2
$\begingroup$

I encounter to the question in reading the following Excise:

Let $(M,g)$ be a $m$-dimensional Riemannian manifold, and $(r,\theta^1,\theta^2,\dotsc,\theta^{m-1})$ be the (geodesic) polar coordinate. Prove that, if $M$ is a space form, i.e., with the constant (sectional) curvature $K$, then $g$ has the following expression (by Gauss's Lemma): $$ \newcommand{\rd}{\mathrm d} \rd s^2=(\rd r)^2+(f(r))^2h_{ij}(\theta)\rd\theta^i\rd\theta^j, $$ where the $m-1$-dimensional metric $(\rd\sigma)^2=h_{ij}(\theta)\rd\theta^i\rd\theta^j$ has constant sectional curvature $1$, and $$ f(r)=\begin{cases} \sin(\sqrt{Kr^2})/\sqrt{K},&\text{if }K>0; \\ r,&\text{if }K=0; \\ \sinh(\sqrt{-Kr^2})/\sqrt{-K},&\text{if }K<0. \end{cases} $$

I already know that the Jacobi-field $J$ ($J(0)=0$, $\lvert\dot J(0)\rvert=1$ and $J$ is perpendicular to the base geodesic curve $\gamma(t)$) for a manifold with constant sectional curvature $K$ is $$ J(t)=f(t)W(t), $$ where $f(t)$ is defined as above and $W(t)$ is a unit parallel vector filed along $\gamma$ with $W(t)\perp\dot\gamma(t)=T(t)$.

So my question is that: can we give a proof that based on the Jacobi-field? If not, what's the relation between the two problem?

In fact, I don't know how the solve the first problem, and try to use Jacobi-field to give a proof without no results.

$\endgroup$
8
  • 2
    $\begingroup$ This question is firstly post on StackExchange:math.stackexchange.com/questions/111504/…, after I read the site meta, I think is more prefer to post it here. $\endgroup$
    – van abel
    Commented Feb 21, 2012 at 0:41
  • 6
    $\begingroup$ Actually, this is foundational material, from a first course on Riemannian manifolds. MSE is better. $\endgroup$
    – Will Jagy
    Commented Feb 21, 2012 at 0:50
  • 2
    $\begingroup$ Will, I am not so sure that the question will get a good answer on MSE $\endgroup$
    – Yemon Choi
    Commented Feb 21, 2012 at 1:23
  • 3
    $\begingroup$ I suggest consulting more than one textbook on Riemannian geometry. There are different ways to derive these formulas, and you might find one approach easier to understand than others. You can do this using moving frames and differential forms; you can also just do everything using local co-ordinates. Just fix a choice of local co-ordinates on the unit sphere and use them along with the radial co-ordinate $r$. $\endgroup$
    – Deane Yang
    Commented Feb 21, 2012 at 11:41
  • 4
    $\begingroup$ Actually, using Jacobi fields to do this exercise is not such a bad idea. It's actually one of my preferred approaches. But it seems to me that when you are learning this stuff for the first time, you shouldn't worry about doing it the "right" way or the "best" way or the "least messiest" way first. Just do it one way or another, even if it involves pages and pages of awful calculations with Christoffel symbols. Then slowly learn how to do it more elegantly. My advice is to keep struggling with it! $\endgroup$
    – Deane Yang
    Commented Feb 21, 2012 at 21:53

1 Answer 1

3
$\begingroup$

By Gauss lemma, $$ \newcommand{\rd}{\mathrm d} \rd s^2=(\rd r)^2+\tilde h_{ij}(r)(\theta)\cdot \rd\theta^i\cdot\rd\theta^j, $$ where $\tilde h_{ij}(r)$ is a Riemannian metric on the sphere that depends on the radius $r$.

Note that $\tilde h_{ij}(r)$ is rotationally symmetric. Therefore $h_{ij}(r)=f(r)^2\cdot h_{ij}(\theta)\cdot\rd\theta^i\cdot\rd\theta^j$ for some function $f$. It remains to check that $f''+K\cdot f=0$.

Note that $f(r)=|J(r)|$, where $J=\tfrac{\partial}{\partial \theta_1}$ is a field on the geodesic $t\mapsto (t,0,\dots,0)$. Note that $J$ is a Jacobi field --- it is a field of variation of radial geodesics. It remains to write Jacobi equation and get $f''+K\cdot f=0$ (you know it already).

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .