## Tag Archives: symplectic

### Symplectic and contact manifolds

Let $(M,\omega)$ be a symplectic manifold. It said to be exact if $\omega=d\lambda$ for some one-form $\lambda$ on $M$.

(1) If $\omega=d\lambda$ is exact, then there is a canonical isomorphism between the v.f. and 1-forms. In particular, there exists a v.f. $X$ such that $\lambda=i_X\omega$. Then we have $\lambda(X)=\omega(X,X)=0$, and $L_X\lambda=i_X d\lambda+d i_X\lambda=i_X\omega +0=\lambda$, and $L_X\omega=d i_X\omega=d\lambda=\omega$.

(2) Suppose there exists a vector field $X$ on $M$ such that its Lie-derivative $L_X\omega=\omega$ (notice the difference with $L_X\omega=0$). Then Cartan’s formula says that $\omega=i_X d\omega+ di_X\omega=d\lambda$, where $\lambda=i_X\omega$. So $\omega=d\lambda$ is exact, and $L_X\lambda=i_Xd\lambda+di_X\lambda=i_X\omega+0=\lambda$.

### Collections

10. Let $f_a:S^1\to S^1$, $a\in[0,1]$ be a strictly increasing family of homeomorphisms on the unit circle, $\rho(a)$ be the rotation number of $f_a$. Poincare observed that $\rho(a)=p/q$ if and only if $f_a$ admits some periodic points of period $q$. In this case $f_a^q$ admits fixed points.

Note that $a\mapsto \rho(a)$ is continuous, and non-decreasing. However, $\rho$ may not be strictly increasing. In fact, if $\rho(a_0)=p/q$ and $f^q\neq Id$, then $\rho$ is locked at $p/q$ for a closed interval $I_{p/q}\ni a_0$. More precisely, if $f^q(x) > x$ for some $x$, then $\rho(a)=p/q$ on $[a_0-\epsilon,a_0]$ for some $\epsilon > 0$; if $f^q(x) 0$; while $a_0\in \text{Int}(I_{p/q})$ if both happen.

Also oberve that if $r=\rho(a)\notin \mathbb{Q}$, then $I_r$ is a singelton. So assuming $f_a$ is not unipotent for each $a\in[0,1]$, the function $a\mapsto \rho(a)$ is a Devil’s staircase: it is constant on closed intervals $I_{p/q}$, whose union $\bigcup I_{p/q}$ is dense in $I$.

9. Let $X:M\to TM$ be a vector field on $M$, $\phi_t:M\to M$ be the flow induced by $X$ on $M$. That is, $\frac{d}{dt}\phi_t(x)=X(\phi_t(x))$. Then we take a curve $s\mapsto x_s\in M$, and consider the solutions $\phi_t(x_s)$. There are two ways to take derivative:

(1) $\displaystyle \frac{d}{dt}\phi_t(x_s)=X(\phi_t(x_s))$.

(2) $\displaystyle \frac{d}{ds}\phi_t(x_s)=D\phi_t(\frac{d}{ds}x_s))$, which induces the tangent flow $D\phi_t:TM\to TM$ of $\phi_t:M\to M$.

Combine these two derivatives together:

$\displaystyle \frac{d}{dt}D_x\phi_t(x_s')=\frac{d}{dt}\frac{d}{ds}\phi_t(x_s) =\frac{d}{ds}\frac{d}{dt}\phi_t(x_s)=\frac{d}{ds}X(\phi_t(x_s)) =D_{\phi_t(x)}X\circ D_x\phi_t(x_s').$

This gives rise to an equation $\displaystyle \frac{d}{dt}D_x\phi_t=D_{\phi_t(x)}X\circ D_x\phi_t.$

Formally, one can consider the differential equation along a solution $x(t)$:
$\displaystyle \frac{d}{dt}D(t)=D_{\phi_t(x)}X\circ D(t)$, $D(0)=Id$. Then $D(t)$ is called the linear Poincare map along $x(t)$. Suppose $x(T)=x(0)$. Then $D(T)$ determines if the periodic orbit is hyperbolic or elliptic. Note that the path $D(t)$, $0\le t\le T$ contains more information than the above characterization.

### Area of the symmetric difference of two disks

This post goes back to high school: the area $\delta_d$ of the symmetric difference of two $d$-dimensional disks when one center is shifted a little bit. Let’s start with $d=1$. So we have two intervals $[-r,r]$ and $[x-r,x+r]$. It is easy to see the symmetric difference is of length $\delta_1(x)=2x$.

Then we move to $d=2$: two disks $L$ and $R$ of radius $r$ and center distance $x=2a. So the angle $\theta(x)$ satisfies $\cos\theta=\frac{a}{r}$.

The symmetric difference is the union of $R\backslash L$ and $L\backslash R$, which have the same area: $\displaystyle (\pi-\theta)r^2+2x\sqrt{r^2-x^2}-\theta r^2=2(\frac{\pi}{2}-\arccos\frac{x}{r})r^2+2x\sqrt{r^2-x^2}$. Note that the limit
$\displaystyle \lim_{x\to0}\frac{\text{area}(\triangle)}{2a}=\lim_{a\to0}2\left(\frac{r^2}{\sqrt{1-\frac{a^2}{r^2}}}\cdot\frac{1}{r}+\sqrt{r^2-a^2}\right)=4r$.
So $\delta_2(x)\sim 4rx$.

I didn’t try for $d\ge3$. Looks like it will start with a linear term $2d r^{d-1}x$.

—————–

Now let ${\bf r}(t)=(a\cos t,\sin t)$ be an ellipse with $a>1$, and ${\bf r}'(t)=(-a\sin t,\cos t)$ be the tangent vector at ${\bf r}(t)$. Let $\omega$ be the angle from ${\bf j}=(0,1)$ to ${\bf r}'(t)$.
Let $s(t)=\int_0^t |{\bf r}'(u)|du$ be the arc-length parameter and $K(s)=|{\bf l}''(s)|$ be the curvature at ${\bf l}(s)={\bf r}(t(s))$. Alternatively we have $\displaystyle K(t)=\frac{a}{|{\bf r}'(t)|^{3}}$.

The following explains the geometric meaning of curvature:

$\displaystyle K(s)=\frac{d\omega}{ds}$, or equivalently, $K(s)\cdot ds=d\omega$. $(\star)$.

Proof. Viewed as functions of $t$, it is easy to see that $(\star)$ is equivalent to $K(t)\cdot \frac{ds}{dt}=\frac{d\omega}{dt}$.

Note that $\displaystyle \cos\omega=\frac{{\bf r}'(t)\cdot {\bf j}}{|{\bf r}'(t)|}=\frac{\cos t}{|{\bf r}'(t)|}$. Taking derivatives with respect to $t$, we get
$\displaystyle -\sin\omega\cdot\frac{d\omega}{dt}=-\frac{a^2\sin t}{|{\bf r}'(t)|^3}$. Then $(\star)$ is equivalent to

$\displaystyle \frac{a}{|{\bf r}'(t)|^{3}}\cdot |{\bf r}'(t)|=\frac{a^2\sin t}{\sin\omega\cdot |{\bf r}'(t)|^3}$, or
$\displaystyle \sin\omega\cdot |{\bf r}'(t)|=a\sin t$. Note that $\displaystyle \sin^2\omega=1-\cos^2\omega=1-\frac{\cos^2 t}{|{\bf r}'(t)|^2}$. Therefore $\displaystyle \sin^2\omega\cdot |{\bf r}'(t)|^2= |{\bf r}'(t)|^2-\cos^2 t=a^2\sin^2 t$, which completes the verification.

### Some notes

Let $(M^{2n},\omega))$ be a symplectic manifold, $S^{2p+q}\subset M$ be a $C^r$ submanifold such that when $\omega$ is restricted to $S$, it has constant rank $2p$. Then for each $x\in S$ there exist a neighborhood $U\ni x$ in $M$ and $C^{r-2}$ symplectic coordinates on $(U,x_l,\cdots, x_{2n})$, such that
$S\cap U=\{p\in U: x_i(p)=0, i=p+q+1,\cdots,n,n+p+l,\cdots,2n\}$,

i. e., there exist symplectic coordinates on $M$ that give $S$ as a vector subspace.

Suppose $(X, \mathcal{X},\mu)$ is a probability space. The classical Borel–Cantelli lemmas state that
(1) if $\{A_n:n\ge0\}$ is a sequence of measurable sets in $X$ and
$\sum_{n\ge0}\mu(A_n)<\infty$, then
$\mu(x\in X: x\in A_n\text{ for infinitely often times})=0$.

(2) moreover if $\{A_n:n\ge0\}$ is independent and $\sum_{n\ge0}\mu(A_n)=\infty$, then
$\frac{S_n(x)}{E_n}\to 1$ for $\mu-$a.e. $x\in X$.
where $S_n(x) =\sum_{j=0}^n 1_{A_j}(x)$ and $E_n=\sum_{j=0}^n \mu(A_j)$.

Suppose $(X,\mu,T)$ is a measure-preserving transformation. If $\{A_n\}$ is a sequence of sets such that $\sum_n \mu(A_n)=\infty$, is $T^n(x)\in A_n$ for infinitely many times for
$\mu-$a.e. $x\in X$ and, if so, is there a quantitative estimate of the asymptotic number
of entry times?

The shrinking target problem: $A_n=B(x,r_n)$.

Let $f:X\to X$ be a homeomorphism and $\Omega_1(f)=\Omega(X,f)$ be the set of nonwandering points. Similarly we can define $\Omega_2(f)=\Omega(\Omega_1(f),f)$ and inductively $\Omega_{\alpha'}(f)=\Omega(\Omega_\alpha(f),f)$ for all $\alpha\in\mathbb{N}$. Let $\Omega_\omega(f)=\bigcap_{\alpha\in\mathbb{N}}\Omega_\alpha(f)$. Once again $\Omega_\omega(f)$ is a compact, invariant subset so we can restart our inductive process for all ordinals $\alpha\in\mathbb{O}$. If the process stabilizes at some ordinal $\alpha$, then we call it the center depth of $(X,f)$ and $\Omega_\alpha(f)$ the Birkhoff center of $(X,f)$. There do exists examples that realize the countable ordinal $\alpha\in\mathbb{O}$ (note there are uncountably many countable ordinals).

A point $x\in X$ is said to be chain recurrent if for each $\epsilon>0$ there exists a nontrivial $\epsilon-$chain $C_\epsilon$ initiated and terminated at $x$. Let $\mathrm{CR}(f)$ be the set of chain recurrent points of $(X,f)$. Unlike the nonwandering set, chain recurrent set always has depth 1:

Proposition. Let $(X,f)$ be continuous. Then $\mathrm{CR}(\mathrm{CR}(f),f)=\mathrm{CR}(f)$.
Proof. Let $x\in \mathrm{CR}(f)$. Then for each $\epsilon>0$ there exists a $\epsilon-$chain $C_\epsilon$ initiated and terminated at $x$. Let $C_x=\limsup_{n\to\infty}C_{1/n}$. If we can show that $C_x\subset \mathrm{CR}(f)$, then for each $\delta>0$, pick $m,n\ge1/\delta$ such that $B(y,1/m)\subset B(fy,\delta)$ for all $y\in X$ and $C_{1/n}\subset B(\mathrm{CR}(f),1/m)$. Hence we $\frac{1}{m}-$perturb $C_{1/n}=[x_k]$ to a new chain $D_{1/n}=[y_k]\subset\mathrm{CR}(f)$. So $d(fy_k,y_{k+1})\le d(fy_k,fx_k)+d(fx_k,x_{k+1})+d(x_{k+1},y_{k+1})<\delta+1/n+1/m<3\delta$.

Now we are left to prove the claim. Let $y\in C_x$. For each $\delta>0$ we pick $m,n\ge1/\delta$ such that $B(y,1/m)\subset B(fy,\delta)$ for all $y\in X$ and $x_k\in C_{1/n}\cap B(y,1/m)$ for some $k$. We replace $x_k$ by $y$ to get a new chain ($3\delta$)
$x=x_0\to\cdots\to x_{k-1}\to y\to x_{k+1}\to \cdots\to x_{K}=x$, or equivalently
$y\to x_{k+1}\to \cdots\to x_{K}=x=x_0\to\cdots\to x_{k-1}\to y$. Since $\delta$ can be arbitrarily small, we see that $y$ is also chain recurrent. QED