Components of Riemann Tensor: 4 Indexes, 16x16 Matrix

[Z3kr0m]

Hello, Riemann tensor $R^i_{jkl}$ 4 indexes, and it should be matrix 16x16 in spacetime if we have time coirdinate - 0 and space coordinates -1,2,3. But how should I write the components to matrix? For example $\begin{pmatrix}R^0_{000} & R^1_{000} & R^2_{000} ... \\ R^0_{100} & R^1_{100} & R^2_{100} ...\\ R^0_{200} & R^1_{200} ... \\ R^0_{300} ... \end{pmatrix}$?

 

[Nugatory]

Hello, Riemann tensor $R^i_{jkl}$ 4 indexes, and it should be matrix 16x16 in spacetime if we have time coirdinate - 0 and space coordinates -1,2,3. But how should I write the components to matrix? For example $\begin{pmatrix}R^0_{000} & R^1_{000} & R^2_{000} ... \\ R^0_{100} & R^1_{100} & R^2_{100} ...\\ R^0_{200} & R^1_{200} ... \\ R^0_{300} ... \end{pmatrix}$?

You can't. The two-dimensional matrix representation only works for rank-two tensors (those with only two indices). Even then you have to be alert to whether you are representing the components of the tensor with contravariant or covariant (upper or lower) indices - this information is lost in the matrix representation, so it's best to avoid it altogether.

 

[Z3kr0m]

So I must write out the components one by one?

 

[Nugatory]

So I must write out the components one by one?

Yes, but you have to do that whether you're arranging the values you're writing down in a matrix or not.

 

[pervect]

Hello, Riemann tensor $R^i_{jkl}$ 4 indexes, and it should be matrix 16x16 in spacetime if we have time coirdinate - 0 and space coordinates -1,2,3. But how should I write the components to matrix? For example $\begin{pmatrix}R^0_{000} & R^1_{000} & R^2_{000} ... \\ R^0_{100} & R^1_{100} & R^2_{100} ...\\ R^0_{200} & R^1_{200} ... \\ R^0_{300} ... \end{pmatrix}$?

It's not really a matrix, its a 4x4x4x4 tensor. But if you group the first two componetns together, and the last two, you can write the 256 element tensor as a 16x16 matrix.

I've only seen discussions (in MTW's "Gravitation") of $R_{ijkl}$ and NOT $R^i{}_{jkl}$ however. To take advantage of the skew symmetries $R_{ijkl} = -R_{jikl} = -R_{ijlk}$ https://en.wikipedia.org/wiki/Riemann_curvature_tensor#Symmetries_and_identities, I believe one needs the indices to be all lower, or all upper, not mixed. It is possible I am mistaken.

If one does consider the all-lower index $R_{ijkl}$, the skew symmetries imply that $R_{00**} = R_{11**} = R_{22**} = R_{33**} = 0$, i.e. any repeated index in the first pair must be zero. A similar argument leads to the same conclusion for repeated indices in the last pair. This lowers the Riemann to a 12x12 matrix by dropping the 4 zero diagional elelents, , and a further reduction to a 6x6 matrix by only specifying $R_{ij**}$ where i<j, as $R_{ji**} = -R_{ij**}$.

The natural pair-groupings are (0,1), (0,2), (0,3) and their duals *(0,1) = (2,3), *(0,2) = (1,3), *(0,3)=(1,2) in a coordinate or orthonormal basis.

Using this approach, the 256 element tensor becomes a 6x6 symmetric matrix, where the symmetry comes from the "interchange symmetry", $R_{ijkl} = R_{klij}$.

This can be further decomposed into three (in GR) 3x3 matrices Wiki calls this the Bel decomposition, https://en.wikipedia.org/wiki/Bel_decomposition, MTW does a similar decomposition but doesn't use this name to describe it.

Two of the 3x3 matrices from this decomposition are symmetric, the so-called electrogravitc and topogravitc tensors. This is a total of 21 degree of freedom, there is one additional constraint due to the Bianchi identity that reduces the degrees of freedom of the Riemann to 20.