Proofs involving covariant derivatives

Start with the Bianchi identity[1]

${\displaystyle R_{abmn;\ell }+R_{ab\ell m;n}+R_{abn\ell ;m}=0.}$

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

${\displaystyle g^{bn}g^{am}(R_{abmn;\ell }+R_{ab\ell m;n}+R_{abn\ell ;m})=0,}$

${\displaystyle g^{bn}(R^{m}{}_{bmn;\ell }-R^{m}{}_{bm\ell ;n}+R^{m}{}_{bn\ell ;m})=0,}$

${\displaystyle g^{bn}(R_{bn;\ell }-R_{b\ell ;n}-R_{b}{}^{m}{}_{n\ell ;m})=0,}$

${\displaystyle R^{n}{}_{n;\ell }-R^{n}{}_{\ell ;n}-R^{nm}{}_{n\ell ;m}=0.}$

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

${\displaystyle R_{;\ell }-R^{n}{}_{\ell ;n}-R^{m}{}_{\ell ;m}=0.}$

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,

${\displaystyle R_{;\ell }=2R^{m}{}_{\ell ;m},}$

which is the same as

${\displaystyle \nabla _{m}R^{m}{}_{\ell }={1 \over 2}\nabla _{\ell }R.}$

Swapping the index labels l and m yields

${\displaystyle \nabla _{\ell }R^{\ell }{}_{m}={1 \over 2}\nabla _{m}R,}$      Q.E.D.     (return to article)

The last equation in the proof above can be expressed as

${\displaystyle \nabla _{\ell }R^{\ell }{}_{m}-{1 \over 2}\delta ^{\ell }{}_{m}\nabla _{\ell }R=0}$

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

${\displaystyle \delta ^{\ell }{}_{m}=g^{\ell }{}_{m},}$

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

${\displaystyle \nabla _{\ell }R^{\ell }{}_{m}-{1 \over 2}\nabla _{\ell }g^{\ell }{}_{m}R=0.}$

Factor out the covariant derivative

${\displaystyle \nabla _{\ell }\left(R^{\ell }{}_{m}-{1 \over 2}g^{\ell }{}_{m}R\right)=0,}$

then raise the index m throughout

${\displaystyle \nabla _{\ell }\left(R^{\ell m}-{1 \over 2}g^{\ell m}R\right)=0.}$

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

${\displaystyle \nabla _{\ell }G^{\ell m}=0,}$     Q.E.D.    (return to article)

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

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

${\displaystyle {\begin{aligned}{\mathcal {L}}_{X}g_{ab}&=X^{c}\partial _{c}g_{ab}+g_{cb}\partial _{a}X^{c}+g_{ca}\partial _{b}X^{c}\\&=X^{c}\partial _{c}g_{ab}+g_{cb}{\bigl (}\partial _{a}X^{c}\pm \Gamma _{da}^{c}X^{d}{\bigr )}+g_{ca}{\bigl (}\partial _{b}X^{c}\pm \Gamma _{db}^{c}X^{d}{\bigr )}\\&={\bigl (}X^{c}\partial _{c}g_{ab}-g_{cb}\Gamma _{da}^{c}X^{d}-g_{ca}\Gamma _{db}^{c}X^{d}{\bigr )}+{\bigl [}g_{cb}{\bigl (}\partial _{a}X^{c}+\Gamma _{da}^{c}X^{d}{\bigr )}+g_{ca}{\bigl (}\partial _{b}X^{c}+\Gamma _{db}^{c}X^{d}{\bigr )}{\bigr ]}\\&=X^{c}\nabla _{c}g_{ab}+g_{cb}\nabla _{a}X^{c}+g_{ca}\nabla _{b}X^{c}\\&=0+g_{cb}\nabla _{a}X^{c}+g_{ca}\nabla _{b}X^{c}\\&=g_{cb}\nabla _{a}X^{c}+g_{ca}\nabla _{b}X^{c}\\&=\nabla _{a}X_{b}+\nabla _{b}X_{a}\end{aligned}}}$

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