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# Christoffel symbols/Proofs

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

Proof 1

Start with the Bianchi identity:$R_\left\{abmn;l\right\} + R_\left\{ablm;n\right\} + R_\left\{abnl;m\right\} = 0,!$.

Contract both sides of the above equation with a pair of metric tensors::$g^\left\{bn\right\} g^\left\{am\right\} \left(R_\left\{abmn;l\right\} + R_\left\{ablm;n\right\} + R_\left\{abnl;m\right\}\right) = 0,,!$

:$g^\left\{bn\right\} \left(R^m \left\{\right\}_\left\{bmn;l\right\} - R^m \left\{\right\}_\left\{bml;n\right\} + R^m \left\{\right\}_\left\{bnl;m\right\}\right) = 0,,!$

:$g^\left\{bn\right\} \left(R_\left\{bn;l\right\} - R_\left\{bl;n\right\} - R_b \left\{\right\}^m \left\{\right\}_\left\{nl;m\right\}\right) = 0,,!$

:$R^n \left\{\right\}_\left\{n;l\right\} - R^n \left\{\right\}_\left\{l;n\right\} - R^\left\{nm\right\} \left\{\right\}_\left\{nl;m\right\} = 0.,!$The first term on the left contracts to yield a Ricci scalar, while the third term contracts to yield a mixed Ricci tensor,:$R_\left\{;l\right\} - R^n \left\{\right\}_\left\{l;n\right\} - R^m \left\{\right\}_\left\{l;m\right\} = 0.,!$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,:$R_\left\{;l\right\} = 2 R^m \left\{\right\}_\left\{l;m\right\},,!$which is the same as:$abla_m R^m \left\{\right\}_l = \left\{1 over 2\right\} abla_l R,!$.Swapping the index labels "l" and "m" yields:$abla_l R^l \left\{\right\}_m = \left\{1 over 2\right\} abla_m R,!$, "Q.E.D." ("return to article")

Proof 2

The last equation in Proof 1 above can be expressed as:$abla_l R^l \left\{\right\}_m - \left\{1 over 2\right\} delta^l \left\{\right\}_m abla_l R = 0,!$where δ is the Kronecker delta. Since the mixed Kronecker delta is equivalent to the mixed metric tensor,:$delta^l \left\{\right\}_m = g^l \left\{\right\}_m,,!$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:$abla_l R^l \left\{\right\}_m - \left\{1 over 2\right\} abla_l g^l \left\{\right\}_m R = 0.,!$Factor out the covariant derivative:$abla_l \left(R^l \left\{\right\}_m - \left\{1 over 2\right\} g^l \left\{\right\}_m R\right) = 0,,!$then raise the index "m" throughout:$abla_l \left(R^\left\{lm\right\} - \left\{1 over 2\right\} g^\left\{lm\right\} R\right) = 0.,!$The expression in parentheses is the Einstein tensor, so:$abla_l G^\left\{lm\right\} = 0,,!$ "Q.E.D." ("return to article")

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