Time has dimension $2$ with respect to the Ricci flow scaling 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
 A: 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 $\lambda$.
