TITLE: The relations between the Perelman's entropy functional and notions of entropy from statistical mechanics
QUESTION [9 upvotes]: I am looking for the relations and analogies between the Perelman's entropy functional,$\mathcal{W}(g,f,\tau)=\int_M [\tau(|\nabla f|^2+R)+f-n] (4\pi\tau)^{-\frac{n}{2}}e^{-f}dV$, and notions of entropy from statistical mechanics. Would you please explain it in details?

REPLY [10 votes]: For metrics on $S^{2}$ with positive curvature, Hamilton introduced the
entropy $N\left(  g\right)  =-\int\ln(R\operatorname{Area})Rd\mu.$ If the
initial metric has $R>0,$ he proved that this is nondecreasing under the Ricci
flow on surfaces; note that $Rd\mu$ satisfies $(\frac{\partial}{\partial t}-\Delta
)(Rd\mu)=0.$ Let $T$ be the singular time; then
$$
\frac{d}{dt}N\left(  g\left(  t\right)  \right)  =2\int\left\vert
\operatorname{Ric}+\nabla^{2}f-\frac{1}{2\tau}g\right\vert ^{2}d\mu+4\int%
\frac{\left\vert \operatorname{div}(\operatorname{Ric}+\nabla^{2}f-\frac
{1}{2\tau}g)\right\vert ^{2}}{R}d\mu,
$$
where $\tau=T-t$ and $\Delta f=r-R.$ ($f$ satisfies $\frac{\partial
f}{\partial t}=\Delta f+rf$; since $n=2$, $\operatorname{Ric}=\frac{1}{2}Rg$)
Perelman's entropy has the main term: $\int fe^{-f}d\mu,$ which is the
classical entropy with $u=e^{-f}$ as Deane Yang wrote. (Besides Section 5 of
Perelman, further discussion of entropy appeared later in some of Lei Ni's
papers as well as elsewhere.) Even though this term is lower order (in terms
of derivatives), geometrically it is the most significant as can be seen by
taking the test function to be the characteristic function of a ball
(multiplied by a constant for it to satisfy the constraint); technically, one
chooses a cutoff function. Thus Perelman proved finite time no local
collapsing below any given scale only assuming a local upper bound for $R,$
since the local lower Ricci curvature bound (control of volume growth is
needed to handle the cutoff function) can be removed by passing to the
appropriate smaller scale.
Heuristically (ignoring the cutoff issue), since the constraint is $\int(4\pi\tau)^{-n/2}e^{-f}d\mu=1,$ if
we take $\tau=r^{2}$ and $e^{-f}=c\chi_{B_{r}},$ then $c\approx\frac{r^{n}%
}{\operatorname{Vol}B_r}.$ So, if the time and scale are bounded from above, by Perelman's monotonicity, we have $$-C\leq\mathcal{W}(g,f,r^{2})\lessapprox r^{2}%
\max_{B_{r}}R+\ln\frac{\operatorname{Vol}B_r}{r^{n}},$$ yielding the volume ratio lower bound.
Added December 12, 2013:
For all of the following, see Perelman, Ni, Topping, etal. Let $\mathcal{N}
=\int fe^{-f}d\mu$ be the classical entropy. Then, under $\frac{\partial
}{\partial t}g=-2(\operatorname{Ric}+\nabla^{2}f)$ and $\frac{\partial
f}{\partial t}=-\Delta f-R$, we have $-\frac{d\mathcal{N}}{dt}=\mathcal{F}
(g,f)\doteqdot\int(R+\left\vert \nabla f\right\vert ^{2})e^{-f}d\mu$
(Perelman's energy). If the solutions are on $[0,T)$, then $\mathcal{F}
(t)\leq\frac{n}{2\left(  T-t\right)  }\int e^{-f}d\mu$, which implies that
$\frac{d}{dt}(\mathcal{N}-(\frac{n}{2}\int e^{-f}d\mu)\log(T-t))\geq0$. Let
$\mathcal{W}(g,f,\tau)=(4\pi\tau)^{-n/2}\left(  \tau\mathcal{F}+\mathcal{N}
\right)  -n\int(4\pi\tau)^{-n/2}e^{-f}d\mu$ (Perelman's entropy). Under
$\frac{\partial}{\partial t}g=-2\operatorname{Ric}$ and $\frac{\partial
f}{\partial t}=-\Delta f+|\nabla f|^{2}-R+\frac{n}{2\tau}$, we have have
$\frac{d\mathcal{F}}{dt}=2\int|\operatorname{Ric}+\nabla^{2}f|^{2}e^{-f}
d\mu-\frac{n}{2\tau}\mathcal{F}$ and $\frac{d\mathcal{N}}{dt}=-\mathcal{F}
+\frac{n}{2\tau}\int e^{-f}d\mu-\frac{n}{2\tau}\mathcal{N}$. So, by coupling
with $\frac{d\tau}{dt}=-1$, we obtain (Perelman's entropy formula)
\begin{align*}
(4\pi\tau)^{n/2}\frac{d\mathcal{W}}{dt}  & =\frac{n}{2\tau}\left(
\tau\mathcal{F}+\mathcal{N}\right)  -\mathcal{F}+\tau\frac{d\mathcal{F}}
{dt}+\frac{d\mathcal{N}}{dt}\\
& =2\tau\int|\operatorname{Ric}+\nabla^{2}f-\frac{1}{2\tau}g|^{2}e^{-f}d\mu.
\end{align*}