# Differential rotations in Chebyshev net

A Chebyshev net obeying Sine-Gordon equation is drawn on a surface of constant negative Gauss curvature $$K$$ so that the asymptotic differential rhombic element corners lie on lines of maximum/minimum normal curvature.

Show that principal rotation of surface normals across diagonals ($$\phi_1$$ = const, $$\phi_2$$ = const.) of rhombus are related as:

$$d \phi_1^2 + d \phi_2^2 = - K ds^2 \tag{0}$$

which is a hyperbolic metric with rotation $$\phi_1,\phi_2$$ as parameters.

EDIT1:

The following fully reinforces original view of hyperbolic geometry where the parameters $$(\phi_1, \phi_2 )$$ take the place of $$(u,v)$$ in the Euclidean metric up to an invariant constant $$a$$ :

$$ds^2= du^2 + dv^2 \tag{1}$$

Let $$K= -1/a^2$$ , primes with respect to hyperbolic geodesic asymptotic arcs

$$\phi_1^{\prime2} +\phi_2^{\prime2} = \frac{1}{a^2} \tag{2}$$

Taking components of arc along direction inclined at $$\psi$$ to fiber along maximum and minimum curvature directions we have by definition of constant $$K$$:

$$\frac{d \phi_1}{ds \cos \psi} \cdot \frac{d \phi_2}{ds \sin \psi} = \frac{ \phi_1 ^{\prime}}{ \cos \psi} \cdot \frac{ \phi_2 ^{\prime}}{ \sin \psi} =\frac{1}{a^2} \tag{3}$$

Solving (2),(3) we obtain derivatives of rotation w.r.t. arc in each direction as:

$$\phi_1^{\prime} = \frac { \cos \psi} {a}, \;\phi_2^{\prime} = \frac { \sin \psi} {a} \tag{4}$$

as one solution taken out of two interchangeable solutions. Squaring and adding,

$$\boxed{ds^2 = a^2( d \phi_1^2 + d \phi_2^2) } \tag{5}$$

What prompts me to post this is: Recognition of this observed identity between Euclidean and Hyperbolic parameters to hopefully remove vagueness (in my mind at least) while recognizing these rotations as hyperbolic parameters:

Thus the above is the curvilinear hyperbolic geodesic Pythagoras theorem. Hypotenuse is allowed only along hyperbolic geodesics and components only along maximum/minimum normal curvature lines.

I have no access to good literature references/ sources but had held this view within myself that... this was Beltrami's original conceptualization. Request enlightened members to please make corrections and give comments on my view.

$$(u,v) \leftrightarrow a (\phi_1 , \phi_2 ) \tag{6}$$

In this connection I quote from the text book authored by DJ Struik, Lectures on Classical Differential Geometry, Second edition pp 153 left bottom:

The whole geometry of Lobachevski-Bolyai could thus be interpreted on a surface of constant negative curvature , parallel lines becoming geodesics (Emphasis mine). Beltrami proved that the consistency of implied consistency of Lobachevski-Bolyai geometry, since an inconsistency in the latter could be interpreted as an inconsistency in the theory of surfaces of constant negative (Gauss) curvature which itself is based on Euclidean postulates.

Above image is made on Mathematica based on the metric correspondence (6). The discussion is for any surface, not necessarily that of revolution as pictured.

EDIT2:

Derivation:

$$\kappa_{1,2}$$ principal curvatures. Euler's normal curvature relation:

$$\kappa_n =\kappa_1 \cos^2 \psi + \kappa_2 \sin ^2 \psi =0 ;\, \kappa_1 \kappa_2 = -1/a^2 \, \rightarrow \kappa_{1,2}= (-\tan\psi/a, \cot\psi/a) \tag{7}$$

Line segment components along curvature extrema directions :

$$2 \, d \phi_1 = 2 \, ds\, \cos \psi \, \kappa_1,\, 2\,d\phi_2 = 2 \, ds \, \sin \psi \, \kappa_2,\ \tag{8}$$

Combining (7),(8) to eliminate $$\kappa_{1,2}$$ we get (5) or (0).

EDIT3:

Constant negative hyperbolic lines $$K$$ in the large for orthogonal lines should accordingly be given by:

$$\cosh (ds/a) = \cosh d \phi_1\, \cosh d \phi_2\$$

and in the small by

$$1 + (ds/a)^2/2 \approx (1 + d\phi_1^2/2)\,(1 + d\phi_2^2/2)$$

so that we have same:

$$ds^2 = a^2(d\phi_1^2 + d\phi_2^2 ). \tag{10=5}$$

EDIT 4:

Further on I bring out a simple but an equally important isometric identity of rotation product $$( d \phi_1.d \phi_2)$$ and second differential of $$\psi$$ which appears to be implied or hidden.

From (4)

$$d \phi_1 =\frac {ds \cos \psi}{R_1}= \kappa_1 {ds \cos \psi};\quad d \phi_2 =\frac {ds \sin \psi}{R_2}= \kappa_2 {ds \sin\psi};\quad \tag {11 }$$

$$d \phi_1 . d \phi_2 =\kappa_1 \kappa_2 \, ds \cos \psi \,ds \sin \psi = K ds^2 \sin \psi\, \cos \psi \tag{12}$$

By Sine-Gordon theorem for half the scissor angle $$2 \psi$$

$$\psi^{''}= \frac{d^2\psi}{ds^2}=K \sin\psi \cos\psi =\kappa_1 \kappa_2 \sin\psi \cos\psi \tag {13}$$

From (13) and (12) the scalar differential rotation product is:

$$\boxed{ d^2\psi = d \phi_1 . d \phi_2} \tag{14}$$

• Something is wrong with your equation. The metric on the left, $d\phi_1^2+d\phi_2^2$, is obviously flat, but the metric on the right, $-K ds^2$, has constant negative Gauss curvature $-1$. Or do you not mean $ds^2$ to be the induced metric? – Robert Bryant Jul 16 '16 at 9:13
• These rotations are calculated and culminate in correct equations (4). Does taking as induced metric help? – Narasimham Jul 16 '16 at 15:17
• @ Robert Bryant At the end is included a short derivation for the apparently " flat metric".I wish to know if there are more fundamental ways to prove it...however.... – Narasimham Jul 16 '16 at 18:53
• Replacing dimensionless lengths by angles this way is fundamental to non-Euclidean geometries as it appears to me.. generating lengths via small angular arguments and then compensatingly back again with a dimensional invariant , as in: $a\, \sinh\, s/a \approx s .$ Shallow first order approximations result when $s<< a$ working out to distances/ arc lengths in Euclidean geometry. – Narasimham Jul 16 '16 at 21:30
• Please be aware that each edit bumps the question to the front of the active questions list, at the expense of other questions which are vying for attention. Better is to do only a few edits to correct many infelicities within the same edit, rather than many edits to do one or two at a time. – Todd Trimble Jul 17 '16 at 21:41