I'm reading the wikipedia pages on some differential geometry topics, e.g. Gaussian curvature. Let's draw our attention to surfaces $z=f(x,y)$ in 3D Euclidian space.
The text states the following:
We represent the surface as the graph of a function, f, of two variables, in such a way that the point $\vec{p}$ is a critical point, i.e., the gradient of f vanishes (this can always be attained by a suitable rigid motion).
Then the Gaussian curvature of the surface at $\vec{p}$ is the determinant of the Hessian matrix of $f(x,y)$.
I was wondering how you can get $\bigtriangledown f(\vec{p})=0$ by a rigid motion. Can anyone give an example?
I can't even imagine the truth of this statement in 2D ($f'(x_0)=0$ after a suitable transformation of the $x$ and $y$-axes.). An example here is welcome too!
Any help is appreciated! Thanks in advance.
We can always represent a regular surface $\Sigma$ in $\mathbb{R}^3$ as the zero set of some function $F:\mathbb{R}^3\to\mathbb{R}$ such that $\nabla F$ is nowhere zero on $\Sigma$. (Take, e.g., the square of the distance function from $\Sigma$.)
Choose a point $p\in\Sigma$. By a rigid motion we may arrange so that $p=0$ and $\nabla F_p$ is vertical. Now by the implicit function theorem, there is a neighborhood $U$ of $p(=0)$ and function $f:\mathbb{R}^2\to\mathbb{R}$ such that in $U$, $$ F(x,y,f(x,y)) = 0. $$
That is, in $U$, the surface $\Sigma\cap U$ coincides with the graph of $f$.