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Last edited by hbghlyj 2025-5-20 17:25Hyperbolic metric geodesically complete
Consider the upper half plane model of the hyperbolic space ($\mathbb{H}$ with the riemannian metric $g=\frac{dx^2+dy^2}{y^2}$). It is known that $(\mathbb{H},g)$ is geodesically complete, which means that no geodesic can reach the border $\partial \mathbb{H}$ in a finite time.
Why is that?
Because $\mathbb{H}$ has many symmetries doing it for one geodesic is enough. You can use an appropriate Mobius transformation, $g$, to realize any geodesic $\gamma$ of $\mathbb{H}$ as
$$\gamma(t)=g\cdot ie^{-t}$$Edit: I saw in your question you didn't want to use explicit geodesics, so you could also do something like this (to see why the denominator forces infinite length)
Let $\gamma(t)=x(t)+iy(t)\subset\mathbb{H}$ be a curve with $\gamma(0)=x_0+iy_0\in\mathbb{H}$ and $\gamma(1)\in\partial \mathbb{H}$ (and say $\dot y<0$).
Then $\ell(\gamma)\ge \int_0^1\frac{|\dot y|}{y} dt=-\int_{y_0}^0 \frac{dy}{y}=\log y|_{0}^{y_0}=\infty$. |
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Czhang271828
Posted 2024-2-6 13:01
