\(
\def\CC{{\mathbb C}}
\def\RR{{\mathbb R}}
\def\NN{{\mathbb N}}
\def\ZZ{{\mathbb Z}}
\def\TT{{\mathbb T}}
\def\CF{{\operatorname{CF}^\bullet}}
\def\HF{{\operatorname{HF}^\bullet}}
\def\SH{{\operatorname{SH}^\bullet}}
\def\ot{{\leftarrow}}
\def\st{\;:\;}
\def\Fuk{{\operatorname{Fuk}}}
\def\emprod{m}
\def\cone{\operatorname{Cone}}
\def\Flux{\operatorname{Flux}}
\def\li{i}
\def\ev{\operatorname{ev}}
\def\id{\operatorname{id}}
\def\grad{\operatorname{grad}}
\def\ind{\operatorname{ind}}
\def\weight{\operatorname{wt}}
\def\Sym{\operatorname{Sym}}
\def\HeF{\widehat{CHF}^\bullet}
\def\HHeF{\widehat{HHF}^\bullet}
\def\Spinc{\operatorname{Spin}^c}
\def\min{\operatorname{min}}
\def\div{\operatorname{div}}
\def\SH{{\operatorname{SH}^\bullet}}
\def\CF{{\operatorname{CF}^\bullet}}
\def\Tw{{\operatorname{Tw}}}
\def\Log{{\operatorname{Log}}}
\def\TropB{{\operatorname{TropB}}}
\def\wt{{\operatorname{wt}}}
\def\Span{{\operatorname{span}}}
\def\Crit{\operatorname{Crit}}
\def\CritVal{\operatorname{CritVal}}
\def\FS{\operatorname{FS}}
\def\Sing{\operatorname{Sing}}
\def\Coh{\operatorname{Coh}}
\def\Vect{\operatorname{Vect}}
\def\into{\hookrightarrow}
\def\tensor{\otimes}
\def\CP{\mathbb{CP}}
\def\eps{\varepsilon}
\)
SympSnip:
- 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
\[V_H= \sum_{i=0}^n Ax_i \partial_{y_i} - (1-A)y_i \partial_{x_i}\]
-
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 \(X\cap \{(\vec x, \vec y) \st |\vec y|\leq 1\}\).