完备 黎曼流形

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

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.

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

hbghlyj

https://gr.inc/question/hopf-rinow-theorem-let-m-g-be-a-connected-riemannian-manifold-the-fol
Step 1: Existence of Minimizing Geodesics
Assume \(\exp_p\) is defined on all of \(T_p M\). We first show that for any \(q \in M\), there exists a minimizing geodesic from \(p\) to \(q\). Let \(d(p, q) = L\). Choose \(\delta > 0\) such that \(B_\delta(p)\) is a normal ball. The geodesic sphere \(S_\delta(p) = \partial B_\delta(p)\) is compact because it is the image of the compact Euclidean sphere \(\{v \in T_p M : |v| = \delta\}\) under the continuous map \(\exp_p\). The function \(x \mapsto d(q, x)\) is continuous on \(S_\delta(p)\), so it attains a minimum at some \(x_0 = \exp_p(\delta X)\), where \(|X| = 1\). Define \(\gamma(s) = \exp_p(sX)\), which is a geodesic defined for all \(s \in \mathbb{R}\) by assumption (b).
Step 2: The Set \(A\) is Open and Closed
Let \(A = \{s \in [0, L] : d(\gamma(s), q) = L - s\}\).
Non-emptiness: \(0 \in A\) since \(d(p, q) = L\).
Closedness: \(A\) is closed because \(d\) is continuous.
Openness: For \(s_0 \in A\) with \(s_0 < L\), let \(\delta_0 > 0\) be such that \(B_{\delta_0}(\gamma(s_0))\) is a normal ball. Let \(y_0 \in S_{\delta_0}(\gamma(s_0))\) minimize \(d(y, q)\). By the triangle inequality:
\[  L - s_0 = d(\gamma(s_0), q) = \delta_0 + d(y_0, q) \implies d(y_0, q) = L - (s_0 + \delta_0).  \]
The curve \(\alpha\) formed by \(\gamma|_{[0, s_0]}\) and the radial geodesic from \(\gamma(s_0)\) to \(y_0\) has length \(s_0 + \delta_0\). Since \(d(p, y_0) \geq s_0 + \delta_0\) (reverse triangle inequality) and \(L(\alpha) = s_0 + \delta_0\), \(\alpha\) is minimizing. By uniqueness of geodesics in normal balls, \(y_0 = \gamma(s_0 + \delta_0)\). Thus, \(s_0 + \delta_0 \in A\), proving \(A\) is open.
Since \(A\) is non-empty, open, and closed in \([0, L]\), \(A = [0, L]\). Hence, \(\gamma(L) = q\) and \(L(\gamma) = d(p, q)\).
Step 3: Heine-Borel Property
Let \(K \subset M\) be closed and bounded. Then \(K \subset \overline{B_R(p)}\) for some \(R > 0\). For any sequence \(\{x_n\} \subset K\), write \(x_n = \exp_p(v_n)\) with \(v_n \in T_p M\) and \(|v_n| \leq R\). The closed Euclidean ball \(\overline{B_R(0)} \subset T_p M\) is compact. Pass to a convergent subsequence \(v_{n_k} \to v\). By continuity of \(\exp_p\), \(x_{n_k} = \exp_p(v_{n_k}) \to \exp_p(v)\). Since \(K\) is closed, \(\exp_p(v) \in K\). Thus, \(K\) is compact.