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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: $$ d(z,C)\leq d(z,\overline{z})\leq \frac{d(x,\overline x)+d(y,\overline y)}{2}=\frac{d(x,C)+d(y,C)}{2}\cdot $$

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 \ref{prop.distance.convex}, $d(\gamma_1(t),\gamma_2(t))$ 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, 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 $\alpha,\beta_t,\theta_t$. 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 $\alpha+\beta\leq \pi$. 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.