# 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 ${\omega}_{0}={\sum}_{i=1}^{n}d{x}_{i}\wedge d{y}_{i}$. 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
${T}_{0}{\mathbb{R}}^{2n}$, 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 $\mathrm{\Psi}$ so that $\mathrm{\Psi}(0)=0$ and ${\mathrm{\Psi}}^{*}\omega ={\omega}_{0}$.
We will obtain this diffeomorphism as the time–$1$ map of 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 ${\mathrm{\Psi}}_{t}$ 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 respect
to the vector field ${X}_{t}$.)

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

$$0={\mathrm{\Psi}}_{t}^{*}(d{\iota}_{{X}_{t}}{\omega}_{t}+\omega -{\omega}_{0})$$ |

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 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) |
---|---|

Canonical name | ProofOfDarbouxsTheoremsymplecticGeometry |

Date of creation | 2013-03-22 14:09:55 |

Last modified on | 2013-03-22 14:09:55 |

Owner | rspuzio (6075) |

Last modified by | rspuzio (6075) |

Numerical id | 8 |

Author | rspuzio (6075) |

Entry type | Proof |

Classification | msc 53D05 |