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\DeclareMathOperator{\sgn}{sgn} \DeclareMathOperator{\Ext}{Ext} \DeclareMathOperator{\TropB}{TropB} \DeclareMathOperator{\weight}{wt} \DeclareMathOperator{\Span}{span} \DeclareMathOperator{\Coh}{Coh} \DeclareMathOperator{\Pic}{Pic} \DeclareMathOperator{\Fuk}{Fuk} \DeclareMathOperator{\str}{star} \DeclareMathOperator{\Ob}{Ob} \DeclareMathOperator{\grad}{grad} \DeclareMathOperator{\Supp}{Supp} \DeclareMathOperator{\Bl}{Bl} \DeclareMathOperator{\Spec}{Spec} \DeclareMathOperator{\Tw}{Tw} \DeclareMathOperator{\Int}{Int} \DeclareMathOperator{\Arg}{\mathbf{M}}\begin{filecontents}{references.bib} @article{ballard2012hochschild, title={Hochschild dimensions of tilting objects}, author={Ballard, Matthew and Favero, David}, journal={International Mathematics Research Notices}, volume={2012}, number={11}, pages={2607--2645}, year={2012}, publisher={OUP} } @article{craw2007explicit, title={Explicit methods for derived categories of sheaves}, author={Craw, Alastair}, publisher={Citeseer} } 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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 \[V_H= \sum_{i=0}^n Ax_i \partial_{y_i} - (1-A)y_i \partial_{x_i}\] \item 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 $X\cap \{(\vec x, \vec y) \st |\vec y|\leq 1\}$. \end{enumerate}\printbibliography \end{document}