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.

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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.

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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<r. So the angle \theta(x) satisfies \cos\theta=\frac{a}{r}.

difference

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.

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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}}.

ellipse

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.

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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