I was wondering, have you ever seen a formula in the Riemannian (more specially Kahlerian but not essential) setting for the derivative $X \cdot |Ric|^2 = 2 g(\nabla\_X Ric, Ric)$ for a vector field $X$?


Or more in general $\nabla\_X Ric$.


Thank you


David



Derivative of (the length of) the Ricci tensor
The comments section was getting unwieldy, so I'll answer here. Hopefully this is helpful.




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What I was trying to say is as follows: suppose that $(M,g,X)$ is a steady gradient soliton, i.e.
$$
\mathcal{L}\_X(g) = 2 Ric\_g
$$
for $X=\nabla f$ for some function $f$. Then, let $\Phi\_t$ denote the flow of $-X$. You may check that $h(t) :=\Phi^\*\_tg$ is a solution to the Ricci flow. Then, the Ricci curvature of $h(t)$ is 
$$
Ric\_{h(t)} = Ric\_{\Phi\_t^\*g} = \Phi\_t^\*Ric\_g.
$$
Thus,
$$
\frac{d}{dt} Ric\_{h(t)} = -\mathcal{L}\_X Ric\_g
$$
Now, relating $\mathcal{L}\_X Ric\_g$ to $D\_X Ric\_g$ in the usual way:
\begin{align\*}
\mathcal{L}\_X Ric(V,W) & = X(Ric(V,W))-Ric([X,V],W)-Ric(V,[X,W])\\
& = D\_X Ric(V,W) +Ric(D\_XV-[X,V],W)+Ric(V,D\_X W -[X,W])\\
& = D\_X Ric(V,W) +Ric(D\_VX,W)+Ric(V,D\_WX)
\end{align\*}
and using the formula for $\frac{d}{dt} Ric\_{h(t)}$ under the Ricci flow, you may find an equation for $D\_X Ric$ (for example, it is the equation found in the paper of Brendle's I linked above; I've matched his sign conventions with $X$). 




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EDIT: I'm going to work out the computation: I have specialized to gradient solitons (I'm not sure what happens if you drop this assumption). 


From Topping's notes <http://homepages.warwick.ac.uk/~maseq/topping_RF_mar06.pdf>, we know that
\begin{align\*}
\frac{d}{dt} Ric\_{h(t)}(V,W) & = \Delta Ric\_{h(t)} (V,W) - 2<Ric\_{h(t)}(V),Ric\_{h(t)}(W)>\_{h(t)} \\
& \qquad + 2<Rm\_{h(t)}(X,\cdot,V,\cdot),Ric\_{h(t)}>\_{h(t)}.
\end{align\*}
If we specialize to $t=0$, then this gives us an equation for $\frac {d}{dt}Ric\_{h(t)}|\_{t=0}$ in terms of $g$-quantities.


On the other hand, notice that (here, we're using the gradient assumption)
$$
DX = D^2f = \frac 12 \mathcal{L}\_X g = Ric\_g.
$$
Thus 
$$
D\_V X= Ric\_g(V).
$$


Now, you can put the $\frac{d}{dt}Ric\_{h(t)}|\_t=0$ computation together with the $\mathcal{L}\_XRic\_g$ computation to see
\begin{align\*}
D\_X Ric\_g (V,W) & = \mathcal{L}\_X Ric(V,W) -Ric\_g(D\_VX,W)-Ric\_g(V,D\_WX)\\
& = -\frac{d}{dt} Ric\_{h(t)}|\_{t=0}(V,W)-2<Ric\_g(V),Ric\_g(W)>\\
& = \Delta Ric\_{g} (V,W) + 2<Rm\_{g}(X,\cdot,V,\cdot),Ric\_{g}>\_{g}.
\end{align\*}


This is the equation that Brendle uses in the linked article. 




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By the way, you can simplify the above argument by using the Uhlenbeck trick (if you work through the details of what I did above, you'll see that there is a good deal of cancellation, which you can exploit by using $D\_{\frac{d}{dt}}$ instead of the time derivative above. See, e.g. ch 6 of this book: <http://download.springer.com/static/pdf/28/bok%253A978-3-642-16286-2.pdf?auth66=1385658655_8c41399e85b51a3addd04347ad203f0d&ext=.pdf> among many other places.


