# Converse of Beltrami-Enneper theorems

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.