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 The Ricci Theorem in Tensor Analysis

The Ricci Theorem in tensor analysis is that the covariant derivative of the metric tensor or its inverse are zero; i.e., all components are zero. Let gij be the metric tensor for some coordinate system (x1,…,xn) for n dimensional space. Then formally,

### Ricci's Theorem (First part): gij, k = 0

where 0 is an n×n×n× array of zeroes.

Proof:

• The covariant derivative of a second rank covariant tensor Aij is given by the formula

#### Aij, k = ∂Aij/∂xk − {ik,p}Apj − {kj,p}Aip

where the symbol {ij,k} is the Christoffel 3-index symbol of the second kind.

• The 3-index symbol of the second kind is defined in terms of the 3-index symbol of the first kind, which has the definition

#### [ij,k] = ½[∂gik/∂xj + ∂gik/∂xi − ∂gij/∂xk]

From this definition it is obvious that [ij,k]=[ji,k]. From the definition it is easily established that

#### ∂gij/∂xk = [ik,j] + [kj,i]

• The 3-index symbols of the second kind are derived from the first kind by multiplying the components of the first kind by the components of the inverse of the metric tensor and summing over the third index; i.e.,

#### {ij,k} = gkp[ij,p]

But from this relation it then follows that the symbols of the first kind can be obtained from those of the second kind by this relation

#### [ij,k] = gkp{ij,p}

Therefore the equation previously derived for ∂gij/∂xk can be expressed as

#### ∂gij/∂xk = gpj{ik,p} + gip{kj,p} or, equivalently as ∂gij/∂xk − gpj{ik,p} − gip{kj,p} = 0

• When the metric tensor gij is substituted for Aij in the general formula for the covariant derivative of a second rank covariant tensor the result is

#### ∂gij,k = ∂gij/∂xk − gpj{ik,p} − gip{kj,p}

But from the previous equation the right-hand side of this equation is identically zero. Thus

#### ∂gij,k = 0

This is the first part of Ricci's Theorem.

The second part of Ricci's Theorem is that

### gij, k = 0

Proof:
• The general formula for the covariant derivative of a second rank contravariant tensor Aij is

#### Aij, k, k = ∂Aij/∂xk + {ik,p}Apj + {kj,p}Aip

From the definitions of the 3-index symbols of the first and second kind and from the inverse relation of gij and gij the following formula can be derived

#### ∂gij/∂xk = −gpj{ik,p} − gip{kj,p} or, equivalently as ∂gij/∂xk + gpj{ik,p} + gip{kj,p} = 0

• When the contravariant metric tensor gij is substituted in the general formula for the covariant derivative of a contravariant tensor of the second rank the result is:

#### gij, k, k = ∂gij/∂xk + {ik,p}gpj + {kj,p}gip

From the previously derived equation the right-hand side of the above equation is identically zero. Thus h4>
gij, k = 0

This is the second part of Ricci's Theorem

The second part could have been derived from the first part by noting that

#### gipgpj = δij

where δij is the Kronecker delta; i.e., δij=0 if i≠j and is equal to 1 if i=j.

Covariant differentiation of the above equation results in

#### gip, kgpj + gipgpj, k= δij, k = 0

Since gij, k=0 the above equation reduces to

#### gip, kgpjgpj = 0

This is a system of linear equations in the unknowns gij, k with all the constants in the equations equal to zero. Since the coefficients matrix of the equation is the metric tensor and the metric tensor has an inverse the only solution to the equations is

#### gij, k = 0

as was previously proven.