# Riemann Curvature Tensor Components

To calculate the number of unique components of the Riemann curvature tensor, we first exploit the symmetry and antisymmetry of the first and last two indices. Before that, a quick review of combinatorics is essential in the process. $\def\multiset#1#2{\left(\kern-.3em\left(\genfrac{}{}{0pt}{}{#1}{#2}\right)\kern-.3em\right)}$

where $n$ is the number of dimensions (eg $4$), and $k$ is the group rank. For example, the metric tensor $g_{\mu \nu}$ has $\multiset{4}{2} = 10$ unique components in $4$ dimensions.

$R_{\alpha \beta \mu \nu}$ has $4^4 = 256$ total components without considering any symmetry or antisymmetry. However, $R_{\alpha \beta \mu \nu}$ is antisymmteric on the first and second pair of indices (eg $R_{\alpha \beta \mu \nu} = - R_{\beta \alpha \mu \nu}$) and is symmetric on the two pairs (eg $R_{\alpha \beta \mu \nu} = R_{\mu \nu \alpha \beta}$) and thus $$$\tag{1} R_{\alpha \beta \mu \nu} = -R_{\beta \alpha \mu \nu} = -R_{\alpha \beta \nu \mu} = R_{\mu \nu \alpha \beta} \label{1}$$$

Each of the antisymmetries in $\alpha \beta$ and $\mu \nu$ has $\binom{n}{2}=6$ unique components in $4$ dimensions, and they are $10$, $12$, $13$, $23$, $03$ and $02$. The symmetry in the two pairs, $R_{\alpha \beta \mu \nu} = R_{\mu \nu \alpha \beta}$ (eg $R_{1012} = R_{1210}$) combined with the antisymmetry reduce the number of components from $256$ to

$\multiset{\binom{n}{2}}{2} = \multiset{\frac{n(n-1)}{2}}{2} = \frac{1}{8}(n^4-2n^3 +3n^2 -2n)$ which is $21$ in $4$ dimensions.

Equation \ref{2} can be re-written as $$$\tag{2*} \label{eq:2*} R_{\alpha [\beta \mu \nu]} = \frac{1}{3}(R_{\alpha \beta \mu \nu} + R_{\alpha \nu \beta \nu} + R_{\alpha \mu \nu \beta}) = 0$$$

I used symmetry and antisymmetry to reduce $\frac{1}{6}(…)$ to \eqref{eq:2*}. Now $[\beta \mu \nu]$ has $\binom{n}{3} = 4$ equations in $4$ dimensions, but $\alpha[\beta \mu \nu]$ are not $4 \cdot4$ because if $\alpha$ is equal to any of ${\beta, \mu, \nu}$ then $\alpha[\beta \mu \nu]$ becomes equation \ref{1}. For example, if $\alpha=\mu$,

$R_{\mu[\beta \mu \nu]} = R_{\mu \beta \mu \nu} + R_{\mu \mu \nu \beta} + R_{\mu \nu \beta \mu} = 0 \quad \text{(Used symmetry and antisymmetry of \ref{1})}$ The only cases missing are when $\alpha \ne { \beta, \mu, \nu }$, but these equations are not independent either. For example, in 4 dimensions, $R_{\alpha[\beta \mu \nu]}$ is

% These 4 equations are all linearly dependent, for example, for $\eqref{eq3}$ and $\eqref{eq6}$:

All the values of $\alpha, \beta, \mu$ and $\nu$ must be different. In $4$ dimensions, these equations are only $1$.

Also, $\alpha$ is redundant in $R_{\alpha[\beta \mu \nu]}$. In fact, $R_{\alpha[\beta \mu \nu]} = R_{[\alpha \beta \mu \nu]} = 0$.

With the aid of Mathematica:

Array[Subscript[R, #1, #2, #3, #4] &, {4, 4, 4, 4}] /. {1 -> \[Alpha],
2 -> \[Beta], 3 -> \[Mu], 4 -> \[Nu]}

proj = Symmetrize[%,
Antisymmetric[{1, 2, 3, 4}]]; SymmetrizedArrayRules[proj]


Using symmetry and antisymmetry of the first and last two indices: $R_{[\alpha \beta \mu \nu]} =\frac{1}{3}(R_{\alpha \beta \mu \nu} + R_{\alpha \nu \beta \nu} + R_{\alpha \mu \nu \beta}) = R_{\alpha [\beta \mu \nu]} = 0 \quad \text{eq. \eqref{eq:2*}}$

The total number of unique components in $n$ dimensions is: $\frac{1}{8}(n^4-2n^3 +3n^2 -2n) - \binom{n}{4} = \boxed{ \frac{n^2(n^2-1)}{12} }$