Reference: Moore, Thomas A., *A General Relativity Workbook*, University Science Books (2013) – Chapter 19; a-b.

We can derive a few useful symmetries of the Riemann tensor by looking at its form in a locally inertial frame (LIF). At the origin of such a frame, all first derivatives of are zero, which means the Christoffel symbols are all zero there. However, the second derivatives of are not, in general, zero, so the derivatives of the Christoffel symbols will not, in general, be zero either.

Using the definition of the Riemann tensor:

we can write it at the origin of a LIF:

The Christoffel symbols are

The symmetries of the Riemann tensor are easiest to write if we look at its form with all indices lowered, that is:

First, we calculate the derivative:

At the origin of a LIF, the first term is zero since all first derivatives of are zero, so we’re left with

Multiplying this by and using , we have

By substituting indices, we can get the second term in 5:

Subtracting 10 from 9 we see that the middle terms cancel, so we’re left with

This equation is valid only at the origin on a LIF.

From this we can get some symmetry properties. First, if we interchange the first two indices and in the tensor we see that the first and third terms on the RHS in 11 swap, as do the second and fourth, so we end up with the negative of what we started with. That is

If we interchange the last two indices and , again the first term swaps with the fourth, and the second with the third, so we get the same result:

If we swap the first and third indices, and also the second and fourth, we get

Thus

the Riemann tensor is symmetric under interchange of its first two indices with its last two:

A final symmetry

property is a bit more subtle. If we cyclically permute the last 3 indices , and and add up the 3 terms, we get

Using the symmetry of and the fact that partial derivatives commute, we find that the first two terms in the first line cancel with the last two terms in the second line, the first two in the second line cancel with the last two in the third line, and the first two in the third line cancel with the last two in the first line, giving the result:

We’ve derived these results for the special case at the origin of 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. Further, in our discussion of LIFs, we showed that we could define a LIF with its origin at *any* point in spacetime, provided that point is locally flat (that is, that there is no singularity at that point). So the argument shows that these symmetries are true for all non-singular points in spacetime.

Incidentally, it might be confusing that we can say that these symmetries are universally valid at all points in all coordinate systems just because they are tensor equations, while we say that 11 is valid only at the origin of a LIF. The difference is that 11 is written explicitly in terms of a particular metric and that metric is defined precisely so that all its first derivatives are zero at the origin of the LIF. If we wanted an equation for at some *other* point in spacetime, we could write it in the same form, but we’d need to find a *different* metric whose first derivatives are zero at this other point. If we wanted to use the original metric, then since this other point is not at the origin of the original LIF, the would not be zero at this point since the derivatives of wouldn’t be zero there, and the expression for would be more complicated in terms of the original metric.

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MichaelThe paragraph after Eq 11 should be the 1st and 3rd terms swap and the 2nd and 4th terms swap, yes ?

growescienceQuite right – fixed now. Thanks.

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