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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). $$ \nabla R(,,,,) + \nabla R(,,,,) + \nabla R(,,,,)=0. $$ 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?

Didier
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Integral fan
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  • I fixed the latex formatting but I think there was a typo in the final line and think you meant "Is $\nabla R (X,Y,Z,W,T) = \nabla_T\langle R(X,Y)Z,W\rangle$?" right? – Didier Dec 05 '23 at 17:55
  • Thank you. Yes that's correct. And I realize now the notational confusion. The covariant derivative of R is a series of (0 ,5) tensors beginning with what's suggested as equality. Of course the following subtractions where the nabla goes inside will vanish when I use normal coordinates. So using metric compatibility on the proposed widget it seems one can reduce to the Bianchi identity where one differentiates outputs of the curvature endomorphism. – Integral fan Dec 05 '23 at 19:12
  • Or actually one can skip that last suggestion. So I get $R_{ijkl;m}+R_{ijlm;k}+R_{ijmk;l} = 0$ and this indeed how wikipedia lists the identity. So I guess my issue is just showing this via Clairaut now. – Integral fan Dec 05 '23 at 19:36
  • It seems one must use some algebraic symmetries of the Riemann tensor now to make the indices play right. So I flipped the indices on each Riemann tensor component using the first pair switches with second pair symmetry. – Integral fan Dec 05 '23 at 20:00

1 Answers1

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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.

Ivo Terek
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  • Thank you for this nice answer. I was also able to do it for the zero-four case using this strategy. – Integral fan Dec 06 '23 at 02:30
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    Yes! Since for normal coordinates we have that $g_{ij}(p) = \delta_{ij}$ and $(\partial_kg_{ij})(p) = 0$ for all indices, you can lower the index $\ell$ on every single equality I wrote (except for the very first one) for free. – Ivo Terek Dec 06 '23 at 02:34