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In Peter Topping's text, Lectures on Ricci Flow, he's establishing an identity regarding the Laplacian of the curvature tensor. He quotes the differential Bianchi identity

$$\nabla_iR_{jkla}+\nabla_jR_{kila}+\nabla_kR_{ijla}=0$$

and then say that by taking another derivative and tracing we get

$$\Delta R_{jkla}+\nabla_i\nabla_jR_{kila}-\nabla_i\nabla_k R_{jila}=0$$ where

$$\Delta R_{jkla}=\Delta RM(\partial_j,\partial_k,\partial_l,\partial_a)$$

I don't see how he gets this identity. It seems that if he's tracing over the elements then there should be some Ricci coefficients. Can someone provide some details with regards to the computation. Thanks.

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Just apply $\nabla_i$ and sum over $i$ to get $$\nabla_i \nabla_iR_{jkla}+\nabla_i \nabla_jR_{kila}+\nabla_i \nabla_kR_{ijla}=0.$$ Recognising $\nabla_i \nabla_i = \Delta$ (since Topping must be working in an orthonormal frame) and $R_{ijla} = -R_{jila}$ gives you your result. There are no Ricci terms because at least one of the indices $i$ we are tracing over appears as a derivative.

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