This is problem7, chapter 5 of Do Carmo's Riemannian geometry.
Let $M$ be a Riemannian two manifold, $p\in M$ , $exp_p$ is a diffeomorphism on a neighbourhood of origin $V\in T_pM$. Let $S_r(0)\subset V$ be a circle of radius $r$ centered at the origin, and let $L_r$ be the length of the curve $exp_p(S_r)$ in $M$. Prove that the sectional curvature at $p\in M$ is
$$K(p)=\lim_{r\to 0}\frac{3}{\pi}\frac {2\pi r-L_r}{r^3}$$
from a previous exercise we already have $$\lim_{r\to 0}\frac{\sqrt{g_{22}}_{rr}}{\sqrt{g_{22}}}=-K(p)$$
where $g_{22}=\big|{\frac{\partial f}{\partial \theta}}\big|^2, f(r,\theta)=exp_p rv(\theta), v $ the unit circle.
I try to use $\frac{\partial f}{\partial \theta}=(dexp_p)_{rv}rv'(\theta)$ , but this is not the case where Gauss lemma is applicable. Also, I cannot figure out where $L_r$ come up. Any suggetion will be apprecaited .
Best Answer
(a) As we are working in a normal ball about $p$, $\exp_p$ furnishes a diffeomorphism from a disc about the origin in $T_pM$ to $B_\delta(p)$. Moreover, the coordinates $(\rho,\theta)$ are given by the composition of the inverse of $\exp_p$ with $\psi_x^2+\psi_y^2$ and $\tan^{-1}\left( \psi_y/\psi_x \right)$, where $\psi_x$ and $\psi_y$ are the coordinates on $T_pM$. These are diffeomorphisms away from the ray $\exp_p(\rho v(0))$ for $0<\rho<\delta$.
(b) Let us compute the components $g_{ij}$ of the metric in polar coordinates $(\rho,\theta)$. First, we have, by Gauss's lemma, \begin{align*} g_{11}&=g\left( \frac{\partial}{\partial \rho},\frac{\partial}{\partial\rho}\right) = g\left( \frac{\partial f}{\partial \rho},\frac{\partial f}{\partial\rho} \right) =\left|\frac{\partial f}{\partial\rho}\right|^2 \end{align*} since, by definition, $\partial f/\partial\rho=f_*(\partial/\partial\rho)$. Moreover, \begin{align*} \left|(f_*)_{\rho v(\theta)}\left( \frac{\partial}{\partial\rho} \right)\right|^2&=\left|\frac{d}{dt}\bigg|_{t=0}\left( \exp_p\left( (\rho +t)v(\theta) \right) \right)\right|^2 =\left|\frac{d}{dt}\bigg|_{t=0}\gamma_p(\rho+t,v(\theta))\right|^2 \\ & =\left|\frac{d}{dt}\bigg|_{t=0}\gamma_p(t,v(\theta))\right|^2 =\left|v(\theta)\right|^2=1, \end{align*} where we have used the fact that $\gamma$ is a geodesic, and hence the magnitude of its tangent vectors is constant. Next, again by Gauss's lemma, we find that \begin{align*} g_{22} &= g\left( \frac{\partial}{\partial\theta}, \frac{\partial}{\partial\theta} \right) =g\left( \frac{\partial f}{\partial\theta},\frac{\partial f}{\partial\theta} \right) =\left|\frac{\partial f}{\partial\theta}\right|^2 \end{align*} Finally, it follows from the proof of Gauss's lemma, c.f. page 70 of do Carmo, that $$g_{12}=g_{21}=g\left( \frac{\partial}{\partial\rho},\frac{\partial}{\partial\theta} \right)=0.$$(c) Along the geodesic $f(\rho,0)$, it is clear that $\partial f/\partial\theta$ is a Jacobi field and so we obtain via a Taylor expansion, see page 115 of do Carmo, that $$\sqrt{g_{22}}=\left|\frac{\partial f}{\partial \theta}\right|=\rho-\frac{1}{6}K(p,\sigma)\rho^3+\tilde R(\rho),$$ where $\lim_{\rho\to 0}\tilde R(\rho)/\rho^3=0$. If we differentiate twice with respect to $\rho$, we obtain $$(\sqrt{g_{22}})_{\rho\rho}=-K(p,\sigma)\rho+R(\rho),$$ where $\lim_{\rho\to 0}R(\rho)/\rho=0$, as desired.
(d) Finally, note that \begin{align*} \lim_{\rho\to0}\frac{(\sqrt{g_{22}})_{\rho\rho}}{\sqrt{g_{22}}} = \lim_{\rho\to 0}\frac{-K(p,\sigma)\rho+R(\rho)}{\rho-K(p,\sigma)\rho^3/6+\tilde R(\rho)} =-K(p,\sigma). \end{align*}
Recall that the arc length of the curve $\text{exp}_p(S_r)$ is given by the integral$$L_r = \int_{-\pi}^\pi \sqrt{g\left({\partial\over{\partial\theta}}, {\partial\over{\partial\theta}}\right)}\,d\theta = \int_{-\pi}^\pi \sqrt{g_{22}}\,d\theta.$$The previous exercise showed that along $\text{exp}_p(rv(0))$, we have the identity$$\sqrt{g_{22}} = r - {1\over6}K(p, \sigma)r^3 + \tilde{R}(r).$$This in fact holds for all $\theta$, as $\sqrt{g_{22}}$ is independent of $\theta$. To see this, we can simply change coordinates to $(r, \theta - \theta_0)$, and the computation above yields the same expression for $\sqrt{g_{22}}(r, 0)$, which is now actually along $\text{exp}_p(rv(\theta_0))$. Hence, we obtain$$L_r = \int_{-\pi}^\pi \left(r - {1\over6}K(p, \sigma)r^3 + \tilde{R}(r)\right)d\theta = 2\pi r - {\pi\over3}K(p, \sigma)r^3 + 2\pi\tilde{R}(r).$$Dividing both sides by $r^3$ and taking the limit as $r \to 0$, we find that$$K(p, \sigma) = \lim_{r \to 0} {3\over\pi} {{2\pi r - L_r}\over{r^3}}.$$