Id like to show this form of the Bianchi identity from do Carmo using normal coordinates. (I am aware one can do this with properties of the curvature tensor and connection by reasoning with operators or via a geodesic frame. But for my own practice I am trying to work through local coordinates).
$$
\nabla R(𝑋,𝑌,𝑍,𝑊,𝑇) + \nabla R(𝑋,𝑌,𝑊,𝑇,𝑍) + \nabla R(𝑋,𝑌,𝑇,𝑍,𝑊)=0.
$$
But it doesn't seem to be working as since $X$ and $Y$ are in the first slots, when I try to reduce everything to second derivatives of Christoffel symbols, it seems I don't have the correct indices to make cancellations via Clairaut's theorem.
I am confused as I remember the 2nd Bianchi identity as an identity about the 1 4 Riemann tensor was fairly easy to prove this way.
Any tips?
Also, is $R(X,Y,Z,W,T) = \nabla(T)\langle R(X,Y)Z,W\rangle$ here?
Best Answer
First of all, we need to know that $$R_{ijk}^{\phantom{ijk}\ell} = \partial_i\Gamma_{jk}^\ell - \partial_j\Gamma_{ik}^\ell + \Gamma\Gamma + \Gamma\Gamma.$$We will not bother with the placement of indices on the $\Gamma\Gamma$ terms. What matters is that if we're dealing with normal coordinates centered at the point $p$, then $\Gamma(p)= 0$ and, at $p$ only, covariant derivatives reduce to partial derivatives. So $R_{ijk}^{\phantom{ijk}\ell}(p) = (\partial_i\Gamma_{jk}^\ell)(p) - (\partial_j\Gamma_{ik}^\ell)(p)$ implies that $$\begin{split} R_{ijk}^{\phantom{ijk}\ell}(p) + R_{jki}^{\phantom{ijk}\ell}(p) + R_{kij}^{\phantom{kij}\ell}(p) &= (\partial_i\Gamma_{jk}^\ell)(p) - (\partial_j\Gamma_{ik}^\ell)(p) \\ &\quad+ (\partial_j\Gamma_{ki}^\ell)(p) - (\partial_k\Gamma_{ji}^\ell)(p) \\ &\quad+ (\partial_k\Gamma_{ij}^\ell)(p) - (\partial_i\Gamma_{kj}^\ell)(p) \\ &= 0\end{split}$$since the six terms cancel in pairs (first and last terms, second and third terms, fourth and sixth ones). As $p$ was arbitrary, this proves the first Bianchi identity, and the exact same argument proves the second Bianchi identity. Namely, we have that $$(\nabla_mR_{ijk}^{\phantom{ijk}\ell})(p) = (\partial_mR_{ijk}^{\phantom{ijk}\ell})(p) = (\partial_m\partial_i\Gamma_{jk}^\ell)(p) - (\partial_m\partial_j\Gamma_{ik}^\ell)(p),$$and so
$$\begin{split} (\nabla_mR_{ijk}^{\phantom{ijk}\ell})(p) + (\nabla_iR_{jmk}^{\phantom{ijk}\ell})(p) + (\nabla_jR_{mik}^{\phantom{ijk}\ell})(p) &= (\partial_m\partial_i\Gamma_{jk}^\ell)(p) - (\partial_m\partial_j\Gamma_{ik}^\ell)(p) \\ &\quad + (\partial_i\partial_j\Gamma_{mk}^\ell)(p) - (\partial_i\partial_m\Gamma_{jk}^\ell)(p) \\ &\quad+ (\partial_j\partial_m\Gamma_{ik}^\ell)(p) - (\partial_j\partial_i\Gamma_{mk}^\ell)(p) \\ &=0,\end{split}$$ since the first and fourth terms cancel each other, and similarly for second and fifth, and for the third and last terms.
TL;DR: The power of normal coordinates is to ignore Christoffel symbols, but this should be done carefully and one point at a time.