Hyperbolic metric geodesically complete

hbghlyj

\begin{aligned} \Bbb H&=\{(x, y) \mid y>0\} \\ \text{first fundamental form}&=\frac{\mathrm{d} x^2+\mathrm{d} y^2}{y^2} . \end{aligned}在Poincaré半平面模型的度量下的柯西列一定收敛到$\Bbb H$里面的点吗？有可能收敛到$\Bbb H$边界吗，那样就不完备了？

Czhang271828

任何内点到边界的距离是无穷远. 测地线是可以无限延申的.

hbghlyj

Czhang271828 发表于 2024-2-6 05:01
任何内点到边界的距离是无穷远. 测地线是可以无限延申的.

找到了：Hopf–Rinow theorem是关于黎曼流形的测地完备性的一套等价命题。定理如下：

设M是黎曼流形，则下列命题等价：

• $ M $的有界闭子集是紧的。
• $ M $是完备度量空间。
• $ M $是测地完备：对$ M $中任意点$ p $，指數映射$ \exp _{p} $可定义在整个切空间$ T_{p}M $.

（math.stackexchange.com/questions/1578321）不能對$[0,1]$使用該定理，因為Hopf-Rinow applies to connected boundaryless Riemannian manifolds. The closed interval $[0,1]$ can only be given its usual topology as a manifold with boundary.

That is: the closed interval is not geodesically complete, despite being Cauchy complete.

hbghlyj

Hyperbolic 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 $\ell(\gamma)\ge \int_0^1\frac{|\dot y|}{y} dt=-\int_{y_0}^0 \frac{dy}{y}=\log y|_{0}^{y_0}=\infty$.

Czhang271828

hbghlyj 发表于 2024-2-6 16:59
找到了：Hopf–Rinow theorem是关于黎曼流形的测地完备性的一套等价命题。定理如下：

设M是黎曼流形，则 ...

用得比较多的等价命题: 存在 $p_0$ 使得 $\exp_{p_0}(v)$ 对任意切向量 $v$ 均有定义.

hbghlyj

Czhang271828 发表于 2024-2-6 05:01
任何内点到边界的距离是无穷远. 测地线是可以无限延申的.

讲解证明的视频：Hopf–Rinow theorem

其它视频：The Cartan-Hadamard theorem