Category: General Relativity
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In our two previous articles, we have deduced the rather complicated expression of the Riemann curvature tensor, a glorious mixture of derivatives and products of connection coefficients, with 256 (=4^4) components in four-dimensional spacetime.

But we have also demonstrated in our article Local Flatness or Local Inertial Frames and SpaceTime curvature that any arbitrary coordinate system could nullify all but 20 second derivatives of a given metric in a curved spacetime. Our aim in this article is to demonstrate that the Riemann tensor has only 20 independant components and that these component are precisely a combination of these second not null derivatives.

The methodology to adopt there is to study the Riemann tensor symmetries  in a Local Inertial Frame (LIF) - where as we know all the Christoffel symbols are null - and to generalize these symmetries to any reference frame, as by definition a tensor equation valid in a given referential will hold true in any other referential frame.

Using the definition of the Riemann tensor as seen in the precedent articles:

and knowing that all the Christoffel symbols are null at the origin of Local Inertial Frame, this expression get simplified to:

By applying the contraction mechanism as exposed in Introduction to Tensors, we can rewrite the Riemann tensor with all indices lowered:

We remember from our article Christoffel symbols in terms of the metric tensor how to write the Christoffel symbol with respect to the metric derivatives:

So that we can write

By substituting indices μ and ν, we get the second term of the Riemann tensor expression:

By substracting the two expressions we see that the middle terms cancel, so we’re left with:

This could obviously be written as:

Bear in mind that this equation is valid only at the origin on a LIF. However, the origin of a LIF defines one particular event in spacetime and since all these symmetries are tensor equations, they must be true for that particular event, regardless of which coordinate system we’re using.

In the same way, we can show easily that the Riemann tensor is symmetric under interchange of its first two indices:

If we swap the first and third indices (α <-> μ), and also the second and fourth (β <-> ν), we get:

A final symmetry property is a bit more subtle. If we cyclically permute the last 3 indices β, μ and ν and add up the 3 terms, we get



The symmetries of the Riemann tensor mean that only some of its 256 components are actually independant.

The two following symmetries (deduced in the previous tab)

show that if α=β or μ=ν then the tensor component Rαβμν is necessarily null as it is equal to its opposite. In four dimensional spacetime, this means that at most 6x6 = 36 components can be non-zero, since we can choose two distinct values for both the first (α,β) and last (μ,ν) pairs of indices.

Indeed, if we were to choose 0,1,2 and 3 the coordinates in four-dimensional spacetime, then the only pairs of indices for which the Riemann tensor components are not null are the six following ones: 01,02,03,12,13 and 23.

We are therefore left with the following 6x6 dimensional matrix as the set of all the not null independant Riemann tensor components:

But using the symmetry

then the lower triangle of this matrix is the mirror image of the upper triangle, meaning we can eliminate 15 more components, leaving 21 possibly independent components: [1]

Is it the end? No, as we know that each indice should be distinct among a given pair, but can an indice of the first pair be equal to an indice of the last pair, say α=μ? There is one final symmetry condition for the Riemann tensor which might be useful for us to test:

So if we choose one indice equal to an indice of another pair,  then the last symmetry  follows from the other three conditions and tells us nothing new; therefore the last symmetry gives us new information only if all the four indices are different.

Looking at our blue right triangle above, it means that the only new information for us is:

 So R1302 can be deduced from R1203 and R2301

And the total number of independent components in four-dimensional spacetime is therefore 21-1 = 20 independant components.

[1] We recall that the number of independant components of a n-dimensional symmetric matrix is n(n+1)/2, here 6x7/2 = 21.