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Terence Tao in his lecture notes on Ricci flow has written:

If we are to find a scale-invariant (and diffeomorphism-invariant) monotone quantity for Ricci flow, it had better be constant on the gradient shrinking soliton. In analogy with $\frac{d}{dt}\mathcal{F}_m=2\int_M |Ric+Hess(f)|^2dm$, we would therefore like the variation of this monotone quantity with respect to Ricci flow to look something like

$$2\int_M|Ric+Hess(f)-\frac{1}{2\tau}g|^2dm (*)$$

where $\tau$ is some quantity decreasing at the constant rate $\dot{\tau}=-1.$

But the scaling is wrong; time has dimension $2$ with respect to the Ricci flow scaling, and so the dimension of a variation of a scale-invariant quantity should be $-2$, while the expression $(*)$ has dimension $-4$. (Note that $f$ should be dimensionless (up to logarithms), $\tau$ has the same dimension of time, i.e. $2$, and $\int_Mdm=1$ is of course dimensionless.

Question:

In the note what is the meaning of this sentence: time has dimension $2$ with respect to the Ricci flow scaling?

Thanks

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For the Ricci flow, as for any second order parabolic equation, time scales like distance squared. In your formula a factor tau is missing. –  Robert Haslhofer May 24 '13 at 12:33
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Btw, I don't think this question is very suitable for Mathoverflow. –  Robert Haslhofer May 24 '13 at 12:34
    
crosspost: math.stackexchange.com/questions/401201/… –  Sepideh Bakhoda May 24 '13 at 13:51

1 Answer 1

Let $u(x,t)$ be a solution to the heat equation $\frac{\partial u}{\partial t}=\Delta u=\sum_{i=1}^{n}\frac{\partial^{2}u}{\partial x_{i}^{2}}$. We have the scaling of space-time (domain parabolic homothety) symmetry that $u_{\lambda}\doteqdot u\circ\lambda^{\lozenge}$, where $\lambda^{\lozenge }(x,t)=(\lambda^{-1}x,\lambda^{-2}t)\doteqdot(x_{\lambda},t_{\lambda})$, is also a solution for any $\lambda>0$ since $\Delta u_{\lambda}(x,t)=\lambda ^{-2}\Delta u(x_{\lambda},t_{\lambda})$ and $\frac{\partial u_{\lambda} }{\partial t}(x,t)=\lambda^{-2}\frac{\partial u}{\partial t}(x_{\lambda },t_{\lambda})$. We have the range scaling symmetry that $u^{\lambda} \doteqdot\lambda u$ is a solution.

Consider the gradient quantity $Q_{1}[u]=t|\nabla u|^{2}+\frac{1}{2}u^{2}$, which satisfies $(\frac{\partial}{\partial t}-\Delta)Q_{1}=-2t|D^{2}u|^{2} \leq0$. If the maximum principle holds, then $\left\vert \nabla u\right\vert ^{2}\leq\frac{1}{2t}(u^{2})_{\max}(0)$ for $t>0$ (Bernstein estimate). Using $t|\nabla u_{\lambda}|^{2}(x,t)=t_{\lambda}|\nabla u|^{2}(x_{\lambda },t_{\lambda})$ and $u_{\lambda}^{2}(x,t)=u^{2}(x_{\lambda},t_{\lambda})$ (scale-invariance), we have $Q_{1}[u_{\lambda}](x,t)=Q_{1}[u](x_{\lambda },t_{\lambda})$. \ Note also that $Q_{1}[u^{\lambda}]=\lambda^{2}Q_{1}[u]$. The two terms $t|\nabla u|^{2}$ and $\frac{1}{2}u^{2}$ comprising $Q_{1}[u]$ scale the same.

Consider the Ricci flow $\frac{\partial}{\partial t}g=-2\operatorname{Ric} _{g}$ on $M^{n}$. Note that $\operatorname{Ric}_{\lambda^{2}g} =\operatorname{Ric}_{g}$, so that the range scaling symmetry is $\frac {\partial}{\partial(\lambda^{2}t)}(\lambda^{2}g)=-2\operatorname{Ric} _{\lambda^{2}g}$, equivalently, $g_{\lambda}(t)\doteqdot\lambda^{2} g(\lambda^{-2}t)$ is also a solution. The domain scaling symmetry is general covariance: if $\varphi$ is a diffeomorphism of $M$, then $\frac{\partial }{\partial t}\varphi^{\ast}g=-2\operatorname{Ric}_{\varphi^{\ast}g}$. For a frame $\{e_{i}(t)\}_{i=1}^{n}$ at a point $x$, if $\frac{\partial}{\partial t}e_{i}=\operatorname{Ric}^{g}(e_{i})$, where $\operatorname{Ric}^{g} :T_{x}M\rightarrow T_{x}M$, then $\frac{\partial}{\partial t}g(e_{i},e_{j} )=0$. Note $\operatorname{Ric}^{\lambda^{2}g}=\lambda^{-2}\operatorname{Ric} ^{g}$. Define $e_{i}^{\lambda}(t)=\lambda^{-1}e_{i}(\lambda^{-2}t)$. Then $\frac{\partial}{\partial t}g_{\lambda}(e_{i}^{\lambda},e_{j}^{\lambda})=0$ and $\frac{\partial}{\partial t}e_{i}^{\lambda}=\operatorname{Ric} ^{g_{\lambda}}(e_{i}^{\lambda})$.

The associated evolution of the scalar curvature is $\frac{\partial R_{g} }{\partial t}=\Delta_{g}R_{g}+2|\operatorname{Ric}_{g}|_{g}^{2}$. Under $g\mapsto\lambda^{2}g$ and $t\mapsto\lambda^{2}t$, each term scales the same: $\frac{\partial R_{\lambda^{2}g}}{\partial(\lambda^{2}t)}=\lambda^{-4} \frac{\partial R_{g}}{\partial t}$, $\Delta_{\lambda^{2}g}R_{\lambda^{2} g}=\lambda^{-4}\Delta_{g}R_{g}$ and $|\operatorname{Ric}_{\lambda^{2} g}|_{\lambda^{2}g}^{2}=\lambda^{-4}\left\vert \operatorname{Ric} _{g}\right\vert _{g}^{2}$. A special case of Hamilton's trace Harnack is $Q_{2}[g(t)]=\frac{\partial\,R}{\partial\,t}+\frac{R}{t}-\frac{1} {2}\operatorname{Ric}^{-1}(\nabla R,\nabla R)\geq0$ assuming $g$ is complete, $\operatorname{Rc}>0$ and $0\leq\operatorname{Rm}$ bounded. Under $g\mapsto\lambda^{2}g$ and $t\mapsto\lambda^{2}t$, each term in the expression for $Q_{2}$ scales by multiplication by $\lambda^{-4}$.

In practice, we can use the following mnemonic: If $g\propto\lambda^{2}$, then $t\propto\lambda^{2}$, $g^{-1}\propto\lambda^{-2}$, $\operatorname{Ric} _{g}\propto\lambda^{0}$, $R_{g}\propto\lambda^{-2}$, $\left\vert \operatorname{Ric}_{g}\right\vert _{g}\propto\lambda^{-2}$, $\nabla_{g} \propto\lambda^{0}$, $\Delta_{g}=\operatorname{tr}_{g}\nabla_{g}^{2} \propto\lambda^{-2}$, etc.; if $T$ is a rank $k$ tensor and $T\propto \lambda^{s}$, then $|T|_{g}\propto\lambda^{s-k}$. For example, $\operatorname{Rm}_{g}$ has rank $4$ and $\operatorname{Rm}_{g}\propto \lambda^{2}$, so $|\operatorname{Rm}_{g}|_{g}\propto\lambda^{-2}$ and $t^{m}| \nabla_{g}^{m}\operatorname{Rm}_{g}|_{g} ^{2}\propto\lambda^{-4}$ (used in the Bernstein-Bando-Shi estimates). See Section 17 of Hamilton's 'Three-manifolds with positive Ricci curvature' paper for a discussion of scaling and the normalized Ricci flow; there, the scaling degree of a tensor is $\frac{1}{2}$ the power of $\lambda$.

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