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Let $\mathbb{H} = \{(u,v) \in \mathbb{R}^2| v>0\}$ be the upper half hyperbolic plane with metric $\mu = \frac{du +dv}{v^2}$, and $\sigma : \mathbb{H} \rightarrow S \subset \mathbb{R}^3$ a parametrization of a regular surface $S$ which inherits the metric $\mu$, and suppose $f(u,v) = 0$ (with $e,f,g$ being the standard notation for the coefficients of the second fundamental form). Find $e \cdot g $ as function of coordinates.

My attempt at the solution was the following: since $S$ inherits the metric $\mu$, the first fundamental form is $F=0, E=G = 1/v^2 =\sqrt{EG - F^2}$. Hence $$eg = \frac{1}{v^4} \langle\sigma_u \wedge \sigma_v,\sigma_{uu}\rangle \langle\sigma_u \wedge \sigma_v,\sigma_{vv}\rangle$$ And I am not sure how to continue from here. Am I missing something trivial about orthogonality between derivatives of vectors? Also I think I have somehow to use the fact that $\sigma_{uv}=0$.

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    $\begingroup$ First, you should write $\mu = \frac{du^2+dv^2}{v^2}$. Second, here's a major hint. The Theorema Egregium tells you that the Gaussian curvature of your surface must be the same as the curvature of $\mathbb H$. $\endgroup$ – Ted Shifrin Sep 8 '18 at 17:49
  • $\begingroup$ Thank you for the hint, I didn't think about that. Now the solution is crystal clear. $\endgroup$ – Siupa Sep 9 '18 at 0:41
  • $\begingroup$ That just means I should have given less of a hint. Drat. $\endgroup$ – Ted Shifrin Sep 9 '18 at 0:47
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HINT: Consider the Theorema Egregium.

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