Finding the dimensions of a spacetime given the Riemann tensor

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.

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

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

and

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

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).