Trying to prove the second Bianchi identity

Question

I'm trying to prove the second Bianchi identity: $$ \nabla_m R^i_{jkl} + \nabla_k R^i_{jlm} + \nabla_l R^i_{jmk} = 0 $$

I tried to prove it using Ricci identity, that relates the Riemann tensor $R$ with $\nabla$ as $$ \left[ \nabla _k ,\nabla _i \right]V^j = R^j_{mki}V^m, $$ if we suppose that the torsion equals zero.

Thanks to it, I get $$ \nabla_m \nabla_k \nabla_l - \nabla_m \nabla_l \nabla_k + \nabla_k \nabla_l \nabla_m - \nabla_k \nabla_m \nabla_l + \nabla_l \nabla_m \nabla_k - \nabla_l \nabla_k \nabla_m $$ and here is where I get stuck. I tried to relate it with the Jacobi identity but I couldn't and I can't see what is the next step.

Thanks.

Answer 1

Recall that in coordinates, we have that $$R_{ijk}^{\phantom{ijk}\ell} = \partial_i\Gamma_{jk}^\ell - \partial_j\Gamma_{ik}^\ell + \Gamma_{\bullet\bullet}^{\bullet}\Gamma_{\bullet\bullet}^{\bullet} - \Gamma_{\bullet\bullet}^{\bullet}\Gamma_{\bullet\bullet}^{\bullet},\tag{1}$$where the placement of indices on each $\Gamma_{\bullet\bullet}^{\bullet}$ is irrelevant. Now, the second Bianchi identity is a tensor identity, so it may established with the aid of normal coordinates centered at arbitrary points $p$. The power of normal coordinates is that, at $p$ (and in general nowhere else), the Christoffel symbols vanish and covariant derivatives reduce to partial derivatives.

With normal coordinates centered at $p$ in place, evaluating $(1)$ at $p$ gives us that $$R_{ijk}^{\phantom{ijk}\ell}(p) = \partial_i\Gamma_{jk}^\ell(p) - \partial_j\Gamma_{ik}^\ell(p),\tag{2}$$from where the first Bianchi identity follows (six terms cancel in pairs). The second Bianchi identity is dealt with in a similar manner: apply $\partial_m$ to $(1)$ to obtain $$\begin{split}\partial_mR_{ijk}^{\phantom{ijk}\ell} = \partial_m&\partial_i\Gamma_{jk}^\ell - \partial_m\partial_j\Gamma_{ik}^\ell \\ &+ (\partial_m\Gamma_{\bullet\bullet}^{\bullet})\Gamma_{\bullet\bullet}^{\bullet} + \Gamma_{\bullet\bullet}^{\bullet}(\partial_m\Gamma_{\bullet\bullet}^{\bullet}) -(\partial_m\Gamma_{\bullet\bullet}^{\bullet})\Gamma_{\bullet\bullet}^{\bullet} - \Gamma_{\bullet\bullet}^{\bullet}(\partial_m\Gamma_{\bullet\bullet}^{\bullet}),\end{split}\tag{3}$$so evaluating $(3)$ at $p$ gives us that $$\nabla_mR_{ijk}^{\phantom{ijk}\ell}(p) = (\partial_m\partial_i\Gamma_{jk}^\ell)(p) - (\partial_m\partial_j\Gamma_{ik}^\ell)(p) \tag{4}$$Hence the second Bianchi identity follows because, again, six terms cancel in pairs (this time using that second order partial derivatives commute).

This strategy also illustrates why there isn't something like a "third Bianchi identity".