The second member of the expression on the left-hand side of (52a) leads first to
- ½ ∂²/∂xα∂xμ (gλβΓλβα) or
to 1/4 ∂²/∂xα∂xμ [gλβgαδ( ∂gδλ/∂xβ
+ ∂gδβ/∂xλ - ∂gλβ/∂xδ)].
The expression arising out of the last member within the round bracket vanishes according to (29) on account of the choice of axes. The two others can be taken together and give us on account of (31), the expression
-½ ∂³gαβ/∂xα∂xβ∂xμ
So that remembering (54) we have
(55) ∂²/∂xα∂xσ (gσβΓμβα
- ½ δμσ gλβ Γλβα) = 0.
identically.