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")