Understanding the symmetries of the Riemann tensor (Part II) The answered first part of this topic is Understanding the symmetries of the Riemann tensor. In my question I suppose that $a\neq b, a\neq c, a\neq d$.
My question is: Consider the components of the Riemann tensor that have all its indices different: 


*

*Independent indices of this type has the form
$$ R_{abcd}, \qquad a<b<c<d \quad (*)\quad \text{or}\quad a<c<b<d\quad (**)$$
and there are $$2\begin{pmatrix} n \\ 4 \end{pmatrix}.$$



  
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*How many dependent components of the Riemann tensor of the form $R_{abcd}$ ($a\neq b, a\neq c, a\neq d$) are there? 2. Explain the result. 3. Show that there's no more.
  

My hypothesis: $$ (4!-2) \begin{pmatrix} n \\ 4 \end{pmatrix},$$
the 4! is for the number of ways of setting $a,b,c,d$ in $(*)$, and $-2\begin{pmatrix} n \\ 4 \end{pmatrix}$ the independent components. The demostration: all posible components are $(n-3)_4$ where $(z)_n$ is the Pochhammer symbol, so $$ (n-3)_4 = n(n-1)(n-2)(n-3)\overset{?}{=} 4! \begin{pmatrix} n \\ 4 \end{pmatrix}.$$
I don't know if this is correct. Thank you for the answers!
 A: From the above I assume you mean not only : $a\neq b, a\neq c, a\neq d$ , but also : $  b\neq c, b\neq d , c \neq d $. 
Indeed there are ${n \choose 4} $ ways to select $4$ different indices from $n$ indices. Now because all indices are different we can arrange them in $4!$ different ways. That gives us a total of $4!{n \choose 4} $ components of the Riemann tensor with  $4$ different indices.
If $2{n \choose 4}$ components are independent then that must mean that $(4!-2){n \choose 4}=22{n \choose 4}$ are dependent components.
To show we have $2{n \choose 4}$ independent components we first show we can put the smallest index first using three of the symmetries of the Riemann tensor, namely :
$R_{\alpha\beta\gamma\lambda}=R_{\gamma\lambda\alpha\beta}$ 
 ; $R_{\alpha\beta\gamma\lambda}=-R_{\beta\alpha\gamma\lambda}$ and $R_{\alpha\beta\gamma\lambda}=-R_{\alpha\beta\lambda\gamma}$
Let $x$ be the smallest index . We have $4$ different possibilities for a component :
1) $R_{x\lambda\beta\gamma}$
2) $R_{\lambda x \beta\gamma}=-R_{x\lambda\beta\gamma}$
3) $R_{\lambda\beta x \gamma}=R_{x \gamma \lambda\beta }$
4) $R_{\lambda\beta  \gamma x}=-R_{x \gamma   \lambda\beta  }$
From this we can see that every component can be made dependent on another component that has its smallest index in the first position. That leaves us with only $3!{n\choose 4}$ possible independent components to investigate if we take the smallest index to be the first.
Now $\frac{1}{2}$ of the remaining components can be made dependent because of $R_{\alpha\beta\gamma\lambda}=-R_{\alpha\beta\lambda\gamma}$ (we always make the 4th index greater than the 3rd).
So now we have only $\frac{1}{2}3!{n \choose 4}=3{n \choose 4}$ components to investigate. If we take $a<b<c<d$ the last possible independent candidates have the form : $R_{abcd} \lor R_{acbd} \lor R_{adbc} $
Because of the symmetry : $R_{\alpha \beta \gamma \lambda} +R_{\alpha\lambda\beta\gamma} + R_{\alpha\gamma\lambda \beta }=0$ we see we can eliminate $\frac{1}{3}$ of the remaining candidates . So $\frac{2}{3}3{n \choose 4}=2{n \choose 4}$ remain independent. 
If we take the forms : $R_{abcd} \lor R_{acbd}$ as independent we can easily check we can not eliminate further using the above symmetries . For example if we try to 'make' $R_{acbd}$ out of combinations of other independent components we only have $R_{abcd}$ at our disposal. All other components are either  dependent or they have one or more indices that are the same. The symmetries of the Riemann tensor all have the same indices on both sides of the equal sign so we can never produce a component with $4$ different indices out of components with one or more identical indices.
$R_{abcd}$ with the symmetry and antisymmetry relations produces only dependent components. The cyclic relation : $R_{abcd} +R_{adbc} = R_{acbd } $ is the only remaining candidate. But for that of course we need two independent components on the left hand side. But $R_{adbc}$ is the one we have taken to be dependent. 
 So there is no way to produce $R_{abcd}$ and $R_{acbd}$ out of other independent components.
