Suppose $M$ is a Riemannian manifold. Consider flow $\frac{\partial}{\partial t}g_{ij}=-2(R_{ij}+\nabla_i \nabla_j f)$, where $f$ is a time-dependent function. I would like to prove that flows of this form are equivalent, up to diffeomorphism, to the Ricci flow $\frac{\partial}{\partial t}g_{ij}=-2R_{ij}$, that is:

By defining a 1-parameter family of diffeomorphism $\Psi(t):M\to M$ by

$$\frac{d}{dt}\Psi(t)=\nabla_{g(t)}f(t),$$ $$\Psi(0)=id_M$$

I want to show that $\bar{g}(t):=\Psi(t)^*g(t)$ satisfy $$\frac{\partial}{\partial t}\bar{g}_{ij}=-2 \bar{R}_{ij}.$$

My problem is that I don't know how to calculate $\frac{\partial}{\partial t}\Psi(t)^*g(t)$. I know that $\frac{\partial}{\partial t}\Psi(t)^* \alpha=\mathcal{L}_{\nabla f} \alpha$, where $\mathcal{L}$ is Lie derivative and $\alpha$ is a time-independent object, but I faced problem when $\alpha$ is a time-dependent object.

Can someone point me in the right direction? Thanks in advance for your time.


You can get the correct answer by applying the formal Leibniz/product rule to the expression:

\begin{eqnarray*} \frac{\partial}{\partial t}\left(\Psi\left(t\right)^{*}g\left(t\right))\right) & = & \frac{\partial\Psi\left(t\right)^{*}}{\partial t}g\left(t\right)+\Psi\left(t\right)^{*}\frac{\partial g\left(t\right)}{\partial t}\\ & = & \Psi^{*}\left(\mathcal{L}_{\nabla f}g\right)-2\Psi^{*}\left(\textrm{Rc}+\nabla\nabla f\right)\\ & = & \Psi^{*}\left(-2\textrm{Rc}\right). \end{eqnarray*}

This calculation is written out in slightly more detail in Chapter 9.4 of http://maths-people.anu.edu.au/~andrews/book.pdf.

As to why this rule is valid: from the definition of the derivative it's easy enough to arrive at $$ \begin{eqnarray*} \frac{\partial}{\partial t}\left(\Psi\left(t\right)^{*}g\left(t\right)\right) & = & \lim_{h\to0}\frac{1}{h}\left(\Psi\left(t+h\right)^{*}\left(g\left(t+h\right)-g\left(t\right)\right)+\left(\Psi\left(t+h\right)^{*}-\Psi\left(t\right)^{*}\right)g\left(t\right)\right).\\ & = & \lim_{h\to0}\frac{1}{h}\left(\Psi\left(t+h\right)^{*}\left(g\left(t+h\right)-g\left(t\right)\right)\right)+\left(\frac{\partial}{\partial t}\Psi\left(t\right)^{*}\right)g\left(t\right); \end{eqnarray*} $$

so it remains to show that the limit we have left is equal to $\Psi(t)^{*}\left({\partial g}/{\partial t}\right)$; i.e. we want to separate the limit of the "product" into a "product" of limits. We can do this because (vague explanation incoming) a small change in $t$ means $\Psi(t)$ only moves points in the manifold by a small amount, so by continuity of $g$ in space and time and continuous differentiability of $\Psi(t)$ in space there will only be a small change in $\Psi(t)^* g(t)$.

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  • 1
    $\begingroup$ It makes sense. I thought that $\Psi (t)^* g(t)$ has the form of $f(g(t))$ and I must use chain rule for calculating its derivative!! thanks a lot. $\endgroup$ – Sepideh Bakhoda Sep 13 '13 at 11:06
  • $\begingroup$ @AnthonyCarapetis: What is the definition of the derivative? Isn't it $$\frac{\partial}{\partial t}\left(\Psi\left(t\right)^{*}g\left(t\right)\right)=\lim_{h\to0}\frac{1}{h}\biggl(\Psi\left(t+h\right)^{*}g\left(t+h\right)-\Psi\left(t\right)^{*}g\left(t\right)\biggr)?$$ $\endgroup$ – C.F.G Feb 24 at 11:48

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