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?

## Best Answer

I think there is a method that I believe is rather simple. Take a look:

There is a thing called 'normal Riemann coordinates'. In this coordinates the metric is expanded around the origin, and the coefficients of expansion are expressed in terms of the Riemann tensor. I suggest that you read about them and check whether the coordinates described below are normal. All needed information can be found here.

Take the coordinates $x_i,\,x=\sqrt{x_i x_i}$ to be (I write $r$ so it cant be confused with $R$ the contraction of Ricci tensor): $$ \psi=x/r\\ x_1=r\psi\sin\theta\cos\phi\\ x_2=r\psi\sin\theta\sin\phi\\ x_3=r\psi\cos\theta $$ Compare it with usual spherical coordinates in $\mathbb{R}^3$, we know: $$ dx_idx_i=r^2d\psi^2+r^2\psi^2(d\theta^2+\sin^2\theta d\phi^2) $$ So we can rewrite your metric as $$ ds^2=\frac{\sin^2\psi}{\psi^2}dx_i dx_i+r^2d\psi^2\left(1-\frac{\sin^2\psi}{\psi^2}\right)=\\ =r^2\frac{\sin^2(x/r)}{x^2}dx_i dx_i+\frac{(x_idx_i)^2}{x^2}\left(1-r^2\frac{\sin^2(x/r)}{x^2}\right) $$

Now, if we believe that $x_i$ are normal, then we have: $$ g_{ik}(x)=\delta_{ik}-\frac{1}{3}R_{iakb}x^ax^b+... $$ So you can just expand the metric and find the symmetrized form $\frac{1}{2}\left(R_{iakb}+R_{ibka}\right)$ of the Riemann tensor at the origin that is sufficient to contract it to the Ricci tensor. The good point is that it is the same level of complexity (in fact, the 'normal' metric of the same form) in any dimension. As soon as you will find $R_{ik}\sim g_{ik}$ in this coordinate system, it holds in any coordinate system. And, due to the symmetry of the sphere, it also holds at every point.