ebsd2021:tema6 []

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 $\ell(xy)\leq \tfrac{A}{2}$ .

Discussion