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The RiemannChristoffel Tensor

The RiemannChristoffel tensor arises as the difference of cross covariant derivatives. Let
A_{i} be any covariant tensor of rank one. Then
A_{i, jk} − A_{i, kj} = R_{ijk}^{p}A_{p}
Remarkably, in the determination of the tensor R_{ijk}^{p} it does not matter which covariant tensor of rank one is used.
The tensor R_{ijk}^{p} is called the RiemannChristoffel tensor of the second kind. As the
notation indicates it is a mixed tensor, covariant of rank 3 and contravariant of rank 1. This has to be
proven.
The determination of the nature of R_{ijk}^{p} goes as follows.
 The general formula for the covariant derivative of a covariant tensor of rank one, A_{i}, is
A_{i, j} = ∂A_{i}/∂x^{j} − {ij,p}A_{p}
 For a covariant tensor of rank two, B_{ij}, the formula is:
B_{ij, k} = ∂B_{ij}/∂x^{k} − {ik,p}B_{pj} − {kj,p}B_{ip}
 A_{i, j} is such a tensor so the above formula applies to it as well. Therefore
A_{i, jk} = ∂A_{i, j}/∂x^{k} − {ik,p}A_{p, j} − {kj,p}A_{i, p}
where (A_{i, j})_{, k} has been written as A_{i, jk}.
 Replacing A_{i, j} by ∂A_{i}/∂x^{j} − {ij,p}A_{p}
and carrying out the indicated differentiation yields
A_{i, jk} = (∂²A_{i}/∂x^{k}x^{j})
− (∂{ij,p}/∂x^{k})A_{p}
− {ij,p}(∂A_{p}/∂x^{k})
− {ik,p}(∂A_{p}/∂x^{j})
+ {ik,p}{pj,q}A_{q}
− {kj,p}(∂A_{i}/∂x^{p})
+ {kj,p}{ip,r}A_{r}
 If the indices j and k are interchanged the result is A_{i, kj}; i.e.,
A_{i, kj} = (∂²A_{i}/∂x^{j}x^{k})
− (∂{ik,p}/∂x^{j})A_{p}
− {ik,p}(∂A_{p}/∂x^{j})
− {ij,p}(∂A_{p}/∂x^{k})
+ {ij,p}{pk,q}A_{q}
− {jk,p}(∂A_{i}/∂x^{p})
+ {jk,p}{ip,r}A_{r}
 The cross partial derivatives (∂²A_{i}/∂x^{j}x^{k})
and (∂²A_{i}/∂x^{k}x^{j}) are equal. The partial derivatives
of A_{p} with respect to x^{j} and x^{k} appear in both expressions although
in different positions. Thus subtracting the expression for A_{i, kj} from the one for
A_{i, jk} yields
A_{i, jk} − A_{i, kj} = − (∂{ij,p}/∂x^{k})A_{p} +
{ik,p}{pj,q}A_{q}
+ (∂{ik,p}/∂x^{j})A_{p} − {ij,p}{pk,q}A_{q}
 The summation indices p and q in the two terms involving a double summation can be interchanged
without affecting the result. This allows the above result to be expressed as
A_{i, jk} − A_{i, kj} =
− (∂{ij,p}/∂x^{k})A_{p} + {ik,q}{qj,p}A_{p}
(∂{ik,p}/∂x^{j})A_{p} − {ij,q}{qk,p}A_{p}
 The above result can further simplified as
A_{i, jk} − A_{i, kj} =
[∂{ik,p}/∂x^{j}) − ∂{ij,p}/∂x^{k} + {ik,q}{qj,p} − {ij,q}{qk,p}]A_{p}
 Let the expression within the brackets be denoted as R_{ijk}^{p} so the above
is represented as
A_{i, jk} − A_{i, kj} = R_{ijk}^{p}A_{p}
 The expression on the lefthand side of the above equation is the difference of two tensors of
covariant rank 3. Therefore it is a tensor of covariant rank 3. The term R_{ijk}^{p}
on the righthand side when multiplied times the components of an arbitrary covariant tensor of
rank 1 and summed yields a covariant tensor of rank 3. Therefore by the
Tensor Quotient Theorem
R_{ijk}^{p} is a mixed tensor of covariant rank 3 and contravariant rank 1. Thus the
notation is justified.
 Thus the condition for the cross covariant derivatives to be equal is that the RiemannChristoffel
tensor of the second kind be identically equal to zero; i.e.,
A_{i, jk} = A_{i, kj}
if and only if
R_{ijk}^{p} = 0
for all i, j, k and p.
(To be continued.)