# Riemann, Ricci curvature tensor and Ricci scalar of the n dimensional sphere

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. • There's a bunch of relatively easy ways to compute these tensors. You could use the technique of Example 0.7 and Problem 0.8 in chapter 4 of these notes for example: rybu.org/?q=mathmultimedia but perhaps you prefer the Ricci calculus formalism. Jun 2, 2013 at 8:21 • It seems wrong to me that the Riemann curvature tensor should be proportional to the square of the metric. After all, the formula you cite is linear in the metric. Jun 2, 2013 at 8:52 • @celtschk: Its linear in the second order derivative of the metric. The expression for the 2 sphere I got is, (I am 100% sure this is correct) for the Ricci tensor is$R_{ij}=\frac{1}{R^2}g_{ij}$. So, I should be getting a same expression in the n dimensional case, with a different constant. For the 2 sphere.$R_{ijk}=\frac{1}{R^2}(\delta^i_j g_{jm}-\delta^i_m g_{jk})$– user23238 Jun 2, 2013 at 9:04 • Your formula for the curvature tensor is incorrect. The correct formula (which you can find in this Wikipedia article) is not linear in the metric and its derivatives. The formula you wrote is valid only at one point in normal coordinates around that point. Jun 2, 2013 at 13:58 • My answer to a similar question Jun 3, 2013 at 22:58 ## 1 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. • Could you please define what the bracket$<X,Y>$means, and what$R(X,Y)$means? Otherwise, it is almost impossible to follow you. Thank you! Nov 28, 2014 at 20:29 • @Kagaratsch: R is the Riemann Curvature Tensor. – user99914 Nov 29, 2014 at 10:03 • And <X, Y> is the Scalar product in$\mathbb{R}^n\$...
– Kofi
Nov 30, 2014 at 15:50
• Basically, you mean to say that <X,Y> is the contraction of the metric tensor with the arbitrary fields X and Y (result is a scalar). And R(X,Y) again is the contraction of the Ricci tensor with two arbitrary fields X and Y (again the result is a scalar). Dec 9, 2014 at 19:23
• Nevermind, I guess those identities should have been obvious to somebody (not me), who has seen the second fundamental form of the sphere
– DCT
Jun 13, 2015 at 16:53