Understanding the symmetries of the Riemann tensor I know that the Riemann tensor $R_{\alpha \beta \gamma \lambda}$ for $n$ dimensions has $$\dfrac{n^2(n^2-1)}{12} \tag{1}$$ independent components because of its properties:


*

*Symmetry: $R_{\alpha\beta\gamma\lambda}=R_{\gamma\lambda\alpha\beta}$.

*Antisymmetry: $R_{\alpha\beta\gamma\lambda}=-R_{\beta\alpha\gamma\lambda}$ and $R_{\alpha\beta\gamma\lambda}=-R_{\alpha\beta\lambda\gamma}$.

*Cyclic relation: $R_{\alpha \beta \gamma \lambda} +R_{\alpha\lambda\beta\gamma} + R_{\alpha\gamma\lambda \beta }=0$.


This PDF document explains the number (1), but I don't understand the last symmetry on the document (the purple color). Can anybody explain me the number of yellow, blue and purple squares based on the symmetry properties of above?
 A: The idea is categorizing the components in two classes : 'dependent' and 'independent'. The dependent components are zero or their values are determined by other (independent) components. If we count the number of independent components we can find and add to that the number of dependent components that follow from our choice of independent components, we should get as the sum of both : $n^4$. If that's the case we know we are done because we have accounted for all the components.
The yellow , blue and purple squares give the independent components based on the number of different indices :
Yellow, $2$ different indices : Components of the form : $R_{abab}$ with $a<b$. The $a=b$ are zero , and $b<a$ become dependent.
There are ${n\choose 2}$ possible choices for $a$ and $b$ this way.
Blue, $3$ different indices : Components of the form : $R_{abac}$ with $b<c$, $a \neq b$, $a \neq c$. The $c<b$ become dependent. We have ${n\choose 3}$ ways to choose different $a$, $b$ and $c$ and $3$ ways to choose the index that is used twice from that. So $3{n\choose 3}$ choices in total.
Purple , $4$ different indices : Components of the form : $R_{abcd}$ or $R_{acbd}$  with $a<b<c<d$. It's easy to see that using the symmetries we can always arrange $4$ different indices in one of these two orders. For example from $R_{acbd}$ follows $R_{acdb}$ . And $R_{acdb}$ together with $R_{abcd}$ determines $R_{adbc}$ . Also placing the smallest index first is no problem using the symmetries. Obviously there are $2{n\choose 4}$ choices.
Now : ${n\choose 2}+3{n\choose 3}+2{n\choose 4}= {n! \over 2!(n-2)!
}+{3n! \over 3!(n-3)!}+{2n! \over 4!(n-4)!}= {n(n-1) \over 2! 
}+{3n(n-1)(n-2) \over 3!}+{2n(n-1)(n-2)(n-3) \over 4!}= {n(n-1)\left((n-2)(n-3)+6(n-2)+6 \right) \over 12}  = {n(n-1)\left(n^2 +n  \right) \over 12}=\frac{n^2(n^2-1)}{12}$
Counting the dependent components gives : $\frac{n^2(11n^2+1)}{12}$.
Sum of both is $n^4 \space \space \square$
