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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

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  • $\begingroup$ You should add some motivation for what you would like in an answer. Is the manifold under consideration "special"? For example, there is a useful formula in the case of Ricci solitons, if $X$ is the soliton vector field. $\endgroup$
    – Otis Chodosh
    Commented Nov 25, 2013 at 18:36
  • $\begingroup$ You're right, I'm sorry. I am in the compact Kahler setting of real dimension $\geq 4$. What is the soliton formula you are talking about? I am indeed in the Kahler-Ricci soliton setting. Thanks $\endgroup$
    – David P
    Commented Nov 25, 2013 at 18:44
  • $\begingroup$ Sorry, I can't find a proper reference off the top of my head (will try to do so later) but you can derive it yourself: in this case $D_X \Ric$ is the same as the time derivative of $\Ric$ under Ricci flow (because $X$ is the soliton potential), so you can find/derive a formula for $D_X \Ric$ using this point of view. $\endgroup$
    – Otis Chodosh
    Commented Nov 25, 2013 at 19:00
  • $\begingroup$ ok, I found somewhere its used: see the first displayed equation in the proof of Proposition 2.3 of Brendle's high dimensional Bryant soliton uniqueness result arxiv.org/pdf/1203.0270v3.pdf. There's no proof, but the proof is exactly what I described above (you should try to work this out yourself, its a very useful thing to understand if you're studying Ricci solitons). $\endgroup$
    – Otis Chodosh
    Commented Nov 25, 2013 at 20:13
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    $\begingroup$ The question arose because I found on a physics forum the formula $\nabla_X Ric = \frac 1 2 X \cdot R g$ which stinks a little to me and I was wondering about it. Probably they meant $\nabla_a R_{ab}$ and not $\nabla_c R_{ab}$. $\endgroup$
    – David P
    Commented Nov 27, 2013 at 11:13

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The comments section was getting unwieldy, so I'll answer here. Hopefully this is helpful.

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 $X$).

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.

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.

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  • $\begingroup$ Do you mean $\frac{d}{dt}Ric_{h(t)} = \Delta_L Ric_{h(t)}$? $\endgroup$
    – David P
    Commented Nov 27, 2013 at 13:54
  • $\begingroup$ There should be an extra term: see Proposition 2.3.7 in Toppings notes: homepages.warwick.ac.uk/~maseq/topping_RF_mar06.pdf $\endgroup$
    – Otis Chodosh
    Commented Nov 27, 2013 at 14:19
  • $\begingroup$ This proposition is about the derivative on a generic curve of metrics, while proposition 2.5.3 says there along the Ricci Flow there is only the Lichnerowicz laplacian. $\endgroup$
    – David P
    Commented Nov 27, 2013 at 14:23
  • $\begingroup$ Ack, sorry, you're right! Note that the $-2<Ric,Ric>$ term in the middle should disappear when you compute $D_X Ric$ (this happens from changing between the Lie derivative and the covariant derivative) $\endgroup$
    – Otis Chodosh
    Commented Nov 27, 2013 at 14:24
  • $\begingroup$ What do you mean? What I get by adding up the pieces is $\Delta R(h)_{ij} - 2 R(h)_i^a R(h)_j^b h_{ab} + R(h)^{ab} R(h)_{iajb} = - \nabla_X - Ric(\nabla_i V, \partial_j) - Ric(\partial_i, \nabla_j V)$. What is cancelled? $\endgroup$
    – David P
    Commented Nov 27, 2013 at 14:42

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