\( \def\CC{{\mathbb C}} \def\RR{{\mathbb R}} \def\NN{{\mathbb N}} \def\ZZ{{\mathbb Z}} \def\TT{{\mathbb T}} \def\CF{{\operatorname{CF}^\bullet}} \def\HF{{\operatorname{HF}^\bullet}} \def\SH{{\operatorname{SH}^\bullet}} \def\ot{{\leftarrow}} \def\st{\;:\;} \def\Fuk{{\operatorname{Fuk}}} \def\emprod{m} \def\cone{\operatorname{Cone}} \def\Flux{\operatorname{Flux}} \def\li{i} \def\ev{\operatorname{ev}} \def\id{\operatorname{id}} \def\grad{\operatorname{grad}} \def\ind{\operatorname{ind}} \def\weight{\operatorname{wt}} \def\Sym{\operatorname{Sym}} \def\HeF{\widehat{CHF}^\bullet} \def\HHeF{\widehat{HHF}^\bullet} \def\Spinc{\operatorname{Spin}^c} \def\min{\operatorname{min}} \def\div{\operatorname{div}} \def\SH{{\operatorname{SH}^\bullet}} \def\CF{{\operatorname{CF}^\bullet}} \def\Tw{{\operatorname{Tw}}} \def\Log{{\operatorname{Log}}} \def\TropB{{\operatorname{TropB}}} \def\wt{{\operatorname{wt}}} \def\Span{{\operatorname{span}}} \def\Crit{\operatorname{Crit}} \def\into{\hookrightarrow} \def\tensor{\otimes} \def\CP{\mathbb{CP}} \def\eps{\varepsilon} \) SympSnip:

  1. Here, we write \(X:=T^*S^n=\{(z_0, \ldots, z_n)\st \sum_{i=0}^n z_i^2=1\}\). Let \(f:= \frac{1}{2}\Re\left(\sum_{i=0}^n z_i^2\right)\), and let \(g=\frac{1}{2}\Im\left(\sum_{i=0}^n z_i^2\right)\). Then \begin{align*} \nabla f = \sum_{i=0}^n x_i \partial_{x_i}- y_i \partial_{y_i} && \nabla g = \sum_{i=0}^n y_i \partial_{x_i} + x_i \partial_{y_i}. \end{align*} The tangent space to \(X\) at a point is the orthogonal complement to \(\RR \nabla f + \RR \nabla g\). Consider the Hamiltonian \begin{align*} H: X\to \RR\\ (\vec x, \vec y )\mapsto \frac{1}{2}|\vec y|^2 \end{align*} To compute the Hamiltonian flow of \(H\), we use that \(V_H=J\circ \nabla_{X} H\), where \(\nabla_X H\) is the gradient of \(H\) on \(X\) computed with the metric inherited from \(\CC^{n+1}\). We use the following identities: \begin{align*} |\nabla f |^2 = |\vec x|^2+|\vec y|^2 && \nabla f \cdot \nabla g = 0 \\ \nabla H \cdot \nabla g = 0 && \nabla H \cdot \nabla f = -|\vec y |^2\\ \end{align*} Computing the gradient by orthogonal projection: \begin{align*} \nabla_X H =& \nabla H - \left(\frac{\nabla f \cdot \nabla H}{|\nabla f|^2}\nabla f \right)\\ =& \sum_{i=0}^n\left(y_i \partial_{y_i}- \frac{-|\vec y|^2}{|\vec x|^2+ |\vec y|^2}( x_i \partial_{x_i}- y_i \partial_{y_i})\right) \\ =& \sum_{i=0}^n y_i\left(1- \frac{|\vec y|^2}{|\vec x|^2 +|\vec y|^2}\right) \partial_{y_i}+ \frac{x_i|\vec y|^2}{|\vec x|^2 +|\vec y|^2}\partial_{x_i} \end{align*} Setting \(A= |\vec y|^2/(|x|^2+|\vec y|^2)\) and multiply through by \(J\) gives \[V_H= \sum_{i=0}^n Ax_i \partial_{y_i} - (1-A)y_i \partial_{x_i}\]
  2. Let \((x_i(t), y_i(t))\) be a Hamiltonian flowline. Then \begin{align*} x_i'=(1-A) y_i && y_i'=A x_i \end{align*} from which we obtain \[x_i''=(1-A) A x_i\] Note that \(A\) depends on \(y_i\). For initial conditions \((a_0, \ldots, a_n, b_0, \ldots, b_n)\), set \begin{align*} A_0 = \frac{|\vec b|^2}{|\vec a|^2 +|\vec b|^2} && B_0=(1-A_0)A_0 \end{align*} We guess the solutions \begin{align*}x_i =& \cos(\sqrt{B_0}t ) a_i + \sin(\sqrt{B_0}t)b_i \\ y_i =& -A_0\sin(\sqrt{B_0}t) a_i + A_0\cos(\sqrt{B_0}t) b_i \end{align*} Observe that \(\vec a \cdot \vec b =0\), so \[|\vec y(t)|^2=(|\vec a|^2 + |\vec b|^2)\cdot A_0 = |\vec b|^2\] So, \(A(y(t))\) is constant for these guessed solutions. From this we check that the guessed solutions are flowlines of \(V_H\). Observe that at the boundary \(|\vec b|=1\), and so \begin{align*} A_0=1/3 && B_0=2/9 \end{align*} Therefore the time \(3\pi \sqrt{2}\) flow of \(V_H\) acts by mulitplication by \(-1\) on the boundary of \(X\cap \{(\vec x, \vec y) \st |\vec y|\leq 1\}\).