Ricci tensor for a 3-sphere without Math packets Let's have the metric for a 3-sphere:
$$
dl^{2} = R^{2}\left(d\psi ^{2} + sin^{2}(\psi )(d \theta ^{2} + sin^{2}(\theta ) d \varphi^{2})\right).
$$
I tried to calculate Riemann or Ricci tensor's components, but I got problems with it. 
In the beginning, I got an expressions for Christoffel's symbols:
$$
\Gamma^{i}_{ii} = \frac{1}{2}g^{ii}\partial_{i}g_{ii} = 0,
$$
$$
\Gamma^{i}_{ji} = \frac{1}{2}g^{ii}\partial_{j}g_{ii},
$$
$$
\Gamma^{k}_{ll} = -\frac{1}{2}g^{kk}\partial_{k}g_{ll},
$$
$$
\Gamma^{k}_{lj} = \Gamma^{k}_{lk}\delta^{k}_{j} + \Gamma^{k}_{jk}\delta^{k}_{l} + \Gamma^{k}_{jj}\delta^{j}_{l} = 0.
$$
The Ricci curvature must be
$$
R_{lj}=\frac{2}{R^{2}}g_{lj}.
$$
But when I use definition of Ricci tensor,
$$
R_{lj}^{(3)} = \partial_{k}\Gamma^{k}_{lj} - \partial_{l}\Gamma^{\lambda}_{j \lambda} + \Gamma^{k}_{j l}\Gamma^{\sigma}_{k \sigma} - \Gamma^{k }_{l \sigma}\Gamma^{\sigma}_{jk},
$$
I can't associate an expression (if I didn't make the mistakes)
$$
R_{lj}^{(3)} = \partial_{j}\Gamma^{j}_{lj} + \partial_{l}\Gamma^{l}_{jl} + \partial_{k}\Gamma^{k}_{ll}\delta^{l}_{j} - \partial_{l}\Gamma^{k}_{jk} - \Gamma^{k}_{jk}\Gamma^{j}_{lj} + \Gamma^{k}_{lk}\Gamma^{l}_{jl} + \Gamma^{\sigma}_{k \sigma}\Gamma^{k}_{ll}\delta^{l}_{j} - \Gamma^{k}_{jk}\Gamma^{k}_{lk} - \Gamma^{l}_{jl}\Gamma^{j}_{lj} - \Gamma^{l}_{kl}\Gamma^{k}_{ll}\delta^{l}_{j} - \Gamma^{j}_{ll}\Gamma^{l}_{jj} - \Gamma^{k}_{ll}\Gamma^{l}_{kl} =
$$ 
$$
= \partial_{j}\Gamma^{j}_{lj} + \partial_{l}\Gamma^{l}_{jl} + \partial_{k}\Gamma^{k}_{ll}\delta^{l}_{j} - \partial_{l}\Gamma^{k}_{jk} - \Gamma^{k}_{jk}\Gamma^{j}_{lj} + \Gamma^{k}_{lk}\Gamma^{l}_{jl} + \Gamma^{\sigma}_{k \sigma}\Gamma^{k}_{ll}\delta^{l}_{j} - \Gamma^{k}_{jk}\Gamma^{k}_{lk} - \Gamma^{l}_{jl}\Gamma^{j}_{lj} - 2\Gamma^{l}_{kl}\Gamma^{k}_{ll}\delta^{l}_{j} - \Gamma^{j}_{ll}\Gamma^{l}_{jj},
$$
where there is a summation only on $k, \sigma$, with an expression for the metric tensor.
Maybe, there are some hints, which can help?
 A: Do not, I repeat, do not mix abstract and concrete indices. It confuses people (and in this case, yourself). In any case, you need to explicitly compute! There's not going to be a magic formula that pops out in "general form" since your expressions for the Christoffel symbol uses only the fact that your metric is diagonal in a certain coordinate system with certain independence of coefficients. In particular, it doesn't care that you have $\sin\psi$ instead of $f(\psi)$ for some arbitrary $f$. Since there exists three manifolds with non-constant curvature, you cannot just read off the identity based on what you have computed. You have to actually compute the various components, in coordinates of the Ricci tensor, and show that they are the appropriate multiplicative factor of the corresponding components of the metric tensor. 
Namely, you should compute things like
$$ \Gamma^\psi_{\theta\varphi}, \Gamma^\psi_{\theta\theta}, \Gamma^\varphi_{\varphi\theta}$$
explicitly and plug it into the expression for the Ricci tensor to obtain explicitly the values
$$ R_{\varphi\varphi}, R_{\psi\theta} $$
etc.
