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How to derivate (covariant derivative) the expressions $R\cdot Ric$ and $Ric^2$ where $Ric^2$ means $Ric \circ Ric$? Here, $Ric$ is the Ricci tensor seen as a operator and $R$ is the scalar curvature of a Riemannian manifold.

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We want to compute $$(\nabla(R\cdot Rc))(X)$$ Connections commute with contractions, so we start by considering the expression (where we contract $X$ with $Rc$, obtaining $Rc(X)$) $$\nabla(R\cdot Rc\otimes X) = (\nabla (R\cdot Rc))\otimes X + R\cdot Rc\otimes (\nabla X)$$ Taking the contraction and moving terms around, we get $$(\nabla(R\cdot Rc))(X) = \nabla(R\cdot Rc(X)) - R\cdot Rc(\nabla X)$$ For the second one, notice that $$Rc\circ Rc = Rc\otimes Rc$$ after contraction, so we proceed as before and start with $$\nabla[Rc\otimes Rc\otimes Y] = [\nabla(Rc\otimes Rc)]\otimes Y + Rc\otimes Rc\otimes \nabla Y$$ We contract and move terms to get $$[\nabla(Rc\circ Rc)](Y) = \nabla(Rc(Rc(Y))) - Rc(Rc(\nabla Y))$$

Now notice that using twice the first equality we derived we obtain $$\nabla(Rc(Rc(Y))) = (\nabla Rc)(Rc(Y)) + Rc((\nabla Rc(Y))) =$$ $$= (\nabla Rc)(Rc(Y)) + Rc((\nabla Rc)(Y) +Rc(\nabla Y)))$$ Putting all together we obtain $$\nabla(Rc\circ Rc) = (\nabla Rc)\circ Rc + Rc\circ(\nabla Rc)$$


This was an explicit derivation of everything. You can also use the various rules on the connections on the bundles associated to the tangent bundle to get directly $$\nabla(R\cdot Rc) = dR\otimes Rc + R\nabla Rc$$ (which agrees with the first result, if you work it out a bit) and $$\nabla(Rc\circ Rc) = \nabla(Rc\otimes Rc) = (\nabla Rc)\otimes Rc + Rc\otimes(\nabla Rc) = (\nabla Rc)\circ Rc + Rc\circ(\nabla Rc)$$ where there is a contraction of the tensors in the intermediate steps.

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  • $\begingroup$ Daniel, as I said, $Ric$ is seen as an operator, that is, as an $(1,1)$ tensor. $\endgroup$ – Myself Jul 19 '13 at 18:08
  • $\begingroup$ @user86970 ah, sorry I misread. I will edit my answer. $\endgroup$ – Daniel Robert-Nicoud Jul 19 '13 at 18:15
  • $\begingroup$ @user86970 Is this what you were trying to obtain? If you want I should be able to do the other one, too. $\endgroup$ – Daniel Robert-Nicoud Jul 19 '13 at 18:21
  • $\begingroup$ Ok Daniel, it was exactly this what I was looking for. I was trying to find $\nabla (R Ric)$ but I was having troubles to interpret the expressions. Please show the other part. $\endgroup$ – Myself Jul 19 '13 at 18:32

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