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--- ebsd2021:tema7 [2021/09/09 16:32] – escola
+++ ebsd2021:tema7 [2021/09/09 16:49] (current) – escola
@@ Line 36: / Line 36: @@
 $$
 d(z,C)\leq d(z,\overline{z})\leq \frac{d(x,\overline x)+d(y,\overline y)}{2}=\frac{d(x,C)+d(y,C)}{2}\cdot
+$$
+__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
+$$
+D^2\leq \tfrac{B^2}{8}+\tfrac{A^2}{4}+\tfrac{C^2}{4}-\tfrac{B^2}{8}-\tfrac{C^2}{4}=\tfrac{A^2}{4}\cdot
 $$
@@ Line 48: / Line 81: @@
 plane.
-**Proof.** By Proposition \ref{prop.distance.convex}, $d(\gamma_1(t),\gamma_2(t))$ is constant.
+**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.
-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, as in the figure below.
 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
+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 Lemma \ref{lem.comparison.2}, $\alpha+\beta_t\leq \pi$, and so taking $t\to\infty$ we conclude that
+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 \ref{cor.comparison.2}, $Q$ is flat. Since this holds for all $t_1<t_2$, the proof is complete.
+By Corollary 2 of the last post, $Q$ is flat. Since this holds for all $t_1<t_2$, the proof is complete.