Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

What is the best way to see that the Ricci scalar curvature of $(S^n(r),g_{round})$ is a constant $n(n-1)/r^2$ ? I essentially only see this value stated in the literature, but no computation associated with it, so I assume it is a straightforward calculation. But...

I keep doing the calculation, and it gets messy... Not sure of the best basis to view the round metric in, and whether to go straight through definitions with Riemannian/Ricci curvature or just use sectional curvature. So in particular, what's the best way to see that the sectional/Riemannian curvature is $1/r^2$ ?

share|cite|improve this question
up vote 4 down vote accepted

I would use the sectional curvature. Pick a point $p\in \mathbb S^{n-1}$ and a two-dimensional subspace $V\subset T_p S^{n-1}$. The geodesics starting from $p$ in directions lying in $V$ are all contained in the $3$-dimensional subspace $W\subset \mathbb R^n$ such that $T_p S^{n-1}\subset T_p W$. Therefore, the calculation of sectional curvature amounts to calculating the Gaussian curvature of $S^{n-1}\cap W$, which is just $S^2$. For that, there is a lot of explicit formulas which yield $1/r^2$ in reasonable time.

Once the sectional curvature is known to be $1/r^2$, the Ricci is determined from one version of the definition: average of sectional curvatures through a given vector (times $n-1$).

[Added] I used the definition of sectional curvature involving the exponential map, as in Helgason's Differential geometry. I quote: let $N_0$ be a normal neighborhood of $0$ in $T_pM$. Let $S$ be a two-dimensional subspace of $T_pM$. Then $\exp (N_0\cap S)$ is a 2-dimensional submanifold of $M$ with induced Riemannian structure. Its curvature at $p$ is called the sectional curvature of $M$ at $p$ along the plane section $S$.

Old-fashioned as it may be, this definition makes it as clear as possible where sectional comes from... Incidentally, Helgason defines the curvature of a 2-dimensional manifold by $$K=\lim_{r\to 0} \frac{12}{r^2}\frac{A_0(r)-A(r)}{A_0(r)}$$ where $A_0(r)$ and $A(r)$ stand for the areas of a disk $B_r(p)\subset T_pM$ and of its image under the exponential map. For the 2-sphere of radius $R$ we have $A(r)=2\pi R^2(1-\cos (r/R))=\pi r^2-\pi r^4/(12R^2)+o(r^4)$, hence $$K=\lim_{r\to 0} \frac{12}{r^2}\frac{\pi r^4 /(12R^2)}{\pi r^2}=\frac{1}{R^2}$$

Yeah... I like the exponential map and I hate tensors.

share|cite|improve this answer

There is a compelling way to see this using the Gauss equation. Let us consider the sphere $S^n \subset \mathbb R^{n+1}$. Choose a point $p \in S^n$ and an orthonormal basis $\{e_i\}$ of $T_pS^n$ in which the second fundamental form is diagonalized, thus $$D_{e_i}\nu = \lambda_ie_i,$$ where $\nu$ is the normal vector ($\nu$ is the position vector in this case) and $D_{e_i}$ is the usual directional derivative in $\mathbb R^n$. Then the Gauss equation reads $$\mathrm{sec_{\mathbb R^{n+1}}}(e_i,e_j) = \mathrm{sec_{S^n}}(e_i,e_j) - \lambda_i\lambda_j.$$ On the other hand it is easy to calculate that $$2\mathrm{Hess}r = \frac{2}{r} g_r,$$ where $g_r = r^2ds^2_{n}$ and $g = dr^2 + g_r$ is the Euclidean metric on $\mathbb R^{n+1}$ written in spherical coordinates. Here $ds^2_{n}$ is the round metric on $S^{n}$. Thus $\lambda_i = \frac{1}{r}$ and since $\mathrm{sec_{\mathbb R^n}(e_i,e_j)} = 0$ it follows that $$\mathrm{sec_{S^n}}(e_i,e_j)= \frac{1}{r^2}.$$

Remark: This method can be found in Chapter 3 of Peter Petersen's great book "Riemannian Geometry."

share|cite|improve this answer

I am not sure that this is the best way, but I find it easy: Calculate the Christoffel symbols and its derivative at the north pole $(0,...,0,r)$. Then using the formula, we can find the Riemannian curvature tensor, and hence sectional curvature and Ricci curvature at the north pole $(0,...,0,r)$. Since the calculation is done at the north pole $(0,...,0,r)$, it simplifies things a lot.

Now, note that $(S^n(r),g_{round})$ is homogenous, i.e. for any two points $p, q$ in $S^n(r)$, there exists an isometry $\sigma$ such that $\sigma(p)=q$. So every point on $(S^n(r))$ has the same Riemannian curvature tensor, and hence sectional curvature and Ricci curvature as the north pole.

share|cite|improve this answer
@ChrisGerig Yes, you are right. I made a mistake. I edited it. For your question, to calculate $\Gamma_{ij}^k$ we can use the formula $\Gamma_{ij}^k=\frac{1}{2}g^{kl}\left(\frac{\partial g_{il}{\partial x_j}+\frac{\partial g_{jl}{\partial x_i}-\frac{\partial g_{ij}{\partial x_l}\right)$. And $g_{ij}$ is known in this case. So we can calculate it explicitly. But as you have said, it's still kind of tedious. – Paul Dec 22 '12 at 3:04
I mean $\Gamma_{ij}^k=\frac{1}{2}g^{kl}(\frac{\partial g_{il}}{\partial x_j}+\frac{\partial g_{jl}}{\partial x_i}-\frac{\partial g_{ij}}{\partial x_l})$ – Paul Dec 22 '12 at 3:08
I want to accept this answer, but it rests on actually being able to compute this; I don't know what $g_{ij}$ would be in terms of coordinates on $S^n(r)$ so that I can get a nice answer. – Chris Gerig Dec 22 '12 at 18:46

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.