ a parameterization of $C$, such that $\vec{\tau} = \frac{\delta\gamma}{\delta t}(t_0)$. $\gamma(t) = \xi(u(t), v(t))$. In the basis $(x_u, x_v)$, $\vec{\tau}$ has coordinates $(\frac{\delta u(t)}{\delta t}, \frac{\delta u(t)}{\delta t}) = (\tau_u, \tau_v)$. Then:

$$\begin{array}{rcl} W_x(\vec{\tau}) = \frac{\delta \vec{n}}{\d... ...\delta v}\\ &=&\tau_u \vec{n}_u + \tau_v \vec{n}_v \end{array}$$

This expression shows that $W_x(\vec{\tau})$ is inpendent from the curve $\Gamma$ chosen to evaluate it. Moreover, the Weingarten map is independent from the chosen parameterization: let the coordinates of $\vec{\tau}$ be $(\tau_u, \tau_v)$ in $(x_u, x_v)$ and $(\tau_{u'}, \tau_{v'})$ in $(x_{u'}, x_{v'})$. The change of coordinates can be expressed by $u' = a_u u + a_v v$ and $v' = b_u u + b_v v$ or $u = \frac{b_v u' - a_v v'}{a_u b_v - a_v b_u}$ and $v = \frac{b_u u' - a_u v'}{a_u b_v - a_v b_u}$. Knowing that $\frac{\delta u}{\delta u'} = \frac{b_v}{a_ub_v - a_vb_u}$, $\frac{\delta u}{\delta v'} = \frac{-a_v v'}{a_ub_v - a_vb_u}$, $\frac{\delta v}{\delta v'} = \frac{-a_u}{a_ub_v - a_vb_u}$ and $\frac{\delta v}{\delta u'} = \frac{b_u}{a_ub_v - a_vb_u}$, the following can be verified:

$$\begin{array}{rcl} W_x(\vec{\tau}) &=&\tau_{u'} \vec{n}_{u'} ... ...n}_v)\\ &=& \tau_u \vec{n}_u + \tau_v \vec{n}_v\\ \end{array}$$