## Regularity of center manifold

Let $X:\mathbb{R}^d\to \mathbb{R}^d$ be a $C^\infty$ vector field with $X(o)=0$. Then the origin $o$ is a fixed point of the generated flow on $\mathbb{R}^d$. Let $T_o\mathbb{R}^d=\mathbb{R}^s\oplus\mathbb{R}^c\oplus\mathbb{R}^u$ be the splitting into stable, center and unstable directions. Moreover, there are three invariant manifolds (at least locally) passing through $o$ and tangent to the corresponding subspaces at $o$.

Theorem (Pliss). For any $n\ge 1$, there exists a $C^n$ center manifold $C^n(o)=W^{c,n}(o)$.

Generally speaking, the size of the center manifold given above depends on the pre-fixed regularity requirement. Theoretically, there may not be a $C^\infty$ center manifold, since $C^n(o)$ could shrink to $o$ as $n\to\infty$. An explicit example was given by van Strien (here). He started with a family of vector fields $X_\mu(x,y)=(x^2-\mu^2, y+x^2-\mu^2)$. It is easy to see that $(\mu,0)$ is a fixed point, with $\lambda_1=2\mu<\lambda_2=1$. The center manifold can be represented (locally) as the graph of $y=f_\mu(x)$.

Lemma. For $n\ge 3$, $\mu=\frac{1}{2n}$, $f_\mu$ is at most $C^{n-1}$ at $(\frac{1}{2n},0)$.

Proof. Suppose $f_\mu$ is $C^{k}$ at $(\frac{1}{2n},0)$, and let $\displaystyle f_\mu(x)=\sum_{i=1}^{k}a_i(x-\mu)^i+o(|x-\mu|^{k})$ be the finite Taylor expansion. The vector field direction $(x^2-\mu^2, y+x^2-\mu^2)$ always coincides with the tangent direction $(1,f'_\mu(x))$ along the graph $(x,f_\mu(x))$, which leads to

$(x^2-\mu^2)f_\mu'(x)=y+x^2-\mu^2=f_\mu(x)+x^2-\mu^2$.

Note that $x^2-\mu^2=(x-\mu)^2+2\mu(x-\mu)$. Then up to an error term $o(|x-\mu|^{k})$, the right-hand side in terms of $(x-\mu)$: $(a_1+2\mu)(x-\mu)+(a_2+1)(x-\mu)^2+\sum_{i=3}^{k}a_i(x-\mu)^i$; while the left-hand side in terms of $(x-\mu)$:

$(x-\mu)^2f_\mu'(x)+2\mu(x-\mu)f_\mu'(x)=\sum_{i=1}^{k}ia_i(x-\mu)^{i+1}+\sum_{i=1}^{k}2\mu i a_i(x-\mu)^i$

$=\sum_{i=2}^{k}(i-1)a_{i-1}(x-\mu)^{i}+\sum_{i=1}^{k}2\mu i a_i(x-\mu)^i$.

So for $i=1$: $2\mu a_1=a_1+2\mu$, $a_1=\frac{-2\mu}{1-2\mu}\sim 0$;

$i=2$: $a_2+1=a_1+4\mu a_2$, $a_2=\frac{a_1-1}{1-4\mu}\sim -1$;

$i=3,\cdots,k$: $a_i=(i-1)a_{i-1}+2i\mu a_i$, $(1-2i\mu)a_i=(i-1)a_{i-1}$.

Note that if $k\ge n$, we evaluate the last equation at $i=n$ to conclude that $a_{n-1}=0$. This will force $a_i=0$ for all $i=n-2,\cdots,2$, which contradicts the second estimate that $a_2\sim -1$. Q.E.D.

Consider the 3D vector field $X(x,y,z)=(x^2-z^2, y+x^2-z^2,0)$. Note that the singular set $S$ are two lines $x=\pm z$, $y=0$ (in particular it contains the origin $O=(0,0,0)$). Note that $D_OX=E_{22}$. Hence a cener manifold $W^c(O)$ through $O$ is tangent to plane $y=0$, and can be represented as $y=f(x,z)$. We claim that $f(x,x)=0$ (at least locally).

Proof of the claim. Suppose on the contrary that $c_n=f(x_n,x_n)\neq0$ for some $x_n\to 0$. Note that $p_n=(x_n,c_n,x_n)\in W^c(O)$, and $W^c(O)$ is flow-invariant. However, there is exactly one flow line passing through $p_n$: the line $L_n=\{(x_n,c_nt,x_n):t>0\}$. Therefore $L_n\subset W^c(O)$, which contradicts the fact that $W^c(O)$ is tangent to plane $y=0$ at $O$. This completes the proof of the claim.

The planes $z=\mu$ are also invariant under the flow. Let’s take the intersection $W_\mu=\{z=\mu\}\cap W^c(O)=\{(x,f(x,\mu),\mu)\}$. Then we check that $\{(x,f(x,\mu))\}$ is a (in fact the) center manifold of the restricted vector field in the plane $z=\mu$. We already checked that $f(x,\mu)$ is not $C^\infty$, so is $W^c(O)$.

## The volume of uniform hyperbolic sets

This is a note of some well known results. The argument here may be new, and may be complete.

Proposition 1. Let $f\in\mathrm{Diff}^2_m(M)$. Then $m(\Lambda)=0$ for every closed, invariant hyperbolic set $\Lambda\neq M$.

See Theorem 15 of Bochi–Viana’s paper. Note that Proposition 1 also applies to Anosov case, in the sense that $m(\Lambda)>0$ implies that $\Lambda=M$ and $f$ is Anosov.

Proof. Suppose $m(\Lambda)>0$ for some hyperbolic set. Then the stable and unstable foliations/laminations are absolutely continuous. Hopf argument shows that $\Lambda$ is (essentially) saturated by stable and unstable manifolds. Being a closed subset, $\Lambda$ is in fact saturated by stable and unstable manifolds, and hence open. So $\Lambda=M$.

Proposition 2. There exists a residual subset $\mathcal{R}\subset \mathrm{Diff}_m^1(M)$, such that for every $f\in\mathcal{R}$, $m(\Lambda)=0$ for every closed, invariant hyperbolic set $\Lambda\neq M$.

Proof. Let $U\subset M$ be an open subset such that $\overline{U}\neq M$, $\Lambda_U(f)=\bigcap_{\mathbb{Z}}f^n\overline{U}$, which is always a closed invariant set (maybe empty). Given $\epsilon>0$, let $\mathcal{D}(U,\epsilon)$ be the set of maps $f\in\mathrm{Diff}_m^1(M)$ that either $\Lambda_U(f)$ is not a uniformly hyperbolic set, or it’s hyperbolic but  $m(\Lambda_U(f))<\epsilon$. It follows from Proposition 1 that $\mathcal{D}(U,\epsilon)$ is dense. We only need to show the openness. Pick an $f\in \mathcal{D}(U,\epsilon)$. Since $m(\Lambda_U(f))<\epsilon$, there exists $N\ge 1$ such that $m(\bigcap_{-N}^N f^n\overline{U})<\epsilon$. So there exists $\mathcal{U}\ni f$ such that $m(\bigcap_{-N}^N g^n\overline{U})<\epsilon$. In particular, $m(\Lambda_U(g))<\epsilon$ for every $g\in \mathcal{U}$. The genericity follows by the countable intersection of the open dense subsets $\mathcal{D}(U_n,1/k)$.

The dissipative version has been obtained in Alves–Pinheiro’s paper

Proposition 3. Let $f\in\mathrm{Diff}^2(M)$. Then $m(\Lambda)=0$ for every closed, transitive hyperbolic set $\Lambda\neq M$. In particular, $m(\Lambda)>0$ implies that $\Lambda=M$ and $f$ is Anosov.

See Theorem 4.11 in R. Bowen’s book when $\Lambda$ is a basic set.

## Doubling map on unit circle

1. Let $\tau:x\mapsto 2x$ be the doubling map on the unit torus. We also consider the uneven doubling $f_a(x)=x/a$ for $0\le x \le a$ and $f(x)=(x-a)/(1-a)$ for $a \le x \le 1$. It is easy to see that the Lebesgue measure $m$ is $f_a$-invariant, ergodic and the metric entropy $h(f_a,m)=\lambda(m)=\int \log f_a'(x) dm(x)=-a\log a-(1-a)\log (1-a)$. In particular, $h(f_a,m)\le h(f_{0.5},m)=\log 2 =h_{\text{top}}(f_a)$ and $h(f_a,m)\to 0$ when $a\to 0$.

2. Following is a theorem of Einsiedler–Fish here.

Proposition. Let $\tau:x\mapsto 2x$ be the doubling map on the unit torus, $\mu$ be an $\tau$-invariant measure with zero entropy. Then for any $\epsilon>0$, $\beta>0$, there exist $\delta_0>0$ and a subset $E\subset \mathbb{T}$ with $\mu(E) > 0$, such that for all $x \in E$, and all $\delta<\delta_0$: $\mu(B(x,\delta))\ge \delta^\beta$.

A trivial observation is $\text{HD}(\mu)=0$, which also follows from general entropy-dimension formula.

Proof. Let $\beta$ and $\epsilon$ be fixed. Consider the generating partition $\xi=\{I_0, I_1\}$, and its refinements $\xi_n=\{I_\omega: \omega\in\{0,1\}^n\}$ (separated by $k\cdot 2^{-n}$)….

Furstenberg introduced the following notation in 1967

Definition. A multiplicative semigroup $\Sigma\subset\mathbb{N}$ is lacunary, if $\Sigma\subset \{a^n: n\ge1\}$ for some integer $a$. Otherwise, $\Sigma$ is non-lacunary.

Example. Both $\{2^n: n\ge1\}$ and $\{3^n: n\ge1\}$ are lacunary semigroups. $\{2^m\cdot 3^n: m,n\ge1\}$ is a non-lacunary semigroup.

Theorem. Let $\Sigma\subset\mathbb{N}$ be a non-lacunary semigroup, and enumerated increasingly by $s_i > s_{i+1}\cdot$. Then $\frac{s_{i+1}}{s_i}\to 1$.

Example. $\Sigma=\{2^m\cdot 3^n: m,n\ge1\}$. It is equivalent to show $\{m\log 2+ n\log 3: m,n\ge1\}$ has smaller and smaller steps.

Theorem. Let $\Sigma\subset\mathbb{N}$ be a non-lacunary semigroup, and $A\subset \mathbb{T}$ be $\Sigma$-invariant. If $0$ is not isolated in $A$, then $A=\mathbb{T}$.

Furstenberg Theorem. Let $\Sigma\subset\mathbb{N}$ be a non-lacunary semigroup, and $\alpha\in \mathbb{T}\backslash \mathbb{Q}$. Then $\Sigma\alpha$ is dense in $\mathbb{T}$.

In the same paper, Furstenberg also made the following conjecture: a $\Sigma$-invariant ergodic measure is either supported on a finite orbit, or is the Lebesgue measure.

A countable group $G$ is said to be amenable, if it contains at least one Følner sequence. For example, any abelian countable group is amenable. Note that for amenable group action $G\ni g:X\to X$, there always exists invariant measures and the decomposition into ergodic measures. More importantly, the generic point can be defined by averaging along the Følner sequences, and almost every point is a generic point for an ergodic measure. In a preprint, the author had an interesting idea: to prove Furstenberg conjecture, it suffices to show that every irrational number is a generic point of the Lebesgue measure. Then any other non-atomic ergodic measures, if exist, will be starving to death since there is no generic point for them :)

## Some simple dynamical systems

Dynamical formulation of Prisoner’s dilemma
Originally, consider the two players, each has a set of stratagies, say $\mathcal{A}=\{a_{i}\}$ and $\mathcal{B}=\{b_{j}\}$. The pay-off $P_k=P_k(a_{i},b_{j})$ for player $k$ depends on the choices of both players.

Now consider two dynamical systems $(M_i,f_i)$. The set of stratagies consists of the invariant probability measures, and the pay-off functions can be

$\phi_k(\mu_1,\mu_2)=\int \Phi_k(x,y)d\mu_1 d\mu_2$, where $\mu_i\in\mathcal{M}(f_i)$;

$\psi_k(\mu_1,\mu_2)=\int \Phi_k(x,y)d\mu_1 d\mu_2-h(f_i,\mu_i)$.

The frist one is related to Ergodic optimization. The second one does sound better, since one may want to avoid a complicated (measured by its entropy) stratagy that has the same $\phi$ pay-off.

Gambler’s Ruin Problem
A gambler starts with an initial fortune of $i, and then either wins$1 (with $p$) or loses $1 (with $q=1-p$) on each successive gamble (independent of the past). Let $S_n$ denote the total fortune after the n-th gamble. Given $N>i$, the gambler stops either when $S_n=0$ (broke), or $S_n=N$ (win), whichever happens first. Let $\tau$ be the stopping time and $P_i(N)=P(S_\tau=N)$ be the probability that the gambler wins. It is easy to see that $P_0(N)=0$ and $P_N(N)=1$. We need to figure out $P_i(N)$ for all $i=1,\cdots,N-1$. Let $S_0=i$, and $S_n=S_{n-1}+X_n$. There are two cases according to $X_1$: $X_1=1$ (prob $p$): win eventually with prob $P_{i+1}(N)$; $X_1=-1$ (prob $q$): win eventually with prob $P_{i-1}(N)$. So $P_i(N)=p\cdot P_{i+1}(N)+q\cdot P_{i-1}(N)$, or equivalently, $p\cdot (P_{i+1}(N)-P_i(N))=q\cdot (P_i(N)-P_{i-1}(N))$ (since $p+q=1$), $i=1,\cdots,N-1$. Recall that $P_0(N)=0$ and $P_N(N)=1$. Therefore $P_{i+1}(N)-P_i(N)=\frac{q^i}{p^i}(P_1(N)-P_{0}(N))=\frac{q^i}{p^i}P_1(N)$, $i=1,\cdots,N-1$. Summing over $i$, we get $1-P_1(N)=P_1(N)\cdot\sum_{1}^{N-1}\frac{q^i}{p^i}$, $P_1(N)=\frac{1}{\sum_{0}^{N-1}\frac{q^i}{p^i}}=\frac{1-q/p}{1-q^N/p^N}$ (if $p\neq .5$) and $P_1(N)=\frac{1}{N}$ (if $p= .5$). Generally $P_i(N)=P_1(N)\cdots\sum_{0}^{i-1}\frac{q^j}{p^j}=\frac{1-q^i/p^i}{1-q^N/p^N}$ (if $p\neq .5$) and $P_1(N)=\frac{i}{N}$ (if $p= .5$). Observe that for fixed $i$, the limit $P_i(\infty)=1-q^i/p^i>0$ only when $p>.5$, and $P_i(\infty)=0$ whenever $p\le .5$. Finite Blaschke products Let $f$ be an analytic function on the unit disc $\mathbb{D}=\{z\in\mathbb{C}: |z|<1 \}$ with a continuous extension to $\overline{\mathbb{D}}$ with $f(S^1)\subset S^1$. Then $f$ is of the form $\displaystyle f(z)=\zeta\cdot\prod_{i=1}^n\left({{z-a_i}\over {1-\bar{a_i}z}}\right)^{m_i}$, where $\zeta\in S^1$, and $m_i$ is the multiplicity of the zero $a_i\in \mathbb{D}$ of $f$. Such $f$ is called a finite Blaschke product. Proposition. Let $f$ be a finite Blaschke product. Then the restriction $f:S^1\to S^1$ is measure-preserving if and only if $f(0)=0$. That is, $a_i=0$ for some $i$. Proof. Let $\phi$ be an analytic function on $\overline{\mathbb{D}}$. Then $\int_{S_1}\phi d(\theta)=\phi(0)$ and $\int_{S_1}\phi\circ f d(\theta)=\phi\circ f(0)$. Significance: there are a lot of measure-preserving covering maps on $S^1$. Kalikov’s Random Walk Random Scenery Let $X=\{1,-1\}^{\mathbb{Z}}$, and $\sigma:X\to X$ to the shift $\sigma((x_n))=(x_{n+1})$. More generally, let $A$ be a finite alphabet and $p$ be probability vector on $A$, and $Y=A^{\mathbb{Z}}$, $\nu=p^{\times\mathbb{Z}}$. Consider the skew-product $T:X\times Y\to X\times Y$, $(x,y)\mapsto (\sigma x, \sigma^{x_0}y)$. It is clear that $T$ preserves any $\mu\times \nu$, where $\mu$ is $\sigma$-invariant. Proposition. Let $\mu=(.5,.5)^{\times\mathbb{Z}}$. Then $h(T,\mu\times \nu)=h(\sigma,\mu)=\log 2$ for all $(A,p)$. Proof. Note that $T^n(x,y)\mapsto (\sigma^n x, \sigma^{x_0+\cdot+x_{n-1}}y)$. CLT tells that $\mu(x:|x_0+\cdot+x_{n-1}|\ge\kappa\cdot \sqrt{n})< \delta(\kappa)$ as $n\to\infty$, where $\delta(\kappa)\to 0$ as $\kappa\to\infty$. There are only $2^{n+\kappa \sqrt{n}}$ different $n$-strings (up to an error). Significance: this gives a natural family of examples that are K, but not isomorphic to Bernoulli. Creation of one sink. 1D case. Consider the family $f_t:x\mapsto x^2+2-t$, where $0\le t\le 2$. Let $t_\ast$ the first parameter such that the graph is tangent to the diagonal at $x_\ast=f_{t_\ast}(x_\ast)$. Note that $x_\ast$ is parabolic. Then for $t\in(t_\ast,t_\ast+\epsilon)$, $f_t(x)=x$ has two solutions $x_1(t), where $x_1(t)$ is a sink, and $x_2(t)$ is a source. 2D case. Let $B=[-1,1]\times[-\epsilon,\epsilon]$ be a rectangle, $f$ be a diffeomorphism such that $f(B)$ is a horseshoe lying above $B$ of shape ‘V’. Moreover we assume $|\det Df|<1$. Let $f_t(x,y)=f(x,y)-(0,t)$ such that $f_1(B)$ is the regular horseshoe intersection: V . Clearly there exists a fixed point $p_1$ of $f_1$ in $B$. We assume $\lambda_1(1)<-1<\lambda_2(1)<0$. Then Robinson proved that $f_t$ admits a fixed point in $B$ which is a sink. First note that for any $t$, and any fixed point of $f_t$ (if exists), it is not on the boundary of $B$. Since $p_1$ is a nondegenerate fixed point of $f_1$, the fixed point continues to exist for some open interval $(t_1,1)$ (assume it is maximal, and denote the fixed point by $p_t$). Clearly $t_1>0$. Note that $p_{t_1}$ is also fixed by $f_{t_1}$, since it is a closed property. If there is some moment with $\lambda_1(t)=\lambda_2(t)$ for the fixed point $p_t$ of $f_t$, then it is already a sink, since $\det Df=\lambda_1\cdot\lambda_2<1$. So in the following we consider the case $\lambda_1\neq\lambda_2$ for all $p_t$, $t\in[t_1,1]$. Then the continuous dependence of parameters implies that both are continuous functions of $t$. The fixed point $p_{t_1}$ must be degenerate, since the fixed point ceases to exist beyond $t_1$, which means: $\lambda_i(t_1)=1$ for some $i\in\{1,2\}$. Case 1. $\lambda_1(t_1)=1$. Note that $\lambda_1(1)<-1$. So $\text{Re}\lambda_1(t_\ast)=0$ for some $t_\ast\in(t_1,1)$, which implies that $\lambda_1(t_\ast)=ai$ for some $a\neq 0$. In particular, $\lambda_2(t_\ast)=-ai$, and $a^2=|\det Df|<1$. So $p_{t_\ast}$ is a (complex) sink. Case 2. $\lambda_2(t_1)=1$. Note that $\lambda_2(1)<0$. Similarly $\text{Re}\lambda_2(t_\ast)=0$ for some $t_\ast\in(t_1,1)$. So in the orientation-preserving case there always exists a complex sink. In the orientation-reversing case ($\lambda_2(1)\in(0,1)$), we need modify the argument for case 2: Case 2′. $\lambda_2(t_1)=1$. Note that $\lambda_2(1)\in(0,1)$. So $|\lambda_1(t_\ast)|<1$ for some $t_\ast\in(t_1,1)$. We pick $t_\ast$ close to $t_1$ in the sense that $|\lambda_1(t_\ast)|>|\det Df|$, which implies $|\lambda_1(t_\ast)|\ast<1$, too. So $p_{t_\ast}$ is also a sink. ## Playing pool with pi This is a short note based on the paper Playing pool with π (the number π from a billiard point of view) by G. Galperin in 2003. Let’s start with two hard balls, denoted by $B_1$ and $B_2$, of masses $0 on the positive real axis with position $0, and a rigid wall at the origin. Without loss of generality we assume $m=1$. Then push the ball $B_2$ towards $B_1$, and count the total number $N(M)$ of collisions (ball-ball and ball-wall) till the $B_2$ escapes to $\infty$ faster than $B_1$. Case. $M=1$: first collision at $y(t)=x$, then $B_2$ rests, and $B_1$ move towards the wall; second collision at $x(t)=0$, then $B_1$ gains the opposite velocity and moves back to $B_2$; third collision at $x(t)=x$, then $B_1$ rests, and $B_2$ move towards $\infty$. Total counts $N(1)=3$, which happens to be first integral part of $\pi$. Well, this must be coincidence, one might wonder. However, Galperin proved that, if we set $M=10^{2k}$, then $N(M)$ gives the integral part of $10^k\pi$. For example, $N(10^2)=31$; and $N(10^4)=314$. ## Notes-09-14 4. Borel–Cantelli Lemma(s). Let $(X,\mathcal{X},\mu)$ be a probability space. Then If $\sum_n \mu(A_n)<\infty$, then $\mu(x\in A_n \text{ infinitely often})=0$. If $A_n$ are independent and $\sum_n \mu(A_n)=\infty$, then for $\mu$-a.e. $x$, $\frac{1}{\mu(A_1)+\cdots+\mu(A_n)}\cdot|\{1\le k\le n:x\in A_k\}|\to 1$. The dynamical version often involves the orbits of points, instead of the static points. In particular, let $T$ be a measure-preserving map on $(X,\mathcal{X},\mu)$. Then $\{A_n\}$ is said to be a Borel–Cantelli sequence with respect to $(T,\mu)$ if $\mu(T^n x\in A_n \text{ infinitely often})=1$; $\{A_n\}$ is said to be a strong Borel–Cantelli sequence if $\frac{1}{\mu(A_1)+\cdots+\mu(A_n)}\cdot|\{1\le k\le n:T^k x\in A_k\}|\to 1$ for $\mu$-a.e. $x$. 3. Let $H(q,p,t)$ be a Hamiltonian function, $S(q,t)$ be the generating function in the sense that $\frac{\partial S}{\partial q_i}=p_i$. Then the Hamilton–Jacobi equation is a first-order, non-linear partial differential equation $H + \frac{\partial S}{\partial t}=0$. Note that the total derivative $\frac{dS}{dt}=\sum_i\frac{\partial S}{\partial q_i}\dot q_i+\frac{\partial S}{\partial t}=\sum_i p_i\dot q_i-H=L$. Therefore, $S=\int L$ is the classical action function (up to an undetermined constant). 2. Let $\gamma_s(t)$ be a family of geodesic on a Riemannian manifold $M$. Then $J(t)=\frac{\partial }{\partial s}|_{s=0} \gamma_s(t)$ defines a vector field along $\gamma(t)=\gamma_0(t)$, which is called a Jacobi field. $J(t)$ describes the behavior of the geodesics in an infinitesimal neighborhood of a given geodesic $\gamma$. Alternatively, A vector field $J(t)$ along a geodesic $\gamma$ is said to be a Jacobi field, if it satisfies the Jacobi equation: $\frac{D^2}{dt^2}J(t)+R(J(t),\dot\gamma(t))\dot\gamma(t)=0,$ where $D$ denotes the covariant derivative with respect to the Levi-Civita connection, and $R$ the Riemann curvature tensor on $M$. ## Exponential map on the complex plane Let $f(z)=e^z=e^x(\cos y+\i\sin y)$ (for $z=x+\i y$) be the exponential map. Note that $f(x)>0$ for all real numbers and $f^{n+1}(x):=f(f^nx)$ goes to $\infty$ really fast: the dynamics of $f$ on $\mathbb{R}$ is trivial. But the dynamics of $f$ on $\mathbb{C}$ is completely different. First note that $e^{2k\pi\i}=1$: the map is not a diffeomorphism, but a covering map branching at the origin. The following theorem was conjectured by Fatou (1926) and proved by Misiurewicz (1981). Theorem (Orbits of the complex exponential map). Let $\mathcal{O}_e(z)$ be the orbit of a point $z\in\mathbb{C}$ under the iterates of $f(z)=e^z$. Then each of the following sets is dense in the complex plane: 1. the basin of $\infty$, $B_e(\infty)=\{z\in\mathbb{C}: f^n(z)\to\infty\}$; 2. the set of transitive points, $\text{Tran}(e)=\{z\in\mathbb{C}: \mathcal{O}_e(z)\text{ is dense}\}$; 3. the set of periodic points, $\text{Per}(e)=\{z\in\mathbb{C}: \mathcal{O}_e(z)\text{ is finite}\}$. So the exponential map is chaotic on the complex plane. Reference: The exponential map is chaotic: An invitation to transcendental dynamics, Zhaiming Shen and Lasse Rempe-Gillen arXiv ## Hegselmann-Krause system There is an interesting preprint today on arxiv by Sascha Kurz: How long does it take to consensus in the Hegselmann-Krause model? http://arxiv.org/abs/1405.5757 Consider $k$-particle system with total energy $E$, and let $\delta=\delta(E)$ be the range of interactions. The evolution of energy distribution $x=(x_i):\mathbb{N}\to \mathbb{R}^k$ is defined by the following algorithm: 1. set the initial energies $x_i(0)\ge 0$ with $E=\sum_{i} x_i(0)$. 2. as time develops, $x_i$ interacts (i.e. exchanges energy) only with the particles with similar energies: let $N_i(0)=\{1\le j\le k: |x_j(0)-x_i(0)|\le \delta\}\ni i$ and $n_i(0)=|N_i(0)|\ge 1$. Then set $x_i(1)=\frac{1}{n_i(0)}\sum_{j\in N_i(0)}x_j(0)$. 3. Suppose we have defined $x_i(n)$. Then let $N_i(n)=\{1\le j\le k: |x_j(n)-x_i(n)|\le 1\}\ni i$ and $n_i(n)=|N_i(n)|\ge 1$. Then set $x_i(n+1)=\frac{1}{n_i(n)}\sum_{j\in N_i(n)}x_j(n)$. • Note that the total energy $E$ is preserved, and the system is scaling invariant. So we can assume $\delta=1$. Then the system can be viewed as a map on the simplex $\Delta_k=\{x=(x_i)\in\mathbb{R}^k:x_i\ge0\text{ and }E=\sum_i x_i\}$. • Note that the order is preserved during the process: $x_i(0)\le x_{i+1}(0)$ implies $x_i(n)\le x_{i+1}(n)$ for all $n$. In particular, if $x_i(n)= x_{i+1}(n)$ for some $n$, then these two particles are identical in the system. So we can replace them by a single particle but with weight $2$. In general we can consider the weighted system $\Omega_k=\{(x_i,k_i):x_i\ge0, k=\sum_i k_i\text{ and }E=\sum_i k_ix_i\}$ (mass conservation and energy conservation). How to characterize the dynamics on $\Omega_k$? • A simple test example is $x_i(0)=i$ for $i=1,\cdots k$. 1. $k=1$: trivial system. 2. $k=2$: $x_1(0)=1$ and $x_2(0)=2$. Then $x_i(n)=1.5$ for all $i=1,2$ and all $n\ge 1$. All $x_i(n)$ reach the same energy at time $T(2)=1$. 3. $k=3$: $x_i(0)=i$. Then $x_1(1)=1.5$, $x_2(1)=2$ and $x_3(1)=2.5$. Now note that $N_i(1)=\{1,2,3\}$ for all $i$. So $x_1(n)=2$ for all $i=1,2$ and all $n\ge 2$. Then reach the same energy at time $T(3)=2$. The set $N_i(n)$ plays an important role. One can check that $T(4)=5$ and $T(5)=6$: the small groups can always reach the same energy. But the case $k=6$ is different: • $X(0)=\langle 1,2,3,4,5,6 \rangle$ • $X(1)=\langle \frac{3}{2},2,3,4,5,\frac{11}{2} \rangle$ • $X(2)=\langle 1\frac{3}{4},2\frac{1}{6},3,4,4\frac{5}{6},5\frac{1}{4} \rangle$ • $X(3)=\langle 1\frac{23}{24},2\frac{11}{36},3\frac{1}{18},3\frac{17}{18},4\frac{25}{36},5\frac{1}{24} \rangle$ • Eventually this leads to two different clusters $x_i(6)=A$ for $i=1,2,3$ and $x_i(6)=B$ for $i=4,5,6$ with large energy difference $B-A > 1$. The two groups of particles never intersect with the other group and will stay in this status forever. So we set $T(6)=6$. There is no general formula for $T(k)$. It is proved that $T(k)=O(k^3)$, and conjectured that $T(k)=O(k)$, and the worse scenario is given the initial data $x_i(0)=i$ for $i=1,\cdots, k$. ## Continuous time Markov process Let $\Sigma=\{1,\cdots, d\}^{\mathbb{N}}$, $B:\Sigma\to\mathbb{R}$ be a Lipschitz potential, and $L_B(f):x\mapsto \sum_{\sigma y=x} e^{B(y)}f(y)$. The potential $B$ is said to be normalized, if $\sum_{\sigma y=x} e^{B(y)}=1$ for all $x\in\Sigma$. If $B$ is normalized, then its topological pressure $P(\sigma, B)=0$, and its equilibrium state $\mu_B$ is an $L_B^\ast$-invariant Gibbe measure. This induces a Markov process $(X_t)$ with values on the state space $\Sigma$. That is, suppose $X_t=x\in\Sigma$. Then it stays at this state for a while, waits for $T\sim \text{Exp}(1)$ and jumps to a point $X_{t+T}=y\in\sigma^{-1}x$ with probability $e^{B(y)}$. Then $\mu_B$ is a stationary measure for this Markov process. More generally, we can assign different jump rates (exponential clocks) at different states. That is, let $r:\Sigma\to[c,C]$. Let $(X_t^r)$ be the modified Markov process with clock $r$. That is, suppose $X_t^r=x\in\Sigma$. Then it waits a time $T\sim \text{Exp}(r(x))$ and jumps to a point $X_{t+T}^r=y\in\sigma^{-1}x$ with probability $e^{B(y)}$. Then the naturally related measure is $\mu^r_B:E\mapsto \frac{1}{\mu_B(1/r)}\cdot\int_E \frac{1}{d}d\mu_B$. Another setting is consider a system of $N$ sites, each carrying an energy $x_i$. Assume the neighboring sites $s_i$ and $s_{i+1}$ exchange energy when an exponential clock $\text{Exp}(\lambda_i(x_i+x_{i+1}))$ rings: $(\hat x_i,\hat x_{i+1})=(a,1-a)(x_i+x_{i+1})$, where$\alpha\sim U([0,1])\$.
Now consider a function $f:\Sigma\to\mathbb{R}$, and its evolution $f(X_t)$ with $X_0=x$. Then the generator $L$ is defined by
$Lf:x\mapsto \mathbb{E}_x\lim_{t\to 0+}\frac{f(X_t)}{t} =\sum_{i}\lim_{t\to 0+}\frac{1}{t}\mathbb{P}(T_t(i))\cdot \mathbb{E}_x(f(X_t)-f(x)|T_t(i))$,
where $T_t(i)$ describes the event that only the i-th clock rings during the time $(0,t)$. For independent exponential clocks, we have
$\mathbb{P}(T_t(i))=e^{-\lambda_i t}\lambda_i t\cdot\prod_{j\neq i}e^{-\lambda_j t}$,
and
$\mathbb{E}_x(f(X_t)-f(x)|T_t(i))=\int_I [f(T_{ia}x)-f(x)]\cdot U(da)$. So
$Lf: x\mapsto\sum_i \lambda_i(x_i+x_{i+1})\cdot\int_I [f(T_{ia}x)-f(x)]\cdot U(da)$

## Asymmetry of Bowen’s dimensional entropy

1. Bowen and Dinaburg gave a alternative definition of topological entropy $h_{\text{top}}(f)$ by calculating the exponential growth rate of the $(n,\epsilon)$-covers. This definition resembles the box dimension of Euclidean subset $E\subset\mathbb{R}^k$, and gives the same value while using the definition given by Adler, Konheim, and McAndrew. In particular, the entropy is time-reversal invariant: $h_{\text{top}}(f^{-1})=h_{\text{top}}(f)$.

2. Later Bowen introduced another definition of topological entropy for noncompact subset in 1973, which resembles the Hausdorff dimension.
Let $f:X\to X$ be a homeomorphism on a compact metric space, $E\subset X$ and $h_B(f,E)$ be Bowen’s topological entropy of $E$ (may not be compact).

Bowen proved that, for any ergodic measure $\mu$, $h_B(f,G_{\mu})=h(f,\mu)$, where $G_{\mu}$ is the set of $\mu$-generic points. This identity has been generalized to general invariant measures of transitive Anosov systems:

Theorem 1. (Pfister–Sullivan link) Let $f:M\to M$ be a transitive Anosov diffeomorphism. Then $h_B(f,G_{\mu})=h(f,\mu)$ for any invariant measure $\mu$.

Note that $\mu(G_\mu)=0$ whenever $\mu$ is invariant but non-ergodic.

3. An interesting fact is that $h_B(f,E)$ may not be time-reversal invariant.

Example 2. Let $f:M\to M$ be a transitive Anosov diffeomorphism, $p$ be a periodic point, $D=W^u(x,\epsilon)$. Then $h_B(f,D) > 0$, but $h_B(f^{-1},D)=0$.

Now let $\mu,\nu$ be two different invariant measures of $f$, $W^s(\mu,f)=G_\mu$ be the set of $\mu$-generic points with respect to $f$, and $W^u(\nu,f)=W^s(\nu,f^{-1})$ be the set of $\mu$-generic points with respect to $f^{-1}$. Let $H_f(\mu,\nu)=B^s(\mu,f)\cap B^u(\nu,f)$ (resemble the heteroclinic intersection of different saddles). Then it is proved (Proposition D in here) that

Proposition 3. Let $f:M\to M$ be a transitive Anosov diffeomorphism. Then $h_B(f,H_f(\mu,\nu))=h_\mu(f)$ and $h_B(f^{-1},H_f(\mu,\nu))=h_\nu(f)$.

A well known fact is that, for any $0\le t\le h_{\text{top}}(f)$, there exists some invariant measure $\mu$ with $h_\mu(f)=t$. So a direct corollary of Proposition 3 is:

Corollary. Let $f:M\to M$ be a transitive Anosov diffeomorphism. Then for any $a, b\in [0, h_{\text{top}}(f)]$, there exists an invariant subset $E$ such that $h_B(f,E)=a$ and $h_B(f^{-1},E)=b$.