proof of Darboux’s theorem (symplectic geometry)

We first observe that it suffices to prove the theorem for symplectic formsMathworldPlanetmath defined on an open neighbourhood of 02n.

Indeed, if we have a symplectic manifold (M,η), and a point x0, we can take a (smooth) coordinate chartMathworldPlanetmath about x0. We can then use the coordinatePlanetmathPlanetmath function to push η forward to a symplectic form ω on a neighbourhood of 0 in 2n. If the result holds on 2n, we can compose the coordinate chart with the resulting symplectomorphism to get the theorem in general.

Let ω0=i=1ndxidyi. Our goal is then to find a (local) diffeomorphism Ψ so that Ψ(0)=0 and Ψ*ω0=ω.

Now, we recall that ω is a non–degenerate two–form. Thus, on T02n, it is a non–degenerate anti–symmetric bilinear formMathworldPlanetmath. By a linear change of basis, it can be put in the standard form. So, we may assume that ω(0)=ω0(0).

We will now proceed by the “Moser trick”. Our goal is to find a diffeomorphism Ψ so that Ψ(0)=0 and Ψ*ω=ω0. We will obtain this diffeomorphism as the time–1 map of the flow of an ordinary differential equationMathworldPlanetmath. We will see this as the result of a deformation of ω0.

Let ωt=tω0+(1-t)ω. Let Ψt be the time t map of the differential equation


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

We will make the ansatz


Now, we differentiate this :


(LXtωt denotes the Lie derivativeMathworldPlanetmathPlanetmath of ωt with respect to the vector field Xt.)

By applying Cartan’s identity and recalling that ω is closed, we obtain :


Now, ω-ω0 is closed, and hence, by Poincaré’s Lemma, locally exact. So, we can write ω-ω0=-dλ.



We want to require then


Now, we observe that ω0=ω at 0, so ωt=ω0 at 0. Then, as ω0 is non–degenerate, ωt will be non–degenerate on an open neighbourhood of 0. Thus, on this neighbourhood, we may use this to define Xt (uniquely!).

We also observe that Xt(0)=0. Thus, by choosing a sufficiently small neighbourhood of 0, the flow of Xt 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