[Math] Riemann, Ricci curvature tensor and Ricci scalar of the n dimensional sphere

differential-geometrygeneral-relativityphysics

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.

$$R_{\alpha \beta \mu \nu}=\frac{1}{2}(g_{\alpha \nu, \beta \mu}-g_{\alpha \mu, \beta \nu }+ g_{\beta \mu, \alpha \nu}-g_{\beta \nu, \alpha \mu})$$

Here $g_{\alpha \nu, \beta \mu}$ means $\frac{\partial g_{\alpha \nu}}{\partial \beta \mu}$.

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

I would say that calculating the curvature of the sphere using charts and Christoffel symbols is possibly the most inefficient way of all.

If $x \in S^n$, then the tangent space at the point $x$ is given by $$ T_x S^n = \{ X \in \mathbb{R}^{n+1} \mid \langle x, X \rangle = 0 \}$$

The curvature tensor of the sphere is then given by $$R(X, Y)Z = \langle Y, Z \rangle X - \langle X, Z \rangle Y .$$ To see this, use the Gauss formula, for example. The Ricci curvature is then the trace w.r.t. $X$, i.e. $$ \mathrm{ric}(Y, Z) = \mathrm{tr} \,R(\cdot, Y)Z = \sum_{i=1}^{n-1}\bigl( \langle Y, Z \rangle \langle e_i, e_i \rangle - \langle e_i Z \rangle \langle Y, e_i \rangle\bigr) = (n-1) \langle Y, Z \rangle$$ where $e_1, \dots e_{n-1} \perp \vec{n}$ is an orthonormal basis of $T_x S^n$.

To get the curvature for spheres of different radius, multiply everything with the inverse radius.