The Riemann-Christoffel 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^pA_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 Riemann-Christoffel 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^kx^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^jx^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^jx^k) and (∂²A_i/∂x^kx^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^pA_p
   

• The expression on the left-hand 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 right-hand 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 Riemann-Christoffel 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.)