Darboux coordinates in neighbourhood of a point

Question

Given a bivector field, $(x\partial_x + y\partial_y)\wedge \partial_z$, how does one find a system of Darboux coordinates in a neighbourhood of the point $(x,y,z) = (1,0,0) \in \mathbb{R}^3$?

Answer 1

We wish to find coordinates $(u,v,w)$ near $(1,0,0)$ that put the bivector $\Pi=(x\partial_{x}+y\partial_{y})\wedge\partial_{z}$ in the Darboux normal form. Since the rank of $\Pi$ at $(1,0,0)$ is $2$, we want to obtain that locally $$ \Pi=\partial_{u}\wedge\partial_{v}. $$ So we get the following requirements:

1) $u$ and $v$ should be local coordinates on the symplectic leaf through $(1,0,0)$.

2) $\{u,v\}=1$.

3) $\{u,w\}=\{v,w\}=0$.

Let us first check what the symplectic leaves of $\Pi$ are. They integrate the image of the vector bundle map $$\Pi^{\sharp}:T^{*}\mathbb{R}^{3}\rightarrow T\mathbb{R}^{3}:\alpha\mapsto\iota_{\alpha}\Pi.$$ In our case, we have \begin{align} \text{Im}(\Pi^{\sharp})&=\text{Span}\{\Pi^{\sharp}(dx),\Pi^{\sharp}(dy),\Pi^{\sharp}(dz)\}\\ &=\text{Span}\{x\partial_{z},y\partial_{z},-x\partial_{x}-y\partial_{y}\}. \end{align} So all points $(0,0,z)$ on the $z$-axis are leaves, since the distribution vanishes there. At any point away from the $z$-axis, this distribution is spanned by the vertical vector field $\partial_{z}$ and the radial vector field $-x\partial_{x}-y\partial_{y}$, so that the remaining leaves are half-planes radiating out of the $z$-axis. In particular, the leaf through the point $(1,0,0)$ is given by $\{y=0,x>0\}$.

We now start building the coordinate chart $(u,v,w)$. Let us take $u:=z$, which is a coordinate on the leaf $y=0$. We now construct $v=v(x,z)$ so that it meets requirement $2)$ above: $$ 1=\{z,v\}=X_{z}(v)=(-x\partial_{x}-y\partial_{y})(v), $$ which leads to $v:=-\ln(x)$. At last, we construct $w$ to satisfy $3)$: \begin{align} &0=\{z,w\}=X_{z}(w)=(-x\partial_{x}-y\partial_{y})(w),\\ &0=\{-\ln(x),w\}=X_{-\ln(x)}(w)=-\partial_{z}(w). \end{align} Setting $w:=y/x$, these requirements are met. So we will set $(u,v,w):=(z,-\ln(x),y/x)$. At last, we have to check that $(x,y,z)\mapsto (z,-\ln(x),y/x)$ is really a change of coordinates near $(1,0,0)$. This is the case, since its Jacobian at $(1,0,0)$ is $$ \left.\begin{pmatrix} 0&0&1\\-1/x&0&0\\-y/x^{2}&1/x&0 \end{pmatrix}\right|_{(1,0,0)}=\begin{pmatrix} 0&0&1\\-1&0&0\\0&1&0\end{pmatrix}, $$ which has nonzero determinant.