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--- ebsd2021:tema6 [2021/09/09 16:08] – escola
+++ ebsd2021:tema6 [2021/09/14 09:14] (current) – escola
@@ Line 1: / Line 1: @@
-**Tema 6: Hadamard manifolds**
+**Tema 6: Hadamard manifolds - part 1**
 In this section we study some geometrical properties of Riemannian manifolds $M$ with negative
@@ Line 16: / Line 16: @@
 We have:
   * $f'=\langle J,J'\rangle$, hence $f'(0)=0$.
-  * We have
+  * We have $f''=\langle J',J'\rangle+\langle J,J''\rangle=\|J'\|^2-\langle J,R(\gamma',J)\gamma'\rangle=\|J'\|^2-K(\gamma',J)\|\gamma'\wedge J\|^2\geq 0$, hence $f$ is convex with $f''(0)>0$.
-$$
-f''=\langle J',J'\rangle+\langle J,J''\rangle = \|J'\|^2-\langle J,R(\gamma',J)\gamma'\rangle=
-\|J'\|^2-K(\gamma',J)\|\gamma'\wedge J\|^2\geq 0,
-$$
-hence $f$ is convex with $f''(0)>0$.
 This implies that $f'$ is non-decreasing with $f'(t)>0$ for $t\approx 0$, and so $f'(t)>0$
@@ Line 43: / Line 38: @@
 This is a direct consequence of Lemmas 1 and 2.
-**Theorem.** (Hadamard-Cartan) Let $M$ be a complete Riemannian manifold with nonpositive curvature. For every $p\in M$, the exponential map ${\rm exp}_p:TM_p\to M$ is a covering map.
+**Theorem 1.** (Hadamard-Cartan) Let $M$ be a complete Riemannian manifold with nonpositive curvature. For every $p\in M$, the exponential map ${\rm exp}_p:TM_p\to M$ is a covering map.
 If $M$ is additionally simply connected, then ${\rm exp}_p$ is a diffeomorphism, hence
 $M$ is diffeomorphic to $\mathbb R^n$.
@@ Line 53: / Line 48: @@
 These two properties guarantee that ${\rm exp}_p$ is a covering map. We arrive at the following definition.
-Hadamard manifold: A Riemannian manifold $M$ is called //Hadamard//
+//Hadamard manifold:// A Riemannian manifold $M$ is called //Hadamard//
 if it is complete, simply connected and has nonpositive curvature.
+By the above results, a Hadamard manifold $M$
+has no conjugate points and it is diffeomorphic to $\mathbb R^n$, with
+every ${\rm exp}_p$ being a diffemorphism. In particular, for every $p,q\in M$
+there is a unique geodesic joining them, which we denote by $pq$ and call it
+a //segment//.
+__Comparison triangles__
+The theorem below is a direct consequence of the Rauch comparison theorem.
+**Theorem 2.** Let $M,\widetilde{M}$ be Riemannian manifolds of same dimension, and suppose that
+for all $p\in M,\widetilde{p}\in\widetilde{M}$ and all planes
+$\sigma\in TM_p,\widetilde{\sigma}\in T\widetilde M_{\widetilde p}$ it holds
+$K(\sigma)\leq K(\widetilde{\sigma})$. Fix $p,\widetilde{p}$ s.t. ${\rm exp}_p,{\rm exp}_{\widetilde p}$
+are diffeomorphisms, and fix a linear isometry $i:TM_p\to T\widetilde M_{\widetilde p}$.
+If $\alpha$ is a curve on $M$ and
+$\widetilde{\alpha}=[{\rm exp}_{\widetilde{p}}\circ i\circ{\rm exp}_p^{-1}](\alpha)$
+is the respective curve on $\widetilde{M}$, then $\ell(\widetilde{\alpha})\leq \ell(\alpha)$.
+In other words, an increase of curvature causes a decrease of distances.
+This result has several consequences for Hadamard manifolds, when comparing
+distances with those in the euclidean space. We state these facts in three lemmas.
+Let $M$ be a Hadamard manifold. Every three points $a,b,c\in M$
+define a unique triangle $\Delta=abc$, whose sides are the segments joining
+pairs of points. We denote the angles of $\Delta$ by $\angle a,\angle b,\angle c$.
+The side lengths $A=\ell(bc)$, $B=\ell(ac)$, $C=\ell(ab)$ satisfy the triangle inequality,
+hence there is a unique euclidean plane triangle $\Delta_0$ with same side lengths.
+**Lemma 3.** Let $\Delta=abc$ be a triangle on $M$ with $\angle c=\gamma$.
+(1) If the euclidean plane triangle $a'b'c'$with sides $A,B$ and
+$\angle c'=\gamma$ has third side equal to $C_0$, then $C_0\leq C$.
+(2) Cosine law: $C^2\geq A^2+B^2-2AB\cos\gamma$.
+**Proof.** (1) Let $g=\{g_p\}_{p\in M}$ be the metric on $M$.
+Let $\widetilde M=TM_c$ with the euclidean metric $\widetilde{g}$.
+Since $d({\rm exp}_c)_0$ is an isometry, the identity
+map ${\rm Id}:(TM_c,g_c)\to (TM_c,\widetilde g)$ is a linear isometry.
+Let $p=c,\widetilde p=0$ in Theorem 2. The curves $0a'=({\rm exp}_c)^{-1}(ca)$
+and $0b'=({\rm exp}_c)^{-1}(cb)$ are euclidean segments with angle
+$\gamma$. Let $C_0$ be the third side length of the triangle they form.
+By Theorem 2, the curve $({\rm exp}_c)^{-1}(ab)$ has euclidean length
+$\ell\leq C$. Since it has endpoints $a',b'$, we also have $C_0\leq \ell$.
+Thus $C_0\leq \ell\leq C$.
+(2) By part (1) and the cosine law applied to the euclidean triangle $0a'b'$:
+$$C^2\geq C_0^2=A^2+B^2-2AB\cos\gamma.
+$$
+**Lemma 4.**
+Let $\Delta=abc$ be a triangle on $M$ with angles $\alpha,\beta,\gamma$,
+and let $\alpha',\beta',\gamma'$ be the angles of $\Delta_0$.
+Then $\alpha\leq \alpha'$, $\beta\leq \beta'$ and $\gamma\leq\gamma'$.
+In particular, $\alpha+\beta+\gamma\leq \pi$, with equality iff $\Delta$ is a plane triangle, i.e.
+iff the convex surface bounded by it has zero sectional curvature.
+**Proof.** It is enough to prove that $\gamma\leq\gamma'$.
+By the cosine laws in $\Delta$ and $\Delta_0$, we have:
+$$
+A^2+B^2-2AB\cos\gamma\leq C^2=A^2+B^2-2AB\cos\gamma' \ \Longrightarrow \ \cos\gamma'\leq \cos\gamma
+ \ \Longrightarrow \ \gamma\leq\gamma'.
+$$
+**Corollary 2.** Let $abcd$ be a quadrilateral on $M$, and let $\alpha,\beta,\gamma,\delta$
+be its angles. Then $\alpha+\beta+\gamma+\delta\leq 2\pi$, with equality iff $abcd$ is plane, i.e.
+iff the convex surface bounded by it has zero sectional curvature.
+**Proof.**
+Just divide the quadrilateral into two triangles and apply Lemma 4.
+The last comparison lemma estimates intermediate distances in $\Delta$ and $\Delta_0$.
+**Lemma 5.** Let $\Delta=abc$ be a triangle on $M$, and let $\Delta_0=a'b'c'$ be the respective
+euclidean plane triangle. Given $x\in ab,y\in ac$, let $x'\in a'b',y'\in a'c'$ with
+$\ell(ax)=\ell(a'x')$ and $\ell(ay)=\ell(a'y')$. Then $\ell (xy)\leq \ell(x'y')$.
+**Corollary 3.** Let $\Delta=abc$ be a triangle on $M$. If
+$x,y$ are the midpoints of $ab,ac$ then $\ell(xy)\leq \tfrac{A}{2}$.
 ~~DISCUSSIONS~~