I have worked out all the connection symbols for the 3-sphere using calculus of variations, cf. this Phys.SE post. So to find the Riemann tensor I am trying to find all the nonzero components of: \begin{equation} R^\rho_{\sigma \mu \nu} = \partial_\mu\Gamma^\rho_{\nu\sigma}-\partial_\nu\Gamma^\rho_{\mu\sigma}+\Gamma^e_{\mu\lambda}\Gamma^\lambda_{\nu\sigma}-\Gamma^\rho_{\nu\lambda}\Gamma^\lambda_{\mu\sigma} \end{equation}

The only strategy I can see is trying every possible combination of Christoffel symbols that produces a nonzero component (Which is of course possible). Surely there must be a better approach than this? What is the general strategy for finding the Riemann tensor?

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    $\begingroup$ In principle, simply "plug and chug" is the safe way to go. However, often simplifications can be made by making use of the various symmetries of the Riemann tensor, as well as those of the space you're analyzing. $\endgroup$ – Danu Oct 4 '14 at 19:18
  • $\begingroup$ I'm fine with plug and chug, but only if there is no better way of doing it. $\endgroup$ – Kevin Murray Oct 4 '14 at 19:22
  • $\begingroup$ Like I said, symmetry will usually allow you to get results a bit quicker. $\endgroup$ – Danu Oct 4 '14 at 19:25
  • $\begingroup$ As a side comment, $a$, $b$, $c$, etc. are easier than $\alpha$, $\beta$, $\gamma$, etc. if you're doing typing! $\endgroup$ – HDE 226868 Oct 4 '14 at 19:25
  • $\begingroup$ @HDE226868: While this is true, typically authors use Roman $a,\,b,\,c$ to denote the spatial components (i.e., 1, 2, 3) while the Greek $\alpha,\,\beta,\,\gamma$ to denote all components (i.e., 0, 1, 2, 3). Unless you are confident in what you are doing, swapping them might lead others to confusion. $\endgroup$ – Kyle Kanos Oct 5 '14 at 0:09

Here's a way to find the Riemann tensor of the 3-sphere with a lot of intelligence but no calculations.

At any point $p$ on a sphere, all directions look the same. Therefore there can be no privileged vector at a point $p$. Now consider the eigenvalue problem for the Ricci tensor, $$R^\alpha{}_\beta x^\beta = \lambda x^\alpha.$$ Since no vector is better than any other vector, either no vector is an eigenvector, or every vector is. Since the Ricci tensor is symmetric, it must be the latter. But the only way for every vector to be an eigenvector is if the tensor is proportional to the identity tensor. Hence, $$R^\alpha{}_\beta = \lambda \delta^\alpha{}_\beta \Rightarrow R_{\alpha\beta} = \lambda g_{\alpha\beta}.$$ Now take the trace of both sides to conclude that $\lambda = R/3$ where $R$ is the curvature scalar.

Now we need to be even more clever. A diagonal element $R_{\alpha\beta}x^\alpha x^\beta$ of the Ricci tensor in $n$ dimensions is $(n-1)$ times the average of the sectional curvature over planes containing $x^\alpha$, for $x_\alpha x^\alpha = 1$. The important thing is that the sectional curvature is the Gaussian curvature of the 2-surface generated by geodesics starting at $p$. What can this 2-surface be? It's a 2-surface characterized by constant curvature (because of all the symmetry we have) -- that's (some part of) a 2-sphere! The Gaussian curvature of a 2-sphere is $1/r^2$. Thus $2/r^2 = \lambda g_{\alpha\beta}x^\alpha x^\beta = \lambda$ and we can conclude $$R_{\alpha\beta} = \frac{2}{r^2} g_{\alpha\beta}.$$


Surely there must be a better approach than this?

Certainly; it's the Cartan formalism which employs differential forms. Consider your case of a sphere, with a metric tensor,

$$ds^2 = r^2dr^2+r^2\sin^2 \theta \, d\theta^2$$

We can choose an orthonormal basis $e^a$ such that $ds^2 = \eta_{ab}e^{a}e^{b}$, and $e^a = e^a_\mu dx^\mu$. For our case:

$$e^r = r \, dr, \, \, e^\theta = r\sin \theta \, d\theta$$ We now compute the exterior derivative of each, and re-express them in terms of the basis:

$$de^r = 0, \quad de^\theta = \sin \theta \, dr \wedge d\theta = \frac{1}{r^2} \, e^r \wedge e^\theta$$

We can now use the first structure equation,$^{1}$ namely $de^a + \omega^a_{b} \wedge e^b = 0$, to deduce the connection components $\omega^a_b$. With some experience, you'll be able to read them off:

$$\omega^r_r = \omega^\theta_\theta = 0, \quad \omega^\theta_r = \frac{1}{r^2}e^\theta = \frac{1}{r}\sin \theta \, d\theta$$

The second structure equation allows us to directly compute components of the Ricci tensor $R^{a}_b$ in the orthonormal basis, namely, $R^a_b = d\omega^a_b + \omega^a_c \wedge \omega^c_b$. We find that,

$$R^\theta_\theta = R^r_r = 0, \quad R^\theta_r = \frac{1}{r^4} e^\theta \wedge e^r$$

Given that $R^a_{b} = R^a_{bcd} e^c \wedge e^d$, we can compute the terms of the full Riemann tensor. To convert back from our orthonormal basis, we can use the relation

$$R^\lambda_{\mu \nu \sigma} = ( e^{-1})^{\lambda}_a \, R^a_{bcd} \, e^b_\mu \, e^c_\nu \, e^d_\sigma$$

By this method, we find for example, $R^\theta_{r\theta r} = \sin^2 \theta$ as expected for the sphere. Generally, the Cartan formalism is much easier than direct computation, but be careful in particular when,

  1. Deducing the connections from the first structure equation; it's not always straightforward;
  2. Converting back to the orthonormal basis.

For an excellent exposition of the method, see the gravitational physics lectures by Professor Ruth Gregory available from the Perimeter Scholars website.

$1$ In fact, the structure equation states $de^a + \omega^a_b + e^b = T^a$, where $T^a$ is the torsion, but in general relativity, we may assume $T^a = 0$. See my own question: Why can we assume torsion is zero in GR?

  • $\begingroup$ Ok, so I don't know anything about Cartan formalism and it isn't in Sean caroll or Hobson, but I am definitely interested! Could you point me to some literature that explains the structure equations and exterior derivatives? $\endgroup$ – Kevin Murray Oct 5 '14 at 13:32
  • $\begingroup$ @KevinMurray: A good introduction is found in Misner, Thorne and Wheeler. But check out those lecture videos, they're really better than the explanation in the book. $\endgroup$ – JamalS Oct 6 '14 at 5:36

I have just now finished an article, "Geometry of the 3-sphere", in which at the end of the paper I give a simple derivation of the Riemann curvature bivector for the unit 3-sphere, using (Clifford) geometric algebra. I also discuss the Lie group $SU(2)$ and Lie algebra $SU(2)$ on the unit 3-sphere, using the powerful, but still rather unknown geometric algebra $G(4)$, the Pauli geometric algebra $G(3)$, and the quaternions $H$, isomorphic to the even subalgebra of $G(3)$.

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    $\begingroup$ Thank you very much your contribution to the site! I formatted your post a little bit. I also removed your link, but checked your site and unfortunately I can't find the named article here. I suggest you to use the direct link to your article. Could you please provide the exact link? I would be happy to edit it into your post again. It would be also important if you would actually answer the original question, using your article. You can do this by clicking the "edit" link below your answer. Kindly regards, Peterh $\endgroup$ – peterh - Reinstate Monica Jul 20 '16 at 22:42
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    $\begingroup$ VLQ voters: please check the meta. $\endgroup$ – peterh - Reinstate Monica Jul 20 '16 at 23:21
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    $\begingroup$ @AlfredCentauri How changed the meta post your voting decision? $\endgroup$ – peterh - Reinstate Monica Jul 21 '16 at 0:58
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    $\begingroup$ @AlfredCentauri I don't know if I am correct, I only suspect. But so I already understand your logic. O.k., no problem. I suggest to check also his site. $\endgroup$ – peterh - Reinstate Monica Jul 21 '16 at 1:35
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    $\begingroup$ Does this actually answer the question? It seems to say that there exists a method of finding the Riemann tensor using Clifford algebra, but it doesn't actually explain that method in sufficient detail to enable the asker to use it. $\endgroup$ – David Z Jul 21 '16 at 10:17

In 3 dimensions, all the data in the Riemman tensor is contained in the Ricci tensor as the Weyl tensor vanishes. So it suffices to compute the Ricci tensor and then using the decomposition given here.


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