Let $X$,$Y$,$Z$,$W$ be vector fields on a riemannian manifold, and let $R(X,Y)W$ be the riemannian curvature:
$$ R(X,Y)W = \nabla_{X}\nabla_{Y}W – \nabla_{Y}\nabla_{X}W – \nabla_{[X,Y]}W $$
Let $g$ be the metric tensor and $\tau$ be the torsion (which I assume to be zero as usual):
$$ \tau(X,Y) = \nabla_{X}Y – \nabla_{Y}X – [X,Y] = 0 $$
With this setting, I want to prove the Second Bianchi Identity:
$$ \nabla_{X} R(Y,Z)W + \nabla_{Z}R(X,Y)W + \nabla_{Y}R(Z,X)W = 0 $$
Now, most proofs I have found seem to rely heavily on index notation and/or using some special frame of reference to simplify the computations. However, I'd like to have a more straightforward proof, that:
1) Uses index-free notation for all tensors involved.
2) Depends only in the abstract properties of the covariant derivative, the torsion and the symmetries of the curvature tensor (including the first Bianchi identity)
How can I proceed to obtain such a proof?
Best Answer
Using the product rule for differentiation $$ \nabla_{X} R(Y,Z)W + \nabla_{Y} R(Z,X)W + \nabla_{Z} R(X,Y)W \\= \nabla_X(R(Y,Z)W) + \nabla_Y(R(Z,X)W) + \nabla_Z(R(X,Y)W) \\- R(\nabla_X Y,Z)W - R(Y,\nabla_X Z)W \\- R(\nabla_Y Z,X)W - R(Z,\nabla_Y X)W \\- R(\nabla_Z X,Y)W - R(X,\nabla_Z Y)W \\- R(Y,Z)\nabla_X W - R(Z,X)\nabla_Y W - R(X,Y)\nabla_Z W $$ Also $$R(\nabla_X Y,Z) + R(Y,\nabla_X Z) + R(\nabla_Y Z,X) + R(Z,\nabla_Y X) + R(\nabla_Z X,Y) + R(X,\nabla_Z Y) \\= R(\nabla_X Y - \nabla_Y X,Z) + R(\nabla_Y Z - \nabla_Z Y,X) + R(\nabla_Z X - \nabla_X Z,Y) \\= R(\tau(X,Y) + [X,Y],Z) + R(\tau(Y,Z) + [Y,Z],X) + R(\tau(Z,X) + [Z,X],Y) $$ And $$ \nabla_X(R(Y,Z)W) + \nabla_Y(R(Z,X)W) + \nabla_Z(R(X,Y)W) \\= \nabla_X \nabla_Y \nabla_Z W - \nabla_X \nabla_Z \nabla_Y W + \nabla_Y \nabla_Z \nabla_X W - \nabla_Y \nabla_X \nabla_Z W + \nabla_Z \nabla_X \nabla_Y W - \nabla_Z \nabla_Y \nabla_X W \\ - \nabla_X \nabla_{[Y,Z]} W - \nabla_Y \nabla_{[Z,X]} W - \nabla_Z \nabla_{[X,Y]} W \\= \nabla_Y \nabla_Z \nabla_X W - \nabla_Z \nabla_Y \nabla_X W + \nabla_Z \nabla_X \nabla_Y W - \nabla_X \nabla_Z \nabla_Y W + \nabla_X \nabla_Y \nabla_Z W- \nabla_Y \nabla_X \nabla_Z W \\ - \nabla_{[Y,Z]} \nabla_X W - R(X,[Y,Z]) - \nabla_{[X,[Y,Z]]}W - \nabla_{[Z,X]} \nabla_Y W - R(Y,[Z,X]) - \nabla_{[Y,[Z,X]]}W - \nabla_{[X,Y]} \nabla_Z W - R(Z,[X,Y]) - \nabla_{[Z,[X,Y]]}W \\= R(Y,Z)\nabla_X W + R(Z,X)\nabla_Y W + R(X,Y)\nabla_Z W \\ - R(X,[Y,Z])W - R(Y,[Z,X])W - R(Z,[X,Y])W $$ noting that $\nabla_{[X,[Y,Z]]}W + \nabla_{[Y,[Z,X]]}W + \nabla_{[Z,[X,Y]]}W = \nabla_{[X,[Y,Z]] + [Y,[Z,X]] + [Z,[X,Y]]}W = 0$ by the Jacobi identity.
Now combine to get $$ \nabla_{X} R(Y,Z) + \nabla_{Y} R(Z,X) + \nabla_{Z} R(X,Y) = R(X,\tau(Y,Z)) + R(Y,\tau(Z,X)) + R(Z,\tau(X,Y))$$
Note the argument is simpler if one has that all the Lie brackets are zero, which is a much weaker assumption than normal coordinates, etc. I would say that the core of the proof is this identity: $$ \nabla_X \nabla_Y \nabla_Z W - \nabla_X \nabla_Z \nabla_Y W + \nabla_Y \nabla_Z \nabla_X W - \nabla_Y \nabla_X \nabla_Z W + \nabla_Z \nabla_X \nabla_Y W - \nabla_Z \nabla_Y \nabla_X W \\= \nabla_Y \nabla_Z \nabla_X W - \nabla_Z \nabla_Y \nabla_X W + \nabla_Z \nabla_X \nabla_Y W - \nabla_X \nabla_Z \nabla_Y W + \nabla_X \nabla_Y \nabla_Z W- \nabla_Y \nabla_X \nabla_Z W $$
Finally let me comment that the problem with using normal coordinates is that you implicitly use that the connection comes from a Riemannian metric. The proofs given here are simple manipulations of the definitions.