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In Example 3 of section 5.10 of Do Carmo's book "Differential Geometry of Curves and Surfaces", we have that the curvature of the (Lobachevsky) hyperbolic plane is

$$K=\frac{1}{2\sqrt{EG}}\bigg\{\left(\frac{E_v}{\sqrt{EG}}\right)_v+\left(\frac{G_u}{\sqrt{EG}}\right)_u\bigg\}$$

which is shown to be equal to -1, since $E=G=1/v^2$ and $F=0$. So, we have $\frac{1}{2\sqrt{EG}}=\frac{1}{2E}=\frac{v^2}{2}$, $G_u=0$ and $E_v=-\frac{2}{v^3}$.

The crucial result to prove that when $F=0$, the parametrization is orthogonal, so we can compute the Gaussian curvature as $$K=\frac{1}{2\sqrt{EG}}\bigg\{\frac{\partial}{\partial u}\frac{G_v}{\sqrt{EG}}+\frac{\partial}{\partial v}\frac{E_v}{\sqrt{EG}}\bigg\}.$$

Could someone please tell me why that is the case?

FunnyBuzer
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  • A detailed answer to the question is done in the post: https://math.stackexchange.com/questions/748974/gaussian-curvature-k-of-of-orthogonal-parametrization-x?rq=1 – FunnyBuzer Mar 02 '20 at 16:52

1 Answers1

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You just have to slog through the computations using the Christoffel symbols (which simplify somewhat when $F=0$) and the Gauss equation. This is a standard (not particularly fun) exercise. See section 4-3 of DoCarmo.

(If you know about differential forms and the method of moving frames, then the computation becomes quite simple.)

Ted Shifrin
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  • Thanks for your answer. I'm familiar with differential forms but not with the method of moving frames (or at least I need to recall it). May I ask you for an extra hint on this please? From my experience, whenever Christoffel symbols are involved, computations become quickly quite cumbersome. – FunnyBuzer Feb 24 '20 at 23:43
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    You can see how to do the moving frames computations quite succinctly in section 3.3 of my differential geometry text, linked in my profile. – Ted Shifrin Feb 24 '20 at 23:45