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--- ebsd2021:tema8 [2021/09/13 15:59] – created escola
+++ ebsd2021:tema8 [2021/09/14 09:16] (current) – escola
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-Tema 8
+**Tema 8: Boundary at infinity of Hadamard manifolds**
+In this topic, we provide some (equivalent) ways to compactify a Hadamard manifold $M$.
+**Asymptotic rays**
+A ray is a geodesic $\gamma:[0,\infty)\to M$.
+//Asymptotic rays:// Two rays $\gamma_1,\gamma_2:[0,\infty)\to M$ are //asymptotic// if the function $t\in [0,\infty)\mapsto d(\gamma_1(t),\gamma_2(t))$ is bounded. In this case, we write $\gamma_1\sim\gamma_2$.
+It is easy to check that $\sim$ is an equivalence relation.
+//Boundary at infinity $M(\infty)$:// The //boundary at infinit//y of $M$ is the set of equivalence relations of $\sim$,
+$$
+M(\infty)=\{[\gamma]:\gamma\text{ is a ray}\}.
+$$
+Every $p\in M$ defines a map $f_p:SM_p\to M(\infty)$ by $f_p(v)=[\gamma_v]$.
+**Lemma 1.** The map $f_p$ is a bijection.
+**Proof.** We start proving injectivity. If $f_p(v)=f_p(w)$ then $\gamma_v\sim\gamma_w$ and so
+$d(t)=d(\gamma_v(t),\gamma_w(t))$ is uniformly bounded for $t\geq 0$. By convexity, $d(t)={\rm const}$, and so $d(t)=d(0)=0$. This implies that $v=w$.
+Now we prove that $f_p$ is surjective. Fix a ray $\gamma$. For each $n$,
+let $\gamma_{v_n}=$ ray starting at $p$ and passing through $\gamma(n)$, say
+$\gamma_{v_n}(t_n)=\gamma(n)$. Passing to a subsequence,
+we can assume that $v_n\to v$. We claim that $\gamma_v\sim \gamma$,
+with $d(\gamma_v(t),\gamma(t))\leq d(p,\gamma(0))$ for all $t\geq 0$.
+To see that, start observing that
+$d(\gamma_v(t),\gamma(t))=\lim\limits_{n\to\infty}d(\gamma_{v_n}(t),\gamma(t))$.
+By convexity, the function $d(\gamma_{v_n}(t),\gamma(t))$ is non-increasing in
+$[0,t_n]$. \footnote{The function $f_n(t)=d(\gamma_{v_n}(t),\gamma(t))$ satisfies $f_n''(t)\geq 0$ and
+$f_n'(t_n)=0$. If $f_n'(t_*)>0$ for some $0\leq t_*\leq t_n$, then $f_n'(t)>0$ for all $t\geq t_*$,
+a contradiction.}
+Hence, for $t_n>t$ we have $d(\gamma_{v_n}(t),\gamma(t))\leq d(p,\gamma(0))$.
+**Lemma 2.**
+For every $p,q\in M$, the composition $f_q^{-1}\circ f_p:SM_p\to SM_q$
+is an homeomorphism.
+//Sphere topology on $M(\infty)$:// The //sphere topology// on $M(\infty)$ is the topology that makes any $f_p$ an homeomorphism.
+By Lemma 2, the definition does not depend on $p$. In particular, $M(\infty)$ is homeomorphic to a sphere of $\mathbb R^n$. There is another way to understand this topology, that actually characterizes the topology of the union $\overline{M}=M\cup M(\infty)$. For $p\in M$, let $B_1(p)\subset TM_p$ be the closed unit ball centered at the origin. Define the map $\varphi=\varphi_p:B_1(p)\to \overline{M}$ by
+$$
+\varphi(v)=
+\left\{
+\begin{array}{ll}
+{\rm exp}_p\left(\tfrac{\|v\|}{1-\|v\|}v\right)&,\text{ if }\|v\|<1,\\
+&\\
+f_p(v)&,\text{ if }\|v\|=1.
+\end{array}
+\right.
+$$
+**Theorem 1.** (Eberlein-O'Neil) The map $\varphi$ is an homeomorphism.
+Hence $\overline{M}$ is homeomorphic to a closed ball of $\mathbb R^n$. The topology of $\overline{M}$ is called the //cone topology//.
 ~~DISCUSSIONS~~