I have a question about a proof of the Beltrami-Enneper theorem:
In the following $\nu$ is the surface-normal and $e_1,e_2,e_3$ the Frenet 3-frame.
It states: Every asymptotic curve $c: I \rightarrow S \subset \mathbb{R}^3$ ( $II(c',c')=0$, where $II$ is the second fundamental form) with curvature $\kappa \neq 0$ and torsion $\tau$ satisfies the equation $\tau^2=-K$, where $\tau$ is the torsion of the curve and $K$ the Gauß-curvature of the surface $S$.
Proof: Let $c(s)$ be an aymptotic curve with $II(c',c')=0$. Then the normal curvature of $c$ vanishes. Hence, $e_2$ is tangential to the surface (so far is everything alright), but now it goes on with, then $e_3 = \nu$.
I mean, I see that $e_3$ is then either $+ \nu$ or $- \nu$, but I don't see why the $+$ sign is already clear here.
Then it goes on like $\tau= \langle e_2',e_3 \rangle = \langle e_2', \nu \rangle $ which is just definition and the fact we just proved, but now it is claimed that this is equal to $II(e_1,e_2).$
Then the author claims that
$K = Det(II)/Det(I) = II(e_1,e_1)II(e_2,e_2) – (II(e_1,e_2))^2 = 0 – \tau^2.$
Apparently, the determinant of the metric tensor is supposed to be equal to one here(why?).
You find this proof in this book:
It is theorem 3.19 on page 85: see here
If anything is unclear, please let me know.
Best Answer
Let $c \colon I \rightarrow \mathbb{R}^3$ be a smooth enough curve with arclength parametrization and consider the associated Frenet-Serret frame. That is, $e_1(t) = c'(t)$, $e_2(t) = \frac{c''(t)}{||c''(t)||}$ and $e_3(t) = e_1(t) \times e_2(t)$. How does the frame change if we change the orientation of the curve? Define $\tilde{c}(t) = c(-t)$. Then $\tilde{e_1}(t) = \tilde{c}'(t) = -c'(-t) = -e_1(-t)$, $\tilde{e_2}(t) = e_2(-t)$ and $\tilde{e_3}(t) = \tilde{e_1}(t) \times \tilde{e_2}(t) = -e_3(-t)$. You can check that the curvature and the torsion of the curve are independent of its orientation, but the Frenet-Serret frame does depend on the orientation as $e_1$ and $e_3$ get flipped.
By the discussion above, reversing orientation if necessary, you can guarantee that $e_3$ in your case will be $\nu$ and not $-\nu$.
As to why $\tau = II(e_1,e_2)$, we have
$$ II_{c(t_0)}(e_1,e_2) = -\left< (d_{c(t_0)}\nu)(e_1), e_2(t_0) \right> = -\left< \left. \frac{d\nu(c(t))}{dt} \right|_{t=t_0}, e_2 \right> = -\left< e_3'(t_0), e_2(t_0) \right> = \left< e_2'(t_0), e_3(t_0) \right> = \tau(t_0) $$ where the equality $$ -\left< e_3'(t_0), e_2(t_0) \right> = \left< e_2'(t_0), e_3(t_0) \right> $$ follows by differentiating $$ \left< e_3(t), e_2(t) \right> \equiv 0. $$
Regarding your question about the determinant, note that a bilinear form $f \colon V \otimes V \rightarrow \mathbb{F}$ doesn't have a well-defined notion of determinant as a number. If $A_B(f)$ is the matrix representing $f$ in a basis $B$, then $A_{B'} = S^T A_B S$ where $S$ is an invertible change of basis matrix. Thus, $\det(A_{B'})(f) = \det(S)^2 \det(A_B)(f)$ and so the determinant is well-defined (independent of basis) only up to a square of the field $\mathbb{F}$. For example, if $\mathbb{F} = \mathbb{R}$, then by choosing an appropriate basis, the determinant of the matrix representing the bilinear form can be made $0,1$ or $-1$. However, if you have two bilinear forms $f,g \colon V \otimes V \rightarrow \mathbb{F}$ with $g$ nondegenerate then the ratio $\frac{\det(A_B(f))}{\det(A_B(g))}$ does not depend on the basis $B$ in which $f$ and $g$ are represented simultaneously. The statement that $K = \frac{\det(II)}{\det(I)}$ should be interpreted in this way. You can compute the ratio using a coordinate frame, but you can also compute the ratio using an orthonormal frame in which $\det(I) = 1$.