Proof of Darboux theorem in symplectic geometry using Moser theorem

differential-geometrysmooth-manifoldssymplectic-geometry

Darboux theorem in symplectic geometry says the following:

Let $(M^{2n},\omega)$ be a symplectic manifold. If $p\in M$, there is a chart $(U,x_1,…,x_n,y_1,…,y_n)$ centered at $p$ such that:
$$\omega=\sum_{i=1}^ndx_i\wedge dy_i$$

In Ana Cannas' Lectures on Symplectic Geometry, the proof begins like this:

"Use any symplectic basis for $T_pM$ to construct coordinates $(x'_
1,…, x'_n, y'_1,…,y'_n)$
centered at $p$ and valid on some neighborhood $U_0$, so that $\omega_p=\sum_{i=1}^n(dx'_i\wedge dy'_i)_p$".

Then she uses Moser theorem, which guarantees there is a neighbourhood $U_1$ and a diffeomorphism $\varphi:U_0\to U_1$ with $\varphi(p)=p$ and $\varphi^*\left(\sum_idx'_i\wedge dy'_i\right)=\omega$. Then we only need to define $x_i:=x'_i\circ\varphi$, $y_i:=y'_i\circ\varphi$.

The final part is clear to me, but the beginning is not. If we have a symplectic basis $\{e_1,f_1,…,e_n,f_n\}$ of $T_pM$, how can we construct the coordinates $(x'_1,…,x'_n,y'_1,…,y'_n)$?

Best Answer

Say we have a symplectic basis $\{e_{1},\ldots,e_{n},f_{1},\ldots,f_{n}\}$ for $(T_{p}M,\omega_{p})$. If $\{e_{1}^{*},\ldots,e_{n}^{*},f_{1}^{*},\ldots,f_{n}^{*}\}$ is the dual basis of $\left(T_{p}M\right)^{*}$, then $\omega_{p}\in\wedge^{2}\left(T_{p}M\right)^{*}$ has the following expression: $$ \omega_{p}=\sum_{i=1}^{n}e_{i}^{*}\wedge f_{i}^{*}. $$

So we have to show that there exists a coordinate chart $(x_{1}',\ldots,x_{n}',y_{1}',\ldots,y_{n}')$ around $p$ satisfying $$ (dx_{i}')_{p}=e_i^{*},\\ (dy_{i}')_{p}=f_{i}^{*}. $$ Such a chart can be constructed as follows. Let $(\xi_{1},\ldots,\xi_n,\eta_1,\ldots,\eta_{n})$ be any coordinate chart around $p$. Now both $\{e_{1}^{*},\ldots,e_{n}^{*},f_{1}^{*},\ldots,f_{n}^{*}\}$ and $\{(d\xi_{1})_{p},\ldots,(d\xi_{n})_{p},(d\eta_{1})_{p},\ldots,(d\eta_{n})_{p}\}$ are bases of $(T_{p}M)^{*}$. Let the following block matrix be the change of bases: $$ \begin{bmatrix} A & C\\ B&E \tag{1} \end{bmatrix}, $$ i.e. $$ e_{1}^{*}=\sum_{j=1}^{n}a_{j1}(d\xi_{j})_{p}+\sum_{j=1}^{n}b_{j1}(d\eta_{j})_{p},\\ \vdots\\ e_{n}^{*}=\sum_{j=1}^{n}a_{jn}(d\xi_{j})_{p}+\sum_{j=1}^{n}b_{jn}(d\eta_{j})_{p},\\ f_{1}^{*}=\sum_{j=1}^{n}c_{j1}(d\xi_{j})_{p}+\sum_{j=1}^{n}e_{j1}(d\eta_{j})_{p},\\ \vdots\\ f_{n}^{*}=\sum_{j=1}^{n}c_{jn}(d\xi_{j})_{p}+\sum_{j=1}^{n}e_{jn}(d\eta_{j})_{p}. $$ This suggests that we should define $$ x_{1}'=\sum_{j=1}^{n}a_{j1}\xi_{j}+\sum_{j=1}^{n}b_{j1}\eta_{j},\\ \vdots\\ x_{n}'=\sum_{j=1}^{n}a_{jn}\xi_{j}+\sum_{j=1}^{n}b_{jn}\eta_{j},\\ y_{1}'=\sum_{j=1}^{n}c_{j1}\xi_{j}+\sum_{j=1}^{n}e_{j1}\eta_{j},\\ \vdots\\ y_{n}'=\sum_{j=1}^{n}c_{jn}\xi_{j}+\sum_{j=1}^{n}e_{jn}\eta_{j}. $$ To check that $(x_{1}',\ldots,x_{n}',y_{1}',\ldots,y_{n}')$ are really coordinates around $p$, we should check that the map $$(\xi_{1},\ldots,\xi_n,\eta_1,\ldots,\eta_{n})\mapsto(x_{1}'(\xi,\eta),\ldots,x_{n}'(\xi,\eta),y_{1}'(\xi,\eta),\ldots,y_{n}'(\xi,\eta))$$ is a local diffeomorphism around $p$. This is the case, since its Jacobian is exactly the transpose of the block matrix (1) above, which is non-singular.

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