updated advanced dyn systems + fixed points

Mathematics/2nd/Linear_geometry/Linear_geometry.tex

@@ -732,7 +732,7 @@
     \end{enumerate}
   \end{proposition}
   \begin{proposition}
-    If the set of fixed points of an affinity $f$, $\text{Fix}(f)$, is non-empty, then $\text{Fix}(f)$ is a subvariety.
+    If the set of fixed points of an affinity $f$, $\Fix(f)$, is non-empty, then $\Fix(f)$ is a subvariety.
   \end{proposition}
   \begin{definition}
     Let $f$ be an affinity. We define the \emph{invariance level} of $f$, $\rho(f)$, as: $$\rho(f)=\min\{\dim L:f(L)\subset L\subset\mathbb{A}\}\in\{0,\ldots,\dim\mathbb{A}\}$$
@@ -743,7 +743,7 @@
   \begin{proposition}[Properties of translations]
     Let $T_{\vf{v}}$ be a translation. Then:
     \begin{enumerate}
-      \item $\text{Fix}(T_{\vf{v}})=\varnothing$.
+      \item $\Fix(T_{\vf{v}})=\varnothing$.
       \item Invariant lines are those with director subspace $\langle\vf{v}\rangle$.
       \item If $\mathcal{R}=\{P;(\vf{v}_1,\ldots,\vf{v}_n)\}$ is an affine frame, then: $$\vf{M}_\mathcal{R}(T_{\vf{v}})=\left(\begin{array}{cccc|c}
                   1      & 0      & \cdots & 0      & 1      \\
@@ -762,7 +762,7 @@
   \begin{proposition}[Properties of reflections]
     Let $f$ be a reflection with root $\vf{v}$ and mirror $H=P+E$. Then:
     \begin{enumerate}
-      \item $\text{Fix}(f)=H$.
+      \item $\Fix(f)=H$.
       \item Invariant lines are those contained on $H$ and those with director subspace $\langle\vf{v}\rangle$.
       \item If $\mathcal{R}=\{P;(\vf{v}_1,\ldots,\vf{v}_{n-1},\vf{v})\}$ is  an affine frame such that $P\in H$ and $\vf{v}_1,\ldots,\vf{v}_{n-1}\in E$, then $$\vf{M}_\mathcal{R}(f)=\left(\begin{array}{cccc|c}
                   1      & 0      & \cdots & 0      & 0      \\
@@ -781,7 +781,7 @@
   \begin{proposition}[Properties of projections]
     Let $f$ be a projection over $H=P+E$ in the direction of $\vf{v}$. Then:
     \begin{enumerate}
-      \item $\text{Fix}(f)=H$.
+      \item $\Fix(f)=H$.
       \item Invariant lines are those contained on $H$.
       \item If $\mathcal{R}=\{P;(\vf{v}_1,\ldots,\vf{v}_{n-1},\vf{v})\}$ is  an affine frame such that $P\in H$ and $\vf{v}_1,\ldots,\vf{v}_{n-1}\in E$, then $$\vf{M}_\mathcal{R}(f)=\left(\begin{array}{cccc|c}
                   1      & 0      & \cdots & 0      & 0      \\
@@ -801,7 +801,7 @@
     Let $f$ be a homothety of similitude ratio $\lambda$. Then:
     \begin{enumerate}
       \item $f$ has a unique fixed.
-      \item If $\mathcal{R}=\{P;\mathcal{B}\}$ is an affine frame with $P\in\text{Fix}(f)$ and $\mathcal{B}$ an arbitrary basis, then $$\vf{M}_\mathcal{R}(f)=\left(\begin{array}{ccc|c}
+      \item If $\mathcal{R}=\{P;\mathcal{B}\}$ is an affine frame with $P\in\Fix(f)$ and $\mathcal{B}$ an arbitrary basis, then $$\vf{M}_\mathcal{R}(f)=\left(\begin{array}{ccc|c}
                     &               &   & 0      \\
                     & \lambda\vf{I} &   & \vdots \\
                     &               &   & 0      \\

Mathematics/5th/Advanced_dynamical_systems/Advanced_dynamical_systems.tex

@@ -255,25 +255,25 @@
   \begin{definition}
     We define the set:
     \begin{multline*}
-      D^0(\TT^1):\{f:\RR\to\RR:f\text{ increasing and}\\
+      \mathcal{D}^0(\TT^1):\{f:\RR\to\RR:f\text{ increasing and}\\
       \text{ homeomorphism}, f(x+1)=f(x)+1\}
     \end{multline*}
     Note that we have the projection:
     $$
-      \function{}{D^0(\TT^1)}{\Homeo(\TT^1)}{f}{F}
+      \function{}{\mathcal{D}^0(\TT^1)}{\Homeo(\TT^1)}{f}{F}
     $$
-    We can define a distance in $D^0(\TT^1)$ as:
+    We can define a distance in $\mathcal{D}^0(\TT^1)$ as:
     $$
       d(f,g)=\max\{ \sup_{x\in\RR}\abs{f(x)-g(x)},\sup_{x\in\RR}\abs{f^{-1}(x)-g^{-1}(x)}\}
     $$
   \end{definition}
   \begin{lemma}
-    $D^0(\TT^1)$ is a complete metric space. Moreover:
+    $\mathcal{D}^0(\TT^1)$ is a complete metric space. Moreover:
     \begin{enumerate}
-      \item $f\to f^{-1}$ is continuous, $f\in D^0(\TT^1)$.
-      \item $(f,g)\to f\circ g$ is continuous, $(f,g)\in D^0(\TT^1)\times D^0(\TT^1)$.
+      \item $f\to f^{-1}$ is continuous, $f\in \mathcal{D}^0(\TT^1)$.
+      \item $(f,g)\to f\circ g$ is continuous, $(f,g)\in \mathcal{D}^0(\TT^1)\times \mathcal{D}^0(\TT^1)$.
     \end{enumerate}
-    Thus, $D^0(\TT^1)$ is a topological group with the composition.
+    Thus, $\mathcal{D}^0(\TT^1)$ is a topological group with the composition.
   \end{lemma}
   \begin{definition}
     Let $\varepsilon\geq 0$ and $\alpha\in\RR$. We define the \emph{Arnold family} as:
@@ -282,7 +282,7 @@
     $$
   \end{definition}
   \begin{lemma}
-    If $0\leq \varepsilon<\frac{1}{2\pi}$, then $f_{\alpha,\varepsilon}\in D^0(\TT^1)$.
+    If $0\leq \varepsilon<\frac{1}{2\pi}$, then $f_{\alpha,\varepsilon}\in \mathcal{D}^0(\TT^1)$.
   \end{lemma}
   \begin{proof}
     Note that ${f_{\alpha,\varepsilon}}'>0\iff \varepsilon<\frac{1}{2\pi}$. Thus, $f_{\alpha,\varepsilon}$ is strictly increasing, and therefore it is a homeomorphism.
@@ -296,7 +296,7 @@
     with $\varphi_n$ 1-periodic.
   \end{remark}
   \begin{lemma}\label{ADS:lema1}
-    Let $f\in D^0(\TT^1)$ be such that $f=\id +\varphi$, with $\varphi$ 1-periodic. Let $m:=\min_{x\in\RR}\varphi$ and $M:=\max_{x\in\RR}\varphi$. Then, we have $m\leq M< m+1$.
+    Let $f\in \mathcal{D}^0(\TT^1)$ be such that $f=\id +\varphi$, with $\varphi$ 1-periodic. Let $m:=\min_{x\in\RR}\varphi$ and $M:=\max_{x\in\RR}\varphi$. Then, we have $m\leq M< m+1$.
   \end{lemma}
   \begin{proof}
     Let $0\leq x\leq y<1\leq x+1$. Then, $f(y)<f(x+1)=f(x)+1$. Thus, $f(y)+x< f(x)+1+y$, and so $\varphi(y) < \varphi(x)+1$. Now take supremum in $y$ and infimum in $x$.
@@ -305,7 +305,7 @@
     Let ${(u_n)}\in\RR$ be a sequence. We say that $(u_n)$ is \emph{subadditive} if $u_{n+m}\leq u_n+u_m$ for all $n,m\in\NN$. We say that $(u_n)$ is \emph{superadditive} if $u_{n+m}\geq u_n+u_m$ for all $n,m\in\NN$, that is, if $(-u_n)$ is subadditive.
   \end{definition}
   \begin{lemma}\label{ADS:lema2}
-    Let $f\in D^0(\TT^1)$ be such that $f=\id +\varphi$. We can write $f^n=\id +\varphi_n$ and let $m_n:=\min_{x\in\RR}\varphi_n$ and $M_n:=\max_{x\in\RR}\varphi_n$. Then, $(M_n)$ is subadditive and $(m_n)$ is superadditive.
+    Let $f\in \mathcal{D}^0(\TT^1)$ be such that $f=\id +\varphi$. We can write $f^n=\id +\varphi_n$ and let $m_n:=\min_{x\in\RR}\varphi_n$ and $M_n:=\max_{x\in\RR}\varphi_n$. Then, $(M_n)$ is subadditive and $(m_n)$ is superadditive.
   \end{lemma}
   \begin{proof}
     We have that:
@@ -328,7 +328,7 @@
     $$
   \end{proof}
   \begin{theorem}[Existence of the rotation number]
-    For all $f\in D^0(\TT^1)$, we have that the sequence of functions $\frac{1}{n}(f^n-\id)$ convergence uniformly to constant function $\rho(f)\in\RR$. This number is called the \emph{rotation number} of $f$.
+    For all $f\in \mathcal{D}^0(\TT^1)$, we have that the sequence of functions $\frac{1}{n}(f^n-\id)$ convergence uniformly to constant function $\rho(f)\in\RR$. This number is called the \emph{rotation number} of $f$.
   \end{theorem}
   \begin{proof}
     By \mcref{ADS:lema1} we have that $\frac{m_n}{n}\leq \frac{M_n}{n}< \frac{m_n}{n}+\frac{1}{n}$, where $m_n:= \min_{x\in\RR}\varphi_n$ and $M_n:= \max_{x\in\RR}\varphi_n$. By \mcref{ADS:lema2,ADS:lema3}, we have that $\frac{m_n}{n}$ and $\frac{M_n}{n}$ have the same limit and moreover:
@@ -341,9 +341,9 @@
     The following properties are satisfied:
     \begin{enumerate}
       \item $\rho(R_\alpha)=\alpha$ $\forall\alpha\in\RR$.
-      \item $\rho(f^n)=n\rho(f)$ $\forall f\in D^0(\TT^1)$, $n\in\NN$.
-      \item $f\leq g\implies \rho(f)\leq \rho(g)$ $\forall f,g\in D^0(\TT^1)$.
-      \item $\rho(f+k)=:\rho(R_k\circ f)=\rho(f) + k$ $\forall f\in D^0(\TT^1)$, $k\in\ZZ$.
+      \item $\rho(f^n)=n\rho(f)$ $\forall f\in \mathcal{D}^0(\TT^1)$, $n\in\NN$.
+      \item $f\leq g\implies \rho(f)\leq \rho(g)$ $\forall f,g\in \mathcal{D}^0(\TT^1)$.
+      \item $\rho(f+k)=:\rho(R_k\circ f)=\rho(f) + k$ $\forall f\in \mathcal{D}^0(\TT^1)$, $k\in\ZZ$.
     \end{enumerate}
   \end{proposition}
   \begin{definition}
@@ -422,13 +422,95 @@
     $$
     Dividing by $n$ and taking limits, we have that $\rho(f)=\mu(\varphi)$. This also shows the second point. To prove the first one, note that $\min\psi_n\leq 0$ and so by \mcref{ADS:lema1} we have $\max\psi_n <1$. Moreover, $\min\psi_n =-\max(-\psi_n) >-1$ (using the same argument as before) and so $\norm{\psi_n}_{\mathcal{C}(\TT^1)}<1$.
   \end{proof}
-  \begin{proposition}
-    Let $f\in D^0(\TT^1)$, $p\in\ZZ$ and $q\in\NN$. Then:
+  \subsubsection{Rational rotation number}
+  \begin{proposition}\label{ADS:characterisation_rot_number}
+    Let $f\in \mathcal{D}^0(\TT^1)$, $p\in\ZZ$ and $q\in\NN$. Then:
     \begin{align*}
       \rho(f)=\frac{p}{q} & \iff \exists x\in\RR\text{ such that }f^q(x)=x+p \\
       \rho(f)>\frac{p}{q} & \iff \forall x\in\RR\text{ we have }f^q(x)>x+p   \\
       \rho(f)<\frac{p}{q} & \iff \forall x\in\RR\text{ we have }f^q(x)<x+p
     \end{align*}
   \end{proposition}
+  \begin{proof}
+    Since $\rho(f^q)= q\rho(f)$ and $\rho(f+p)=\rho(f)+p$, we have that if $g=f^q-p$, $\rho(g)=q \rho(f)-p$. Thus, an easy check shows that we can assume that $p=0$ and $q=1$. We will only prove the equivalences to the left, as it is sufficient.
+    \begin{enumerate}
+      \item Assume $f(x)=x$ for some $x\in\RR$. Then, from the definition of $\rho(f)$ applied to the point $x$, we have that $\rho(f)=0$.
+      \item Assume $f(x)>x$ and write $f=\id+\varphi$ with $\varphi\in\mathcal{C}(\TT^1)$ and $\varphi>0$. Since, $\TT^1$ is compact, we have in fact that $\varphi\geq \min\varphi=:\varepsilon>0$. Now:
+            $$
+              f^n-\id = \sum_{i=0}^{n-1}\varphi\circ f^i\geq n\varepsilon
+            $$
+            And so $\rho(f)\geq \varepsilon>0$.
+      \item Proceed as in the previous case.
+    \end{enumerate}
+  \end{proof}
+  \begin{definition}
+    Let $F\in\Homeo(\TT^1)$ and $x\in \TT^1$. We define the \emph{orbit} of $x$ as:
+    $$
+      \mathcal{O}_F(x):=\{F^n(x):n\in\ZZ\}
+    $$
+    We also define the \emph{positive orbit} of $x$  and the \emph{negative orbit} of $x$ as:
+    \begin{align*}
+      \mathcal{O}_F^+(x) & :=\{F^n(x):n\in\ZZ_{\geq 0}\} \\
+      \mathcal{O}_F^-(x) & :=\{F^n(x):n\in\ZZ_{\leq 0}\}
+    \end{align*}
+    If the homeomorphism is not specified, we will omit the subscript.
+  \end{definition}
+  \begin{definition}
+    Let $F\in\Homeo(\TT^1)$ and $X\subset \TT^1$. We say that $X$ is \emph{positively invariant} if $F(X)\subseteq X$ and \emph{negatively invariant} if $F^{-1}(X)\subseteq X$. We say that $X$ is \emph{invariant} if $F(X)=X$.
+  \end{definition}
+  \begin{proposition}
+    Let $X\subset \TT^1$ and $x\in \TT^1$. Then:
+    \begin{enumerate}
+      \item $X$ is invariant $\iff \forall x\in X$, $\mathcal{O}(x)\subseteq X\iff X$ is a union of orbits.
+      \item $\mathcal{O}(x)$ is finite $\iff x$ is periodic.
+      \item The omega limit $\omega(x)$ and the alpha limit $\alpha(x)$ are non-empty compact invariant sets.
+    \end{enumerate}
+  \end{proposition}
+  \begin{definition}
+    Let $F\in \Homeo(\TT^1)$. We define the \emph{positively recurrent points} and \emph{negatively recurrent points} as:
+    \begin{align*}
+      R^+(F):=\{x\in\TT^1:x\in\omega(x)\} \\
+      R^-(F):=\{x\in\TT^1:x\in\alpha(x)\}
+    \end{align*}
+  \end{definition}
+  \begin{proposition}
+    Let $F\in \Homeo(\TT^1)$. Then, $R^\pm(F)$ are invariant non-closed sets.
+  \end{proposition}
+  \begin{definition}
+    Let $F\in \Homeo(\TT^1)$ and $x\in \TT^1$. We say that $x$ is a \emph{wandering point} if there exists a neighborhood $U$ of $x$ such that $\forall n\geq 1$ we have $F^n(U)\cap U=\varnothing$. The orbit $U$ is called a \emph{wandering domain}. We define the set:
+    $$
+      \Omega(F):=\{ x\in\TT^1: x\text{ is not wandering}\}
+    $$
+  \end{definition}
+  \begin{remark}
+    A point $x\in\TT^1$ is \emph{non-wandering} if it is not wandering, i.e.\ if $\forall U$ neighborhood of $x$ $\exists n\geq 1$ such that $F^n(U)\cap U\ne\varnothing$.
+  \end{remark}
+  \begin{proposition}
+    Let $F\in \Homeo(\TT^1)$. Then, $\Omega(F)$ is an invariant closed set.
+  \end{proposition}
+  \begin{remark}
+    Note that:
+    $$
+      \Fix(F)\subseteq \Per(F)\subseteq R^\pm(F)\subseteq \Omega(F)\subseteq \TT^1
+    $$
+  \end{remark}
+  \begin{definition}
+    Let $F\in \Homeo(\TT^1)$ and $X\subseteq \TT^1$ be closed and invariant. We say that $X$ is \emph{minimal} if $\forall x\in X$, $\overline{\mathcal{O}(x)}=X$. If $X=\TT^1$, we say that $F$ is \emph{minimal}.
+  \end{definition}
+  \begin{proposition}
+    Let $F\in \Homeo(\TT^1)$ and $X\subseteq \TT^1$ be a closed and invariant. Then, $X$ is minimal $\iff$ $\forall Y\subseteq X$ closed, invariant and non-empty, $Y=X$.
+  \end{proposition}
+  \begin{theorem}
+    Let $F\in \Homeo(\TT^1)$ with $\rho(F)=\frac{p}{q}\in \quot{\QQ}{\ZZ}$. Then:
+    \begin{enumerate}
+      \item $F$ has periodic points of period $q$, and any periodic point of $F$ has minimal period $q$.
+      \item For any $x\in \TT^1$, $\omega(x)$ and $\alpha(x)$ are periodic orbits.
+    \end{enumerate}
+  \end{theorem}
+  \begin{proof}
+    First we assume $q=1$ and $p=0$. Let $f\in \mathcal{D}^0(\TT^1)$ be a lift of $F$. By \mcref{ADS:characterisation_rot_number}, we have that $\exists x\in \RR$ with $f(x)=x$. So $\Fix(f)\ne \varnothing$, and it is closed and invariant by translations. Now we write $\RR\setminus\Fix(f)$ as union of open intervals. Let $(a,b)$ be one connected component. Inside it, we must have either $f(x)>x$ or $f(x)<x$. In the first case we have that $(f^n(x))$ is strictly increasing $\forall x\in (a,b)$ and so $\omega(x)=\{b\}$ and $\alpha(x)=\{a\}$ $\forall x\in(a,b)$. The second case is exactly the opposite.
+
+    Now we do the general case. Assume $\rho(f)=\frac{p}{q}$. Then, again by \mcref{ADS:characterisation_rot_number}, we have that $\exists x\in \RR$ with $f^q(x)=x+p$. Assume we have $x'\in\RR$ and $p',q'\in\ZZ$ with $q'\geq 1$ such that $f^{q'}(x')=x'+p'$. By \mcref{ADS:characterisation_rot_number}, we have that $\frac{p}{q}=\frac{p'}{q'}$ and so $\exists k\in\NN$ such that $q'=kq$ and $p'=kp'$ because $\frac{p}{q}$ is irreducible. Now let $g=f^q-p$. Then, an easy calculation shows that $g^k(x')=x'$. But $\rho(g)=0$ and in the previous case we have seen that the periodic points are only fixed points, so $k=1$. For the second part, we proceed as in the previous case with the function $g=f^q-p$.
+  \end{proof}
 \end{multicols}
 \end{document}

preamble_formulas.sty

@@ -410,6 +410,9 @@
 \newcommand{\CPP}{\mathrm{CPP}} % Compound Poisson Process
 \newcommand{\RPM}{\mathrm{RPM}} % Random Poisson Measure
 
+%%% ADVANCED DYNAMICAL SYSTEMS
+\newcommand{\Fix}{\mathrm{Fix}} % Fixed points
+
 %%% vectors and matrices
 \newcommand{\vf}[1]{\boldsymbol{\mathrm{#1}}} % math style for vectors and matrices and vector-values functions (previously it was \*vb{#1} but this does not apply to greek letters)
 \newcommand{\transpose}[1]{{#1}^{\mathrm{T}}} % math style for vectors and matrices