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For surfaces in $\mathbb R^3$ given that geodesic torsion $\tau_g$ and Gauss negative curvature $K$ are constant along a line on a surface show that the line must be asymptotic, i.e., must have a vanishing normal curvature $\kappa_n$. This is Beltrami-Enneper converse theorem Number 1.

Given that if along a curve geodesic torsion $\tau_g$ is constant and also that it has zero normal curvature $\kappa_n$ as an asymptotic line show that the surface has constant negative Gauss curvature $K$. This is Beltrami-Enneper converse theorem Number 2.

Important Remark:

If $ |K| \ne \tau_g$, $|K| = 1/b$, $b \ne a$ then the asymptotics would be different, do not have vanishing Euler normal curvature, given by roots of normal curvature equation:

$$ k_n^2 -2 k_n \cot 2 \psi/b + (1/a^2 + 1/b^2) = 0 $$

$$ k_{n1} = \cot (2\psi) /b - \sqrt{{(\csc (2 \psi)/b )}^2 - 1/a^2} $$

For $ K = -1/a^2$, $b =a $, its roots are $ k_n = [0, 2 \cot( 2 \psi)/a] $ which are asymptotic and non-asymptotic normal curvatures respectively.

As scalar curvature expressions are readily available, no need to visit original derivations of Enneper again, only algebraic manipulations are required for me to answer.

Notation

$K$ Gauss curvature, $k_n $ normal curvature, $ k_1$, $k_2$ principal curvatures, $\tau_g $ geodesic torsion, $ k_1 >0 $ because $K<0$. $\psi $ angle between asymptotic and principal curvature lines

First Converse theorem:

Given const. $K$, $ \tau_g $, show that $ k_n =0$.

$$ K = k_1 k_2 =\tau_g^2 \tag{1*} $$

$$ (k_1 +k_2)\sin\psi \cos\psi = \tau_g \tag{2*} $$

Eliminate $\tau_g $ between (1*) (2*) by squaring (2*) and equating,

$$ k_1^2 + 2 k_1 k_2 \left[ 1- {\dfrac{1/2}{(\sin \psi \cdot \cos \psi )^2}} \right] + k_2^2 = 0 \tag{3*} $$

Let $ \tan \psi = t $, so the square bracket term equals $ -(t^2 + 1/t^2))/2 $.

Letting $ R= \dfrac{k_1}{k_2} $

$$ R^2 - R (t^2 + 1/t^2) +1 =0 \tag{4*} $$

$$ R = t^2, R = 1/t^2 \tag{5*} $$

These give (Euler) normal curvatures zero brought back to classical forms :

$$ k_n = k_1 \cos ^2\psi + k_2 \sin^2 \psi = 0 \tag{6*}$$

$$ k_n = k_2 \cos^2 \psi + k_1 \sin^2 \psi = 0 \tag{7*}$$

as required to be shown. The second result was not expected but should not be surprising as prevails between conjugate pairs of warped surface parts around saddle point/asymptotic lines for $K <0$ .

Second Converse theorem:

Given $ k_n =0 $ and $ \tau_g = const.$, show $ K = -\tau_g^2. $

$$ k_n = -k_1 \cos ^2\psi + k_2 \sin^2 \psi = 0 \tag{8*}$$

From (8*) and (2*)

$$ \sin\psi = \sqrt{ k_1 a} ;\, \cos\psi= \sqrt{ k_1 a} \tag{9*}$$

where $a$ is an arbitrary constant for scaling in trig. triangles.

$$ \cos^2\psi + \sin^2\psi = 1 = a (k_1 + k_2) \tag{10*} $$

Plug in (9*) and (10*) into (2*) and squaring,

$$ 1/a \cdot \sqrt{k_1 k_2 } \cdot a = \tau_g \rightarrow K = k_1 k_2 = \tau_g^2 = 1/a^2 \tag{11*} $$

whose signs need to be changed to get into classical form.

Actually my aim/motivation for this post has been:

1) To verify converse theorems of Beltrami-Enneper as stated for constant $K$, which I have already done as above and also

2) To demonstrate/verify general validity in converse theorems even when $K$ is variable, which is still an open question now.

It is here perhaps that the machinery of differential geometry would be brought to work.

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