I am reading the John Lee's Introduction to Riemannian manifolds, proof of proposition 11.3. and stuck at some statement. I think that it is first time to see this kind of pattern (?) ..
Proposition 11.3. Suppose $(M,g)$ is a Riemannian manifold, $U\subseteq M$ is 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 $U – \{p\}$ :
$$ \mathcal{H}_r = \frac{s_c'(r)}{s_c(r)}\pi_r, \tag{11.6}$$
where $s_c$ is defined by $(10.8)$, and 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 $\partial|_q$).
Proof. $ \Rightarrow)$ (Omitted..) We accept this direction.
$ \Leftarrow)$ Conversely, suppose $\mathcal{H}_r$ is given by $(11.6)$. Let $\gamma$ be a radial geodesic starting at $p$, and let $J$ be a normal Jacobi field along $\gamma$ that vanishes at $t=0$. By Proposition 11.2, $D_tJ(t) = \mathcal{H}_rJ(t)=s_c'(t)J(t)/s_c(t)$. A straightforward computation then shows that $s_c(t)^{-1}J(t)$ is parallel along $\gamma$. Thus we can write every such Jacobi field in the form $J(t)=ks_c(t)E(t)$ for some constant $k$ and some parallel unit normal vector field $E$ along $\gamma$. Proceeding exactly as in the proof of Theorem $10.14$, we conclude that $g$ is given by formula $(10.17)$ in these coordinates, and therefore has constant sectional curvature $c$. QED.
If needed, I will upload more information about involving notations, theorems etc.. more in detail. I don't understand the bold statement. How can we find such constant $k$ and parallel unit normal $E$ along $\gamma$? It seems like it could work, but I'm not sure how can we find that precisely
Can anyone help?
Best Answer
The vector field $V(t) = s_c(t)^{-1}J(t)$ satisfies $D_t V(t) = 0$ (this is the definition of being parallel along the underlying geodesic). In particular, it has constant norm: indeed, $\frac{d}{dt}\|V(t)\|^2 = 2 \langle D_tV(t),V(t)\rangle = 0$. Let $k = \|V(t)\|\geqslant 0$ be this constant, and assume that $k >0$. Define $E(t) = \frac{1}{k}V(t)$, which is then a unit vector field, parallel along the underlying geodesic. Then by definition, $V(t) = kE(t)$. This is still true if $k = 0$. Hence, $J(t) = ks_c(t)E(t)$ for some constant $k \geqslant 0$ and $E(t)$ unit parallel vector field along the underlying geodesic.