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?

  • 6
    $\begingroup$ Why don't you do this explicitly instead of these symbolic manipulations? You have a metric (which is even diagonal!), so start by just calculating the Christoffel symbols. There is no magic way to turn an arbitrary metric into a Riemann tensor! $\endgroup$
    – Vibert
    May 1 '13 at 11:40
  • $\begingroup$ Possible duplicate by OP: physics.stackexchange.com/q/62717/2451 . Please do not post the same question in a new entry. Re-edit the old entry instead. $\endgroup$
    – Qmechanic
    May 1 '13 at 13:23
  • 4
    $\begingroup$ If you can't use a software package (because this is homework), then the best thing to do is just work step by step from the definitions instead of trying anything fancy right off the bat. Once you've done it once this way, learn a nicer way (Cartan) and do that once or twice, then just use the software. :) So just work out the components one by one. One thing I can say though: you have repeated indices all over the place. Be extra super careful about what is being summed over and what isn't! $\endgroup$
    – Michael
    May 1 '13 at 13:37
  • 1
    $\begingroup$ It's not a homework, I need it for improving my abilities. [:)]. Thank you! $\endgroup$
    – user8817
    May 1 '13 at 14:31

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.


Given that your metric is diagonal, it simplifies a lot these calculations. However, the Riemman tensor is such an object...

First, start with the Christoffel Symbols

$$ \Gamma^i{}_{k\ell}= {1 \over 2} g^{im} (g_{mk,\ell} + g_{m\ell,k} - g_{k\ell,m})$$

Note that $g_{im}=0$ for $i \neq m$ so it simplifies to

$$ \Gamma^i{}_{k\ell}= {1 \over 2} g^{ii} (g_{ik,\ell} + g_{i\ell,k} - g_{k\ell,i})$$

and the metric does not depend on $\varphi$ so $g_{\mu \nu,\varphi}=0$ Because the 3-sphere is a manifold without torsion, the following symmetry happens:

$$ \Gamma^i{}_{k\ell}= \Gamma^i{}_{\ell k} $$

Plug all of that into the Riemman tensor

$$R^\rho{}_{\sigma\mu\nu} = \partial_\mu\Gamma^\rho{}_{\nu\sigma} - \partial_\nu\Gamma^\rho{}_{\mu\sigma} + \Gamma^\rho{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\sigma} - \Gamma^\rho{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\sigma}$$

and then

$$R_{ab}=R^{c}_{acb} $$

If you don't want to calculate the Riemman Tensor, just do the following

$$ R_{\alpha\beta} = {R^\rho}_{\alpha\rho\beta} = \partial_{\rho}{\Gamma^\rho_{\beta\alpha}} - \partial_{\beta}\Gamma^\rho_{\rho\alpha} + \Gamma^\rho_{\rho\lambda} \Gamma^\lambda_{\beta\alpha} - \Gamma^\rho_{\beta\lambda}\Gamma^\lambda_{\rho\alpha} $$

  • $\begingroup$ Did you read the body of my answer? I derived the general expression for a Ricci tensor of sphere (and, also, for hyperboloid). I only needed to simplify it. $\endgroup$
    – user8817
    May 2 '13 at 0:05

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