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**Tema 6: Hadamard manifolds - part 1**

In this section we study some geometrical properties of Riemannian manifolds $M$ with negative
sectional curvature. For simplicity, we will just write "$M$ has nonpositive curvature".
The crucial property under nonpositive curvature is convexity.


__Preliminaries in nonpositive curvature__


We begin proving that nonpositive curvature is stronger than no conjugate points.

**Lemma 1.** If $M$ has nonpositive sectional curvature, then $M$ has no conjugate points. 

**Proof.** Let $J$ be a Jacobi field with $J(0)=0$ and $J'(0)\neq 0$, and consider the differentiable function $f(t)=\tfrac{1}{2}\|J(t)\|^2=\tfrac{1}{2}\langle J(t),J(t)\rangle$.
We have:
  * $f'=\langle J,J'\rangle$, hence $f'(0)=0$.
  * 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 $f''(0)>0$.

This implies that $f'$ is non-decreasing with $f'(t)>0$ for $t\approx 0$, and so $f'(t)>0$
for all $t>0$. Theferore $f$ is increasing, and so $f(t)\neq 0$ for all $t>0$. 


Next, we collect some known facts in differential geometry. For proofs, 
see do Carmo's book //Riemannian Geometry//. The first represents
Jacobi fields that vanish at $t=0$ by the exponential map.

**Lemma 2.** 
Let $\gamma:[0,a]\to M$ be a geodesic with $\gamma(0)=p$. If $J$ is a Jacobi field
along $\gamma$ with $J(0)=0$, then
$$
J(t)=(d{\rm exp}_p)_{t\gamma'(0)}(tJ'(0)),\ \ t\in[0,a].
$$

**Corollary 1.** If $M$ has nonpositive curvature, then ${\rm exp}_p$ is a local diffeomorphism
for every $p\in M$.

This is a direct consequence of Lemmas 1 and 2.

**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 $\mathbb R^n$.

The proof of this theorem consists on the following steps:
  * By Corollary 1, ${\rm exp}_p$ is a local diffeomorphism.
  * ${\rm exp}_p$ lifts paths: to see this, observe that $TM_p$ with the induced metric of $M$ is complete, and so we can lift any path $\alpha:[0,1]\to M$ to $\overline{\alpha}:[0,1]\to TM_p$ s.t. ${\rm exp}_p\circ \overline{\alpha}=\alpha$.

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//
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~~