Hamiltonian phase, complete-domain flow and reference measure — Examples

Choose the normalized three-cut one-spacer Chacon transformation supplied by [F2], on its completed Lebesgue probability space. The exact convention in [F1] matches the supplier's absolute-Cesàro definition, so this system is weakly mixing.

By [F3], the correlations for the fixed pair $(E,E)$ stay at least $2\mathrm{/}81$ above zero along $h_r\to \mathrm{\infty }$. Strong mixing in [F1] would force these same correlations to tend to zero, which they do not. Hence this probability-preserving system is not strongly mixing and refutes the statement.

Define $H(q,p)={\sum }_{i=1}^2(p_i^2\mathrm{/}(2m_i)+m_i{\omega }_i^2{q}_i^2\mathrm{/}2)$ on ${\mathbb{R}}^4$. The Hamilton equations are ${\dot{q}}_i=p_i\mathrm{/}{m}_i$ and ${\dot{p}}_i=-m_i{\omega }_i^2{q}_i$, whose explicit oscillatory solutions exist for every real time. For $J=(0,\mathrm{\infty })$, $\mathrm{\nabla }H$ vanishes only at the origin and hence nowhere on $H^{-1}(J)$. For every compact $K\subset J$, $H^{-1}(K)$ is closed and bounded because $H$ is a positive-definite quadratic form, hence compact in ${\mathbb{R}}^4$. The level ${\mathrm{\Sigma }}_E=H^{-1}(\{E\})$ is therefore a nonempty compact regular shell with the exact hypotheses of [F2].

Let $E_1(q_1,p_1)=p_1^2\mathrm{/}(2m_1)+m_1{\omega }_1^2{q}_1^2\mathrm{/}2$ and put $S=\{z\in {\mathrm{\Sigma }}_E:E_1(z)<E\mathrm{/}2\}$. The equations in step 1.1 conserve $E_1$ because $\frac{d}{dt}{E}_1=(p_1\mathrm{/}{m}_1)(-m_1{\omega }_1^2{q}_1)+(m_1{\omega }_1^2{q}_1)(p_1\mathrm{/}{m}_1)=0$. Hence every time map satisfies ${\mathrm{\Phi }}_{\mathrm{\Delta }}^{-1}S=S$. The set $S$ is a nonempty relatively open subset of ${\mathrm{\Sigma }}_E$; for instance points with $(q_1,p_1)=(0,0)$ lie in it. Its complement contains a nonempty relatively open subset with $E_1>E\mathrm{/}2$. The regular-shell graph density $\mathrm{\mid }\mathrm{\nabla }H{\mathrm{\mid }}^{-1}dA$ is positive on every graph chart, so both open subsets have positive shell measure. Since the whole shell measure is finite and positive by [F2], the normalized probability of $S$ lies strictly between zero and one.

The invariant measurable set $S$ has intermediate probability, so [F4] rules out ergodicity of ${\mathrm{\Phi }}_{\mathrm{\Delta }}$ for every $\mathrm{\Delta }>0$. By [F2], the normalized shell law is a finite invariant probability; apply [F3] to that same law and sampled map to obtain almost-everywhere recurrence. This is a recurrent Hamiltonian system that is not ergodic, refuting the statement.

Use the Hamiltonian $H(\theta ,I)=\nu I$, with $\theta$ dimensionless, $I$ in J s, and $\nu$ in s${}^{-1}$, so $H$ is in J. The convention in [F1] gives $\dot{\theta }=\nu$ and $\dot{I}=0$. Thus ${\mathrm{\Phi }}_t(\theta ,I)=(\theta +\nu t\mathrm{m}\mathrm{o}\mathrm{d}1,I)$ is complete. The smooth involution $\mathcal{J}(\theta ,I)=(-\theta \mathrm{m}\mathrm{o}\mathrm{d}1,I)$ satisfies ${\mathcal{J}}^{\ast }(d\theta \wedge dI)=-d\theta \wedge dI$ and $\mathcal{J}{\mathrm{\Phi }}_t\mathcal{J}(\theta ,I)=(\theta -\nu t\mathrm{m}\mathrm{o}\mathrm{d}1,I)={\mathrm{\Phi }}_{-t}(\theta ,I)$, proving reversibility in [A2]. Use the open invariant domain $R=S^1\times (I_0,I_0+a_0)\subset T^{\ast }{S}^1$; it is preserved by $\mathcal{J}$, complete under the flow, and its reference measure has total mass $a_0^{-1}{\int }_0^{1}d\theta {\int }_{I_0}^{I_0+a_0}dI=1$. Define on this open domain the density ${\rho }_0=4{\mathbf{1}}_{[0,1\mathrm{/}4)\times (I_0,I_0+a_0)}$ relative to $m_0{\mathrm{\mid }}_R$. Its integral is $4\cdot (1\mathrm{/}4)\cdot (a_0\mathrm{/}{a}_0)=1$; excluding the two action endpoints changes no reference-measure mass.

Partition the open domain into $C_1=[0,1\mathrm{/}2)\times (I_0,I_0+a_0)$ and $C_2=[1\mathrm{/}2,1)\times (I_0,I_0+a_0)$. The transported density has angular support $[\nu t,\nu t+1\mathrm{/}4)$ modulo one. At $t=0$ its cell probabilities are $(1,0)$. At $t=3\mathrm{/}(8\nu )$ its support is $[3\mathrm{/}8,5\mathrm{/}8)$, so each cell receives angular length $1\mathrm{/}8$ and the probabilities are $(1\mathrm{/}2,1\mathrm{/}2)$. At $t=1\mathrm{/}(2\nu )$ its support is $[1\mathrm{/}2,3\mathrm{/}4)$, so the probabilities are $(0,1)$.

For any fixed $k>0$, the coarse Shannon entropy $S_C(t)=-k{\sum }_{j=1}^2{p}_j(t)\log p_j(t)$, with $0\log 0=0$, takes the values $0$, $k\log 2$, and $0$ at those three times. It therefore decreases between $3\mathrm{/}(8\nu )$ and $1\mathrm{/}(2\nu )$, contradicting the proposed monotonicity. Here $k$ is a mathematical scale; choosing $k=k_B$ gives the usual entropy units J/K without changing the calculation.

The initial density satisfies ${\int }_R\mathrm{\mid }{\rho }_0\log {\rho }_0\mathrm{\mid }d{m}_0=4\log 4\cdot m_0(\mathrm{supp}{\rho }_0)=\log 4<\mathrm{\infty }$. By [F2] its fine reference-measure entropy remains constant, even though the displayed coarse partition entropy changes.