I am calculating the Riemann curvature tensor, Ricci curvature tensor, and Ricci scalar of the $n$ sphere $$x_0^2 + x_1^2 + ….+x_n^2=R^2,$$ whose metric is
$$ds^2=R^2(d\phi_1^2 + \sin{\phi_1}^2 d\phi_1^2+\sin{\phi_2}^2\sin{\phi_1}^2 d\phi_2^2+……).$$
I have done the same exercise for the 2 sphere, and found that the Riemann curvature wih a lowered index is proportional to the product of two $g \cdot g $, and the Ricci tensor to $g$. I am hoping to see the same result for the n dimensional sphere.
I am using the following expression which can be found in Schutz.
$$R_{\alpha \beta \mu \nu}=\frac{1}{2}(g_{\alpha \nu, \beta \mu}-g_{\alpha \mu, \beta \nu }+ g_{\beta \mu, \alpha \nu}-g_{\beta \nu, \alpha \mu}).$$
It can be seen clearly that only derivatives of the form $g_{\phi_p \phi_p, \phi_m \phi_n}$ where $m,n < p$ are non zero. How do I show that the Riemann curvature tensor is proportional to the product of the metric tensors? The Ricci tensor is $R_{ij}=R^l_{ilj}=g^{lm}R_{milj}$, as the metric tensor is diagonal $m=l, and setting first and third index equal in the expression for the RIemann curvature which would cause the first and third term to be zero as the derivatives are zero by the above argument, I get
$$R_{ij}=-\frac{1}{2}g^{il}(g_{ll,ij}+g_{ij,ll})=-\frac{1}{2}g^{ll}g_{ll,ij}.$$
I was predicting, that from the 2 sphere case to get a term proportional the the metric tensor. (For the 2-sphere case $R_{ij}=\frac{1}{R^2}R_{ij}$). Where am I going wrong? How do I convert it to a form which reduced to the 2-sphere on substituting, $n =2$.
The Ricci scalar expression is much more horrible, $R=g^{im}R_{mi}=g^{ii}R_{ii}$. I moronically expanded this and got a terrrible expression in $\cot{\phi}$ which cant be summed easily, while I need a number inversely proportional to $R$.
Any help is appreciated.
Also, (maybe this should be posted as a separate question), could I get the same tensors using some other method, e.g. tetrads. I have heard of this but don't know much. So it would be great if somebody could show it specifically for this case.
Best Answer
The direct calculation of the derivatives isn't that hard. But one can also quickly see the values of the Riemann tensor for a sphere – and similar simple shapes – by using the definition of the Riemann tensor via the parallel transport of vectors. $$\delta V^\alpha = R_{\alpha\beta\gamma\delta}V^\beta d\Sigma^{\gamma\delta}$$ Around a point of the sphere $S^d$, the transport around an area given by $d\Sigma^{\gamma\delta}$ for fixed values of the indices (locally orthonormal basis) allows you to see that all of this is happening on an $S^2$ only. The other dimensions are unaffected. That's why you get $R_{\alpha\beta\gamma\delta}$ equal to $(1/a^2)$ times $g_{\alpha\gamma}g_{\beta\delta}-g_{\alpha\delta}g_{\beta\gamma}$. Effectively, the antisymmetrized pair of indices $\alpha\beta$ has to be the same as the pair $\gamma\delta$. I didn't assume anything special about the point; all points on a sphere are equally good by a symmetry. So the Ansatz for the Riemann tensor has to hold everywhere.
Note that it's multiplied by $1/a^2$, the inverse squared radius of the sphere. Many of your formulae omit it; moreover, you are using a confusing symbol $R$ for the radius which looks like the Ricci scalar – a different thing.
In $d=2$, the Riemann tensor only has one independent component and the formula for the Riemann tensor in terms of the metric tensor above actually holds for any surface if $1/a^2$ is replaced by $R/2$. Note that the two-sphere has (Ricci scalar) $R=2/a^2$. Also, the Ricci tensor is $R_{ij}=Rg_{ij}/2$ in $d=2$ so that the vacuum Einstein equations are obeyed identically.
For $d=3$, the Riemann tensor has 3 independent components, just like the Ricci tensor, so the Riemann tensor may be written in terms of the Ricci tensor. That's not true for higher dimensions, either.