ricci tensor of 2-sphere $S^2$ Hi could someone show me explicitly how to compute the ricci tensor $g_{ij}$?

 A: This might not be the most straightforward way, but it's a nice trick that bypasses a lot of the calculation. For the Ricci tensor we use this formula:
$$R_{ij}=\frac{\partial { \Gamma^l}_{ij}}{\partial x^l}-\frac{\partial { \Gamma^l}_{il}}{\partial x^j}+{ \Gamma^l}_{ij}{ \Gamma^m}_{lm}-{ \Gamma^m}_{il}{ \Gamma^l}_{jm}$$
Where we sum over repeated indices, and instead of lets say ${ \Gamma^1}_{21}$ I will write explicitly ${ \Gamma^\theta}_{\phi\theta}$. 
We could just use a formula and calculate everything directly, but here's a faster approach. It's known that geodesics extremize arclength. That is, given a curve $\gamma:[0,1]\to M$, with its tangent (or velocity) vector $T^a (t_0) =\gamma_*( \frac{\partial}{\partial t}|_{t=t_0})\in T_{\gamma(t_0)}M$ and given the functional 
 $$ S[\gamma ]:=\int_{t=0}^{t=1} \left( g_{ab} T^a T^b \right)^{1/2}\mathrm d t$$
called arc-length, $\gamma$ is a geodesic iff it extremizes $S$. 
Let's call the integrand $\left( g_{ab} T^a T^b \right)^{1/2}=(2L)^{1/2}$. It can be shown explicitly that the resulting equations are the same as if we had the integrand (in coordinates) $$L=\frac{1}{2}g_{\mu\nu}\frac{d x^{\mu}}{d t}\frac{d x^{\nu}}{d t}$$
which isn't strange since one integrand is a monotonic function of the other. 
So if we compare the Euler-Lagrange equation with the geodesic equation, $$\begin{align}\text{Euler-Lagrange: }\: \frac{d}{dt}\frac{\partial L}{\partial \dot x^i}&=\frac{\partial L}{\partial x^i}\\
\text{Geodesic:} \phantom{{}=1}\frac{d^2 x^{i}}{d t^2}&= -{ \Gamma^i}_{jk}\frac{d x^{j}}{d t}\frac{d x^{k}}{d t}\end{align}$$
we see that if we find the E-L equations, we find the Christoffel symbols, and the Ricci tensor. 
In our case 
$$L=\frac{1}{2}\left(\dot \theta^2 + \sin^2\theta\:\dot\phi^2\right)$$
where the dots indicate derivatives wrt the parameter $t$. 
For the $\theta$ coordinate we find: 
$$\ddot\theta=\sin\theta\cos\theta\:\dot\phi^2$$
and we conclude that 
$${\Gamma^{\theta}}_{\phi\phi}=-\sin\theta\cos\theta,\:{\Gamma^{\theta}}_{\theta\phi}={\Gamma^{\theta}}_{\phi\theta}={\Gamma^{\theta}}_{\theta\theta}=0$$
Similarly, the other equation is $\frac{d}{dt}(\sin\theta\:\dot\phi):$
$$\ddot\phi = -\cot\theta\:\dot\theta\dot\phi$$
so that
$${\Gamma^{\phi}}_{\theta\phi}={\Gamma^{\phi}}_{\phi\theta}=\cot\theta,\:{\Gamma^{\phi}}_{\theta\theta}={\Gamma^{\phi}}_{\phi\phi}=0.$$
So we calculate the Ricci tensor components:
$$R_{\phi\phi}=\partial_{\phi}{\Gamma^{\phi}}_{\phi\phi}+\partial_{\theta}{\Gamma^{\theta}}_{\phi\phi}-\partial_{\phi}{\Gamma^{\phi}}_{\phi\phi}-\partial_{\phi}{\Gamma^{\phi}}_{\theta\phi} + 
{\Gamma^{\phi}}_{\phi\phi}{\Gamma^{\phi}}_{\phi\phi}
+
{\Gamma^{\phi}}_{\phi\phi}{\Gamma^{\theta}}_{\phi\theta}
+
{\Gamma^{\theta}}_{\phi\phi}{\Gamma^{\phi}}_{\theta\phi}
+
{\Gamma^{\theta}}_{\phi\phi}{\Gamma^{\theta}}_{\theta\theta}
- 
{\Gamma^{\phi}}_{\phi\phi}{\Gamma^{\phi}}_{\phi\phi}
-
{\Gamma^{\phi}}_{\phi\theta}{\Gamma^{\theta}}_{\phi\phi}
- 
{\Gamma^{\theta}}_{\phi\phi}{\Gamma^{\phi}}_{\phi\theta}
-
{\Gamma^{\theta}}_{\phi\theta}{\Gamma^{\theta}}_{\phi\theta}
= 0 +(\sin^2\theta- \cos^2\theta) - 0 - 0 +0 +0 +(-\sin\theta\cos\theta\cot\theta)+0-0-(-\sin\theta\cos\theta\cot\theta)-(-\sin\theta\cos\theta\cot\theta)-0 = \sin^2\theta $$
$$R_{\phi\theta}=R_{\theta\phi}=\partial_{\phi}{\Gamma^{\phi}}_{\phi\theta}+\partial_{\theta}{\Gamma^{\theta}}_{\phi\theta}-\partial_{\theta}{\Gamma^{\phi}}_{\phi\phi}-\partial_{\theta}{\Gamma^{\phi}}_{\phi\phi} + 
{\Gamma^{\phi}}_{\phi\theta}{\Gamma^{\phi}}_{\phi\phi}
+
{\Gamma^{\phi}}_{\phi\theta}{\Gamma^{\theta}}_{\phi\theta}
+
{\Gamma^{\theta}}_{\phi\theta}{\Gamma^{\phi}}_{\theta\phi}
+
{\Gamma^{\theta}}_{\phi\theta}{\Gamma^{\theta}}_{\theta\theta}
- 
{\Gamma^{\phi}}_{\phi\phi}{\Gamma^{\phi}}_{\theta\phi}
-
{\Gamma^{\phi}}_{\phi\theta}{\Gamma^{\theta}}_{\theta\phi}
- 
{\Gamma^{\theta}}_{\phi\phi}{\Gamma^{\phi}}_{\theta\theta}
-
{\Gamma^{\theta}}_{\phi\theta}{\Gamma^{\theta}}_{\theta\theta} = 0 + 0 - 0 - 0 + 0 + 0 +0 +0 -0 -0 -0 -0 = 0 $$
$$R_{\theta\theta}=\partial_{\phi}{\Gamma^{\phi}}_{\theta\theta}+\partial_{\theta}{\Gamma^{\theta}}_{\theta\theta}-\partial_{\theta}{\Gamma^{\phi}}_{\theta\phi}-\partial_{\theta}{\Gamma^{\theta}}_{\theta\theta} + 
{\Gamma^{\phi}}_{\theta\theta}{\Gamma^{\phi}}_{\phi\phi}
+
{\Gamma^{\phi}}_{\theta\theta}{\Gamma^{\theta}}_{\phi\theta}
+
{\Gamma^{\theta}}_{\theta\theta}{\Gamma^{\phi}}_{\theta\phi}
+
{\Gamma^{\theta}}_{\theta\theta}{\Gamma^{\theta}}_{\theta\theta}
- 
{\Gamma^{\phi}}_{\theta\phi}{\Gamma^{\phi}}_{\theta\phi}
-
{\Gamma^{\phi}}_{\theta\theta}{\Gamma^{\theta}}_{\theta\phi}
- 
{\Gamma^{\theta}}_{\theta\phi}{\Gamma^{\phi}}_{\theta\theta}
-
{\Gamma^{\theta}}_{\theta\theta}{\Gamma^{\theta}}_{\theta\theta}= 0 + 0 - (- \csc^2\theta)- 0+0+0+0+0-\cot^2\theta-0-0-0 = 1$$
To get the scalar curvature, we form 
$$R = g^{ij}R_{ij} = g^{\phi\phi}R_{\phi\phi}+g^{\theta\theta}R_{\theta\theta}= \csc^2{\theta}\sin^2\theta + 1=2.$$
