Proofs involving covariant derivatives

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This article contains proof of formulas in Riemannian geometry that involve the Christoffel symbols.

Contracted Bianchi identities

Proof

Start with the Bianchi identity[1]

[math]\displaystyle{ R_{abmn;\ell} + R_{ab\ell m;n} + R_{abn\ell;m} = 0. }[/math]

Contract both sides of the above equation with a pair of metric tensors:

[math]\displaystyle{ g^{bn} g^{am} (R_{abmn;\ell} + R_{ab\ell m;n} + R_{abn\ell;m}) = 0, }[/math]
[math]\displaystyle{ g^{bn} (R^m {}_{bmn;\ell} - R^m {}_{bm\ell;n} + R^m {}_{bn\ell;m}) = 0, }[/math]
[math]\displaystyle{ g^{bn} (R_{bn;\ell} - R_{b\ell;n} - R_b {}^m {}_{n\ell;m}) = 0, }[/math]
[math]\displaystyle{ R^n {}_{n;\ell} - R^n {}_{\ell;n} - R^{nm} {}_{n\ell;m} = 0. }[/math]

The first term on the left contracts to yield a Ricci scalar, while the third term contracts to yield a mixed Ricci tensor,

[math]\displaystyle{ R_{;\ell} - R^n {}_{\ell;n} - R^m {}_{\ell;m} = 0. }[/math]

The last two terms are the same (changing dummy index n to m) and can be combined into a single term which shall be moved to the right,

[math]\displaystyle{ R_{;\ell} = 2 R^m {}_{\ell;m}, }[/math]

which is the same as

[math]\displaystyle{ \nabla_m R^m {}_\ell = {1 \over 2} \nabla_\ell R. }[/math]

Swapping the index labels l and m yields

[math]\displaystyle{ \nabla_\ell R^\ell {}_m = {1 \over 2} \nabla_m R, }[/math]      Q.E.D.     (return to article)

The covariant divergence of the Einstein tensor vanishes

Proof

The last equation in the proof above can be expressed as

[math]\displaystyle{ \nabla_\ell R^\ell {}_m - {1 \over 2} \delta^\ell {}_m \nabla_\ell R = 0 }[/math]

where δ is the Kronecker delta. Since the mixed Kronecker delta is equivalent to the mixed metric tensor,

[math]\displaystyle{ \delta^\ell {}_m = g^\ell {}_m, }[/math]

and since the covariant derivative of the metric tensor is zero (so it can be moved in or out of the scope of any such derivative), then

[math]\displaystyle{ \nabla_\ell R^\ell {}_m - {1 \over 2} \nabla_\ell g^\ell {}_m R = 0. }[/math]

Factor out the covariant derivative

[math]\displaystyle{ \nabla_\ell\left(R^\ell {}_m - {1 \over 2} g^\ell {}_m R\right) = 0, }[/math]

then raise the index m throughout

[math]\displaystyle{ \nabla_\ell \left(R^{\ell m} - {1 \over 2} g^{\ell m} R\right) = 0. }[/math]

The expression in parentheses is the Einstein tensor, so [1]

[math]\displaystyle{ \nabla_\ell G^{\ell m} = 0, }[/math]     Q.E.D.    (return to article)

this means that the covariant divergence of the Einstein tensor vanishes.

The Lie derivative of the metric

Proof

Starting with the local coordinate formula for a covariant symmetric tensor field [math]\displaystyle{ g=g_{ab}(x^c)dx^a\otimes dx^b }[/math], the Lie derivative along a vector field [math]\displaystyle{ X = X^a \partial_a }[/math] is

[math]\displaystyle{ \begin{align} \mathcal{L}_X g_{ab} & = X^c \partial_c g_{ab}+g_{cb}\partial_a X^c + g_{ca}\partial_b X^c \\ & = X^c \partial_c g_{ab} + g_{cb}\bigl(\partial_a X^c\pm \Gamma^{c}_{da}X^d\bigr) + g_{ca}\bigl(\partial_b X^c \pm \Gamma^{c}_{db}X^d\bigr) \\ & = \bigl( X^c \partial_c g_{ab} - g_{cb}\Gamma^{c}_{da}X^d - g_{ca}\Gamma^{c}_{db}X^d\bigr)+ \bigl[ g_{cb}\bigl(\partial_a X^c + \Gamma^{c}_{da}X^d\bigr) + g_{ca}\bigl(\partial_b X^c + \Gamma^{c}_{db}X^d\bigr) \bigr] \\ & = X^c\nabla_c g_{ab} + g_{cb}\nabla_a X^c + g_{ca}\nabla_b X^c \\ & = 0 + g_{cb}\nabla_a X^c + g_{ca}\nabla_b X^c \\ & = g_{cb}\nabla_a X^c + g_{ca}\nabla_b X^c \\ & = \nabla_a X_b + \nabla_b X_a \end{align} }[/math]

here, the notation [math]\displaystyle{ \partial_a = \frac{\partial}{\partial x^a} }[/math] means taking the partial derivative with respect to the coordinate [math]\displaystyle{ x^a }[/math].      Q.E.D.     (return to article)

See also

References

  1. 1.0 1.1 Synge J.L., Schild A. (1949). Tensor Calculus. pp. 87–89–90. 

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