The question is:

For a spacetime the Riemann tensor is given below:

$$R_{\mu \nu \rho \sigma} = \frac{R}{6} (g_{\mu \rho} g_{\nu \sigma} - g_{\mu \sigma} g_{ \nu \rho} )$$

What is the dimension of this spacetime?

My thoughts (what I think is the correct method but not 100% sure how to do):

  • Contract the indexes on the left hand side until you obtain the Ricci scalar which in turn will automatically give a constraint on the number of dimensions.

Ricci tensor $ R_{ \nu \sigma} = g^{\mu \rho} R_{\mu \nu \rho \sigma}$

Ricci scalar $ R = g^{\nu\sigma} R_{\nu \sigma}$

The Ricci scalar itself should be a number, this number I think should be related to the number of dimensions.

So for us we have, $$R = g^{\nu\sigma} g^{\mu \rho} R_{\mu \nu \rho \sigma}$$

Not sure if plugging in our spacetime into this and working through would yield the correct answer. $$R = \frac{R}{6} g^{\nu\sigma} g^{\mu \rho}(g_{\mu \rho} g_{\nu \sigma} - g_{\mu \sigma} g_{ \nu \rho} )$$

Perhaps something like this? Not 100% sure though?

  • $\begingroup$ be careful: that relation is valid only for maximally symmetric spaces! That said the denominator 6 is encoding info about dimensions! $\endgroup$ – AoZora May 10 '19 at 20:36
  • 3
    $\begingroup$ Just evaluate the tensor contraction in your $gg(gg-gg)$ formula... $\endgroup$ – Hans Moleman May 10 '19 at 20:44
  • $\begingroup$ We also require $R\ne0$ to infer the dimension. $\endgroup$ – J.G. May 11 '19 at 12:33

As already commented to finish up the conclusion one has to compute:

$$g^{\mu\rho}g_{\mu\rho} = \delta^\rho_\rho = n$$


$$g^{\nu\sigma}g_{\nu\sigma} = \delta^\sigma_\sigma = n$$

and finally:

$$g^{\nu\sigma}g_{\mu\sigma}g^{\mu\rho}g_{\nu\rho} = \delta^\nu_\mu \delta^\mu_\nu = \delta^\nu_\nu = n$$ from which were conclude:

$$R = \frac{R}{6}g^{\nu\sigma}g^{\mu\rho}(g_{\mu\rho}g_{\nu\sigma} -g_{\mu\sigma}g_{\nu\rho}) = \frac{R}{6}(n^2-n)$$

From this result we can guess that the dimension of the given space is $n=3$.

The following contractions were used (note that $\delta^\nu_\mu$ is the Kronecker symbol):

$$g^{\nu\sigma}g_{\mu\sigma} =\delta^\nu_\mu$$

It actually expresses that the contravariant metric tensor is the inverse of the covariant metric tensor. And the trace of the Kronecker symbol:

$$\delta^\mu_\mu = n$$.

Indices summation is done over double appearing indices (Einstein convention).

  • $\begingroup$ Hi, thank you for that, I got exactly the same thing after I contracted too! $\endgroup$ – user61882 May 12 '19 at 10:30
  • $\begingroup$ @user61882: good! $\endgroup$ – Frederic Thomas May 12 '19 at 13:46

The dimensions of the space-time could be determined by looking at the symmetries of the Riemann tensor. The idea is to find the total number of independent parameters required to specify the curvature at any space-time event.

When we choose any locally-inertial frame in a space-time manifold in curved background, the metric at that region is not exactly Minkowski. Rather that region is projected onto a tangent space which approximates an Euclidean space. Thus, in a locally-inertial frame, the space is not exactly flat. However, the extent of non-flatness depends on the derivatives (particularly the second derivatives) of the metric tensor at that region.

For this we need to expand the metric tensor $g_{\alpha \beta}$ in Taylor series. In the locally-inertial frame, the following points are to be noted:

  • The metric corresponds to that of the Minkowski metric i.e. $g_{\alpha \beta}=\eta_{\alpha \beta}$.
  • All the first derivatives of the metric tensor $g^{'}_{\alpha \beta}$ vanish.
  • The second derivatives of the metric tensor $g^{''}_{\alpha \beta}$ don't all vanish.

The number of surviving second derivatives $g^{''}_{\alpha \beta}$ is equal to the number of independent components of the Riemann tensor which equals $\frac{1}{12}D^2(D^2-1)$, where $D$ is the dimension of the space-time. Now, given the Riemann tensor, we can determine the dimension of the space-time from this result.

Reference: This is discussed in details at the end of Chap. 7 of The Feynman Lectures on Gravitation.

  • $\begingroup$ Whilst this answer isn't wrong, it doesn't actually answer the question does it? OPs question can be directly answered in the manner they suggest in their post, they just haven't followed through with all the necessary steps in the calculation. $\endgroup$ – jacob1729 May 11 '19 at 10:35
  • $\begingroup$ Hi, thank you for you answer as Jacob said this is correct but not 100% relevant for my question. The correct answer has been stated by Frederic Thomas $\endgroup$ – user61882 May 12 '19 at 10:31

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