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_{abmn;l} + R_{ablm;n} + R_{abnl;m} = 0,!.

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

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

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

: R^n {}_{n;l} - R^n {}_{l;n} - R^{nm} {}_{nl;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,: R_{;l} - R^n {}_{l;n} - R^m {}_{l;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,: R_{;l} = 2 R^m {}_{l;m},,!which is the same as: abla_m R^m {}_l = {1 over 2} abla_l R,!.Swapping the index labels "l" and "m" yields: abla_l R^l {}_m = {1 over 2} 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 {}_m - {1 over 2} delta^l {}_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 {}_m = g^l {}_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 {}_m - {1 over 2} abla_l g^l {}_m R = 0.,!Factor out the covariant derivative: abla_l (R^l {}_m - {1 over 2} g^l {}_m R) = 0,,!then raise the index "m" throughout: abla_l (R^{lm} - {1 over 2} g^{lm} R) = 0.,!The expression in parentheses is the Einstein tensor, so: abla_l G^{lm} = 0,,! "Q.E.D." ("return to article")


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