15
$\begingroup$

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 $1/r^2$ ?

$\endgroup$
0

4 Answers 4

7
$\begingroup$

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.

$\endgroup$
0
6
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ @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. $\endgroup$
    – Paul
    Dec 22, 2012 at 3:04
6
$\begingroup$

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."

$\endgroup$
1
$\begingroup$

I'll show the direct computation, because I think it is the best way to see this. Work in this map $\phi$, which is either defined on the upper hemisphere $\{x_{n+1} >0\}$, or lower hemisphere $\{x_{n+1}<0\}$, $$ \phi(x_1,\dots,x_{n+1}) = (x_1,\dots,x_n,\sum_{i=1}^{n+1}x_i^2 -R^2) = (y_1,\dots,y_{n+1}) $$ Write $(\vec{e}_1,\dots,\vec{e}_{n+1})$ the canonical basis of $\mathbb{R}^{n+1}$. Then compute the matrices of $d\phi$ and $d\phi^{-1}$ to see that for all $i\leq n$, $$ \partial y_i = \frac{\partial}{\partial y_i} = \vec{e}_i - \frac{x_i}{x_{n+1}}\vec{e}_{n+1} $$ Differentiate the $\partial y_i$ in euclidean $\mathbb{R}^{n+1}$, $$ \tilde{\nabla}_{\partial y_i}\partial y_j = \left(-\frac{x_ix_j}{x_{n+1}^3}-\frac{\delta_{ij}}{x_{n+1}}\right) \vec{e}_{n+1} $$ Then we need to project $\vec{e}_{n+1}$ orthogonally on the $n$-sphere. For that, check that the metric tensor is $g_{ij} = \delta_{ij} + \frac{x_ix_j}{x_{n+1}^2}$ and its inverse is $g^{ij} = \delta_{ij} - \frac{x_ix_j}{R^2}$. The orthogonal projection of $\vec{e}_{n+1}$ is therefore $-\sum_{i=1}^n \frac{x_ix_{n+1}}{R^2}\partial y_i$. This yields the Christoffel symbols of $\nabla$ in map $\phi$ : $$ \nabla_{\partial y_i}\partial y_j = \left(\frac{x_ix_j}{x_{n+1}^2}+\delta_{ij}\right)\sum_{k=1}^n\frac{x_k}{R^2}\partial y_k, \quad\quad \text{i.e.} \quad\quad \Gamma_{ij}^k = \left(\frac{x_ix_j}{x_{n+1}^2}+\delta_{ij}\right)\frac{x_k}{R^2} $$ As you can see, the Christoffels are fairly simple. Continue to compute the right part of the Riemann curvature tensor $$ \sum_{\lambda=1}^n \Gamma^\rho_{\mu\lambda}\Gamma^\lambda_{\nu\sigma} - \Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma} = \frac{x_\rho(x_\mu\delta_{\nu\sigma}-x_\nu\delta_{\mu\sigma})}{R^2 x^2_{n+1}} $$ and the left part $$ \partial y_\mu\Gamma^\rho_{\nu\sigma} - \partial y_\nu\Gamma^\rho_{\mu\sigma} = \left(\frac{x_\nu x_\sigma}{x^2_{n+1}}+\delta_{\nu\sigma} \right)\frac{\delta_{\rho\mu}}{R^2} - \left(\frac{x_\mu x_\sigma}{x^2_{n+1}}+\delta_{\mu\sigma} \right)\frac{\delta_{\rho\nu}}{R^2} +\frac{x_\rho x_\nu\delta_{\sigma\mu} - x_\rho x_\mu\delta_{\sigma\nu}}{R^2 x^2_{n+1}} $$ Finally, compute the traces to see that the Ricci tensor $R_{\sigma\nu}=\sum_{\rho=1}^n R^\rho_{\sigma\rho\nu}$ is equal to $\frac{n-1}{R^2}$ times the metric tensor. That means the scalar curvature is $\frac{n(n-1)}{R^2}$, as requested.

You see there is no magic : just do it. In the process you got a lot more insight into the geometry of the $n$-sphere than just its scalar curvature : its Ricci tensor, Riemann curvature tensor and the Christoffels.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .