proof of Darboux’s theorem (symplectic geometry)

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

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 thenuse the coordinate function to push $\eta$ forward to a symplectic form$\omega$ on a neighbourhood of $0$ in ${ℝ}^{2n}$. If the result holds on ${ℝ}^{2n}$,we can compose the coordinate chart with the resulting symplectomorphism toget the theorem in general.

Let ${\omega }_0={\sum }_{i=1}^{n}d{x}_i\wedge d{y}_i$. Our goal is then to find a (local) diffeomorphism $\mathrm{\Psi }$ so that $\mathrm{\Psi }(0)=0$ and ${\mathrm{\Psi }}^{*}{\omega }_0=\omega$.

Now, we recall that $\omega$ is a non–degenerate two–form. Thus, on$T_0{ℝ}^{2n}$, it is a non–degenerate anti–symmetric bilinear form. By a linear change of basis, it can be put in the standard form. So, wemay assume that $\omega (0)={\omega }_0(0)$.

We will now proceed by the “Moser trick”. Our goal is to find adiffeomorphism $\mathrm{\Psi }$ so that $\mathrm{\Psi }(0)=0$ and ${\mathrm{\Psi }}^{*}\omega ={\omega }_0$.We will obtain this diffeomorphism as the time–$1$ map of theflow of an ordinary differential equation. We will see this as theresult of a deformation of ${\omega }_0$.

Let ${\omega }_t=t{\omega }_0+(1-t)\omega$.Let ${\mathrm{\Psi }}_t$ be the time $t$ map of the differential equation

$$\frac{d}{dt}{\mathrm{\Psi }}_t(x)=X_t({\mathrm{\Psi }}_t(x))$$

in which $X_t$ is a vector field determined by a condition to be stated later.

We will make the ansatz

$${\mathrm{\Psi }}_t^{*}\omega ={\omega }_t.$$

Now, we differentiate this :

$$0=\frac{d}{dt}{\mathrm{\Psi }}_t^{*}{\omega }_t={\mathrm{\Psi }}_t^{*}(L_{X_t}{\omega }_t+\frac{d}{dt}{\omega }_t).$$

($L_{X_t}{\omega }_t$ denotes the Lie derivative of ${\omega }_t$ with respectto the vector field $X_t$.)

By applying Cartan’s identity and recalling that $\omega$ is closed, weobtain :

$$0={\mathrm{\Psi }}_t^{*}(d{\iota }_{X_t}{\omega }_t+\omega -{\omega }_0)$$

Now, $\omega -{\omega }_0$ is closed, and hence, by Poincaré’s Lemma,locally exact. So, we can write $\omega -{\omega }_0=-d\lambda$.

Thus

$$0={\mathrm{\Psi }}_t^{*}(d(i_{X_t}{\omega }_t-\lambda ))$$

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 anopen neighbourhood of $0$. Thus, on this neighbourhood, we may use this todefine $X_t$ (uniquely!).

We also observe that $X_t(0)=0$. Thus, by choosing a sufficientlysmall neighbourhood of $0$, the flow of $X_t$ will be defined for timegreater than $1$.

All that remains now is to check that this resulting flow has the desiredproperties. This follows merely by reading our of the ODE,backwards.