# Ricci flow as a gradient flow and its Lyapunov function

In study of Ricci flow, for making Ricci flow as a gradient flow I faced $\mathcal{F}(g,f)=\int (R+|\nabla f|^2)e^{-f}$. I know that if we suppose $\frac{df}{dt}=-R$, then $\frac{d}{dt}\mathcal{F}(g,f)=\int \langle-Ric-Hess(f),\dot{g}\rangle e^{-f}dV$. So by definition, gradient of $\mathcal{F}$ is given by $\nabla \mathcal{F}=-Ric-Hess(f)$. In this point we define modified Ricci flow as $\dot{g}=-2(Ric+Hess(f))$, then $\dot{g}=2\nabla\mathcal{F}$.

Question: By Monotonicity of $\mathcal{F}$ we know that $\frac{d}{dt}\mathcal{F}(g,f)\ge0.$ Since $\mathcal{F}$ is Lyapunov function of modified Ricci flow, some equilibrium points of the flow may be unstable. Why don't we define modified Ricci flow as $\dot{g}=-2\nabla\mathcal{F}$? In this case $\frac{d}{dt}\mathcal{F}(g,f)\le0$ and all equilibrium points would be stable. Doesn't this make a better definition?

-
Doesn't that change the PDE from a heat equation to a backwards heat equations? If so, that makes it unlikely to have short time existence. – Ben McKay Apr 28 '13 at 17:29

If $\frac{\partial}{\partial s}g=v$, then $\frac{\partial R}{\partial s}=-\Delta V+\operatorname{div}^{2}v-\left\langle v,\operatorname{Ric} \right\rangle$ and $\frac{\partial}{\partial s}d\mu=\frac{1}{2}Vd\mu$, where $V=\operatorname{tr}_{g}v$. So $$\frac{\partial}{\partial s}(Rd\mu)=(-\Delta V+\operatorname{div}^{2}v+\langle v,\tfrac{R}{2}g-\operatorname{Ric}\rangle)d\mu.$$ Integrating this, we see that the Euler-Lagrange equation for $\int Rd\mu$ is $\tfrac{R}{2}g-\operatorname{Ric}=0$ (Einstein-Hilbert). To get Ricci flow, we want to get rid of the $\tfrac{R}{2}g$ term due to the variation of the volume form. Perelman accomplished this by introducing $f$ with $e^{-f}d\mu$ fixed. Imposing $\frac{\partial}{\partial s}\left( e^{-f}d\mu\right) =0$, i.e., $\frac{\partial f}{\partial s}=\frac{V}{2}$, we obtain $$\frac{d}{ds}\int Re^{-f}d\mu=-\int\left\langle v,\operatorname{Ric} \right\rangle e^{-f}d\mu+\int\left( -\Delta V+\operatorname{div}^{2}v\right) e^{-f}d\mu.$$ One pays the price that the divergence terms no longer integrate to zero. Serendipitously, $$\frac{d}{ds}\int\left\vert \nabla f\right\vert ^{2}e^{-f}d\mu=\int\left( -v\left( \nabla f,\nabla f\right) +\Delta V\right) e^{-f}d\mu,$$ so, by combining the above and integrating by parts, one obtains Perelman's formula: $$\frac{d}{ds}\int\left( R+\left\vert \nabla f\right\vert ^{2}\right) e^{-f}d\mu=-\int\left\langle v,\operatorname{Ric}+\nabla^{2}f\right\rangle e^{-f}d\mu.$$ The gradient flow is $\frac{\partial}{\partial t}g=-2(\operatorname{Ric} +\nabla^{2}f)$, $\frac{\partial f}{\partial t}=-R-\Delta f$. One cannot solve this forward, so one makes the gauge change: $\frac{\partial}{\partial t}g=-2\operatorname{Ric},$ $\frac{\partial f}{\partial t}=-R-\Delta f+\left\vert \nabla f\right\vert ^{2}$ (by adding $\mathcal{L}_{\nabla f}$ to both equations). Since we have decoupled the first equation from the second, we can solve it forward in time (Hamilton-DeTurck). For applications, $f$ is solved backward in time. For geometric flows, the idea of using the backward heat kernel to obtain a monotonicity formula was originally used by Gerhard Huisken for the mean curvature flow and by Michael Struwe for the harmonic map heat flow. Hamilton tried to get this to work for Ricci flow; his interest is evident from his paper with Matt Grayson.