What is the local canonical form of a regular surface?

672 Views Asked by At

We know that for any space curve $\alpha:I\rightarrow {\rm I\!R}^3$ parametrized by arc length, there exists a "local canonical form" around $s_0\in I$: $$ \alpha(s) = \begin{bmatrix}s-\frac{k^2 s^3}{6} + o(s^3)\\ \frac{k's^3}{6}+\frac{s^2 k}{2}+o(s^3) \\-\frac{s^3 k \tau}{6}+o(s^3)\end{bmatrix} $$ where $k$ is the curvature and $\tau$ the torsion. If I understood correctly, this can be achieved by setting the origin on $\alpha(s_0)$ and using the Frenet frame $\lbrace t, n, b\rbrace$ as coordinate system.

(see: Differential Geometry of Curve and Surface, Do Carmo, Sec 1-6, p27)

What is the local canonical form of a regular surface ?

I saw, for example on these notes (Sec 2.4, p12), that the canonical form of a regular surface $\gamma:U\rightarrow{\rm I\!R}^3$ is: $$ \gamma(u,v) = \begin{bmatrix} u \\ v \\ \frac{1}{2}(\kappa_1 u^2 + \kappa_2 v^2) + o(u^2 + v^2)\end{bmatrix} $$ with $\kappa_1$ and $\kappa_2$ the two principal curvatures.

  • Is this true?

  • The coefficient multiplying the $uv$ term of the Taylor expansion is null. Why?

  • How to prove it?

  • How to find it? (I assume that taking the two principal curvature directions and the unit normal vector as coordinate system leads to this formula).