Direct proof of the second Bianchi identity

Question

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?

Answer 1

You can find such a proof in my book Riemannian Manifolds: An Introduction to Curvature, pp. 123-124. The computation is hugely simplified by choosing a point and then extending the given vectors to vector fields whose brackets and first covariant derivatives both vanish at the given point. (This can be done, for example, by taking them to be coordinate vectors in Riemannian normal coordinates, but that's not the only way to do it.) From there, it's just a matter of writing down the definition of the covariant derivative of $R$ three times with indices cyclically permuted, and adding.

You can do the whole proof without relying on a special frame, just by keeping track of the commutators and first covariant derivatives, and noting that they all cancel. But why would you want to?

Answer 2

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.

Answer 3

The proof you are looking for is here: http://www.math.ucla.edu/~petersen/Bianchi_Ricci_Identities.pdf

It has the advantage (which is what I suspect you are interested in) of isolating precisely how the various symmetries/algebraic properties are used in the proof.

The proof follows by iterating the Ricci identities and using the first Bianchi identity. The first Bianchi identity is obtained by taking the Lie derivative of the connection - an idea that goes back at least to Yano, Bochner and/or Kostant some time around the 50's ([1,2]). Of course, the Jacobi identity plays a central role and it turns out the by reversing the argument, one could prove the Jacobi identities from the Bianchi identity.

There is no need to choose any frames, or extend vectors in any particular way which is something I find appealing. Any other proof is secretly using identities such as Jacobi by building it in to the chosen local frame and using tensorality.

Edit: See (Kostant is much easier to read!)

1. K. Yano and S. Bochner, Curvature and Bette numbers, Annals of Mathematics Studies, no. 32, Princeton, 1953.
2. Bertram Kostant, Holonomy and the lie algebra of infinitesimal motions of a riemannian manifold, Trans. Amer. Math. Soc. 80 (1955), 528–542.