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. (Sorry if the question is too trivial.)
 A: Let us start with the simple example of heat equation
$$ \partial_t u(t,x) = \triangle u(t,x) $$
we see that if $u(t,x)$ is a solution, then $u_\lambda(t,x) = u(\lambda^2 t,\lambda x)$ is also a solution, as
$$ (\partial_t u_\lambda)(t,x) = \lambda^2 (\partial_t u)(\lambda^2 t,\lambda x) = \lambda^2 (\triangle u)(\lambda^2 t,\lambda x) = (\triangle u_\lambda)(t,x)~.$$
This shows that if we change $x$ by a scale factor $\lambda$, we need to change $t$ by scale factor $\lambda^2$. So if $x$ has unit "Length", the units for $t$ should be "Length$^2$". 
This fact is usually true for parabolic equations (though nonlinearities sometimes throw a wrench into the proceedings; in those cases it is also important to keep track of the solution $u$'s units; note that in the linear case $u$'s units do not matter, since by linearity they must appear in the same "strength" for all terms). 
For the Ricci flow, consider the metric $g_{\mu\nu}$ as a function in local coordinates. Let the "scaled metric" $g^{(\lambda)}_{\mu\nu}(x) = g_{\mu\nu}(\lambda x)$ and compute the corresponding Christoffel symbols, you will see that $\Gamma^{(\lambda)} = \lambda \Gamma$ and the curvature scales like $\mathrm{Riem}^{(\lambda)} = \lambda^2 \mathrm{Riem}$. Taking the trace does not change the scaling, so we have that the same analysis as in the linear case gives that if $g_{\mu\nu}(t,x)$ solve, in a fixed coordinate system, the Ricci flow, then so would $g^{(\lambda)}_{\mu\nu}(t,x) = g_{\mu\nu}(\lambda^2 t,\lambda x)$. 

Response to followup:
Again let us first start with the simple examples. Think back about classical physics. The distance $x$ an object has traveled has unit "Length". The time $t$ it took to travel has unit "Time". So the "variation of the distance with respect to time" $v = \frac{\mathrm{d}x}{\mathrm{d}t}$ has unit "Length / Time". 
Now in the case of a parabolic equation, suppose $\mathcal{F}$ is something that has no units (a pure number, or in SI the unit of "one"), then the variation of $\mathcal{F}$ with respect to "time", which we recall has unit "Length$^{2}$", would have unit 
$$ \frac{\text{Unit of } \mathcal{F}}{\text{Unit of time}} = \frac{1}{\text{Length}^2} = \text{Length}^{-2} $$
