I am reading the John Lee's Introduction to Riemannian manifold ( 2nd Edition ), Theorem 11.7 ( Hessian Comparison ) and stuck at some statement. This is second question.
First I arrange some relavant definitions and theorems.
Definition 1. Given a finite dimensional inner product space $V$ and self-adjoint endomorphism $A,B : V \to V$, the notation $A \le B$ means that $\langle Av ,v \rangle \le \langle Bv ,v \rangle$ for all $v\in V$, or equivalently that $B- A$ is positive defitinte.
Theorem 11.5 ( Riccati Comparison Theorem ). Suppose $(M,g)$ is a Riemannian manifold and $\gamma : [a,b] \to M$ is a unit-speed geodesic segment. Suppose $\eta$, $\tilde{\eta}$ are self-adjoint endomorphism fields along $\gamma_{(a,b]}$ that satisfy the following Riccati equations :
$$ D_t \eta + \eta^2+\sigma=0, D_t\tilde{\eta}+\tilde{\eta}^2+\tilde{\sigma}=0,$$
where $\sigma$ and $\tilde{\sigma}$ are continuous self-adjoint endomorphism fields along $\gamma$ satisfying $$\tilde{\sigma}(t) \ge \sigma(t) \ \operatorname{for all} t\in [a,b] $$
Suppose further that $\lim_{t \searrow a} ( \tilde{\eta}(t) – \eta(t))$ exists and satisfies
$$ \lim_{t \searrow a}(\tilde{\eta}(t)-\eta(t)) \le 0.$$
Then $$\tilde{\eta}(t) \le \eta(t) \ \operatorname{for all} t\in (a,b]. $$
Definiton 2. For each $c\in \mathbb{R}$, let us define a function $s_c : \mathbb{R} \to \mathbb{R}$ by
$$ s_c(t) = \begin{cases}
t, \ \operatorname{if} c=0 ; \\
R \operatorname{sin} \frac{t}{R}, \ \operatorname{if} c= \frac{1}{R^2}>0 ; \\
R \operatorname{sinh} \frac{t}{R} , \ \operatorname{if} c= – \frac{1}{R^2} <0 .
\end{cases} $$
Definition 3. Suppose $(M,g)$ is a Riemannian manifold, $U\subseteq M$ is a normal neighborhood of $p\in M$, and $r$ is the radical distance function on $U$. For each $q\in U-\{p\}$, $\pi_r : T_qM \to T_qM$ is the orthogonal projection onto the tangent space of the level set of $r$ ( equivalently, onto the orthogonal complement of $\partial_r|_q$).
Proposition 11.3. Suppose $(M,g)$ is a Riemannian manifold, $U\subseteq M$ a normal neighborhood of $p\in M$, and $r$ is the radial distance function on $U$. Then $g$ has constant sectional curvature $c$ on $U$ if and only if the following formula holds at all points of $U – \{ p\}$ : $$ \mathcal{H}_r = \frac{s_c'(r)}{s_c(r)}\pi_r$$,
Theorem 11.4 ( The Riccati Equation ). Let $(M,g)$ be a Riemannian manifold; let $U$ be a normal neighborhood of a point $p\in M$ ; let $r : U \to \mathbb{R}$ be the radial distance function ; and let $\gamma : [0,b] \to U$ be a unit-speed radial geodesic. The Hessian operator $\mathcal{H}_r$ satisfies the following equation along $\gamma|_{(0,b]}$, called a Riccati equation :
$$ D_t \mathcal{H}_r + \mathcal{H}_r^{2} + R_{\gamma'}=0,$$
where $\mathcal{H}^2_r$ and $R_{\gamma'}$ are the endomorphism fields along $\gamma$ defined by $\mathcal{H}_r^2(w)=\mathcal{H}_r ( \mathcal{H}_r(w))$ and $R_{\gamma'}(w)=R(w,\gamma')\gamma'$, with $R$ is the curvature endomorphism of $g$.
Now I state Hessian comparison theorem.
Theorem 11.7 ( Hessian Comparison ) Suppose $(M,g)$ is a Riemannian $n$-manifold. $p\in M$. $U$ is a normal neighborhood of $p$, and $r$ is the radial distance function on $U$.
(a) If all sectional curvature of $M$ are bounded above by a constant $c$, then the following inequality holds in $U_0 – \{p\}$ :
$$ \mathcal{H}_r \ge \frac{s_c'(r)}{s_c(r)}\pi_r,$$
where $s_c$ and $\pi_r$ are defined as in Proposition 11.3 (above definitions), and $U_0=U$ if $c\le 0$, while $U_0 = \{q\in U : r(q) <\pi R \}$ if $c= 1/R^2 >0$.
(b) If all sectional curvature of $M$ are bounded below by a constant $c$, then the following inequality holds in all of $U-\{p\}$ : $$ \mathcal{H}_r \le \frac{s'_c(r)}{s_c(r)}\pi_r.$$
Now I upload proof of the hessian comparison. It's little bit long so I upload it as an iamge.
( If there are any questions about each step except the underlined statement, I will upload detailed explanation of my own. )
I don't understand the final (underlined) statement at all. I am struggling with this for hours. How can we prove $U \subseteq U_0 = \{ q\in U : r(q) < \pi R \}$ if $c= 1/R^2 >0$ rigorously? Assume that $U_0$ is a proper subset of $U$. Then there exists $q\in U$ such that $r(q) \ge \pi R$. Then, how can we connect the fact " $s_c'(r)/s_c(r) \to -\infty$ as $r \nearrow \pi R$ " to "$\mathcal{H}_r$ is defined and smooth in all of $U – \{p\}$ and $\mathcal{H}_r \le \mathcal{H}_r^{c}$ on $U_0$" to deduce contradiction?
Can anyone help?
Best Answer
I prefer writing this out differently.
Let $c: [0,\delta) \rightarrow M$ be a unit speed geodesic, $E = (e_1, \dots, e_{n-1},e_n)$ be an orthonormal frame along $c$ such that it is parallel along $c$ and $e_n = c'$. Let $J = (J_1, \dots, J_n)$ be Jacobi fields along $c$ sucht that $J_k(0) = 0$ and $\nabla_{c'(0)}J_k = e_k.$
Then there exists a matrix $M$ along $c$ such that $J = EM$. It is straightfoward to show that the Jacobi equations are equivalent to $$ M'' + MK = 0,$$ where $$K^i_j = g(e_i,R(e_j,e_n)e_n.$$
The second fundamental form of the geodesic sphere with radius $r$ is $A= M^{-1}M'$ and satisfies the Riccati equation $$ A' + A^2 + K = 0. $$ I believe that $A$ is also the Hessian of the distance function from $c(0)$.
You now should be able to get the desired estimate applying either the Sturm comparison theorem to the Jacobi equation or the Riccati comparison theorem to the Riccati equation satisfied by $A$.
I looked for a reference to where I first learned this, but I wasn't able to. I think I first learned this approach from the writings of Jost, Karcher, Heintze, or Eschenburg. The advantage of this approach is that in the end you're just working with plain old matrix-valued functions of 1 variable instead of all the abstract stuff on a Riemannian manifold.