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Let $R'$ be a tensor of order 4 in a riemannian manifold $M$ defined by: $R'(W,Z,X,Y)=\langle W,X \rangle \langle Z,Y\rangle - \langle Z,X\rangle \langle W,Y\rangle $

And let $R$ be the curvature tensor of $M$, if we have $R=KR'$, how to conclude that $\nabla_U R=(UK)R'$?

By definition: $\nabla_U R(W,Z,X,Y)=\nabla R(W,Z,X,Y,U)=U(R(W,Z,X,Y))-R(\nabla_UW,Z,X,Y)- \ldots$ and stuck here.

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I would use the product rule and then try to show that $\nabla_UR' = 0$ instead of mucking with $\nabla_UR$ ... – Neal Jun 6 '12 at 3:03

1 Answer 1

Expanding on my comment ... your recollection of the definition of the covariant derivative of a tensor is correct. However, instead of trying to compute $\nabla_UR$, I would try to work with $\nabla_UR'$: $$\begin{eqnarray}(\nabla_UR')(W,Z,X,Y) &= U(R'(W,Z,X,Y))\\ &- R'(\nabla_UW,Z,X,Y)\\ &- R'(W,\nabla_UZ,X,Y)\\ &- R'(U,Z,\nabla_UX,Y) \\ &- R'(U,Z,X,\nabla_UY).\end{eqnarray}$$ Now use the product rule, the definition of $R'$, and the fact that the LCC is a metric connection to finish the proof ...

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