Counting independent components of Riemann curvature tensor

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I'm having some trouble understanding the counting procedure for the number of independent components of Riemann curvature tensor $R_{iklm}$ in 4D spacetime. (The answer is supposed to be 20, but I'm trying to understand how so!)

Let's do it sequentially:

  • Since we are looking at a rank 4 tensor in 4D, $R_{iklm}$ could bear $4^4 = 256$ components in general. (I follow this argument.)

  • Since the tensor is anti-symmetric w.r.t. first and second pair of indices, i.e. $$R_{iklm} = - R_{ikml} = - R_{kilm}$$ all 256 are not independent. e.g. $R_{1234} = - R_{2134}$, and those where these antisymmetric indices get repeated, vanish. e.g. $R_{1134} = - R_{1134}$, and hence $=0$. Thus, the number of independent components is far lesser. (I follow this qualitative argument.)

Trouble begins when we have to count how many. An attempt towards the answer can be found at this link, from which I quote verbatim:

The two conditions show that all components where either the first and second indices, or the third and fourth indices are equal must be zero. In four dimensional spacetime, this means that at most $\left[\ {}^4C_2\ \right]^2 = 36$ components can be non-zero, since we can choose two distinct values for both the first and last pairs of indices.

I don't get the above argument cleanly: of course with repeated indices, we get a vanishing answer, so the indices have to be distinct, but how does number of independent components amount to "choosing two distinct values for both the first and last pairs of indices"? i.e. what is the origin of the latter half of the sentence?

  • An alternative reasoning is adopted in a comment made below the post in the aforementioned link:

It was stated,…. “The two conditions [(1) & (2)] show that all components where either the first and second indices, or the third and fourth indices are equal must be zero.”

…This is correct as stated, since an interchange of two indices that are equal refers to the same tensor component, and the only number that can be equal to the negative of itself, is zero. However this only accounts for 112 components being zero; with 256 – 112 = 144 still left that are not proven to be zero.

A 4 X 4 anti-symmetric matrix is one where the 6 bottom-left triangle of components are equal to the negative of the 6 upper-right triangle of components, and so dependent, and that the diagonal components are zero (already counted).

So, for the nj-matrix, get rid of the 6 dependent, for each of the lm combinations,.. that’s 6*4*4=96. For the lm-matrix, again get rid of the 6 dependent, for each of the nj combinations,.. that’s 4*4*6=96. That’s 96nj + 96lm – 84 (that are counted twice) = 108. Now, 256 – 112 (above) – 108 = 36.

The spirit underlying this answer looks cleaner, but there are some missing steps:

How do we count that 112 vanish? How do we count that 84 were counted twice as per the former argument?

I'm sure that if I am able to understand this part, I would be able to figure out the rest of the counting myself too (i.e. slashing 36 to 21 because of the symmetry argument, and reducing one more due to the Bianchi identity constraint.)

Any help in clearing the haze would be greatly appreciated.

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(Reference : Spivak's book A comprehensive introduction to Riemannian geometry II 197p.) As far as I understand you count the number of generators : $$ T : V\times V\times V\times V\rightarrow {\bf R} $$ is linear map on $V={\bf R}^4$ which satisfies the following :

(1) $T_{ijkl}=-T_{jikl}$ (2) $T_{ijkl}=-T_{ijlk}$ (3) $T_{ijkl} + T_{jkil} + T_{kijl}=0$ (4) $T_{ijkl}=T_{klij}$

(Note that (4) is followed from the three)

Let $N:=\{i,j,k,l\}$

Case 1 : $N=2$ : By the above comment we have $6$

Case 2 : $N=4$ : Notet that by (1), (2), and (4), we suffice to consider $T_{3421},\ T_{4231},\ T_{2341}$ That is last two places must contain index $1$

In further $(3)$ implies that $$ T_{3421} + T_{4231} + T_{2341}=0$$

Hence we have $2$

Case 3 : Similar to case 2, we have $$ T_{1231},\ T_{2311}=0,\ T_{3121} $$

That is by (3) we have $1$. That is since we do not consider index $4$, we have $4$.

Hence we have total $12$

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3 pages proof of numbers of independent components of Riemann-Christoffel curvature tensor 1st page

2nd page

3rd page