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 upto 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. > $$ (u,v) \leftrightarrow a (\phi_1 , \phi_2 ) \tag{6} $$ [![Hyp_Geods_Pseudosphere/metric][1]][1] 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: $$ \boxed{ d^2\psi = d \phi_1 . d \phi_2} \tag{14} $$ [1]: https://i.sstatic.net/e7aZe.png