# Riemann Curvature Tensor Components

20 Oct 2018 · CommentsTo 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

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

which is $21$ in $4$ dimensions.

Equation \ref{2} can be re-written as

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 $\alpha=\mu$,

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:

The total number of unique components in $n$ dimensions is:

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