Tchebyshef net and Gaussian Curvature $K$

Question

If the coordinate curves form a Tchebyshef net ( Here provides a definition) then $E=G=1$ and $F=\cos(\theta)$. Show that in this case $$K = -\frac{\theta_{uv}}{\sin \theta}$$

When, I calculate Christoffel symbols, I get: $$\Gamma_{11}^2 = - \frac{\theta_u}{\sin \theta} \hspace{10 pt} \Gamma_{11}^1=\tan\theta \theta_u $$ $$\Gamma_{22}^2 = - \frac{\theta_v}{\sin \theta} \hspace{10 pt} \Gamma_{22}^1=\tan\theta \theta_v $$ $$\Gamma_{12}^1 = \Gamma_{12}^2 = 0$$ Using one of the equatións of compatibility : $$ -EK =(\Gamma_{12}^2)_u-(\Gamma_{11}^2)_v+\Gamma_{12}^1\Gamma_{11}^2+\Gamma_{12}^2\Gamma_{12}^2-\Gamma_{11}^2\Gamma_{22}^2-\Gamma_{11}^1\Gamma_{12}^2 $$ From the relations above we get: $$ -K =-(\Gamma_{11}^2)_v+-\Gamma_{11}^2\Gamma_{22}^2 $$ Hence : $$ -K =-(\Gamma_{11}^2)_v+-\Gamma_{11}^2\Gamma_{22}^2 $$ $$ -K =-(-\frac{\theta_{uv}+\theta_u\theta_v\cos \theta}{sin\theta^2})+-\frac{\theta_u \theta_v}{\sin\theta^2} $$ But, this is not the result what we want. My question is. What is wrong with my calculations? Any help is appreciate.

References : [Manfredo Do Carmo] Pages 232,234,237

Answer 1

Christoffel symbols:

$$\Gamma_{11}^2=-\frac{\theta_u}{\sin\theta} \hspace{10 pt} \Gamma_{11}^1=\frac{\theta_u}{\tan\theta} $$

$$\Gamma_{22}^2=\frac{\theta_v}{\tan\theta} \hspace{10 pt} \Gamma_{22}^1=-\frac{\theta_v}{\sin\theta}$$

$$\Gamma_{12}^1 = \Gamma_{12}^2 = 0$$ Using one of the equatións of compatibility :

$$ -EK =(\Gamma_{12}^2)_u-(\Gamma_{11}^2)_v+\Gamma_{12}^1\Gamma_{11}^2+\Gamma_{12}^2\Gamma_{12}^2-\Gamma_{11}^2\Gamma_{22}^2-\Gamma_{11}^1\Gamma_{12}^2 $$

Then $$-K =-(\Gamma_{11}^2)_v-\Gamma_{11}^2\Gamma_{22}^2 $$ Imply

$$-K =-\left(-\frac{\theta_u}{\sin\theta}\right)_v-\left(-\frac{\theta_u}{\sin\theta} \right)\left(\frac{\theta_v}{\tan\theta}\right) $$

$$-K =\left(\frac{\theta_{uv}\sin\theta-\theta_u\theta_v\cos\theta}{\sin^2\theta}\right)+\left(\frac{\theta_u\theta_v\cos\theta}{\sin^2\theta}\right) $$

$$K = -\frac{\theta_{uv}}{\sin \theta}$$