## You are here

Homeproof of Darboux's theorem (symplectic geometry)

## Primary tabs

# 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 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.

## Mathematics Subject Classification

53D05*no label found*

- Forums
- Planetary Bugs
- HS/Secondary
- University/Tertiary
- Graduate/Advanced
- Industry/Practice
- Research Topics
- LaTeX help
- Math Comptetitions
- Math History
- Math Humor
- PlanetMath Comments
- PlanetMath System Updates and News
- PlanetMath help
- PlanetMath.ORG
- Strategic Communications Development
- The Math Pub
- Testing messages (ignore)

- Other useful stuff