proof of Darboux’s theorem (symplectic geometry)

We first observe that it suffices to prove the theorem for symplectic forms defined on an open neighbourhood of $0\in\mathbb{R}^{2n}$.

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 $\mathbb{R}^{2n}$. If the result holds on $\mathbb{R}^{2n}$, 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 $T_{0}\mathbb{R}^{2n}$, 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 :

 $0=\frac{d}{dt}\Psi_{t}^{*}\omega_{t}=\Psi_{t}^{*}(L_{X_{t}}\omega_{t}+\frac{d}% {dt}\omega_{t}).$

($L_{X_{t}}\omega_{t}$ 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 Poincaré’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 of the ODE, backwards.

Title proof of Darboux’s theorem (symplectic geometry) ProofOfDarbouxsTheoremsymplecticGeometry 2013-03-22 14:09:55 2013-03-22 14:09:55 rspuzio (6075) rspuzio (6075) 8 rspuzio (6075) Proof msc 53D05