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




__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~~