ProofOfDarbouxsTheoremSymplecticGeometry 2004-02-16 14:22:15 2007-06-24 00:18:34 Proof Darboux's theorem (symplectic geometry) 0 Moser trick % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \newcommand{\R}{\mathbb{R}^{2n}} We first observe that it suffices to prove the theorem for symplectic forms defined on an open neighbourhood of $0 \in \R$. Indeed, if we have a symplectic manifold $(M, \eta)$, and a point $x_0$, we can take a (smooth) coordinate chart about $x_0$. We can then use the coordinate function to push $\eta$ forward to a symplectic form $\omega$ on a neighbourhood of $0$ in $\R$. If the result holds on $\R$, we can compose the coordinate chart with the resulting symplectomorphism to get the theorem in general. Let $\omega_0 = \sum_{i=1}^{n} dx_i \wedge dy_i$. Our goal is then to find a (local) diffeomorphism $\Psi$ so that $\Psi(0) = 0$ and $\Psi^*\omega_0 = \omega$. Now, we recall that $\omega$ is a non--degenerate two--form. Thus, on $T_{0} \R$, it is a non--degenerate anti--symmetric bilinear form. By a linear change of basis, it can be put in the standard form. So, we may assume that $\omega(0) = \omega_0(0)$. We will now proceed by the ``Moser trick''. Our goal is to find a diffeomorphism $\Psi$ so that $\Psi(0) = 0$ and $\Psi^*\omega = \omega_0$. We will obtain this diffeomorphism as the time--$1$ map of the flow of an ordinary differential equation. We will see this as the result of a deformation of $\omega_0$. Let $\omega_t = t \omega_0 + (1-t) \omega$. Let $\Psi_{t}$ be the time $t$ map of the differential equation $$\frac{d}{dt} \Psi_t(x) = X_t(\Psi_t(x))$$ in which $X_t$ is a vector field determined by a condition to be stated later. We will make the ansatz $$\Psi_{t}^* \omega = \omega_t.$$ Now, we differentiate this \PMlinkescapetext{identity}: $$ 0 = \frac{d}{dt} \Psi_t^*\omega_t = \Psi_t^*( L_{X_t}\omega_t + \frac{d}{dt} \omega_t). $$ ($L_{X_t}\omega_t$ denotes the Lie derivative of $\omega_t$ with respect to the vector field $X_t$.) By applying Cartan's identity and recalling that $\omega$ is closed, we obtain : $$ 0 = \Psi_t^*(d \iota_{X_t} \omega_t + \omega - \omega_0) $$ Now, $\omega - \omega_0$ is closed, and hence, by Poincar\'e's Lemma, locally exact. So, we can write $\omega - \omega_0 = - d\lambda$. Thus $$ 0 = \Psi_t^*( d ( i_{X_t} \omega_t - \lambda )) $$ We want to require then $$ i_{X_t} \omega_t = \lambda. $$ Now, we observe that $\omega_0 = \omega$ at $0$, so $\omega_t = \omega_0$ at $0$. Then, as $\omega_0$ is non--degenerate, $\omega_t$ will be non--degenerate on an open neighbourhood of $0$. Thus, on this neighbourhood, we may use this to define $X_t$ (uniquely!). We also observe that $X_t(0) = 0$. Thus, by choosing a sufficiently small neighbourhood of $0$, the flow of $X_t$ will be defined for time greater than $1$. All that remains now is to check that this resulting flow has the desired properties. This follows merely by reading our \PMlinkescapetext{derivation} of the ODE, backwards.