Proof of the differential Bianchi identity

Question

I was trying to prove the differential Bianchi identity by applying the covariant derivatives to each of the Riemann tensor terms

$R^{\lambda}_{\sigma\mu\nu;\rho}+R^{\lambda}_{\sigma\nu\rho;\mu}+R^{\lambda}_{\sigma\rho\mu;\nu}=0\space\space\space\space\space\space(1)$

and I got here:

$R^{\lambda}_{\sigma\mu\nu;\rho}=R^{\lambda}_{\sigma\mu\nu,\rho}+\Gamma^{\lambda}_{m\rho}R^{m}_{\sigma\mu\nu}-\Gamma^{m}_{\sigma\rho}R^{\lambda}_{m\mu\nu}-\Gamma^{m}_{\mu\rho}R^{\lambda}_{\sigma m\nu}-\Gamma^{m}_{\nu\rho}R^{\lambda}_{\sigma\mu m}\space\space\space\space\space(2)$

$R^{\lambda}_{\sigma\nu\rho;\mu}=R^{\lambda}_{\sigma\nu\rho,\mu}+\Gamma^{\lambda}_{m\mu}R^{m}_{\sigma\nu\rho}-\Gamma^{m}_{\sigma\mu}R^{\lambda}_{m\nu\rho}-\Gamma^{m}_{\nu\mu}R^{\lambda}_{\sigma m\rho}-\Gamma^{m}_{\rho\mu}R^{\lambda}_{\sigma\nu m}\space\space\space\space\space(3)$

$R^{\lambda}_{\sigma\rho\mu;\nu}=R^{\lambda}_{\sigma\rho\mu,\nu}+\Gamma^{\lambda}_{m\nu}R^{m}_{\sigma\rho\mu}-\Gamma^{m}_{\sigma\nu}R^{\lambda}_{m\rho\mu}-\Gamma^{m}_{\rho\nu}R^{\lambda}_{\sigma m\mu}-\Gamma^{m}_{\mu\nu}R^{\lambda}_{\sigma\rho m}\space\space\space\space\space\space(4)$

I know that using the torsion free property and the symmetries of the Riemann tensor, the last two terms of each of the equations (2),(3) and (4) when summed over cancel each other out. I dont know what to do from here in order to complete the proof.

Answer 1

Here is another proof. The covariant derivatives satisfy the Jacobi identity $$ [\nabla_\mu,[\nabla_\nu,\nabla_\kappa]]+ [\nabla_\nu,[\nabla_\kappa,\nabla_\mu]]+[\nabla_\kappa,[\nabla_\mu,\nabla_\nu]]=0. $$ This can be verified directly, but it is also known that pretty much any associative algebra will satisfy the Jacobi identity, and the elements $\nabla_1,...,\nabla_n$ basically generate a formal associative algebra.

Then letting the Jacobi identity act on any vector field $X^\rho$ we get for one of the terms $$ [\nabla_\kappa,[\nabla_\mu,\nabla_\nu]]X^\rho=\nabla_\kappa[\nabla_\mu,\nabla_\nu]X^\rho-[\nabla_\mu,\nabla_\nu]\nabla_\kappa X^\rho=\nabla_\kappa(R^\rho_{\ \sigma\mu\nu}X^\sigma)-R^\rho_{\ \sigma\mu\nu}\nabla_\kappa X^\sigma + R^\sigma_{\ \kappa\mu\nu}\nabla_\sigma X^\rho \\ =\nabla_\kappa R^\rho_{\ \sigma\mu\nu}X^\sigma+R^\rho_{\ \sigma\mu\nu}\nabla_\kappa X^\sigma-R^\rho_{\ \sigma\mu\nu}\nabla_\kappa X^\sigma+R^\sigma_{\ \kappa\mu\nu}\nabla_\sigma X^\rho \\ = \nabla_\kappa R^\rho_{\ \sigma\mu\nu}X^\sigma+R^\sigma_{\ \kappa\mu\nu}\nabla_\sigma X^\rho. $$

Now writing this into the Jacobi identity gives $$ 0=\left[\nabla_\kappa R^\rho_{\ \sigma\mu\nu}+\text{ cyclic permutations on }\kappa,\mu,\nu\right] X^\sigma+\left[R^\sigma_{\ \kappa\mu\nu}+\text{ cyclic permutations}\right]\nabla_\sigma X^\rho. $$

The second term here vanishes identically because of the algebraic Bianchi identity (cyclic identity), and what we are left with is the differential Bianchi identity.

Alternatively, this can be taken to be a proof of both the differential and algebraic Bianchi identity, since at any point $x$ one may take $$ X^\sigma(x)=\delta^\sigma_\alpha,\quad \nabla_\sigma X^\rho(x)=0, $$ which gives the differential Bianchi identity, and take $$ X^\sigma(x)=0,\quad \nabla_\sigma X^\rho(x)=\delta^\rho_\sigma, $$ which gives the algebraic Bianchi identity.

In this proof I have assumed torsionlessness, but the generalization to the torsionful case is similar, though more laborous.

Answer 2

You can show that in two steps.

1. Show that if a Lorentz tensor vanishes in one Lorentz frame, it vanishes in all Lorentz frames. This is quite simple so I won't do it.

2. This means that you switch to any frame and calculate this identity since what you want to get is zero, which, if true, is valid in all reference frames. Thus, switch to the normal coordinates $g_{\mu\nu} = \eta_{\mu\nu}$ in which case the $\Gamma_{\mu\nu}^\rho$ vanishes but its derivative doesn't. Then, covariant derivative on the Riemann tensor simply read

$\nabla_\lambda R_{\mu\nu\rho\sigma} =\frac12 \partial_\lambda \left(\partial_\mu \partial_\sigma g_{\nu\rho} - \partial_\mu \partial_\rho g_{\nu\sigma} - \partial_\nu \partial_\sigma g_{\mu\rho} + \partial_\nu \partial_\rho g_{\mu\sigma}\right) $

Finally, if you can perform the cycle of the indices and do the math, you get zero. Since the answer is zero in this frame, it holds in general and the Bianchi identity is satisfied.