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--- ebsd2021:tema7 [2021/06/28 16:08] – external edit 127.0.0.1
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-Tema 7
+**Tema 7: Hadamard manifolds - part 2**
+__Convexity properties__
+We continue exploring the convexity properties on Hadamard manifolds.
+//Convex function:// A function $f:M\to\mathbb R$ is //convex// if for every geodesic $\gamma$ the composition $f\circ\gamma:\mathbb R\to\mathbb R$ is convex.
+//Convex set:// A set $C\subset M$ is //convex// if for every $x,y\in C$ the segment $xy\subset C$.
+Here are examples of convex sets: if $f:M\to\mathbb R$ is a convex function then $f^{-1}(-\infty,\alpha]$ is a convex set for all $\alpha\in\mathbb R$. Indeed, let $f(x)\leq \alpha$ and $f(y)\leq\alpha$, and let $\gamma:[0,t]\to M$ be the segment $xy$. Since $f\circ\gamma:[0,t]\to \mathbb R$ is convex, it is bounded by $\alpha$ and so $xy\subset f^{-1}(-\infty,\alpha]$. The main result of this subsection is the following.
+**Proposition 1.** If $\gamma_1,\gamma_2$ are geodesics on $M$, then $t\mapsto d(\gamma_1(t),\gamma_2(t))$
+is convex.
+**Proof.** Fix $t_1,t_2$, let $t=\tfrac{t_1+t_2}{2}$, and let $x$ be the midpoint of the segment
+$\gamma_1(t_1)\gamma_2(t_2)$.
+By the triangle inequality,
+$$
+d(\gamma_1(t),\gamma_2(t))\leq d(\gamma_1(t),x)+d(x,\gamma_2(t)).
+$$
+By Corollary 2 of the previous post, $d(\gamma_1(t),x)\leq \tfrac{d(\gamma_1(t_2),\gamma_2(t_2))}{2}$
+and $d(x,\gamma_2(t))\leq \tfrac{d(\gamma_1(t_1),\gamma_2(t_1))}{2}$, thus
+$$
+d(\gamma_1(t),\gamma_2(t))\leq \frac{d(\gamma_1(t_1),\gamma_2(t_1))+d(\gamma_1(t_2),\gamma_2(t_2))}{2}\cdot
+$$
+**Corollary 1.** If $C$ is a convex set, then $x\mapsto d(x,C)$ is a convex function.
+**Proof.** For simplicity, assume that $C$ is closed (the general case follows by approximating $d(x,C)$
+by $d(x,y)$ for $y\in C$). Fix a geodesic $\gamma$, and $x=\gamma(t_1)$, $y=\gamma(t_2)$.
+Let $\overline{x},\overline{y}$ s.t. $d(x,C)=d(x,\overline{x})$ and $d(y,C)=d(y,\overline{y})$.
+If $z,\overline{z}$ are the midpoints of $xy,\overline{x}\overline{y}$ then by the previous proposition:
+$$
+d(z,C)\leq d(z,\overline{z})\leq \frac{d(x,\overline x)+d(y,\overline y)}{2}=\frac{d(x,C)+d(y,C)}{2}\cdot
+$$
+__Bruhat-Ti ts characterization of nonpositive curvature__
+The next result, due to Bruhat and Ti ts, is an alternate characterization of nonpositive curvature.
+We state it for Riemannian manifolds, but it also works for complete metric spaces.
+**Theorem 1.** (Bruhat-Ti ts) Let $M$ be a complete Riemannian manifold. Then $M$ is Hadamard iff for
+any points $x,y\in M$ there is a point $m\in M$ (the midpoint of $xy$) s.t.
+$$
+d(z,m)^2\leq \tfrac{1}{2}\left[d(z,x)^2+d(z,y)^2\right]-\tfrac{1}{4}d(x,y)^2, \ \ \forall z\in M.
+$$
+Intuitively, the above inequality states that the triangle $xyz$ is thinner than the respective
+euclidean triangle. Indeed, in zero curvature we obtain equality, by Stewart's theorem.
+For a proof, see Ballmann's lecture notes.
+The above theorem gives another proof of Corollary 3 of the last post.
+For that, let $D=\ell(xy)$ and $E=\ell(by)$. By the Bruhat-Ti ts theorem applied to
+the triangle $abc$ with midpoint $y$ and to the triangle $aby$ with midpoint $x$, we have:
+\begin{align*}
+&E^2\leq \tfrac{1}{2}\left[A^2+C^2\right]-\tfrac{1}{4}B^2\\
+&\\
+&D^2\leq \tfrac{1}{2}\left[\left(\tfrac{B}{2}\right)^2+E^2\right]-\tfrac{1}{4}C^2.
+\end{align*}
+Substituting the first inequality into the second, we get that
+$$
+D^2\leq \tfrac{B^2}{8}+\tfrac{A^2}{4}+\tfrac{C^2}{4}-\tfrac{B^2}{8}-\tfrac{C^2}{4}=\tfrac{A^2}{4}\cdot
+$$
+__Flat strip theorem__
+A consequence of the previous results is the so-called flat strip theorem.
+**Theorem 2.** (Flat strip theorem) Let $M$ be a Hadamard manifold, and let $\gamma_1,\gamma_2$ be two geodesics. If the function $t\mapsto d(\gamma_1(t),\gamma_2(t))$ is bounded, then $\gamma_1,\gamma_2$
+bound a flat strip, i.e. a convex region isometric to the convex hull of two parallel lines in the euclidean
+plane.
+**Proof.** By Proposition 1, $d(\gamma_1(t),\gamma_2(t))$ is constant. For $t_1<t_2$, consider the quadrilateral $Q$ with vertices $\gamma_1(t_1),\gamma_2(t_1),\gamma_2(t_2),\gamma_1(t_2)$, and let $\alpha,\beta,\delta,\theta$ be its angles.
+We claim that $\alpha+\beta=\pi$.
+For that, consider the triangle $\gamma_1(t_1)\gamma_2(t_1)\gamma_1(t)$ for $t>t_1$, with angles
+$\alpha,\beta_t,\theta_t$. We have $\beta=\lim\limits_{t\to\infty}\beta_t$.
+By the last post, $\alpha+\beta_t\leq \pi$, and so taking $t\to\infty$ we conclude that
+$\alpha+\beta\leq \pi$. If we apply the same calculations for $t\to-\infty$, we get that
+$(\pi-\alpha)+(\pi-\beta)\leq \pi$, i.e. $\alpha+\beta\geq\pi$, therefore
+$\alpha+\beta=\pi$. Similarly, $\delta+\theta=\pi$, and so $\alpha+\beta+\delta+\theta=2\pi$.
+By Corollary 2 of the last post, $Q$ is flat. Since this holds for all $t_1<t_2$, the proof is complete.
 ~~DISCUSSIONS~~