Constant curvature on a sphere?

Question

$ds^2 = \frac{1}{1- r^2}dr^2 + r^2d\theta^2$ denotes a 2d spherical surface and it should have a constant curvature. The Riemann curvature tensor components are linear in their all 3 inputs. Since the curvature is constant, we expect $R^r_{\theta r \theta} = k|e_\theta|^2|e_r|$, where k is a constant.

However, $|e_r| = \frac{1}{\sqrt{1- r^2}}$, $|e_\theta| = r$ but $R^r_{\theta r \theta} = r^2$ which is not of the form $k|e_\theta|^2|e_r|$. Looks like it's not growing linearly as $e_r$ grows.

So, what am I missing?

Answer 1

In Riemannian geometry, a constant curvature space means that the curvature tensor is of the form $$ R_{ijkl}=k(g_{ik}g_{jl}-g_{il}g_{jk}), $$ where $k$ is a constant. This equivalently means that the sectional curvature is constant (unlike for the Riemann tensor, it is meaningful for the sectional curvature to be constant).

In two dimensions, the curvature tensor is always of the form $$ R_{ijkl}=\frac{R}{2}(g_{ik}g_{jl}-g_{il}g_{jk}), $$ so in this case the space is constant curvature if and only if the scalar curvature $R$ is constant. But this is only in two dimensions.

The condition $\nabla_mR_{ijkl}=0$ is strictly weaker. Such spaces are called locally symmetric.

EDIT:

It appears that OP is rather confused regarding the concepts of symmetry in Riemannian geometry. Let's overview some terms:

• Constant curvature space
• Locally maximally symmetric space
• Maximally symmetric space
• Locally symmetric space
• Symmetric space

These are all a priori different concepts.

Constant curvature spaces: As mentioned, a (pseudo-)Riemannian manifold $(M,g)$ is a constant curvature space if the sectional curvature function $\mathrm{Sec}$ is constant. This is a function on the 2-Grassmannian bundle of $M$, rather than $M$ itself (technically, for pseudo-Riemannian metrics, it is a function on an open subset of the Grassmannian that consists of those 2-planes which are nondegenerate with respect to the metric), i.e. it associates a value $\mathrm{Sec}(\Delta)$ to each $2$-plane $\Delta\le T_pM$.

A completely equivalent characterization is that the Riemann curvature tensor has the form $$ R_{ijkl}=k\left( g_{ik}g_{jl}-g_{il}g_{jk}\right) $$with $k$ a constant.

(Locally) maximally symmetric spaces: Truthfully, this terminology seems to be more "classical" and don't often appear in the modern literature on the subject, so it is possible that I get something wrong in the local-global distinction here.

I think it is relatively safe to say that a pseudo-Riemannian space $(M,g)$ is locally maximally symmetric if each point $p\in M$ has some neighborhood $U\subseteq M$ such that on $U$, there exist the maximal number $m(m+1)/2$ (where $m=\dim M$) of linearly independent Killing vector fields.

Then $(M,g)$ is maximally symmetric if its isometry group (which is necessarily a Lie group; Myers--Steenrod theorem and its pseudo-Riemannian generalization) has the maximum allowed dimension $m(m+1)/2$. I think this is equivalent to $M$ admitting $m(m+1)/2$ global, linearly independent Killing vector fields, but I am not absolutely sure.

Local maximal symmetry is equivalent with being constant curvature. Basically, one can use the Frobenius integrability theorem to show that the completely integrability of the Killing equation implies that the curvature tensor is of the form $R_{ijkl}=k(g_{ik}g_{jl}-g_{il}g_{jk})$ given above (with $k$ a constant), and conversely, if the curvature tensor has this form, then the integrability conditions for the Killing equation are satisfied identically (a somewhat unrigorous version of this argument can be found in Weinberg: Gravitation and Cosmology, but it can easily be rigourified). However the Frobenius theorem concerns itself only with local existence so I think this only establishes equivalence with local maximal symmetry.

(Locally) symmetric spaces: Given a point $p$ in $(M,g)$, a geodesic symmetry is a local diffeomorphism $f:U\rightarrow U$ ($U$ is a neighborhood of $p$) such that $f(p)=p$ (it fixes $p$) and $f(\gamma(t))=\gamma(-t)$ for any geodesic $\gamma$ with $\gamma(0)=p$.

Then $(M,g)$ is locally symmetric if each point has a geodesic symmetry, and each geodesic symmetry is a local isometry (it does not need to be, normally!). The space is (globally) symmetric if in addition, each geodesic symmetry extends to a global isometry of $(M,g)$.

A standard result is that $(M,g)$ is locally symmetric if and only if $\nabla_m R_{ijkl}=0$ ($m$ here is a free index, not the dimension of $M$!), i.e. the curvature tensor is parallel/covariant constant.

This condition is strictly weaker than having constant curvature. Every constant curvature space is locally symmetric, but there are locally symmetric spaces that do not have constant curvature.

Answer 2

It is true that scalar curvature is constant when all covariant derivatives of the Riemann tensor are zero: That assumption $$\tag1 \nabla_\mu R^\nu{}_{\eta\,\rho\,\sigma}=0 $$ implies for the Ricci tensor $$ \nabla_\mu R^\rho{}_{\eta\,\rho\,\sigma}=\nabla_\mu R_{\eta\,\sigma}=0 $$ which implies for the scalar curvature $$ \partial_\mu R=\nabla_\mu R=\nabla_\mu (R_{\rho\,\eta}\,g^{\eta\,\rho})=g^{\eta\,\rho}\,\underbrace{\nabla_\mu R_{\rho\,\eta}}_0+R_{\rho\,\eta}\,\underbrace{\nabla_\mu\,g^{\eta\,\rho}}_0=0\,. $$ provided that the metric has zero covariant derivative (Levi-Civita connection).

I think (1) is a much stronger assumption and another cumbersome condition to verify. In your case you can show directly that the scalar curvature is constant and equals two.

The covariant derivative of the Riemann tensor is \begin{align}\tag2 \nabla_\mu R^\nu{}_{\eta\,\rho\,\sigma}&=\partial_\mu R^\nu{}_{\eta\,\rho\,\sigma}+\underbrace{\Gamma^\nu_{\alpha\,\mu}\,R^{\alpha}{}_{\eta\,\rho\,\sigma}-\Gamma^\alpha_{\eta\,\mu}\,R^{\nu}{}_{\alpha\,\rho\,\sigma}-\Gamma^\alpha_{\rho\,\mu}\,R^{\nu}{}_{\eta\,\alpha\,\sigma}-\Gamma^\alpha_{\sigma\,\mu}\,R^{\nu}{}_{\eta\,\rho\,\alpha}}\,. \end{align} In your case the non zero Christoffel symbols are \begin{align} \Gamma^r_{r\,r}&=\tfrac r{1-r^2}\,,&\Gamma^r_{\theta\,\theta}&=-r(1-r^2)\,,&\Gamma^\theta_{r\,\theta}=\Gamma^\theta_{\theta\,r}&=\tfrac1r \end{align} and the non zero Riemann components are \begin{align} R^r{}_{\theta\,r\,\theta}&=-R^r{}_{\theta\,\theta\,r}=r^2\,,&R^\theta_{r\,\theta\,r}&=-R^\theta_{r\,r\,\theta}=\tfrac 1{1-r^2}\,. \end{align} This leads to zero covariant derivatives. Only in the following cases the underbraced term in (2) is not zero. But it cancels with the partial derivatives term:

\begin{align*} \nabla_r R^r{}_{\theta\,r\,\theta}&=-\nabla_r R^r{}_{\theta\,\theta\,r}=2r-2r=0\,,\\[2mm] \nabla_r R^\theta{}_{r\,\theta\,r}&=-\nabla_r R^\theta{}_{r\,r\,\theta}=\tfrac {2r}{1-r^2}-\tfrac {2r}{1-r^2}=0\,. \end{align*}

Answer 3

I got the answer. For curvature to remain constant the Riemann curvature tensor should be constant, meaning its covariant derivative along any direction must be zero. I ran EinsteinPy code and verified it for r direction.