## Physical entropy production

A simple fact: let $T:X\to X$ be a homeomorphism preserving the measure-class of $\omega$, $J(T^k,x)$ be the Jacobian of $T^k$ at $x$. Then for any sequence $a(k)\to\infty$ with $\sum_k \frac{1}{a(k)}<+\infty$ we have $\limsup_k\frac{J(T^k,x)}{a(k)}=0$ for $\omega$-a.e. $x\in X$. For example $a(k)=e^{t k}$ with $t>0$ or $a(k)=k^{\alpha}$ with $\alpha>1$. This is a direct corollary of Borel-Cantelli property:

Consider the set $E_{k,\delta}=\{x\in X: \frac{J(T^k,x)}{a(k)}\ge\delta\}$. It is easy to see $\omega(E_{k,\delta})\le \frac{1}{\delta\cdot a(k)}$ and hence $\sum\omega(E_{k,\delta})\le \sum\frac{1}{\delta\cdot a(k)}<+\infty$. So $\omega(x\in E_{k,\delta} \text{infinitely often})=0$ for all $\delta>0$

Notes from papers by Jaksic, Pillet, Rey-Bellet, Ruelle and Young.

4. Let $\nu\ll \mu$ be two probability measures on $X$ and $\phi=\frac{d\nu}{d\mu}$ be its density. Then $1=\nu(X)=\mu(\phi)$. Moreover $1/\phi$ is well-defined with respect to $\nu$ and $\nu(1/\phi)=\mu(1)=1$, too.

The relative entropy can be defined as $E(\nu|\mu)=\nu(\log\phi)$ when $\nu\ll\mu$, $+\infty$ otherwise.
Note that $E(\nu|\mu)=\nu(-\log\frac{1}{\phi})\ge-\log\nu(1/\phi)=0$. So the relative entropy is nonnegative.

Convexity: assume $\nu_1,\nu_2\ll\mu$ and $p+q=1$, then $\phi=p\phi_1+q\phi_2$ and $E(p\nu_1+q\nu_2|\mu)\le p\nu_1(\log\phi_1)+q\nu_2(\log\phi_2)$.

3. Let $f:X\to X$ be a homeomorphism, and $\phi$ be a continuous function on $X$. Then a probability $\mu$ on $X$ is said to be a Gibbs measure with respect to $(f,\phi)$, if there exist $C_n(\epsilon)$ with $\displaystyle \lim_{n\to\infty}\frac{\log C_n(\epsilon)}{n}=0$ for all $\epsilon>0$, such that $1/C_n(\epsilon)\le \frac{\mu(B_n(x,\epsilon))}{e^{\phi_n(x)}}\le C_n(\epsilon)$ for all $x\in X$.

Prop. Assume there exists some Gibbs measure $\mu$ for $(f,\phi)$. Then for any continuous function $g$, the moment generating function $\displaystyle e(\theta)=\lim_{n\to\infty}\frac{1}{n}\log\mu(e^{\theta g_n})=P(f,\phi+\theta g)$.
Proof. Let $\epsilon>0$ and $\delta=\text{var}_\epsilon g$. Then $\displaystyle e^{\phi_n(x)+\theta g_n(x)}\overset{C}{\sim} e^{\theta g_n(x)}\cdot\mu(B_n(x,\epsilon))\overset{\delta}{\sim}\int_{B_n(x,\epsilon)}e^{\theta g_n}d\mu$.

Assume $f$ is expansive and $(f,\phi)$ admits a unique equilibrium state, say $\mu_\phi$. Then the pressure function is differentiable at $\phi$ and $\frac{d}{d\theta}|_{\theta=0}P(f,\phi+\theta g)=\mu_\phi(g)$. Then according to Prop., $\displaystyle \mu_\phi(g)=e'(0)=\lim_{n\to\infty}\frac{1}{n}\mu(g_n)=\lim_{n\to\infty}\mu_n(g)$ for all $g$. In particular we see $\mu_n\to\mu_\phi$ for every $(f,\phi)$-Gibbs measure $\mu$.

Remark. The Gartner-Ellis Theorem actually says that for every $(f,\phi)$-Gibbs measure $\mu$, $\nu_{n,x}\to\mu_\phi$ for $\mu$-a.e. $x\in X$.

2. Let $f:(X,\mu)\to (X,\mu)$ be a local homeomorphism preserving the measure-class of $\mu$. Suppose there is an decomposition $X=\bigcup_k X_k$, and the branch inverse $f_k^{-1}:X\to X_k$. Let $J(f)$ be the Jacobian of $f$, and $L=L_0$ be the induced transfer operator: $\displaystyle (L\phi)(x)=\sum_{fy=x}\frac{\phi(y)}{J(f,y)}$. Then for any measurable function $\phi$: $\int L(\phi) d\mu=\int\phi d\mu$.
Proof. $\int L(\phi) d\mu=\sum_k \int\frac{\phi(x_k)}{J(f,x_k)}d\mu(x)=\sum_k \int \phi(x_k) J(f_k^{-1},x)d\mu(x)$
$=\sum_k \int_{X_k}\phi d\mu(x_k)=\int\phi d\mu$.

Moreover for any two functions $\phi,\psi$, we have
$\int (L\phi)\cdot\psi d\mu=\int L(\phi\cdot\psi\circ f)d\mu=\int \phi\cdot\psi\circ f d\mu$.

An $\mu$-abs. cts measure $\nu=h\mu$ is $f$-invariant if and only if $Lh=h$.
Proof. $\int \phi\circ\!f\, d\nu=\int \phi\circ\!f\, h d\mu=\int \phi\cdot (Lh) d\mu=\int\phi\, hd\mu=\int\phi d\nu$.

Then we generalize the transfer operator $L_g:\phi\mapsto (x\mapsto \sum_{fy=x}\frac{e^{g(y)\phi(y)}}{J(f,y)})$. The moment generating function of $g$ with respect to an $f$-invariant measure $\nu=h\mu$ is $e(\theta)=\lim_{n\to\infty}\frac{1}{n}\log\nu(e^{\theta g_n})$. This is related to the transfer operator $L_g$:
$\nu(e^{\theta g_n})=\int e^{\theta g_n}\cdot hd\mu=\int L^n(e^{\theta g_n}\cdot h)d\mu=\mu(L_{\theta g}^n h)$.

In particular, if $L_{\theta g}$ has a spectral gap, then $e(\theta)$ is analytic and leads to large deviation results of $(\nu,g)$.

1. Let $f:M\to M$ be a $C^2$ diffeomorphism, $m$ a pre-chosen smooth probability volume, $J(f,\cdot)$ the Jacobian of $f$ with respect to $m$. The entropy production of $(f,m)$ is $e_f(m)=\int-\log J(f,x)dm(x)\ge-\log\int J(f,x)dm(x)=0$ (that is, entropy is non-decreasing). Moreover, $e_f(m)=0$ if and only if $m$ is $f$-invariant (that is, an equilibrium state).

Ruelle introduced the entropy production for general  $f$-invariant measure $\nu$ as $e_f(\nu)=\int-\log J(f,x)d\nu(x)$. Clearly $e_{f^{-1}}(\mu)=\int-\log J(f^{-1},x)d\nu(x)=\int\log J(f,f^{-1}x)d\nu(x)$ $=\int\log J(f,x)d\nu(x)=-e_f(f)$, since $\mu$ is $f$-invariant. Compare with $e_{f^{-1}}(m)=\int-\log J(f^{-1},x)dm(x)=\int\log J(f,f^{-1}x)dm(x)$ $=\int\log J(f,x)dm(fx)=\int J(f,x)\log J(f,x)dm(x)$.

It is observed in his paper that $e_f(\nu)=-\int\sum_i\lambda_i(x)d\nu(x)=-\sum_i\lambda_i(\nu)$ be Oseledec Multiplicative Ergodic Theorem, and hence independent of the choice of volumes.

Theorem 1.2. (1). Suppose $h_\mu(f)=\sum_i\lambda^+_i(\mu)$, then $e_f(\mu)\ge 0$. (He called such measures SRB there)
(2). Assume $\mu$ is hyperbolic SRB. Then $e_f(\mu)>0$ if and only if $\mu$ is singular.
(3). $m(B(\mu,f))=0$ if $e_f(\mu)<0$.

Proof. (1). $e_f(\mu)=-\Lambda^u(\mu)-\Lambda^s(\mu)=-(h_\mu(f)+\Lambda^s(\mu))\ge0$ by Ruelle entropy inequality.
(2). Note that SRB is equivalent to $\mu^u_x\ll m^u_x$. So $\mu\ll m$ if and only if $\mu^s_x\ll m^s_x$, which also equivalent to (by Ledrappier-Young) $h_\mu(f)=-\Lambda^s(\mu)$ and hence $e_f(\mu)=-\Lambda^u(\mu)-\Lambda^s(\mu)=-h_\mu(f)-\Lambda^s(\mu)=0$.
(3). Pick $N\ge1$. For each $x\in B(\mu,f)$, there exists $n_x\ge N$ with $\displaystyle \frac{1}{n}\sum_{0\le k\le n-1}-\log J(f,f^kx)\le -c=:e_f(\mu)/2<0$. So $B(\mu,f)\subset\bigcup_{n\ge N}A_n$, where $A_n=\{x:J(f^n,x)\ge e^{nc}\}$. Clearly $m(A_n)\le e^{-nc}$ and hence $m(B(\mu,f))\le\sum_{n\ge N}e^{-nc}\to 0$ as $N\to\infty$. In particular $m(B(\mu,f))=0$.