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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 .
Indeed, if we have a symplectic manifold , and a point , we can take a (smooth) coordinate chart about . We can then use the coordinate function to push forward to a symplectic form on a neighbourhood of in . If the result holds on , we can compose the coordinate chart with the resulting symplectomorphism to get the theorem in general.
Let . Our goal is then to find a (local) diffeomorphism so that and .
Now, we recall that is a non–degenerate two–form. Thus, on , 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 .
We will now proceed by the “Moser trick”. Our goal is to find a diffeomorphism so that and . We will obtain this diffeomorphism as the time– map of the flow of an ordinary differential equation. We will see this as the result of a deformation of .
Let . Let be the time map of the differential equation
in which is a vector field determined by a condition to be stated later.
We will make the ansatz
Now, we differentiate this identity:
( denotes the Lie derivative of with respect to the vector field .)
By applying Cartan’s identity and recalling that is closed, we obtain :
Now, is closed, and hence, by Poincaré’s Lemma, locally exact. So, we can write .
Thus
We want to require then
Now, we observe that at , so at . Then, as is non–degenerate, will be non–degenerate on an open neighbourhood of . Thus, on this neighbourhood, we may use this to define (uniquely!).
We also observe that . Thus, by choosing a sufficiently small neighbourhood of , the flow of will be defined for time greater than .
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 Symplectic manifolds, general- Forums
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