16
$\begingroup$

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?

$\endgroup$

3 Answers 3

24
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ Why is it you are able to use product rule on $R$ and its input vectors? I would have expected chain rule to be used here, since $R$ is being used like a function. $\endgroup$
    – eigenchris
    Jul 27, 2019 at 21:36
  • $\begingroup$ @eigenchris The notation is like that of a function, but we know that it is multiplication by a tensor. $\endgroup$ Jul 27, 2019 at 21:38
17
$\begingroup$

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?

$\endgroup$
7
  • $\begingroup$ Thank you very much for your answer. I'm interested in a proof of this kind for the following reason: much of what one does with tensors and the covariant derivative is strictly algebraic (as it depends mostly on the algebraic properties required of the objects involved, and not of their specific construction in terms of the manifold itself). Therefore, I'd like to flesh out as clearly as possible those aspects of the covariant derivative and of the curvature tensor that are strictly dependent on algebraic properties (and can thus be generalized to more abstract settings). $\endgroup$ Oct 25, 2015 at 18:51
  • $\begingroup$ in your book, pp. 124, i'm confused that $\nabla _{W}Rm\left( Z,V,X,Y\right)=\nabla _{W}⟨R(Z,V) X,Y⟩=⟨ \nabla _{W}\nabla _{Z}\nabla _{V}X-\nabla _{W}\nabla _{V}\nabla _{Z}X,Y⟩=0$ due to $\nabla _{V}X$, $\nabla _{Z}X=0$, but $\nabla Rm$ is not a zero tensor, what is wrong here? $\endgroup$
    – user360777
    Feb 1, 2017 at 11:58
  • $\begingroup$ @user360777: I'm not sure I understand the question. $Rm$ is not a zero tensor, but $Rm(Z,V,X,Y)$ is; but what does that have to do with it? The first equality above is just the definition of $Rm$, and the second is a few steps combined: replace $R(Z,V)X$ by its definition, using the fact that $[Z,V]\equiv 0$; expand $\nabla_W\langle \cdot,\cdot\rangle$ using compatibility with the metric, and finally use the fact that $\nabla_W Y = 0$ at $p$. The equations $\nabla_V X =0$ and $\nabla_Z X=0$ don't enter into it at this stage. (I probably should expand on this computation in the next edition.) $\endgroup$
    – Jack Lee
    Feb 1, 2017 at 15:09
  • 1
    $\begingroup$ @JackLee Thanks a lot! I understand your proof, but why can't $\nabla_V X=0$ and $\nabla_Z X=0$ enter into $⟨\nabla _{W}\nabla _{Z}\nabla _{V}X-\nabla _{W}\nabla _{V}\nabla _{Z}X,Y⟩$ and $\nabla _{W}Rm(Z,V,X,Y)$ is not necessarily 0? $\endgroup$
    – user360777
    Feb 1, 2017 at 16:16
  • 1
    $\begingroup$ $\nabla_V X$ and $\nabla_Z X$ are zero only at the point $p$. Their derivatives may not be zero. $\endgroup$
    – Jack Lee
    Feb 1, 2017 at 20:16
5
$\begingroup$

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.
$\endgroup$
3
  • $\begingroup$ Thank you very much. Indeed, what I wanted was a "lean" proof, using nothing more than the symmetries and the definitions. I do believe that going for such proofs is the better way of understanding any mathematical subject. $\endgroup$ Oct 11, 2017 at 17:28
  • $\begingroup$ I really like the link by Petersen, but can you explain how they concluded that $$\nabla^2_{X,Y}s= (\nabla_X\nabla)_Ys+\nabla_Y(\nabla_X s)$$? I don't follow how this comes about? $\endgroup$ Mar 5, 2022 at 13:50
  • $\begingroup$ Just after the sentence, "The two concepts are related by"? Combine the two previous equations. $\endgroup$
    – Paul Bryan
    Mar 5, 2022 at 22:33

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .