# Is there any additional symmetry in Riemann curvature tensor to tell which components are zero?

It is known that $R^{\alpha}_{\beta \gamma \delta} = -R^{\alpha}_{\beta \delta \gamma}$. I.e it's skew-symmetric in its last two indices. So if the last two indices are the same one can just say by looking at the component that it would be zero. So work reduces. Is there any other symmetry to tell which other terms vanish ? There certainly should be such relations because the number of the independent terms are $\dfrac{n^2(n^2-1)}{12}.$

• There are several other such symmetries. Check the Wikipedia page on the Riemann tensor, for instance. – gj255 Oct 16 '17 at 11:21

The Riemann tensor is defined to be that tensor for which $$(\nabla_\mu \nabla_\nu - \nabla_\nu \nabla_\mu) \omega_\rho = R_{\mu \nu \rho} {}^\sigma \omega_\sigma$$ for any one-form $\omega_\sigma$. Various other conventions exist; consult the endpapers of MTW for an exhaustive list. This is the one I'm most familiar with, so it's what I'll use. I'll also assume that we're using a torsion-free derivative operator throughout, i.e., $(\nabla_\mu \nabla_\nu - \nabla_\nu \nabla_\mu)f = 0$ for any scalar function $f$. See these notes if you're interested in how this generalizes when this condition is relaxed.
1. From the definition, it is immediately evident that $$R_{\mu \nu \rho} {}^{\sigma} = - R_{\nu \mu \rho} {}^\sigma.$$
2. It is not too hard to show that for any one-form $\omega_\rho$, we must have $$\nabla_{[\mu} \nabla_{\nu} \omega_{\rho]} = 0.$$ You can show this by writing the covariant derivatives in terms of Christoffel symbols and using their symmetry, assuming a torsion-free derivative operator. Alternately, you can use the fact that $\mathrm{d}^2 \pmb{\omega} = 0$ for any one-form $\pmb{\omega}$. In terms of the Riemann tensor, this implies that $$R_{[\mu \nu \rho]}{}^{\sigma} = 0,$$ since $R_{[\mu \nu \rho]}{}^{\sigma} \omega_\sigma$ must vanish for any $\omega_\sigma$.
3. Using the fact that the derivative operator is torsion-free, it is possible to show that for any rank-2 covariant tensor field (such as the metric $g_{\rho \sigma}$), we have $$(\nabla_\mu \nabla_\nu - \nabla_\nu \nabla_\mu) g_{\rho \sigma} = R_{\mu \nu \rho} {}^\lambda g_{\lambda \sigma} + R_{\mu \nu \sigma} {}^\lambda g_{\rho \lambda} = R_{\mu \nu \sigma \rho} + R_{\mu \nu \rho \sigma}.$$ But if $\nabla$ is the covariant (i.e., metric-compatible) derivative operator, then $\nabla_\mu g_{\rho \sigma} = 0$, which implies that $$R_{\mu \nu \rho \sigma} = - R_{\mu \nu \sigma \rho}.$$
With these three symmetries, one can show that there are $n^2(n^2 - 1)/12$ independent components of the Riemann tensor. Note, however, that this is not the same as saying that there are $n^2(n^2 - 1)/12$ non-vanishing components; properties (1) and (2) together impose relations of the form $$R_{\mu \nu \rho} {}^\sigma + R_{\nu \rho\mu } {}^\sigma + R_{\rho \mu \nu } {}^\sigma = 0,$$ but this does not necessarily imply that any of these components vanish, particularly when all three of $\mu$, $\nu$, and $\rho$ are distinct.