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 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 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 derivation of the ODE, backwards.