Perron numbers

Consider a monic polynomial with integer coefficients: $p(x)=x^d + a_1 x^{d-1} + \cdots + a_{d-1}x + a_d$, $a_j \in \mathbb{Z}$.
The complex roots of such polynomials are called algebraic integers. For example, integers and the roots of integers are algebraic integers. Note that the Galois conjugates of an algebraic integer are also algebraic integers.

Consider a square matrix $A$ with positive integer entries. By Perron-Frobenius Theorem, there is a unique positive eigenvalue $\lambda >0$. Moreover, it admits an eigenvector of all positive entries, and satisfies $\lambda > |\lambda'|$ for any other eigenvalue $\lambda'$ of $A$. In particular, $\lambda > |\lambda_{\sigma}|$ for any of its Galois conjugates. Such a number is called a Perron number. More generally, a weak Perron number is a real algebraic integer whose modulus is greater than or equal to that of all of its Galois conjugates.

Let $f$ be a continuous map on the interval $I=[0,1]$, and $h_{\text{top}}(f)$ the topological entropy. Assume $f$ is postcritically finite: $\text{PC}(f)=\{f^n c: c\in \text{Crit}(f)\}$ is a finite set. Then the partition $\alpha$ of $I$ along the postcritical set $\text{PC}(f)$ is a Markov partition for $f$, since (1) the endpoints are sent to endpoints, (2) every folding corresponds to a critical point of $f$. Therefore, $\lambda(f)=e^{h_{\text{top}}(f)}$ is the leading eigenvalue of the incidence matrix $A_{f,\alpha}$ associated to the Markov partition $\alpha$. It follows that $\lambda(f)$ is a weak Perron number.

Surfaces with boundaries

Let $M$ be a compact surface with boundary. Then its 2nd homology group $H_2(M,\mathbb{R})=0$, since there is no nontrivial 2-cycle: for any triangulation $\{c_j\}$ of $M$, $\partial (\sum a_j c_j)=0$ if and only if $a_j=0$ for all $j$. It follows that the 2nd de Rham cohomology group $H^2_{dR}(M,\mathbb{R}) \simeq H_2(M,\mathbb{R})=0$. Therefore, every 2-form on $M$ (automatically closed) is exact. In particular, the area form $\omega$ on $M$ satisfies $\omega = d\lambda$ for some 1-form $\lambda$ on $M$: any symplectic surface with boundary is exact.

Recall that for a general symplectic manifold $(M,\omega)$, the one forms $\alpha$ are in 1-1 correspondence to vector fields $V:M \to TM$ via $\alpha=i_V\omega$. Then $\omega=d\lambda$ being exact is equivalent to the existence of a Liouville vector field — a vector field $V$ satisfying $L_V\omega = d\circ i_V\omega = d\lambda =\omega$.

Topological groups

Let $G$ be a group and $\mathscr{T}$ be a topology on $G$. Then $G$ is said to be a topological group if the two structures on $G$ are compatible: both $G\times G \to G, (x, y) \mapsto x\cdot y$ and $G\to G, x\mapsto x^{-1}$ are continuous.
(1) any topological group can be made Hausdorff by taking an appropriate canonical quotient.

(2) any group with the discrete topology is a topological group: it is just an ordinary group.

We will assume all group topologies are Hausdorff and non-discrete. More generally, a group is topologizable if it admits a non-discrete Hausdorff topology. The following is based on the article.

[Markov 1944] A subset $S\subset G$ is unconditionally closed (unc-closed) if it is closed in every Hausdorff group topology on $G$. For example, if $S=G\backslash \{e\}$ is unc-closed, then for any group topology on $G$, $\{ e\}$ is an open set. Then the topology must be discrete. So $S=G\backslash \{e\}$ is unc-closed if and only if $G$ is non-topologizable.

The collection of unc-closed sets form a topology on $G$, which is called the Markov topology $\mathscr{M}_G$. Note that $\mathscr{M}_G$ is $T_1$, but not necessarily Hausdorff. We can reformulate the above observation as: $G$ is non-topologizable if and only if $\mathscr{M}_G$ is the discrete topology.

[Markov] A subset $E\subset G$ is elementary algebraic if there exist $n\ge 1$, $a_j\in G$ and $\epsilon_j \in \{-1, 1\}$ such that $E=\{x\in G| x^{\epsilon_1}a_1\cdots x^{\epsilon_n}a_n =1\}$. Then a subset $F \subset G$ is algebraic if it can be written as an (arbitrary) intersection of finite unions of elementary algebraic subsets of $G$. This resembles the property of being closed sets. [1976, 1999] The Zariski topology $\mathscr{Z}_G$ is the topology in which closed sets are precisely the algebraic subsets of $G$. For example $c(g)=\{x\in G| xgx^{-1}g^{-1}=1\}$ is an ele-alg subset. So the center $Z(G)=\bigcap_G c(g)$ is an algebraic subset. It is easy to see that algebraic sets are unc-closed. Therefore, $\mathscr{Z}_G$ is coarser than $\mathscr{M}_G$.

A natural question is: when does $\mathscr{Z}_G=\mathscr{M}_G$? Markov proved it for countable groups, and Perel’man proved it for abelian groups. There are groups for which $\mathscr{Z}_G \subsetneq\mathscr{M}_G$.

Hausdorff topology

Let $X$ be a set, $\mathscr{T}$ be a topology on $X$. Then the topology is $T_1$ if the singleton $\{x\}$ is closed for every $x\in X$. It follows that every finite subset of $X$ is closed. In particular, for a finite set, the only $T_1$ topology is the discrete topology. From now on we will consider infinite sets. For example, on $\mathbb{N}$: the smallest $T_1$ topology is the cofinite topology.

A topology on $X$ is Hausdorff if it is $T_1$ and for any two points $x\neq y$, there are open sets $U\ni x$ and $V\ni y$ such that $U\cap V =\varnothing$.

For $\mathbb{N}$, we know that the cofinite topology is not Hausdorff, since any two nonempty open sets intersect and their intersection is again cofinite. The discrete topology on $\mathbb{N}$ is Hausdorff. Is it the only Hausdorff topology on $\mathbb{N}$?

(1). Consider the map $f:\mathbb{N} \to [0, 1]$, $f(1)=0$ and $f(n)=\frac{1}{n-1}$ for every $n\ge 2$. Let $\mathscr{T}_f$ be the induced topology on $\mathbb{N}$ as a subspace of $[0, 1]$. So it is Hausdorff since $[0, 1]$ is. Note that $\{n\}$ is open for every $n\ge 2$, while $\{1\}$ is not open: the neighborhoods are $U_n={k\in \mathbb{N}: k\ge n}$ for all $n\ge 2$.

(2). We know that $|\mathbb{N}| = |\mathbb{Q}|$. So there is an bijection $g: \mathbb{N} \to \mathbb{Q} \subset \mathbb{R}$. Let $\mathscr{T}_g$ be the induced topology on $\mathbb{N}$ as a labelling of $\mathbb{Q}$. So it is Hausdorff since $\mathbb{R}$ is. Note that none of the singletons $\{n\}$ is open: every nonempty open set is infinite with infinite complements.

So one can embed $\mathbb{N}$ into different spaces, and different embeddings induce different topologies.

Consider the collection $\mathscr{H}$ of all Hausdorff topologies on $X$. We can give a parital order on $\mathscr{H}$ by inclusion. Note that every totally ordered subset $\mathscr{A} \subset \mathscr{H}$ admits a minimal element $\bigcap \mathscr{A}$ (unfortunately not necessarily Hausdorff). So one has to definite minimal Hausdorff topology directly: $\mathscr{T}$ is a minimal Hausdorff topology if it is Hausdorff and minimal with respect to the inclusion relation.

One can ask: does minimal topology exist? In particular, what can we say about the minimal Hausdorff topologies on $\mathbb{N}$?

An interesting result is that if $X$ is compact Hausdorff, then its topology is minimal. Therefore, minimal topologies do exist. As of $\mathbb{N}$, our topology $\mathscr{T}_f$ is minimal since it is Hausdorff and compact. Again compactness is sufficient but not necessary: Katetov constructed a minimal Hausdorff space which is not compact.

Notes on Lyapunov exponents

Alena Erchenko gave a talk on Flexibility of Lyapunov exponents. Let $f:M\to M$ be a diffeomorphism, $\mu$ be an $f$-invariant and ergodic measure. Let $\chi^+(\mu,f)$ be the sum of positive Lyapunov exponents of $\mu$ under $f$. Then Ruelle’s entropy inequlity states that $h_{\mu}(f) \le \chi^+(\mu,f)$.

Now if $f$ is a $C^2$ diffeomorphism, then the Ledrappier-Young showed that Pesin’s entropy formula $h_\mu(f) = \chi^+(\mu,f)$ holds if and only if $\mu$ is SRB: the conditional measures of $\mu$ along the (measurable) unstable foliations are absolutely continuous with respect to the Lebesgue measure.

Now assume $f$ has a unique measure of maximal entropy $\mu_{mme}$ and a unique SRB measure $\mu_{SRB}$. For example, we can take $f$ as a transitive Anosov diffeomorphism on $\mathbb{T}^n$. It follows from their definitions that

$\chi^+(\mu_{SRB},f) = h_{\mu_{SRB}}(f) \le h_{\text{top}}(f) = h_{\mu_{mme}}(f) \le \chi^+(\mu_{mme},f)$.

An interesting observation she made was the following dichotomy:

(1) either $\mu_{SRB} = \mu_{mme}$: then $\chi^+(\mu_{SRB},f) = \chi^+(\mu_{mme},f)= h_{\text{top}}(f)$;

(2) or $\mu_{SRB} \neq \mu_{mme}$: then $\chi^+(\mu_{SRB},f) < h_{\text{top}}(f) < \chi^+(\mu_{mme},f)$.

In the first case, one wants to show rigidity results: unless there are some resonance, the map $f$ should be smoothly conjugtate to a linear model. The second alternative is an open set. The first question we have in our mind is how flexible the inequalities can be: could one push $\chi^+(\mu_{SRB},f)$ to $0$ and/or $\chi^+(\mu_{mme},f)$ to $\infty$? Note that the first will bring the entropy $h_{\mu_{SRB}}(f) \to 0$, while the second will bring the dimension $\dim(\mu_{mme}) \to 0$ (by Young’s entropy formula).

On $\mathbb{T}^2$ it is very flexible. For higher dimensions, one can study the vector of postive Lyapunov exponents. How flexible can the Lyapunov vector can be? It turns out to be very flexible, too.

The Lyapunov exponents also encode some geometric information about the pressure function of the geometric potential. Let $\phi_t= -t\cdot \log DF|_{E^u}$, $p(t)= P(f, \phi_t)$, and $\mu_t$ be the unique equilibrium state of $\phi_t$. Then

(1) $\mu_0 = \mu_{mme}$, and $\mu_1= \mu_{SRB}$;

(2) $p(0)=h_{\mu_{mme}}(f) = h_{\text{top}}(f)$, $p(1)= h_{\mu_{SRB}}(f) -\mu_{SRB}(\log DF|_{E^u}) =0$;

(3) $p'(0) = - \chi^+(\mu_{mme},f)$, and $p'(1)= -\chi^+(\mu_{SRB},f)$

Multiplicative functions

A function $f: \mathbb{N} \to S^1$ is called completely multiplicative if $f(m\cdot n) = f(m) \cdot f(n)$. An example is the function $\Omega(n)=q_1 + \cdots + q_k$ if $n=p_1^{q_1}\cdots p_k^{q_k}$. Note that $\lambda(n)=(-1)^{\Omega(n)}$ is the Liouville function.

In link the authors introduced multiplicative dynamical systems. Given a continuous map $T:X\to X$, it induces an additive dynamical system $(X, T)$, in the sense that $T^{m+n} = T^m\circ T^n$. It also induces a multiplicative dynamical system $(X, T^{\Omega})$, in the sense that $T^{\Omega(m\cdot n)} = T^{\Omega(m)}\circ T^{\Omega(n)}$.

More generally, a multiplicative dynamical system is a map $S: \mathbb{N} \to C(X, X)$ such that $S(m\cdot n) = S(m) \circ S(n)$. It is easy to see that a multiplicative dynamical system is determined by the generators $S(p)$, $p$ prime. Then $(Y, S)$ is said to be finitely generated if $\{S(p)| p \text{ prime}\}$ is a finite set. In the case $S(n)=T^{\Omega(n)}$, we see that $S(p)= T$ for every prime $p$. In particular, it is finitely generated.

Motivated by the Mean Value Theorem for prime numbers, the authors formulated the following (much more generalized) version of Sarnak Conjecture: let $(X, T)$ be an additive dynamical system, $(Y, S)$ a multiplicative dynamical system. Suppose both have low complexity, and there is no local obstruction (that is, aperiodic). Then the two systems are disjoint: for any continuous functions $f$ on $X$ and $g$ on $Y$, and any points $x\in X$ and $g\in Y$, the sequences $(f(T^n x))$ and $(g(S_ny))$ are asymptotically independent.

In the particular case that $Y=\{0, 1\}$, $S_n(k)= k+ \Omega(n) \mod 2$, $g(k)=(-1)^k$, the above conjecture reduces to that the average of $(-1)^{\Omega(n)}f(T^n x)$ is $0$.

The reason that they need to put the low complexity assumption on the dynamical systems is that there are counterexamples. For example, let $T$ be the shift map on $X=\{0,1\}^{\mathbb{N}}$, $f(x)=(-1)^{x_0}$, and $\hat x\in X$ be the point generated by $\Omega$. Then $f(T^n \hat x) =(-1)^{\Omega(n)}$ and hence $(-1)^{\Omega(n)}f(T^n \hat x) =1$ for all $n$. In particular, the two sequences are the opposite of asymptotically independent.

They also mentioned a fundamental conjecture in this direction: the subshift generated by $\Omega$ has zero entropy. This must be true for the above conjecture to make sense. This might not be an easy question since it is really about the properties of all natural numbers.

Let $\displaystyle \zeta(s)= \sum_{n\ge 1}\frac{1}{n^s}, z\in \mathbb{C}$ be the Riemann zeta function. Euler (before Riemann) observed the product formula for $s>1$: $\displaystyle \zeta(s)= \prod_{p \text{ prime}}\frac{1}{1- p^{-s}}$. Then the quotient

$\displaystyle \frac{\zeta(2s)}{\zeta(s)} = \prod_{p} \frac{1- p^{-s}}{1- p^{-2s}} =\prod_{p} \frac{1}{1+p^{-s}} =\sum_{n\ge 1}\frac{\lambda(n)}{n^s}$, the later being the Dirichlet series for the Liouville function.

The space of characteristics

Let $(M, \omega)$ be a symplectic manifold. Given a smooth function $H: M \to \mathbb{R}$, the level set $E_h=\{x\in M| H(x) =h\}$ is a hypersurface possibly with singularities. Assume $N=E_h$ is connected and has no singularity. Then the kernal $L=\ker(\omega|_{N})$ is a one-dimensional subbundle of the tangent bundle $TN$, which is called the direction of characteristics. The integral foliation of the characteristic directions is called the characteristic foliation. We are interested in the space $X$ of characteristics. It is a symplectic manifold: its symplectic form $\mu$ is the quotient form $(\omega|_{N})/\sim$ of the restriction $\omega|_N$ of $\omega$ to $N$: $\mu$ is closed since $\omega$ is; $\mu$ is non-degenerate since $X$ is the quotient of $N$ by the kernel of $\omega_N$.

In the case that $(M,g)$ is a Riemannian manifold, the cotangent bundle $T^{\ast}M$ is a symplectic manifold with the standard symplectic structure $\omega = d\lambda$. A special function is the kinetic energy $H(q,p)= \frac{1}{2}|p|^2_q$. Then $N=E_2$ is the unit cotangent bundle, and the characteristics are geodesics (with their tangent vectors).

Example. Consider the standard Euclidean space $M=\mathbb{R}^n$. The unit cotangent bundle $N=\mathbb{R}^n \times S^{n-1}$, and the characteristics are just lines (with their tangent vectors) $[q,p]=\{(q+ tp, p) \in \mathbb{R}\}$. To find an explicit description of the space of characteristics, we consider the projection $\pi: N \to S^{n-1}$, $(q,p) \to p$. Note that the geodesic flows factors through this projection: $\pi(q+ tp, p) =p$ for every $t\in\mathbb{R}$. Therefore, the fiber of the space $X$ of characteristics over $p$ is $X_p= \mathbb{R}^n/\sim_p$, where $q_1 \sim_p q_2$ if $q_1 - q_2 \in \langle p \rangle$. Note that we can identify $X_p$ with $T_p^{\ast}S^{n-1}=\{\eta\in \mathbb{R}^{n}| \eta(p) =0\}$. Therefore, $X_p = T_p^{\ast}S^{n-1}$ for every $p\in S^{n-1}$, and hence $X= T^{\ast}S^{n-1}$. One can identify the standard symplectic form on $T^{\ast}S^{n-1}$ with the induced symplectic form on $X$ from the one on $T^{\ast}\mathbb{R}^n$.

The cross-ratio

Given three distinct points $A, B,C$ on a line $L$, we define their cross-ratio $(A, B, C)= AC/BC$ (directed displacements). Given another point $P$ outside the line $L$, one can draw three lines $\alpha=PA$, $\beta=PB$ and $\xi= PC$. We can define the cross-ratio of these three lines $(\alpha, \beta, \xi)= \frac{\sin(\alpha,\xi)}{\sin(\beta,\xi)}$. These two cross-ratios may not be the same. But they are closely related:

$\frac{(A, B, C)}{(\alpha, \beta, \xi)} = \frac{|PA|}{|PB|}$ (absolute displacements).

Now given a fourth point $D$ on the line $L$, we define their cross-ratio $(A, B, C, D)= \frac{(A, B, C)}{(A, B, D)}$. Similarly, we define $(\alpha, \beta, \xi, \eta) = \frac{(\alpha, \beta, \xi)}{(\alpha, \beta, \eta)}$, where $\eta=PD$.

It follows from the definitions that $\frac{(A, B, C, D)}{(\alpha, \beta, \xi, \eta)}=\frac{|PA|/|PB|}{|PA|/|PB|}=1$. It has the following consequences:

(1) fixing $L$ and the four points $(A, B, C, D)$, while letting $P$ move in the plane containing $L$, the cross-ratio $(\alpha, \beta, \xi, \eta)=(A, B, C, D)$ is independent of the choice of the point $P$.

(2) fixing $P$ and the four lines $(\alpha, \beta, \xi, \eta)$, while letting $L$ slide on the plane, the cross-ratio $(A, B, C, D)=(\alpha, \beta, \xi, \eta)$ is independent of the choice of the line $L$.

Random maps

Consider a collection of maps $T_j: X\to X$ and a collection of probability function $p_j: X\to [0, 1]$ with $\sum_{j} p_j(x) =1$ for every $x\in X$. It induces the following random walk on $X$. Given $x\in X$, let $T(x) = T_j(x)$ with probability $p_j(x)$. Given a probability measure $\nu$ on $X$, we can define the push-forward $T(\nu)$ as the distribution of $T(x)$ with $x \sim \nu$. More precisely, $(T\nu)(A)=P(T(x)\in A)$.

The transition kernel $K(x, \cdot)$. We pick $x$ according to the measure $\nu$.
Then $P(T(x)\in A)= E(I_A(T(x)))= E(E(I_A(T(x))|x))=E(\sum_j p_j(x)I_A(T_j(x)))=\int_X \sum_j p_j(x)I_A(T_j(x)) d\nu(x)$.
Therefore, $K(x, A) = \sum_j p_j(x)I_A(T_j(x))$.

Then a probability measure $\mu$ on $X$ is said to be stationary if $T\mu=\mu$. That is, $\mu(A) = \sum_j \int_X p_j(x)I_A(T_j(x)) d\nu(x)$
for every $A$.

Index of fixed points

The topological index of an isolated equilibrium point of the flow (or vector field) on a manifold $M^d$. Suppose $X(p)=0$ for some $p\in M$, and $X(q) \neq 0$ for any $q$ with $d(q, p) =\epsilon$. Consider the map $\phi_p: \partial B(p, \epsilon) \to S^{d-1}$, $q\mapsto \frac{1}{|X(q)|}X(q)$, and let $i(p,f)$ be the degree of the induced map $[\phi_p]$ on $S^{d-1}$.

Similarly, one defines the topological index of an isolated fixed point of a homeomorphism. Consider a homeomorphism $f$ on a closed manifold $M^d$. Let $p$ be an isolated fixed point of $f$. That is, there exists $\epsilon > 0$ such that $\{x\in B(p, \epsilon) | f(x)=x \}=\{p\}$. Consider the map $\phi_p: \partial B(x,\epsilon) \to S^{d-1}$, $x\mapsto \frac{f(x) - x}{|f(x) - x|}$, and let $i(p,f)$ be the degree of the induced map $[\phi_p]$ on $S^{d-1}$.

Suppose all fixed points of $f$ are isolated (hence there are finitely many of them), say $F(f)=\{p_1, \dots, p_n\}$. Then the Lefschetz Fixed Point Theorem states that $\sum_{p\in F(f)} i(p, f) = L(f):=\sum_{k=0}^d (-1)^k tr(f_{\ast}: H_k(M) \to H_k(M))$.
In the particular case that $f$ is homotopic (or homologous) to identity, then $f_{\ast}=Id$ on $H_k(M) \to H_k(M)$, and hence $L(f)=\chi(M)$, the Euler characteristic number on the manifold $M$. Then we have $\sum_{p\in F(f)} i(p, f) =\chi(M)$. In the case $M=S$ is a surface, we have $L(f) =2- tr(f_{\ast}: H_1(S) \to H_1(S))$.

Example. Consider the map $f$ on $S^2=\mathbb{C} \cup\{\infty\}$ induced by $f(z) = z + 1$ with the extension $f(\infty) = \infty$. Then $\infty$ is only one fixed point of $f$. We want to compute the index $i(\infty, f)$. Note that $S^2$ can be covered by two charts: $(\mathbb{C},z)$ around $0$, and $(\mathbb{C},w)$ around $\infty$. The transfer map between two charts is $w=\frac{1}{z}$. Then in the chart $(\mathbb{C},w)$, we have $f(w)= f(\frac{1}{z}) = \frac{1}{z+1}=\frac{w}{1+w}$. Therefore, $f(w) - w = \frac{w^2}{1+w}$. Then for sufficiently small $\epsilon$, the map $\phi_p(\epsilon e^{i\theta})$ is a small perturbation of the map $\epsilon e^{i\theta} \mapsto e^{i2\theta}$. Therefore, $i(\infty,f) =2$. On the other hand, it is easy to see that $L(f) = 2$ since $H_1(S^2) =0$.

Note that the above function $f$ is the time-1 map of the flow $\psi_t: z\mapsto z+t$. Around $\infty$, we have $\psi_t: w\mapsto \frac{w}{1+ tw}$. So the vector field $X(w)=\frac{d}{dt}\Big|_{t=0}\frac{w}{1+ tw}=-w^2$ and $X(\epsilon e^{i\theta}) = \epsilon^2 e^{i(2\theta+\pi)}$. So the two definations are compatible if the map is a time-1 map of a flow.